1 The Classical Groups

1.1 The groups

Let F denote either the real numbers, R, or the complex numbers, C. In thissection we will describe the main players in the rest of this book the ClassicalGroups as designated by Hermann Weyl..This section should be treated as adictionary. The groups as named here will appear throughout the book.

1.1.1 The general linear group.

Let V be a finite dimensional vector space over F. The set of all invertibletransformations of V to V will be denoted GL(V ). This set has a group struc-ture under composition of transformations with identity element the identitytransformation Id(x) = x for all x ∈ V . We will consider this family of groupsin this subsection. We will be recalling some standard terminology related tolinear transformations and their matrices.Let V and W be finite dimensional vector spaces over F with bases v1, ..., vn

and w1, ..., wm respectively. If T : V →W is a linear map then we have

Tvj =mPi=1

aijwi

with aij ∈ F. The aij are called the matrix coefficients or entries of the lineartransformation T with respect to the two bases and the m× n array

A =

a11 a12 · · · a1ma21 a22 · · · a2m...

.... . .

...am1 am2 · · · amn

is the matrix of T with respect to the two bases. If S : W → U is a linearmap with U an l dimensional vector space with basis u1, ..., ul and if B is thematrix of S with respect to the bases w1, ..., wm and u1, ..., ul then the matrixof S ◦T with respect to v1, ..., vn and u1, ..., ul is given by BA the product beingthe usual product of matrices.We denote by Mn(F) the space of all n× n matrices over F. Let V be an n

dimensional vector space over F with basis v1, .., vn. If T : V → V is a linearmap we will (for the moment) write µ(T ) for the matrix of T with respect tothis basis. If T, S ∈ GL(V ) then the observations above imply that

µ(S ◦ T ) = µ(S)µ(T ).

Furthermore, if T ∈ GL(V ) then µ(T ◦ T−1) = µ(T−1 ◦ T ) = µ(Id) = I theidentity matrix (the entries being δij = 1 if i = j and 0 otherwise. A matrixA ∈ Mn(F) is said to be invertible if there is a matrix B ∈ Mn(F) such thatAB = BA = I. We note that a linear map T : V → V is in GL(V ) if and only

1

if µ(T ) is invertible. We also recall that a matrix A ∈Mn(F) is invertible if andonly if its determinant is non-zero.We will use the notation GL(n,F) for the set of n×n invertible matrices with

coefficients in F. GL(n,F) is a group under matrix multiplication with identitythe identity matrix. We note that if V is an n-dimensional vector space over Fwith basis v1, ..., vn then the map µ : GL(V )→ GL(n,F) corresponding to thisbasis is a group isomorphism. The group GL(n,F) is called the general lineargroup of rank n. If w1, ..., wn is another basis of V and if

wj =Pigijvi

thenvj =

Phijwi

with [hij ] the inverse matrix to [gij ]. If T is a linear transformation of V to Vand if A = [aij ] is the matrix of T with respect to v1, ..., vn and if B = [bij ] isthe matrix with respect to w1, ..., wn then

Twj = T

µPigijvi

¶=PigijTvi =P

igij(

Pk

akivk) =Pigij(

Pk

aki(Pl

hlkwk).

Thus, if g = [gij ] then B = g−1Ag.

1.1.2 The special linear group.

The special linear group of n×n matrices is the set of all elements, A, ofMn(F)such that det(A) = 1. Since det(AB) = det(A) det(B) and det(I) = 1 we seethat the special linear group is a subgroup of GL(n,F) with will be denotedSL(n,F).We note that if V is an n-dimensional vector space with basis v1, ..., vn and

if µ : GL(V )→ GL(n,F) is as above then the set {T ∈ GL(V )|det(µ(T )) = 1}is independent of the choice of basis. (See the change of basis formula in theprevious subsection. You will be called upon to prove this in Exercise 1 at theend of this section). This subset us the subgroup µ−1(SL(n,F)) and we willdenote it by SL(V ).

1.1.3 Isometry groups of bilinear forms.

Let V be an n dimensional vector space over F and let B : V × V → F bea bilinear map Then we denote by O(B) the set of all T ∈ GL(V ) such thatB(Tv, Tw) = B(v,w) for all v, w ∈ V . We note that O(B) is a subgroupof GL(V ). Let v1, ..., vn be a basis of V relative to this basis we may writeB(vi, vj) = bij. Assume that T has matrix A = [aij ] then we have

B(Tvi, Tvj) =Pk,l

akialjB(vk, vl) =Pk,l

akialjbkl.

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Thus if At is the matrix [cij ] with cij = aji (the usual transpose of A) then thecondition that T ∈ O(B) is that

[bij ] = At[bij ]A.

If B is non-degenerate (B(v, w) = 0 for all w implies v = 0 and B(v, w) = 0for all v implies w = 0) then we have det([bij ]) 6= 0. Hence using the formulaabove we see that if B is nondegenerate and if T : V → V is linear and satisfiesB(Tv, Tw) = B(v, w) for all v, w ∈ V then T ∈ O(B). The next two subsectionswill discuss the most important special cases of this class of groups.

1.1.4 Orthogonal groups.

We will start this section by first introducing the matrix groups and then iden-tifying them with the corresponding isometry groups. Let O(n,F) denote theset of all g ∈ GL(n,F) such that ggt = I. That is

gt = g−1.

We note that (AB)t = BtAt and if A,B ∈ GL(n,F) then (AB)−1 = B−1A−1.It is therefore obvious that O(n,F) is a subgroup of GL(n,F). This group iscalled the orthogonal group of n×n matrices over F. If F = R we introduce theindefinite orthogonal groups, O(p, q), with p+ q = n with p, q ∈ N. Let

Ip,q =

·Ip 00 −Iq

¸with Ir denoting the r × r identity matrix. Then we define

O(p, q) = {g ∈Mn(R)|gtIp,qg = Ip,q}.

We note that O(n, 0) = O(0, n) = O(n,R). Also, if

σ =

0 0 · · · 1......

. . ....

0 1 0 01 0 0 0

that is the matrix with all entries zero except for ones on the skew diagonal (thei, n+1− i entries are 1). Then σIp,qσ

−1 = σIp,qσ = σIp,qσt = −Iq,p. Thus the

mapφ : O(p, q)→ GL(n,R)

given by φ(g) = σgσ defines an isomorphism of O(p, q) onto O(q, p).We will now describe these groups in terms of bilinear forms let V be an

n-dimensional vector space over F and let B be a symmetric non-degeneratebilinear form over F (either R or C). We have

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Proposition 1 If F = C then there exists a basis v1, ..., vn of V such thatB(vi, vj) = δij. If F = R then there exist p, q ≥ 0 with p + q = n and a basisv1, ..., vn of V such that B(vi, vj) = εiδij with εi = 1 for i ≤ p and εi = −1 fori > p. Furthermore, if we have another such basis then the corresponding p, qare the same.

Remark 2 The number p− q in the case when F = R is called the signature ofthe form B. We will also say that (p, q) is the signature of B.

We preface the proof of this result with a simple lemma.

Lemma 3 Let M be a symmetric bilinear form on V . Then if M(v, v) = 0 forall v ∈ V then M = 0.

Proof. Using the symmetry and bilinearity we have the identity

4B(v, w) = B(v + w, v + w)−B(v − w, v − w)

for all v,w ∈ V . This implies the lemma.We will now prove the proposition. We will first show that there exists a

basis w1, ..., wn of V such that

B(wi, wj) = 0 for i 6= j

andB(wi, wi) 6= 0

by induction on n (such a basis is called an orthogonal basis with respect toB). Since B is non-degenerate the previous lemma implies that there exists avector wn ∈ V with B(wn, wn) 6= 0. If n = 1 then this is what we set out toprove. Assume for n − 1 ≥ 1. Set V 0 = {v ∈ V |B(wn, v) = 0}. If v ∈ V thenv0 = v − B(v,wn)

B(wn,wn)wn ∈ V 0. This implies that V = V 0 + Fwn. So, in particular,

dimV 0 = n−1. We assert that the form B0 = B|V 0×V 0 is non-degenerate on V 0.Indeed if v ∈ V 0 satisfies B(v0, w) = 0 for all w ∈ V 0 then since B(v0, wn) = 0we see that B(v0, w) = 0 for all w ∈ V so v0 = 0. We can now apply theinductive hypothesis to find a desired basis w1, ..., wn−1 for V 0. Now it is clearthat w1, ..., wn has the desired properties.If F = C we can now complete the proof. Let w1, ..., wn be an orthogonal

basis of V with respect to B. Let zi be a choice of square root of B(wi, wi) thenif vi = 1

ziwi the B(vi, vj) = δij . At this point the reader should have noticed

why we need the p when F = R.We now consider the case when F = R. We rearrange (if necessary) so that

B(wi, wi) ≥ B(wi+1, wi+1) for i = 1, ..., n − 1. Let p = 0 if B(w1, w1) < 0.Otherwise let p = max{i|B(wi, wi) > 0}. Then B(wi, wi) < 0 for i > p. Definezi to be a square root of B(wi, wi) for i ≤ p. Define zi to be a square root of−B(wi, wi) if i > p. Then if vi = 1

ziwi we have B(vi, vj) = εiδij . We are left

with proving that the p is intrinsic to B.To complete the proof for F = R we will use a terminology that will appear

later in this opus.

