Linear groups and ergodic theory

Uri Bader

(Instytut Technologii Technion w Hajfie, Izrael)

(Wrocław, 01 – 12.09.2014)

Notes From Wroclaw University Course About

Algebraic Representations of Ergodic Actions

October 1, 2014

Abstract

These notes are in preliminary form and might contain typos and in-

accuracies. Proceed with caution.

Part I

Lectures

1 Lecture 1 - An Introduction

The Basic Problem

Let us consider groups of matrices, e.g. GL2(R), which is the group of all 2 × 2invertible matrices with entries in R. We have the determinant homomorphismdet ∶ GL2(R) → R× = R/0, and the kernel of this map is denoted SL2(R) =A ∈ GL2(R) ∶ det(A) = 1.

Let G = SL2(R). It acts on R2 by matrix multiplication, and there are twoorbits - one is the xed point 0 and one is R2/0. We can identify R2/0 withG/U where U is the stabilizer of some point in R2/0, say (1

0). The matrices in U

must be of the form⎛⎝

1 t0 s

⎞⎠in order to x (1

0), and the determinant condition

implies that the only matrices in U are of the form⎛⎝

1 t0 1

⎞⎠. U is called a

unipotent group.

Note. The group U is isomorphic to the additive group of the reals, (R,+).Check this.

Problem 1.1. Find all maps φ ∶ R2 → R2 which commute with G = SL2(R),that is:

∀x ∈X, g ∈ G ∶ φ(gx) = gφ(x)

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We observe that if φ is such a map, then φ(0) = 0. Indeed, this commutingcondition implies that φ(0) is also a point xed by G, but the only such pointis 0. We ask ourselves where can (1

0). We know that (1

0) is xed by U , so its

image under φ must also be xed by U , meaning that it must be of the form(α

0), as the U orbits on R2 look like:

If we now take a general point x ∈ R2/0, we nd some g ∈ G such thatx = g ⋅ (1

0), meaning that:

φ(x) = φ(g ⋅ (1

0)) = g ⋅ φ((1

0)) = g ⋅ (α

0)gcommutes with

scalar multiplication= αg ⋅ (1

0) = αx

this also works for x = 0, so we get that the only maps R2 → R2 commutingwith G = SL2(R) are given by scalar multiplication.

Remark. Let G be a group and let H ≤ G. It can be seen that any mapG/H → G/H which commutes with the G-action must be of the form gH ↦ gnHfor some xed n ∈ NG(H), the normalizer of H in G. This gives a canonicalidentication of the collection of automorphisms of G/H as a G-space, denotedAutG(G/H), to the group N/H. Check this.

In our example, H = U and the normalizer is NG(U) = P =⎧⎪⎪⎨⎪⎪⎩

⎛⎝a b0 a−1

⎞⎠∶

a ∈ R×, b ∈ R⎫⎪⎪⎬⎪⎪⎭≅ R× ⋉ R, and P/U ≅ R×, which can be identied (as above) to

the collection of non-zero scalars (the non-zero demand comes into play becauseone requires the maps G/H → G/H to be an automorphism, and in particularbijective.

We now ask a dierent question:

Problem 1.2. Find all maps φ ∶ R2 → R2 which commute with the action ofΓ = SL2(Z), the collection of matrices in SL2(R) with integer entries.

The diculty in this case is that the orbit structure is wild - the stabilizerof a generic point is trivial, so we have in R2 lots and lots of copies of SL2(Z).

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One can dene a function R2 → R2 by multiplying the elements of each suchorbit by an element of Γ, which will depend on the orbit itself. This is whyunderstanding all such maps is very hard.

In our previous example G = SL2(R) R2, the orbit structure was muchnice - all orbits were locally closed, i.e. open in their closure. The dynamicshere are what we call tame. This is also what we see for U R2, in which theorbits are even closed.

Example 1.3. More examples of wild dynamics:

Irrational rotation on this circle - the orbits are far from locally closed,and they even equidistribute.

Γ = SL2(R) G/U and U G/Γ (as we'll see later).

We reformulate the problem and seek only (Borel) measurable maps R2 → R2

commuting with Γ = SL2(Z).

Theorem 1.4. Every such map is equal to a multiplication be a scalar Lebesgue-a.e.

What is special about Γ in this case? It is a lattice, i.e. Γ ≤ G = SL2(R) isdiscrete and G/Γ has a G-invariant Borel probability measure. Note that Z ≤ Ris a lattice by denition, and R/Z is even compact. As we'll see later, this in notthe case form SL2(Z) ≤ SL2(R).

In order to answer this question, we turn our heads to a seemingly unrelatedtopic.

Invariant Metrics on Homogenous Spaces

Denition 1.5. Let G be a group acting on a set X. A G-invariant metric onX is a metric d on X satisfying:

d(gx, gy) = d(x, y), ∀x, y ∈X,g ∈ G

Theorem 1.6. Let G = SL2(R) and X = R2/0 = G/U. There are no continu-ous G-invariant semi-metrics on X but the trivial semi-metric: d(x, y) = 0 forall x, y ∈X.

Remark. Continuous here means that the map d ∶ R2/0 ×R2/0 → [0,∞) iscontinuous, where the topology on R2/0 is the quotient topology on G/U. Itcoincides with the Euclidean topology on R2/0.

Before proving the theorem, let us note another useful subgroup of G. Let

A =⎧⎪⎪⎨⎪⎪⎩

⎛⎝λ

λ−1

⎞⎠∶ λ ∈ R×

⎫⎪⎪⎬⎪⎪⎭. The orbits of A R2 can be depicted:

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each red orbit is two-sheeted - it contains two connected components in oppositequadrants. We are now ready to prove the theorem:

Proof. Take a semi-metric d and let ε > 0. We'll show that for any two pointsx, y ∈ R2/0, one has that d(x, y) < ε. This will show that d is trivial.

Let x0 = (10), and denote the ball in the d-semi-metric of radius ε around x0

by B. By continuity of d, the set B contains a small Euclidean ball (the insideof the red ball in the picture):

Take a point x = (sr) inside this Euclidean ball (where r ≠ 0) and apply any

element u ∈ U . Because x0 is U -xed and d is G-invariant:

d(x0, ux) = d(ux0, ux) = d(x0, x) < ε

but by applying an element of U , we can get to any point in the horizontal linethrough x. Thus B contains the blue strip in the picture. To be exact, we didnot show that it contains the x-axis (whose points are U -xed), but B ⊆ R2/0is closed in the Euclidean topology (as d is continuous) and therefore B alsocontains the x-axis.

Now, take any two points x, y ∈ R2/0 (the purple points in the picture).By applying an element a of A, we can make them both go simultaneously into

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the blue strip (the brown points in the picture are the images of x and y undera). Thus:

d(x, y) = d(ax, ay) ≤ d(ax, x0) + d(x0, ay) < ε + ε = 2ε

as ε > 0 was arbitrary, we get that d(x, y) = 0 for all x, y ∈ R2/0.

Denition 1.7. The action G X is called ergodic if whenever A ⊆ X ismeasurable G-invariant, it's either null or full (i.e., it's complement X/A isnull). Equivalently, the collection of all G-invariant bounded function X → R,L∞(X)G, consists only of constant functions. If X has a G-invariant probabilitymeasure, this is equivalent to saying that the collection of G-invariant squareintegrable function, L2(X)G, consists only of constant functions.

Example 1.8. The action G G/Γ is ergodic, as there are no G-invariant setsin X = G/Γ but ∅ and X itself.

Corollary 1.9. The action U G/Γ is ergodic.

Proof. We need to show that L2(G/Γ)U = consts, which is ne as there is aG-invariant probability measure on G/Γ.

If G acts on a space X, then there is a one-to-one correspondence betweenthe U -invariant points XU and the collection of measurable functions from G/Uto X which respect the action of G, L(G/U,X)G, given by the orbit maps - ifx ∈ XU then gU ↦ gx is a well dened measurable function which respects theaction of G. Plugging X = L2(G/Γ), we nd that:the

L2(G/Γ)U ≅ L(G/U, L2(G/Γ))G

Note that the orbit maps are not only measurable, but also continuous.Given such a map φ ∈ L(G/U, L2(G/Γ)G, we can dene a semi-metric d on G/Ugiven by

d(xU, yU) = ∣∣φ(xU) − φ(yU)∣∣L2

beacuse the map φ is continuous, d is a continuous semi-metric and because φrespects the action of G, d will be G-invariant. The theorem implies that d ≡ 0,meaning that the image of φ consists of a single element ψ ∈ L2(G/Γ). Becauseφ is G-equivariant, it's image ψ needs to be a G-xed point. Thus:

L2(G/Γ)U ≅ L(G/U, L2(G/Γ))G ≅ L2(G/Γ)G

and the latter consists only of constants as the action G G/Γ is ergodic.

We can now conclude:

Theorem 1.10. Every measurable map φ ∶ R2 → R2 which is Γ = SL2(Z)-equivariant is essentially multiplication by a scalar.

We've already seen that U G/Γ is ergodic and that G/Γ has a G-invariantprobability measure. However, we are interested in the spae G/U ≅ R2/0, sowe look for a duality principle which bonds U G/Γ and Γ G/U together.

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Proof. (of the theorem) Let φ ∶ G/U ≅ R2/0 → R2 be a Γ-equivariant map.We build a map Φ ∶ G/Γ → R2 which is U -equivariant. First extend φ to a map

ψ ∶ G→ R2/0 by ψ(x) = φ(xU), which is exactly the same as ψ(x) = φ(x⋅(10)).

As it stands now, Ψ is not U -equivariant, but only Γ-equivariant. Let us deformit a bit - let Ψ ∶ G→ R2 by dened by

Ψ(g) = g ⋅ ψ(g−1) = g ⋅ φ(g−1(1

0))

if g = hγ when γ ∈ Γ and h ∈ G, then:

Ψ(γh) = hγ ⋅φ(γ−1h−1 ⋅(1

0))

φ is Γ−equivariant= hγγ−1 ⋅φ(h−1 ⋅(1

0)) = h ⋅φ(h−1 ⋅(1

0)) = Ψ(h)

meanign that Ψ is Γ-invariant. Let Φ ∶ G/Γ → R2 be dened by Φ(xΓ) = Ψ(x).Φ is U -equivariant as if x ∈ G and u ∈ U then:

Φ(uxΓ) = Ψ(ux) = ux⋅φ(x−1u−1(1

0))

U xes (10)

= u⋅x⋅φ(x−1⋅(1

0)) = u⋅Ψ(x) = u⋅Φ(xΓ)

To conclude the above, Φ ∶ G/Γ→ R2 is a U -equivariant map. If we push theG-invariant ergodic probability measure from U G/Γ to U R2 using Φ, weget a U -invariant probability measure µ which is ergodic.

Push the U -invariant probability ergodic probability measure using the pro-jection on the y-coordinate projy ∶ R2 → R and get an ergodic probability mea-sure on R which is U -invariant - but U does not act on this space as points inthe same U -orbit have the same y-coordinate. By ergodicity we conclude thatthe image measure projy⋆(µ) must be a delta measure at some point y0.Weclaim that y0 must be equal to 0.

Indeed, if projy⋆(µ) is supported as the single point y0, the original mea-sure µ on R2 must be supported on y = y0. If y0 ≠ 0, the action of U onthis line is by translations, and one can translate left and right by any amount.However, the only translation invariant measure on the real line is the Lebesgue

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measure (up to scaling), which is not a probability measure, although µ is suchtranslation invariant (as it is U -invariant) probability measure on a copy of thereal line, which is impossible. Thus y0 = 0.

We got an ergodic U -invariant probability measure µ which is supported onR × 0, but the action of U on this space is trivial. By ergodicity we deduceagain that µ is supported on one point, say (α

0). µ was the push forward of

the G-invariant probability measure on G/Γ by Φ. This means that the essentialimage of Φ is (α

0). Reading backwards, for essentially every x = g−1(1

0) in R2/0

we have that:

φ(x) = φ⎛⎝g−1(1

0)⎞⎠= g−1 ⋅ gφ(g−1(1

0)) = g−1Φ((1

0)) = g−1(α

0)

g−1 acts linearly, so it commutes with scalar multiplication. As a result:

φ(x) = g−1(α0) = α ⋅ g−1(1

0) = αx

so φ is essentially multiplication by the scalar α.

Up to now, we introduced a number of dierent topics which one shouldelaborate on:

Lattices in groups.

Metrics on coset spaces (And a related topic: the Howe-Moore theorem).

Ergodic theory.

Algerbaic groups (as SL2(R) and U) and dynamics.

Measures on algebraic varieties.

Algebraic representation of ergodic actions and applications.

Lattices in R2

We saw a fact in the argument above - Γ = SL2(Z) is a lattice inside SL2(R) = G.It is clear that Γ ≤ G is a discrete subgroup, and we will see in the rst tutorialthat there is a G-invariant measure on G/Γ (see corollary 11.12 in the tutorialssection). We need to explain why it is nite. We want to understand X = G/Γ.

Consider a lattice inside R2. One can take Z2 as an example. Anotherexample can be g ⋅ Z2 when g ∈ GL2(R). If Λ is a lattice and g ∈ GL2(R) theng ⋅Λ is also a lattice.

Exercise 1.11. (Baby Mostow Rigidity) Show that if Λ ≤ R2 is a lattice thenΛ = g ⋅Z2 for some g ∈ GL2(R). In particular, the action of GL2(R) on the spaceof lattices is transitive.

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We stabilizer of Z2 is exactly GL2(Z), so the space of all lattices can beidentied by GL2(R)/GL2(Z). To restrict ourselves to SL2 instead of GL2, wedene the covolume of a lattice to be:

covol(gZ2) = ∣det(g)∣ = vol( fundamental domain )

let us set consider the space of all lattices of covolume 1, which will be calledunimodular lattices.

Exercise 1.12. Show that X = SL2(R)/SL2(Z) can be identied with the spaceof all unimodular lattices.

We now present a few useful denitions:

Denition 1.13. Let Λ be a unimodular lattice.

A primitive basis for Λ is a pair u.v of points in Λ so that det(u, v) = 1.These represent edges of a parallelogram fundamental domain.

A primitive vector in Λ is a vector u ∈ Λ which is a part of a primitivebasis.

Exercise 1.14. Show that x ∈ Λ is primitive if and only if there is no y ∈ Λ andn ∈ Z/±1 such that x = ny. Hint: show it for Λ = Z2 rst.

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2 Lecture 2 - SL2(Z) and SL2(R)We want to nd a fundamental domain for the right action of Γ = SL2(Z) onG = SL2(R), i.e. to nd a nice subset F of SL2(R) so that:

G = F ⋅ Γ ≅←Ðproductmap

F × Γ

pictorially,

We may view the action G G/Γ as G F : if x ∈ F ⊆ G and g ∈ G, we canwrite g ⋅ x = gx uniquely as fγ when f ∈ F and γ ∈ G. We then dene:

g ⋅ x = gxγ−1 = f ∈ F

this element γ is usually denoted α(g, x) and satises some cocycle condition.Pictorially:

If the group G is unimodular (as SL2(R)), the right multiplication by γpreserves the Haar measure on G, meaning that the action G F is mea-sure preserving. Recall we denoted the space of unimodular lattices by X ≅

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SL2(R)/SL2(Z). Denoting G = SL2(R), Γ = SL2(Z) and U =⎧⎪⎪⎨⎪⎪⎩

⎛⎝

1 t1

⎞⎠∶ t ∈ R

⎫⎪⎪⎬⎪⎪⎭,

we have seen that G/U ≅ R2/0 and we've also seen a form of duality betweenG/Γ (space of lattices) and G/U (space of vectors). We wish to nd a commonpredecessor to these two.

Dene Y = (Λ, v) ∶ Λ ∈ X, v a primitive vector in Λ. We have the forget-ting maps:

Y

R2/0 X

writing these in terms of G, X ≅ G/Γ and R2/0 ≅ G/U. What is Y ? One can seethat the action of G on Y is transitive (why?), and the stabilizer of the point

(Z2, (10)) consists exactly of the integer points in U :

U(Z) =⎧⎪⎪⎨⎪⎪⎩

⎛⎝

1 n1) ∶ n ∈ Z

⎫⎪⎪⎬⎪⎪⎭= U ∩ Γ

writing U = U(R), we have the picture:G/U(Z)

G/U(R) G/Γ

The Siegel Transform

We wish to take a function on R2/0 and transform it to be a function of X.Recall that whenever G acts on the coset space G/H, the normalizer N =

NG(H) acts on G/H from the right: n ⋅ (gH) = gn−1H. The actions of G andof N commute with each other. Modding the N -action out, G still acts on thequotient space G/N by g ⋅ hN = ghN . In our case, G = SL2(R) acts on G/U(Z),and the normalizer of H = U(Z) is N = U(R) (exercise), so we have a rightaction of U(R)/U(Z) ≅ R/Z, which is compact. Modding the N -action out, we getthe projection map G/U(Z) ≅ Y → R2/0 ≅ G/U(R), so the bre above each pointis a circle R/Z. Thus the projection Y → R2/0 is proper.

Given f ∶ R2/0 → R continuous with compact support, we compose itwith the projection Y → R2/0 to get a continuous map f ∶ Y → R withcompact support (as projection is proper). We notice that the bre of the mapY = G/U(Z) → G/Γ = X above every point is countable, so we can push f to get

a map f ∶ X → R which is continuous with compact support by summing overbres. In coordinates, it is dened by:

f(Λ) = ∑x∈Λ primitive

f(x)

the map taking f ∈ Cc(R2/0) and returning f ∈ Cc(X) is called the Siegeltransform.

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Let us now dualize. Given a linear functional on Cc(X), we precompose itwith the Siegel transform to get a functional on Cc(R2/0). It is known thatthe dual space of Cc(Z) (if Z is a nice enough topological space) if the space ofall Radon measures on Z, denoted M(Z). We thus get a map:

M(X)→M(R2/0)

take the G-invariant measure µ on X ≅ G/Γ and push it forward to get aG-invariant measure on R2/0. There is a unique G-invariant measure onR2/0 ≅ G/U, up to scale, which is the Lebesgue measure λ. Let us pickthe scaling of the G-invariant measure on X so that the pushed measure onR2/0 is the Lebesgue measure. By dention of the pushed measure, givenf ∈ Cc(R2/0) we get the identity:

∫R2/0

f = ∫Xf

by completion, this identity holds for all f ∈ L1(R2/0) = L1(R2). This identityis called Siegel's identity.

Claim 2.1. Let f = χ[−2,2]2 be the characteristic function of [−2,2]2. Then f ≥ 1everywhere. As a result, the total measure of X = G/Γ is bounded by 16, henceΓ ≤ G is a lattice.

We rst explaing why the latter part is true. Indeed, if f ≥ 1 then:

µ(X) = ∫X

1dµ ≤ ∫Xχ[−2,2]2dµ = ∫

R2χ[−2,2]2 = λ([−2,2]2) = 16

In order to prove the rst part of the claim, we state (and prove later) atheorem by Minkowski:

Theorem 2.2. Let C ⊆ R2 be a convex set which is symmetric (i.e., −C = C).Suppose that λ(C) > 4. Then C ∩Z2 ≠ 0.

We now explain why the theorem implies the rst part of the claim. Indeed,we denote f = χ[−2,2]2 . GivenΛ = gZ2 a lattice, we write:

f(Λ) = ∑x∈Λ primitive

f(x) = ∑x∈Z2 primitive

f(gx) = ∑x∈Z2 primitive

χg−1[−2,2]2(x)

because g−1 acts by measure preserving linear transformations, C = g−1[−2,2]2is a convex symmetric set, and λ(C) = λ([−2,2]2) > 4. Thus C ∩ Z2 is containssome non-zero vector x0. Note that because C is convex and symmetric, if x ∈ Cand t ∈ [−1,1] then tx ∈ C. If 0 ≠ x0 ∈ Z2 is not primitive, there is some primitvey0 ∈ Z2 and n ∈ Z/±1,0 such that x0 = ny0, so that y0 = 1

nx0 is also inside C.

To conclude, C = g−1[−2,2]2 contains a primitive vector from Z2, hence:

f(Λ) = ∑x∈Z2 primitive

χg−1[−2,2]2(x) ≥ 1

To conclude the proof that Γ = SL2(Z) is a lattice inside G = SL2(R), weprove the theorem of Minkowski:

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Proof. Consider the map 12C → [0,1]2 which takes a point (x, y) to (x− [x], y −

[y]), when [z] is the integer part of z. This transformation preserves theLebesgue measure (as on each square [m,m+1)×[n,n+1) it acts by translation),and:

λ(1

2C) = 1

4λ(C) > 1 = λ([0,1]2)

therefore there are two dierent points x, y ∈ 12C which has the same image, i.e.

x − y ∈ Z2/0. We claim that x − y ∈ C. Indeed, 2x,2y ∈ C, so −2y ∈ C (as C issymmetric) and because C is convex:

x − y = 1

2(2x) + 1

2(−2y) ∈ C

which completes the proof.

Remark 2.3. Firstly, a similar proof gives that SLn(Z) ≤ SLn(R) is a lattice.However, for n > 2, the map from the projection Yn → Rn/0 will not beproper - the bre over each point will be isomorphic to the non-compact space

SLn−1(R)⋉Rn−1/SLn−1(Z)⋉Zn−1, as U is now of the form⎛⎝

1 Rn−1

SLn−1(R)⎞⎠. But not

all hope is lost - by induction on n, these have nite measure. Integrating overbres turns the map Yn → Rn/0 into a map L1(Rn) = L1(Rn/0)→ L1(Yn).The Siegel transform becomes L1(Rn) → L1(Xn) and Minkowski (in higherdimension) can be applied.

Secondly, replacing Minkowski's theorem by a Poisson summation argumentyields an exact formula for the volume:

vol(SLn(R)/SLn(Z)) = ζ(2)ζ(3)⋯ζ(n)

when the normalization is taken so that f ↦ (Λ↦ ∑x∈Λ/0 f(x)) is an isometry

of L1(Rn) with L1(Xn), when Rn is equipped with the Lebesgue measure.Lastly, another proof for n = 2 could be given using hyperbolic geometry.

A fundamental domain for the action of Γ on the hyperbolic plane H2 can begiven by the black part of the following picture:

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when evaluating the measure of this domain, the integrand at height y turnsout to be 1

y2 , so even adding the green part (making it easier to integrate), the

total area turns out to be nite, meanign that the G-invariant measure on G/Γis nite. When writing a KAN decomposition for SL2(R), the domain withthe green part corresponds to KA

≥√

32

N[− 12 ,

12 ]. Borel and Harish-Chandra used

similar techniques to show that for a general semisimple algebraic group G overR, G(Z) is a lattice inside G(R). The exact area of the fundamental domaincould be calculated by calculating the two lower triangles of the fundamentaldomain (and of course, it is possible to calculate the exact integral by hand).

The Group SL2(Z)

We want to nd a nice fundamental domain for the action of SL2(Z) on SL2(R).We have the map G = SL2(R) → SL2(R)/SL2(Z) = X taking a unimodular basisto its Z-span. We wish to nd a section, associating a primitive basis to eachunimodular lattice Λ. Here is an algorithmic way to nd one:

Algorithm 2.4. Let Λ be a unimodular lattice in R2:

1. Let ` = minw∈Λ/0 ∣∣w∣∣. Out of all the vectors in Λ which has length `,let x be the one which has the smallest angle with the positive part of thex-axis.

2. Look at the set I = w ∶ det(x,w) = 1. This is a line of the formw0 + tx ∶ t ∈ R where w0 is the clockwise rotation of x by π/2, dividedby ∣∣x∥2. w0 is the closest point to the origin on this line.

3. Draw a line segment J on the line I, centered at w0, which has length`. Because x is a primitive vector, the line I must have a lattice pointon it, call it v. By the choice of x, there is at least one point out ofv +nx ∶ n ∈ Z which lies on J . By the minimality of `, either that there

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is a unique lattice point inside J which is in its interior, or that there areexactly two lattice points which form its boundary. Pick a lattice point z inthis segment J . Then x, z is a primitive basis for Λ, which is of positiveorientation.

By rotating and scaling, x can be mapped to (10). The point z must be

mapped to the black area in the picture above - it cannot be in the lowerhalf-plane because the orientation must be preserved even after the scaling androtation. It must lie outside the unit circle because of the minimality of `, andit's x value is in [− 1

2, 1

2] because of z being in the segment J , whose length is `.

More formally, we can identify all of the hyperbolic plane H2 with:

H2 =⎧⎪⎪⎨⎪⎪⎩f∈Hom(Z2,C)∶

orientation

³ ¹¹¹¹¹¹¹¹¹¹¹¹¹¹· ¹¹¹¹¹¹¹¹¹¹¹¹¹¹µdet(f)>0

⎫⎪⎪⎬⎪⎪⎭/ C×

°Rotation and Scaling

SL2(Z) acts on H2 by precomposition. The matrix (−1−1

) acts trivially

(as it just multiplies f by −1, which is insignicant as we divide by all complexscalars). This gives us an action PSL2(Z) H2. We identify H2 with the upperhalf-plane in the following manner - given f ∈ Hom(Z2,C) with det(f) > 0, we

multiply it by some element α of C× making αf((10)) equal to (1

0). We note

that maps in the same equivalnce class in H2 acts the same after multiplication

by this map-depending scalar, and we are allowed to identify [f] with αf((01)).

