sign in sign up
<<
Home , Additive group, Classical group, Class of groups, Degenerate bilinear form, General linear group, Group (mathematics)

1

1.1 The groups

Let F denote either the real , R, or the complex numbers, C.Inthis section we will describe the main players in the rest of this book the Classical Groups as designated by ..This section should be treated as a dictionary. The groups as named here will appear throughout the book.

1.1.1 The general linear .

Let V be a finite dimensional vector over F. The of all invertible transformations of V to V will be denoted GL(V ).Thissethasagroupstruc- ture under composition of transformations with the identity transformation Id(x)=x for all x V . We will consider this family of groups in this subsection. We will be recalling∈ some standard terminology related to linear transformations and their matrices. Let V and W be finite dimensional vector over F with bases v1, ..., vn and w1, ..., wm respectively. If T : V W is a linear then we have → m Tvj = aijwi i=1 P with aij F.Theaij are called the coefficients or entries of the linear transformation∈ T with respect to the two bases and the m n array × a11 a12 a1m ··· a21 a22 a2m A =  . . ···. .  . . .. .    am1 am2 amn   ···    is the matrix of T withrespecttothetwobases.IfS : W U is a linear → map with U an l dimensional with u1, ..., ul and if B is the matrix of S with respect to the bases w1, ..., wm and u1,...,ul then the matrix of S T with respect to v1,...,vn and u1, ..., ul is given by BA the product being the usual◦ product of matrices. We denote by Mn(F) the space of all n n matrices over F.LetV be an n × dimensional vector space over F with basis v1, .., vn.IfT : V V is a we will (for the moment) write µ(T ) for the matrix of T →with respect to this basis. If T,S GL(V ) then the observations above imply that ∈ µ(S T )=µ(S)µ(T ). ◦ 1 1 Furthermore, if T GL(V ) then µ(T T − )=µ(T − T )=µ(Id)=I the ∈ ◦ ◦ (the entries being δij =1if i = j and 0 otherwise. A matrix A Mn(F) is said to be invertible if there is a matrix B Mn(F) such that AB∈= BA = I. WenotethatalinearmapT : 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 and only if its 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 with identity the identity matrix. We note that if V is an n-dimensional vector space over F with basis v1, ..., vn then the map µ : GL(V ) GL(n, F) corresponding to this → basis is a group . The group GL(n, F) is called the general of n. If w1, ..., wn is another basis of V and if

wj = gijvi i P then vj = hijwi with [hij] the inverse matrix to [gij].IfPT is a linear transformation of V to V and if A =[aij] is the matrix of T with respect to v1, ..., vn and if B =[bij] is the matrix with respect to w1, ..., wn then

Twj = T gijvi = gijTvi = µ i ¶ i gij( akivk)= gPij( aki( Phlkwk). i k i k l P P P P P 1 Thus, if g =[gij] then B = g− Ag.

1.1.2 The .

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)=1we see that the special linear group is a of GL(n, F) with will be denoted SL(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 the} previous subsection. You will be called upon to prove this in Exercise 1 at the 1 of this section). This us the subgroup µ− (SL(n, F)) and we will denote it by SL(V ).

1.1.3 groups of bilinear forms.

Let V be an n dimensional vector space over F and let B : V V F be a Then we denote by O(B) the set of all T GL(×V ) such→ that B(Tv,Tw)=B(v,w) for all v, w V .WenotethatO∈(B) is a subgroup ∈ of GL(V ).Letv1, ..., vn be a basis of V relative to this basis we may write B(vi,vj)=bij. Assume that T has matrix A =[aij] then we have

B(Tvi,Tvj)= akialjB(vk,vl)= akialjbkl. k,l k,l P P 2 t Thus if A is the matrix [cij] with cij = aji (the usual of A) then the condition that T O(B) is that ∈ t [bij]=A [bij]A.

If B is non-degenerate (B(v,w)=0for all w implies v =0and B(v,w)=0 for all v implies w =0)thenwehavedet([bij]) =0. Hence using the formula aboveweseethatifB is nondegenerate and if T6 : V V is linear and satisfies B(Tv,Tw)=B(v, w) for all v, w V then T O(B).→ The next two subsections will discuss the most important special∈ cases∈ of this 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 the set of all g GL(n, F) such that ggt = I.Thatis ∈ t 1 g = g− .

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

Ip 0 Ip,q = 0 Iq · − ¸ with Ir denoting the r r identity matrix. Then we define × t O(p, q)= g Mn(R) g Ip,qg = Ip,q . { ∈ | } We note that O(n, 0) = O(0,n)=O(n, R). Also, if 00 1 ··· . . .. . σ =  . . . .  0100    1000     that is the matrix with all entries zero except for ones on the skew (the 1 t i, n +1 i entries are 1). Then σIp,qσ− = σIp,qσ = σIp,qσ = Iq,p.Thusthe map − − φ : 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-degenerate over F (either R or C). We have

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

Remark 2 The p q inthecasewhenF = R is called the signature of the 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 on V .ThenifM(v,v)=0for all v V then M =0. ∈ Proof. Using the 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 firstshowthatthereexistsa basis w1,...,wn of V such that