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Definition 4 Let V be a vector space over R and let M be a symmetric bilinearform on V then M is said to be positive definite if M(v, v) > 0 for everyv ∈ V, v 6= 0.

With this concept in hand we will complete the proof. Fix a basis v1, ..., vnsuch that B(vi, vj) = εiδij with εi = 1 for i ≤ p and εi = −1 for i > p.Set V+ equal to the linear span of {v1, ..., vp} and V− equal to the linear spanof {vp+1, ..., vn}. Then V = V+ + V− direct sum. Let π : V → V+ be theprojection onto the first factor. We note that B|V+ is positive definite. Let Wbe a subspace of V such that B|W is positive definite. Suppose that w ∈W andπ(w) = 0. Then w ∈ V− thus w =

Pi>p

aivi. Hence

B(w,w) =Pi,j>p

aiajB(vi, vj) = −Pi>p

a2i ≤ 0.

Thus since B|W has been assumed to be positive definite w = 0. This impliesthat π : W → V+ is injective. Hence dimW ≤ dimV+ = p. If we had anotherbasis satisfying the desired conditions yielding p0 as above then this argumentimplies that p0 ≤ p and p ≤ p0 so p = p0. This completes the proof.

In the case when F = R and B is a non-degenerate symmetric bilinear formon a finite dimensional vector space V we will call a basis v1, ..., vn of V suchthat there exists n > p > 0 such that B(vi, vj) = εiδij with εi = 1 for i ≤ p andεi = −1 for i > p a pseudo orthonormal basis. We have seen that the numberp is an invariant of B.We now come to the crux of the matter. If B is a non-degerate symmetric

bilinear form on the ndimensional vector space V over F and if v1, ..., vn is basisof V then we will use the notation µ(g) for the matrix of the linear operator gwith respect to this basis.The following result is now a direct consequence of the above and the dis-

cussion in subsection 1.1.3.

Proposition 5 Let the notation be as above. If v1, ..., vn is an orthonormalbasis for V with respect to B or −B then µ : O(B) → O(n,F) defines a groupisomorphism. If F = R and if v1, ..., vn is a pseudo-orthonormal basis of V withsignature (p, n− p) then µ : O(B)→ O(p, n− p) is a group isomorphism.

The special orthogonal group over F is the subgroup SO(n,F) = O(n,F) ∩SL(n,F) of O(n,F). The indefinite special orthogonal groups are the groupsSO(p, q) = O(p, q) ∩ SL(p+ q,R).

1.1.5 The symplectic group.

We set

J =

·0 I−I 0

¸

5

with I the n × n identity matrix. The symplectic group of rank n over F isdefined to be

Sp(n,F) = {g ∈M2n(R)|gtJg = J}.As in the case of the orthogonal groups one sees without difficulty that Sp(n,F)is a subgroup of GL(2n,R).We will now look at the coordinate free version of these groups. For this we

need

Proposition 6 Let V be an m dimensional vector space over F and let B bea non-degenerate, skew symmetric (B(v, w) = −B(w, v)), bilinear form on V .Then m must be even. If m = 2n then there exists a basis v1, ..., v2n such that[B(vi, vj)] = J.

Proof. Assume that B is non-degerate and dimV = m > 0. Let v be a non-zero element of V . Then there exists w ∈ V with B(v,w) 6= 0. Replacing wwith 1

B(v,w)w we may assume that B(v, w) = 1. Let W = {x ∈ V |B(v, x) =0, B(w, x) = 0}. If x ∈ V then set x0 = x−B(v, x)w −B(x,w)v. Then

B(v, x0) = B(v, x)−B(v, x)B(v, w)−B(w, x)B(v, v) = 0

since B(v, w) = 1 and B(v, v) = 0. Similarly

B(w, x0) = B(w, x)−B(v, x)B(w,w) +B(w, x)B(w, v) = 0

since B(w, v) = −1, B(w,w) = 0. Thus V = U +W where U is the span of vand w. We note that B|U is non-degenerate so U ∩W = {0}. This implies thatdimW = m − 2. We also note that B|W is non-degenerate. We will leave thisto the reader (use the same technique as we used for the orthogonal groups seeexercise 3 at the end of this section). We will now prove the first part of theproposition. We have seen that if dimV > 0 then dimV ≥ 2. Thus the resultis true for dimV ≤ 2. Assume that the result is true for 2 ≤ dimV < k. We willprove the result for dimV = k. If we argue as above we have V = U+W a directsum decomposition. With dimW = dimV − 2. Since B|W is non-degeneratethe inductive hypothesis implies that dimW is even, Hence dimV is even.We will now prove the second part of the result by induction on n where

m = 2n. If n = 1 then we take v1 = v and v2 = w with v, w as above.Assume the result for n− 1 ≥ 1. Then set vn = v, v2n = w. Then since B|W isnon-degenerate we have a basis w1, ..., w2n−2 of W as in the second part of theassertion. Take vi = wi and vn+1−i = wn−i for i ≤ n− 1. The basis v1, ..., v2nis a desired basis of V .Now let V be an m dimensional vector space over F and let B be a non-

degerate, bilinear, skew symmetric form on V . Then we have seen that m iseven m = 2n. Let v1, ..., v2n be a basis of V as in the previous result and letµ(g) be the matrix of the linear transformation g : V → V with respect to thisbasis.

Proposition 7 Let the notation be as above. Then µ : O(B) → Sp(n,F) is agroup isomorphism.

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1.1.6 The unitary groups.

In this subsection we will look at an important class of subgroups ofGL(n,C) theunitary groups and special unitary groups for definite and indefinite Hermitianforms. If A ∈ Mn(C) then we will use the standard notation for its adjointmatrix

A∗ = At

where A is the matrix obtained from A by complex conjugating all of the entries.The unitary group of rank n is the group U(n) = {A ∈Mn(C)|A∗A = I}. Thespecial unitary group is U(n) ∩ SL(n,C). Let Ip,q be as in 1.1.4. We define theindefinite unitary group of signature (p, q) to be

U(p, q) = {g ∈Mn(C)|g∗Ip,qg = Ip,q}.

The special indefinite unitary group of signature (p, q) is SU(p, q) = U(p, q) ∩SL(n,C).We will now consider a coordinate free description of these groups. Let V be

an n-dimensional vector space over C then a bilinear map of B : V × V → C asvector spaces over R is said to be a Hermitian form if it satisfies the followingtwo conditions:1. B(av,w) = aB(v,w) for all a ∈ C, v, w ∈ V .2. B(w, v) = B(v, w) for all v, w ∈ V.A Hermitian form is said to be nondegenerate for v ∈ V if B(v,w) = 0

for all w ∈ V then v = 0.. It is said to be positive definite if B(v, v) > 0 ifv ∈ V and v 6= 0 (note that if M is a Hermitian form then M(v, v) ∈ R for allv ∈ V ). We define U(B) to be the group of all elements, g, of GL(V ) such thatB(gv, gw) = B(v, w) for all v,w ∈ V . U(B) is called the unitary group of B.

Proposition 8 Let V be an n-dimensional vector space over C and let B be anon-degenerate Hermitian form on V . Then there exists an integer n ≥ p ≥ 0and a basis v1, ..., vn of V over C such that

B(vi, vj) = εiδij

with εi = 1 for i ≤ p and εi = −1 for i > p. The number depends on p and noton the choice of basis.