We can get any point in this way which lies in the upper half-plane (but not inthe lower half-plane, because f preserves orientation).

How does SL2(Z) act on H2 ≅ z ∈ C ∶ Im(z) > 0?

Exercise 2.5. Show that (a bc d

) acts by sending z to az+bcz+d

. The scaling 1cz+d

is the scalar α above.

By our discussion on primitive bases, the fundamental domain for the actionSL2(Z) H2 is the black domain:

14

Theorem 2.6. Γ = SL2(Z) is generated by S = (0 −11 0

) and T = (1 10 1

).

We will prove this fact based on the geometric picture above.

Exercise 2.7. Prove this fact by using the Euclidean algorithm for nding agreatest common divisor.

Proof. Because the black domain F in the picture is a fundamental domain forΓ H2, for each z ∈ H2 there is a unique γ ∈ Γ such that γz falls in F (Actually,there are two unique elements ±γ, but S2 = −Id, so we can always move by⟨S,T ⟩ from one to the other). Therefore it's enough to show that γ can bechosen to be inside ⟨S,T ⟩. Let us x z.

T acts on H2 by translating along horizontal lines: T (z) = z+1. We act on zby T,T −1 until it falls inside the strip Re ∈ [− 1

2, 1

2], and denote the resulting

point by z1. If z1 is in F , we won. Otherwise, we act on it by S, taking it toz1 = − 1

z1, i.e. we invert it with respect to the unit circle. We can now act on z1

by T,T −1 until it falls in this strip again, resulting in the point z2. Now repeatthis argument again and again. We need to show that this halts after nitelymany steps. Suppose that this is not the case.

To see this, we denote the b = Im(z) > 0. It's easy to see that for any n ≥ 1,Im(zn), Im(zn) ≥ b, meaning that all points zn lie inside the square z ∶∣Re(z)∣ ≤ 1

2, Im(z) ∈ [b,1]. This means that there are innitely many points

in the same Γ-orbit. However, if we look at the domains γFγ∈Γ intersectingthis square, one gets the picture

15

It's easy to see that there are only nitely many domains γFγ∈Γ whichintersect the square, so because F is a fundamental domain, there are onlynitely many points in each orbit which lie in this square, contradicting the factthat all zn lie in this square. Thus zn will eventually fall in F .

Exercise 2.8. Make this more explicit - Show that if y0 <√

32

then there is

some n ≥ 1 such that Im(zn) ≥ y0. Now show that there is some y0 <√

32

so that the domains F,S ⋅ F,T −1STS ⋅ F and TST −1S ⋅ F cover the squarez ∶ ∣Re(z)∣ ≤ 1

2, Im(z) ∈ [y0,1]. Identify these domains in the picture above.

Corollary 2.9. PSL2(Z) is generated by S and U = (0 −11 1

) = ST . Note that

U3 = 1 in PSL2(Z).

Theorem 2.10. PSL2(Z) = ⟨S⟩⋆ ⟨U⟩ ≅ C2 ⋆C3, meaning that there is no otherrelation but U3 = S2 = 1.

Proof. One needs to show that given any word in S,U which is of the formw = UεSU±SU±⋯SUη where ε, η ∈ ±1,0 which is non-trivial, it represents anon-trivial elements in PSL2(Z). To do this, consider the action of PSL2(Z)on P1(R):

16

S acts by z ↦ − 1z, so it sends maps 0 to ∞ and vice versa. Also, −1 is

mapped to 1. Thus (by continuity), the left half-circle is mapped by S to theright arc (S hits right).

U acts by z ↦ − 11+z

, making it act by 0 ↦ −1,∞ ↦ 0,1 ↦ − 12, so U sends

the right half-circle to the arc between −1 and 0 (in particular, it hits left).Similarly, U−1 takes the right half-circle to the arc between ∞ and −1, meaningthat it also hits left. Thus U,U−1 both send the right half-circle to the lefthalf-circle.

Now, suppose our word w starts and ends by U or U−1. We have the structurew = U±SU±SU±⋯SU±. If we look at w ⋅ 1, the U± hits it left, then S hits hitright, then U± hits it left, and this goes on. It ends with U±, meaning thatthe last hit was left, i.e. w ⋅ 1 is in the left half circle, unlike 1. Thus w ≠ 1in PSL2(Z). The other cases can be shown by conjugating the word to thisform.

Exercise 2.11. Use this to show that:

1. SL2(Z) ≅ C4 ⋆C2

C6.

2. [PSL2(Z), PSL2(Z)] is a free group of rank 2, generated by [S,U], [S,U−1],whose index in PSL2(Z) is 6.

3. [SL2(Z), SL2(Z)] is a free group of rank 2 whose index in SL2(Z) is 12.

17

3 Lecture 3 - Isometric Actions of Semisimple

Groups

Example 3.1. There is no continuous metric on R2/0 ≅ SL2(R)/U, invariantunder the SL2(R)-action.

Also, there is an essentially unique SL2(R)-invariant measure on H2 ≅SL2(R)/SO(2). Note that SO(2) is compact, but U is not.

Consider A ≤ G = SL2(R) to be the subgroup of all diagonal matrices. Thenthere is an A-invariant point in G/A, and:

Claim 3.2. (Mautner's phenomenon) Suppose G = SL2(R) acts on a metricspace X by isometries and that the map (g, x) ↦ gx is continuous. If x ∈ X isA-xed, then x is G-xed.

Proof. We'll show that x is U xed . We have that:

⎡⎢⎢⎢⎢⎣

λλ−1

⎤⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎣

1 t0 1

⎤⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎣

λ−1

λ

⎤⎥⎥⎥⎥⎦=⎡⎢⎢⎢⎢⎣

1 λ2t1

⎤⎥⎥⎥⎥⎦

λ→0→ Id

if u =⎡⎢⎢⎢⎢⎣

1 t0

⎤⎥⎥⎥⎥⎦∈ U , we take aλ =

⎡⎢⎢⎢⎢⎣

λ−1

λ

⎤⎥⎥⎥⎥⎦. Then ux = x because:

d(ux,x)x is Axed= d(ua−1

λ x, a−1λ x)

G acts by

isometries= d(aλua−1λ x,x)

λ→0→ d(x,x) = 0

so d(ux,x) = 0, meaning that x is U -xed. We therefore get a continuousG-invariant semi-metric on G/U dened by:

δ(gU,hU) = d(gx, hx)

by what we already saw in the rst lecture, this semi-metric must be trivial,implying that for any g ∈ G,

d(gx, x) = δ(gU, eU) = 0

meaning that x is g-xed. Therefore x is G-xed.

In general, the actions of SLn(R) must behave well:

Denition 3.3. A continuous map f ∶ X → Y between topological spaces iscalled proper if whenever K ⊆ Y is compact, f−1(K) is compact.

The continuous action GX is said to be proper if the map G×X →X ×Xtaking (g, x) to (x, gx) is proper.

Exercise 3.4. Show that if G X is a continuous action and G × X → Xdened by (g, x)↦ gx is proper, then G is compact.

We will prove that:

18

Theorem 3.5. Suppose that G = SLn(R) acts continuously by isometries on ametric space (X,d), without xed points. Then the action is proper.

Note that this result says something about general continuous actions byisometries, as one can just throw out the G-xed points XG.

Exercise 3.6. Let GX be a continuous action, and suppose both G and Xare metrizable. The following are equivalent:

1. The action is proper.

2. If K ⊆X is compact, the set g ∈ G∣gK ∩K ≠ ∅ ⊆ G is compact.

3. If xn → x in X and gn → ∞ in G, then gnxn → ∞ in X (we say thatαn →∞ if it has no converging subsequence).

Remark 3.7. The assumption that G and X are metrizable is not really needed,but then the third conditions needs to be expressed in the language of nets.

A Scientic Approach to the Mautner Phenomenon

Let G be a locally compact second countable group (from now on, this collectionof adjactives will be abbreviated to lcsc), and let an∞1 ⊆ G. We dene:

U+ = u ∈ G ∶ a−1n uan → e ⊆ G

U− = u ∈ G ∶ anua−1n → e ⊆ G

and

U0 =⎧⎪⎪⎨⎪⎪⎩u ∈ G ∶ both anua−1

n anda−1n uanare pre-compact

⎫⎪⎪⎬⎪⎪⎭⊆ G

Exercise 3.8. Show that U+, U− and U0 are all subgroups of G. Also, show thatU0 normalizes both U+ and U−, so ⟨U+, U−⟩ is a normal subgroup of ⟨U0, U+, U−⟩.

Also, if G = SLn(R), A is the diagonal subgroup of G and an ∈ A then⟨U0, U+, U−⟩ = G. Hint: let Eij be the n × n matrix dened by:

(Eij)k` =⎧⎪⎪⎨⎪⎪⎩

1 k = `, or (k, `) = (i, j)0 otherwise

compute anEija−1n and show that one can take a subsequence so that Eij ∈

U+ ∪ U− ∪ U0. By repeating this act n2 − n times, one sees that all of the Eij-sare in U+∪U−∪U0. Now show that the collection Eij generates all of SLn(R).

Actually, if an is bounded then U0 = G. Otherwise, by passing to a propersubsequence, we will get that all the Eij-s are in U+ ∪ U−. Because the o-diagonal entry in Eij can be replaced by any non-zero number without changingthe asymptotics of the an-dynamics, U+ ∪U− is non-compact. Thus ⟨U+, U−⟩ isa non-compact normal subgroup of ⟨U+, U−, U0⟩ = SLn(R), so because SLn(R)is nearly simple (see remark 12.12 on page 68), ⟨U+, U−⟩ = SLn(R) = G.

19

Claim 3.9. (The Moutner Phenomenon for SLn(R)): Suppose that G = SLn(R)acts continuously by isometries on a metric space (X,d). If x ∈X and anx→ x,then x is xed by ⟨U+, U−⟩.

Proof. if u ∈ U±,

d(ux,x)← d(ua±nx, a±nx) = d(a∓nua±nx,x)u∈U±

→ d(ex, x) = d(x,x) = 0

note that we have used that anx → y if and only if a−1n y → x (for y = x), which

only works because G acts by isometries. It is not true for a general topologicalspace.

This can be improved slightly:

Claim 3.10. If an → ∞ and anx → y, then there are bn ∈ ⟨aj∞j=1⟩ such thatbn →∞ and bnx→ x.

Proof. Note that a−1n y → x. Let us choose indices nk∞k=1 such that bk =

a−1k ank →∞, and observe that a−1

k ankx→ x.

Dynamics at Innity and the KAK decomposition

Claim 3.11. Every g ∈ GLn(R) can be written as kak′, when k, k′ ∈ O(n) and

a ∈⎧⎪⎪⎨⎪⎪⎩

⎛⎝

λ1

⋱λn

⎞⎠∶ λi > 0

⎫⎪⎪⎬⎪⎪⎭.

Proof. For g ∈ GLn(R), let g be the transpose of g. The matrix gg is positivedenite, so there is some k ∈ O(n) so that:

kggk−1 = positive diagonal =⎛⎝

µ1

⋱µn

⎞⎠

let us denote by a the diagonal matrix with entriesõ1, ...,

õn. Thus kgg

k−1 =a2 = aa, implying that gg = kaak. Let k′ = gk−1a−1. Then:

(k′) = (a−1)(k−1)ggg=kaak

= ((g)−1ka)

= akg−1 = (gk−1a−1)−1 = (k′)−1

meaning that k′ ∈ O(n) and that g = k′ak.

Corollary 3.12. Let G = SLn(R), K = SO(n) and A =⎧⎪⎪⎨⎪⎪⎩

⎛⎝

λ1

⋱λn

⎞⎠∶ λi >

0, ∏n1 λi = 1

⎫⎪⎪⎬⎪⎪⎭. Then G =KAK.

20

Proof. Take g ∈ G and write it as g = kak′ when k, k′ ∈ O(n) and a is somediagonal matrix. Taking determinants, we nd that:

1 = det(g) = det(k)det(a)det(k′)

because k, k′ ∈ O(n), we nd that det(a) ∈ ±1. But a is a diagonal ma-trix with positive diagonal entries, so only det(a) = 1 is possible. If det(k) =

det(k′) = 1 then we are done, otherwise replace k be k⎛⎝

−11

⋱1

⎞⎠and k′

by⎛⎝

−11

⋱1

⎞⎠k′. This will work as the diagonal matrix will commute with

a.

We are now ready to prove the theorem we stated before:

Theorem. Suppose that G = SLn(R) acts continuously by isometries on ametric space (X,d), without xed points. Then the action is proper.

Proof. Assume G×X →X×X is not proper. By one of the equivalent conditionswe had before, we nd some gn ∈ G and xn, x, y ∈X so that gn →∞, xn → x butgnxn → y. Writing gn = knank′n, by passing to a subsequence we may assumethat both kn, k

′n converge in K, say to k, k′ and that an →∞.

Thus ank′x→ k−1y. By a claim we had before, we can nd bn ∈ ⟨aj∞j=1⟩ ⊆ A

such that bn → ∞ and bnz → z for z = k′x. But then Mautner's phenomenongives that z is ⟨U+, U−⟩-xed, and the latter group is all of G, contradictingabsence of xed points in X.

Corollary 3.13. For every continuous action by isometries of G = SLn(R)with no xed points on a metric space X, all point stabilizers are compact (usethe second formulation of properness for K = x0). Actually, proper mapsare closed (exercise), so the orbits are closed. Moreover, the orbit space X/G isHausdor (exercise).

Remark 3.14. We acually need less than metric structure, but only uniformstructure. If (X,d) is a metric space, then the collection (x, y) ∈X2 ∶ d(x, y) ≤ε

ε≥0allows us to talk about notions as uniform continuity. A uniform structure

generalizes this. It is a collection of subsets of X ×X which allows intersection(taking minimum of nitely many ε-s) and if U is in the collection then there issome V in the collection so that V ⋅V = (x, z) ∈X2 ∶ ∃y, (x, y), (y, z) ∈ V ⊆ U(the ε/2 condition).

As an example, one can take a topological group G and dene the followinguniform structure:

⎧⎪⎪⎨⎪⎪⎩(x, y) ∈ G ×G ∶ xy−1 ∈ Ω

RRRRRRRRRRRe ∈ Ω ⊆ G is open

⎫⎪⎪⎬⎪⎪⎭

21

when we are given a uniform structure, we can talk about uniform continuity andequicontinuity. We actually proved that if G = SLn(R) acts equicontinuouslyon a uniform space X with no xed points, then the action is proper.

Corollary 3.15. If H is a Hausdor topological group and φ ∶ SLn(R) → His a continuous homomorphism, then the image is closed (as proper maps areclosed). This is done by consider the action of SLn(R) on the uniform space Hby g ⋅ h = hφ(g), which has no xed points if φ /≡ e.

Remark 3.16. Let us examine the proof (of the properness of actions theorem)once again. We only used that fact that there is some A ≤ G such that:

G =KAK for some compact set K ⊆ G.

If an∞1 ⊆ A, the by passing to a subsequence we can assure that U+ ∪U−is not compact, and ⟨U+, U0, U−⟩ = G.

In this generality, we get:

Theorem. If G is as above acts continuously and by isometries on a metric space(X,d), and that there is not point x ∈X which is xed by a non-compact normalsubgroup, then the action is proper.

These conditions are satised by many groups - all semisimple groups satisfythem, all tree group1 which act 2-transitively on the boundary, and all groupsacting strongly transitively on an ane building. All such groups also satisfythe corollary above.

1i.e., groups acting on regular trees

22

4 Lecture 4 - Metric Ergodicity

We start with an exercise:

Exercise 4.1. Prove the Howe-Moore theorem for G = SLn(R): if G → U(V )is a unitary representation of G and V G = 0 then for every v, v′ ∈ V ,

⟨gv, v′⟩ g→∞→ 0

Equivalent, gv converges weakly to 0 as g →∞. Hints: We know that stabilizersin V /V G = V /0 are compact. Show that if gn →∞ and gnx converges weaklyto y then the stabilizer of y′ is non-compact, when y′ = k−1y for some k ∈ K.Actually, if gn = knank′n then k is some partial limit of the kn-s, and the stabilizer

will be U(an)+ . You can use the fact that if xn is a bounded sequence in V , it

has a weakly converging subsequence.

Metric Ergodicity

We have basically seen:

Theorem 4.2. Let G = SLn(R) and let H ≤ G be non-compact. Any continuousG-invariant semimetric d on G/H is the trivial one.

Proof. Indeed, consider the equivalence relation on G/H dene by xH ∼ yH ifd(xH, yH). G acts on the metric space G/H/∼ continuously by isometries andtransitively, so this space can be identied with G/H for some H ≥ H. Thestabilizer of the basepoint eH is non-compact (as H isn't), meaning that thisaction has a xed point. Thus H = G and hence all points of G/H are identiedby ∼, meaning that d ≡ 0.

Corollary 4.3. Same H and G as above. Then every (Borel) measurable sep-arable2 G-invariant semi-metric d on G/H is trivial.

Proof. Let us assume WLOG that d is a metric (by divinding out 0-distantpoints). We'll show that d is actually continuous, implying the corollary. Denoteby τ the natural topology on G/H coming from G. We'll show that id ∶ (G/H, τ)→(G/H, d) is continuous, implying that d is continuous as it is the composition ofthe continuous maps:

(G/H × G/H, τ × τ)→ (G/H × G/H, d × d) d→ R

To show this, it's enough to show that the projection map π ∶ G → (G/H, d)is continuous, as it is H-invariant and the universal propery of (G/H, τ). It'senough to show that for any ε > 0, π−1(Bd(eH, ε)) contains an open set arounde in G, where Bd(x, δ) is ball around x of radius δ in the metric d.

2meaning that the semi-metric space (G/H, d) is separable

23

To show it, we cover G/H by countably many d-balls of radius ε2(which is

possible due to separability). Because the preimages of these balls under π coverG, one of these preimages will have positive (Haar) measure. Because π is G-equivariant and d is aG-invariant metric, g⋅π−1(Bd(xH, ε2)) = π

−1(Bd(gxH, ε2)),so the set A = π−1(Bd(eH, ε2)) has positive Haar measure.

Exercise 4.4. A = A−1, A⋅A ⊆ π−1(Bd(eH, ε)) and AA−1 = A2 contains an iden-tity neighborhood (see theorem 13.1 for proof of the latter part). In particular,π−1(Bd(eH, ε)) contains an open neighborhood of e.

Denition 4.5. Let X be a Lebesgue space (a separable metric space witha measure class), and suppose that G acts on X measurably, preserving themeasure class. We say that the action is metrically ergodic (and sometimesabbriviate to ME) if whenever U is a separable complete metric space on whichG acts continuously by isometries, every G-equivariant mapX → U is essentiallyconstant (and its image lies inside UG).

The previous corollary can be seen as:

Corollary 4.6. Let G = SLn(R) and H ≤ G be non-compact. Then G G/H isME.

Remark 4.7. Firstly, by taking U = 0,1 with the trivial action and the discretemetric, we get that ME implies ergodicity.

Secondly, one can restrict himself to considering metric spaces U with boundedmetric by replacing d by mind,1.Proposition 4.8. If GX is doubly ergodic (meaning that the diagonal actionGX ×X is ergodic), then X is ME.

Proof. Let φ ∶X → U be a measurable G-equivariant map to a separable metricspace on which G acts continuously by isometries. Consider the map:

X ×X φ×φ→ U ×U d→ [0,∞)

becuase d is G-invariant, this map is G-invariant, so by double ergodicity weconclude that it is essentially constant. Let α ∈ [0,∞) the essential image ofthis map. If α = 0 then the map φ is essentially constant, concluding the proof.

Else, α > 0, so the essential image φ(X) ⊆ U is discrete. U is separable, soφ(X) must be at most countable. We claim that in this case φ(X) must be asingle point, contradicting α > 0. Indeed, let µ be a probability measure on Xof the appropriate measure class, and let ν be φ⋆µ, which must be atomic as ameasure on a countbale set. If one pushes µ×µ along φ×φ he gets the measureν × ν on φ(X) × φ(X).

If we let ∆ ⊆ φ(X) × φ(X) be the diagonal, then ν × ν(∆) is positive as νis atomic. But ν × ν(∆) is equal to µ × µ((φ × φ)−1(∆)), and (φ × φ)−1(∆) isG-invariant. Thus its µ × µ measure is 1, so ν × ν(∆) = 1. Because ν is atomic,this is only possible if ν is a delta measure at some point, meaning that theessential image of φ is that point.

24

This is not a two sided implication (see exercise 13.10 on page 74 for an ex-plicit example). However, if the action GX is probability measure preserving(pmp for short), then GX is doubly ergodic if and only if it is ME. In fact,

Theorem 4.9. For a probability measure preserving action GX, the follow-ing are equivalent:

1. If G Y is ergodic and probability measure preserving then GX ×Y isergodic.

2. GX2 is ergodic.

3. GX is ME.

Remark. It will not be proved, but these are equivalent to G L2(X) havingno nite-dimensional invariant subspaces but const.s.

Denition 4.10. G X is called weakly mixing if the conditions above aresatised.

It reamins to prove (3)⇒ (1):

Proof. Suppose G X be ME and let G Y be ergodic and probabilitymeasure preserving. By Fubini:

L2(X × Y )G = L(X,L2(Y ))G

the action of G on L2(Y ) is by isometries, and it is separable (as Y is aLebesgue space, i.e. a separable metrizable space), so because G X is ME,L(X,L2(Y ))G is exactly L2(Y )G, which consists only of constants by ergodicityof G Y .

Theorem 4.11. Let G = SLn(R), and suppose G acts ergodically on X whilepreserving a probability measure. Then GX is ME.

Remark 4.12. Howe-Moore theorem actually implies that such actions are mix-ing.

Proof. (of the theorem) let φ ∶ X → U be a measurable G-equivariant mapfrom X to a separable metric space on which G acts continuously by isometries.We rst note that if φ hits UG, then its image must be contained by UG.Indeed, φ−1(UG) is G-invariant, meaning that if it has positive measure thenthe essential image of φ is contained inside UG (by ergodicity). If φ only hitspoints in UG, then the pushed forward measure on UG by φ is a 0,1-measure(by ergodicity), so because UG is nice enough (separable metric), this is a deltameasure somewhere, meaning that φ is essentially constant.

We now assume that φ misses UG. We can completely remove UG from U ,allowing the assumption that UG = ∅. Hence the action G U is proper. We'llshow that cannot be. Indeed, push the measure from X via φ to get a measureµ on U . We nd a compact subset K of U such that µ(K) > 1

2. For each g ∈ G,

25

the µ-measure of gK is > 12, and the total measure of U is 1, meaning that

gK ∩K ≠ ∅. Thus g ∈ G ∶ gK ∩K ≠ ∅ = G is not compact, contradictingproperness of the action G U .

Corollary 4.13. Let Γ ≤ G = SLn(R) be a lattice, e.g. Γ = SLn(Z). Theaction G G/Γ is ergodic and probability measure preserving, so it is ME.Taking Y = G/Γ, the action G Y is ergodic and probability measure preserving,meaning that GX ×Y = (G/Γ)2 is ergodic and probability measure preserving.Iterating this procedure gives that G (G/Γ)n is ergodic for all n.

Proposition 4.14. If G = SLn(R) acts ergodically on X while preserving aprobability measure, and if H ≤ G is a non-compact subgroup, then H actsergodically on X. In particular, replacing X by X2, gives that G X beingmetrically ergodic implies that H X is metrically ergodic.

Proof. We have that:

L2(X)Horbitmaps= L(G/H, L2(X))

G GG/His ME= L2(X)G = cosnt.s

Corollary 4.15. If Γ ≤ SLn(R) is a lattice and H ≤ SLn(R) is a non-compactsubgroup then H (G/Γ)n is ergodic for all n.

Can one reverse the roles of H and Γ?

Theorem 4.16. If SLn(R) = GX is ME, then ΓX is ME. In particular,Γ G/H is ME.

Proof. Suppose Γ acts on a metric space U continuously by isometries, and thatφ ∶ X → U is Γ-equivariant. We dene Φ ∶ G ×X → U by Φ(g, x) = φ(g1 ⋅ x).By Fubini, Φ is a map in X → LΓ−equiv(G,U) when the action on G is by rightmultiplication (check!). Also, Φ ∶ X → LΓ−equiv(G,U) is a G-equivariant map(when the action on G is from the left):

Φ(g′x)(g) = φ(g−1 ⋅ (g′x)) = φ((g′−1g)−1 ⋅ x) = Φ(x)(g′−1g)

Assuming the metric on U is bounded, there is a G-invariant metric onLΓ−equiv(G,U) - given two Γ-equivariant maps α,β ∶ G→ U , we have:

d(α(γx), β(γy)) = d(γα(x), γβ(y))Γ acts by

isometries= d(α(x), β(y))

,so the expression

D(α,β) = ∫G/Γ

d(α(x), β(x))dx

denes a metric (the L1-metric) on LΓ−equiv(G,U), which is G-invariant as themeasure on G/Γ is G-invariant. This makes LΓ−equiv(G,U) a metric space.