B(wi,wj)=0for i = j 6 and B(wi,wi) =0 6 by induction on n (such a basis is called an orthogonal basis with respect to B). Since B is non-degenerate the previous lemma implies that there exists a vector wn V with B(wn,wn) =0.Ifn =1then this is what we set out to ∈ 6 prove. Assume for n 1 1.SetV 0 = v V B(wn,v)=0 .Ifv V then B(v,wn) − ≥ { ∈ | } ∈ v0 = v wn V 0.ThisimpliesthatV = V 0 + Fwn. So, in particular, − B(wn,wn) ∈ dim V 0 = n 1. WeassertthattheformB0 = B V V is non-degenerate on V 0. − | 0× 0 Indeed if v V 0 satisfies B(v0,w)=0for all w V 0 then since B(v0,wn)=0 ∈ ∈ we see that B(v0,w)=0for all w V so v0 =0. We can now apply the ∈ inductive hypothesis to find a desired basis w1, ..., wn 1 for V 0. Now it is clear − that 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.Letzi beachoiceofsquarerootofB(wi,wi) then 1 if vi = wi the B(vi,vj)=δij. At this point the reader should have noticed zi 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.Letp =0if B(w1,w1) < 0. ≥ − Otherwise let p =max i B(wi,wi) > 0 .ThenB(wi,wi) < 0 for i>p.Define { | } zi to be a root of B(wi,wi) for i p.Define zi to be a of 1 ≤ B(wi,wi) if i>p.Thenifvi = wi we have B(vi,vj)=εiδij.Weareleft zi −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.

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

With this concept in hand we will complete the proof. Fix a basis v1, ..., vn such that B(vi,vj)=εiδij with εi =1for i p and εi = 1 for i>p. ≤ − Set V+ equal to the of v1, ..., vp and V equal to the linear span { } − of vp+1, ..., vn .ThenV = V+ + V . Let π : V V+ be the { } − → onto the first factor. We note that B V+ is positive definite. Let W | be a subspace of V such that B W is positive definite. Suppose that w W and | ∈ π(w)=0.Thenw V thus w = aivi. Hence ∈ − i>p P 2 B(w, w)= aiajB(vi,vj)= ai 0. i,j>p − i>p ≤ P P Thus since B W has been assumed to be positive definite w =0. This implies | that π : W V+ is injective. Hence dim W dim V+ = p. If we had another → ≤ basis satisfying the desired conditions yielding p0 as above then this implies that p0 p and p p0 so p = p0. This completes the proof. ≤ ≤ InthecasewhenF = R and B is a non-degenerate symmetric bilinear form on a finite dimensional vector space V we will call a basis v1, ..., vn of V such that there exists n>p>0 such that B(vi,vj)=εiδij with εi =1for i p and ≤ εi = 1 for i>pa pseudo . We have seen that the number p is an− 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 basis of V then we will use the notation µ(g) for the matrix of the linear g with 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 orthonormal basis for V with respect to B or B then µ : O(B) O(n, F) defines a group − → isomorphism. If F = R and if v1, ..., vn is a pseudo-orthonormal basis of V with signature (p, n p) then µ : O(B) O(p, n p) is a . − → − The special orthogonal group over F is the subgroup SO(n, F)=O(n, F) ∩ SL(n, F) of O(n, F).Theindefinite special orthogonal groups are the groups SO(p, q)=O(p, q) SL(p + q, R). ∩ 1.1.5 The . We set 0 I J = I 0 · − ¸

5 with I the n n identity matrix. The symplectic group of rank n over F is defined to be × t Sp(n, F)= g M2n(R) g Jg = 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 be a 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 dim V = m>0.Letv be a non- zero element of V .Thenthereexistsw V with B(v,w) =0. Replacing w with 1 w we may assume that B(v, w∈)=1.LetW = 6 x V B(v, x)= B(v,w) { ∈ | 0,B(w, x)=0 .Ifx 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)=1and 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.ThusV = U + W where U is the span of v − and w.WenotethatB U is non-degenerate so U W = 0 .Thisimpliesthat | ∩ { } dim W = m 2. WealsonotethatB W is non-degenerate. We will leave this to the reader− (use the same technique| as we used for the orthogonal groups see exercise 3 at the end of this section). We will now prove the firstpartofthe proposition. We have seen that if dim V>0 then dim V 2.Thustheresult is true for dim V 2. Assume that the result is true for 2 ≥dim V

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

6 1.1.6 The unitary groups.

In this subsection we will look at an important class of of GL(n, C) the unitary groups and special unitary groups for definite and indefinite Hermitian forms. If A Mn(C) then we will use the standard notation for its adjoint matrix ∈ t A∗ = A where A is the matrix obtained from A by complex conjugating all of the entries. The of rank n is the group U(n)= A Mn(C) A∗A = I .The { ∈ | } is U(n) SL(n, C).LetIp,q be as in 1.1.4. We define the indefinite 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 as × → vector spaces over R is said to be a Hermitian form if it satisfies the following two 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 if ∈ v V and v =0(note that if M is a Hermitian form then M(v, v) R for all v ∈ V ). We de6 fine U(B) to be the group of all elements, g,ofGL(V )∈such that B∈(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 a non-degenerate Hermitian form on V . Then there exists an n p 0 ≥ ≥ and a basis v1, ..., vn of V over C such that

B(vi,vj)=εiδij with εi =1for i p and εi = 1 for i>p. The number depends on p and not on the choice of basis.≤ −

The proof of this Proposition is almost identical to that of Proposition 4 and willbelefttothereader(seeexercise5foraguidetoaproof). 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 called a pseudo-orthonormal basis (if p = n then it will be called an orthonormal basis.Thepair(p, n p) will be called the signature of B. The following result is proved in exactly− the same way as the corresponding result for orthogonal groups.