The proof of this Proposition is almost identical to that of Proposition 4 andwill be left to the reader (see exercise 5 for a guide to a proof).If V is an n-dimensional vector space over C and B is a non-degenerate

Hermitian form on V then a basis as in the above proposition will be calleda pseudo-orthonormal basis (if p = n then it will be called an orthonormalbasis. The pair (p, n− p) will be called the signature of B. The following resultis proved in exactly the same way as the corresponding result for orthogonalgroups.

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Lemma 9 Let V be a finite dimensional vector space over C and let B be anon-degenerate Hermitian form on V of signature (p, q) and let v1, ..., vp+q be apseudo-orthonormal basis of V relative to B, Then if µ : GL(V )→ GL(n,C) isthe assignment of a linear transformation of V to its matrix then µ(g) ∈ U(p, q)for all g ∈ U(B) and µ : U(B)→ U(p, q) defines a group homomorphism.

1.1.7 The quaternionic general linear group.

We will first recall some basic properties of the quaternions. We consider thevector space, H, over R given by

H =½·

x y−y x

¸|x, y ∈ C

¾.

One checks directly that H is closed under multiplication in Mn(C), that theidentity matrix is in H and that since·

x y−y x

¸ ·x −yy x

¸= (|x|2 + |y|2)I

every non-zero element of H is invertible. Thus H is a division algebra (or skewfield) over R. This division algebra is a realization of the quaternions. The moreusual way of introducing the quaternions is to consider the vector space, H, overR with basis 1, i, j,k and define a multiplication so that 1 is the identity and

i2 = j2 = k2 = −1,

ij = k, ji = −k, ik = −j,ki = j, jk = i,kj = −iand extended to H by linearity. We note that this gives an isomorphic divisionalgebra to the one described if we take

1 = I, i =

·i 00 −i

¸, j =

·0 1−1 0

¸,k =

·0 ii 0

¸.

We define GL(n,H) to be the group of all invertible n×n matrices over H. Herethe definition of matrix multiplication is the usual one but one must be carefulabout the order of multiplication.If we look at the real vector space Hn then we can define multiplication by

H by

a · (u1, ..., un) = (u1a, ..., una).We note that 1 · u = u and (ab) · u = a · (b · u). We can therefore think of Hn

as a vector space over H. We will think of Hn as n× 1 columns and define

Au

8

for u ∈ Hn and A ∈ GL(n,H) by matrix multiplication. Then A(a ·u) = aAu soA defines a quaternionic linear map. Now has 3 different structures as a vectorspace over C. By looking at the subfields

R1+Ri,R1+Rj,R1+Rj.

Relative to each of these complex structures GL(n,H) acts by complex lineartransformations. We leave it to the reader to prove that the determinant ofA ∈ GL(n,H) as a complex linear transformation with respect to any of thesethree structures is the same (see exercise 10). We can thus define SL(n,H) tobe the elements of determinant 1 with respect to any of these structures. Thisgroup is also denoted classically as SU∗(2n).

1.1.8 The quaternionic unitary groups.

If w ∈ H, withw =

·a b−b a

¸then we note that w∗w = ww∗ = |a|2 + |b|2. If X ∈Mn(H) then set X∗ = [x∗ji]if X = [xij ]. The indefinite quaternionic unitary groups are the groups

Sp(p, q) = {g ∈ GL(p+ q,H)|g∗Ip,qg = Ip,q}.

We leave it to the reader to prove that this set is indeed a subgroup of GL(p+q,H). One can define quaternionic Hermitian forms and prove a result analo-gous to Proposition 8 but we will have no need of such material in the rest ofthis book.

1.1.9 The group SO∗(2n).

If we include covering groups (to be defined later) we will have described up allof the classical groups over C and R except for one infinite family that is relatedto the orthogonal groups. We will describe that family now.Let

J =

·0 I−I 0

¸as usual with I the n×n identity matrix. We note that J2 = −I2n the 2n×2nidentity matrix. Thus the map GL(2n,C) to itself given by θ(g) = −JgJ definesan automorphism and θ2 is the identity automorphism. Our last classical groupis

SO∗(2n) = {g ∈ SO(2n,C)|θ(g) = g}here, as usual, g is the matrix whose entries are the complex conjugates of g.

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1.1.10 Exercises.

1. Show that if v1, ..., vn and w1, ..., wn are bases of V , a vector space over F,and if T : V → V is a linear map with matrices A and B respectively relativeto these bases then detA = detB.2. What is signature of the form B(x, y) =

Pixiyn+1−i on Rn?

3. Let V be a vector space over F and B a skew symmetric or symmetricnon-degenerate bilinear form on V . Assume that W is a subspace of V onwhich B restricts to a nondegenerate form. Prove that the restriction of B toW⊥ = {v ∈ V |B(v, w) = 0 for all w ∈W} is non-degenerate.4. Let V denote the vector space of symmetric 2× 2 matrices over F = R or

C. If x, y ∈ V define B(x, y) = 12 (det(x+ y)− det(x− y)) show that B(x, y) =

B(y, x) for all x, y ∈ V . Show that if F = R then the signature of the form Bis (2,1). If g ∈ SL(2,F) define ϕ(g) ∈ GL(V ) by ϕ(g)(v) = gvgt. Then showthat ϕ : SL(2,F) → O(V,B) is a group homomorphism with image SO(V,B)and kernel {±I}.5. The purpose of this exercise is to prove Proposition 8 by the method of

proof of Proposition 1. First prove the analogue of Lemma 3. That is, if M isHermitian form such that M(v, v) = 0 for all v then M = 0. To do this notethat if the condition is satisfied then M(v+ tw, v+ tw) = tM(w, v)+ tM(v, w).Now substitute values for t to see that M(v, w) = 0. The rest of the argumentsare the same as for Proposition 4 once one observes that ifM is Hermitian thenM(v, v) ∈ R.6. Let V be a 2n-dimensional vector space over F = R or C. We consider

the space W =V

nV . Fix a basis ω of the one dimensional vector spaceV2nV .

Consider the bilinear form B(u, v) defined by u ∧ v = B(u, v)ω. Show that theform is nondegenerate and skew symmetric if n is odd and symmetric if n iseven. If n is even and F = R then determine the signature of B.7. In the notation of problem 6. we consider the case when V = F4 and if

e1, e2, e3, e4 is the standard basis then ω = e1 ∧ e2 ∧ e3 ∧ e4. Consider ϕ(g) =V2 g for g ∈ SL(4,F). Show that ϕ : SL(4,F) → O(V2 F4, B) is a group

homomorphism with kernel {±I}. If F = C and we choose an orthonormal basisthen ϕ defines a group homomorphism of SL(4,C) to SO(6,C). If F = R and wechoose a pseudo-orthonormal basis show that ϕ defines a group homomorphismto SO(3, 3).8. In the notation of problem 7 let φ be the restriction of ϕ to Sp(4,F). Let

ν = e1 ∧ e3 + e2 ∧ e4. Show that φ(g)ν = ν. Show that B(v, v) = −2. LetW = {w ∈ V2 F4|B(ν, w) = 0}. Let ρ(g) = φ(g)|W . If R show that the signatureof the restriction of B to W is (3,2). Show that ρ is a group homomorphismfrom Sp(4,F) to SO(W,B|W ) with kernel {±1}.9. Let G = SL(2,F)×SL(2,F) and define ϕ : G→ GL(F2⊗F2) be given by

ϕ(a, b) = a⊗b. Let B be the bilinear form on F2⊗F2 given by B(ei⊗ej , ek⊗el) =εikεjl with εij = −εji for i, j = 1, 2 and ε12 = 1. Show that B defines asymmetric nondegenerate form on F2 ⊗ F2 and calculate the signature of B ifF = R. Show that if g ∈ G then ϕ(g) ∈ SO(F2 ⊗ F2, B). What is the kernel of

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ϕ?10. Show that if X ∈Mn(H) and we consider the three versions of C in the

quaternions as in 1.1.8. These give 3 ways of thinking of X as a 2n× 2n matrixover C. Show that the determinants of these three matrices are equal.