Now, Φ ∶ X → LΓ−equiv(G,U) is a G-equivariant map to a G-metric space,so it is essentially constant. This means φ is essentially constant.

26

Foundations of Ergodic and Measure Theory

Denition 4.17. A few denitions regarding Borel spaces:

A Borel space is a set equipped with a σ-algebra. A Borel action is anaction of a Borel group G (i.e. a Borel space with the structure of a groupon which multiplication and inversion are measurable maps. Usually, lo-cally compact topological groups) on a Borel space X, so that the mapsG ×X →X ×X is a Borel map, and G acts by elements of AutBorel(X).

A Borel space is called countably generated if the σ-algebra is generatedby a countable collection which separated to points, i.e., if Ann is thiscollection and x, y are two points in the underlying set, then:

n ∶ x ∈ An = n ∶ y ∈ An Ô⇒ x = y

A Borel space is called countbaly separated if there is a countable sepa-rating collection of measurable sets.

A Borel space is called standard if it is isomorphic (as Borel spaces) toa polish space (i.e., a separable completely metrizable topological space)with the Borel σ-algebra, i.e. the σ-algebra generated by the open sets.

Example 4.18. IfX = R and B = Borel(X) then the space is of course standard(as R is a polish space). It is also countably generated by (q1, q2)q1,q2∈Q. Thelatter is true when X is replaced by a general second countable topologicalspace. This shows that a standard Borel space is countably generated, and acountably generated Borel space is countably separated.

Fact 4.19. All uncountable polish spaces are Borel-isomorphic.

We will show something a bit weaker:

Theorem 4.20. Every countably generated Borel space with an atomless mea-sure class is isomorphic as such to [0,1] with the Lebesgue measure class.

Proof. (Sketch)Step 1: Map X to the Cantor set C = 0,1N in the following way - Let

Ann be a countable generating set for the σ-algebra. Dene φ ∶ X → 0,1Nby φ(x) = (χAn(x))∞n=1, which is measurable and also injective by separation.Push he measure class from X to C, so that the inverse map φ is dened almosteverywhere.

Step 2: Map C to [0,1] by passing from a trinary basis to a binary one:ξ((xn)∞n=1) = ∑∞

n=1xn2n. This is a measurable bijection, and one can push the

measure class to [0,1].Step 3: Normalize the measure class to be the Lebesgue one: Take a

probability measure µ on [0,1] of the appropriate measure class, and deneψ ∶ [0,1] → [0,1] by ψ(t) = µ([0, t]), which is measurable and invertible whendropping a null set.

Composing all of those together give the sought for isomorphism.

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5 Lecture 5 - Foundations of Ergodic Theory and

Measure Theory

Tame and Wild Dynamics

Denition 5.1. Let X be a Borel space.

Let ∼ be a measurable equivalence relation on X (measurability meansthat (x, y) ∶ x ∼ y ⊆ X ×X is measurable with respect to the productσ-algebra). We dene the quotient Borel space X/∼ to be the similarly-denoted quotient set, with the maximal σ-algebra so that the projectionmap X → X/∼, sending each x ∈X to its equivalence class, is measurable.Inparticular, if GX is a Borel action then one can dene X/G as X/∼, wherex ∼ y if x and y are in the same G-orbit of X.

A Borel action GX is called tame if the quotient space X/G is countablyseparated. Otherwise, is it called tame.

Note 5.2. If G X is a Borel action, we may map X/G into [0,1] injectively(using the process above), which gives a G-invariant map X → [0,1] for whichthe bre over each point of [0,1] is a G-orbit in X.

Example 5.3. A few examples of tame and wild dynamics:

Let U =⎧⎪⎪⎨⎪⎪⎩(1 t

1) ∶ t ∈ R

⎫⎪⎪⎬⎪⎪⎭and let X = R2. The action U X is tame.

Indeed, the space X/G is isomorphic (as Borel spaces) to R⊔R× - the copyof R coming from the x-axis and the copy of R× coming from everythingelse. Note that this is not an isomorphism as topological spaces, andactually X/G (as a topological space) is not even Hausdor. However, it isT1.

Let A =⎧⎪⎪⎨⎪⎪⎩(λ

λ−1) ∶ λ ∈ R⎫⎪⎪⎬⎪⎪⎭and let X = R2. The action A X is tame.

Indeed, the space X/G ≅ ⊔ ⊔ ⊔ (0,∞). Indeed, the rst point isthe origin, the second point is the image of the x-axis without the origin,the third point is the image of the y-axis when forgetting about the origin,and the copy of (0,∞) comes from the everything but the axes.

Let G = Q and X = R. The action GX by g ⋅ x = g + x is wild.

Let G = Z and X = S1. Take α ∈ R/Q and dene an action of G on X byn ⋅ x = e2πinαx, known as the irrational rotation. Then X/G is wild.

Let G = SL2(Z) and X = R2. The action of G on X is wild.

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Some Topology

Denition 5.4. Let X be a topological space.

A subset of X is called constructible if it is in the algebra generated bythe open (and closed) sets of X.

A subset of X is called locally closed if it is open in its closure.

Exercise 5.5. Let X be a topological space. Show the following facts aboutconstructible and locally closed sets:

1. For a subset A ⊆X, the following are equivalent:

(a) A is locally closed.

(b) A is the intersection of a closed an and open set in X.

(c) For every a ∈ A, there is some neighborhood Ua ⊆ A such that A∩Ua =A ∩Ua.

2. A subset of X is constructible if and only if it is the union of locally closedsets.

3. Show that if A ⊆ X is constructible then there is subset U ⊆ A which isopen and dense in A.

Remark 5.6. One can dene the notion of a constructible map between topo-logical spaces - A continuous map φ ∶ X → Y is constructible if the image of aconstructible set in X is a constructible set in Y .

An action G X is constructible if the map G ×X → X ×X, taking (g, x)to (x, gx), is constructible.

Exercise 5.7. Show that the orbits of a constructible action GX are actuallylocally compact.

Claim 5.8. Let X be a topological space and suppose that we have a continuousaction G X . Suppose that X is second countable and T0. If the orbits ofGX are locally closed, then the action GX is tame.

To see why this is true, we take a countable basis for X, so it's image underthe projection X → X/G separates points in X/G (exercise).

Proposition 5.9. Let X be countably separated. Every probability measure µon X which only takes the values 0 and 1 is a dirac mass at some point.

Proof. Take µ to be such a measure. Let An∞n=1 ⊆X be a countable collectionof measurable sets which separates the points of X. Let I = n ∈ N ∶ µ(An) = 1.Let B = ⋂n∈I An. Then:

B is non-empty as I is countable, implying that µ(B) = 1.

B cannot contain more than one point, as An∞n=1 separates the pointsof X.

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Thus B consists of one point, call it x0. We deduce by the rst point thatµ(x0) = 1. Because µ is a probability measure, we deduce that µ(X/x0) = 0.Therefore µ is a dirac mass at x0.

Corollary 5.10. Let X be a countably separated Borel space. Every probabilitymeasure is either a convex combination of other probability measures supportedon disjoint sets, or a dirac mass at some point.

Proof. If µ is measure which attains some value other than 0 and 1, we may takea subset A of X to be of measure 0 < µ(A) < 1. We can condition on A and on

X/A, i.e. dene µ1(E) = µ(E∩A)

µ(A)and µ2(E) = µ(E/A)

µ(X/A), both will be probability

measures on X, and:

µ = µ(A) ⋅ µ1 + µ(X/A) ⋅ µ2 = αµ1 + (1 − α)µ2

for α = µ(A).

Corollary 5.11. Let GX be a tame action. Then every ergodic proababilitymeasure on X is supported on a single G-orbit.

Proof. It's enough to show that every ergodic (with respect to the trivial actionG X/G) probability measure on X/G is a dirac mass somewhere. To show this,we note that every subset of X/G is now G-invariant, so ergodicity implies thatthe measure of every set is either 0 or 1. Now use the proposition above.

Note 5.12. Let X be a second countable T0 topological space, and suppose thatG X is a proper action. Then the orbits are closed, making G X a tameaction. In particular, any continuous action of a compact group is tame.

We will see later:

Theorem 5.13. (Chevalley) Let k be an algebraically closed eld. Every alge-braic action (over k) of an (ane) algebraic group on an algebraic variety istame. Actually, it is constructible.

Remark 5.14. This is also true for general local elds k, but requires heaviermachinery. In this setting, the theorem is called the Borel-serre theorem, andrequires the use of Galois cohomology. See the book by Platonov and Rapinchukfor more details.

Lebesgue Spaces

Denition 5.15. A Lebesgue space is a countably generated Borel space, en-dowed with a measure class (of a σ-nite measure, or equivalently, of a proba-bility measure).

Example. The basic example of such is R with the Lebesgue measure class onit.

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Morphisms between Lebesgue spaces are measure class-preserving Borel maps,which are dened up to null sets. The dual category to the category of allLebesgue spaces can be viewed as the category of commutative weak-* separa-ble von-Neumann algebras by the correspondence X L∞(X).

The collection of all morphisms between two Lebesgue spaces X and Y willbe denoted Maps(X,Y ) or MapsLeb(X,Y ).

For X a Lebesgue space and U a Borel space, we denote by L(X,U) thecollection of all Borel maps f ∶X → U which are dened up to null sets.

Roughly, ergodic theory is the study of Lebesgue actions: G is a locallycompact second countable group, X is a Lebesgue space, there is a homomo-prhism G→ AutLeb(X) so that G ×X →X ×X dened by (g, x)↦ (x, gx) is amorphism. One can consider the left action of G on G/H as an example.

Fact 5.16. Every Lebesgue space is isomorphic (as such) to one of the following:1, ..., n, N, [0,1], 1, ..., n ∪ [0,1] or N ∪ [0,1]. This will not be proved.

Proposition 5.17. Note that any map φ ∶X → Y between two Lebesgue spacescan be viewed as a continuous map between compact metrizable spaces.

Proof. Let us pick a countable generating collection Bn∞n=1 for Y , and considerthe embedding of Y into the cantor set Y → 0,1N dened by x↦ (χBn(x))∞n=1,where χE is the characteristic function of E.

Now, pick such a collection for X, and extend it by adding all φ−1(Bk)-s.Let An∞n=1 be the resulting collection. Embed X into the cantor set in thesame manner. The map φ now induces a map 0,1N → 0,1N which is theprojection on the indices n for which An is of the form φ−1(Bk), so the thefollowing diagram commutes:

X → 0,1N↓ φ ↓Y → 0,1N

Note that the same basic argument will work for countably many maps φ.

Example 5.18. The map φ ∶ [0,1]→ 0,1 dened by φ = χ[0,1/2] can be madecontinuous by cutting [0,1] at the point 1/2.

Remark 5.19. A compact metrizable model for a Lebesgue space X could bemade by choosing a weak-* dense norm-closed separable subalgebraA of L∞(X).Indeed, the weak-* denisty assures that A sees everything, and the separabilityassures that the spectrum of A is metrizable.

Roughly, if we have a map φ ∶ X → Y which needs to be realized as acontinuous map between metrizable spaces, we choose such a subalgebra forL∞(X), and then take its preimage under φ⋆ ∶ L∞(Y )→ L∞(X) to get one forY . Now taking the induced map between the spectrum does the job.

Theorem 5.20. (A version of Mackey's point realization theorem) Every Leb-segue action has a compact metrizable model.

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Proof. (Sketch)If G is countable, use the cantor set construction above with the φ-s being

the elements of G.Generally, we want to nd a G-invariant subalgerba A of L∞(X) which has

all the good properties above. We pick some subalgbera A′ which has theseproperties and dene A = L1(G) ⋆A′.

Corollary 5.21. In a Lebesgue action, stabilizers are always closed and if G isσ-compact, then orbits are measurable (as a countable union of compact sets).

We end this part with a theorem about integrationSuppose that π ∶ X → Y is any measurable map, and that we are given a

probability measure ν on Y . Suppose that there are probability measures µy on(ν-almost) every bre Xy = π−1(y). These measures together give a notionof integrating on X- if f ∶ X → R is a measurable function, we can dene it'sintegral as:

∫Y

⎡⎢⎢⎢⎢⎣∫Xyf(x)dµy(x)

⎤⎥⎥⎥⎥⎦dν(y)

this denes a measure on X just by integrating 1A for measurable sets A. Themeasure is denoted ∫ µydν(y).

Now, suppose we are given a measure µ on X, and we ask ourselves wetherit is possible to write it in this form, i.e. ∫ µydν(y).

Theorem 5.22. (Disintegration of measures) Let π ∶ X → Y be a continuousmap between compact metrizable spaces. Let µ ∈ Prob(X), and let ν = π⋆µ bethe pushed forward measure under π.

Then there is a map in L(Y,Prob(X)), taking y to µy ∈ Prob(X) such that:

µy is supported on Xy = π−1(y), i.e. µy(Xy) = 1.

µ can be represented as ∫ µydν(y)

Corollary 5.23. By existance of compact metrizable models, we conclude thatthis is true for any map π ∶X → Y between two Lebesgue spaces.

Proof. (Sketch)We have a map L1(X,µ)→ L1(Y, ν) which is dened by f ↦ g if π⋆(fdµ) =

gdν. We denote the image of f ∈ L1(X,µ) by f ∈ L1(Y, ν).Let fn∞n=1 ⊆ L1(X) be a dense countable collection of functions. Push it

forward to get a collection of functions fn∞n=1 in L1(Y ). After removing a null

set from Y , each y dened a functional on this collection by Iy(fn) = fn(y).This is to be thought of as the integral of fn on Xy with respect to the requestedmeasure µy.

Now one extends Iy from spanfn∞n=1 to the whole space using continuityand uses Riesz's theorem to get actual measures µy. The identity ∫ fdµ =

∫Y⎡⎢⎢⎢⎢⎣f(x)dµy(x)

⎤⎥⎥⎥⎥⎦dν(y) can be checked by hand, and the condition that µy(Xy) =

1 can also be easily checked, as the value of Iy(f) depend only on f(y).

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6 Lecture 6 - Ane Algebraic Geometry

Let k be an algebraically closed eld, which has a complete separable absolutevalue, the latter is required for talking about measures (and general analysis)on varieties over k. As examples, one can take k = C or k = Cp, the completionof the algebraic closure of Qp.

Denition 6.1. An ane k-algebra is is a nitely generated k-algebra withno nilpotents. Morphisms between such are just k-homomorphisms. An anevariety over k is an object in the dual category to the category of all anek-algebras. Morphisms between such are duals of morphisms between ane k-algebras. One can use the classical denition of an ane algberaic variety asthe zeros of some polynomials in the ane space An(k), and this identicationis given by Hilbert's nullstellensatz.

Given an ane k-algebra A, we dene X ∶= specm(A) =Homk−alg(A,k).

By the nullstellensatz, there is an injection3 A Func(X,k) dened inthe following way: if a ∈ A and x ∈ X, we dene a(x) = x(a) (recall that x is afunction A→ k). The image of A in Func(X,k) is denoted O(X) (or sometimesk[X]), and it is called the coordinate ring on X.

Endow X with two topologies:

The weakest making O(X) a collection of regular functions on X. This iscalled the Zariski topology, or Z-topology.

The weakest making O(X) a collection of continuous maps on X (whenk is taken with the absolute value topology). This is called the Hausdortopology, or H-topology.

of course, the Z-topology is weaker than the H-topology.

Remark 6.2. A restatement of the dention above - a map φ ∶ X → Y betweenvarieties is a morphism if φ⋆ ∶ O(Y )→ O(X) is a k-algebra homomorphism, i.e.pulled back regular functions are regular).

Exercise 6.3. Check that morphisms are continuous under both topologies.

So, why is A Func(specm(A), k) an injection?

We denote X = specm(A) = Homk−alg(A,k) for convenience. We need to showthat given some a ∈ A such that x(a) = 0 for every x ∈ Homk−alg(A,k), a mustbe the zero element in A.

To do this, pick some a ∈ A/0 and pick some maximal ideal m in A[ 1a].

Condsider the k-algebra homomorphism φ ∶ A→ A[ 1a ]/m sending some b to b+m.

The image is a eld in which a is invertible (as 1aexists), and therefore φ(a) ≠ 0.

If we somehow knew a priori that this eld A[ 1a ]/m was k, then we would have

done, as φ was in Homk−alg(A,k). All we know about this eld A[ 1a ]/m is that

it is a k-algebra, which is nitely generated as such (as A is nitely generated,

3The proof that this is indeed an injection will be displayed later

33

and we only throw in 1a). We claim that in this situation, A[ 1

a ]/m must be a eld.Indeed:

Lemma 6.4. (Zariski's lemma) an ane k-algebra which is a eld must beisomorphic (as elds) to k.

Remark 6.5. A similar situation arises in the theory of C⋆-algebras:

Theorem. (Gelfand) A commutative C⋆-algebra which is a eld must be C.

Corollary 6.6. There is a one-to-one correspondence between maximal idealsin an ane k-algebra A to k-algebra homomorphisms Homk−alg(A,k). Thesignicance of this statement is that the homomorphisms are into k and not toa general eld.

In order to prove Zariski's lemma, we'll need to work a bit. We start bydening the notion of integral extensions:

Denition 6.7. Let A B be an injection of ane k-algebras. It is called anintegral extension if B is nitely generated as an A-module.

Dually (for varieties), we say that the morphism specm(B) → specm(A) isnite.

We now prove a lemma:

Lemma 6.8. (Going up lemma) If A B is an integral extension and a is aproper ideal of A then aB, the ideal generated by a in B, is a proper ideal of B.

Proof. We assume that aB = B and show that 1 ∈ a, meaning that a = A.Because the extension is integral, we may nd some b1 = 1, b2, b3, ..., bn ∈ B suchthat B = ∑n1 Abi. By the condition aB = B, we may write each bi as a sum

∑n1 aijbj when aij ∈ a. Writing this in matrix notation:

(bi) = (aij)(bj)

if we denote the vector of the bi-s by b, we get that (aij − δij)b = 0.We want to conclude that det(aij − δij) = 0, as then we could expand using

the denition of the determinant, resulting in products between δ-s and a-s,only on of which excludes all elements from a. Therefore we get an equation ofthe form 1 + a = 0 for some a ∈ a, implying that 1 ∈ a. To indeed, show thatdet(aij − δij) = 0, we multiply the equation (aij − δij)b = 0 by the adjoint matrixto (aij − δij). We get that:

0 = adj(aij − δij) ⋅ (aij − δij) ⋅ b = det(aij − δij)b

but det(aij − δij) is some scalar in A, so reading the rst entry (and recallingthat b1 = 1), we conclude that det(aij − δij) = 0, and deduce that 1 ∈ a.

We can now prove Zariski's lemma:

34

Proof. Let A be a nitely generated k-algebra with no nilpotents. We can writeA = k[⋯]. Let X = x1, ..., xn be a maximal algebraically independent set(over k), and let Y = y1, ..., ym be algberaic over k[X] so that A = K[X,Y ](formally, we use Noether's normalization theorem).

Each yi satises some polynomial in k[X]. Multiplying by the least commonmultiple of the leading coecientsm we nd some 0 ≠ f ∈ k[X] such that all theyi-s satisfy monic polynomials over k[X][ 1

f]. Because A = k[X,Y ] we conclude

that A[ 1f] is nitely generated as a k[X][ 1

f]-module.

However, A is a eld and f ∈ k[X] ⊆ A is non-zero, so 1falready exists in A.

Hence the extension k[X][ 1f] A is integral. Because a commutative ring is a

eld if and only if its ideals are all trivial, the lemma implies that k[X][ 1f] must

be a eld itself. We'll show that it is impossible unless n = 0, i.e. that X = ∅and therefore A = k[Y ] is a nitely generated module over k[X] = k. Thus Ais a nite (hence algebraic) eld extension of k, but k was already algebraicallyclosed, therefore A = k.

Indeed, assume that n ≥ 1. Let a = ⟨f + 1⟩ be an ideal in k[X][ 1f]. If f ∉ k,

then f +1 has no inverse, making a a non-trivial ideal in k[X][ 1f], which implies

that latter cannot be a eld. However, if f ∈ k then a is a trivial ideal - it isequal to k[X][ 1

f]. However, in this case 1

f∈ k so k[X][ 1

f] = k[X], and k[X]

is not a eld as x1 ∈ k[X] has no inverse (x1 exists in k[X] as n ≥ 1). Thusk[X][ 1

f] is not a eld, and one proceeds as above to conclude that A = k.

Corollary 6.9. Every nite morphism Y → X of ane algebraic varieties issurjective. Note that this is not trivial, as the injection of the dual k-algebrahomomorpshim A B only implies that the image of Y →X is (Zariski) dense.

Proof. Assume φ ∶ Y → X is nite, i.e. that O(X) O(Y ) is integral. Letx ∈X be a point and we show that x has a preimage under φ.

Let m ⊲ O(X) be the maimal ideal associated with x. Let mO(Y ) be theideal generated in O(Y ). By the lemma, mO(Y ) is not all of O(Y ), so wecan nd some maximal ideal m′ ≤ O(Y ) containing mO(Y ). We pull it back toO(X) to get the ideal O(X) ∩m′this is a maximal ideal of O(X) (which is noteverything as 1 is not inside - it is outside m′), so it corresponds to some x ∈X.Thus φ(x) = y.

Exercise 6.10. Show that in any ane k-algebra:

⋂m maximal

m = ⋂p prime

p = 0

(as there are no nilpotents). Conclude whatever version of the nullstellensatzthat you know.

Products and Morphisms

When dening the notion of an algebraic group, we will need to as the multi-plication map G ×G → G to be a morphism. Before doing this, we must denea product in the category of ane algebraic varieties.

35

Denition 6.11. The category of ane k-algebras has a coproduct - namelythe tensor product:

A↓

A⊗B C↑ B

Exercise. Check that the tensor product of ane k-algebras is ane.

Dualizing this gives the product for ane varieties.

Exercise 6.12. Check that as sets,

specm(A⊗B) ≅ specm(A) × specm(B)

this is actually a homeomorphism for the Hausdor topology. However, this isfar from true when we turn our heades to the Zariski topology - for example, theZariski topology on C2 is very far from the product topology on (C, Zariski)×(C, Zariski).

Check that projectionsX×Y →X,Y are both open maps, when the topologyon both X and Y is either the Hausdor topology or the Zariski topology, whenthe topology on X × Y is the product topology.

This denition allows us to prove:

Theorem 6.13. Let φ ∶ X → Y be a morphism of ane varieties, both consid-ered with the Zariski topology. Then there exists a non-empty open subset W ofφ(X) which is open in φ(X).

Proof. First of all, we restrict the range of φ to assure that Y = φ(X). Henceφ⋆ ∶ O(Y ) O(X) is an injection.

As before, we nd U = (u1, ...., un) in O(X) which are algebraically indepen-dent over O(Y ), and V = (v1, ..., vm) in O(X) which are algebraic over O(Y )[U]so that O(X) = O(Y )[U,V ].

As before, we nd some f ∈ O(Y )[U] ⊆ O(X) such that O(X)[ 1f] is an

integral extension of O(Y )[U][ 1f]. In order to consider O(X)[ 1

f], we use the

process of localization (see page 86 for more on localization) and consider Xf ⊆X to be the set of points on which f does not vanish, i.e., Xf = f−1(k/0).This gives the following diagram:

Xopen

⊇ Xf

localization

= specm(O(X)[ 1

f])

nite

→φ⋆

specm(O(Y )[U][ 1

f]) = (Y ×An)

f

open

⊆ Y ×An

Let W be the projection of φ⋆(Xf) to Y . Then we have by a previouscorollary, and because the projection is an open map,

Xfsurjective→ (Y ×An)f

open

⊆ Y ×Ansurjective

↓ ↓W

open

⊆ Y

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but it is clear that W = φ(Xf) ⊆ φ(X), so we found a non-empy subset of φ(X)which is open in φ(X) = Y .

Denition 6.14. A topological space is Noetherian if every non-empty collec-tion of closed subsets has a minimum.

Fact 6.15. (Hilbert's Basis theorem) An ane k-algebra is Noetherian - everynon-empty collection of ideals has a maximum.

Exercise 6.16. Use the duality to show that every ane variety is a Noetheriantopological space, when equipped with the Zariski topology. Show that this isnot the case for the Hausdor topology.

We now state and prove a very potent theorem that will help us later. Wewill use the notion of constructible sets, on which we elaborate in page 83.

Theorem 6.17. (Chevalley-Tarski) The image of a morphism between alge-braic varieties is constructible in the Zariski topology (hence for the Hausdortopology).

Remark 6.18. Actually, every morphism is a constructible map (as a continuousmap between topological spaces). For further reading, view page 16.10.

Proof. Let φ ∶ X → Y be a morphism of ane varieties. Assume that φ(X) isnot constructible, and consider the collection:

Λ =⎧⎪⎪⎨⎪⎪⎩X ′ ⊆X closed

RRRRRRRRRRRφ(X ′) is not constructible

⎫⎪⎪⎬⎪⎪⎭

by assumption, Λ is a non-empty collection of closed sets in X. By the Noethe-rian property of X, we conclude that there is some minimum in this collection.Let X0 be this minimum.

