Introduction To Lie Groups Michael Taylor

21. First look at Lie groups1.1. Definition and first examplesA Lie group G is a group that is also a smooth manifold, such that the groupoperations G G G and G G given by (g, h) 7 gh and g 7 g 1 aresmooth maps.We consider some examples, starting with(1.1.1)Gl(n, R) {A M(n, R) : A 1 exists},where M(n, R) consists of n n real matrices.Proposition 1.1.1. The set Gl(n, R) is open in M(n, R).Proof. One way to see this is to note that Gl(n, R) {A M(n, R) :det A ̸ 0}, and det : M(n, R) R is continuous. Here is another.Given A Gl(n, R), we have A B A(I A 1 B), which is invertibleprovided I A 1 B is invertible. Now if C M(n, R) we have the operatornorm(1.1.2) C sup { Cv : v Rn , v 1},and we see that C k C k , and hence(1.1.3) C 1 (I C) 1 ( C)k ,k 0with absolute convergence, so A 1 B 1 implies A B is invertible. The group Gl(n, R) inherits a manifold structure from the vector spaceM(n, R). Since (A, B) 7 AB is bilinear, it is clearly smooth. Furthermore,κ(A) A 1 gives a smooth map on Gl(n, R), withDκ(A)X A 1 XA 1 .(1.1.4)In fact, for X small,(1.1.5)(A X) 1 (A(I A 1 X)) 1 (I A 1 X) 1 A 1 A 1 ( 1)k (A 1 X)k A 1 ,k 1which yields (1.1.4).Similar considerations apply to(1.1.6)Gl(n, C) {A M(n, C) : A 1 exists},where M(n, C) consists of n n complex matrices.Many other basic examples of Lie groups arise as subgroups of Gl(n, R)and Gl(n, C). For example, we have(1.1.7) Sl(n, F) {A M(n, F) : det A 1} Gl(n, F),F R or C.

1.1. Definition and first examples3Other examples areO(n) {A M(n, R) : A A I},(1.1.8)U(n) {A M(n, C) : A A I},whereA (ajk ) A (akj ).(1.1.9)Also we have(1.1.10)SO(n) {A O(n) : det A 1},SU(n) {A U(n) : det A 1}.The proof that (1.1.7)–(1.1.10) define Lie groups follows from the fact thesegroups are all smooth submanifolds of M(n, F). This fact in turn can bededuced from the following result, which is a consequence of the inversefunction theorem.Theorem 1.1.2. (Submersion mapping theorem.) Let V and W befinite-dimensional vector spaces, and F : V W a smooth map. Fix p W ,and considerS {x V : F (x) p}.(1.1.11)Assume that, for each x S, DF (x) : V W is surjective. Then S is asmooth submanifold of V . Furthermore, for each x S,(1.1.12)Tx S ker DF (x).For a proof of this result, see §A.1. The proof takes the following approach. Given q S, defineGq : V W ker DF (q),Gq (x) (F (x), Pq (x q)),where Pq : V ker DF (q) is a projection. Then the inverse functiontheorem can be applied to Gq .We show how Theorem 1.1.2 can be applied to show that the groupsdescribed in (1.1.7)–(1.1.10) are smooth submanifolds of M(n, F). We startwith (1.1.7). Here we take(1.1.13)V M(n, F), W F, F : V W, F (A) det A.Now given A invertible,(1.1.14)F (A B) det(A B) (det A) det(I A 1 B),and inspection shows that, for X M(n, F),(1.1.15)det(I X) 1 Tr X O( X 2 ),

41. First look at Lie groupssoDF (A)B (det A) Tr(A 1 B).(1.1.16)Now, given A Sl(n, F), or even A Gl(n, F), it is readily verified thatτA : M(n, F) F,(1.1.17)τA (B) Tr(A 1 B),is nonzero, hence surjective, and Theorem 1.1.2 applies.We turn to O(n), defined in (1.1.8). In this case,V M(n, R),(1.1.18)W {X M(n, R) : X X },F : V W,F (A) A A.Now, given A V ,F (A B) A A A B B A O( B 2 ),(1.1.19)soDF (A)B A B B A A B (A B) .(1.1.20)We claim that(1.1.21)A O(n) DF (A) : M(n, R) W is surjective.Indeed, given X W , i.e., X M(n, R), X X , we have1(1.1.22)B AX DF (A)B X.2Again, Theorem 1.1.2 applies.Similar arguments apply to U(n) in (1.1.8) and to the groups in (1.1.10).For SU(n) we takeV M (n, C),W {X M (n, C) : X X } R,F : V W,F (A) (A A, ℑ det A).Note that A U(n) implies det A 1, so ℑ det A 0 det A 1.As a further comment on O(n), we note that, given A M(n, R), definingA : Rn Rn ,(1.1.23)A O(n) (Au, Av) (u, v), u, v Rn ,where (u, v) is the Euclidean inner product on Rn : (1.1.24)(u, v) uj vj ,jwhere u (u1 , . . . , un ), v (v1 , . . . , vn ). Similarly, given A M(n, C),defining A : Cn Cn ,(1.1.25)A U(n) (Au, Av) (u, v), u, v Cn ,