1.2 The Lie algebras.

Let V be a vector space over F. Let End(V ) denote the algebra (undercomposition) of F-linear maps of V to V . If X,Y ∈ End(V ) then we set[X,Y ] = XY − Y X. This defines a new product on End(V ) that satisfies twoproperties (see exercise 1 at the end of this section) :1. [X,Y ] = −[Y,X] for all X,Y (skew symmetry).2. For all X,Y,Z we have [X, [Y.Z]] = [[X,Y ], Z] + [Y, [X,Z]].The latter condition is a substitute for the associative rule.

Definition 10 A pair of a vector space, g, over F and a bilinear binary oper-ation [. . . , . . .] is said to be a Lie algebra if for each X,Y,Z ∈ g the conditions1. and 2. are satisfied.

Thus, in particular, we see that End(V ) is a Lie algebra under the binaryoperation [X,Y ] = XY − Y X.We also note the obvious fact that if g is a Lie algebra and if h is a subspace

such that [X,Y ] ∈ h implies that [X,Y ] in h then h is a Lie algebra under therestriction of [..., ...]. We will call h a sub-Lie algebra or a subalgebra.Suppose that g and h are Lie algebras over F then a Lie algebra homomor-

phism of g to h is an F-linear map T : g → h such that T [X,Y ] = [TX,TY ].We will say that a Lie algebra homomorphism is an isomorphism if it is bijective(you will be asked in Exercise 2, at the end of this section, to prove that the in-verse of a bijective Lie algebra homomorphism is a Lie algebra homomorphism).

1.2.1 The general linear and special linear Lie algebras.

Let V be as in the previous section. We will use the notation gl(V ) for End(V )looked upon as a Lie algebra under [X,Y ] = XY − Y X. We note that if wechoose a basis for V and if V is n-dimensional then the matrices of the elementsof End(V ) form a Lie algebra which we will denote by gl(n,F).If V is a finite dimensional vector space over F and if v1, ..., vn is a basis then

each T ∈ End(.has a matrix. A, with respect to this basis. If A ∈ Mn(F) andA = [aij ] then tr(A) =

Piaii. We note that

tr(AB) = tr(BA).

This implies that if A is the matrix of T with respect to v1, ..., vn then tr(A) isindependent of the choice of basis. We will write tr(T ) = tr(A). This impliesthat we can define

sl(V ) = {T ∈ End(V )|tr(T ) = 0}.

11

Choosing a basis we may look upon this Lie algebra as

sl(n,F) = {A ∈ gl(n,F)|trA = 0}.

These Lie algebras will be called the special linear Lie algebras.

1.2.2 The Lie algebras associated with binary forms.

Let V be a vector space over F and let B : V × V → F be a bilinear map. Wedefine

so(B) = {X ∈ End(V )|B(Xv,w) = −B(v,Xw)}.We leave it to the reader to check that so(B) is a Lie subalgebra of gl(V ) bythe obvious calculation (see Exercise 3 at the end of this section). If B isnondegenerate you will also be asked to prove in Exercise 3 that trX = 0 forX ∈ so(B). For this the following discussion should be useful.Let the notation be as in section 1.1.3. That is we assume that V is finite

dimensional and we choose a basis v1, ..., vn of V . Then bij = B(vi, vj(. By acalculation completely analogous to that in 1.1.3 we see that if T ∈ so(B) thenits matrix, A, relative to v1, ..., vn satisfies

At[bij ] + [bij]A = 0.

1.2.3 The orthogonal Lie algebras.

We defineso(n,F) = {X ∈Mn(F)|Xt = −X}.

Since trXt = trX this implies that so(n,F) is a subspace of sl(n,F). It is alsoeasily checked (by direct calculation) that so(n,F) is a Lie subalgebra of sl(n,F).By analogy with the definition of O(p, q) let p, q ≥ 0 be integers such that

p+ q = n. Set (as in 1.1.4)

Ip,q =

·Ip 00 −Iq

¸.

We setso(p, q) = {X ∈Mn(R)|XtIp,q = −Ip,qX}.

As observed in the previous section so(p, q) ⊂ sl(n,R) and it is easily checkedto be a Lie subalgebra.We can now apply the material in section 1.1.4. Let B denote a non-

degenerate symmetric bilinear form on V a finite dimensional vector space overF. If F = C then we have seen in Proposition 1 (1.1.4) that there is a basisv1, ..., vn of V such that B(vi, vj) = δij . If µ(T ) is the matrix of T ∈ End(V )relative to this matrix then µ defines a Lie algebra isomorphism of so(B) ontoso(n,C). If F = R and B has signature p, q and v1, ..., vn is a correspondingpseudo-orthonormal basis then the analog of µ defines a Lie algebra isomorphismof so(B) onto so(p, q).

12

1.2.4 The symplectic Lie algebra.

In this case we take

J =

·0 I−1 0

¸with I the n× n identity matrix. We define

sp(n,F) = {X ∈M2n(F)|XtJ = −JX}.

Then this subspace of the n× n matrices is a Lie subalgebra which we call thesymplectic Lie algebra of rank n.Let V be a vector space over F of dimensionm <∞ and B a skew symmetric

bilinear form on V . Then m is even m = 2n and applying Proposition 6 section1.5 we see that there exists a basis v1, ..., v2n of V such that [B(vi, vj)] = J .The map that assigns to an endomorphism of V its matrix relative to this basisdefines an isomorphism of so(B) onto sp(n,F).

1.2.5 The unitary Lie algebras.

We will use the notation and results of section 1.1.6. Let be integers withp, q ≥ 0, We define

u(p, q) = {X ∈Mp+q(C)|X∗Ip,q + Ip,qX = 0}.

We note that this space is a real subspace ofMp+q(C). One checks directly thatit is a Lie subalgebra of Mp+q(C) thought of as a Lie algebra over R.We define su(p, q) = u(p, q) ∩ sl(p+ q,C).Let n = p+ q, let V be an n-dimensional vector space over C and let B be

a non-degenerate Hermitian form on V . Then we define

u(B) = {T ∈ EndC(V )|B(Tv,w) +B(v, Tw) = 0, v, w ∈ V }.

and su(B) = u(B) ∩ sl(V ). If B has signature (p, q) and if v1, ..., vn is a pseudoorthogonal basis of V then the assignment T 7→ µ(T ) of T to its matrix relativeto this basis defines a Lie algebra isomorphism of u(B) with u(p, q) and su(B)with su(p, q).

1.2.6 The quaternion general linear Lie algebra.

In this subsection we will follow the notation of section 1.1.7. The Lie algebra inquestion consists of the n× n matrices over the quaternions, H, with the usualmatrix commutator. We will denote the Lie algebra gl(n,H). This Lie algebrais to be thought of as a Lie algebra over R (we have not defined Lie algebras overskew-fields). If we consider Hn to be C2n relative to any one of the isomorphiccopies of C: R1 + Ri, R1 + Rj,R1 + Rk. Then we can take sl(n,H) to be theelements of trace zero in any of these structures. This real Lie algebra is usuallydenoted su∗(2n).

13

1.2.7 The quaternionic unitary Lie algebras.

Here we will use the notation of section 1.1.8 (especially the quaternionic ad-joint). Then we define for n = p+ q with p, q non-negative integers

sp(p, q) = {X ∈ gl(n,H)|X∗Ip,q + Ip,qX = 0}.

1.2.8 The Lie algebra so∗(2n).

Here we will be using the notation of section 1.1.9 especially the definition ofθ. The Lie algebra in question is defined to be space of all X ∈ so(2n,C) suchthat θ(X) = X. This real vector subspace of so(2n,C) defines a Lie subalgebraif we consider so(2n,C) to be a Lie algebra over R.

1.2.9 The list.

The Lie algebras described in the preceding sections constitute the list of clas-sical Lie algebras over R and C. These Lie algebras will be the main subjectof study throughout the remainder of this book. We will find that the choicesof bases for the matrix forms of these Lie algebras are not the most convenient.An observant reader should have noticed by now that the Lie algebras thathave a name that is "frakturized" form of a name in the previous section canbe obtained by “differentiating” the definition as a group. That is we look atthe rule for inclusion in the corresponding group, G. Let us call it R. We willdenote the Lie algebra with the “frakturized” designation g. Consider a curveσ : (−ε, ε) → GL(V ) such that σ(0) = I and σ is differentiable at 0. Thenone should observe that if R(σ(t)) is satisfied for all t ∈ (−ε, ε) then σ0(0) ∈ g.For example, if G is the subgroup O(B) of GL(V ) defined by a binary formB (i.e. G = {g ∈ GL(V )|B(gv, gw) = B(v, w) for all v,w ∈ V }. Then theindicated conditions on the curve are B(σ(t)v, σ(t)w) = B(v, w) for all v, w ∈ Vand t ∈ (−ε, ε). If we differentiate these relations we have

0 =d

dtB(σ(t)v, σ(t)w)|t=0 = B(σ0(0)v, σ(0)w) +B(σ(0)v, σ0(0)w).

