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Moduli Theory and Invariants

Lectures by Tom GarrityPCMI 2015 Undergraduate Summer School

Notes by David [email protected]

Contents

Lecture 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

Lecture 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

Lecture 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

Lecture 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

Lecture 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

Lecture 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

Lecture 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

Lecture 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

Lecture 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

Lecture 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

Lecture 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

Lecture 12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

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mailto:[email protected]

Lecture 1 June 29, 2015

As kids, we learn that “functions describe the world”. This is true in otherforms of math too. For instance in algebraic geometry we study polynomials,and in complex analysis we study holomorphic functions.

Another big theme of math is equivalence problems, i.e. when are thingsthe same? In topology, things are “the same” up to bending and twisting; inalgebraic geometry, things are the same up to change of coordinates.

This is the idea behind moduli spaces. We classify some sorts of objectsup to equivalence, and the equivalence classes are the moduli space. Ideally,we want to define our moduli spaces such that they have nice geometric oralgebraic structure. This is HARD!

Example 1. We’ll focus on the moduli space of triangles in an extended exam-ple. The equivalence relation here is similarity, as in the sense of basic high-

school geometry. For instance, is similar to but not to .For the moduli space of triangles, we may assume that one vertex is at the

origin, with the shortest side the segment from p0, 0q to p0, 1q. We place thetriangle entirely in the first quadrant with the longest side having one of itsendpoints at the origin. Because we only care about triangles up to similarity,all of these moves are okay.

px, yq

Due to our clever maneuvering of these triangles, any triangle is describeduniquely by the point px, yqwith a few caveats. If the triangle is isosceles, thenby moving the point px, yq either vertically up or vertically down we get thesame triangle. So our moduli space is a bit redundant in that regard. In fact,we want to remove the redundant points in our moduli space to make it nicer.

2

This set also has some geometric structure. Namely, it’s locally a manifold!If we wiggle the point px, yq in the first picture in any direction, we get a dif-ferent, equally valid scalene triangle. So there is some open neighborhood ofpx, yqwhere this looks like R2.

Problem 2. What does the moduli space of triangles up to similarity properlylook like (i.e. without redundancies)? What about quadrilaterals up to similar-ity? What about n-gons? (This last one is an open problem!)

Now for some algebraic geometry. Classical algebraic geometry sought toclassify conics up to a change of coordinates (where by change of coordinateswe mean a linear transformation with some translation). There are seven typesof conics, of which four are degenerate.

• Ellipses ax2 ` by2 “ 1;

• Hyperbolae ax2 ´ by2 “ 1;

• Parabolae y “ ax2 ` bx` c;

• Double lines px´ yq2 “ 0;

• Two lines px` yqpx´ yq “ 0;

• Points x2 ` y2 “ 0;

• Empty set x2 ` y2 “ ´1.

Note that all ellipses are equivalent up to change of coordinates. For exam-ple, changing a circle x2`y2 “ 1 to a ellipse 4u2`9v2 “ 1 is simply done by thechange of coordinates x “ 2u, y “ 3v. Likewise, all hyperbolae are equivalentup to real change of coordinates, and all parabolae are equivalent, all doublelines, etc. But a hyperbola cannot become an ellipse under any real change ofcoordinates.

It’s different in the complex numbers. Here, by the change of coordinatesx “ u and y “ iv, we can turn a circle into a hyperbola. Moreover, certaindegeneracies vanish. For example, x2 ` y2 “ ´1 is no longer empty, and x2 `

y2 “ 0 is more than just a point.However, we still cannot make a circle into a parabola just by working over

C. Ideally, the three nondegenerate classes would all join together and thedegeneracies would go away. So we work in projective space.

Definition 3. n-dimensional Projective space over the complex numbers isdefined as

PnC “ Cn`1ztp0, 0, 0qu{ „

where „ is the equivalence relation λpx, y, zq „ px, y, zq for complex λ ‰ 0.

3

We can think of P2C as lines through the origin in C3, because each equiva-lence class of a point in PnC determines a line; two points are on the same lineif and only if they are scalar multiples of each other.

And in P2C, px, y, zq „ px{y, x{z, 1q for z ‰ 0, so we think of the plane z “ 1

in P2C as a copy of C2. Each line through the origin passes through this planein exactly one point, and hence gives a point px{y, x{zq in C2. Therefore, wethink of P2C as C2 with some extra “points at infinity” where z “ 0.

We can take any conic in C2 and homogenize it to get a conic in P2C. Tohomogenize, replace x by x{z and y by y{z and then clear denominators.

x2 ` y2 “ 1 ù x2 ` y2 “ z2

Our change of coordinates is slightly different in projective space too.

x “ au` bv ` cw

y “ du` ev ` fw

z “ gu` hv ` jw

det

»

–

a b c

d e f

g h j

fi

fl ‰ 0

Problem 4.

In P2C, there are only three conics up to change of coordinates: circles, twodistinct lines, and double lines. Each is described by a homogeneous equation

ax2 ` 2bxy ` 2cxz ` dy2 ` 2eyz ` fz2 “ 0,

or equivalently a vector in C6zt~0u

»

—

—

—

—

—

—

–

a

2b

2c

d

2e

f

fi

ffi

ffi

ffi

ffi

ffi

ffi

fl

.

As a final note, finding all higher degree curves is very very hard. Degreethree are elliptic curves, and that topic is a large portion of the whole field ofnumber theory!

Lecture 2 June 30, 2015

Today we’ll talk about groups acting on objects we care about. Yesterday, wesaw that the projective line was the right space to think about in algebraic ge-ometry. The notion of “sameness” in algebraic geometry is equivalence under

4

change of coordinates. We saw that a projective change of coordinates is givenby

»

–

x

y

z

fi

fl “

»

–

a11 a12 a13a21 a22 a23a31 a32 a33

fi

fl

»

–

u

v

w

fi

fl

This matrix also defines a transformation on conics in P2C. This is an exampleof a group representation!

Definition 5. A group representation of a group G is a group homomorphismρ : GÑ GLpV q for some vector space V .

Example 6. The cyclic group of order 3, C3 may be realized as either Z{3Z orxt | t3 “ 1y, among many other ways. These are examples of different ways torepresent a group, but not representations. This group can be represented asmatrices that show symmetries of a triangle in a map T : xt | t3 “ 1y Ñ GL2C.

T p1q “

„

1 0

0 1

T ptq “

„

cosp2π{3q ´ sinp2π{3q

sinp2π{3q cosp2π{3q

“

«

´12

´?3

2?32

´12

ff

T pt2q “

„



Problem 7. Why is„

cos θ ´ sin θ

sin θ cos θ

rotation by θ? Why does rotation by angles (rotation matrices) form a group?

Change of coordinates relevant for conics in P2 corresponds to a represen-tation ρ : GLp3q Ñ GLp6q. A conic in P2 is

ax2 ` bxy ` cxz ` dy2 ` eyz ` fz2 “ 0.

5

This can be thought of as just the coefficients.

»

—

—

—

—

—

—

–

a

b

c

d

e

f

fi

ffi

ffi

ffi

ffi

ffi

ffi

fl

P C6ztp0, 0, 0, 0, 0, 0qu.

If our change of coordinates is given by the matrix equation»

–

x

y

z

fi

fl “

»

–

a11 a12 a13a21 a22 a23a31 a32 a33

fi

fl

»

–

u

v

w

fi

fl ,

where A “ paijq is the matrix with detA ‰ 0. Changing coordinates is a repre-sentation of GL3C, and in particular a map ρ : GL3CÑ GL6C.

Example 8. Let’s think about cubics in P1C, with an equation

ax3 ` bx2y ` cxy2 ` dy3 “ 0 or

»

—

—

–

a

b

c

d

fi

ffi

ffi

fl

P C4zt~0u.

The change of coordinates x “ u` v and y “ v is represented by the followingmatrix equation:

„

x

y

“

„

1 1

0 1

„

u

v

.

After change of coordinates, we get

ax3`bx2y`cxy2`dy3 ÞÑ au3`p3a`bqu2v`p3a`3b`cquv2`pa`b`c`dqv3.