1.1. Definition and first examples5where (u, v) denotes the Hermitian inner product on Cn : (1.1.26)(u, v) uj v j .jNote that⟨u, v⟩ ℜ(u, v)(1.1.27)defines the Euclidean inner product on Cn R2n , and we have(1.1.28)U(n) , O(2n).Analogues of O(n) and U(n), with R and C replaced by the ring H of quaternions, will be discussed in §1.2.Having defined several matrix groups, we now define a family of Liegroups that are not a priori subgroups of Gl(N, F). Namely we define theEuclidean group E(n) as a group of isometries of Rn . As a set, E(n) O(n) Rn , and the action of (A, v) on Rn is given by(1.1.29)(A, v)x Ax v,A O(n), v, x Rn .The group law is seen to be(1.1.30)(A, v) · (B, w) (AB, Aw v).Actually, E(n) is isomorphic to a matrix group, via()A v(1.1.31)(A, v) 7 ,0 1as one verifies that)() ()(A vB wAB Aw v .(1.1.32)0 10 101There are Lie groups that are not isomorphic to matrix groups, but itis a fact (not established here) that every connected Lie group is locallyisomorphic to a matrix group. This is a consequence of a result known asAdo’s theorem.

61. First look at Lie groups1.2. Quaternions and the groups Sp(n)The space H of quaternions is a four-dimensional real vector space, identifiedwith R4 , with basis elements 1, i, j, k, the element 1 identified with the realnumber 1. Elements of H are represented as follows:(1.2.1)ξ a bi cj dk,with a, b, c, d R. We call a the real part of ξ (a Re ξ) and bi cj dkthe vector part. We also have a multiplication on H, an R-bilinear mapH H H coinciding with the standard product on the real part, andotherwise governed by the rules(1.2.2)ij k ji,jk i kj,ki j ik,andi2 j 2 k 2 1.(1.2.3)Otherwise stated, if we write(1.2.4)ξ a u,a R,u R3 ,and similarly write η b v, b R, v R3 , the product is given by(1.2.5)ξη (a u)(b v) (ab u · v) av bu u v.Here u · v is the dot product in R3 and u v is the cross product of vectorsin R3 . The quantity ab u · v is the real part of ξη and av bu u v isthe vector part.We also have a conjugation operation on H:(1.2.6)ξ a bi cj dk a u.A calculation gives(1.2.7)ξη (ab u · v) av bu u v.In particular,(1.2.8)Re(ξη) Re(ηξ) (ξ, η),the right side denoting the Euclidean inner product on R4 . Setting η ξ in(1.2.7) gives(1.2.9)ξξ ξ 2 ,the Euclidean square-norm of ξ. In particular, whenever ξ H is nonzero,it has a multiplicative inverse:(1.2.10)ξ 1 ξ 2 ξ.A routine calculation gives(1.2.11)ξη η ξ.

1.2. Quaternions and the groups Sp(n)7Hence(1.2.12) ξη 2 (ξη)(ξη) ξηηξ η 2 ξξ ξ 2 η 2 ,or ξη ξ η .(1.2.13)Note that C {a bi : a, b R} sits in H as a commutative subring,for which the properties (1.2.9) and (1.2.13) are familiar.We consider the set of unit quaternions:Sp(1) {ξ H : ξ 1}.(1.2.14)Using (1.2.10) and (1.2.13) it is clear that Sp(1) is a group under multiplication. It sits in R4 as the unit sphere S 3 . We compare Sp(1) with the groupSU(2), consisting of 2 2 complex matrices of the form()ξ η(1.2.15)U , ξ, η C, ξ 2 η 2 1.xη ξThe group SU(2) is also diffeomorphic to S 3 . Furthermore we have:Proposition 1.2.1. The groups SU(2) and Sp(1) are isomorphic under thecorrespondenceU 7 ξ jη,(1.2.16)for U as in (3.15).Proof. The correspondence (1.2.16) is clearly bijective. To see that it is ahomomorphism of groups, we calculate:)()( ′) ( ′′ξ η ′ξ ηξξ ηη ′ ξη ′ ηξ(1.2.17) ′′ ,η ξη′ ξηξ ′ ξη ′ ηη ′ ξξgiven ξ, η C. Noting that, for a, b R, j(a bi) (a bi)j, we have(ξ jη)(ξ ′ jη ′ ) ξξ ′ ξjη ′ jηξ ′ jηjη ′(1.2.18) ξξ ′ ηη ′ j(ηξ ′ ξη ′ ).Comparison of (1.2.17) and (1.2.18) verifies that (1.2.16) yields a homomorphism of groups. To proceed, we consider n n matrices of quaternions:(1.2.19)A (ajk ) M(n, H),ajk H.