Since σ(0) = I we see that σ0(0) ∈ so(B) as asserted. This is the reason whythese Lie algebras are usually called the infinitesimal form of these groups.

1.2.10 Exercises

1. Prove the identities 1) and 2) in the beginning of this section.2. Prove that the inverse of a bijective Lie algebra homomorphism is a Lie

algebra homomorphism.3. Prove that B is a bilinear form on V then so(B) is a Lie subalgebra of

gl(V ). Also show that if B is non-degenerate then trX = 0 for all X ∈ so(B).4. Prove that the indicated spaces that were not proved to be Lie algebras in

each of the previous sections is indeed a Lie subalgebra of the indicated generallinear Lie algebra over R or C.

14

5. Let X ∈ Mn(H) for each of the choices of a copy of C in H write outthe corresponding matrix of X as an element of M2n(C). Use this formula toobserve that the trace condition is indeed independent of the choice.6. Show that sp(p, q) ⊂ sl(p+ q,H).7. Carry out the infinitesimal calculation indicated in section 1.2.9 for each

of the examples. For the case of SL(n,F) you will need to prove that if σ :(−ε, ε)→ GL(n,F) is differentiable and σ(0) = I then

d

dtdet(σ(t))|t=0 = tr(σ0(0)).

1.3 Closed subgroups of GL(n,R).In this section we will give an introduction to some Lie theoretic ideas that willmotivate the later developments in this book. We will begin with the definitionof a topological group and then emphasize the topological groups that are closedsubgroups of GL(n,F). Our main tool will be the exponential map which wewill deal with explicitly.

1.3.1 Topological Groups.

Let G be a group with a topology such that the maps

G×G→ G, g, h 7→ gh

andG→ G, g 7→ g−1

are continuous is called a topological group. Here the set G × G is given theproduct topology above. We note that GL(n,F) is a topological group when en-dowed with the topology of an open subset ofMn(F) looked upon as Fn

2

by lay-ing out the matrix entries as, say, (x11, x12, ..., xn1, x21, ..., x2n, ..., xn1, ..., xnn).The multiplication is continuous and Cramer’s rule implies that the inverse iscontinuous.We note that if G is a topological group then we have for each g ∈ G maps

Lg : G→ G and Rg : G→ G given by Lg(x) = gx and Rg(x) = xg. The groupproperties and continuity imply that Rg and Lg are homeomorphisms.If G is a topological group and H is a subgroup that is closed as a topological

space then H is also a topological group called a topological subgroup. Thus allof the examples in section 1 are topological groups.A topological group homomorphism will mean a continuous topological group

homomorphism. A topological group homomorphism is said to be a topologicalgroup isomorphism if it is bijective and its inverse is also a topological grouphomomorphism.In this section we will be studying closed subgroups of GL(N.R) for all

N = 1, 2, ...We first show how we will be looking at GL(n,F) for F = C or H as

15

a closed subgroup of GL(dn,R) with d = dimR F = 2, 4 for C and H respectively.If we consider Cn to be R2n and multiplication by i to be the action of the matrix

J =

·0 I−I 0

¸with I the n×n identity matrix thenMn(C) consists of the matrices inM2n(R)that commute with J and GL(n,C) consists of the elements of GL(2n,R) thatcommute with J . The case of the quaternionic groups is handled similarly wherewe look at the standard n-dimensional vector space over the quaternions, Hn, asa R4n with three operators J1, J2, J3 such that JkJl = −JlJk for k 6= l, J2i = −Iand J1J2 = J3, J1J3 = −J2, J2J3 = J1. Given as follows

J1 =

·J 00 −J

¸, J2 =

·0 I−I 0

¸, J3 =

·0 JJ 0

¸.

Then Mn(H) is the set of X ∈ M4n(R) such that XJi = JiX for i = 1, 2, 3.Furthermore, GL(n,H) is with this identification the subgroup of all elementsof GL(4n,R) that commute with J1, J2 and J3.Before we study our main examples we will prove a few general results about

topological groups.

Lemma 11 Let H be an open subgroup of a topological group, G, then H isclosed in G.

Proof. We note that G is a disjoint union of left cosets. If g ∈ G then sinceLg is a homeomorphism gH = Lg(H) is open. Hence every left coset is open.Since H the complement of the left cosets other than the identity coset is H,we see that the complement of H is open. Thus H is closed.

Lemma 12 Let G be a topological group then the identity component of G (thatis the connected subgroup that contains the identity element, e)is a normal sub-group.

Proof. Let X be the identity component of G. If g ∈ X then since e ∈ X wesee that g ∈ LgX. Since Lg is a homeomorphism we see that LgX = X. HenceX is closed under multiplication. Since e ∈ LgX we see that g−1 ∈ X. So X isa subgroup. If g ∈ G then define τ(g)(x) = gxg−1 for x ∈ G. Then τ(g) is ahomeomorphism of G and τ(g)(e) = e. Thus τ(g) maps the identity componentof G to itself. Hence X is a normal subgroup.

1.3.2 The exponential map.

On Mn(R) we define the inner product hX,Y i = tr(XY t). The correspondingnorm

kXk = hX,Xi 12 = (Xij

x2ij)12

has the following properties.

16

1) It is a norm that is kX + Y k ≤ kXk+kY k, kcXk = |c| kXk and kXk = 0if and only if X = 0.

2) kXY k ≤ kXk kY k.We will now show how to derive 2) from the usual Cauchy-Schwarz inequal-

ity.

kXY k2 =Xij

ÃXk

xikykj

!2.

Now |Pk xikykj |2 ≤¡P

k x2ik

¢ ³Pk y

2kj

´by the usual Cauchy-Schwarz inequal-

ity. Hence

kXY k2 ≤Xij

ÃXk

x2ik

!ÃXk

y2kj

!=

ÃXik

x2ik

!Xkj

y2kj

= kXk2 kY k2 .

Taking the square root of both sides completes the proof.IfA ∈Mn(R) and r > 0 we will use the notationBr(A) = {X ∈Mn(R)| kX −Ak <

r}.We can use this norm in order to define some matrix valued analytic func-

tions. We note that if we have a power seriesP∞

m=0 amzm and if r > 0 is such

thatP∞

m=0 |am|rm < ∞ then the given series defines a function f(z) on thedisc of radius r. We note that for the same series if k ≥ l and kXk < r then°°°°°°

X0≤m≤k

amXm −

X0≤m≤l

amXm

°°°°°° =

°°°°°°X

l≤m≤kamX

m

°°°°°° ≤X

l≤m≤k|am| kXmk

≤X

l<m≤k|am| kXkm ≤

∞Xm>l

|am|rm.

This goes to 0 as l→∞. Thus the seriesPm≥0 amXm defines a function from

the open ball of radius r in Mn(R) to Mn(R). The absolute convergence of thisseries implies that we can interchange the order of summation without changingthe value. In this section we will be concentrating on two functions. The firstis defined by the power series

exp(x) =∞Xn=0

xn

n!

and the second

log(1 + x) =∞Xn=1

(−1)n+1xn

n.

The first converges on every disc and for the other the best choice is r = 1. Wenote that we have the formal identities

exp(x+ y) = exp(x) exp(y),

log(1 + (exp(x)− 1)) = x.

17

We have therefore defined two matrix valued functions functions

exp :Mn(R)→Mn(R)

andlog : {X ∈Mn(R)| kX − Ik < 1}→Mn(R).

Both of these functions are given by power series so are real analytic functions.We note that

exp(0) = I

andexp(X + Y ) = exp(X) exp(Y )

due to the formal identity. This implies that exp(X) exp(−X) = I thus

exp :Mn(R)→ GL(n,R).