By the previous theorem, using φ∣X0 , we can nd some non-empty subset W

of φ(X0) which is open in φ(X0). In particular, W is locally closed. But wecan write:

φ(X0) =W ∪ φ(X0/φ−1(W ))

when W is locally closed, and φ(X0/φ−1(W )) is constructible (as X0/φ−1(W )

is a closed set which is smaller than X0), making φ(X0) a union of two con-structible sets, hence itself constructible. This is a contradiction to the choiceof X0, implying that φ(X) must be constructible.

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Algebraic Groups

Denition 6.19. An (ane) algebraic group is a group object in the categoryof all ane varieties, that is, an ane variety G which has the structure of agroup so that the maps m ∶ G ×G → G dened by m(x, y) = xy and j ∶ G → Gdened by j(g) = g−1 are morphisms. We will omit the adjactive ane fromthe expression ane algebraic group from time to time, as we will not beinterested in non-ane ones.

An algebraic homomorphism between two algebraic groups is a group homo-morphism which is also a morphism between the underlying varieties.

Exercise 6.20. Let G be an algebraic group and let H ≤ G. Show that the

Zariski closure of H, HZ, is an algebraic group.

Corollary 6.21. If G is an algebraic group and H ≤ G is constructible, then itis closed.

Proof. By constructibility, there is some non-empty open set W of H which is

open in HZ. Thus H = ⋃h∈H hU is open in H

Z.

This implies that H is closed in HZ, as the complement is ⋃h′∉H h′H, which

is open. Thus H =HZand H is closed.

Corollary 6.22. If φ ∶ G → G′ is an algebraic homomorphism, then its imageis closed (as it is constructible.

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7 General Varieties

Up to now, we studied only ane varieties. However, it is not enough to fullylearn the theory of ane algebraic groups. For example, one would like toconsider coset spaces of algebraic groups. It turns out that if G is an algebraicgroup and H ≤ G is an algebraic subgroup then G/H can be given the structureof a variety, but it will not be ane in general. For example, let G = SL2(C)

and H =⎧⎪⎪⎨⎪⎪⎩

⎛⎝a b

a−1) ∶ a, b ∈ C, a ≠ 0

⎫⎪⎪⎬⎪⎪⎭, so that G/H can be identied with P1(C),

which is not an ane variety, although it locally looks like one.To generalize the theory of ane varieties, we study sheaves:

Sheaves of functions

The idea behind sheaves is to stitch local structure to a global phenomenon.

Example 7.1. Let X be a (locally compact, metrizable) topological space.For each open set U ⊆ X we associate the space C(U) of continuous functionsU → C.

Usually, C(U) is richer then C(X) itself - for example, x↦ 1xcan be found

in C(R/0) but not in C(R).Observe that if V U then we have the restriction operation ∣UV ∶ C(U) →

C(V ). Furthermore, if W V U then the restriction f ∣UW of a functionU → C is equal to f ∣UV ∣VW . We'll denote f ∣UV by f ∣V .

Furthermore, we require local behavior - if Uii is an open cover of someU ⊆ X and fi ∈ C(Ui) is a collection of function which are compatible whichother - fi∣Ui∩Uj = fj ∣Ui∩Uj for all i, j, then these functions can be glued together- there is some f ∈ C(U) such that f ∣Ui = fi for all i. This can be summarizedusing the following exact diagram:

C(U)(⋅∣Uk )→ ∏

k

C(Uk)(fi∣Ui∩Uj )

(fj ∣Ui∩Uj )

∏i,j

C(Ui ∩Uj)

satisfying this diagram is called the sheaf axiom.

This process can be thought of in a more general manner:

Denition 7.2. Let X be a topological space, and let Top(X) be the categoryof all open sets in X, morphisms being inclusions.

A pre-sheaf on X is a contravariant functor on Top(X).A sheaf is a pre-sheaf which satises that sheaf axiom.

We want to consider the space of regular functions on an ane variety, O(X),as some prototype of a sheaf. More specically, we need to talk about rationalfunctions on subsets of X (like 1

xon R/0). In order to do this properly, we

rst need to gure out when is O(X) a domain, to avoid expressions of the form⋆

0.

39

Denition 7.3. A topological space X is irreducible if whenever X =X1 ∪X2

for two closed sets X1,X2, either X =X1 or X =X2.

Exercise 7.4. Show that the following is true:

An irreducible Hausdor space is a singleton.

Every Noetherian space (i.e., a space in which every non-empty collectionof closed sets has a minimal element) is the union of nitely many closedirreducible subsets. Hint: take a minimal closed set which is not such aunion.

Show that irreducible spaces must be connected, but the other implicationis false.

Show that an algebraic group is connected if and only if it is irreducible(for the Zariski topology).

Conclude from the previous parts that if G is an algebraic group and G0

is the connected component of the identity, then G0 P G and that G/G0 isnite.

Show that for an ane variety X, X is irreducible if and only if O(X) isa domain.

From now on, unless stated otherwise, we assume our varieties are irre-ducible.

Now, let X be an ane variety, and consider O(X) = k[X], the space ofregular functions on X. It is a domain, so it has a eld of fractions, usuallydenoted k(X). We think of its elements as partially-dened rational functionson X.

Denition 7.5. Let X be irreducible, and let x ∈ X. Dene Ox =⎧⎪⎪⎨⎪⎪⎩

fg∈ k(X) ∶

g(x) ≠ 0

⎫⎪⎪⎬⎪⎪⎭⊆ k(X). These are the functions which are dened on x.

Theorem 7.6. Let X be an ane irreducible variety. O(X) = ⋂x∈X Ox, i.e. arational function which can be dened everywhere must be regular functions.

Proof. It's clear that O(X) ⊆ ⋂x∈X Ox.As for the other inclusion, let h ∈ ⋂x∈X Ox, meaning that for every point

x ∈ X, there is a representation h = fxgx

when gx(x) ≠ 0. Let I be the ideal

in O(X) which is generated by gxx∈X . We claim that I = O(X). If this isthe case, 1 ∈ I, so one can nd x1, ..., xn ∈ X and t1, ..., tn ∈ O(X) such that1 = ∑n1 tigxi . As a result,

h = h ⋅ 1 = h ⋅ (n

∑1

tigxi) =n

∑1

ti ⋅ (hgxi)h=fxgx=

n

∑1

tifxi ∈ O(X)

40

We now show that I = O(X). Assume this is not the case, meaning thatthere is some maximal ideal m in O(X) such that I ⊆ m. The ideal m isassociated with some point x0 ∈ X, meaning that f(x0) = 0 for all elements fof m. In particular, gx(x0) = 0 for all x ∈ X. This is absurd, as we have chosengx0 ∈ O(X) satisfying gx0(x0) ≠ 0.

Denition 7.7. We now dene the functor O on an ane variety X - if U ⊆X,we dene O(U) = ⋂x∈U Ox.

Exercise 7.8. Show that O ∶ Top(X)→ k − alg is a sheaf.

We want to dene morphisms between spaces with sheaves.

Denition 7.9. Let (X,CX) and (Y,CY ) be spaces with sheaves, both takingvalues in the same category. A morphism between them is a continuous mapφ ∶X → Y so that the induced map φ⋆ gives a natural transformation OY ⇒ OX .More concretely, for any open sets V ⊆ U ⊆ Y , we require that the followingdiagram will commute:

OX(φ−1(U)) ← OY (U)↓ ↓

OX(φ−1(V )) ← φY (V )

where the left arrows are induced by composition with φ and the down arrowsare restrictions.

Exercise 7.10. Show that a morphism of ane varieties is a morphism ofspaces with sheaves, when we equip an ane variety with the sheaf O.

Remark 7.11. A space with a sheaf is sometimes called a ringed space.

We are nally ready to dene the notion of a general variety:

Denition 7.12. A pre-variety X is a topological space with a sheaf C ∶Top(X) → k − alg, which is covered by nitely many by nitely many aneopen sets, i.e. open sets U which are isomorphic as ringed spaces to an anevariety, when endowed with the restriction of the sheaf.

Exercise 7.13. Show the following:

A pre-variety is a Noetherian topological space.

Products exist in the category of pre-varieties.

A topological space of X is Hausdor if and only if the diagonal ∆ =(x,x) ∶ x ∈ X is closed in X ×X, when the latter is endowed with theproduct topology

Dene an analouge for the Hausdor topology on a pre-variety. It need notbe Hausdor, but we shall continue to name it the Hausdor topology.

41

Denition 7.14. A variety X is a prevariety such that ∆ ⊆ X ×X is closed,when the latter is endowed with the Zariski topology.

Exercise 7.15. Show that if X is a variety, the Hausdor topology is actuallyHausdor.

Example 7.16. The following are examples of variety. The proof is left as anexercise:

Pn.

A closed subset of a variety.

Notation. Closed subsets of An are called ane varieties. Closed subsetsof Pn are called projective varieties.

Open sets of varieties are varieties.

Notation. Open sets of An are called quasi-ane varieties. Open sets ofPn are called quasi-projective varieties.

Locally closed subsets of varieties.

Denition 7.17. Let G be an ane algberaic group and let V be a variety.We say that G acts on V algebraically, or that V is a G-variety, if there is ahomomorphism G→ Autalg(V ), and the action map G×V → V is a morphism.

Example 7.18. The following are examples of algberaic actions:

GLn An.

GLn An/0.

GLn Pn.Exercise 7.19. Show that a morphism between general varieties is constructible.Use the fact that you already know this fact for ane varieties.

Corollary 7.20. If G acts on V algebraically, the action is constructible. Inparticular, the orbits are locally closed, and when considering the Borel structurearising from the Hausdor topology, the action is tame.

We now state a highly useful theorem about quotient of algberiac groups

Theorem 7.21. (Chevalley)Let G be an ane algebraic group, and let H ≤ G be an ane algbraic

subgroup. There exists a G-variety V and a basepoint x ∈ V with the followingproperties:

1. Set theortically, V ≅ G/H.

2. x is H-xed.

3. V has the quotient universal property - whenever G acts on a variety Uand y ∈ U is H-xed, there is a unique G-morphism V → U which maps xto y.

V is usually denoted G/H, and x is identied with eH.Furthermore, if H is normal in G then the variety G/H is ane.

42

Ergodic Theory on Varieties

Let G =⎧⎪⎪⎨⎪⎪⎩(1 t

1) ∶ t ∈ C

⎫⎪⎪⎬⎪⎪⎭, and let L = G(R) ≅ (R,+). Suppose G acts on a

variety V and let µ ∈ Prob(V )L, where V is equipped with the Borel structurearising from the Hausdor topology.

Claim 7.22. µ is supported on V G.

Note 7.23. The Zariski closure of L, LZ, is equal to G. Thus V L = V G.

Proof. Assume that µ(V /V G) is non-zero toward contradiction. V G is closed,so V /V G is also a G-variety. By conditioning on it, we get another measure(which we also denote by µ, by abuse of notation) to get that µ is supported onV /V G. That is, we reduce ourselves to the case in which V G = ∅, and need toshow that there is no L-invariant probability measure on V .

Let us disintegrate the measure µ over the (Borel) map V → V/G. By takinga generic bre, we nd a measure (which we again denote by µ), which issupported on a single orbit G/H. However, G ≅ C, meaning that it has noalgebraic subgroups but 0 and C itself. H cannote be the later, as V has notG-xed points, meaning that G/H ≅ C. The measure µ is an L ≅ R invariantmeasure on C, where the action is by addition. By again distinegrating (thistime with respect to the R-action), we may change µ to get a measure which issupported on a single orbit, R + iα for some xed real α. The action of R onthis line is by translations.

To conclude, we got a translation-invariant probability measure on a copy ofthe real line. It is well known that there are no such measures, meaning that wearrived to a contradiction. Thus µ(V /V G) = 0, and µ is supported on V G.

Example 7.24. Let L = SO(2) and G = SL2(C). There is an L-invariantmeasure on the G-variety C2/0 (the Lebesgue measure on some circle), eventhough there are not L-xed points in the variety. The reason for this is that Lis compact, and we'll elaborate on this subject next time.

43

8 Lecture 8 - Measures on Varieties

From now on, unless stated otherwise, we consider the Hausdor topology onvarieties.

Theorem 8.1. Let G be an ane algebraic group, V an algebraic variety andG V algberaically. Let µ ∈ Prob(V ), and let L = stabG(µ) ≤ G.

Then L is almost algebraic - there exists a normal subgroup H of L whichis algebraic in G such that L/H is compact. Moreover, µ is supported on V H .

Colloquially speaking, stabilizers of measures must be some compact averg-ing of a measure supported on some invariant points..

In order to prove the theorem, we rst state (and later prove) a proposition:

Proposition 8.2. Let G be an ane algebraic group, L ≤ G xed. Then one

can nd some normal algebraic subgroup H in the Zariski closure of L, LZ,

such that L has a precompact image in LZ/H. Moreover, if V is a G-variety and

µ ∈ Prob(V ) is L-invariant, then µ is supported on V H .

We now explain why does the proposition imply the theorem:

Proof. Fix an algebraic action G V and µ ∈ Prob(V ). Take L = stabG(µ).We nd a normal algberaic subgroup H ≤ L

Zas in the proposition, so that

µ(V H) = 1. Then µ is H-invariant, meaning that H ≤ L. Becuase H is normal

in LZ, it is also normal in the smaller subgroup L.

Finally, L is pre-compact modulo H, and L/H is closed in LZ/H (as L is closed

in G, and H ≤ L). Thus L/H is actually compact.

Before proving the theorem, let us make some corollaries of the proposition:

Corollary 8.3. Suppose SLn(C) acts algebraically on V , and let µ ∈ Prob(V )SLn(C).Then µ is supported on SLn(C)-xed points.

Proof. Let L = SLn(R), so that LZ = G = SLn(C). Let H P L

Z = G = SLn(C)be the subgroup given by the proposition. By almost simplicitly of SLn(C)(PSLn(C) is simple), we nd that H is either nite or H = G.

If H is nite, H is actually composed from elements ine 2πikn Id ∶ k =

0,1, ..., n − 1, so L/H = SLn(R)/H is not pre-compact. Thus H = G, and µ issupported on V H = V SLn(C).

Corollary 8.4. (Borel density theorem) If Γ ≤ SLn(R) is a lattice, then ΓZ =

SLn(C).Proof. Consider the action of SLn(C) on SLn(C)/ΓZ , which we want to show isa point. Consider the SLn(R)-equivariant map SLn(R)/Γ → SLn(C)/ΓZ inducedby inclusion. Push the SLn(R)-invariant probability measure from SLn(R)/Γ toSLn(C)/ΓZ = V , and denote the outcome by µ ∈ Prob(V ). Then µ is SLn(R)-invariant

By the previous corollary, the support of this measure is inside V SLn(C). But

V = SLn(C)/ΓZ has no SLn(C)-xed points unless ΓZis equal to SLn(C).

44

We now prove the proposition:

Proof. Let us x G an ane algebraic group and L ≤ G. Without loss of

generality, we may assume that LZ = G (as the proposition lives inside L

Z

anyway). Consider the collection:

Λ = H ≤ G algebraic∣ ∃L-invariant probability measure on G/H

Note that G ∈ Λ, meaning that Λ is a non-empty collection of closed sets (for theZariski topology, which is Noetherian) of G. Thus, by the Noetherian propertywe deduce that there is some minimal H0 ∈ Λ. Let ν be some L-invariantprobability measure on G/H0.

Part 1: We claim that H0 is normal in LZ = G.

Assume not, and let N = NG(H) be the normalizer of H inside G, which isa proper subgroup of G. Let φ ∶ G ×G → G be dened by φ(x, y) = y−1x, andlet U ⊆ G/H0 × G/H0 be the image of φ−1(G/N) under the projection to H0.

Exercise 8.5. Show that if (xH0, yH0) is an element of U , then y−1x ∉ N .Conclude that the stabilizer of (xH0, yH0) in G/H0 × G/H0 is a proper subgroupof some conjugate of H0.

Now consider the L-invariant probability measure ν×ν on Prob(G/H0×G/H0).The set supp(ν × ν)

Zis L×L invariant, and Z-closed (i.e., closed in the Zariski

topology), meaning that the set must be L ×LZ = G ×G invariant, which is anon-empty subset of G/H0 × G/H0. Thus ν × ν is supported on a Z-dense (i.e.,dense in the Zariski topology) set inside G/H0×G/H0. In particular, U has positivemeasure as it is open.

By disintegrating the G action we nd some ν ∈ Prob(G/H) which is sup-ported on some orbit coming from U , i.e. H is a proper subgroup of a conjugateof H0. This measure is of course L-invariant, as the original measure was L-invariant and the G-orbit is preserved under the L-action. This contradicts the

minimality of H0 in the collection Λ. This implies that H0 is normal in G = LZ .Part 2: We claim that L is precompact modulo H0.Let us recall where we arrived - we had ν ∈ Prob(G/H0) which is L-invariant.

Let L be the closure (in the Hausdor topology) of the image of L in G/H0

(We basically replace G by G/H0). Because the action of G on Prob(G/H0)is continuous (when equipping G and G/H0 with the Hausdor topology andProb(G/H0) with the weak-* topology), stabilizers are closed. Thus ν is L-invariant.

Consider the action of L on G/H0 from the left. This action of the closedsubgroup L ≤ G/H0 on the topological group G/H0 is tame - orbits are copies of L,hence closed. Therefore, by disintegrating the L-action, we get an L-invariantprobability measure ν which is supported on an L-orbit on G/H0, i.e. a coset of

45

L. We identify the latter with L itself, giving an L-invariant probability measureon L. Thus L is compact.4, which means that L/H0 is precompact.

Part 3: We need to show that if G acts algebraically on some variety V andµ ∈ Prob(V ) is L-invariant, then supp(µ) ⊆ V H0 . To do this, we rst show thisin a specic case:

Claim. If V H0 = ∅, then there are no L-invariant probability measures on V .

Proof. Suppose there are such measures, and let µ be an L-invairant probabilitymeasure on V . Consider the L-invariant probability measure µ× ν on V ×G/H0.By disintegrating it with respect to with respect to the G-action, we nd someL-invariant probability measure on one orbit G/H, where H is the stabilizer ofsome point in V × G/H0.

Exercise 8.6. Use the fact that V H0 = ∅ to show that the stabilizer H is aproper subgroup of a conjugate of H0.

This contradicts the minimality of H0. Thus there are no L-invariant prob-ability measure on V .

We now return to our general setting. Suppose G acts algebraically on avariety V , and let µ ∈ Prob(V ) be L-invariant. If µ(V /V H0) is non-zero, we cancondition the measure µ on V /V H0 to get an L-invariant probability µ measure

on V /V H0 . Note that because H0 is normal in LZ = G, V H0 is G-invariant, so

U = V /V H0 is a G-variety without any H0-xed points. The previous claim nowgives that there are no L-invariant measures on U = V /V H0 , meaning that wearrived at a contradiction. Thus µ(V /V H0) is zero, and µ is supported on V H0

. This concludes the proof of the proposition.

We now claim:

Claim 8.7. If V is a G-variety, L ≤ G and µ ∈ Prob(V ) is L-invariant so thatµ × µ is L-ergodic, then µ is a dirac mass at some point in V L.

The main theorem will basically help us reduce to the case in which L iscompact. This calls for a lemma:

Lemma 8.8. Let K be a compact group which has a Lebesgue action on aLebesgue space X. If µ is a K-invariant probability measure on X, and µ×µ isergodic on X2, then µ is a dirac mass at some point in XK .

Note 8.9. Every ergodic probability measure under the action of K is supportedon a single orbit - this is because the pushed measure on X/K is a 0,1-measure,and the space X/K is nice enough - it is countably separated. See 5.9 for moredetails.

We now prove the lemma:

4We know this if L was locally compact. A theorem by Weyl says that a group whichsupports a measure invariant under the left action can be given a locally-compact structure.This is needed in the case in which the eld is not locally compact, as Cp.

46

Proof. By the note, it's enought to consider the homogenous case - X = K/M forsome M ≤ K. Suppose that µ ∈ Prob(K/M) is K-invariant. There is a uniqueK-invariant measure on K/M, namely the Haar measure. Thus µ = HaarK/M ,and µ is fully supported.

It follows that µ×µ is fully supported on K/M ×K/M, but ergodcity (and thenote), it is supported on a single orbit. Thus M =K, and µ is a dirac mass.

We are now ready to prove our claim - if V is a G-variety, L ≤ G , µ is anL-invariant probability measure on V which is doubly L-ergodic, then µ is adirac mass at some point in V L.

Proof. Without loss of generality:

1. LZ = G, as the claim focuses on what happens inside L

Z.

2. V = V H0 , where H0 is in the previous theorem. This is ne as µ issupported only there.

3. H0 = e. This can be done by replacing G by G/H0, which is ne as theaction of H0 on V is now trivial.

4. L is compact. Indeed, L is pre-compact in G/H0 = G, and the closure of L

also stabilizes µ and acts doubly ergodic (and V L = V L).

This means we arrived exactly to the setting of the lemma, meaning that weare done.

47

9 Lecture 9 - Algebraic Repesentations of Er-

godic Actions

Denition 9.1. Let S be a locally compact second countable group, and sup-pose G acts ergodically on a Lebesgue space X. Let G be some ane algebraicgroup (over a xed eld, which is algebraically closed and complete with respect

to some absolute value), and x some Sρ→ G which is an algebraic representation

of S, i.e. a continuous homomorphism whose image is Z-dense.An algebraic representation of X (or to be more exact, of the action S X

given ρ) is composed of three components:

An algebraic variety V .

An algebraic action G V .

A ρ-equivariant Borel map Xφ→ V which is dened up to null sets.

The resulting commuting diagram is:

Sρ→ G

ÿ ÿX

φ→ V

where only V , the action G V and the map φ are metamorphosing.

A morphism of two algebraic representations Xφ→ V and X

ψ→ U is a G-morphism V → U so that the following diagram commutes (up to null sets):

Xφ→ Vψ

↓U

Basically, this is a way to represent the ergodic theoretical being S X asan algebraic creature.

Exercise 9.2. Let S be as above and let T ≤ S be a closed subgroup. DeneX = S/T , and consider the left action of S on X. Given G and ρ, show that for

any algebraic representation Xφ→ V , the following diagram commutes:

X → G/ρ(T )Z

φ

↓ ∃!V

Theorem 9.3. Let G = SLn(C) and S = SLn(R). Pick some lattice Γ ≤ S,and consider the inclusion Γ

ρ G, which has a Z-dense image by Borel densitytheorem.

48

Let T ≤ S be some non-compact closed subgroup, and consider the left action

of Γ on X = S/T . Then any algebraic representation S/T =X φ→ V is actually S-equivariant.. In particular, the exercise implies that for any such representation,the following diagram holds:

X → G/ρ(T )Z

φ

↓ ∃!V

Proof. Let us recall claim 8.7 - if G V is an algebraic action and L ≤ Gstabilizes some probability measure µ ∈ Prob(V ), and µ × µ is L-ergodic, then

µ is a dirac mass at some point in V L = V LZ

. This gives a corollary, which isof similar, but indepedent interest:

Corollary 9.4. Let S = SLn(R) and T ≤ S be closed and non-compact. Pick

a lattice Γ ≤ S. If there is an algebraic representation Tρ→ G, then every ρ-

algebraic representation of the T -space X = S/Γ is essentially constant:

X → ⋆ ;

V

This corollary is true as on can push forward the S-invariant probabilitymeasure from S/Γ to V , giving a T -invariant probability measure which is doublyT -ergodic (as T (S/Γ)n is ergodic for all n).

To now prove the theorem, we use the duality trick that was employedin the rst chapter - if we had a Γ-equivarinat map φ ∶ S/T → V , we x a T -invariant base-point x in S/T and dene Φ ∶ S/Γ → V by Φ(sΓ) = s ⋅ φ(s−1x). Itturns out that Φ is actually well dened, and furthermore T -equivariant. Forexplicit calculations showing this, retrace to the proof of theorem 1.10.

Now, given a Γ-equivariant map φ ∶ S/T → V we dualize and get a mapΦ ∶ S/Γ→ V which is T−equivariant. But the latter must be essentially constantby the corollary, meaning that the former must also be essentially constantbecause of the relation Φ(sΓ) = s ⋅ φ(s−1x).

In the last three few examples we saw a reoccuring theme - the category ofall representation has an initial element5. Moreover, it is of the form G/H forsome algebraic subgroup H of G (in the case of a single point, H = G). As onewould might expect, this repeating phenomenon is more than just chance:

Theorem 9.5. Let S X be an ergodic action, G is some algebraic group, andρ is an algebraic representation of S into G.

5An initial element in a category C is an object X in the category, so that for any givenobject Y in the category, there is a unique morphism X → Y . For example, this is the emptyset in the category of sets and the trivial group in the category of groups.

49

Then there exists some algebraic subgroup H0 ≤ G and an algebraic repre-

sentation Xφ0→ G/H0 which is an inital object in the category of algebraic repre-

sentations of X, meaning that for any algebraic representation Xφ→ V , we have

the diagram:

Xφ0→ G/H0

φ

↓ ∃!V

The algebraic representation Xφ0→ G/H0 is called the (algebraic) gate of X.