We can rewrite it as a transformation of the coefficients in a matrix equation»

—

—

–

a

3a` b

3a` 3b` c

a` b` c` d

fi

ffi

ffi

fl

“

»

—

—

–

1 0 0 0

3 1 0 0

3 2 1 0

1 1 1 1

fi

ffi

ffi

fl

»

—

—

–

a

b

c

d

fi

ffi

ffi

fl

Here, ρ : GL2pCq Ñ GL4C, so we get that

ρ

ˆ„

1 1

0 1

˙

“

»

—

—

–

1 0 0 0

3 1 0 0

3 2 1 0

1 1 1 1

fi

ffi

ffi

fl

6

Problem 9. It’s not immediately clear that this defines a group homomor-phism. Show it!

Problem 10. Do this for quartics and quintics. Do it in general for an arbitrarychange of coordinates A “ paijq1ďi,jď2 for cubics in P1 (ρ : GL2CÑ GL4C). Doit in general for an arbitrary change of coordinates A “ paijq1ďi,jď3 for conicsin P2 (ρ : GL3CÑ GL6C). Use Mathematica!

Definition 11. Let G be a group. Let ρ1 : G Ñ GLpV q and ρ2 : G Ñ GLnC betwo representations. We say ρ1 is equivalent to ρ2 if they are the same underchange of basis, i.e. there is some change of basis matrix D such that for allg P G, Dρ1pgqD´1 “ ρ2pgq. ρ1 and ρ2 are really the same representation, butthey looked different because we chose bases for V .

Now let’s talk about irreducible representations. For our representationρ : G Ñ GLpV q, we can make a new one. This new representation is calledρ : G Ñ GLpV q ‘ GLpV q and given by ρpgq “ ρpgq ‘ ρpgq. This is an exampleof a reducible representation.

Definition 12. A representation ρ : GÑ GLpV q is reducible if ρ is equivalent toa representation ρ that is a block diagonal matrix for every g. That is, for eachg P G, ρpgq looks like

ρpgq “

„

A 0

0 B

for matrices A,B.

Since we can build every representation from irreducible representations,then we want to find all of the irreducible representations for a given group G.

Example 13. Let’s find some representations of the first interesting symmetricgroup.

S3 “ tid, p1 2q, p1 3q, p1 2 3q, p1 3 2qu

There’s the boring representation ρ1pσq “ 1 for all σ P S3. (called the trivialrepresentation).

Another representation is the sign representation, ρ2pσq “ sgnpσq for allσ P S3.

The last one is the representation given by S3

œ

C2 by rotations and reflec-tions.

ρ3pidq “„

1 0

0 1

ρ3p1 2 3q “

„



ρ3p1 3 2q “

„

1 0

0 ´1

7

Why do we care? Representations of the Poincare group, which describessymmetries of the universe, are fundamental to physics. From the irreduciblerepresentations we can rediscover Maxwell’s laws and the Dirac equation, whichis amazing.

Lecture 3 July 2, 2015

We’ve been talking about equivalence problems, in which we want to classifyobjects up to equivalence. The notion of equivalence we’ve been using is somegroup acting on these objects, and if they lie in the same orbit, they’re the same.

Last time, we were thinking about the group GLnC acting by change ofvariables on Pn. We got representations of GLnC by

GLnCÑ GLpV q

for a much larger space V . We ended up with the definition of a representation

Definition 14. A representation of a group G is a group homomorphism GÑ

GLpV q for some vector space V .

Essentially, we represent elements of the group as matrices. We want to un-derstand the irreducible representations of a group, which are building blocksfor all representations.

Now we’ll look at the corresponding problem for finite groups. The outlinefor the next few days is as follows:

1. there are only a finite number of irreducible representations for any finitegroup G;

2. the number of irreducible representations of G is the same as the numberof conjugacy classes of the group G;

3. there is a function χT : G Ñ C associated to each representation T : G Ñ

GLpV q, called its character, that contains lots of information;

4. under suitable inner products, these characters for the irreducible repre-sentations form an orthonormal basis for all such functions.

The first three are kind of reasonable. The last one is in the realm of mir-acles. We encode the characters for a group in something called a charactertable.

First, conjugacy classes.

Definition 15. LetG be a group. We say that x is conjugate to y if there is someg P G such that gxg´1 “ y.

8

If G is a matrix group, this is a change of basis. So we’re trying to cap-ture the notion of change of basis in an abstract group. An important fact isthat conjugacy is an equivalence relation between elements of the group. Theequivalence classes under this relation are called conjugacy classes.

Example 16. In S3, there are three conjugacy classes.

tidu tp1 2q, p1 3q, p2 3qu tp1 2 3q, p1 3 2qu.

Example 17. In any abelian group, x is conjugate to y if and only if for someg P G, gxg´1 “ y if and only if gg´1x “ y if and only if x “ y. So each elementis in its own conjugacy class.

As a bit of a preview, here’s the definition of the character of a representa-tion.

Definition 18. Let T : GÑ GLpV q be a representation. Then the character of Tis the function χT : GÑ C given by

χT pgq :“ trpT pgqq.

Problem 19. Why is the character defined by the trace? Why not determinant?It’s totally reasonable to define the character this way, but it seems reasonableto define it using the determinant as well. (Note: the answer “because it works”is not acceptable).

Example 20. The characters corresponding to the three representations of S3

are as follows

S3 1 p1 2q p1 3q p2 3q p1 2 3q p1 3 2q

χT11 1 1 1 1 1

χT2 1 ´1 ´1 ´1 1 1

χT32 0 0 0 ´1 ´1

this is a primitive form of what we call a character table.

Notice that the characters are constant on character classes. This is in factalways true, not just rigged for S3! Real character tables will lump all conjugacyclasses together in the columns.

Theorem 21. If x is conjugate to y, then χT pxq “ χT pyq for any representationT .

Proof. There is some g P G such that gxg´1 “ y.

χT pyq “ trpT pyqq

“ trpT pgxg´1qq

“ trpT pgqT pxqT pg´1qq

“ trpT pgqT pxqT pgq´1q

“ trpT pgqT pgq´1T pxqq

“ trpT pxqq “ χT pxq

9

This means we can rewrite the character table for S3 lumping together theconjugacy classes in the columns, as follows.

tidu tp1 2q, p1 3q, p2 3qu tp1 2 3q, p1 3 2qu

χT11 1 1

χT21 ´1 1

χT32 0 ´1

This is usually how we write character tables. There’s a huge amount of struc-ture here, and again it’s not by coincidence. The sum of the squares of entriesin the first column is always the size of the group. The rows are orthonormalwhen weighted by the size of each of the conjugacy classes. The columns areorthonormal.

For the symmetric group Sn, the conjugacy class depends on cycle type,which is the same as the number of partitions of n, as defined below.

Definition 22. A partitions of n is the number of ways to summing

n “ k1 ` k2 ` . . .` k`

with k1 ě k2 ě . . . ě k`. We denote by P pnq the number of partitions of n.

Example 23.

n P pnq how to sum to n1 1 1 “ 1

2 2 2 “ 2 “ 1` 1

3 3 3 “ 3 “ 2` 1 “ 1` 1` 1

4 5 4 “ 4 “ 3` 1 “ 2` 2 “ 2` 1` 1 “ 1` 1` 1` 1

We should make that fact we said earlier a theorem.

Theorem 24. The number of partitions of n is equal to the number of conjugacyclasses of Sn.

A diagrammatic way of representing the partitions of n is Young Diagrams.These are little box diagrams corresponding to the partitions. The number ofdiagrams with n boxes is equal to P pnq.

10

4 “ 4

4 “ 3` 1

4` 2` 2

4 “ 2` 1` 1

4 “ 1` 1` 1` 1

We will use these Young diagrams later to understand the irreducible rep-resentations of Sn.


As always, we’re thinking about equivalence problems where equivalent ismoving things around with groups. We talked about how to represent groups,and then about the irreducible representations of a group. Each irreduciblerepresentation comes with a character, and the character table encodes all theinformation about representations of a group.

Let G be a finite group. Say we have k-many conjugacy classes, C1, . . . , Ck.This means, as last lecture, that there are necessarily k-many irreducible repre-sentations. Let ci “ #Ci be the number of elements in a conjugacy class Ci. Letgi be the Then the inner product on characters (or really, any class function, i.e.a function constant on each Ci) is as follows.

Definition 25. The inner product on two class function ν, µ : GÑ C is

xν, µy :“1

|G|

kÿ

i“1

ciνpgiqµpgiq,

where gi P Ci Ă G is a representative of the conjugacy class.

This inner product is defined by a matrix»

—

—

—

—

–

c1|G| 0 . . . 0

0 c2|G| . . . 0

......

. . ....