7 Lemma 9 Let V be a finite dimensional vector space over C and let B be a non-degenerate Hermitian form on V of signature (p, q) and let v1,...,vp+q be a pseudo-orthonormal basis of V relative to B,Thenifµ : GL(V ) GL(n, C) is the 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 .∈ ∈ → 1.1.7 The quaternionic . We will first recall some basic properties of the . We consider the vector space, H,overR given by xy H = x, y C . y x | ∈ ½· − ¸ ¾ One checks directly that H is closed under multiplication in Mn(C), that the identity matrix is in H and that since xy x y − =(x 2 + y 2)I y x yx | | | | · − ¸· ¸ every non-zero element of H is invertible. Thus H is a (or skew field) over R. This is a realization of the quaternions. The more usual way of introducing the quaternions is to consider the vector space, H,over R 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 = i − − − and extended to H by linearity. We note that this gives an isomorphic division algebra to the one described if we take

i 0 01 0 i 1 = I,i = , j = , k = . 0 i 10 i 0 · − ¸ · − ¸ · ¸ We define GL(n, H) to be the group of all invertible n n matrices over H.Here the definition of is the usual one× but one must be careful about the 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 so A defi∈nes a quaternionic∈ linear map. Now has 3 different structures· as a vector space over C.Bylookingatthesubfields

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

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

1.1.8 The quaternionic unitary groups.

If w H,with ∈ ab w = b a · − ¸ 2 2 then we note that w∗w = ww∗ = a + b .IfX Mn(H) then set X∗ =[xji∗ ] | | | | ∈ 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).Onecandefine quaternionic Hermitian forms and prove a result analo- gous to Proposition 8 but we will have no need of such material in the rest of this book.

1.1.9 The group SO∗(2n). If we include covering groups (to be definedlater)wewillhavedescribedupall of the classical groups over C and R except for one infinite family that is related to the orthogonal groups. We will describe that family now. Let 0 I J = I 0 · − ¸ 2 as usual with I the n n identity matrix. We note that J = I2n the 2n 2n × − × identity matrix. Thus the map GL(2n, C) to itself given by θ(g)= JgJ defines an and θ2 is the identity automorphism. Our last classical− group is SO∗(2n)= g SO(2n, C) θ(g)=g { ∈ | } here, as usual, g is the matrix whose entries are the complex conjugates of g.

9 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 relative to these bases→ then det A =detB. n 2. What is signature of the form B(x, y)= xiyn+1 i on R ? i − 3. Let V be a vector space over F and B aP skew symmetric or symmetric non- on V .AssumethatW is a subspace of V on which B restricts to a nondegenerate form. Prove that the restriction of B to W ⊥ = v V B(v, w)=0for all w W is non-degenerate. { ∈ | ∈ } 4. Let V denote the vector space of symmetric 2 2 matrices over F = R or 1 × C.Ifx, y V define B(x, y)= (det(x + y) det(x y)) show that B(x, y)= ∈ 2 − − B(y,x) for all x, y V . Show that if F = R then the signature of the form B ∈ is (2,1). If g SL(2, F) define ϕ(g) GL(V ) by ϕ(g)(v)=gvgt.Thenshow ∈ ∈ that ϕ : SL(2, F) O(V,B) is a with SO(V,B) and 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 is Hermitian form such that M(v, v)=0for all v then M =0. Todothisnote that 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 are the same as for Proposition 4 once one observes that if M is Hermitian then M(v,v) R. ∈ 6. Let V be a 2n-dimensional vector space over F = R or C.Weconsider the space W = nV . Fix a basis ω of the one dimensional vector space 2nV . Consider the bilinear form B(u, v) defined by u v = B(u, v)ω. Show that the form is nondegenerateV and skew symmetric if n∧is odd and symmetricV if n is even. 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 = and if e1,e2,e3,e4 is the then ω = e1 e2 e3 e4.Considerϕ(g)= 2 ∧ ∧ ∧ 2 g for g SL(4, F). Show that ϕ : SL(4, F) O( F4,B) is a group ∈ → homomorphism with kernel I .IfF = C and we choose an orthonormal basis V {± } V then ϕ defines a group homomorphism of SL(4, C) to SO(6, C).IfF = R and we choose a pseudo-orthonormal basis show that ϕ defines a group homomorphism to 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.Let ∧ 2 4 ∧ − W = w F B(ν,w)=0 .Letρ(g)=φ(g) W .IfR show that the signature of the{ restriction∈ | of B to W }is (3,2). Show that| ρ is a group homomorphism V from 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 × 2 2 → ⊗ ϕ(a, b)=a b.LetB be the bilinear form on F F given by B(ei ej,ek el)= ⊗ ⊗ ⊗ ⊗ εikεjl with εij = εji for i, j =1, 2 and ε12 =1. Show that B defines a − symmetric nondegenerate form on F2 F2 and calculate the signature of B if ⊗ F = R. Show that if g G then ϕ(g) SO(F2 F2,B). What is the kernel of ∈ ∈ ⊗

10 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 matrix × over C. Show that the of these three matrices are equal.

1.2 The Lie .

Let V be a vector space over F.LetEnd(V ) denote the algebra (under composition) of F-linear maps of V to V .IfX, Y End(V ) then we set [X, Y ]=XY YX.Thisdefines a new product on End∈ (V ) that satisfies two properties (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 if for each X, Y, Z g the conditions 1. and 2. are satisfied. ∈

Thus, in particular, we see that End(V ) is a Lie algebra under the binary [X, Y ]=XY YX. 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 the restriction of [...,∈ ...].Wewillcallh a sub-Lie algebra or a . 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 → 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 YX. Wenotethatifwe choose a basis for V and if V is n-dimensional then the− matrices of the elements of 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) and ∈ ∈ A =[aij] then tr(A)= aii.Wenotethat i P tr(AB)=tr(BA).