Since exp is, in particular, continuous there exists r > 0 such that if X ∈Br(0) = {X ∈ Mn(R)| kXk < r} then exp(X) ∈ B1(I). We also note that theformal identity relating exp and log implies that

log(I + (expX − I)) = X

for X ∈ Br(0).We calculate the differential of exp at 0 we we expand the series to order 1

and haveexp(tv) = I + tv +O(t2).

This implies thatd exp0(v) = v.

The inverse function theorem implies that there exists a ball Bs(0) with s > 0and s ≤ r (defined above) such that exp(Bs(0)) = U is open in GL(n,R) and

exp : Bs(0)→ U

is a homeomorphism with analytic inverse. We note that U ⊂ B1(I) so, in fact,³exp|Bs(0)

´−1= log|U .

To give an idea of the power of what we have done so far we will prove

Theorem 13 Let φ : R → GL(n,R) be a continuous group homomorphismof the additive group of real numbers then there exists X ∈ Mn(R) such theφ(t) = exp(tX) for all t ∈ R. Furthermore, X is uniquely determined by φ.

18

Proof. If we set φε(t) = φ(εt) then φε is also a continuous homomorphism ofR into GL(n,R). Since φ is continuous we may assume that φ(t) ∈ exp(B s

2(0))

for |t| ≤ 2. Thus φ(1) = expX for some X ∈ B s2(0). We also note that

φ( 12) = expZ with Z ∈ B s2(0). Thus φ(1) = exp(2Z). Since exp : Bs(0) → U

is injective this implies that 2Z = X so Z = 12X. If we do the same argument

with φ( 14) = exp(W ) with W ∈ B s2(0) we have exp(2W ) = expZ and since

both 2W and Z are in Bs(0) we have W = 12Z = 1

4X. We can continue thisargument and find that

φ(1

2k) = exp(

1

2kX).

If a = a12 +

a222 + ...+ ak

2k+ ... with aj ∈ {0, 1} is the dyadic expansion of the real

number 0 ≤ a < 1 then by continuity and the the assumption that φ is a grouphomomorphism we have

φ(a) = limk→∞

φ(a12+

a222+ ...+

ak2k) = lim

k→∞φ(1

2)a1φ(

1

4)a2 · · ·φ( 1

2k)ak

= limk→∞

exp(1

2X)a1 · · · exp( 1

2kX)ak

= limk→∞

exp((a12+

a222+ ...+

ak2k)X) = exp(aX).

Now if 0 ≤ a < 1 then φ(−a) = φ(a)−1 = exp(aX)−1 = exp(−aX). Finally,if a ∈ R there exists k > 0 with k an integer such that |ak | < 1. Then φ(a) =φ(ak )

k = exp(akX)k = exp(aX).

If exp(tX) = exp(tY ) for all t then 0 = exp(tX)−exp(tY )t = X − Y +O(t) as

t→ 0. Thus X = Y .

1.3.3 The Lie algebra of a closed subgroup of GL(n,R).

Let G be a closed subgroup of GL(n,R) then we will denote the set {X ∈Mn(R)| exp(tX) ∈ G, t ∈ R} by Lie(G). We will show that this set is a Liesubalgebra of Mn(R) relative to the bracket [X,Y ] = XY − Y X. This will bethe first of four apparently different definitions of Lie algebra of a class of groupsthat will be used in this book.For this we need more information about the exponential and logarithm. We

consider

log(I + (exp tX exp tY − I))

19

we will expand the series in t to second order. We have

log(I + (exp tX exp tY − I))

= log((I + tX +t2

2X2 +O(t3))(I + tY +

t2

2Y 2 +O(t3)))

= log(I + tX + tY + t2XY +t2

2X2 +

t2

2Y 2 +O(t3))

= t(X + Y ) + t2XY +t2

2X2 +

t2

2Y 2 − 1

2(tX + tY + t2XY +

t2

2X2 +

t2

2Y 2)2 +O(t3)

= t(X + Y ) + t2XY +t2

2X2 +

t2

2Y 2 − 1

2(tX + tY )2 +O(t3)

since all the rest of the terms in the last expression involve at least t3. This lastexpression is

tX + tY + t2XY +t2

2X2 +

t2

2Y 2 − 1

2t2(X2 +XY + Y X + Y 2) +O(t3)

= tX + tY +t2

2[X,Y ] +O(t3).

If we do exactly the same argument with

log(I + (exp sX exp tY − I))

neglecting any terms of order t2 or s2 then we have

log(I + sX + tY + stXY +O(s2) +O(t2))

= sX + tY + stXY − 12(sX + tY + stXY )2 +O(s2) +O(t2)

= sX + tY + stXY − 12(stXY + stY X) +O(t2) +O(s2)

= sX + tY +st

2[X,Y ] +O(t2) +O(s2).

Lemma 14 As t→ 0 we have

exp tX exp tY = exp(tX + tY +t2

2[X,Y ] +O(t3))

and as t→ 0, s→ 0 we have

exp sX exp tY = exp(sX + tY +ts

2[X,Y ] +O(t2) +O(s2)).

This result implies a few (standard) limit identities

Lemma 15 Let X,Y ∈Mn(R) then1) limk→∞(exp( 1kX) exp(

1kY ))

k = exp(X + Y ).

2)limk→∞(exp( 1kX) exp(1kY ) exp(− 1

kX) exp(− 1kY ))

k2 = exp([X,Y ]).

20

Proof. We have (using the first asymptotic result in the previous lemma

(exp(1

kX) exp(

1

kY ))k = exp(

1

kX +

1

kY +O(

1

k2))k = exp(X + Y + kO(

1

k2)).

This implies the first limit formula. As for the second we use the same asymp-totic information and get

(exp(1

kX) exp(

1

kY )) = exp(

1

k(X + Y ) +

1

2k2[X,Y ] +O(

1

k3))

hence

(exp(1

kX) exp(

1

kY ))(exp(−1

kX) exp(−1

kY ))

= exp(1

k(X + Y ) +

1

2k2[X,Y ] +O(

1

k3)) exp(−1

k(X + Y ) +

1

2k2[X,Y ] +O(

1

k3))

= exp(1

k2[X,Y ] +O(

1

k3)).

We therefore have

(exp(1

kX) exp(

1

kY ))(exp(−1

kX) exp(−1

kY ))k

2

= (exp(1

k2[X,Y ] +O(

1

k3)))k

2

.

This implies the second limit formula.This result implies

Proposition 16 If G is a closed subgroup of GL(n,R) then Lie(G) is a Liesubalgebra of Mn(R).

Proof. If X ∈ Lie(G) then tX ∈ Lie(G) for all t ∈ R. If X,Y ∈ Lie(G), t ∈ Rthen

exp(t(X + Y )) = limk→∞

(exp(t

kX) exp(

t

kY ))k

is in G since G is a closed subgroup. Similarly

exp(t[X,Y ]) = limk→∞

(exp(1

kX) exp(

1

kY ))(exp(−1

kX) exp(−1

kY ))k

2

which is in G.We will use the identification of the groups GL(n,C) and GL(n,H) with the

corresponding subgroups of GL(2n,R) and GL(4n,R) respectively as in the endof subsection 1.3.1 and use the notation therein. With these identifications theLie algebras in section 1.2 line up perfectly with the groups in section 1.1 inthe sense that one replaces several of the capital letters in the group name withfraktur letters. We will do a few examples and leave the rest as exercises.That Lie(GL(n,R)) = Mn(R) = gl(n,R) is obvious. We now look at

Lie(GL(n,C)). This Lie algebra is the space of all X ∈ M2n(R) such thatexp(tX)J = J exp(tX). This says that J−1 exp(tX)J = exp(tX). SinceA−1 exp(X)A = exp(A−1XA) we see that X ∈ Lie(GL(n,C)) if and only

21

if exp(tJ−1XJ) = exp(tX) for all t ∈ R. But the uniqueness in Theorem11 now implies that J−1XJ = X so X ∈ Mn(C) = gl(n,C). As aboveLie(GL(n,H)) = gl(n,H).We now look at SL(n,R). We will first calculate det(exp(tX)) for X ∈

Mn(R) this calculation uses a method that we will use in general in the nextsubsection. We note that we can apply Theorem 11 to the case of n = 1.Thus we have det(exp(tX)) = exp(tµ(X)) with µ(X) ∈ R. Now exp(tµ(X)) =1 + tµ(X) + O(t2)). We will now give an expansion of the other side of theequation (here I = [δij ])

det(exp(tX)) = det(I + tX +O(t2)) =

det(I + tX) +O(t2) =P

σ∈Snsgn(σ)(δσ1,1 + txσ1,1)(δσ2,2 + txσ2,2) · · · (δσn,n + txσn,n)

We note that if σi 6= i for some i there must be a j 6= i with σj ± j thus ifσ is not the identity permutation then the corresponding term in the sum isof order at least t2. Thus the only terms contributing to order 0 and 1 comefrom the identity element. That term is (1 + tx11) · · · (1 + txnn) = 1 + t(x11 +... + xnn) + O(t2). We conclude that µ(X) = trX. We therefore see thatLie(SL(n,R)) = {X ∈Mn(R)|trX = 0} = sl(n,R).For other classical groups it is convenient to use the following simple result.