0 0 . . . ck|G|

fi

ffi

ffi

ffi

ffi

fl

11

The goal of this lecture is to see why, for two representations, ρ and ψ, thecharacters are orthogonal.

xχρ, χψy “

#

1 if i “ j

0 if i ‰ j

The vital piece of this theorem is Schur’s Lemma, which says the following.

Lemma 26 (Schur’s Lemma). Let G be a finite group and let T1 : G Ñ GLpV1qand let T2 : G Ñ GLpV2q be irreducible representations. Let S : V1 Ñ V2 be alinear map. Assume for all g P G, ST1pgq “ T2pgqS.

(a) If T1 and T2 are inequivalent, then S “ 0;

(b) if V1 “ V2, and T1 “ T2, then S “ λid.

The condition that we assume is that the following diagram commutes. Wecall S an intertwining operator.

V1 V1

V2 V2

T1pgq

S S

T2pgq

Recall that

Definition 27. T : G Ñ GLpV q is reducible if T is equivalent to a represen-tation A ‘ B for representations A : G Ñ GLpV1q, B : G Ñ GLpV2q such thatV1 ‘ V2 “ V . More concretely, T pgq is equivalent to a matrix

ˆ

Apgq 0

0 Bpgq

˙

under some change of basis.

Notice that the subspace V1 of V is invariant under the action of T pgq foreach g P G, because of the block matrix form of T pgq. Therefore, we can makean alternative (and much neater) definition of reducible.

Definition 28. T : G Ñ GLpV q is reducible if there is a nontrivial subspaceW Ď V such that for all g P G and w PW , T pgqpwq PW .

Now we can prove Schur’s Lemma.

Proof of Schur’s Lemma 26. We first prove part (a). Assume that T1 is not equiv-alent to T2, which means there is no matrix P such that PT1pgqP´1 “ T2pgq. IfS “ 0, then we’re done!

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So now suppose that S ‰ 0. We will show that S is both injective andsurjective, which means that it’s bijective and therefore ST1pgqS´1 “ T2pgq.This is a contradiction, because we assumed that T1 and T2 were not equivalent.

Let W “ kerS Ď V . S is injective if and only if kerS “ 0. We will showthat W is an invariant subspace, i.e. for all g P G and w P W , T1pgqpwq P W .If true, then dimW “ 0 or dimW “ dimV1. The latter implies S “ 0, whichcontradicts our assumption that S ‰ 0.

So now we show that for all g P G, w P W , T1pgqpwq P W . Because W “

kerS, we now show that SpT1pgqpwqq “ 0.

SpT1pgqpwqq “ T2pgqpSpwqq “ T2pgqp0q “ 0.

So T1pgqpwq P kerS “ W . Hence W is an invariant subspace. Therefore,either dimW “ 0 or dimW “ dimV1. In the latter case, kerS “ W “ V1, andso S “ 0, which we assumed was not the case. In the former case, dimW “ 0,so kerS “ 0 and S must be injective.

Now we show that S is surjective. Let W “ imS. The strategy here is thesame. We will show that for all W P W , T2pgqpwq P W , which implies by asimilar argument to the above that either W “ 0 or W “ V2.

To that end, let v2 P W Ď V2. Since W “ imS, there is some v1 P V1 withSpv1q “ v2. Then

T2pgqpv2q “ T2pgqpSpv1qq “ SpT pgqpv1qq P imS

Hence, T2pgqpv2q P imS “ W . So W is an invariant subspace under the actionof T2pgq for each g P G. Because T2 is an irreducible representation, this meansthat either W “ V2 or W “ 0. If W “ 0, then S “ 0, which is a contradictionbecause we assumed S ‰ 0. This leaves the former possibility, namely thatW “ V2, then S is surjective, as desired.

So S is both surjective and injective, so it is an isomorphism. Hence, ST1pgq “T2pgqS implies that ST1pgqS´1 “ T2pgq, so T1 and T2 are equivalent. This con-tradicts our assumption that they were inequivalent!

Now we prove part (b). Here, V1 “ V2 “ V , and T1 “ T2 “ T . We want toshow that S “ λidV for some λ P C.

Let λ be an eigenvalue of S : V Ñ V . Hence, λ is a root of detpxidV ´Sq “ 0.Let w be an eigenvector for λ, so Sw “ λw. Define S “ S´λidV . We will showthat for all v P V , v P kerpSq, and therefore Sv “ λv for all v P V .

This will follow if we can show for all g P G, ST pgq “ T pgqS and then applypart (a).

ST pgqpvq “ ST pgqv ´ λidV T pgqv

“ T pgqSv ´ T pgqpλidV vq

“ T pgq

ˆ

Sv ´ λidV v˙

“ T pgqSv

13

Now we can apply part (a) to claim that S “ 0. Expanding the definition ofS, we see that S ´ λidV “ 0, so therefore S “ λidV .

This proof doesn’t use the fact that G is a finite group. In fact, it holds inmuch more generality.


Last week, we talked about sets of objects up to equivalence, which are calledModuli spaces. The equivalence relations we care about are the orbits of agroup action.

Let G be a finite group. Today, we will show that characters of irreduciblerepresentations form an orthonormal basis for the space of class functions sub-ject to the inner product

xχT , χSy “1

|G|

ÿ

gPG

χT pgqχSpgq.

Here, T, S are irreducible representations. Recall that

χT psq “ trpT pgqq “ sum of diagonal of matrix “ sum of eigenvalues .

Let g P G, and let T : GÑ GLnC. Let λ be an eigenvalue of T pgq. We claimthat λ lies on the unit circle. Let ~v be an eigenvector for T pgq corresponding toλ. Then,

T pg2q~v “ T pgqT pgq~v “ T pgqλ~v “ λ2~v

Thus, we can concludeT pgkq~v “ λk~v.

Since G is a finite group, g has a finite order, say gn “ 1. So,

~v “ T p1q~v “ T pgnq~v “ λn~v,

and therefore λn “ 1. Doing some more manipulations, we find that

1

λ“ λn´1.

Now if λ “ a` bi, then1

λ“ a´ bi “ λ.

What does this tell us about characters? Well, the trace is the sum of theeigenvalues, so

χT pgq “ trpT pgqq “ λ1 ` . . .` λn

14

Taking conjugates,χT pgq “ λ1 ` . . .` λn.

Using that 1λ “ λ, then

χT pgq “1

λ1` . . .`

1

λn“ χT pg

´1q,

because the eigenvalues of the inverse of a matrix are the reciprocals of theeigenvalues.

That was just the first step. Let T and S be inequivalent irreducible repre-sentations.

T : GÑ GLnC

S : GÑ GLmC

Our goal is to show that

0 “1

|G|

ÿ

gPG

χT pgqχSpgq. (1)

Notice first that

1

|G|

ÿ

gPG

χT pgqχSpgq “1

|G|

ÿ

gPG

χT pgqχSpg´1q.

We’re going to rewrite this in terms of not eigenvalues, but instead as thesums of diagonals of matrices. Let

T pgq “ ptijpgqq1ďi,jďn Spgq “ psijpgqq1ďi,jďm,

so for example

χT pgq “mÿ

i“1

tiipgq.

χSpg´1q “

mÿ

i“1

siipg´1q.

Let’s plug these into the equation 1 from earlier. We now want to show that

0 “1

|G|

ÿ

gPG

˜

mÿ

i“1

tiipgq

¸˜

mÿ

i“1

siipg´1q

¸

The strategy here is to plug in things to a polynomial equation and get zero,and from there conclude that all coefficients (these nasty sums) have to be zero.This is like saying that the polynomial ax2 ` bx ` c “ 0 for all x implies thata “ b “ c “ 0.

15

Let X : Cn Ñ Cm be a matrix X “ pxijq for 1 ď i ď m, 1 ď j ď n. X is ourmatrix of variables. Define X0 : Cn Ñ Cm by

X0 “ÿ

gPG

Spgq´1XT pgq.

This new function is the composition

X0 : Cn T pgqÝÝÝÑ Cn X

ÝÑ Cm SpgqÝÝÝÑ Cm.

The key point here is that X0 is an intertwining operator! This point is high-lighted in the following fact.

Fact 29. For all g P G, X0T pgq “ SpgqX0.

If we know this, then because S and T are inequivalent, X0 must be zero.

Proof. Compute both sides.