This implies that if A is the matrix of T with respect to v1, ..., vn then tr(A) is independent of the choice of basis. We will write tr(T )=tr(A). This implies that 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. We define × → 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 ) by the obvious calculation (see Exercise 3 at the end of this section). If B is nondegenerate you will also be asked to prove in Exercise 3 that trX =0for X 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 .Thenbij = B(vi,vj(.Bya calculation completely analogous to that in 1.1.3 we see that if T so(B) then ∈ its matrix, A,relativetov1, ..., vn satisfies

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

1.2.3 The orthogonal Lie algebras. We define t so(n, F)= X Mn(F) X = X . { ∈ | − } Since trXt = trX this implies that so(n, F) is a subspace of sl(n, F).Itisalso easily checked (by direct calculation) that so(n, F) is a Lie subalgebra of sl(n, F). By with the definition of O(p, q) let p, q 0 be such that p + q = n. Set (as in 1.1.4) ≥

Ip 0 Ip,q = . 0 Iq · − ¸ We set t so(p, q)= X Mn(R) X Ip,q = Ip,qX . { ∈ | − } As observed in the previous section so(p, q) sl(n, R) and it is easily checked to 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 over F.IfF = C then we have seen in Proposition 1 (1.1.4) that there is a basis v1, ..., vn of V such that B(vi,vj)=δij.Ifµ(T ) is the matrix of T End(V ) relative to this matrix then µ defines a Lie algebra isomorphism of so∈(B) onto so(n, C).IfF = R and B has signature p, q and v1, ..., vn is a corresponding pseudo-orthonormal basis then the analog of µ defines a Lie algebra isomorphism of so(B) onto so(p, q).

12 1.2.4 The symplectic Lie algebra. In this case we take 0 I J = 10 · − ¸ with I the n n identity matrix. We define × t sp(n, F)= X M2n(F) X J = JX . { ∈ | − } Then this subspace of the n n matrices is a Lie subalgebra which we call the symplectic Lie algebra of rank× n. Let V be a vector space over F of m< and B a skew symmetric bilinear form on V .Thenm is even m =2n and applying∞ Proposition 6 section 1.5 we see that there exists a basis v1,...,v2n of V such that [B(vi,vj)] = J. The map that assigns to an of V its matrix relative to this basis defines 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 with p, q 0,Wedefine ≥ u(p, q)= X Mp+q(C) X∗Ip,q + Ip,qX =0 . { ∈ | } We note that this space is a real subspace of Mp+q(C). One checks directly that it 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,letV be an n-dimensional vector space over C and let B be a non-degenerate Hermitian form on V .Thenwedefine

u(B)= T End (V ) B(Tv,w)+B(v,Tw)=0,v,w V . { ∈ C | ∈ } and su(B)=u(B) sl(V ).IfB has signature (p, q) and if v1, ..., vn is a pseudo orthogonal basis of∩V then the assignment T µ(T ) of T to its matrix relative to this basis defines a Lie algebra isomorphism7→ of u(B) with u(p, q) and su(B) with su(p, q).

1.2.6 The general linear Lie algebra. In this subsection we will follow the notation of section 1.1.7. The Lie algebra in question consists of the n n matrices over the quaternions, H, with the usual × matrix . We will denote the Lie algebra gl(n, H). This Lie algebra is to be thought of as a Lie algebra over R (wehavenotdefined Lie algebras over skew-fields). If we consider Hn to be C2n relative to any one of the isomorphic copies of C: R1 + Ri, R1 + Rj,R1 + Rk.Thenwecantakesl(n, H) to be the elements of zero in any of these structures. This real Lie algebra is usually denoted 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 of θ. The Lie algebra in question is defined to be space of all X so(2n, C) such ∈ that θ(X)=X. This real vector subspace of so(2n, C) defines a Lie subalgebra if 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 subject of study throughout the remainder of this book. We will find that the choices of 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 that have a name that is "frakturized" form of a name in the previous section can be obtained by “differentiating” the definition as a group. That is we look at the rule for inclusion in the corresponding group, G. Let us call it . We will denote the Lie algebra with the “frakturized” designation g.ConsideracurveR σ :( ε, ε) GL(V ) such that σ(0) = I and σ is differentiable at 0.Then − → one should observe that if (σ(t)) is satisfied for all t ( ε, ε) then σ0(0) g. For example, if G is the subgroupR O(B) of GL(V ) de∈fined− by a binary form∈ B (i.e. G = g GL(V ) B(gv, gw)=B(v, w) for all v,w V .Thenthe indicated conditions{ ∈ on the| are B(σ(t)v, σ(t)w)=B(v,w∈) for} all v, w V and t ( ε, ε).Ifwedifferentiate these relations we have ∈ ∈ − d 0= B(σ(t)v,σ(t)w) t=0 = B(σ0(0)v, σ(0)w)+B(σ(0)v, σ0(0)w). dt |

Since σ(0) = I we see that σ0(0) so(B) as asserted. This is the reason why these 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 =0for 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 general linear 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 out ∈ the corresponding matrix of X as an element of M2n(C).Usethisformulato observe 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 det(σ(t)) t=0 = tr(σ0(0)). dt |