Before proving this theorem, we show a simple but sometimes powerful ap-plication:

Proposition 9.6. Let S,X,G be as above, and denote by AutLebS (X) the group

of all S-equivariant Lebesgue automorphisms of X. Denote by AutAlgG (G/H0) thegroup of all G-algebraic automorphisms of G/H0.

Exercise. Show that AutAlgG (G/H0) is isomorphic to NG(H0)/H0, where NG(H0)is the normalizer of H0 in G.

Then there is a homomorphism ρ⋆ ∶ AutLebS (X) → AutAlgG (G/H0) with thefollowing property - if ω ∶ X → X is a Lebesgue automorphism, then φ0ω =ρ⋆(ω)φ0.

Proof. Given such ω, we consider the representation φ0ω ∶ X → G/H0. By ini-

tiality of φ0 ∶X → G/H0, there is a unique ξ ∈ AutAlgG (G/H0) such that φ0ω = ξφ0.The uniqueness assures that ρ⋆(ω) ∶= ξ denes a homomorphism (check!). Dia-gramatically, we got:

Xφ0→ G/H0

ω ↓ Ç ∃!

Xφ0→ G/H0

We now prove the theorem:

Proof. We recall an ace we have up our sleeve - the minimality trick. Let

Λ =⎧⎪⎪⎨⎪⎪⎩H ≤ G algebraic

RRRRRRRRRRR∃ an algebraic representation X → G/H

⎫⎪⎪⎬⎪⎪⎭

the collection Λ is non-empty, asH = G is inside. By the Noetherian property, wend some minimal element H0 in Λ. Let us pick some algebraic representationφ0 ∶ X → G/H0. We claim that this is an initial object in the category. We

need to show that given any algebraic representation Xφ→ V , there is a unique

morphism G/H0 → V in the category of representations.

50

Exercise. Show uniqueness. For this part, you do not need the minimality ofH0, but only the fact that two G-equivariant maps on a G-homogenous spacewhich agree on a single point must coincide.

We now show existance. Given φ and φ0, we can dene the product repre-

sentation Xφ×φ0→ V × G/H0:

X ↓

V ← V × G/H0 → G/H0

by S-ergodicity of X and the tameness of the action G V ×G/H0, we concludethat the support of the image of X in V × G/H0 must consist of a single orbit,call it G/H, meaning that we have a representation φ ∶ X → G/H through whichX → V × G/H0. By composing with the projection V × G/H0 → G/H0, we get aG-morphism G/H → G/H0.

However, even when forgetting about the algebraic structure for a moment,this must imply that H is a subgroup of some conjugation of H0 (by consideringstablizers - check!). But minimality of H0 must imply equality, so the mapG/H → G/H0 was actually an isomorphism. By considering the inverse of it, weget the following partial diagram:

G/H↓

V ← V × G/H0G/H0

which gives a morphism G/H0 → V in the category of representations.

Theorem 9.7. Let S = SLn(R), Γ ≤ S a lattice, and T ≤ S an amenable non-

compact subgroup (e.g., T =⎧⎪⎪⎨⎪⎪⎩

⎛⎝

1 t⋱

1

⎞⎠

⎫⎪⎪⎬⎪⎪⎭≅ (R,+)). Let G = SLm(R). Assume

ρ ∶ S → G is an algebraic representation of S and that ρ(Γ) is unbounded.Consider the left action ΓX ∶= S/T . Then the gate of X is non-trivial.

Proof. Consider the compact G-variety Pm−1(C) ≅ G/Q, where Q =⎧⎪⎪⎨⎪⎪⎩

⎛⎝λ v0 A

⎞⎠

RRRRRRRRRRRλ ≠

0, v ∈ Cm−1, A ∈ GLm−1(C), det(A) = λ−1

⎫⎪⎪⎬⎪⎪⎭.

Let C = Prob(G/Q). This is a compact convex space (with the weak-* topol-ogy, coming from C(G/Q)) on which Γ acts by ρ. We claim that there is aΓ-equivariant map X → C6

Indeed, we rst nd a Γ-equivariant map S → C. To do this, we take afundamental domain F ⊆ S for the left Γ-action on G, meaning that S = ΓF .

6The following couple of paragraph are exactly the proof that the action Γ S/T isamenable whenever Γ and T are closed and T is amenable.

51

We map F arbitrarily to C (say, by a constant map) and extend the map toall of S by forcing it to be Γ-equivariant. This shows that MapsΓ(S,C), thecollection of all Γ-equivariant (Borel, dened up to null sets) maps S → C, isnon-empty.

Consider the action of T on MapsΓ(S,C) by the right action of S. Thecollection MapsΓ(S,C) is compact and convex (it lies inside L∞(S,C), as C isbounded, and we can take the weak-* topology from L1), so amenability of Timplies that there is a T -invariant function inMapsΓ(S,C), i.e. a Γ-equivariantmap X = S/T → C.

The theorem now follows from the following proposition:

Proposition 9.8. Suppose Γ X is metrically ergodic, G is simple, and ρ ∶Γ→ G is Z-dense and unbounded. If Q is a proper algebraic subgroup of G andthere is a Γ-equivariant map X → Prob(G/Q), then there exists some algebraicsubgroup N ≨ G and a Γ-equivariant map (an algebraic representation) X → G/N.

Proof. Let H be minimal in the collection

Λ =⎧⎪⎪⎨⎪⎪⎩H ≤ G algebraic

RRRRRRRRRRR∃ a Γ-equivariant map X → Prob(G/H)

⎫⎪⎪⎬⎪⎪⎭

Q ∈ Λ and Q is a proper subgroup of G, so H ≠ G. Let φ ∶ X → Prob(G/H) beΓ-equivariant.

Exercise. Show that H ≠ e. To apply this theorem for G = SLm(C), showthat H cannot be nite (as only PSLm(C) is simple). You can argue in thefollowing way:

Build aG-invariant metric on Prob(G/H) so that the actionG Prob(G/H)is continuous. Look for the Wasserstein metric.

Assuming that there is a Γ-equivariant map X → Prob(G/H), use metricergodicity to deduce that Prob(G/H)Γ is non-empty.

Use the techniques from yesterday (specically, proposition 8.2 or anothertheorem of similar essence) to show that Prob(G/H)Γ is empty if H isnite.

Let N = NG(H), so that N ≨ G (by simplicity). Consider the Z-open set Uin G/H × G/H dened by

U = (xH, yH) ∶ y−1x ∉ N

which is the collection of all the elements in G/H×G/H whose stabilizer is strictlysmaller than some conjugate of H. Denote φ(x) = µx.Claim 9.9. for almost every x ∈X, the set U is µx × µx-null

52

Proof. Colloquially speaking, this is true by disintegrating and the minimalityof H. However, we need to push the envelope a bit.

The map x ↦ µx × µx(U) is Γ-equivariant (as x ↦ µx is), so the ergodicityof X implies that this is equal to a constant. Assume this constant is non-zero.By conditioning on U , we have a map X → Prob(U), denoted x ↦ λx. Wecan look at the map Prob(U) → Prob(U/G). Because X is ergodic and U/G iscountably separated, the image of X in Prob(U/G) is a single measure ν (as thecomposition X → Prob(U/G) is Γ-equivairant).

There is a map X → Maps((U/G, ν), P rob(U)) which is dened by disinte-gration of λx with repsect to the G-action. By Fubini, it's the same as a map

X × U/G → Prob(U). Each y ∈ U/G gives a map Xy→ Prob(U). Actually, the

process of disintegration assures that the image of this map is inside Uy, whichis the bre over y in U → U/G.

But that's just a G-orbit in U , which is isomorphic to G/H′ where H ′ is

strictly smaller than some conjugate of H. We thus get a map Xy→ Prob(G/H′),

contradicting the minimality of H.

We now complete the proof of our earlier proposition. We want to build aΓ-equivariant map X → G/N, where N is the normalizer of H.

Note. A measure µ on a Lebesgue space Z is a delta measure at some point ifand only if µ × µ() = 1, when ∆ is the diagonal. Actually, the measure of thediagonal is non-zero if and only if µ has an atom.

The image of X → Prob(G/H × G/H) misses U . Considering the map G/H ×G/H → G/N × G/N dened by dividing out N , the preimage of ∆ ⊆ G/N × G/N is(xH, yH) ∶ y−1x ∈ N - the complement of U . Composing with this map, weget a map X → Prob(G/N ×G/N) which only hits ∆, meaning that the measuresof the image are delta measures at points in G/N. This gives a map X → G/N.This map is Γ-equivariant as the original map X → Prob(G/H × G/H) was Γ-equivariant, and all the actions we took preserved this fact.

The proof is nally complete.

53

10 Algebraic Bi-Representations of Bi-Actions and

Margulis Super-Rigidity

Our ultimate goal is:

Theorem 10.1. (A special case of Margulis Super-Rigidity)Let S = SL3(R), Γ ≤ S a lattice and let G = PSLn(C) be a simple group.

Take some Zariski-dense unbounded continuous homomorphism ρ ∶ Γ→ G.Then there is a continuous homomorphism ρ ∶ S → G such that ρ∣Γ = ρ. (Ac-

tually, the demand ρ(S)Z= G implies that n = 3. Furthremore, ρ is continuous,

which turns out to force ρ to be algebraic).

Remark 10.2. The high-rank property of SL3(R) will be reected by havingthe following subgroups:

T1 =⎧⎪⎪⎨⎪⎪⎩

⎛⎝

1 t1

1

⎞⎠

⎫⎪⎪⎬⎪⎪⎭, T2 =

⎧⎪⎪⎨⎪⎪⎩

⎛⎝

1 t1

1

⎞⎠

⎫⎪⎪⎬⎪⎪⎭, T3 =

⎧⎪⎪⎨⎪⎪⎩

⎛⎝

11 t

1

⎞⎠

⎫⎪⎪⎬⎪⎪⎭

T4 =⎧⎪⎪⎨⎪⎪⎩

⎛⎝

1t 1

1

⎞⎠

⎫⎪⎪⎬⎪⎪⎭, T5 =

⎧⎪⎪⎨⎪⎪⎩

⎛⎝

11

t 1

⎞⎠

⎫⎪⎪⎬⎪⎪⎭, T6 =

⎧⎪⎪⎨⎪⎪⎩

⎛⎝

11t 1

⎞⎠

⎫⎪⎪⎬⎪⎪⎭each of them is non-compact, closed and amenable, and all togehter they gen-erate SL3(R) and commute in successive pairs.

We've seen:

Theorem 10.3. Using the notations above, and for every i = 1, ...,6, there is aproper algebraic subgroup Hi =H ≤ G and an algebraic representation

Γρ→ G

ÿ ÿS/Ti

φ→ G/H

Equivalently, we have a repesentation

Γ × Tiρ×τ→ G × e

ÿ ÿ

Sφ′→ G/H

where Ti acts from the right and τ is the trivial homomorphism. This is, how-ever, not the most eective way to represent the Γ × Ti action - it is possibleto act on G/H from the right by the group NG(H)/H. In general, replacing thegroup e by something larger will make H become smaller and smaller, andwhen it'll become miniscule, or even better, trivial, the map φ′ will become anapproximation for ρ.

Note that in this discussion Γ is xed, but Ti might vary, so their roles areasymmetric.

54

Denition 10.4. (A bi-representation of a bi-action)Suppose Γ, T are locally compact second countable groups acting commuta-

tively on a Lebesgue space X, so that the action Γ × T X is ergodic. Let Gbe a (xed) algebraic group and let ρ ∶ Γ → G be a (xed) Z-dense continuoushomomorphism. A bi-representation of the bi-action Γ × T X is composed ofthe following:

An algebraic group L and a continuous Z-dense homomorphism θ ∶ T → L.

A G ×L-variety V .

A ρ × θ-equivariant map Xφ→ V .

For convenience, we often denote the bi-representation alone by V and denotethe cooresponding accessories by LV , φV and θV .

We claim that these form a category, so we need to dene a morphismbetween two such bi-representations. Let U,V be two bi-representations of the

bi-action Γ×T X. A morphism between the two is a G×L-morphism Uχ→ V

where L = θU × θV (T )Z≤ LU × LV acts on U and V by the projections on LU

and LV .

Remark 10.5. This denition of a morphism might seem unorthodox, in thesense that it will be more intuitive to seek for a homomorphism λ ∶ LU → LVand demand a G × λ-equivariant morphism betweem U and V . The reason wedon't use this dention (beside relative eectiveness of the former) is that onewould sometimes like to dilate LU , resulting in more freedom when acting onV .

Theorem 10.6. Let S = SL3(R), Γ ≤ S a lattice and T = Ti for some i = 1, ...,6.Consider the action of Γ × T on X = SL3(R) by left and right multiplication,correspondingly. Let ρ ∶ Γ → G = PGLn(C) be a Z-dense continuous homomor-phism.

There exist an algebriac subgroup H ≤ G and some L ≤ NG(H)/H alge-braic, some Z-dense continuous homomorphism θ ∶ T → L and an algebraic

bi-representation Xφ0→ G/H0 which is the initial object in the category of bi-

representations of Γ × T X.

Proof. The proof will be extremely similar to the one we had for algberaicrepresentations (instead of bi-representations), but will require an extra step.

We shall use the minimality trick. Let

Λ =⎧⎪⎪⎨⎪⎪⎩H ≤ G algebraic

RRRRRRRRRRR∃ an algebraic bi-representation X → G/H

⎫⎪⎪⎬⎪⎪⎭

the collection Λ is non-empty, asH = G is inside. By the Noetherian property, wend some minimal elementH0 in Λ. Let us pick some algebraic bi-representationφ0 ∶ X → G/H0, and let L0 and θ0 be the corresponding algebraic group and

55

morphism. L0 can be mapped into NG(H0)/H0 so that only the kernel of theaction on G/H0 is mapped trivially, as the action of L0 on G/H0 commutes withthe left G action, hence comes from a right action. We thus can assume thatL0 ≤ NG(H0)/H0 We claim that this is an initial object in the category. We need to

show that given any algebraic representationXφ→ V , there is a unique morphism

G/H0 → V in the category of bi-representations. We head for existance.

Given φ and φ0, we can dene the product bi-representation Xφ×φ0→ V ×G/H0:

X ↓

V ← V × G/H0 → G/H0

G ×LV ← G ×L → G ×L0

where L = θV × θ0(T )Z≤ LV × L0. By S-ergodicity of X and the tameness of

the action G×L V ×G/H0, we conclude that the support of the image of X inV × G/H0 must consist of a single orbit, call it G×L/M.

Here we stumble upon the main issue with this argument as it stood before- we need to get a G-homogenous variety to use the minimality of H0, whichallows us to reverse the arrow G×L/M → G/H0 and conclude the existance of amorphism as we have done in the past.

Lemma 10.7. If we have a bi-representation

Γ × T ρ×θ→ G ×Lÿ ÿX

φ→ G×L/M

of the bi-action Γ × T X, then G×L/M is G-transitive.

Exercise 10.8. Complete the proof of the theorem given the lemma.

We now prove the lemma:

Proof. Let us mod out the G-action:

S =X φ→ G×L/M¿ ↓ /G

L/projL(M)

the resulting map X → L/projL(M) is Γ-left equivariant, meaning that we have amap Γ/S → L/projL(M). This means that we found a representation of the actionT Γ/S:

Tθ→ L

ÿ ÿΓ/S

φ→ L/projL(M)

56

If we push the invariant probability measure on Γ/S to L/projL(M), we get andoubly L-ergodic (and even doubly T -ergodic) measure on L/projL(M), so claim8.7 implies that the image of Γ/S is a single point in (L/projL(M))T , so it must

be θ(T )Z= L-xed. But L/projL(M) contains L-invariant points if and only if it

is a single point.To conclude, we shouwed that G×L/M/G = L/projL(M) is a single point, hence

G×L/M is G-transitive.

We are now ready to prove Margulis Super-Rigidity:

Proof. Recall that S = SL3(R), Γ ≤ S is a lattice, G = PSLn(C) and ρ ∶ Γ → Gis a continuous Z-dense homomorphism.

We take T = Ti for i = 1, ...,6. Fix an initial element in the category ofbi-representations of Γ × Ti X = S:

Xφi→ G/Hi

Γ × Ti

ρ×θi→ G ×Li

where Hi is a proper subgroup of G and Li ≤ NG(Hi)/Hi. Let us denote NG(Hi)by Ni.

Observe that T1 and T2 commute, so Γ×T1×T2 X - Γ acts from the left andT1, T2 both act from the right. Thus, for each t2 ∈ T2 we get a Γ×T1-equivariant

Lebesgue automorphism of X. Thus, as we saw yesterday, by initiality of Xφ1→

G/H1, we get:

Xφ1→ G/H1

t2 ↓ Ç ∃!

Xφ1→ G/H1

let us denote the G-morphism G/H1 → G/H1 by θ′2(t2). This gives us a homomor-

phism T2

θ′2→ AutAlgG (G/H1) ≅ N1/H1. Let L′2 be the Z-closed of its image.Note that

Xφ1→ G/H1

Γ × T2

ρ×θ′2→ G ×L′2is a bi-representation of the bi-action Γ × T2 X. We conclude (by initiality)that there is a morphism G/H2 → G/H1 in the category of bi-representations:

Xφ2→ G/H2

φ1

∃!G/H1

57

in particular, this is a G-morphism of varieties G/H2 → G/H1. Note that theroles of the indices 1 and 2 here are symmetric (all we used is the facts thatT1 and T2 commutes), so we have G-morphisms G/H1 G/H2. This impliesthat G/H1 ≅ G/H2 as G-spaces. The roles of the indices 1 and 2 can be replacedby any two successive indices, so we nd that all of the G-spaces G/Hi areisomorphic. This implies that up to conjugation all of the subgroupsHi coincide.Because conjugation does not change G/H, we can choose all Hi-s to be equalto each other, and denote them all by H, and similarly, which in turn makesthe G-morphisms G/H1 G/H2 to be the identity, implying that all the mapsφi ∶X → G/H are the same:

Xφ2→ G/H2

φ1

∃!G/H1

we denote them all by φ.We denote the (identical) groups Ni = NG(Hi) by N . We end up with the

following diagram:

Xφ→ G/H

Γ × Ti

ρ×θi→ G ×Liwhere Li ≤ NG(H)/H = N/H. Consider the right action of N/H on G/H and mod itout:

Γ × Tiρ×θi→ G ×Li

ÿ ÿX

φ→ G/HÇ ↓

G/N

the resulting map X → G/N is both Γ-equivariant and Ti-invariant, so it mustsplit through X/⟨Ti⟩ = S/⟨Ti⟩ = ⋆:

Γ × Tiρ×θi→ G ×Li

ÿ ÿX

φ→ G/H↓ ↓

⋆ Γ-map G/N

the image of the point in G/N must be a Γ-invariant point, hence ρ(Γ)Z-invariant.

But Borel density theorem implies that ρ(Γ)Z= G, so G/N contains a G-invariant

point. Thus N = G.However, N = NG(H) is the normalizer of H. As we said in the begining,

H = Hi is a proper subgroup of the simple group G. Thus N = G must imply

58

that H = e, and N/H = G. We end up with the following diagram:

Xφ→ G

Γ × Ti

ρ×θi→ G ×Li

where Li ≤ G. In some sense, each θi is a fragment of the homomorphism

Sρ→ G that we look for. We would like to glue all of the shards to get this

actual homomoprhism. There is more than one way to do it, and we choose touse function algebras.

Consider the following diagram:

L(X) φ⋆← O(G)

Γ × Tiρ×θi→ G ×Li

the map φ⋆ is injective as φ(X)Z= G. To see why this is true, we note that the

essential image φ(X) is ρ(Γ)-invariant, so the Z-closure φ(X)Zis a non-empty,

ρ(Γ)Z= G-invariant subset of G, so it must equal to G itself. The injectivity

of φ⋆ allows us to consider O(G) as a sub-algebra of L(X), invariant under theaction of all Ti-s. This means that it is a ⟨Ti⟩ = S-invariant subalgebra of L(S).

This means that G × S acts on O(G) by algebra automorphisms, where Gacts on G the left and S acts on the right. This means that G × S acts on thespectrum specm(O(G)) ≅ G in the same manner. The right action of S gives a

homomorphism Sθ→ G, and we have the following diagram:

S =X φ→ G

Γ × S ρ×θ→ G ×G

now one sets ρ = θ, and checks that ρ∣Γ = ρ. This completes the proof.

59

Part II

Tutorials

11 The Haar Measure

The Haar Measure

Denition. Let G be a topological group and let B the Borel σ-algebra on G.A (non-zero) measure µ on (G,B) is called a (left) Haar measure if:

1. µ is left invariant - if g ∈ G and E ∈ B then µ(gE) = µ(E).

2. µ is nite on compact sets - if K ⊆ G is compact then µ(K) <∞.

3. µ is inner regular - If U ⊆ G is open then:

µ(U) = supµ(K) ∶ K ⊆ U compact

4. µ is outer regular - if E ∈ B then:

µ(E) = infµ(U) ∶ U ⊇ E open

(I will call properties 2-4 the regularity properties below). The reasonone wants such regularity properties is that it makes is possible to integratecontinuous functions (with compact support) against µ. Another way to thinkof these is by functionals. We seek for a linear functional I ∶ Cc(G) → R whichis invariant under the left action of G, and is continuous with respect to thetopological structure on Cc(G), that is, for any compact set K ⊆ G there issome constant C = CK > 0 such that for any f ∈ Cc(G) such that supp(f) ⊆K,∣I(f)∣ ≤ C ∣∣f ∣∣. The reason for this equivalence is a theorem by Riesz:

Fact 11.1. (Riesz) Let X be a Hausdor LC topological space. There is a 1-1 correspondence between positive measures on X which satisfy the regularityproperties and linear functionals on Cc(X) which are positive (that is, if f is anon-negative function then I(f) ≥ 0). This correspondence takes a measure µto the functional I(f) = ∫ fdµ.

Theorem 11.2. Let G be a LC group. Then there is a left Haar measure on G.Furthermore, if µ,µ′ are two such measures then µ′ = aµ for some a > 0 (Theuniqueness will be shown later for the general case of G-invariant measures onG/H).

Example 11.3. .

If G = Rn then the n-dimensional Lebesgue measure is a left Haar measureon G.

60

If G is a discrete group then µ(A) = ∣A∣ (the cardinality of A) is a leftHaar measure on G.

If G = Q with the topology coming from R (Making it not LC!), thereare no left Haar measures on G. Indeed, if we suppose such measure µexists then by G-invariance µ(x) = µ(y) for any x, y ∈ G. Now, forany E ∈ B we have:

µ(E) = µ⎛⎝ ⋃x∈E

x⎞⎠

Eis at mostcountbale= ∑

x∈E

µ(x) = ∑x∈E

µ(0) = µ(0) ⋅ ∣E∣

this means that up to a scalar multiplication, µ is the counting measure.However, this measure in this situation is not outer regular - every openset in Q is an intersection of an open set in R with Q and therefore innite,meaning that:

µ(0) = infµ(U) ∶ U ⊇ E open = infµ(0)∣U ∣, U ⊇ 0 open =∞

contradicting the compactness of 0.

In the exercise sheet there will be more examples of such.

Remark 11.4. One can dene a Right Haar measure using the same demands,but replacing left-invariance by right-invarince: µ(Eg) = µ(E) for all E mea-surable and g ∈ G.

Exercise 11.5. Let µ be a measure on G. µ is a left Haar measure if and onlyif ν dened by ν(E) ≜ µ(E−1) = µ(x−1 ∶ x ∈ E) is a right Haar measure.

The Modular Character

Fix a LC group G and a left Haar measure µ on it. We know that µ is invariantunder left translations, but what about right translations?

Denition 11.6. Take some g ∈ G and look at ν dened by ν(E) = µ(Eg).One notices that ν is also a left Haar measure - all the regularity demands arenear obvious to check (one only uses that F is open/compact i Eg is), andonly left invariance remains. For that we notice that the right multiplicationand left multiplication commute - h(xg) = (hx)g - meaning that (hE)g = h(Eg)and thus:

ν(hE) = µ((hE)g⎞⎠= µ

⎛⎝h(Eg)

⎞⎠

µ is left

invariant= µ(Eg) = ν(E)

this means by the theorem above that for some scalar α we have ν = αµ. Wedenote α = ∆(g). We get a map ∆ ∶ G → (0,∞), which is a homomorphismbasically by:

∆(g1)∆(g2)µ(E)R is commutative

= ∆(g2)∆(g1)µ(E) = ∆(g1)µ(Eg1) =

61

= µ((Eg1)g2) = µ(E(g1g2)) = ∆(g1g2)µ(E)

the map ∆ is called the modular character or modular function.Note that by the denition, for all E measurable such that µ(E) > 0 and all

g ∈ G,

∆(g) = µ(Eg)µ(E)

Proposition 11.7. ∆ is a continuous homomorphism G→ (0,∞) and for everyf ∈ L1(G) = L1(µ),

∫ f(xy)dµ(x) = ∆(y)−1 ∫ f(x)dµ(x)

Proof. We rst prove the integral formula - for f = χE we have:

∫ f(xy)dµ(x) = ∫ χE(xy)dµ(x) = ∫ χEy−1(x)dµ(x) = µ(Ey−1) = ∆(y−1)µ(E) = ∆(y−1)∫ f(x)dµ(x)

so our formula works for characteristic function. But every function in L1 can beapproximated (in that sense) by linear combinations of characteristic functions,thus the formula is valid for all f .