X0T phq “ÿ

gPG

Spg´1qXT pgqT phq

“ÿ

gPG

Spg´1qXT pghq

“ÿ

k“gh

Sphk´1qXT pkq

“ÿ

kPG

SphqSpk´1qXT pgq

“ Sphqÿ

kPG

Spk´1qXT pgq “ SphqX0

Now we apply Schur’s Lemma1 and conclude X0 “ 0. Therefore, examin-ing each individual entry of the matrix X0,

0 “ x0ij “ÿ

gPG

mÿ

a“1

nÿ

b“1

siapg´1qxabtbjpgq

Now let a “ i, b “ j. The coefficient of xij in the above sum is

0 “ÿ

gPG

siipg´1qtjjpgq (2)

We’re not quite there yet, but this looks a lot like the trace formula! It’s enoughto use it.

1without which we are nothing

16

xχT , χSy “1

|G|

ÿ

gPG

trT pgqtrSpg´1q

“1

|G|

ÿ

gPG

pt11pgq ` t22pgq ` . . .` tnnpgqq`

s11pg´1q ` s22pg

´1q ` . . .` smmpg´1q

˘

“1

|G|

ÿ

gPG

nÿ

i“1

mÿ

j“1

tiipgqsjjpg´1q

“1

|G|

nÿ

i“1

mÿ

j“1

˜

ÿ

gPG

tiipgqsjjpg´1q

¸

“1

|G|

nÿ

i“1

mÿ

j“1

ˆ

coefficientof xij as in (2)

˙

“ 0


Last time we talked about the Schur orthogonality relations among charactersof a finite group. This says that the rows of the character table form an or-thonormal basis for the space of class functions on a group G, that is,

tf : GÑ C | fpxgx´1q “ fpgq for all g P Gu.

The inner product on this space is given by

xα, βy “1

|G|

ÿ

gPG

αpgqβpgq.

If χS , χT are characters of irreducible representations, then

xχT , χSy “

#

1 T – S

0 otherwise.

There is a similar orthogonality relation among the columns of the charactertable, but there is a slightly different inner product.

Proposition 30. Let g, h P G, and let χi for i “ 1, . . . , k index the irreduciblerepresentations of G. Then if g is in conjugacy class C,

1

|G|

kÿ

i“1

χipgqχiphq “

#

|C| if g and h are conjugate0 otherwise

.

Example 31. Let’s go back to S3. Before we have anything in the charactertable, we can determine what the dimensions of the representations are. Notethat

χV pidq “ trpρV pidqq “ tridV “ dimV.

17

Therefore, the first column is the dimensions. And by the column orthogonal-ity, we know that the sum of squares of dimensions is |S3| “ 6. The only wayto write 6 as the sum of three squares 6 “ a2` b2` c2 is 6 “ 12` 12` 22, so thedimensions must be 1, 1 and 2.

S3 tidu tp1 2q, p1 3q, p2 3qu tp1 2 3q, p1 3 2qu

trivial a “ 1

sgn b “ 1

χ c “ 2

Of course, we know that the first row is p1, 1, 1q and the last row is p2, 0,´1q.We can write the inner product on characters as a bilinear form:

xtrivial, χy “

C

»

–

1

1

1

fi

fl ,

»

–

2

0

´1

fi

fl

G

“1

6r2, 0,´1s

»

–

1 0 0

0 3 0

0 0 2

fi

fl

»

–

1

1

1

fi

fl (3)

Notice that the matrix has sizes of conjugacy classes on the diagonal!Now let T : S3 Ñ GLpnq be any representation, with character χT pidq “ 11,

χT p1 2q “ ´3, and χT p1 2 3q “ ´1. We want to decompose T into irreps. Weknow that T pgq is similar to a matrix

»

—

—

—

–

T1pgq

T2pgq. . .

T`pgq

fi

ffi

ffi

ffi

fl

where each Ti is an irrep. So form this, we know that

χT “ α trivial` β sgn` γχ,

and we want to find α, β, γ. So we just take some inner products to find thatα “ xχT , trivialy, β “ xχT , sgny, and γ “ xχT , χy. Let’s compute it using thebilinear form from (3).

xχT , trivialy “1

6r1, 1, 1s

»

–

1 0 0

0 3 0

0 0 2

fi

fl

»

–

11

´3

´1

fi

fl “1

6r1, 1, 1s

»

–

11

´9

´2

fi

fl “ 0

xχT , sgny “1

6r1,´1, 1s

»

–

1 0 0

0 3 0

0 0 2

fi

fl

»

–

11

´3

´1

fi

fl “1

6r1,´1, 1s

»

–

11

´9

´2

fi

fl “ 3

xχT , χy “1

6r2, 0,´1s

»

–

1 0 0

0 3 0

0 0 2

fi

fl

»

–

11

´3

´1

fi

fl “1

6r2, 0,´1s

»

–

11

´9

´2

fi

fl “ 4

18

From this, we know that the representation T is composed of 3 copies of thesign representation and 4 copies of the 2-dimensional representation.

Remark 32. Fourier analysis is the representation theory of an abelian infinitegroup, namely the unit circle S1. This is an abelian group, so even though thegroup is infinite, each of the irreducible representations is one-dimensional.These representations are just eiθ. This is related to a construction on groupscalled the Pontryagin Dual.

For now, we’ll be done with character tables and finite group representa-tions. Instead, we’ll focus on explicitly finding irreducible representations ofthe symmetric groups Sn. This is important to our eventual goal of going backto algebraic geometry, because the irreducible representations of Sn are veryclosely related to the irreducible representations of GLpnq. And that’s what wereally want to know in algebraic geometry.

The first thing we need to know are the conjugacy classes of Sn. These cor-respond to the cycle types of elements, and are represented by Young diagrams.For instance, for n “ 3, we get

identity

two cycles

three-cycles

Definition 33. A Young tabloid is a Young diagram filled in with numbersfrom 1 to n, where n is the number of boxes in the diagram. They are consid-ered equivalent up to permutation of the rows.

Example 34. For instance, with the Young diagram

there are 3 possibilities,

1 23

1 32

2 31

.

These correspond to the permutations p1 2qp3q, p1 3qp2q, and p2 3qp1q, respec-tively.

Definition 35. A partition of n is a sequence λ “ pλ1, λ2, . . . , λkq with λ1 ě

λ2 ě . . . ě λk such that λ1 ` λ2 ` . . .` λk “ n. A partition of n is often writtenλ % n.

19

Young diagrams correspond to partitions, as described in a previous lec-ture. Briefly, the partition p2, 1q of 3 corresponds to . We say a Young dia-gram is of shape λ where λ is the corresponding partition of n.

The irreps of Sn correspond to conjugacy classes, which correspond to Youngdiagrams. So it’s natural to search for irreps in the Young diagrams.

To that end, define a representation space for Sn for λ a partition of n by

F pMλq “à

Young tabloids Yof shape λ

CY.

It’s the C-vector space with basis all of the Young tabloids of shape λ.For example, when λ “ 2 ` 1 a partition of 3, we have an element of

F pM2`1q given by

p3, 2, 6q “ 3 1 23

` 2 1 32

` 6 2 31

.

What is the action of S3 on this space? It just permutes the basis vectorsby permuting the corresponding boxes in the Young tabloid. For instance, ifσ “ p1 2q P S3,

σ ¨ 1 23

“2 13

.


Last time, we talked about representations of the symmetric group. Our goalthis time is to introduce an algorithm to explicitly find all irreducible represen-tations for any Sn. We’re going to start with an example and do it for S3 andon your problem set you will do it for S4.

The first thing we should know for our algorithm is the termination condi-tion. How many irreducible representations should we have? We know thatthe number of irreducible representations is equal to the number of conjugacyclasses. For Sn, the conjugacy classes are given by cycle type, which is capturedby the Young diagrams. This number is equal to P pnq, the partition number ofn. This is equal to the number of Young diagrams.

Example 36. For S3, we need to find the partitions of 3. There are three ofthem: 3 “ 3 “ 2` 1 “ 1` 1` 1. These correspond to Young diagrams.

p3q

p2, 1q

p1, 1, 1q

20

Our goal here is to match each Young diagram with the corresponding ir-reducible representation of Sn. First, we introduce a partial ordering on Youngdiagrams.

Definition 37. The partial ordering on partitions / Young diagrams is givenfor two partitions of n, called µ “ pµ1, . . . , µnq and λ “ pλ1, . . . , λnq as follows.We say λ ě µ if and only if

λ1 ě µ1

λ1 ` λ2 ě µ1 ` µ2

λ1 ` λ2 ` λ3 ě µ1 ` µ2 ` µ3

λ1 ` . . .` λn ě µ1 ` . . .` µn

Example 38. For example, among partitions of 3, the partition p3q ě p2 ` 1q

and p2` 1q ě p1` 1` 1q.