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

1.3.1 Topological Groups. Let G be a group with a such that the maps

G G G, g, h gh × → 7→ and 1 G G, g g− → 7→ are continuous is called a topological group.HerethesetG G is given the × above. We note that GL(n, F) is a topological group when - n2 dowed with the topology of an open subset of Mn(F) looked upon as F 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 is continuous. 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 group → → properties and continuity imply that Rg and Lg are . If G is a topological group and H is a subgroup that is closed as a then H is also a topological group called a topological subgroup. Thus all of 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 topological group isomorphism if it is bijective and its inverse is also a topological group homomorphism. 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 0 I J = I 0 · − ¸ with I the n n identity matrix then Mn(C) consists of the matrices in M2n(R) × that commute with J and GL(n, C) consists of the elements of GL(2n, R) that commute with J. The case of the quaternionic groups is handled similarly where we look at the standard n-dimensional vector space over the quaternions, Hn,as 4n 2 a R with three operators J1,J2,J3 such that JkJl = JlJk for k = l, Ji = I − 6 − and J1J2 = J3,J1J3 = J2,J2J3 = J1.Givenasfollows − J 0 0 I 0 J J = ,J = ,J = . 1 0 J 2 I 0 3 J 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 elements of 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,thenH is closed in G.

Proof. We note that G is a disjoint of left . If g G then since ∈ Lg is a gH = Lg(H) is open. Hence every left 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 of G (that is the connected subgroup that contains the , e)is a normal sub- group.

Proof. Let X be the of G.Ifg X then since e X we ∈ ∈ see that g LgX.SinceLg is a homeomorphism we see that LgX = X. Hence ∈ 1 X is closed under multiplication. Since e LgX we see that g− X.SoX is ∈ 1 ∈ a subgroup. If g G then define τ(g)(x)=gxg− for x G.Thenτ(g) is a homeomorphism of∈ G and τ(g)(e)=e.Thusτ(g) maps the∈ identity component of G to itself. Hence X is a .

1.3.2 The exponential map.

t On Mn(R) we define the inner product X, Y = tr(XY ). The corresponding h i 1 2 1 X = X, X 2 =( x ) 2 k k h i ij ij X has the following properties.

16 1) It is a norm that is X + Y X + Y , cX = c X and X =0 if and only if X =0. k k ≤ k k k k k k | |k k k k

2) XY X Y . Wek willk now≤ k showkk howk to derive 2) from the usual Cauchy-Schwarz inequal- ity. 2 2 XY = xikykj . k k ij à k ! X X 2 2 2 Now xikykj x y by the usual Cauchy-Schwarz inequal- | k | ≤ k ik k kj ity. Hence P ¡P ¢ ³P ´

XY 2 x2 y2 = x2 y2 = X 2 Y 2 . k k ≤ ik kj ik  kj k k k k ij à k !à k ! à ik ! kj X X X X X Takingthesquarerootofbothsidescompletestheproof.  If A Mn(R) and r>0 we will use the notation Br(A)= X Mn(R) X A < r . ∈ { ∈ |k − k } We can use this norm in order to define some matrix valued analytic func- m tions. We note that if we have a power m∞=0 amz and if r>0 is such m that m∞=0 am r < then the given series defines a function f(z) on the disc of radius| r.Wenotethatforthesameseriesif| ∞ P k l and X

m m m m amX amX = amX am X ° − ° ° ° ≤ | |k k °0 m k 0 m l ° °l m k ° l m k ° ≤X≤ ≤X≤ ° ° ≤X≤ ° ≤X≤ ° ° ° ° ° ° ° °m ∞ m ° ° ° am X ° am r . ≤ | |k k ≤ | | ll X≤ X m This goes to 0 as l .Thustheseries m 0 amX defines a from →∞ ≥ the open ball of radius r in Mn(R) to Mn(R). The absolute convergence of this series implies that we can interchange theP order of without changing the value. In this section we will be concentrating on two functions. The first is defined by the

∞ xn exp(x)= n! n=0 X and the second ∞ xn log(1 + x)= ( 1)n+1 . n n=1 − X The first converges on every disc and for the other the best choice is r =1.We note 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) → and log : X Mn(R) X I < 1 Mn(R). { ∈ |k − k } → Both of these functions are given by power series so are real analytic functions. We note that exp(0) = I and exp(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) X

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 have exp(tv)=I + tv + O(t2). This implies that d exp0(v)=v.

The implies that there exists a ball Bs(0) with s>0 and 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, ⊂ 1 − exp Bs(0) =logU . | | ³ ´ 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 homomorphism → of the of real numbers then there exists X Mn(R) such the ∈ φ(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 of R into GL(n, R).Sinceφ is continuous we may assume that φ(t) exp(B s (0)) ∈ 2 for t 2.Thusφ(1) = exp X for some X B s (0). Wealsonotethat 2 1 | | ≤ ∈ φ( )=expZ with Z B s (0).Thusφ(1) = exp(2Z).Sinceexp : B (0) U 2 2 s ∈ 1 → is injective this implies that 2Z = X so Z = 2 X. If we do the same argument 1 with φ( )=exp(W) with W B s (0) we have exp(2W )=expZ and since 4 2 ∈ 1 1 both 2W and Z are in Bs(0) we have W = 2 Z = 4 X.Wecancontinuethis argument and find that 1 1 φ( )=exp( X). 2k 2k

a1 a2 ak If a = 2 + 22 + ... + 2k + ... with aj 0, 1 is the dyadic expansion of the 0 a<1 then by continuity∈ and{ the} the assumption that φ is a group homomorphism≤ we have