The reason for continuity is that the map y ↦ Ry−1f = f(xy−1) is continuousin L1, and if ∫ f ≠ 0 then

∆(g) = ∫f(xg−1)dµ(x)∫ fdµ

As a conclusion, we have:

Theorem 11.8. Let G be a locally compact group. Suppose that one of thefollowing is true:

1. G is abelian.

2. G is compact.

3. G is countable

4. G = [G,G] (e.g., G is simple).

Then G is unimodular

Proof. 1 is clear as left and right multiplication are the same. 3 is also clear asthe Haar measure in that case is just the counting measure. 4 is true as ∆ is ahomomorphism into an abelian group, making [G,G] be in its kernel.

As for 2, Consider ∆(G) = ∆(g) ∶ g ∈ G ⊆ (0,∞). Because ∆ is ahomomorphism, this is a subgroup of (0,∞). Because ∆ is continuous andG is compact, ∆(G) must also be compact. The only compact subgroup of(0,∞) is 1, meaning that ∆(g) = 1 for all g, and the left Haar measure rightinvariant.

62

Measures on Homogenous Spaces

Let H ≤ G be a subgroup. We ask ourselves when do we have a G-invariantmeasure on G/H. We rst show:

Theorem 11.9. If there are two G-invariant measures on G/H, then they arethe same up to a multiplicative constant.

Proof. Take two such measures µ and ν0, and dene ν = µ + ν0. Obviously,µ≪ ν, so by the Radon-Nikodym theorem we nd f = dµ

dνsuch that fdµ = dν.

Fix g ∈ G and push both measures by the map g ∶ x ↦ gx. By assumption,they are invariant, and we have that:

fdµ = dν = g⋆(dν) = g⋆(fdµ) = f g−1 ⋅ g⋆(dµ) = f g−1dµ

but the Radon Nikodym derivative is unique (up to measure zero), so f = f g−1,making f a G-invariant function on G/H. Such function must be a constant, andtherefore fdµ = dν means that ν = αµ for some constant α > 0 (as both µ, ν arenon-zero measures), resulting in ν0 = (α − 1)µ, proving the claim.

It turns out that it is not always true that there is a G-invariant measure onG/H.

Theorem 11.10. (Unproven) Let H ≤ G be a closed subgroup. There is a G-invaraint measure on G/H if and only if ∆G∣H = ∆H , where ∆G,∆H are themodular characters of G,H. In this case, we have the following Fubini-esqueequality: for every f ∶ G→ R which is continuous and has compact support:

∫Gf(x)dµG(x) = ∫

G/H

⎡⎢⎢⎢⎢⎣∫Hf(aξ)dµH(ξ)

⎤⎥⎥⎥⎥⎦dµG/H(aH)

up to a multiplicative factor which does not depend on f .

Remark 11.11. This condition on the modular characters is essential, as onesees by taking x ↦ f(xy) for y ∈ G and f ∈ Cc(G) xed and using change ofvariables.

Corollary 11.12. Let G = SL2(R) and Γ = SL2(Z). Then there is a G-invariant measure on G/Γ. This is true as both ∆G and ∆Γ are identically one.

Exercise 11.13. .

1. Use this theorem to show that if K ≤ G is compact then G/K has a G-invariant measure on it.

2. Let G = SL2(R) and P =⎧⎪⎪⎨⎪⎪⎩

⎡⎢⎢⎢⎢⎣

a ba−1

⎤⎥⎥⎥⎥⎦

RRRRRRRRRRRa ∈ R/0, b ∈ R

⎫⎪⎪⎬⎪⎪⎭. Show, either by

using the theorem or arguing directly, that G/P has no invariant measureson it.

63

We conclude this part with another useful Fubini-esque Theorem:

Theorem 11.14. Let L ≤ H ≤ G be LC groups and suppose that there areinvariant measures on G/H,G/L,H/L. Denote by µG/H , µG/L and µH/L these mea-sures. Then the equality:

∫G/L

f(xL)dµG/L(xL) = ∫G/H

⎡⎢⎢⎢⎢⎣∫H/L

f(abL)dµH/L(bL)⎤⎥⎥⎥⎥⎦dµG/H(aH)

is true for all f ∈ Cc(G/H) up to a multiplicative factor which is independent off , but only on the choice of the measures.

Proof. It's enough to show that

f ↦ I(f) = ∫G/H

⎡⎢⎢⎢⎢⎣∫H/L

f(abL)dµH/L(bL)⎤⎥⎥⎥⎥⎦dµG/H(aH)

gives a G-invariant linear functional on G/H which is positive - that is, if fis a non-negative function then I(f) ≥ 0 (as by Riesz's theorem, every suchfunctional gives a measure on G/H, which will again be G-invariant, so it's ascalar multiplication of G/L).First, I is obviously a linear functional and if f ≥ 0then for any aH ∈ G/H,

∫H/L

f(abL)dµH/L(bL) ≥ 0

, so integrating this over G/H gives I(f) ≥ 0. By Riesz's theorem, I is given byintegration against some measure ν on G/H, and we would like to show that it isa G-invariant measure. To show this, it's enough to show that I is G-invariant.

Let g ∈ G, and let ψ(xL) = f(gx)L. Then:

I(ψ) = ∫G/H

⎡⎢⎢⎢⎢⎣∫H/L

ψ(abL)dµH/L(bL)⎤⎥⎥⎥⎥⎦dµG/H(aH) = ∫

G/H

⎡⎢⎢⎢⎢⎣∫H/L

f(gabL)dµH/L(bL)⎤⎥⎥⎥⎥⎦dµG/H(aH)

change variables a′H = gaH, which is allowed as µG/H is G-invariant, to get that:

= ∫G/H

⎡⎢⎢⎢⎢⎣∫H/L

f(a′bL)dµH/L(bL)⎤⎥⎥⎥⎥⎦dµG/H(a′H) ≜ I(f)

thus I is G-invariant, and the measure ν is a G-invariant measure on G/L. Thismeans that µG/L = ν (up to a multiplicative constant), implyig that for allf ∈ Cc(G/L):

∫G/L

fdµG/L = ∫ fdν ≜ ∫G/H

⎡⎢⎢⎢⎢⎣∫H/L

f(abL)dµH/L(bL)⎤⎥⎥⎥⎥⎦dµG/H(aH)

up to the same multiplicative constant.

64

A Word on Measure Classes

Denition 11.15. Let X be a set and B a σ-algebra on it. A σ-ideal J ⊆ B isa subcollection satisfying:

1. ∅ ∈ J .

2. If A1,A2, ... ∈ J then ⋃∞1 Ai ∈ J .

3. If A ∈ J and B ∈ B then A ∩B ∈ J .

Example 11.16. Given a measure µ on (X,B), we dene the σ-ideal Jµ = A ∈B ∶ µ(A) = 0. This is called the σ-ideal of null sets for µ.

Denition. Two measures on the same Borel space (X,B) are said to be of thesame measure class if they share the same null sets. The measure class [µ] of ameasure µ is the collection of all measures which are in the same class of µ.

Note 11.17. We might not have a general way to dene a G-invariant measureon G/H, but one can always dene a measure class on G/H which is G-invariant.To do this, one takes the measure class given by the Haar measure on G andpushes it by the projection map G → G/H. That is, E ⊆ G/H will be null if andonly if EH ⊆ G is null.

Remark 11.18. We cannot do something similar to build a G-inv. measure onG/H as the Haar measure pushed by the projection map would result in a G-measure on G/H but the regularity assumptions might not be fullled. As anexample, take G = R2 and H = R, so G/H ≅ R. Denoting the 2-D Lebesguemeasure by V ol, we get that the measure µ on G/H will satisfy:

µ([0,1]) = V ol(R × [0,1]) =∞

although [0,1] is compact.

65

12 Simplicity of PSLn and embedding free groups

Simplicity of PSLn(k)Let k be a eld and V be a vector space over V . Let P(V ) be the projectivizationof V :

P(V ) = V /0/k× = V /0/v∼λv

for n ≥ 1, we dene the n-dimensional projective space over k to be:

Pn(k) = P(kn+1)

Exercise 12.1. Convince yourselves that P1(R) can be identied with S1, andthat P1(C) can be identied with S2.

We will denote the equivalence class of v ∈ V /0 by [v].Now, GLn(k) acts on kn, and it preserves this relation (as these are linear

maps). Thus the action descends to GLn(k) Pn−1(k). Note that scalarmatrices in GLn(k) act trivially: if g = λId then:

g([v]) = [λv] = [v]

we quotient out the scalars from GLn(k) to get a group PGLn(k), which actson Pn−1(k).

Denition 12.2. Let PSLn(k) be the image of SLn(k) under GLn(k) →PGLn(k).

Theorem 12.3. PSLn(k) is a simple group in each of the following cases:

1. n ≥ 3.

2. n = 2 and k ≠ F2,F3.

Remark 12.4. One can show computationally that PSL2(F2) ≅ A3 and PSL2(F3) ≅A4. These are not simple groups.

We now prove the theorem:

First Part - Actions on Pn−1(k).

Proposition 12.5. The group G = PSLn(k) acts 2-transitively on X = Pn−1(k).

Proof. We prove for n = 2, and leave the general case as an exercise. Let[u], [v] ∈ Pn−1(k) be dierent points. We pick the representatives u and v.Because u, v are not colinear, α = det(u, v) ≠ 0. Let g be the matrix ( 1

αu, v).

It's determinnt is 1, and:

g ⋅ (1

0) = 1

αu ∼ u, g ⋅ (0

1) = v

meaning that [g] ∈ PSL2(R) takes the standard basis to ([u], [v]). This provesthe claim.

66

Claim 12.6. The action of G = PSLn(k) on X = Pn−1(k) is faithful (i.e., onlythe group identity acts trivially).

Proof. Let g ∈ SLn(k) act trivially on X. Let e1, ..., en be the standard basiselements. We know that g ⋅ej ∼ ej , as g xes ej , so g ⋅ej = λj ⋅ej for some λj ∈ k.As a result, g must be a diagonal matrix with diagonal entries λ1, ..., λn. Butthen

⎛⎝

1⋮1

⎞⎠∼ g ⋅

⎛⎝

1⋮1

⎞⎠=⎛⎝

λ1

⋮λn

⎞⎠

meaning that all λ-s are equal. As a result, g is a scalar matrix. But we alreadykilled all scalars when going to PSLn(k). Thus no element in PSLn(k) but thegroup identity acts trivially.

We now take N ≠ e be a normal subgroup of G. We wish to show thatN = G.

Proposition 12.7. N acts transitively on X = Pn−1(k).

Proof. N ≠ e and the action has no kernel, so there is at least one N -orbitcontaining two points, and let x, y ∈ X be diernt points in the same orbit.Suppose that the N action is not transitive, i.e. that there is some z ∈X whichis not in the orbit of y. By 2-transitiveness of G, we nd some g ∈ G so thatgx = x, gy = z. But this cannot hold, as if y = nx for n ∈ N then:

z = g ⋅ y = gn ⋅ x∃n1 ∈N, as N PG

= n1g ⋅ x = n1x

which cannot hold as x and z are in dierent N -orbits. Thus N acts transitively.

Note 12.8. Note that G = PSL2(k) acts transitively on Pn−1(k), so the lattercan be identied as G/Q when Q is the stabilizer of some point. The stabilizerof [e1] is given by:

Q =⎧⎪⎪⎨⎪⎪⎩

⎛⎝

1 uA

⎞⎠∶ u ∈ kn−1, A ∈ GLn−1(k)

⎫⎪⎪⎬⎪⎪⎭≅ U ⋊GLn−1(k)

where U ≅ kn−1. Note that beacuse N acts transitively on X = G/Q, we havethat G = NQ.

Second Part - Some Algebraic Stu

Claim 12.9. NU is a normal subgroup of G.

Proof. Indeed, U is a normal subgroup of Q, and G = NQ implies that Q→ G/Nis surjective, hence the image of U is normal in G/N. But then the preimage ofU in G, NU , is a normal subgroup of G, as there is a general 1-1 correspondencebetween normal subgroups of G/H to normal subgroups of G containing H.

67

Exercise 12.10. Let i0, j0 ∈ 1, ..., n be dierent and let Ei0,j0 be the unipotentmatrix whose (i0, j0)-entry is 1 and all other non-diagonal entries are zero, i.e.(Ei0,j0)i,j = δi,j + δi0,j0 . Show that the normal subgroup generated by Ei0,j0 ,which we denoted by ⟨⟨Ei0,j0⟩⟩, is all of G = PSLn(k). Hint: For σ ∈ Sn apermutation, we dene (Aσ)ij = δi,σ(j). Show that AσEi0,j0A

−1σ = Eσ(i0),σ(j0).

Corollary 12.11. NU = G. Indeed, E1,2 ∈ U , and NU is a normal subgroup ofG, thus

G = ⟨⟨Ei0,j0⟩⟩ ⊆ NU

Furthermore, we have that:

G/N = NU/N ≅ U/N∩U

but U is abelian, so U/N∩U is abelian. This means that G/N is abelian, andtherefore N ≥ [G,G].

Therefore, in order to show that G is simple, it's enough to show that G =[G,G].

Third Part - Hands on Computations and arguing by cases

We rst assume that n ≥ 3. In this case, one can check that [E1,2,E2,3] = E1,3.Thus E1,3 ∈ [G,G], so:

G = ⟨⟨E1,3⟩⟩ ⊆ [G,G]

as [G,G] is a normal subgroup of G. This implies what we want.We now assume that n = 2, and see why do the demands on k arise. We take

some a ≠ 0 and b ∈ k. We compute:

⎡⎢⎢⎢⎢⎣

⎛⎝a

a−1

⎞⎠,⎛⎝

1 b1

⎞⎠

⎤⎥⎥⎥⎥⎦=⎛⎝

1 (a2 − 1)b1

⎞⎠

if the eld k has an element a such that a2 ≠ 1, a ≠ 0, then we pick it and varyb to get E1,2 as a commutator, and argue as above. This can be done only if khas at least four elements. This completes the proof of the theorem.

Remark 12.12. This shows that PSLn(R) is simple. It's easy to see that allnormal subgroups of SLn(R) must be contained in the set of scalar matrices. Ifn is odd, then one has that SLn(R) is actually simple, and otherwise the onlynormal subgroups are e, SLn(R) and ±Id.

Embedding free groups into SO(3)In the Banach-Tarski paradox, it's useful to embed the free group F2 into SO(3).Usually it is done by considering a case of the ping-pong lemma for complicatedelements and sets on which we act. However, in the case of PSL2(Z), it is easyto nd a free subgroup (as we did in the lecture). We now explain how one canuse this to show that there is a free subgroup of SO(3). The idea is to show

68

that SO(3) (or to be precise, SU(2)) is Zariski dense in SL2(C), in which wealready know that there is a free subgroup. Then Baire category theorem willgive us a free subgroup of SO(3). In order to show that SU(2) is Zariski densein SL2(C), we will work locally.

Let a, b be elements of PSL2(Z) generating F2. Take a, b ∈ SL2(Z) which lifta, b. They also generate a free subgroup, now of SL2(C). We pass to SL2(C)because we want to apply to following denition:

Denition 12.13. Let h be a complex Lie algebra and let g a real Lie subal-gerba. We say that g is a real form of h if h ≅ g⊗RC, i.e. h = g⊕ig. We say thata real Lie subgroup G of a complex Lie group H is a real form of it if Lie(G) isa real form of Lie(H).

Exercise 12.14. SL2(R) and SU(2) are both real forms of SL2(C). Also, ifG is a real form of H then G ×G is a real form of H ×H.

Proposition 12.15. If G is a real form of H, then G is Zariski dense in H,i.e. if a polynomial P on H vanishes on G, it vanishes on all of H.

Proof. It's enough to show that the derivative of P at each point is zero, whichis preferable as we understand polynomials on Cn or Rn better then on othersurfaces. Pick a point and let p be the dierential of P at that point. If weidentify the tangent space of H as Cn, then we have a real subspace of dimensionn, namely g, on which p vanishes. By applying an orthogonal transformationA ∈ O(2n), it's enough to prove this result for g = Rn. To show this, one inductson n. This is left as an exercise.

Corollary 12.16. SU(2) contains a free group, and therefore SO(3) ≅ SU(2)/±1

contains a free group.

Proof. Consider the polynomial in two variables (A,B) ↦ w(A,B) − Id wherew ∈ F2. It does not identically vanish on H = SL2(C) × SL2(C), as SL2(C)contains a free group. By the previous exercise, it does not vanish on G =SU(2) × SU(2), so its zero set is of lower dimension in G. Because there areonly countably many words in F2, we conclude (by Baire category) that thereis some (A,B) ∈ G×G at which all of these polynomials do not vanish. Thus Aand B generate a free subgroup of G. This shows SU(2) contains a free group.

To show that SO(3) contains a free subgroup, we need to show that ifw ∈ ⟨A,B⟩ ⊆ SU(2), then w ≠ ±Id, as this is the kernel of SU(2) → SO(3). Wealready know that w ≠ Id, and w = −Id would imply w2 = Id, which is againimpossible. Thus SO(3) contains a free subgroup of rank 2.

69

13 A Bit on Convolution and Some Ergodic The-

ory Examples

Convolution - why does AA−1 contain an open set around e.

As it arised in the lecture, we will show the following fact:

Theorem 13.1. Let G be any LC group and suppose that A ⊆ G has positivemeasure. Then AA−1 ⊆ G contains an open ball around e.

Before proving this, we shall talk about convolution:

Denition 13.2. Let f, h ∈ L1(G). We dene a function f ⋆ h ∈ L1(G) by:

f ⋆ h(x) = ∫Gf(y)h(y−1x)dy

(A use of Fubini shows that ∣∣f ⋆ h∣∣1 ≤ ∣∣f ∣∣1 ⋅ ∣∣h∣∣1).

Let us prove a few properties of this action:

Proposition 13.3. .

1. If f and h are both compactly supported, then supp(f) ⋅supp(h) ⊇ supp(f ⋆h).

2. If f ∈ L1(G) and h ∈ L∞(G) (or vice versa) then f ⋆ h ∈ L∞(G) and∣∣f ⋆ h∣∣∞ ≤ ∣∣f ∣∣1∣∣h∣∣∞ (or ∣∣f ⋆ h∣∣∞ ≤ ∣∣f ∣∣∞∣∣h∣∣1).

3. The convolution L1 ×L1 → L1 is a bilinear map

4. If f ∈ L1(G) and h ∈ Cc(G) then f ⋆ h is continuous (or vice versa)

5. If f ∈ L1(G) ∩L∞(G) and h ∈ L1(G) (or vice versa) then f ⋆ h is contin-uous.

Proof. (1) and (3) are obvious. For (2):

RRRRRRRRRRR∫Gf(y)h(y−1x)dy

RRRRRRRRRRR≤ ∫

G∣f(y)h(y−1x)∣dy ≤ ∫

G∣f(y)∣∣∣h∣∣∞dy = ∣∣f ∣∣1∣∣h∣∣∞

For (4) - if f was also in Cc(G), then f ⋆ h would have been continuousas (x, y) ↦ f(y)h(y−1x) is continuous. But if we let fn∞1 ⊆ Cc(G) so that

fnL1

→ f then:

∣∣fn ⋆ h − f ⋆ h∣∣∞(3)

= ∣∣(fn − f) ⋆ h∣∣∞(2)

≤ ∣∣fn − f ∣∣1 ⋅ ∣∣h∣∣∞ → 0

What does that mean? it means that the continuous function fn ⋆ h convergeuniformly to f ⋆ h - hence f ⋆ h is continuous. The proof of (5) is the same

70

We are now ready to prove the theorem:

Proof. Shrink A to be smaller so that A,A−1 both have nite Haar measure 7

Now χA ∈ L∞(G) ∩L1(G) and χA−1 ∈ L1(G), so by part (5) of the proposi-tion, χA ⋆ χA−1 is continuous. Also, by part (1) of the proposition, the supportof χA ⋆ χA−1 is contained in AA−1.

The value of χA ⋆ χA−1 at e is:

χA ⋆ χA−1(e) = ∫ χA(y)χA−1(y−1e)dyχA(y)=χA−1 (y

−1)

= ∫ χA(y)χA(y)dy =

χA=χAχA

= ∫ χA(y)dy = µ(A) > 0

by continuity, there is an open set around e in which χA ⋆χA−1 > 0, and this setmust be contained in supp(χA ⋆ χA−1) ⊆ AA−1.

The Bernoulli Shift

Denition 13.4. Consider the spaceX = 0,1Z, which consists of all two-sidedinnite binary sequences. On each copy of 0,1 we take the 50-50 measure,and take the product probability measure on X - if i1, ..., ik are some indicesand a1, ..., ak ∈ 0,1 then:

µ⎛⎝

⎧⎪⎪⎨⎪⎪⎩(xn)n∈Z = x ∈X

RRRRRRRRRRR∀j = 1, ..., k, xij = aj

⎫⎪⎪⎬⎪⎪⎭

⎞⎠= 1

2k

basically meaning that the coordinates are independent of each other - specifyinga coordinate costs us 1

2of the measure no matter what coordinates we already

specied. The topology on X, the product topology, is such that all open sets

are unions of sets of the form

⎧⎪⎪⎨⎪⎪⎩(xn)n∈Z = x ∈ X

RRRRRRRRRRR∀j = 1, ..., k, xij = aj

⎫⎪⎪⎬⎪⎪⎭, i.e.

specifying only nitely many coordinates.Consider the shift map σ on X - taking a sequence and shifting it one place

left - if x = (xn)n∈Z ∈ X then (σx)i = xi+1. This denes an action Z X byk ⋅ x = σkx.

Claim 13.5. This action is ergodic. Actually, it's metrically ergodic.

Proof. We actually show something stronger, known as mixing, that is, if A,Bare two measurable sets then:

limn→∞

µ(σnA ∩B) = µ(A)µ(B) (⋆)

7(The measure of A−1 will not be zero, as it is the right Haar measure of A and the Haarmeasures are absolutely continuous with respect to one another. However, it's not importantfor our cause, as even then we will have that χA−1 ∈ L1

(G)).

71

.We rst explain why mixing implies ergodicity. If A is an invariant set, plugB = A and get that:

µ(A) = limn→∞

µ(A) = limn→∞

µ(A ∩A) = limn→∞

µ(σnA ∩A) = µ(A)µ(A)

which implies that µ(A) ∈ 0,1. Therefore the action is ergodic.is true if A and B are such cylinder sets, i.e. sets that specify only nitely

many coordinates. Suppose that A species mA coordinates and B species mB

coordinates. The action of σ on A makes the specied coordinates by shiftedone place to the left (if A species the 7th coordinate, σA specices the 6th).Thus, if n is large enough then σnA and B do not both specify a coordinate.This means that σnA ∩B is a cylinder set specifying mA +mB coordinates if nis large enough. Thus:

µ(σnA ∩B) = 1

2mA+mB= 1

2mA⋅ 1

2mB= µ(A)µ(B)

for all n large enough. Taking n → ∞ gives (⋆). The same argument aboveworks if A and B are nite unions of cylinder sets.

In order to get (⋆), one uses approximation of general measurable sets by(nite unions) of cylinder sets.

What about metric ergodicity?

Exercise 13.6. Repeat this argument for X = ΩZ, when Ω is some measurespace, to show that the shift action ZX is mixing.

Meaning that GX2 is ergodic, as X2 = (ΩZ)2 ≅ (Ω2)Z.

An Example of a Metrically Ergodic System Whose PowersAre Not Ergodic

Consider the action of GL2(R) on P1(R) by Mobius transformations: We denote

by

⎡⎢⎢⎢⎢⎣

xy

⎤⎥⎥⎥⎥⎦the equivalence class of (x

y). Each element (but

⎡⎢⎢⎢⎢⎣

10

⎤⎥⎥⎥⎥⎦) can be written as

⎡⎢⎢⎢⎢⎣

x1

⎤⎥⎥⎥⎥⎦, and we identify it with x ∈ R. In that case,

⎛⎝a bc d

⎞⎠

⎡⎢⎢⎢⎢⎣

x1

⎤⎥⎥⎥⎥⎦=⎡⎢⎢⎢⎢⎣

ax + bcx + d

⎤⎥⎥⎥⎥⎦=⎡⎢⎢⎢⎢⎣

ax+bcx+d

1

⎤⎥⎥⎥⎥⎦

this needs to be explained when cx + d = 0, and we are interested in the case of⎡⎢⎢⎢⎢⎣

10

⎤⎥⎥⎥⎥⎦as well (which we identify with ∞). One gets that the action of this matrix

72

is:

x↦

⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩

ax+bcx+d

x ≠∞, cx + d ≠ 0

∞ x ≠∞, cx + d = 0

a/c x =∞, c ≠ 0

∞ x =∞, c = 0

as we already saw (for SL2, but the reasonin is the same), this action splitsthrough PGL2(R) (as scalar matrices act trivially), and the action of PGL2(R)is faithful - there are no stabilizers at any point by e. One can see (in a sim-ilar way to the proof of 2-transitiveness of PSL2) that G = PGL2(R) acts3-transitively on X = P1(R). Furthermore, if an element of PGL2(R) xesthree distinct points, it must be trivial. Indeed, it's enough to look what is

xing 0,1,∞ (by 3-transitiveness). If a matrix A ∈ GL2(R) xes ∞ ↔⎡⎢⎢⎢⎢⎣

10

⎤⎥⎥⎥⎥⎦, it

must be of the form

⎡⎢⎢⎢⎢⎣

a b0 d

⎤⎥⎥⎥⎥⎦. Assuming that it xes 0 ↔

⎡⎢⎢⎢⎢⎣

01

⎤⎥⎥⎥⎥⎦yields that b = 0,

and xing 1↔⎡⎢⎢⎢⎢⎣

11

⎤⎥⎥⎥⎥⎦gives that a = d, i.e. the matrix A is scalar, so it's image in

PGL2(R) is the trivial elemnt. We conlcude that the action of G on triples ofdistinct points in P1(R) is simply transitive.