This is only a partial ordering. There may be some incomparable partitions.For example, the partitions p3, 3q and p4, 1, 1q are not compatible. Although4 ě 3, we also have 4` 1 ď 3` 3.

Problem 39. Describe this ordering in terms of the Young diagrams.

Last time, we introduced Young diagrams and Young tabloids. The Youngdiagrams are the boxes, and the Young tabloids are the boxes with numbers 1

through n inside of them, but among the rows it doesn’t matter in what orderthe numbers appear.

Example 40. For the Young diagram corresponding to the partition p2, 1q, thereare three associated Young tabloids

ù1 23

1 32

2 31

.

For each Young diagram, we want an irreducible representation – a vectorspace with an action of Sn. We have some number of Young tabloids for eachpartition λ of n. For notation, let’s list the different Young tabloids correspond-ing to λ as δ1, . . . , δk. These are going to be the basis elements for a vector spaceon which Sn will acts.

Mλ “

kà

i“1

Cδi.

This will not be the final vector space on which Sn will act to give an irreduciblerepresentation, but it will contain the irreducible representation.

Example 41. Back to S3. For the partition 3 “ 2` 1, we get the vector space

Mp2,1q “

"

α 1 23

` β 1 32

` γ 2 31

ˇ

ˇ

ˇ

ˇ

α, β, γ P C*

– C3

21

For each such vector space, we get an action of Sn. Let’s illustrate with anexample.

Example 42. There is a natural action of S3 on Mp2,1q. Let σ “ p1 2q. There is anatural action on Mp2,1q given by

σ ¨ i jk

“i1 j1

k1,

where i1 “ σpiq, j1 “ σpjq and k1 “ σpkq.The action of σ permutes the basis vectors as

σ ¨ 1 23

“1 23

σ ¨ 1 32

“2 31

σ ¨ 2 31

“1 32

Therefore, σ acts on any vector in Mp2,1q by

σ ¨

ˆ

α 1 23

` β 1 32

` γ 2 31

˙

“ β 1 23

` α 1 32

` γ 2 31

.

So the matrix corresponding to σ is

σ “

»

–

0 1 0

1 0 0

0 0 1

fi

fl .

In general, we get a representation Sn Ñ GLpMλq in this way. However,this representation is not irreducible, but it has an irreducible representationcorresponding to λ and all other irreducible representations making up thisrepresentation correspond to Young diagrams µ ě λ. This means that by start-ing with the largest partition under the partial order, we can find one irrep.And at each next step, the representation decomposes as only irreps we havealready found and exactly one irrep we have not yet seen.

Example 43. So the steps of our algorithm are as follows, illustrated for S3.Here we have

p3q ě p2, 1q ě p1, 1, 1q.

So we begin with the representation for the largest partition, which is p3q. Wefirst start by finding S3

œ

Mp3q, and then we find S3

œ

Mp2,1q and strip outparts corresponding to S3

œ

Mp3q. And then we move on to S3

œ

Mp1,1,1q butremove parts equivalent to S3

œ

Mp3q and S3

œ

Mp2,1q.

22

Now, we need to talk about the Young Tableaux. To distinguish the Youngtabloids from the Young tableaux, set

1 23

“ δ31 32

“ δ22 31

“ δ1

Definition 44. The Young Tableaux are the different ways to put the numbers1 through n in the different boxes of a Young diagram of size n.

Example 45. For the partition p2, 1q of 3, there are six Young Tableaux.

1 23

2 13

1 32

3 12

2 31

3 21

These project by a map π onto the Young tabloids, as follows:

1 23

2 13

1 32

3 12

2 31

3 21

1 23

“ δ31 32

“ δ22 31

“ δ1

π π π π π π

Let t be a Young Tableaux for some λ a partition of n. Let

Ct “ tσ P Sn | σ permutes elements in the same column u.

Example 46.C 1 2

3

“ tid, p1 3qu Ă S3

We also define elements et PMλ via

et :“ÿ

σPCt

sgnpσqπpσptqq.

Let’s see what this is in the case of S3.

Example 47.

e 1 23

“ π

ˆ

1 23

˙

´ π

ˆ

3 21

˙

“1 23

´2 31

“ δ3 ´ δ1

Our irreducible representation corresponding to a Young diagram λ will be

GLˆ

Spanpet | t Young tableaux of λq˙

23

Example 48. Let’s do it again for S3. Set

t “ 1 32

.

Then Ct “ tid, p1 2qu. We find

id ¨ t “ 1 32

p1 2q ¨ t “ 2 31

.

Therefore, et “ δ2 ´ δ1. Now set

t “ 2 13

.

Then Ct “ tid, p2 3qu. So

id ¨ t “ 2 12

p2 3q ¨ t “ 3 12

.

Therefore, et “ δ3 ´ δ2.So the three vectors we get are δ3 ´ δ1, δ2 ´ δ1, and δ3 ´ δ2. These are not

linearly independent, so we get a representation of S3 of dimension 2.


Today we’ll explicitly find all of the irreducible representations of the symmet-ric group for the case of S3, and along the way illustrate the general method.

Young diagrams index the conjugacy classes of Sn, and therefore the irre-ducible representations. Let λ be a Young diagram, and let δ1, . . . , δk denoteYoung Tabloids for λ. Set

Mλ “ Cδ1 ‘ Cδ2 ‘ . . .‘ Cδk.

The representation Sn Ñ GLpMλq contains one new irreducible representation.Let t be a Young Tableaux for λ. Set

Ct “ tσ P Sn | σpermutes columns of tu

for example,C 1 2

3

“ tid, p1 3qu.

The irreducible representation is GLpSpantetuq, where

et “ÿ

σPCt

sgnpσqπpσptqq.

The first half of this lecture is basically contained in the following example.

24

Example 49. For S3 and λ “ , the Young Tableaux are

1 23

2 13

1 32

3 12

2 31

3 21

These project by a map π onto the Young tabloids, as follows:

1 23

2 13

1 32

3 12

2 31

3 21

1 23

“ δ31 32

“ δ22 31

“ δ1

π π π π π π

We found last time that

Spanptet | t Young Tableaux for λuq “ Spanpδ1 ´ δ3, δ1 ´ δ2q.

There is an irreducible representation T : S3 Ñ GLp2,Cq with character givenby

χT pidq “ 2, χT p1 2q “ 0, χT p1 2 3q “ ´1.

Let’s write

δ1 ´ δ3 “ v1 „

ˆ

1

0

˙

δ1 ´ δ2 “ v2 „

ˆ

0

1

˙

.

How do the elements of S3 act on the vectors v1 and v2?

element of S3 v1 “ δ1 ´ δ3 v2 “ δ1 ´ δ2id v1 v2p1 2q v1 ´ v2 ´v2p1 3q

p2 3q

p1 2 3q ´v2 v1 ´ v2p1 3 2q

This shows that the matrix of the element p1 2q is given by

T p1 2q “

„

1 0

´1 ´1

,

so we can confirm that χT p1 2q “ trpT p1 2qq “ 0. Similarly, the matrix for p1 2 3q

is

T p1 2 3q “

„

0 1

´1 ´1

,

and as expected, χT p1 2 3q “ trpT p1 2 3qq “ ´1.

25

Remember that λ “ . The representation Mλ was three dimensional! Sowe found a two-dimensional irreducible representation, so can we decomposeMλ into irreducibles?

Well, there’s a subspace Cpδ1 ` δ2 ` δ3q, which we didn’t look at before. Itis invariant because for any α P C,

σpαpδ1 ` δ2 ` δ3qq “ αpδσp1q ` δσp2q ` δσp3qq

Switching gears – invariant theory.

We are interested about equivalence questions. Given a space X , we oftenconsider elements of X equivalent up to action of some group G. Elementsx, y P X are equivalent if there is some g P G such that g ¨ x “ y. For example,in the discussion of conics, we found that there were three equivalence classescorresponding to double lines, two distinct lines, and circles.

Definition 50. Let G

œ

X . The Orbit of x P X is the set

tg ¨ x | g P Gu.

Definition 51. An invariant of a space X is a function f : X Ñ C such that ifx „ y, then fpxq “ fpyq.

Example 52. The genus of a surface is a topological invariant. We can classifycompact two-dimensional surfaces, for example, by their genus.

