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Notes on Integration on Lie Groups Michael Taylor Contents 1 Notes on Integration on Lie Groups Michael Taylor Contents 1. Construction of Haar measure 2. Integrating a representation 3. Weyl orthogonality 4. The adjoint representation 5. Haar measure in exponential coordinates 6. The Weyl integration formula 7. Ensembles of Hermitian matrices 8. The discriminant of a matrix 9. The integral of j Tr M jj2 10. The integral of j Tr Mj2j Abstract. This is a very informal set of notes on integration on Lie groups and connections with basic representation theory. We give some constructions of the Lie group integral, show how some integrals can be computed by using simple symmetry considerations, and present some cases where more earnest e®orts are required to compute integrals. 1 2 1. Construction of Haar measure For our ¯rst construction, assume G is a compact subgroup of the unitary group U(n), sitting in Mn(C), the space of complex n £ n matrices. The space Mn(C) has a Hermitian inner product, (1.1) (A; B) = Tr AB¤ = Tr B¤A; giving a real inner product hA; Bi = Re (A; B). This induces a Riemannian metric on G. Let us de¯ne, for g 2 G, (1.2) Lg;Rg : Mn(C) ¡! Mn(C);LgX = gX; RgX = Xg: Clearly each such map is a linear isometry on Mn(C), and we have isometries Lg and Rg on G. A Riemannian metric tensor on a smooth manifold induces a volume element on M, as follows. In local coordinates (x1; : : : ; xN ) on U ½ M, say the metric tensor has components hjk(x). Then, on U, q (1.3) dV (x) = det(hjk) dx1 ¢ ¢ ¢ dxN : In such a way we get a volume element on a compact group G ½ U(n), and since Lg and Rg are isometries, they also preserve the volume element. We normalize this volume element to de¯ne normaluized Haar measure on G: Z Z 1 (1.4) f(g) dg = f dV: V (G) G G We have left invariance Z Z (1.5) f(hg) dg = f(g) dg G G and right invariance Z Z (1.6) f(gh) dg = f(g) dg; G G for all h 2 G, in such a situation. We now give a second construction of Haar measure, valid in much greater generality. Let G be any Lie group, say of dimension N. Pick any nonzero !e 2 3 N ¤ ¤ Te G, where e denotes the identity element of G. Then there is a unique N form !` on G such that ¤ (1.7) !`(e) = !e;Lg!` = !`; 8 g 2 G; and a unique N-form !r on G such that ¤ (1.8) !r(e) = !e;Rg!r = !r; 8 g 2 G: ¤ ¤ In fact !e = Lg!`(g) and !e = Rg!r(g). If we use !` (or !r) to de¯ne an orientation on G, then we have volume elements, which we denote dV` and dVr. Note that, for all h 2 G, Z Z Z Z (1.9) f(hg) dV`(g) = f(g) dV`(g); f(gh) dVr(g) = f(g) dVr(g): G G G G N ¤ Since ¤ Te G is 1-dimensional, it is clear that both dV` and dVr are unique, up to a constant positive multiple. ¤ ¤ ¤ Note that Lg and Rh commute for each g; h 2 G. Hence Rg!` is left-invariant ¤ and Lg!r is right-invariant for each g; h 2 G. The uniqueness mentioned above implies ¤ ¤ (1.10) Rh!` = ®(h)!`;Lg!r = ¯(g)!r; for all g; h 2 G, with ®; ¯ : G ! (0; 1). It is clear that ® and ¯ are homomorphisms: (1.11) ®(gh) = ®(g)®(h); ¯(gh) = ¯(g)¯(h): We say G is unimodular if ® ´ 1 (equivalently, ¯ ´ 1). In such a case, the left invariant Haar measure is also right invariant; we say Haar measure is bi-invariant on G, and that G is unimodular. The Haar measure constructed on a compact group G ½ U(n) at the begining of this section is bi-invariant. From another perspective, note that the image of G under ® is a subgroup of (0; 1); if G is compact this must be a compact subgroup, hence f1g. Lots of noncompact Lie groups are also unimodular, but some are not unimod- ular. 4 2. Integrating a representation Let G be a compact Lie group, ¼ a unitary representation of G on V , a ¯nite- dimensional vector space with an inner product. We set Z P v = ¼(g)v dg: G Claim. P is the orthogonal projection of V on the space where ¼ acts trivially. The