a1 a2 ak 1 1 1 φ(a) = lim φ( + + ... + ) = lim φ( )a1 φ( )a2 φ( )ak k 2 22 2k k 2 4 ··· 2k →∞ →∞ 1 1 = lim exp( X)a1 exp( X)ak k 2 ··· 2k →∞ a a a = lim exp(( 1 + 2 + ... + k )X)=exp(aX). k 2 22 2k →∞ 1 1 Now if 0 a<1 then φ( a)=φ(a)− =exp(aX)− =exp( aX). Finally, ≤ − a − if a R there exists k>0 with k an integer such that k < 1.Thenφ(a)= a ∈k a k | | φ( k ) =exp(k X) =exp(aX). exp(tX) exp(tY ) If exp(tX)=exp(tY ) for all t then 0= −t = X Y + O(t) as t 0.ThusX = 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 Lie | ∈ ∈ } subalgebra of Mn(R) relative to the [X, Y ]=XY YX.Thiswillbe the first of four apparently different definitions of Lie algebra− of a that will be used in this book. For this we need more information about the exponential and . We consider

log(I +(exptX exp tY I)) −

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

log(I +(exptX exp tY I)) − t2 t2 =log((I + tX + X2 + O(t3))(I + tY + Y 2 + O(t3))) 2 2 t2 t2 =log(I + tX + tY + t2XY + X2 + Y 2 + O(t3)) 2 2 t2 t2 1 t2 t2 = t(X + Y )+t2XY + X2 + Y 2 (tX + tY + t2XY + X2 + Y 2)2 + O(t3) 2 2 − 2 2 2 t2 t2 1 = t(X + Y )+t2XY + X2 + Y 2 (tX + tY )2 + O(t3) 2 2 − 2 since all the rest of the terms in the last expression involve at least t3.Thislast expression is t2 t2 1 tX + tY + t2XY + X2 + Y 2 t2(X2 + XY + YX+ Y 2)+O(t3) 2 2 − 2 t2 = tX + tY + [X, Y ]+O(t3). 2 If we do exactly the same argument with

log(I +(expsX exp tY I)) − neglecting any terms of order t2 or s2 then we have

log(I + sX + tY + stXY + O(s2)+O(t2)) 1 = sX + tY + stXY (sX + tY + stXY )2 + O(s2)+O(t2) − 2 1 = sX + tY + stXY (stXY + stY X)+O(t2)+O(s2) − 2 st = sX + tY + [X, Y ]+O(t2)+O(s2). 2 Lemma 14 As t 0 we have → t2 exp tX exp tY =exp(tX + tY + [X, Y ]+O(t3)) 2 and as t 0,s 0 we have → → ts exp sX exp tY =exp(sX + tY + [X, Y ]+O(t2)+O(s2)). 2

This result implies a few (standard) identities

Lemma 15 Let X, Y Mn(R) then 1 ∈ 1 k 1) limk (exp( X)exp( Y )) =exp(X + Y ). →∞ k k 1 1 1 1 k2 2)limk (exp( X)exp( Y )exp( X)exp( Y )) =exp([X, Y ]). →∞ k k − k − k

20 Proof. We have (using the first asymptotic result in the previous lemma 1 1 1 1 1 1 (exp( X)exp( Y ))k =exp( X + Y + O( ))k =exp(X + Y + kO( )). k k k k k2 k2 This implies the first limit formula. As for the second we use the same asymp- totic information and get 1 1 1 1 1 (exp( X)exp( Y )) = exp( (X + Y )+ [X, Y ]+O( )) k k k 2k2 k3 hence 1 1 1 1 (exp( X)exp( Y ))(exp( X)exp( Y )) k k −k −k 1 1 1 1 1 1 =exp((X + Y )+ [X, Y ]+O( )) exp( (X + Y )+ [X, Y ]+O( )) k 2k2 k3 −k 2k2 k3 1 1 =exp( [X, Y ]+O( )). k2 k3 We therefore have

1 1 1 1 2 1 1 2 (exp( X)exp( Y ))(exp( X)exp( Y ))k =(exp( [X, Y ]+O( )))k . k k −k −k k2 k3 This implies the second limit formula. This result implies

Proposition 16 If G isaclosedsubgroupofGL(n, R) then Lie(G) is a Lie subalgebra of Mn(R).

Proof. If X Lie(G) then tX Lie(G) for all t R.IfX, Y Lie(G), t R then ∈ ∈ ∈ ∈ ∈ t t exp(t(X + Y )) = lim (exp( X)exp( Y ))k k k k →∞ is in G since G is a closed subgroup. Similarly

1 1 1 1 2 exp(t[X, Y ]) = lim (exp( X)exp( Y ))(exp( X)exp( Y ))k k k k k k →∞ − − 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 end of subsection 1.3.1 and use the notation therein. With these identifications the Lie algebras in section 1.2 line up perfectly with the groups in section 1.1 in the sense that one replaces several of the capital letters in the group name with fraktur 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 that 1 ∈ exp(tX)J = J exp(tX).ThissaysthatJ − exp(tX)J =exp(tX).Since 1 1 A− exp(X)A =exp(A− XA) we see that X Lie(GL(n, C)) if and only ∈