We can instead think about invariants in terms of category theory too. In-stead of a map to C, invariants can take values in a different category. Givenany category C, we can have as an invariant some functor F : C Ñ D for someother category D.

Example 53. The fundamental group π1 : Top Ñ Groups is an invariant ontopological spaces taking values in the category of groups.

Given some spaceX and an actionG

œ

X , we want to find all the functionswhich are constant on orbits, that is, all the possible invariants of the space. Tothat end, let R be the ring of all functions on X .

Definition 54. Let G

œ

X , and let R be the ring of all functions on X . Then thering of invariant functions on X is

RG :“

"

f P R | fpg ¨ xq “ fpxq for all x P X, g P G*

.

In general, there may be infinitely many invariant functions in RG – if f isinvariant, so is f ˝f , and f ˝f ˝f , and so on. So instead of asking ifRG is finite,we ask if it’s finitely generated. And even if RG is finitely generated, we wantto know if there are finitely many relations between generators? Or can we getall of the relations from finitely many? So we want to know the answer to thefollowing questions

26

(1) is RG finitely generated?

(2) is the ring of relations between generators finitely generated?

In invariant theory, there are two important theorems which answer thesequestions, respectively. They are given the boring names

(1) The First Fundamental Theorem of Invariant Theory;

(2) The Second Fundamental Theorem of Invariant Theory.

They are really more schema for questions to ask rather than honest-to-godtheorems. Often called syzygies2.

Example 55. Let R “ Crx1, . . . , xns and let Sn

œ

R by permuting the coordi-nates.

σ ¨ fpx1, . . . , xnq “ fpxσp1q, . . . , xσpnqq

For example, with σ “ p1 2q and R “ Crx1, x2, x3s,

f “ x21 ` x22 ` x2 ÞÑ σ ¨ f “ x22 ` x

21 ` x3.

What is the ring of invariants? It is generated by the elementary symmetricpolynomials

σ1 “ x1 ` x2 ` . . .` xn

σ2 “ x1x2 ` x2x3 ` . . .` xn´1xn

......

σn “ x1x2 ¨ ¨ ¨xn

This answers the First Fundamental Theorem of Invariant Theory for Sn

œ

R.The Second Fundamental Theorem is much easier to answer – no relations.

Why do we care? Let’s switch gears for a while and talk about science (ismath not science? Are we not doing science now?). To a mathematician, thefollowing completely describes science.

Joke 56.Science = Numbers in boxes

For any physical phenomena in an experiment, you should get the samenumbers (up to units). This is an equivalence relation. The invariants are thesymmetries of a system. These symmetries are very important to science, be-cause they can help us determine the structure of physical phenomena. We candescribe the relations among these ideas.

2worth 24 points in Scrabble!

27

Lagrangian ofyour system

AllowableSymmetries

Invariants(Conservation Laws)

These are connected by Noether’s Theorem, which is possibly the ultimate in-variant theorem in Physics.


Problem 57. Make a pop-up book about Schur’s lemma.

Last time we talked about invariants and invariant theory. Given a space Xand a finite group G acting on X , we want to find objects up to equivalence.An invariant for the action of G on X is a function f : X Ñ C that is constanton each orbit of G

œ

X .Given a spaceX , we consider the ringR of functions onX . The subringRG

is the ring of invariant functions, which is constant on each orbit (alternatively,functions on the orbit space X {{G).

There are two fundamental theorems of invariant theory. The first describesif RG is finitely generated, and the second describes the number of relationsamong the generators. These are not really theorems, but questions that youask of a particular action of a group on some space.

Today we’re going to talk about the alternating group acting on polynomi-als and find the first two fundamental theorems.

An “ tσ P Sn | sgnpσq “ 1u

We let the alternating group act on

R “ Crx1, . . . , xns.

Of course An Ď Sn, so if some polynomial is invariant under the action of Sn,then it is also invariant under action of An.

Hence, the ring of invariants RAn still contains all of the elementary sym-metric polynomials,

σ1 “ x1 ` x2 ` . . .` xn

σ2 “ x1x2 ` x2x3 ` . . .` xn´1xn

......

σn “ x1x2 ¨ ¨ ¨xn.

28

But it also contains a new invariant. Claim that the new invariant is

∆px1, . . . , xnq “ź

iăj

pxi ´ xjq.

Example 58. When n “ 3, we have

∆ “ px1 ´ x2qpx1 ´ x3qpx2 ´ x3q

Notice that ∆ is not invariant under S3, because

p1 2q ¨∆ “ px2 ´ x1qpx2 ´ x3qpx1 ´ x3q “ ´∆,

but it is invariant under A3. For instance,

p1 2 3q¨∆ “ p1 2 3q¨px1´x2qpx1´x3qpx2´x3q “ px2´x3qpx2´x1qpx3´x1q “ ∆

Notice in the above example that the action of σ P S3 on ∆ picks up afactor of sgnpσq. This is true in general, and gives us a clue as that there isan algebraic relation between ∆ and the symmetric polynomials. In particular,∆2 is a symmetric polynomial in RSn , because acting on ∆ might pick up anegative sign, but squaring it eliminates the sign difficulties.

Example 59. For n “ 2,

∆2 “ px2 ´ x1q2

“ x22 ´ 2x1x2 ` x21

“ px1 ` x2q2 ´ 4x1x2 “ σ2

1 ´ 4σ2

Something similar happens for n “ 3, 4, . . .

Are there any other polynomial generators needed for RAn? The answer isno! This is the first fundamental theorem of invariant for An

œ

Crx1, . . . , xns.

Claim 59.1.RAn “ Crσ1, . . . , σn,∆s

Proof. Let f P RAn , which means that for all τ P An, τ ¨ f “ f . Our goal is towrite f as

f “ symmetric polynomial `∆ p symmetric polynomial q ,

and we know that we don’t need any other powers of ∆ because ∆2 is sym-metric.

Let f “ p1 2q ¨ f . Then we claim that f “ pi jqf for any transposition pi jq.To show this, notice that

pi jq “ p1 2qp1 2qpi jq,

29

and moreover p1 2qpi jq is an even permutation. So p1 2qpi jq ¨ f “ f because fis invariant under the action of the alternating group, and p1 2qpi jq P An.

pi jqf “ p1 2qp1 2qpi jqf

“ p1 2qf

“ f .

Now we can think about f ` f and f ´ f and rewrite

f “1

2pf ` fq `

1

2pf ´ fq.

This is closer to what we want if one of those two terms is symmetric. Let’sclaim that f ` f is symmetric. For σ P Sn an even permutation,

σpf ` fq “ σf ` σp1 2qf

“ f ` p1 2qσf σ even

“ f ` p1 2qf

“ f ` f .

Now that we’ve shown that f ` f is invariant under An, we just need to showthat it’s invariant under transpositions to show that f ` f is invariant underSn. To that end, let pi, jq P Sn.

pi jqpf ` fq “ pi jqpf ` pi jqfq

“ pi jqf ` pi jqpi jqf

“ f ` f.

Therefore, f ` f is a symmetric polynomial, that is, an element of RSn .Now let’s think about f ´ f . What happens when we act on it by p1 2q?

p1 2qpf ´ fq “ p1 2qpf ´ p1 2qfq

“ p1 2qf ´ p1 2qp1 2qf

“ p1 2qf ´ f

“ f ´ f “ ´pf ´ fq

Of course, there was nothing special about p1 2q, so this works in fact for anytransposition in Sn. Acting on f ´ f by a transposition pi jq picks up a sign.We can think of f ´ f as a one-variable polynomial in xi, and when we plug inxi “ xj , the polynomial is equal to its own opposite and therefore identicallyzero. Since xj is a root of pf ´ fq as a polynomial in xi, there must be a factorof pxi ´ xjq in f ´ f .

pxi ´ xjq | pf ´ fq

30

Since this is true for all transpositions, and the pxi ´ xjq are relatively prime(by degree considerations), we conclude that

∆ | pf ´ fq

Hence, we writef ´ f “ ∆g

for some polynomial g.Finally, claim that g is symmetric. For σ P Sn, write σ “ p1 2qτ where τ is

even. We know that

σ ¨ pf ´ fq “ ´pf ´ fq,

but also,

σ ¨ pf ´ fq “ σ ¨ p∆gq

“ σ ¨∆σ ¨ g

“ ´∆pσ ¨ gq,

so we conclude that

´p∆gq “ ´pf ´ fq “ ´∆pσ ¨ gq,

so σ ¨ g “ g and g is symmetric.Hence, we write

f “1

2pf ` fq `

1

2pf ´ fq “

1

2pf ` fq `

1

2∆g,

and pf ` fq, g P RSn are symmetric.