Dene a function CR on 4-tuples of distinct points in P1(R) with values inR in the following way: given a 4-tuple of such points (w,x, y, z), there is aunique g ∈ PGL2(R) such that g(w) = 0, g(x) = 1 and g(y) = ∞. We deneCR(w,x, y, z) = (w,x, y, z) = g(z).

Exercise 13.7. Show that if z1, z2, z3 ∈ P1(R) ≅ R ∪ ∞, the unique Mobius

transformation taking z1, z2, z3 to 0,1,∞ is f(z) = (z−z1)(z2−z3)(z−z3)(z2−z1)

.

Claim 13.8. The action of G = SL2(R) on X = P1(R) is metrically ergodic, butnot mixing.

Remark 13.9. The action G X is transitive, and the stabilizer of

⎡⎢⎢⎢⎢⎣

10

⎤⎥⎥⎥⎥⎦is

Q =⎧⎪⎪⎨⎪⎪⎩

⎛⎝a b

a−1

⎞⎠

RRRRRRRRRRRa ≠ 0, b ∈ R

⎫⎪⎪⎬⎪⎪⎭, so X ≅ G/Q. As we saw in the rst day, ∆G ≡ 1 but

∆Q ≠ 1, so there is no invariant G-measure on G/Q, and in particular the actionis not pmp.

Proof. (of claim)The action is not mixing as GX4 has a G-invariant function, namely the

cross ratio CR. Indeed, if w,x, y, z are dierent points in X and h ∈ SL2(R) andg is such that g(w) = 0, g(x) = 1 and g(y) =∞ then gh−1(hw) = 0, gh−1(hx) = 1and gh−1(hy) =∞, meaning that:

CR(hw,hx,hy, hz) = gh−1(hz) = g(z) = CR(w,x, y, z)

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To be exact, CR is dened on the set of distinct 4-tuples, and on its com-plement in X we dene CR ≡ 0. This is a G-invariant measurable function, soGX4 is not ergodic.

However, we saw in class that for G = SLn(R) and H any non-compactsubgroup, the action G G/H is metrically ergodic. Taking n = 2 and H = Qgives us our result.

Exercise 13.10. Show directly that G X2 is ergodic by noting that theaction of G X is 2-transitive and showing that the diagonal inside X2 is anull set.

Actually, this shows that the action G X2 is essentially transitive. Usethis to show directly that GX2 is metrically ergodic, although it is not doublyergodic.

74

14 A Classical Introduction to Ane Algberaic

Geometry

For this tutorial, k is always a eld.

Ane Algberaic Geometry

Denition 14.1. Let n ≥ 0. The ane space of dimension n over k is denedto by An(k) = kn. The reason for this change of notation is that we wantto treat kn merely as a set of points, partially forgeting about the linear andmutltiplicative structure hidden there. We also denote the space of polynomialsover k in n variables (denoted by x1, ..., xn) by k[x1, ..., xn].

We are interested in polynomials over k in n variables. If we had some topo-logical space X (which is nice enough, say metrizable) and we were interestedin continuous functions on X, we notice that a set F ⊆ X is closed if and onlyif there is a continuous map f ∶ X → R such that F = f−1(0). When giving atopology to An(k), we want that the role of continuous maps will be replacedby polynomials p ∈ k[x1, ..., xn].

Denition 14.2. The Zariski topology on An(k) is dened in the followingway - let S ⊆ k[x1, ..., xn] be a collection of polynomials polynomials. We denethe vanishing set of these polynomials to be:

V (S) =⎧⎪⎪⎨⎪⎪⎩y = (yi)ni=1 ∈ An(k) ∶ ∀f ∈ S, f(y1, ..., yn) = 0

⎫⎪⎪⎬⎪⎪⎭

a set is in An(k) is closed if and only if it is of the form V (S) for some set S.These sets are sometimes called algebraic sets.

Exercise 14.3. Let S,T ⊆ k[x1, ..., xn]. Show that:

1. V (S) ∩ V (T ) = V (S ∪ T )

2. V (S) ∪ V (T ) = V (ST ) where ST = pq ∶ p ∈ S, q ∈ T.

3. If ⟨S⟩ is the ideal generated by S (that is, all polynomials of the formf1s1 + ... + fksk for some s1, ..., sk ∈ S and f1, ..., fk ∈ k[x1, ..., xk]), thenV (S) = V (⟨S⟩).

4. Conclude from parts 1 and 2 that this actually denes a topology onAn(k), in a way that all the axioms are satised.

Denition 14.4. A subset V ⊆ An(k) is called an (ane) algebraic variety ifit is Zariski-closed, that is V = V (S) for some S ⊆ k[x1, .., xn].

Example 14.5. The following are examples (and non-exmaples) of (ane)algebraic varieties:

75

If (ai)ni=1 = a ∈ An(k), then a is an algebraic variety. Indeed, it is equalto V (S) for S = x1 − a1, ..., xn − an.

The unit circle is an algebraic variety over k = R. Indeed, S1 = (x, y) ∈A2(R) ∶ x2 + y2 − 1 = 0.

The set SLn(R) ⊂ Rn2

is an algberaic variety. Indeed, SLn(R) = (gij)ni,j=1 =g ∈ An

2

(R) ∶ det(g)−1 = 0, and the determinant map is just a polynomialin the entries of the matrix.

The set of upper triangular matrices is an algebraic varietry. Indeed, ifwe take our coordinates to be xijni,j=1, then the set of upper triangularmatrices is just the vanishing set of the collection S = xiji>j of polyno-mials

Let us consider k = R and V = R× = R/0. A priori, this is not an algebraicvariety, as it is dened by non-vanishing of the polynomial p(x) = x.However, let us consider a set in R2:

X = (x, y) ∈ R2 ∶ xy = 1

this is obviously an algebraic variety, and there is an identication of Vwith X by x ↦ (x, 1

x). This means that R× can be given the structure of

an algebraic variety

Let's try and generalize the previous example - take GLn(k) to be thecollection of all invertible matrices. A priori, it is not an ane algebraicvariety under this denition, as it is given by det ≠ 0. However, considerthe following set in kn

2+1:

X = (g, y) ∈ kn2

× k ∶ det(g) ⋅ y = 1

this is an algebraic variety in Rn2+1, as (g, y)↦ det(g) ⋅ y is a polynomial.

Obviously, the map g ↦ (g, 1detg

) is an identication of GLn(k) with X,

so GLn(k) has the structure of an algebraic variety.

The set Z ⊆ R is not algebraic. Indeed, if Z = V (S) for some S ⊆ R[x]then every polynomial p ∈ S has innitely many zeros - the points of Z.Therefore p must be the zero polynomial (as a polynomial of degree ncannot have more than n zeros). Thus S ⊆ 0 and therefore V (S) = R, acontradiction.

Exercise 14.6. Show that the Zariski topology on A1(k) coincides with theconite topology, that is, as set U ⊆ A1(k) is Zariski-open if and only if U = ∅or that A1(k)/U is nite.

Denition 14.7. Let V ⊆ An(k) be an algebraic variety over k . Given such V ,we dene I(V ) to be the set of all polynomials vanishing on V . This is an ideal

76

of k[x1, ..., xn], and we can look at the quotient ring k[x1,...,xn]/I(V ), meaningthat we identify polynomials which attain the same values on V . This ring iscalled the coordinate ring of V , and denoted k[V ]. Elements of this ring arecalled regular functions on V . Note that k[V ] has the structure of a k-algebra..

Exercise 14.8. Use the construction of an algberaic structure on R× to showthat the map f(x) = 1

xis a regular map on R×. Also show that f(g) = 1

det(g)is

a regular map on GLn(k).Actually, show that the coordinate ring on R× is isomorphic to k[x, 1

x] and

that the coordinate ring on GLn(k) is isomorphic to k[xij , 1det

].

Denition 14.9. Amorphism of (ane) algebraic varieties is dened as follows:let V ⊆ An(k) and U ⊆ Am(k). A morphism φ ∶ V → U is given by m regularmaps on V by φ(v) = (f1(v), ..., fm(v)), where the image of V under this mapsatises the polynomial equations dening U . A morphism φ ∶ V → U is anisomorphism if there is a morphism ψ ∶ U → V such that φψ = idU and ψφ = idV ,where idU , idV are the identity maps on U and V , respectively.

Exercise 14.10. Show that if V ⊆ An(k) and U ⊆ Am(k) are isomorphic, thenk[V ] ≅ k[U] as k-algebras. Hint: show that V ↦ k[V ] will dene a function(a functor) from the category of ane algebraic varieties to the category of k-algerbas, and that morphisms of varieties translate into morphisms of k-algebras.

Ane Algebraic Groups

Denition 14.11. Let V ⊆ An(k) be an ane algebraic variety. V is called analgebraic group if it has the structure of a group, where both maps

m ∶ V × V → V, j ∶ V → V

m(g, h) = g ⋅ h, j(g) = g−1

are given by polynomials (or more accurately, are regular, when V ×V is thoughtof in A2n(k)).

Example 14.12. Examples of ane algerbaic groups:

Let Ga(k) be the additive group of the eld k, Ga(k) = (k,+). ObviouslyGa(k) ⊆ A1(k) is an algebraic variety (zeros of p(x) = 0). The groupmultiplication is given by m(x, y) = x + y, which a polynomial, and theinverse is given by j(x) = −x, which is also a polynomial. This Ga(k) isan algebraic group (over k).

More generally, one can take G = (kN ,+) be the additive group of kN .Obviously, G ⊆ AN(k) is an algebraic variety (zeros of p(x1, ..., xN) = 0),and the group multiplication and inverse are given by polynomials. Thus(kN ,+) is an algebraic group (over k).

77

As another example, one can take G = SLn(k). We already saw that

G ⊆ An2

(k) is an algebraic variety. The multiplication in the group isgiven by polynomials: the i, j entry of AB is:

(AB)ij =n

∑k=1

AikBkj

which is a polynomial in the entries of A and B. As for the inverse, werecall a formula from linear algebra:

A−1 = 1

det(A)⋅ adj(A)

where adj(A) is the adjoint matrix to A. The denition of A is not ofhigh importance right now, but we will remind that the entries of adj(A)are given by certain minors of the matrix A, and in particular they arepolynomials A's entries. Thus, if A ∈ SLn(k) then:

j(A) = A−1 = 1

det(A)adj(A) = adj(A)

is also given by polynomials, thus SLn(k) is an algebraic group (over k).

Similarly, we claim that GLn(k) is an algebraic group. We already gaveit the structure of an algberaic variety and the multiplication is just thesame as in SLn(k). As for the inverse, it is given by:

j(A) = A−1 = 1

det(A)adj(A)

recall that 1det(A)

is a regular map on GLn(k), and that it is in adj(A)is given by polynomials. Thus j ∶ GLn(k) → GLn(k) is a regular map onGLn(k) and therefore GLn(k) is an algebraic group (over k).

Consider the multiplicative group of the eld k, denote Gm(k) = (k× =k/0, ⋅). One can argue as above to show that it is an algebraic group(over k) , but we notice that Gm(k) is the same as GL1(k), as there is anobvious identication of an element x ∈ k with the matrix (x).

More generally, the group ((k×)N , ⋅) is an algebraic group. The proof isgiven as an exercise.

Suppose that G is an algberaic group and that H ≤ G is a subgroup.Suppose that H ⊆ G is a subvariety (meaning that it is the zero set ofadditional polynomials). Then H is an algebraic group. This is beacusethe multiplication and inversion maps on H are restrictions of the mapson G, which are already given by polynomials.

78

Consider the set of invertible upper-triangular matrices P =⎧⎪⎪⎨⎪⎪⎩

⎛⎝

x11 ⋯ x1n

⋱ ⋮xnn

⎞⎠∶

xii ∈ k×, xij ∈ k if i < j⎫⎪⎪⎬⎪⎪⎭. This is a subgroup of GLn(k). We show that it

is a subalgebraic variety. Indeed, P is given as the zeros of the additionalpolynomials xiji>j . Thus P is an algebraic group.

Consider the set of invertible diagonal matricesD =⎧⎪⎪⎨⎪⎪⎩

⎛⎝

x11

x22

⋱xnn

⎞⎠∶

xii ∈ k×⎫⎪⎪⎬⎪⎪⎭. This is a subgroup of GLn(k), which is also a subvariety. In-

deed, it is given as the zeros of the additional polynomials xiji≠j

Denition 14.13. A homomorphism of algebraic groups is a morphism of theunderlying algebraic varieties, which is also a group homomorphism. An iso-morphism of such is a homomorphism which has an inverse, which is also analgebraic group homomorphism.

Example 14.14. Show that the algebraic groups (k×)n and the group of in-vertible diagonal matrices are isomorphic as algebraic groups over k.

Remark 14.15. When we talked about embedding the free group inside SO(3),the main tool was to show that SU(2) is a real form of SL2(C), and thenshow that every polynomial which vanishes on SU(2) must vanish on SL2(C).What we exactly shown is that if H is an algebraic group over C, and if G isan algberaic subgroup over R (we now consider H as a real algebraic group),then a polynomial which vanishes on G must vanish on H, i.e. that G is Zariskidense in H.

Algebraic actions

Denition 14.16. Let G be an algebraic group and V an algebraic variety,both are dened over k. An algebraic action of G on V is an usual group actionG V such that the action map G×V → V , dened by (g, v)↦ g ⋅v, is denedby polynomials in the entries of g and v

Example 14.17. Let G = SL2(R) and V = R2 = A2(R). The action G Vdened by A ⋅ v = Av (i.e. matrix multiplication) is algebraic, as it is dened bypolynomials:

(a bc d

) ⋅ (xy) = (ax + by

cx + dy)

Proposition 14.18. Let G V be an algebraic action, and let v ∈ V . LetStabG(v) be the stabilizer of the point v. Then it is an algebraic subgroup of G.

79

Proof. Suppose that V ⊆ An(k). Fix v as above, and consider the followingfunction on G: f ∶ G→ An(k), f(g) = g ⋅ v − v.

Because the action map is given by polynomials and v is xed, f is givenby polynomials. But StabG(v) is the subgroup of G which is dened by theaddition demand f(g) = 0, which is given by polynomials. Thus StabG(v) isa dened by additional polynomials, making it a subvariety of G and thus analgebraic subgroup.

Exercise 14.19. Consider the action of R on S1 ⊂ R2 by rotations. Show thatthis action is not algebraic without writing an explicit formula for the actionmap.

Basics of Commutative Algebra

Denition 14.20. Let R be a ring (commutative with 1). We say R is Noethe-rian if any ascending sequence of ideals in R

I1 ≤ I2 ≤ I3 ≤ I4 ≤ ...

eventually stabilizes - if n is large enough then In = In+1.

Exercise 14.21. Show that a ring R is Noetherian if and only if every ideal inR is nitely generated.

An important theorem (which will not be proved) about such rings is:

Theorem 14.22. (Hilbert Bais Theorem) If R is Noetherian, then R[x1, ..., xn]is Noetherian.

Exercise 14.23. Use the theorem and the previous exercise to deduce thatwhenever V ⊆ An(k) is an algberaic variety, there are polynomials f1, ..., fm ∈k[x1, ..., xn] such that V = V (f1, ..., fm).

We know that every variety V can be written as V = V (J) when J is anideal in R = k[x1, ..., xn], but there might be redundancies. For example, in onevariable, V (⟨x⟩) = V (⟨x2⟩).

Denition 14.24. For an ideal J in a ringR, dene√J = x ∈ R ∶ ∃m, xm ∈ J.

This is an ideal of R (the hard part is showing that it's closed under addition,and this is true by the binomial formula). The ideal J is called radical if J =

√J .

Note that√J is a radical ideal.

Note that if p ∈√J then p = 0 ⊆ V (J), as pm ∈ J and pm vanishes at a

point if and only if p vanishes there. This means that V (√J) = V (J), so every

variety can be written as V (J) for some radical ideal J in R = k[x1, ..., xn]. Isthis the only way for two vanishing sets to be the same, i.e. does V (J) = V (J ′)for two radical ideals J, J ′ imply that J = J ′?

80

Theorem 14.25. (Hilbert's Nullstellensatz) Yes!, but only if k is algebraicallyclosed. Suppose that k is algebraically closed. Then for a general ideal J of R,

I(V (J)) = p ∈ R ∶ p∣V (J) = 0 =√J

i.e., if p vanishes whenever all elements of J vanish, then pm ∈ J for some powerm ∈ N.

Equivalently, if k is algebraically closed, f1, ..., fN , g ∈ R and g vanisheswhenever all fi-s vanish, then g = ∑N1 hifi for some other polynomials h1, ..., hN ∈R.

Of course, this theorem has a weaker form, when g = 1:

Theorem 14.26. (Nullstellemstaz's weak form)If k is algebraically closed, If f1, ...., fN ∈ R do not have a common zero, then

1 = ∑N1 hifi for some other polynomials hi ∈ R.

Exercise 14.27. .

1. Show that even the weak form is incorrect if k is not algberaically closed.

2. Show that the weak form implies the stronger form. Hint: try adding adummy variable xn+1 and look at f1, ..., fN ,1 − xn+1g. This is called theRabinowitsch trick.

Corollary 14.28. (From Nullstellensatz) There is a one-to-one correspondencebetween radical ideals and ane varieties by J ↔ V (J).

This correspondence gives a one-to-one correspondence between maximal ide-als and points in An(k) by a = (a1)n1 ↔ ⟨x1 − a1, ..., xn − an⟩.

We found what happens with radical and maximal ideals, but there's alsothe collection of primes ideal in the middle. We would like to nd out whathappens with them:

Denition 14.29. Let V be a variety. It's called irreducible if whenever V =⋃finite Vi for Vi being Zariski closed subsets of V , for at least one i we haveVi = V .

Proposition 14.30. The correspondence above gives a one-to-one correspon-dence between irreducible varieities and prime ideals.

Proof. Let V = V (J) where J is radical.⇒: Assume V is irreducible, and let p, q ∈ R = k[x1, ...., xn] such that pq ∈ J .

Consider V1 = V ∩ p = 0 and V2 = V ∩ q = 0. These are Zariski closed setsin V which cover it (as pq = 0 on V ), meaning that at least one of them is Vitself. Assuming WLOG that V1 = V , we get that p = 0 on V and thereforep ∈ I(V (J)) =

√J = J . Hence J is prime.

⇐: Assume J is prime. Let V = ⋃N1 Vi be a Zariski closed cover of V . WriteVi = V (Ji). Then:

V (J) = V =N

⋃1

Vi =N

⋃1

V (Ji) = V (N

∏1

Ji)

81

so by Nullstellensatz, J =√J =

√∏N

1 Ji ⊇ ∏N1 Ji. Because J is prime, there is

some i so that Ji ⊆ J , implying that:

V = V (J) ⊆ V (Ji) = Vi ⊆N

⋃1

Vi = V

meaining that Vi = V , and therefore V is irreducible.

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15 Constructible Sets

Proposition 15.1. For a subset A of a topological space X, the following areequivalent:

1. A = F ∩U for some closed set F and open set U .

2. A = A ∩U for some open set U .

3. If a ∈ A, there is some open neighborhood Ua of a such that A∩Ua = A∩Ua.

Denition 15.2. If A satises the conditions above, it is called locally closed.

Proof. (1) Ô⇒ (2): Given A = F ∩U , we note that F is a closed set containingA, so it contains A. Therefore intersecting both sides of A = F ∩ U with A, weget:

A = A ∩A = (F ∩U) ∩A = (F ∩A) ∩U = A ∩U(2) Ô⇒ (3): Obvious, as one can take Ua = U for any a ∈ A.(3) Ô⇒ (1): Let U = ⋃a∈AUa be the open set and F = A be the closed set.

Then A ⊆ U and:

A = A ∩U = A ∩ (⋃a

Ua) =⋃a

(A ∩Ua) =⋃a

(A ∩Ua) = A ∩ (⋃a

Ua) = A ∩U

Denition 15.3. Let X be a topological space. The collection of all con-structible sets is dened as the algebra generated by open and closed sets inX.

Proposition 15.4. A set C ⊆ X is constructible if and only if it is a union ofnitely many locally closed sets.

Proof. Let A = Alg(open, closed) be the algebra generated by open and closedsets. Because every locally closed set is the intersection of an open and a closedset, locally closed sets are in A, hence nite unions of locally closed sets are alsoin A. To show that other inclusion, we show that nite unions of locally closedsets form an algebra themselves.

Indeed, the intersection of two locally closed sets is locally closed - if F,Eare closed and U,V are open then:

(F ∩U) ∩ (E ∩ V ) = (F ∩E) ∩ (U ∩ V )

where F ∩E is closed and U ∩V is open. Thus, if we are given two nite unionsof locally closed sets, their intersection is again such as:

(N

⋃1

Pi) ∩ (M

⋃1

Qj) =N

⋃1

M

⋃1

(Pi ∩Qj)

it's also clear that the collection of nite unions of locally closed sets is closedunder union (of two such). It is left to show that this collection is closed undertaking complements.

83

Exercise 15.5. Show that the collection of all nite unions of locally closedsets is closed under taking complements, completing the proof.

Theorem 15.6. Let X be a topological space, C ⊆X constructible. Then thereis a subset O ⊆ C which is open and dense in C.

Proof. Write C = ⋃N1 Ei where each Ei is locally closed. Dene Zi = Ei/Ei,Z = ⋃N1 Zi and O = C/Z.

Step 1: We show that O ⊆ C. Indeed,

C ∪Z = (N

⋃1

Ei) ∪ (N

⋃1

(Ei/Ei)) =N

⋃1

Ei =N

⋃1

Ei = C

removing Z from both sides, we get:

O = A/Z = (C ∪Z)/Z = C/Z ⊆ C

Step 2: We explain why O ⊆ C is closed. Indeed, we need to show thatZ ⊆ C is closed, meaning that it suces to show that each Zi is closed in C. Eiis locally closed, meaning that Ei ⊆ Ei is open. Therefore Zi = Ei/Ei is a closedset of Ei, which is itself a closed set of C, so Zi is closed in C.

Step 3: We explain why O is dense in C. Assume not, meaning that thereis some non-empty open set V inside C/O = ⋃N1 Zi. Let i0 ∈ 1,2, ..., i0 be theminimal index so that there is some non-empty open set inside ⋃i01 Zi, and takesuch open set U . We claim that it is not possible that i0 = 1 nor that U ⊆ Zi0 .Both are implied by:

Claim 15.7. There is no i ∈ 1, ...,N such that U ⊆ Zi.

Proof. Indeed, Zi = Ei/Ei, so if U ⊆ Ei was disjoint from Ei then Ei was notdesne in Ei, which is absurd.

returning to the proof, we have that i0 ≥ 2 and that U /⊆ Zi0 . Because Zi0 ⊆ Cis closed, U/Zi0 is open and non-empty in C, and:

∅ ≠ U/Zi0 ⊆i0−1

⋃1

Zi

contradicting the minimality of i0.

We now explain an alternative way to prove such a fact.

Denition 15.8. Let A be a subset of a topological space X. We dene:

U(A) =⎧⎪⎪⎨⎪⎪⎩x ∈ A ∶ ∃a neighborhoodUa of a

such that A ∩Ua = A ∩Ua

⎫⎪⎪⎬⎪⎪⎭

Exercise 15.9. Check that:

84

U(U(A)) = U(A)

U(A) = A if and only if A is locally closed (in particular, U(A) is locallyclosed).

Dene A1 = A/U(A), A2 = A1/U(A1) and so on. Show that A is con-structible if and only if An = ∅ for some n. In particular, if A is con-structible then U(A) ≠ ∅.

We recall the notion of a constructible action:

Denition 15.10. A continuous action G X is constructible if the mapG ×X →X ×X, taking (g, x) to (x, gx) is constructible.

Claim 15.11. If GX is constructible, then orbits are locally closed.

Proof. We rst explain why orbits are constructible. Indeed, x x ∈X and lookat G × x, which is a closed set in G × X. Therefore its image x × Gx isconstructible in X ×X. We can write it as a nite union of locally closed sets,all must be of the form x × Ai. Thus x × Gx is actually constructible inx×X (as x×Ai is open in its closure, x×Ai). But because x×X ≅X,we get that Gx is constructible in X.