81. First look at Lie groupsIf Hn denotes the space of column vectors of length n, whose entries arequaternions, then A M(n, H) acts on Hn by the usual formula. If ξ (ξj ), ξj H, we have (1.2.20)(Aξ)j ajk ξk .kNote thatA : Hn Hn(1.2.21)is R-linear, and commutes with the right action of H on Hn , defined by(1.2.22)(ξb)j ξj b,ξ Hn , b H.Composition of such matrix operations on Hn is given by the usual matrixproduct. If B (bjk ), then (1.2.23)(AB)jk ajℓ bℓk .ℓWe define a conjugation on M(n, H); with A given by (1.2.19),A (akj ).(1.2.24)A calculation using (1.2.11) gives(AB) B A .(1.2.25)We are ready to define the groups Sp(n) for n 1:(1.2.26)Sp(n) {A M(n, H) : A A I}.Note that A is a left inverse of the R-linear map A : Hn Hn if andonly if it is a right inverse (by real linear algebra). In other words, givenA M(n, H),A A I AA I.(1.2.27)In particular,(1.2.28)A Sp(n) A Sp(n) A 1 Sp(n).Also, given A, B Sp(n),(1.2.29)(AB) AB B A AB B B I.Hence Sp(n), defined by (1.2.26), is a group. We claim that (1.2.26) defines asmooth, compact submanifold of M(n, H), so Sp(n) is a compact Lie group.We omit the check of smoothness, which goes along the lines of (1.1.18)–(1.1.22), but we will establish compactness, using a construction of separateinterest.

1.2. Quaternions and the groups Sp(n)9We define a quaternionic inner product on Hn as follows. If ξ (ξj ), η (ηj ) Hn , set (1.2.30)⟨ξ, η⟩ η j ξj .jFrom (1.2.8) we have(1.2.31)Re⟨ξ, η⟩ (ξ, η),where the right side denotes the Euclidean inner product on Hn R4n .Now, if A M(n, H), A (ajk ), then ⟨Aξ, η⟩ η j ajk ξkj,k(1.2.32) ajk ηj ξkj,k ⟨ξ, A η⟩.Hence(1.2.33)⟨Aξ, Aη⟩ ⟨ξ, A Aη⟩.In particular, given A M(n, H), we have A Sp(n) if and only if A :Hn Hn preserves the quaternionic inner product (1.2.30). Given (1.2.31),we have(1.2.34)Sp(n) , O(4n).From here it is easy to show that Sp(n) is closed in O(4n), and hence compact.Remark. Further results on quaternions are given in §10.1 and in §D.2,where there is also a refinement of (1.2.34).

101. First look at Lie groups1.3. The matrix exponential and other functions of matricesIf A M(n, C), we defineetA (1.3.1) k tk 0k!Ak .We also denote this by Exp(tA). Making use of the operator norm (1.2) andnoting that Ak A k , we see that (1.3.1) is absolutely convergent forall A and all t. The power series (1.3.1) can be differentiated term by term,and we obtaind tA(1.3.2)e AetA etA A.dtUsing this we can establish the identitye(s t)A esA etA .(1.3.3)To get this, we can first computed [ (s t)A tA ](1.3.4)ee e(s t)A Ae tA e(s t)A Ae tA 0,dtusing the product rule; hence e(s t)A e tA is independent of t. Evaluating att 0 givese(s t)A e tA esA .(1.3.5)Setting s 0 givesetA e tA I.(1.3.6)Thus e tA is the multiplicative inverse of etA . Using this, we can multiplyboth sides of (1.3.5) on the right by etA and obtain (1.3.3).A similar argument, which we leave to the reader, gives(1.3.7)AB BA esA tB esA etB ,though such an identity fails when A and B do not commute.We note a few easy identities:(1.3.8)etX 1 AX X 1 etA X,( ) etA etA ,given X invertible, t R. If A is diagonal, etA is obtained by exponentiatingthe diagonal entries. Also one has(1.3.9)det etA et Tr A .If A is diagonal this is checked by the remarks above; it then follows for Adiagonalizable, by (1.3.8). It can be shown that the set of diagonalizablematrices is dense in M(n, C), and then (1.3.9) holds for all A, by continuity.Alternatively, it is quite easy to show that there exists an open subset of