Definition 60. Let GLpnq act onX with ring of functionsR. Let g P GLpnq, andlet f P RGLpnq. Then f is covariant with weight k if

gpfq “ pdet gqkf.

Example 61. The action of GLpnq on nˆ n matrices given by

φpBq “ ATBA

has a covariant of weight 2 given by det : Mn Ñ C, because

detpσpBqq “ detpATBAq “ detpAT qdetpBqdetpAq “ detpAq2 detB

The invariants here are coefficients of detpλI ´Bq.

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Given a homogeneous polynomial of degree two, P px, yq “ ax2`bxy`cy2,a covariant under change of coordinates is fpa, b, cq “ b2 ´ 4ac. This is the dis-criminant of the quadratic formula! A big problem in the 19th century amongGerman mathematicians was the following. This is what algebraic geometrywas all about in that era.

Problem 62 (Gordon’s problem). Solve the first fundamental problem of in-variant theory for change of coordinates of homogeneous polynomials of eachdegree and number of variables.

After many years of people trying to find algorithms to find the generatorsof the rings of invariants, David Hilbert solved the whole thing in about tenpages: all invariants are finitely generated, but he didn’t produce an algorithm.Gordon said of this proof “Das ist nicht Mathematik, das ist Theologie.” Twoyears later, Hilbert found an algorithm to compute all of the generators of thering of invariants.

Much of this theory leads to moduli spaces and David Mumford’s work inalgebraic geometry, which we will talk about for the rest of this week.


Today we’re going to give an intro to algebraic geometry in haunting similaritywith the morning lecture. We will see that the notion of objects up to equiva-lence is very closely related to invariant functions on some space.

In terms of algebraic geometry, there is a close relationship between geo-metric objects called algebraic sets and algebraic objects: namely, commutativerings. Functions describe the world, and these commutative rings are indeedrings of functions.

We will use the ring of invariants to create a new algebraic object, whichwill capture the orbits of the objects of some space under action of a group.

Definition 63. Let S Ă Crx1, . . . , xns. An algebraic set is the set of commonzeros of polynomials in S.

tpa1, . . . , anq P Cn | P pa1, . . . , anq “ 0 for all P P Su

Example 64. The quintessential example is the zeros of x2 ` y2 ´ 1 in C2. This

32

is the circle.

For our algebraic sets, we may assume that S “ I is an ideal of Crx1, . . . , xns.The notation is

VpIq :“ tpa1, . . . , anq | P pa1, . . . , anq “ 0, @ P P Iu

The next question we want to consider is: “What are the (polynomial) func-tions defined on VpIq?”, that is, functions VpIq Ñ C.

Example 65. What are the polynomial functions on the circle? For example,Qpx, yq “ x3 is an example. This is a function on all of C2, but also on the circle.Additionally, P px, yq “ x3`x2`y2´1 is a polynomial function on all of C2. Ingeneral, P ‰ Q in C2. But on the other hand, for points px, yq P Vpx2 ` y2 ´ 1q,P px, yq “ Qpx, yq. They’re the same at all values on the circle! So logically, weshould say that P “ Q on the circle. Indeed, P,Q represent the same coset in

Crx,ys{xx2`y2´1y

In general, the functions on VpIqwill be described by

Crx1,...,xns{I .

Technically, we want I to be a radical ideal, that is, if fn P I , then f P I .

Definition 66. Given an algebraic set V associated to an ideal IpV q, the ring ofregular functions on V is

OV “Crx1,...,xns{IpV q.

So we know how to go one direction: given an algebraic set V , we canproduce a ring associated to it. We want to be able to do the converse as well:how can we go from knowing a commutative ring to knowing the algebraic setassociated to it? This theorem is a partial answer.

Theorem 67. LetR be a commutative ring (with some additional assumptions).The points in the corresponding algebraic set are in bijection with the maximalideals of R.

33

Example 68. Let R “ Crxs. What is the algebraic set associated to it? Well, themaximal ideals of R are xx ´ ay for a P C, so for each maximal ideal we get apoint for each a P C.

For Crx, ys, the maximal ideals are xx´ a, y´ by, so correspond to points inC2.

Definition 69. Given a ring R, the associated geometric object is the spectrumof R,

SpecpRq “ tI CR | I primeu.

The maximal ideals are called geometric points.

The spectrum of a ring R has an associated topology, called the Zariskitopology, and we can do much much more with these spaces. But we’ll leaveit here for now.

Example 70.SpecZ “

x0y, x2y, x3y, x5y, x7y, . . .(

To summarize, for any algebraic set V , there is an associated ring OV . Tothis ring OV , we associate it’s field of fractions

KV “

"

f

g

ˇ

ˇ

ˇ

ˇ

f, g P OV , g ‰ 0

*

.

This field is the set of rational functions on the algebraic set.

Definition 71. An algebraic set V is a variety if

IpV q “ tP | @ ~a P V, P p~aq “ 0u

is a prime ideal.

Now let’s go back to equivalence problems.

Definition 72. There are two notions of equivalence between two algebraicvarieties V and W .

(a) V is isomorphic to W if and only if OV is isomorphic to OW ;

(b) V is birationally equivalent to W if and only if KV is isomorphic to KW

as a field.

While we may naıvely think that these are the same, but they are very muchnot the same. The first is a very rigid notion of equivalence, while birationalequivalence isn’t so rigid. In recent years, there have been a number of break-throughs in the area of birational geometry called the minimal model program,which aims to find a canonical element in each birational equivalence class.

Example 73. P2C is birationally equivalent to P1C ˆ P1C, yet these are notisomorphic.

34

Projective algebraic geometry

So far, everything we’ve done is in affine space. But we want to work overprojective space, which is where we’ve been thinking about equivalences. Forexample, there are lots of conics in affine space, but only three types in projec-tive space.

So just replace every instance of Spec in the above with Proj, right?Okay, we’ll do it carefully. In PnC, px0, . . . , xnq “ pλx0, . . . , λxnq for any

λ ‰ 0. So how does it make sense to define a polynomial on PnC?

Example 74. Let P px, y, zq “ x2 ` y2 ´ z2. This polynomial is poorly behavedin projective space, where p1, 1, 1q and p2, 2, 2q are the same point.

P p1, 1, 1q “ 12 ` 12 ´ 12 “ 1

P p2, 2, 2q “ 22 ` 22 ´ 22 “ 2

So instead we work with homogeneous polynomials (every monomial hasthe same degree), where at least the zero sets are well-defined.

Definition 75. Let I be an ideal of Crx0, x1, . . . , xns. I is homogeneous if it isgenerated by homogeneous polynomials.

Here, we think of Crx0, x1, . . . , xns as a graded ring

Crx0, x1, . . . , xns “8à

n“0

Crx0, x1, . . . , xnsdeg“n,

where Crx0, x1, . . . , xnsdeg“n is the polynomials of degree n.

Definition 76. Let I be a homogeneous ideal. Then the projective algebraicset defined by P is

VpIq “ tpa0, . . . , anq | P pa0, . . . , anq “ 0 for all P P Iu

Definition 77. Let R be a graded ring

ProjpRq “

homogeneous, prime ideals of R(


Let’s recall what’s been going on here. Let X be a geometric object with anaction of a groupG. We consider x „ y if there is some g P G such that g ¨x “ y.Let R be the ring of functions on X (in algebraic geometry, OX ). Then the ringof invariants is RG “ tf P R | fpg ¨ xq “ fpxq for all x P X, g P Gu. We willrelate the space of orbits X {{G to SpecpRGq.

35

Suppose that RG is generated by f1, . . . , fk. There’s a map X Ñ Ck definedby

p ÞÑ pf1ppq, f2ppq, . . . , fkppqq.

We hope that for each orbit of G

œ

X , there is a corresponding single point inCk under the image of this map.

Let’s look at some concrete examples. Sometimes this works out really well,but sometimes it fails horribly.

Example 78. Let G “ pC,`q. Let X “ C2 and let t P C. Define an action of Gon X by t ÞÑ σt : X Ñ X ,

σtpa, bq “ pa, b` tq.

For concreteness,σ3p8, 11q “ p8, 14q.