21 1 if exp(tJ − XJ)=exp(tX) for all t R. But the uniqueness in Theorem 1 ∈ 11 now implies that J − XJ = X so X Mn(C)=gl(n, C).Asabove ∈ Lie(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 next subsection. 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.Nowexp(tµ(X)) = 1+tµ(X)+O(t2)). We will now give an expansion∈ of the other side of the equation (here I =[δij]) det(exp(tX)) = det(I + tX + O(t2)) = 2 det(I + tX)+O(t )= sgn(σ)(δσ1,1 + txσ1,1)(δσ2,2 + txσ2,2) (δσn,n + txσn,n) σ Sn ··· ∈ P We note that if σi = i for some i theremustbeaj = i with σj j thus if σ is not the identity6 then the corresponding6 term in± the sum is of order at least t2. Thus the only terms contributing to order 0 and 1 come from the identity element. That term is (1 + tx11) (1 + txnn)=1+t(x11 + 2 ··· ... + xnn)+O(t ). Weconcludethatµ(X)=trX. We therefore see that Lie(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 isaclosedsubgroupofG ⊂ ⊂ and G isaclosedsubgroupofGL(n, R) then H isaclosedsubgroupofGL(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).IfX Lie(H) ∈ then exp(tX) H G for all t R.ThusX Lie(G). Thisprovesthelemma. ∈ ⊂ ∈ ∈ 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 1.2.9 ∈ ∈ complete 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 . { ∈ | } t2 2 t2 2 We note that exp(tX)∗ =(I +tX + 2 X +...)∗ = I +tX∗ + 2 (X∗) +....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. k Hence Lie(U(p, q)) is contained in u(p, q).IfX u(p, q) then (X∗) Ip,q = k k ∈ ( 1) Ip,qX thus − exp(tX∗)Ip.q = Ip,q exp( tX). −

22 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 , their differentials and a theorem of Von Neumann.

Let G GL(n, R) and let H GL(m, R) be closed subgroups. Let φ : H G be a continuous⊂ group homomorphism.⊂ If X Lie(H) then t φ(exp→tX) ∈ 7→ defines a continuous homomorphism of R into GL(n, R). Hence Theorem 11 implies 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 ∈ t t φ(exp(t(X + Y )) = φ(lim(exp( X)exp( Y ))k) k k k →∞ t t = lim (φ(exp( X))φ(exp( Y )))k k k k →∞ t t = lim (exp( µ(X)))φ(exp( µ(Y ))))k k k k →∞ =exp(tµ(X + Y )).

Hence the uniqueness in Lemma 11 implies that µ(X + Y )=µ(X)+µ(Y ).Itis also obvious that µ(tX)=tµ(X). Using the second limit formula in Lemma 13 inexactlythesamewayaswedidthefirst 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 linear groups. By the end of this subsection we will have made a link to the material in appendix D on Lie groups. We will now record what we have proved this point in this subsection.

Proposition 18 Let G GL(n, R) and H GL(m, R) be closed subgroups and let ϕ : H G be⊂ a continuous homomorphism.⊂ Then there exists a unique 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 algebra of a closed subgroup, G,ofGL(n, R) the group structure. We first note that since the map t exp tX from R to G the Lie algebra of G isthesameasthe Lie algebra of its7−→ identity component. It is therefore reasonable to confine our attention to connected groups in this discussion. The key result in this direction is due to Von Neumann.

23 Theorem 19 Let G beaclosedsubgroupofGL(n, R) then there exists an open neighborhood, V ,of0 in Lie(G) so that exp(V ) is open in G and exp : V exp(V ) is a homeomorphism. →

t Proof. Let W = X Mn(R) trX Lie(G)= 0 .ThenMn(R)=Lie(G) W { ∈ | { }} and orthogonal direct sum. If we take orthonormal basis of Mn(R) that starts L with a basis of Lie(G) then continues with W then we can think of Mn(R) 2 as Rn and of Lie(G) as the subspace consisting of those vectors with the last w =dimW entries 0 and W as those with the first d =dimLie(G) entries 0.We consider 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 .Wenotethatϕ(tX)= 2 ∈ 2 ∈ 2 (I + tX1 + O(t ))(I + tX2 + O(t )) = I + t(X1 + X2)+O(t )).Wethere- 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 that

ϕ : Bs1 (0) GL(n, R) has an open image U1 with the map ϕ : Bs1 (0) U1 a 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 element that outside of exp(Lie(G)).Thusthereexistsforeach≥ ∩ n 1 an element in ≥ s1 Zn B 1 (0) such that exp(Zn) G but Zn / Lie(G).WewriteZn = Xn + Yn ∈ n ∈ ∈ with Xn Lie(G) and Yn W.Then ∈ ∈

ϕ(Zn)=exp(Xn)exp(Yn).

1 Since exp(Xn) G we see that exp(Yn) G.Wealsoobservethat Yn n . ∈ 1 ∈ k k ≤ Let εn = Yn then εn s1. There exists a positive integer mn such that k k ≤ n ≤ s1 mnεn < 2s1. Hence ≤ s1 mnYn < 2s1. ≤ k k m Now exp(mnYn)=exp(Yn) n G. Since the mnYn is bounded we can replace it with a subsequence∈ that converges. We may therefore assume that there exists Y W with limn mnYn = Y .SinceG is closed exp(Y ) G. We will now use (variant∈ of) the classic→∞ argument of Von Neumann to show∈ that if 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.

Hence an exp(tmnYn)=exp(Yn) exp(bnYn).