Lemma 17 If H ⊂ G ⊂ GL(n,R) are such that H is a closed subgroup of Gand G is a closed subgroup of GL(n,R) then H is a closed subgroup of GL(n,R)and Lie(H) = {X ∈ Lie(G)| exp(tX) ∈ H, t ∈ R}.Proof. It is obvious that H is a closed subgroup of GL(n,R). If X ∈ Lie(H)then exp(tX) ∈ H ⊂ G for all t ∈ R. Thus X ∈ Lie(G). This proves the lemma.

We consider Lie(Sp(n,C)). We can consider Sp(n,C) ⊂ GL(2n,C) ⊂GL(2n,R). We can thus look upon Lie(Sp(n,C)) as the set of X ∈ Mn(C)such that exp tX ∈ Sp(n,C) for all t ∈ R. The remarks in subsection 1.2.9complete the argument to show that Lie(Sp(n,C)) = sp(n,C).We will do one more family of examples. We consider G = U(p, q) ⊂ GL(p+

q,C). Here

Lie(G) = {X ∈Mn(C)| exp(tX)∗Ip,q exp(tX) = Ip,q}.We note that exp(tX)∗ = (I+ tX+ t2

2 X2+ ...)∗ = I+ tX∗+ t2

2 (X∗)2+ .... Thus

if we expand the two sides of

exp(tX)∗Ip,q exp(tX) = Ip,q

to first order in t we have

Ip.q + t(X∗Ip,q + Ip,qX) = Ip,q.

Hence Lie(U(p, q)) is contained in u(p, q). If X ∈ u(p, q) then (X∗)kIp,q =(−1)kIp,qXk thus

exp(tX∗)Ip.q = Ip,q exp(−tX).

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That is,exp(tX)∗Ip.q exp(tX) = Ip.q.

Hence exp(tX) ∈ U(p, q) for all t ∈ R. We therefore see that Lie(U(p, q)) =u(p, q).

1.3.4 Continuous homomorphisms, their differentials and a theoremof Von Neumann.

Let G ⊂ GL(n,R) and let H ⊂ GL(m,R) be closed subgroups. Let φ : H → Gbe a continuous group homomorphism. If X ∈ Lie(H) then t 7→ φ(exp tX)defines a continuous homomorphism of R into GL(n,R). Hence Theorem 11implies that there exists µ(X) ∈Mn(R) such that

φ(exp(tX)) = exp(tµ(X)).

We will now prove that µ : Lie(H) → Lie(G) is a Lie algebra homomorphism.We will use Lemma 13. If X,Y ∈ Lie(H) then by continuity we have

φ(exp(t(X + Y )) = φ( limk→∞

(exp(t

kX) exp(

t

kY ))k)

= limk→∞

(φ(exp(t

kX))φ(exp(

t

kY )))k

= limk→∞

(exp(t

kµ(X)))φ(exp(

t

kµ(Y ))))k

= exp(tµ(X + Y )).

Hence the uniqueness in Lemma 11 implies that µ(X+Y ) = µ(X)+µ(Y ). It isalso obvious that µ(tX) = tµ(X). Using the second limit formula in Lemma 13in exactly the same way as we did the first we find that µ([X,Y ]) = [µ(X), µ(Y )].We will use the notation dφ = µ. This will be our preliminary definition of

differential of a continuous homomorphism of closed subgroups of general lineargroups. By the end of this subsection we will have made a link to the materialin appendix D on Lie groups.We will now record what we have proved up to this point in this subsection.

Proposition 18 Let G ⊂ GL(n,R) and H ⊂ GL(m,R) be closed subgroupsand let ϕ : H → G be a continuous homomorphism. Then there exists aunique Lie algebra homomorphism µ : Lie(H)→ Lie(G) such that ϕ(exp(X)) =exp(µ(X)). We will use the notation dϕ for µ.

We will now study in more detail the relationship between the Lie algebraof a closed subgroup, G, of GL(n,R) the group structure. We first note thatsince the map t 7−→ exp tX from R to G the Lie algebra of G is the same as theLie algebra of its identity component. It is therefore reasonable to confine ourattention to connected groups in this discussion. The key result in this directionis due to Von Neumann.

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Theorem 19 Let G be a closed subgroup of GL(n,R) then there exists an openneighborhood, V , of 0 in Lie(G) so that exp(V ) is open in G and exp : V →exp(V ) is a homeomorphism.

Proof. LetW = {X ∈Mn(R)|trXtLie(G) = {0}}. ThenMn(R) = Lie(G)L

Wand orthogonal direct sum. If we take orthonormal basis of Mn(R) that startswith a basis of Lie(G) then continues with W then we can think of Mn(R)as Rn2 and of Lie(G) as the subspace consisting of those vectors with the lastw = dimW entries 0 andW as those with the first d = dimLie(G) entries 0. Weconsider the map ϕ : Mn(R) → GL(n,R) given by ϕ(X) = exp(X1) exp(X2)with X = X1 + X2 and X1 ∈ Lie(G), X2 ∈ W . We note that ϕ(tX) =(I + tX1 + O(t2))(I + tX2 + O(t2)) = I + t(X1 + X2) + O(t2)). We there-fore see that the differential of ϕ at 0 is the identity map. We may there-fore apply the inverse function theorem and find there exists s1 > 0 such thatϕ : Bs1(0) → GL(n,R) has an open image U1 with the map ϕ : Bs1(0) → U1a homomorphism. With these preliminaries we can begin the argument. Sup-pose that for any ε > 0 with s1 ≥ ε the set ϕ(Bε(0)) ∩ G contains an elementthat outside of exp(Lie(G)). Thus there exists for each n ≥ 1

s1an element in

Zn ∈ B 1n(0) such that exp(Zn) ∈ G but Zn /∈ Lie(G). We write Zn = Xn + Yn

with Xn ∈ Lie(G) and Yn ∈W . Then

ϕ(Zn) = exp(Xn) exp(Yn).

Since exp(Xn) ∈ G we see that exp(Yn) ∈ G. We also observe that kYnk ≤ 1n .

Let εn = kYnk then εn ≤ 1n ≤ s1. There exists a positive integer mn such that

s1 ≤ mnεn < 2s1. Hences1 ≤ kmnYnk < 2s1.

Now exp(mnYn) = exp(Yn)mn ∈ G. Since the sequence mnYn is bounded we

can replace it with a subsequence that converges. We may therefore assumethat there exists Y ∈W with limn→∞mnYn = Y . Since G is closed exp(Y ) ∈ G.We will now use (variant of) the classic argument of Von Neumann to show thatif t ∈ R then exp(tY ) ∈ G. Indeed, let mnt = an + bn with an ∈ Z and bn ∈ R,0 ≤ bn < 1 (i.e. an is the integer part of mnt). Then

tmnYn = anYn + bnYn.

Henceexp(tmnYn) = exp(Yn)

an exp(bnYn).

Now 0 ≤ bb < 1 so since limn→∞ Yn = 0 we see that limn→∞ exp(bnYn) = I.We therefore see that

exp(tY ) = limn→∞ exp(tmnYn) = lim

n→∞ exp(Yn)an ∈ G.

this completes the proof that exp(tY ) ∈ G for all t. We have thus come uponthe contradiction Y ∈ Lie(G)∩W = {0} since Y 6= 0. We have therefore shownthat there exists an ε > 0 such that ϕ(Bε(0)) ∩G ⊂ exp(Lie(G)).