1.3. The matrix exponential and other functions of matrices11M(n, C) consisting of diagonalizable matrices. Since both sides of (1.3.9)are holomorphic on M(n, C), this suffices.We remark on the behavior of the exponential map on the tangent spaceat the identity to the groups described in (1.1.7)–(1.1.10). Making use ofthe criterion (1.1.12), one can calculate the following:TI Sl(n, F) {A M(n, F) : Tr A 0},(1.3.10)TI O(n) {A M(n, R) : A A} TI SO(n),TI U(n) {A M(n, C) : A A},TI SU(n) {A M(n, C) : A A, Tr A 0}.For the first two, take A I in (1.1.16) and (1.1.20), respectively, yieldingDF (I)A Tr A and DF (I)A A A , respectively. Having (1.3.10) andmaking use of (1.3.8)–(1.3.9), one readily verifies the following.Proposition 1.3.1. For each Lie group listed above,(1.3.11)Exp : TI G G.We will discuss how this result fits in a more general framework in §§3.1–3.2.We next want to calculate the derivative of the map Exp : M(n, R) Gl(n, R). Equivalently, if A, B M(n, R), we calculated A tB(1.3.12)e.t 0dtWhen A and B commute, this is easily calculated via (1.3.7). Otherwise,matters are more complicated. To calculate (1.3.12), it is useful to look at(1.3.13)U (s, t) es(A tB) ,which satisfies U (A tB)U (s, t), U (0, t) I. sThen Ut U/ t satisfies Ut (s, t) (A tB)Ut (s, t) BU (s, t), Ut (0, t) 0,(1.3.15) sand in particular (1.3.16)Ut (s, 0) AUt (s, 0) BU (s, 0), Ut (0, 0) 0. sThis is an inhomogeneous linear ODE, whose solution is se(s σ)A BU (σ, 0) dσUt (s, 0) 0 s(1.3.17) e(s σ)A BeσA dσ.(1.3.14)0

121. First look at Lie groupsWe get (1.3.12) by setting s 1:(1.3.18)d A tBedtt 0 1e(1 σ)A BeσA dσ, 0so (1.3.19)1D Exp(A)B eAe σA BeσA dσ.0The method (1.3.1) of defining the matrix exponential extends to othercases. Suppose F (z) is a holomorphic function with a power series expansion(1.3.20)F (z) ak z k .k 0If (1.3.20) converges on the disk DR {z C : z R}, and if A M(n, C), A R, then we can define(1.3.21)F (A) a k Ak ,k 0and this power series is absolutely convergent. Power series manipulationsshow that if also G(z) is holomorphic on DR , and we set H(z) F (z)G(z),then, for A R,(1.3.22)F (A)G(A) H(A).We will see more examples of (1.3.21) in subsequent sections.Here we look into one other example, namely, for tA 1, set (1.3.23)log(I tA) tA ( 1)k 1t2 2 t3 3A A ··· tk Ak .23kk 1We aim to prove thatelog(I tA) I tA.(1.3.24)To see this, note that for tA 1,(1.3.25)X(t) log(I tA) X ′ (t) A(I tA t2 A2 · · · ) A(I tA) 1 ,as follows from (1.3.23) by differentiating term by term. For such X(t),we see that X(t) and X(s) always commute, so it follows from (1.3.19) (orotherwise) that(1.3.26)d X(t)e X ′ (t)eX(t) .dtConsequently, if we set(1.3.27)V (t) (I tA) 1 elog(I tA) ,