proof consists of four easy pieces: (2.1) ¼(g)P v = P v; 8 g 2 G; Z (2.2) P ¤ = ¼(g¡1) dg = P; ZZ ZZ (2.3) P 2 = ¼(g)¼(h) dg dh = ¼(gh) dg dh = P; (2.4) ¼(g)v = v 8 g =) P v = v: 5 3. Weyl orthogonality Let G be a compact Lie group. Assume ¼ is an irreducible unitary representation of G on V and ¸ an irreducible unitary representation of G on W . De¯ne P acting on Hom(V; W ) as follows. If A : V ! W , set Z (3.1) P (A) = ¸(g)A¼(g)¡1 dg: G It is readily veri¯ed that (3.2) ¸(g)P (A)¼(g)¡1 = P (A); 8 g 2 G: In other words, P (A) intertwines ¼ and ¸. Now Schur's lemma gives the following: ¼ not ¼ ¸ =) P (A) = 0; 8A; (3.3) ¼ = ¸ =) P (A) = c¼(A)I; where c¼(A) is scalar and I the identity operator on V = W . In the latter case, taking the trace yields d¼ c¼(A) = Tr A (where d¼ = dim V ), hence c¼(A) = ¡1 d¼ Tr A, so Z ¤ ¡1 (3.4) ¼(g)A¼(g) dg = d¼ (Tr A)I: G If matrix entries are denoted ¼(g)jk;Ajk, etc., we have X Z ¡1 ¼(g)jkAk`¼(g)m` dg = d¼ ±jm Tr A k;` G (3.5) X ¡1 = d¼ ±jm ±k`Ak`; k;` hence Z ¡1 (3.6) ¼(g)jk¼(g)m` dg = d¼ ±jm ±k`: G These are Weyl orthogonality relations. They are complemented by Z (3.7) ¼(g)jk¸(g)m` dg = 0; ¼ not ¼ ¸; G which follows from the ¯rst part of (3.3). 6 4. The adjoint representation The adjoint representation of a Lie group G is a representation of G on its Lie algebra g. We recall that g consists of left invariant vector ¯elds on G. Such X 2 g is uniquely determined by X(e) 2 TeG, so g ¼ TeG. A vector ¯eld X on G is left t invariant if and only if the flow FX it generates commutes with Lg for all g 2 G, t t that is, g(FX h) = FX (gh) for all g; h 2 G. If we set t (4.1) γX (t) = FX e; s t t s we obtain γX (t + s) = FX (FX e) ¢ e = (FX e)(FX e), and hence γX (s + t) = 0 γX (s)γX (t), for s; t 2 R. Clearly γX (0) = X(e). The exponential map (4.2) Exp : g ¡! G is de¯ned by (4.3) Exp(X) = γX (1): If G is a Lie subgroup of Gl(n; C), then TeG is a subspace of Mn(C), and (4.2) coincides with the matrix exponential eX . To de¯ne the adjoint representation of G on g, consider ¡1 (4.4) Kg : G ¡! G; Kg(h) = ghg : Then Kg(e) = e, and we set (4.5) Ad(g) = DKg(e): TeG ¡! TeG; identifying TeG ¼ g. Since Kgh = Kg ± Kh, the chain rule implies (4.6) Ad(gh) = Ad(g) Ad(h): Note that γ(t) = g Exp(tX) g¡1 is a 1-parameter subgroup of G satisfying γ0(0) = Ad(g)X. Hence (4.7) Exp(t Ad(g)X) = g Exp(tX) g¡1: In particular, ¡ ¢ (4.8) Exp Ad(Exp sY )tX = Exp(sY ) Exp(tX) Exp(¡sY ): s t ¡s The right side of (4.8) is equal to FY ± FX ± FY (e). 7 In general a representation ¼ of G on V yields a representation d¼ of g on V by (4.9) d¼(X) = D¼(e)X; D¼(e): TeG ! V: One shows that, for X; Y 2 g, (4.10) [d¼(X); d¼(Y )] = d¼([X; Y ]); where [X; Y ] denotes the Lie bracket of vector ¯elds. (See [T2], Appendix B, for more details on this, and on the material below.) From (4.8) it can be deduced that D Ad(X) = ad X, given by (4.11) ad X(Y ) = [X; Y ]: We mention another useful identity: (4.12) ead X = Ad(Exp X); a special case of the more general identity (4.13) et d¼(X) = ¼(Exp tX); valid when ¼ is a representation of G on V . Finally we tie in with the question of whether G is unimodular. A comparison of (1.10) and (4.4) shows that (4.14) ®(g) = det Ad(g): In other words, G is unimodular if and only if det Ad(g) ´ 1. 8 5. Haar measure in exponential coordinates Let G be a Lie group, with Lie algebra g. We assume G ½ Gl(Cn), so g ½ End(Cn), and Exp : g ! G is given by Exp(X) = eX . We have 1 ¡ e¡z (5.1) D Exp(X)Y = eX ¥(ad X)Y; ¥(z) = : z X Here D Exp(X): g ! TgG; g = Exp(X), and also left multiplication by e n of ¥(ad X)Y 2 g yields an element of TgG ½ End(C ). Using the left-invariant volume form on G, we have (5.2) det D Exp(X) = det ¥(ad X): Thus Haar measure pulled back to g is given by H(X) dX, with dX Lebesgue measure on g and H(X) = j det ¥(ad X)j.
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