What do the orbits of this action look like?

pa, bq

Here, R “ Crx, ys is the polynomial functions on X . The functions whichare invariant are those that don’t depend on y, so RG “ Crxs. This ring isgenerated by fpxq “ x.

We have the map X “ C2 Ñ C defined by

pa, bq ÞÑ fpa, bq “ a,

and for each orbit we have one point on the X-axis. We see this from X {{G “

C2{C “ C.Alternatively, we could have thought about SpecpRGq “ SpecCrxs which

corresponds to C.

Here’s another example that doesn’t work out so well.

Example 79. Let G “ Cˆ, and let X “ C2. Let G act on X by t ÞÑ σt : X Ñ X ,

σtpa, bq “`

ta, b{t˘

.

There are three types of orbits.

36

(1) If ab ‰ 0, then we get a hyperbola centered at the origin.

(2) if exactly one of a, b is zero (that is, 0 P ta, bu), then we get one of the axeswithout the origin. For example, if a ‰ 0, b “ 0, then we get the followingpicture. Note that there are two distinct versions of this orbit: the x-axisand the y-axis. These are kind of degenerate hyperbolae.

(3) If a “ b “ 0, then we get the origin.

The ring of functions here is R “ Crx, ys (this is also OX “ R). An example ofan invariant function on a hyperbola orbit xy “ 1 is the function fpx, yq “ xy.These are the only types of invariant functions.

So RG “ Crxys and is generated by fpx, yq “ xy. Our function X Ñ C ispa, bq ÞÑ ab. Again, SpecpRGq is just the complex numbers.

37

Unfortunately, this fails to distinguish all of the orbits. It distinguishes all ofthe hyperbola orbits as xy “constant, but fails to distinguish the other orbits:each gets mapped to zero.

What’s the failure here? It turns out to be the fact that the orbits of thesecond type are not closed, yet the hyperbola orbits are closed.

An invariant function on the orbit of p0, 1q has to be constant, and maintainthat value at the origin too because any invariant function is continuous. Thus,this invariant function has to be the same on the orbit of p1, 0q – if it was not,then the invariant function would be discontinuous on the orbit of p1, 0q.

In general, orbits whose closures intersect will be identified.

Example 80. Let G “ Cˆ, let X “ Cn and let t P Cˆ. Let σt : Cn Ñ Cn bedefined by σtpa1, . . . , anq “ pta1, . . . , tanq.

What are the orbits? Lines through the origin skipping the origin or theorigin itself.

None of these orbits are closed, and yet all of the orbit closures intersect at theorigin. So the only invariant functions are the constants!

In general, given a space X (generally algebraic), we consider the ring ofregular functions on X , called R “ OX . We have a group G acting on X .We want to find all invariants of the orbit space X {{ G, and identify the “badpoints” onX . These are called the nonstable points. There are also semistablepoints, and stable points (which are a subset of the semistable ones).

The set of stable pointsXstable ofX is a big open set ofX , and the orbit spaceXstable {{G is well-behaved – we can distinguish all orbits within there. The setof semistable points is not too much worse, and in addition it’s the closure ofthe stable points.

Here’s a formal definition of stable.

Definition 81. A point p P X is stable if the orbit of p is closed and the stabilizer

StabGppq “ tg P G | g ¨ p “ pu

is finite.

38

The condition that the stabilizer is finite means that we can distinguishthings nearby a point p from the point p itself.

Example 82. Going back to example 79, the stable points of X “ C2 is the set

Xstable “ tpa, bq | ab ‰ 0u.


We’ve been studying equivalence problems: when are things the same?For a space X , we have an action of G on X . We consider things equiva-

lent if they lie in the same orbit. When does the orbit space X {{ G have goodstructure? We split this up into stable, semi-stable, and unstable points. To-day, we’re going to give a numerical criterion for the stable, semi-stable, andunstable points of X {{G.

Definition 83.

Vn,d “ t polynomials, homogeneous of degree d in n variables u

Example 84.

V3,2 “ tax2 ` bxy ` cxz ` dy2 ` eyz ` fz2u “

$

’

’

’

’

’

’

&

’

’

’

’

’

’

%

»

—

—

—

—

—

—

–

a

b

c

d

e

f

fi

ffi

ffi

ffi

ffi

ffi

ffi

fl

,

/

/

/

/

/

/

.

/

/

/

/

/

/

-

– C6

We saw a few weeks ago that the change of variables on x, y, z is given by a3ˆ 3 matrix. But changing variables also changes coefficients in V3,2, so we geta representation

ρ : GLp3q Ñ GLp6q.

Now we hope to understand this for any n and any d. Note that

dimVn,d “

ˆ

d` n´ 1

n´ 1

˙

.

Let’s begin with V2,d, which are single variable polynomials of the form

adxd ` ad´1x

d´1 ` . . .` a0

The zero loci of these polynomials are just points on P1C. Our space is X , andwe’ll restrict attention to the action of G “ SLpn,Cq by change of coordinates.We want to find Vn,d {{SLpn,Cq. This is a representation SLpnq Ñ SL

´

`

d`n´1n´1

˘

¯

.

39

Theorem 85 (Numerical Stability Criterion). Mumford found criteria for pointsin the orbit space V {{ SLpn,Cq to be unstable, semistable, or stable, for anyaction of SLpn,Cq on the vector space V . Let x P V .

x unstable x semistable x stable0 P closure of x-orbit 0 R closure of x-orbit 0 R closure of x-orbit

and Stabpxq finitefor all nonconstant there is nonconstant there is nonconstant

SLpn,Cq-invariant f SLpnq-invariant f , SLpnq-invariant f , fpxq ‰ 0 andfpxq “ 0 fpxq ‰ 0 trdegpV SLpnqq “ dimV ´ dim SLpnq

1. x unstable: there is a 1-parameter subgroup λ : Cˆ Ñ SLpnq such thatthe weights of x with respect to λ are all positive

2. x semistable: for all 1-parameter subgroups λ, not all weights are posi-tive;

3. x stable: for all 1-parameter subgroups λ : Cˆ Ñ SLpnq, there is a nega-tive weight

Definition 86. A 1-parameter subgroup λ : Cˆ Ñ SLpnq is a map such that, upto change of basis,

λptq “

»

—

—

—

–

tr1

tr2

. . .trn

fi

ffi

ffi

ffi

fl

Example 87. In SLp2q, a one parameter subgroup is

t ÞÑ

„

t 0

0 1{t

.

Definition 88. A weight of λ with respect to ~x P V is

µpx, λq “ weight “ ´maxt´ri : xi ‰ 0u,

where

λptq “

»

—

—

—

–

tr1

tr2

. . .trn

fi

ffi

ffi

ffi

fl

~x “

»

—

—

—

–

x1x2...xn

fi

ffi

ffi

ffi

fl

.

40

Example 89. Let’s go back to our action of SLp2q by change of coordinates inV2,4. A vector is, for example,

a4x4 ` a3x

3y ` a2x2y2 ` a1xy

3 ` a0y4

and we have a change of coordinates defined by the matrix λ

λ

„

x

y

“

„

t 0

0 1{t

„

x

y

“

„

txy{t

.

This polynomial transforms as

a4t4x4 ` axt

2x3y ` a2x2y2 `

a1t2xy3 `

a0t4y4,

so the representation of the matrix is

»

—

—

—

—

–

t4 0 0 0 0

0 t2 0 0 0

0 0 1 0 0

0 0 0 t´2 0

0 0 0 0 t´4

fi

ffi

ffi

ffi

ffi

fl

This is still in SLp5q. What are the weights?

µpp1, t, 1, 1, 1q, λq “ ´4

µpp1, 1, 1, 1, 0q, λq “ ´2

µpp1, 1, 1, 0, 0q, λq “ 0

Using the criteria for stability, we can figure out when things are stable, semi-stable, or unstable. It turns out to be unstable when there is a root of very highmultiplicity. The cutoff for “very high” is half of the degree of the polynomial.Using the theorem, we can calculate exactly which points are stable and whichare unstable, which is what made Hilbert famous.

What to take from these lectures

This is all very classical stuff. It was known to Hilbert and his contemporaries.If we had another week, we would go on to talk about the modern version ofthis stuff, which leads into Lie groups, Lie algebras, and much more algebraicgeometry. This is the beginnings of

We began with equivalence problems, and then moved on to representationtheory, invariant theory, some algebraic geometry, and now finally we havetalked about moduli spaces. Hopefully you’ve got some idea of what’s goingon in the general picture.

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moduli theory and invariants - cornell...

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