1.3. The matrix exponential and other functions of matrices13we have V (0) I and(1.3.28)V ′ (t) A(I tA) 2 elog(I tA) A(I tA) 2 elog(I tA) 0,so (2.24) is established.It follows directly from (1.3.1) thatExp(0 B) I B O( B 2 ),and hence(1.3.29)D Exp(0)B B,i.e., D Exp(0) is the identity operator on M(n, R). (This is of course also aspecial case of (1.3.19).) It follows from the inverse function theorem thatthere are neighborhoods O of 0 M(n, R) and Ω of I Gl(n, R) such that(1.3.30)Exp : O Ω, diffeomorphically,hence there is a smooth inverse from Ω to O. The results (1.3.23)–(1.3.24)provide an explicit formula for this inverse. Putting this together withProposition 1.3.1 yields the following.Proposition 1.3.2. For each Lie group G listed in (1.3.10), there existsa neighborhood O of 0 in TI G and a neighborhood Ω of I in G such that(1.3.30) holds.

141. First look at Lie groups1.4. Integration on a Lie groupFor our first construction, assume G is a compact subgroup of the unitarygroup U(n), sitting in M(n, C), the space of complex n n matrices. Thespace M(n, C) has a Hermitian inner product,(A, B) Tr AB Tr B A,(1.4.1)giving a real inner product ⟨A, B⟩ Re (A, B). This induces a Riemannianmetric on G. Let us define, for g G,(1.4.2)Lg , Rg : M(n, C) M(n, C),Lg X gX,Rg X Xg.Clearly each such map is a linear isometry on M(n, C) (given that g U (n)),and we have isometries Lg and Rg on G.A Riemannian metric tensor on a smooth manifold induces a volumeelement on M , as follows. In local coordinates (x1 , . . . , xN ) on U M , saythe metric tensor has components hjk (x). Then, on U , (1.4.3)dV (x) det(hjk ) dx1 · · · dxN .See §A.2 for a demonstration that dV is well defined, independent of thechoice of coordinates.In such a way we get a volume element on a compact group G U(n),and since Lg and Rg are isometries, they also preserve the volume element.We normalize this volume element to define normalized Haar measure on G: 1f (g) dg V (G)(1.4.4)GWe have left invariancef dV.G (1.4.5)and right invariance f (hg) dg f (g) dgGG (1.4.6)f (gh) dg Gf (g) dg,Gfor all h G, in such a situation.We give a more general construction of Haar measure, working on anyLie group G. To start, we fix some Euclidean inner product on Te G g;call it ⟨ , ⟩g . Here e denotes the identity element of G. Defining Lg and Rgon G as in (1.4.2), we have(1.4.7)DLg 1 , DRg 1 : Tg G Te G g.

1.4. Integration on a Lie group15We define two metric tensors on G as follows. Given U, V Tg G, we defineinner products⟨U, V ⟩ℓ ⟨DLg 1 U, DLg 1 V ⟩g ,(1.4.8)⟨U, V ⟩r ⟨DRg 1 U, DRg 1 V ⟩g .A straightforward computation shows that, for each g G, Lg : G Gis an isometry for ⟨ , ⟩ℓ and Rg : G G is an isometry for ⟨ , ⟩r . Nowthe procedure (1.4.3) yields two volume elements on G, which we denotedVℓ and dVr . As noted above, isometries of Riemannian manifolds naturallypreserve the induced volume elements, so we have, for all h G,(1.4.9) f (hg) dVℓ (g) Gf (g) dVℓ (g),Gf (gh) dVr (g) Gf (g) dVr (g).GThus dVℓ is left-invariant and dVr is right-invariant. We call these Haarmeasures.We discuss the extent to which dVℓ is unique. If dVℓ′ is another leftinvariant measure, given in local coordinates by a smooth multiple of Lebesguemeasure, then dVℓ′ φ(g) dVℓ for a smooth positive function φ, and fromthe left invariance of both measures one can deduce that φ(hg) φ(g) forall g, h G, so φ must be constant. A similar remark holds for dVr .We consider the effect of a right translation on dVℓ . For convenience set (1.4.10)Iℓ (f ) f (g) dVℓ (g),Gso right translation by h yieldsIℓh (f )(1.4.11) f (gh) dVℓ (g).GIt is easy to check thatabove we have(1.4.12)Iℓhis left-invariant, so by the uniqueness describedIℓh

Chapter 1. First look at Lie groups 1 x1.1. De nition and rst examples 2 x1.2. Quaternions and the groups Sp(n) 6 x1.3. The matrix exponential and other functions of matrices 10 x1.4. Integration on a Lie group 14 Chapter 2. Lie groups and representations 19 x2.1. Basic notions of representation theory 20 x2.2. Weyl orthogonality 23 x2.3. The .