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

an exp(tY ) = lim exp(tmnYn) = lim exp(Yn) G. n n →∞ →∞ ∈ this completes the proof that exp(tY ) G for all t. Wehavethuscomeupon the contradiction Y Lie(G) W = 0∈ since Y =0. Wehavethereforeshown ∈ ∩ { } 6 that there exists an ε>0 such that ϕ(Bε(0)) G exp(Lie(G)). ∩ ⊂

24 This result implies that a closed subgroup of GL(n, R) hasanopenneighbor- 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 structure. We also note that Proposition 16 be reformulated as

Proposition 20 Let G GL(n, R) and H GL(m, R) be closed subgroups ⊂ ⊂ and 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 have now arisen. Let xij be the standard coordinates of Mn(R),thatiswewriteX Mn(R) ∈ as X =[xii]. In other words we take the standard basis Eij (the matrix with exactly one non-zero entry that is a 1 in the i, j position. Then the xij are the coordinates 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 ∂ d d f(I)= f(I + tEij) t=0 = f(exp(tEij)) t=0. ∂xij dt | dt | Using the chain rule we see that d f(X(I + tEij)) t=0 = Dijf(X) dt |

∂ with Dij = xki . We therefore see that if A Mn( ) then ∂xkj R k ∈ P d f(X(I + tA)) t=0 = aijDijf(X). dt | ij P Our observations about the expansion of exp(tA) in powers of t imply that d f(X exp(tA)) t=0 = aijDijf(X). dt | ij P ∂ We will write XA for aijDij.Then(XA)I = aij TI (Mn( )).We ∂xij R ij ij ∈ note that a direct calculationP implies that [XA,XPB]=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 into → GL(n, R). Weassertthat(theLie(G) in the statement is in the sense of this section)

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 ∈ d vAf = f(exp(tA)) t=0 dt |

25 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) .Sincedim Lie(A)=dimTI (G) we see that the two spaces⊃ { are the| same.∈ } We also note that if A Lie(G) then we can define ∈ G d XA f(g)= f(g exp(tA)) t=0. dt | Then by the very definition and the observation that the map R G G, t, g G × → 7→ g exp tA is smooth we see that XA is a left invariant vector field. We note Lemma 22 We have if A, B Lie(G) then [XG,XG]=XG . ∈ A B [A,B] Proof. If A, B Lie(G) then ∈ G 2 f(exp(sA)exp(tB)) = f(exp sA)+tXB f(exp(sA)) + O(t ) 2 G G G 2 = f(I)+sXAf(I)+O(s )+tXB f(I)+stXA XB f(I)+O(t ). We therefore see that since f(exp(sA)exp(tB)) f(exp(tB)exp(sA)) = st(XGXGf(I) XGXGf(I))+O(s2)+O(t2) − A B − B A G G ∂2 we have [XA ,XB ]f(I)= ∂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 ∈ GL(n,R) [XA,XB]=X[A,B] and XA = XA) ∂2 (f(exp(sA)exp(tB)) f(exp(tB)exp(sA))) s=0,t=0 = X[A,B]f(I). ∂s∂t − | G G G This implies that (diG)I ([XA ,XB ]I )=(X[A,B])I =(diG)I (X[A,B])I . Hence, G G G G G G since [XA ,XB ] is completely determined by [XA ,XB ]I we see that [XA ,XB ]= G X[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 ϕ → is of class C∞. We will now calculate dϕ : TI (H) TI (G) we have (using I → the notation in the proof of the preceding lemma) the map Lie(H) TI (G), → A vA.Nowiff C∞(G) then 7→ ∈ d dϕI (vA)f = f(ϕ(exp(tA))) t=0 . dt |

But ϕ(exp(tA)) = exp(tµ(A)) (as in Proposition 16). Thus dϕI (vA)=vµ(A). H H From this we see that if we define the vector field XA on H by XA f(h)= d H G dt f(h exp(tA)) t=0 then we find that dϕh(XA )h =(Xµ(A))ϕ(h). We therefore| see that the abstract definition of Lie algebra as left invariant vector fields and the concrete notion here of a space of matrices yield the same G Lie algebras under the correspondence A XA . Furthermore the definition of differential in Proposition 16 yields the same7→ differential as the abstract one under this correspondence.

26 1.3.5 Covering and groups. 1.3.6 Exercises. 1. Complete the proof that the Lie algebras of the groups described in section 1.1 “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 to prove 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 to prove that the image of ϕ is open and closed in the corresponding orthogonal group. 4. In the notation of exercises 8 and 9 of 1.1.10 prove that the appropriate differentials are Lie algebra .

2 Addendum for Appendix D.

In this addendum we add the following result which is for the most part proved in 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 1 open neighborhood U of e (the identity of G) such that if u U then u− U m ∈ ∈ and a surjective homeomorphism Φ : U Br(0) R for some r>0,ans 1 → ⊂ with 0

We also observe that this combined with the following discussion gives a simple proof of the fact that a connected of a connected Lie group is a Lie group. If G is a connected and locally arcwise connected topological group we will 1 now show that if we have a covering space π : H G and we choose eo π− (e) → ∈ then H has a structure of a toplogical group with identity eo such that π is a group homomorphism. We will now begin that task. Let Lg : G G be given (as usual) by Lg(x)=gx.Thenforeachh H → ∈ there exists a unique homeomorphism L˜h : H H such that L˜h(eo)=h ˜ → ˜ and π(Lh(x)) = Lπ(h)(π(x)). We assert that if we set m(u, v)=Lu(v) then with this multiplication H is a group. The identity map has the same property ˜ assumed for Leo thus m(eo,u)=u and by definition m(u, eo)=u thus eo is and identity element for the multiplication m.Ifu H then there exists a ∈ unique v H such that L˜u(v)=eo Thus if we show that the associative rule ∈ is satsified we will see that H with multiplication m and identity eo is a group.

27 ˜ We note Lx Ly = Lxy and π(m(x, y)) = xy and since m(x, y)=Lm(x,y)(eo)= ˜ ˜ ◦ ˜ ˜ ˜ Lx Ly(eo) we have Lm(x,y)=Lx Ly. This is the content of the associative rule.◦ ◦ 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. µ˜ : H H H of µ ∈ × → such that µ˜(eo,eo)=eo.Sincem is another such lift we see that m =˜µ.

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

28