Now, consider the set U(Gx) of points in which Gx is locally closed. As Gxis constructible, it is non-empty, and it is clear (as G acts on X by homeomor-phisms, preserving Gx) that U(Gx) is also G-invariant. But Gx is a G-transitivespace, which means that Gx = U(Gx), implying that Gx is locally closed.

A Little Extra About Irreducible Varieties

Theorem 15.12. Let V be any (ane) variety. Then V is the union of nitelymany irreducible varieties.

Proof. First, by the one-to-one correpsondence we had between radical idealsand varieties, and also between prime ideals and irreducible varieties, it's enoughto prove that every radical ideal is the intersection of nitely many prime ideals.

Exercise 15.13. Let J be a radical ideal, and suppose that ab ∈ J . Show thatJ =

√J + (a) ∩

√J + (b).

Now, let J be any radical ideal. If it is not prime itself, we write it (asabove) as the intersection of two larger radical ideals, J1 and J2. Do the samewith J1, J2 to (possibly) get J11, J12, J21 and J22. We need to explain why thisprocess eventually halts, but that follows from the Noetherian property.

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16 Localization and Some Absolute Values

Localization

We want to study rational functions, i.e. of the form f/g, on a variety X, whenf, g ∈ k[X]. This is a problem to study these globally, as they are not denedwhen g = 0. However, they are near points in which g ≠ 0.

Example 16.1. Firstly, if a ∈X, we would like to restrict our attention to a.Doing this, we invert all functions g ∈ k[X] such that g(a) ≠ 0. The function f

gcan be thought of as dened in an arbitrarily small neighborhood of a.

Secondly, if g ∈ k[V ] and we would like to restrict out attention to g ≠ 0,we invert g, g2, g3, ..., and get functions f

gkwhich are dened on g ≠ 0. This

is exactly what we did on Friday when considering V = R/0 ⊆ A1(R), anddening a structure of a variety by looking at:

X = (x, y) ∈ A2(R) ∶ xy = 1

we forced the function y = 1xto be a regular function on V , i.e. we inverted by

x,x2, x3, ....

To be exact:

Denition 16.2. Let R be a ring. A set S ⊆ R is called multiplicative if0 ∉ S,1 ∈ S and whenever a, b ∈ S, also ab is in S.

Given S ⊆ R multiplicative, we dene S−1R to be the ring of all fractions rs∶ r ∈ R,s ∈ S, where addition and multiplication are as usual, and equality

is given by rs= r′

s′if and only if there's some u ∈ S such that u(rs′ − sr′) = 0 in

R.

Remark 16.3. Why u is needed? consider the variety V = V (xy) ⊆ A2(R), andlet a = (1,0) ∈ V . At the point a, the functions 0 and y are the same, so wewould like that when localizing at a, y

1= 0

1. If we would not have added to u,

this would be equivalent to saying that y ⋅1−0 ⋅1 = 0 in R = R[x,y]⟨xy⟩

, which is false.

If we pick u ∈ S = p ∈ R[x, y] ∶ p(a) ≠ 0 as u = x, we get that u(y ⋅1−0 ⋅1) = xywhich is indeed zero in R.

Notation 16.4. If S = R/p when p ⊲ R is a prime ideal, one writes S−1R = Rp.If a ∈ S is not nilpotent and S = 1, a, a2, a3, ... then S−1R is denoted by Ra

Exercise 16.5. Let R = R[x,y]/⟨xy⟩ and let R′ = R[x,y,z]/⟨xy.yz.zx⟩. Show thatR⟨x−1⟩ ≅ R[x]⟨x−1⟩, and explain this geometrically. What about R′

⟨x−1⟩?

Recall that we dened principal open sets in a variety X as sets of the formf ≠ 0 for f ∈ k[X].Claim 16.6. These form a basis of topology for the Zariski topology

Proof. A general open set in X is given by X/f1 = ... = fm = 0 = f1 = 0 ∨ f2 =0 ∨⋯∨ fm = 0 = ⋃m1 fi ≠ 0, and the intersection of two principal closed setsf ≠ 0,g ≠ 0 is also one, fg ≠ 0.

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Denition 16.7. A set E in X is called good if it is of the form:

f1 ≠ 0, ...., fm ≠ 0, g1 = ... = g` = 0

These sets have the structure of ane varieties, as they are closed sets off1 ≠ 0, ...., fm ≠ 0, which are ane themselves by the process of localizationby a = f1f2⋯fn ∈ R = k[X].

Claim 16.8. Any locally closed set K in X is a nite union of good sets. Asa result, every consturctible set is a nite union of such.

Proof. Write K = U ∩ F when U is open and F is closed. Write F = g1 = .... =g` = 0 and use the above to write U as a nite union of principle open sets.This completes the proof.

Recall that we have seen in class:

Proposition 16.9. The image of a morphism between ane varieties is con-structible. (See Theorem 6.17 on page 37)

We now conclude:

Corollary 16.10. A morphism φ ∶X → Y of ane varieties is constructible

Proof. It's enough to show that good sets in X map to constructible sets inY , as any constructible set is the nite union of such. Take C ⊆X a good setand consider the inclusion map i ∶ C → X, which is a morphism. It gives bycomposition the map φ i ∶ C → Y , which is a morphism of ane varieties, soits image φ(i(C)) = φ(C) is constructible.

PGLn(C) is ane

We saw in class that if G is an ane algebraic group and H ≤ G is a normalane algebraic subgroup, then G/H is an ane algebraic subgroup. This is abit surprising, as one can take G = GLn(C) and H = Scalars and end up withthe fact that PGLn(C) is ane. We explain why this is true.

Indeed, consider the conjugation action of GLn(C) onMn(C), the collectionof all n×n matrices. Only the scalars act trivially, so this action factors throughPGLn(C) and has no kernel there. Because each element acts by an invertiblelinear transformation on M2(C) ≅ C4, we get an algebraic monomorphism φ ∶PGLn(C) GLn2(C). We want to show that the image is closed.

Exercise 16.11. For n = 2, try writing this map in coordinates and showingdirectly that the image is closed.

Now, the image of φ is the same as the image of ψ ∶ GLn(C) → GLn2(C)given by the action. This is an algberaic morphism of ane algebraic groups,so the image is locally closed. To show that it must be closed, restrict to theclosure of the image Im(ψ), which is an algebraic subgroup. By local closeness,

Im(ψ) is open in Im(ψ), hence closed. Thus Im(ψ) = Im(ψ) and the image ofψ (and thus φ) is closed. This shows that PGLn(C) is actually ane.

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Absolute Values on Fields - The Very Basics

We restricted our attention in class to algebraically closed elds which are com-plete w.r.t. an absolute value. We would like to understand better how eldswith absolute values look like. We know many dierent examples - if we denoteby ∣x∣∞ =

√x2 for x ∈ R and ∣x∣∞ =

√xx, then (Q, ∣ ⋅ ∣∞), (R, ∣ ⋅ ∣∞) and (C, ∣ ⋅ ∣∞)

are all elds with absolute values on them. One can also take some prime p andconsider the p-adic absolute value on Q, dened by:

∣pm ⋅ ab∣p

= p−m

for every m ∈ Z, and a, b ∈ Z coprime to p. One can complete Q with respect tothe metric d(x, y) = ∣x − y∣p and get the p-adics (Qp, ∣ ⋅ ∣p).Denition 16.12. Let k be a eld. An absolute value of k is a function ∣ ⋅ ∣ ∶k → [0,∞) which satises:

∣x∣ = 0 if and only if x = 0.

∣xy∣ = ∣x∣∣y∣

∣x + y∣ ≤ ∣x∣ + ∣y∣Furthermore, the absolute value is called Non-archimedean if:

∣x + y∣ ≤ max∣x∣, ∣y∣.We will classify all absolute values on Q and understand archimedean abso-

lute values tomorrow. For the mean time, we make a lemma:

Lemma 16.13. For an absolute value ∣ ⋅ ∣ on a eld k, the following are equiv-alent:

For all n ∈ N, ∣n∣ ≤ 1.

∣ ⋅ ∣ is non-archimedean.

Proof. If ∣ ⋅ ∣ is non-archimedean, we note that the rst and second axiom implythat ∣1∣ = 1. Thus:

∣n∣ = ∣1 + 1 + ... + 1´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

∣n times

≤ max∣1∣, ∣1∣, ..., ∣1∣ = 1

For the other direction, let x, y ∈ k. We note that for all n and i, ∣(ni)∣ ≤ 1 as

these are integers. Thus:

∣x + y∣ = ∣(x + y)n∣1/n = ∣n

∑i=0

(ni)xiyn−i∣

1/n

≤⎛⎝

n

∑i=0

∣(ni)∣∣x∣i∣y∣n−i

⎞⎠

1/n

≤

≤⎛⎝

n

∑i=0

∣x∣i∣y∣n−i⎞⎠

1/n

≤⎛⎝(n+1)⋅max

i∣x∣i∣y∣n−1

⎞⎠

1/n

= n√n + 1⋅ n

√max∣x∣, ∣y∣ n→∞→ max∣x∣, ∣y∣

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17 Some More on Absolute Values - Ostrowski's

Theorems

Ostrowski's Little Theorem

Last time, we dened the notion of an absolute value on a eld K, and partiallyshowed that:

Lemma 17.1. For a non-trivial absolute value ∣ ⋅ ∣ on Q, the following areequivalent:

1. ∣ ⋅ ∣ is nonarchimedean.

2. ∣n∣ ≤ 1 for all n ∈ N.

3. There is some M > 0 such that for all n ∈ N, ∣n∣ ≤M .

4. There is some n0 ∈ N such that ∣n0∣ < 1.

If Q is exchanged by another eld, the rst three are still equivalent, andthe latter still implies all of them.

Proof. We already showed (1) ⇐⇒ (2). (2) Ô⇒ (3) is clear. For (3) Ô⇒(2), we take a ∈ N and look at ak for some k ∈ N. We have:

∣a∣ = ∣ak ∣1/kak∈N

≤ M1/k k→∞→ 1

implying ∣a∣ ≤ 1 for arbitrary a ∈ N. (2) Ô⇒ (4) is true by non-triviality ofthe absolute value. As for (4) Ô⇒ (3), we take some a ∈ N and write it in thebasis of n0:

a = a0 + a1n0 + ... + arnr0, ai ∈ 0,1, ..., n0 − 1

letting M = max∣0∣, ∣1∣, ..., ∣n0 − 1∣, we conclude using the triangle inequalitythat:

∣a∣ = ∣r

∑0

aini0∣ ≤

r

∑0

∣ai∣∣n0∣i ≤M∞

∑0

∣n0∣i ≤M

1 − ∣n0∣

and a ∈ N was arbitrary.

We are now ready to classify absolute values on Q:

Theorem 17.2. (Ostrowski's Little Theorem) Let ∣ ⋅ ∣ be any non-trivial absolutevalue on Q. Then either:

∣ ⋅ ∣ is given by ∣ ⋅ ∣α∞, where ∣ ⋅ ∣∞ is the strandard absolute value on Q, andα ∈ (0,1].

∣ ⋅ ∣ is given by ∣ ⋅ ∣αp , where ∣ ⋅ ∣p is the p-adic absolute value on Q for p prime,and α > 0.

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Proof. We split into two cases:Case 1: ∣ ⋅ ∣ is nonarchimedean. Dene I ≤ Z to be:

I = a ∈ Z ∶ ∣a∣ < 1

by the lemma above, I ≠ ∅, and 1 ∉ I. Furthermore, I is an ideal (by the lemma,for all b ∈ Z we have ∣b∣ ≤ 1, so if a ∈ I and b ∈ Z then ∣ab∣ = ∣a∣ ⋅ ∣b∣ ≤ ∣a∣ < 1),and furthermore I is a prime ideal (if ∣a∣∣b∣ < 1 then at least one of ∣a∣ and ∣b∣ issmaller than 1). Thus I = pZ for some p prime. Note that by the lemma above(part (2)), if a ∉ pZ = I then ∣a∣ = 1. We claim that ∣ ⋅ ∣ must be some power ofthe p-adic absolute value ∣ ⋅ ∣p. Indeed, for x ∈ Q we write x = pma

bwhere m ∈ Z

and a, b ∈ Z are coprime to p, so:

∣x∣ = ∣pm ab∣ = ∣p∣m ∣a∣

∣b∣a,b∉pZ

= ∣p∣m = pm logp ∣p∣ = p−mlog(p)log(∣p∣) = ∣pm∣

α

p

where α = log(p)log(∣p∣)

. But:

∣x∣p = ∣pm ab∣p

= ∣p∣mp∣a∣p∣b∣p

a,b∉pZ

= ∣p∣mp

meaning that ∣ ⋅ ∣ = ∣ ⋅ ∣αp for this value of α.Case 2: ∣ ⋅ ∣ is archimedean. In this case, the lemma above implies that there

is some n0 ∈ N such that ∣n0∣ > 1.Take any a, b ∈ N bigger than 1, and write bk in basis a:

bk = ∑0≤i<r

ciai

where each ci is in 0,1, ..., a − 1 and r = k log bloga

+ 1. Thus:

∣b∣k = ∣bk ∣ ≤RRRRRRRRRRR∑

0≤i<r

ciaiRRRRRRRRRRR≤ ∑

0≤i<r

∣ci∣∣a∣i

by the triangle inequality, each ∣ci∣ is smaller than ci, which is itself smaller thana. We thus get:

∣b∣k ≤ ∑0≤i<r

∣ci∣∣a∣i ≤ b ∑0≤i<r

∣a∣i ≤ b ⋅ r ⋅ max0≤i<r

∣a∣i = b ⋅ r ⋅max1, ∣a∣r−1

taking k-th root and recalling that r = k log bloga

+ 1, we get that:

∣b∣ ≤ k√b ⋅ k

√k

log b

log a+ 1 ⋅max1, ∣a∣

log(b)log(a) k→1→ max1, ∣a∣

log(b)log(a) (⋆)

this is true for general a, b > 1 in N.

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Fix N ∋ a > 1 arbitrary and plug b = n0 to get that:

1 < ∣n0∣ = ∣b∣ ≤ max1, ∣a∣log(b)log(a)

which can only make sense if ∣a∣log(b)log(a) > 1, i.e. if ∣a∣ > 1. Knowing this information,

we return to (⋆) with arbitrary a and b and get that:

∣b∣ ≤ ∣a∣log(b)/ log(a) Ô⇒ ∣b∣1/ log(b) ≤ ∣a∣1/ log(a)

the roles of a, b are symmetric here, so we get that:

∣a∣1/ log(a) = ∣b∣1/ log(b)

Now take 2 ≤ n ∈ N to be a and b = 2, giving:

∣a∣ = ∣2∣log(a)log(2)

algebra

= ∣a∣log ∣2∣log(2)∞

because ∣1∣ = 1 and ∣0∣ = 0, this will be true for all a ∈ Z. Using quotients givesthis equality for all a ∈ Q. The claim is now proved.

Ostrowski's Big Theorem

Theorem 17.3. Let K be any eld which is complete with respect to somearchimedean absolute value. Then either K = R and K = C, and the absolutevalue is of the form ∣ ⋅ ∣α∞.

The tactics to prove this theorem is as follows. The lemma in the beginningimplies that char(K) = 0 (otherwise there is some n ∈ N such that n = 0, meaningthat ∣n∣ = ∣0∣ = 0 < 1). Thus Q embeds into K, and the restriction of the absolutevalue to Q is archimedean. By the small theorem, we conclude that it is of theform ∣ ⋅ ∣α∞, meaning that the completion of Q with respect to it is R. BecauseK is complete, we conclude that R embeds into K, and the restriction of theabsolute value to the embedded copy of R is ∣ ⋅ ∣α∞.

There are now two cases - one for the case in which K contains an elementsatisfying x2 = −1, i.e. C K, and one for the case in which there is no suchelement. The rst case will be dealt with rst, and to solve the second one wewill just consider K(i), and extend the absolute value to it.

Case 1: C embeds into K. In this case, we claim that the absolute valueon C is indeed ∣ ⋅ ∣α∞.Claim 17.4. Every archimedean absolute value on C is of the form ∣ ⋅ ∣α∞ for someα ∈ (0,1].

Proof. Let ω ∈ C be a root of unity, ωN = 1. Then:

∣ω∣N = ∣ωN ∣ = ∣1∣ = 1

so ∣ω∣ = 1. The set of roots of unity is dense in S1 ⊆ C. We claim:

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Exercise 17.5. The map ∣⋅∣ ∶ C→ R is continuous, when one takes the Euclideantopology on C. Hint: Think of C as R2. ∣ ⋅ ∣ is almost a norm on R2 - it hasall the properties, but it scales wrong - the real scalars go out with some power

(as by what was said in the discussion above: ∣ ⋅ ∣RRRRRRRRRRRR

= ∣ ⋅ ∣α∞ for some α). Repeat

the proof of equivalence of norms on R2, but use the almost-norms ∣ ⋅ ∣ and∣ ⋅ ∣α∞ on C ≅ R2 to show that they dene the same topology. Deduce that ∣ ⋅ ∣ iscontinuous with respect to ∣ ⋅ ∣∞.

It follows that ∣ ⋅ ∣ = 1 on S1. Now take general z = reiθ ∈ C, and write:

∣z∣ = ∣reiθ ∣ = ∣r∣∣eiθ ∣ = ∣r∣ = ∣r∣α∞ = ∣r∣α∞∣eiθ ∣α∞ = ∣z∣α∞

as claimed.

Proposition 17.6. K = C, and the absolute value on K is ∣ ⋅ ∣α∞.

Remark 17.7. An easier way to prove this theorem than what we will show, if

∣ ⋅ ∣RRRRRRRRRRRC

is the usual absolute value on C, is to use the Gelfand-Mazur theorem:

every C⋆-algebra over C in which every element but 0 is invertible must be aeld.

Proof. This proof is a bit tricky. Suppose we have some γ ∈ K/C, and denef(x) = ∣x − γ∣ on C.

Exercise 17.8. Use the triangle inequality and the form of ∣ ⋅ ∣ on C to showthat f is continuous on C.

Also, if x→∞ then f(x)→∞:

∞← ∣x∣ ≤ ∣γ∣const

+ ∣x − γ∣ = ∣γ∣ + f(x)

thus f has a global minimum, say at a ∈ C. We dene β = γ − a, so that∣β∣ = infb∈C ∣β − b∣ = infb∈C ∣γ −(a+ b)∣. Because γ ∉ C and a ∈ C, β cannot be zero.Thus ∣β∣ > 0.

Now comes the tricky part - we will take some c ∈ C such that 0 < ∣c∣ < ∣β∣and show that ∣β − c∣ = ∣β∣. To do this, we notice that identity:

βn − cn

β − c= ∏ε≠1,εn=1

(β − εc)

note that by minimality of ∣β∣, we have that ∣β − εc∣ ≥ ∣β∣ for all ε ∈ C. Inparticular, taking absolute values gives that:

∣βn − cn∣∣β − c∣

= ∏ε≠1,εn=1

∣β − εc∣ ≥ ∣β∣n−1

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therefore:∣β − c∣∣β∣

≤ ∣βn − cn∣∣βn∣

= ∣1 − ( cβ)n

∣ ≤ 1 + ( ∣c∣∣β∣

)nn→∞→ 1

as ∣c∣ < ∣β∣. This implies that ∣β−c∣ ≤ ∣β∣, but the minimality of β forces equality,i.e. that ∣β − c∣ = ∣β∣. In particular ∣c∣ < ∣β∣ = ∣β − c∣. We can repeat this argumentagain and again, replacing β with β − c, to get that for all m ∈ Z, ∣β∣ = ∣β −mc∣,but this is absurd - any element in C can be reached by moving a lot of veryshort steps from 0, meaning that if z ∈ C then we can write z = mc for somem ∈ Z and ∣c∣ < ∣β∣ (recall that the topology given by ∣ ⋅ ∣ on C is the Euclideanone), thus ∣β−z∣ = ∣β∣ for all z ∈ C, so the function f we started with is constant,

which is impossible as f(x) x→∞→ ∞.

This pair of claims completes the proof for the case in which C embeds intoK.

Case 2: There is no solution to x2 = −1 in K. We can just throw in i, andlook at K ′ = K(i). If we show that one can extend the absolute value to K ′,then we could complete K ′ and use the above to deduce that K ′ = C, henceK = R.

We would like to dene ∣a + bi∣ =√

∣a2 + b2∣ or√

∣a∣2 + ∣b∣2, as we wouldfor passing from R to C. The problem is that the second one might not bemultiplicative, so we remain with ∣a + bi∣ =

√∣a2 + b2∣. It is easy to show that

∣ ⋅ ∣ satises ∣x∣ = 0 ⇐⇒ x = 0 (basically because x2 + 1 ≠ 0 for all x ∈ K), andthe check that ∣ ⋅ ∣ is multiplicative is the same as for C. However, showing thetriangle inequality is arduous to impossible. If one tries to imitate the proof forC, he ends up with needing to showing the inequality:

∣aa′ + bb′∣ ≤√

∣a2 + b2∣√

∣a′2 + b′2∣

for a, b, a′, b′ ∈ K, and one cannot use the Cauchy-Schwarz inequality, as ∣ ⋅ ∣ isnot the standard absolute value.

To solve this, we follow Artin:

Denition 17.9. A quasi-absolute value (this is not a standard name) is a amap ∣ ⋅ ∣ ∶K → [0,∞) satisfying:

1. ∣x∣ = 0 if and only if x = 0.

2. ∣xy∣ = ∣x∣ ⋅ ∣y∣.

3. There is some constant C > 0 such that for all b ∈ K, ∣b∣ ≤ 1 implies∣1 + b∣ ≤ C.

Exercise 17.10. Let ∣ ⋅ ∣ be a quasi-absolute value. Show that ∣ ⋅ ∣ is an absolutevalue if and only if C ≤ 2 in the denition.

Conclude that if ∣ ⋅ ∣ is a quasi-absolute value, then ∣ ⋅ ∣λ is an absolute valuefor some value of 1 ≥ λ > 0.

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Our goal is now to show that ∣ ⋅ ∣ is a quasi-absolute value on K(i), so wewill be able to rescale it to get an actual absolute value, and then we can dowhat we wanted to do in at rst, which is to complete it and rely on the casein which C embeds into the eld. To show this, one shows:

Fact 17.11. Let K be a eld in which there is no solution to x2 + 1 = 0. Thenthere is some constant ∆ > 0 such that

∣a2 + b2∣ ≥ ∆ max∣a∣2, ∣b∣2 (#)

This will be a guided exercises in the exercises sheet.We deduce:

Corollary 17.12. ∣ ⋅ ∣ is a quasi absolute value on K(i)

Proof. One needs to show that if ∣a+ib∣ ≤ 1, then ∣1+a+ib∣ ≤ C for some constantC independent of a and b.

Let ∆ be as above (for K). Then ∣a + ib∣ ≤ 1 implies ∣a2 + b2∣, which impliesby (#) that ∣a∣, ∣b∣ ≥ ∆−1/2. Thus:

∣1 + a + ib∣2 = ∣(1 + a)2 + b2∣ ≤ ∣(1 + a)2∣ + ∣b∣2 = ∣1 + 2a + a2∣ + ∣b∣2 ≤

≤ 1 + ∣2∣∣a∣ + ∣a∣2 + ∣b∣2 ≤ 1 + ∣2∣∆−1/2 +∆−1 +∆−1

where the right hand side is some constant independent of a and b. Thus ∣ ⋅ ∣ isa quasi-absolute value, and the proof is complete.

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References

Lecture 1 (Intro):

Rigidity of group actions on homogeneous spaces, III / Bader, Furman,Gorodnik, Weiss.

Lecture 2 (Lattices):

Notes by Burger.

Lecture 3 (Isometric Actions on Coset Spaces of SemisimpleGroups):

Equicontinuous Actions of Semi-Simple groups / Bader, Gelander.

Lecture 4 (Metric Ergodicity):

Boundary Rigidity and Lyapunov Exponents / Bader, Furman.

Weak Mixing Properties For Nonsingular Action / Weiss, Glasner.

Lecture 5 (Basics of Ergodic Theory)

Ergodic Theory and Semisimple Groups / Zimmer.

A survey by Alon Lessel - To Appear.

Lectures 6-7 (Algebraic Geometry and Algebraic Groups,Constructible Actions)

Algebraic Geometry and Algebraic Groups

Linear Algebraic Groups / Borel

Linear Algebraic Groups / Humphries

Constructible Actions

Appendix in Representation of the Group GL(n,F) where F is a Non-Archimedean Local Field / Bernstein, Zelevinsky.

Algebraic Groups and Number Theory / Platonov, Rapinchuk

Lie Groups and Lie Algebras / Serre.

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Lecture 8 (Measures on Varieties)

Ergodic Theory and Semisimple Groups / Zimmer

Smoothness of Algebraic Actions / Bader, Duchesne, Lécureux - To Ap-pear.

Lectures 9-10 (Algebraic Representations and Margulis Super-Rigidity)

Algebraic Representations

Algebraic Representations of Ergodic Actions and Super-Rigidity / Bader,Furman

Classical Approach to Super-Rigidity

Discrete subgroups of semisimple Lie groups / Margulis

Ergodic Theory and Semisimple Groups / Zimmer

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