Bourbaki

Exercise 6, § 6

Tag 01FR, permanent. Cite it as [Bourbaki, Tag 01FR].

Lie Groups and Lie Algebras, IX, 412

Put bi(G)=b_i(G) = dimRHi(G)_{\mathbf{R}}H^i(G) for i≥0i\geq 0 and, if X denotes an indeterminate, PG(X)=∑i≥0bi(G)XiP_G(X) =\sum_{i\geq 0}b_i(G)X^i.

a) Prove that bi(G)=∫Gb_i(G) =\int_G Tr ∧∧i\wedge \wedge^i(Ad gg)dgdg and PG(X)=∫GP_G(X) =\int_G det(1 +X+ X.Ad gg)dgdg (cf. Appendix II, Lemma 1).

b) Deduce the equalities ∑ibi(G)=2r\sum_ib_i(G) = 2^r, and ∑i(−1)ibi(G)=0\sum_i(-1)^ib_i(G) = 0 if dim G>0G>0 (use H. Weyl’s formula, or Exerc. 8 of §2).

c) Take G=U(n,C)G =\mathbf{U}(n,\mathbf{C}). Show that PG(X)P_G(X) is the coefficient of (X1...Xn)2n−2(X_1. . .X_n)^{2n-2} in the polynomial 1n!(1+X)n∏1≤i,j≤n\frac{1}{n!}(1 + X)^n\prod_{_{1\leq i,j\leq n}}(XXi+Xj_i+ X_j)(Xi−Xj)(X_i-X_j) (with coefficients in Z[X]\mathbf{Z}[X]).i≠j.i_{\not=j}

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