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This result implies that a closed subgroup of GL(n,R) has an open neighbor-hood of I that is homeomorphic with an open ball in Rm with m = dimLie(G).Now Theorem A.D.1 implies that it has a Lie group structure.We also note that Proposition 16 be reformulated as

Proposition 20 Let G ⊂ GL(n,R) and H ⊂ GL(m,R) be closed subgroupsand let ϕ : H → G be a continuous homomorphism. Then ϕ is of class C∞.

We will now relate the two notions of Lie algebra and differential that havenow arisen.Let xij be the standard coordinates of Mn(R), that is we write X ∈Mn(R)

as X = [xii]. In other words we take the standard basis Eij (the matrix withexactly one non-zero entry that is a 1 in the i, j position. Then the xij are thecoordinates of X with respect to this basis. We first note that if f ∈ C∞(U)with U an open neighborhood of I in Mn(R) then

∂

∂xijf(I) =

d

dtf(I + tEij)|t=0 =

d

dtf(exp(tEij))|t=0.

Using the chain rule we see that

d

dtf(X(I + tEij))|t=0 = Dijf(X)

with Dij =Pk

xki∂

∂xkj. We therefore see that if A ∈Mn(R) then

d

dtf(X(I + tA))|t=0 =

Pij

aijDijf(X).

Our observations about the expansion of exp(tA) in powers of t imply that

d

dtf(X exp(tA))|t=0 =

PijaijDijf(X).

We will write XA forPijaijDij . Then (XA)I =

Pijaij

∂∂xij

∈ TI(Mn(R)). We

note that a direct calculation implies that [XA,XB] = X[A,B].Now assume that G is a closed subgroup of GL(n,R). Then we have seen

that G is a Lie subgroup. Let iG : G → GL(n,R) be the injection of G intoGL(n,R). We assert that (the Lie(G) in the statement is in the sense of thissection)

Lemma 21 We have (diG)I(TI(G)) = {(XA)I |A ∈ Lie(G)}.Proof. Let A ∈ Lie(G) then the map of R to G given by t→ exp(tA) is a C∞

map. If f ∈ C∞(G) then we define

vAf =d

dtf(exp(tA))|t=0

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an element of TI(G). Now (diG)I(vA)f = (XA)I . We therefore see that(diG)I(TI(G)) ⊃ {(XA)I |A ∈ Lie(G)}. Since dimLie(A) = dimTI(G) we seethat the two spaces are the same.We also note that if A ∈ Lie(G) then we can define

XGA f(g) =

d

dtf(g exp(tA))|t=0.

Then by the very definition and the observation that the map R×G→ G, t, g 7→g exp tA is smooth we see that XG

A is a left invariant vector field. We note

Lemma 22 We have if A,B ∈ Lie(G) then [XGA ,X

GB ] = XG

[A,B].

Proof. If A,B ∈ Lie(G) then

f(exp(sA) exp(tB)) = f(exp sA) + tXGBf(exp(sA)) +O(t2)

= f(I) + sXAf(I) +O(s2) + tXGBf(I) + stXG

AXGBf(I) +O(t2).

We therefore see that since

f(exp(sA) exp(tB))−f(exp(tB) exp(sA)) = st(XGAX

GBf(I)−XG

BXGAf(I))+O(s

2)+O(t2)

we have [XGA ,X

GB ]f(I) =

∂2

∂s∂t (f(exp(sA) exp(tB))− f(exp(tB) exp(sA)))|s=0,t=0.On the other hand if f ∈ C∞(GL(n,R)) then the same argument implies (since[XA,XB] = X[A,B] and X

GL(n,R)A = XA)

∂2

∂s∂t(f(exp(sA) exp(tB))− f(exp(tB) exp(sA)))|s=0,t=0 = X[A,B]f(I).

This implies that (diG)I([XGA ,X

GB ]I) = (X[A,B])I = (diG)I(X

G[A,B])I . Hence,

since [XGA ,X

GB ] is completely determined by [X

GA ,X

GB ]I we see that [X

GA ,X

GB ] =

XG[A,B].We have already observed that if G ⊂ GL(n,R) and H ⊂ GL(m,R) are

closed subgroups and if ϕ : H → G is a continuous homomorphism then ϕis of class C∞. We will now calculate dϕI : TI(H) → TI(G) we have (usingthe notation in the proof of the preceding lemma) the map Lie(H) → TI(G),A 7→ vA. Now if f ∈ C∞(G) then

dϕI(vA)f =d

dtf(ϕ(exp(tA)))|t=0 .

But ϕ(exp(tA)) = exp(tµ(A)) (as in Proposition 16). Thus dϕI(vA) = vµ(A).From this we see that if we define the vector field XH

A on H by XHA f(h) =

ddtf(h exp(tA))|t=0 then we find that dϕh(X

HA )h = (X

Gµ(A))ϕ(h).

We therefore see that the abstract definition of Lie algebra as left invariantvector fields and the concrete notion here of a space of matrices yield the sameLie algebras under the correspondence A 7→ XG

A . Furthermore the definitionof differential in Proposition 16 yields the same differential as the abstract oneunder this correspondence.

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1.3.5 Covering and quotient groups.

1.3.6 Exercises.

1. Complete the proof that the Lie algebras of the groups described in section1.1 “line up” correctly with the Lie algebras described in section 1.2.2. We will use the notation of Exercise 4 in 1.1.10. Observe that ϕ is

continuous and prove that dϕ is a Lie algebra isomorphism. Use this result toprove that the image of ϕ is open in SO(V,B) and hence closed/3. In this exercise we will use the notation of Exercise 7 of 1.1.10. Observe

that ϕ is continuous and prove that dϕ is a Lie algebra isomorphism use this toprove that the image of ϕ is open and closed in the corresponding orthogonalgroup.4. In the notation of exercises 8 and 9 of 1.1.10 prove that the appropriate

differentials are Lie algebra isomorphisms.

2 Addendum for Appendix D.In this addendum we add the following result which is for the most part provedin the end of the proof of Theorem D.2.8.Therem A.D.1. Let G be a topological group and assume that there is an

open neighborhood U of e (the identity of G) such that if u ∈ U then u−1 ∈ Uand a surjective homeomorphism Φ : U → Br(0) ⊂ Rm for some r > 0, an swith 0 < s < r such that if V = Φ−1(Bs(0)) then if u, v ∈ V then uv ∈ U anda C∞ map F : Bs(0)×Bs(0)→ Br(0) such that1) Φ(u−1) = −Φ(u) for u ∈ U ,2) Φ(uv) = F (Φ(u),Φ(v)) for u, v ∈ V .Then there exists a structure of a Lie group on G compatible with its structure

as a topological group.

We also observe that this combined with the following discussion gives asimple proof of the fact that a connected covering space of a connected Liegroup is a Lie group.If G is a connected and locally arcwise connected topological group we will

now show that if we have a covering space π : H → G and we choose eo ∈ π−1(e)then H has a structure of a toplogical group with identity eo such that π is agroup homomorphism. We will now begin that task.Let Lg : G → G be given (as usual) by Lg(x) = gx. Then for each h ∈ H

there exists a unique homeomorphism L̃h : H → H such that L̃h(eo) = hand π(L̃h(x)) = Lπ(h)(π(x)). We assert that if we set m(u, v) = L̃u(v) thenwith this multiplication H is a group. The identity map has the same propertyassumed for L̃eo thus m(eo, u) = u and by definition m(u, eo) = u thus eo isand identity element for the multiplication m. If u ∈ H then there exists aunique v ∈ H such that L̃u(v) = eo Thus if we show that the associative ruleis satsified we will see that H with multiplication m and identity eo is a group.

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We note Lx◦Ly = Lxy and π(m(x, y)) = xy and sincem(x, y) = L̃m(x,y)(eo) =

L̃x ◦ L̃y(eo) we have L̃m(x,y) = L̃x ◦ L̃y. This is the content of the associativerule.We must now prove that m : H ×H → H is continuous. We note that if we

set µ(x, y) = xy for x, y ∈ G then there is a unique lift. µ̃ : H ×H → H of µsuch that µ̃(eo, eo) = eo. Since m is another such lift we see that m = µ̃.

It also implies an easy proof of the fact that if H ⊂ G is a closed subgroupof a Lie group G then G/H has the structure of a Lie group.

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