Bourbaki

Exercise 3, § 5

Lie Groups and Lie Algebras, V

Bourbaki marks this one as harder than the rest.

Let KK be a field of characteristic zero, VV a vector space of finite dimension ll over KK, and GG a subgroup of GL(V)GL(V) generated by pseudo-reflections. Put q=Card(G)q = \mathrm{Card}(G); denote by SS (resp. LL) the symmetric algebra (resp. exterior algebra) of VV. If x∈Vx \in V, denote by xx (resp. x′x') its canonical image in SS (resp. LL).

a) Let E=S⊗LE = S \otimes L be the tensor product of SS and LL. Show that there exists a unique derivation dd on EE such that dx=x′dx = x' and dx′=0dx' = 0 for all x∈Vx \in V.

b) Let SGS^G be the algebra of symmetric invariants of GG and let P1,…,PlP_1, \ldots, P_l be homogeneous elements of SGS^G such that SG=K[P1,…,Pl]S^G = K[P_1, \ldots, P_l]. For every subset I={i1,…,ir}I = \{i_1, \ldots, i_r\} of [1,l][1, l] with i1<⋯<iri_1 < \cdots < i_r, put

ωI=dPi1…dPir. \omega_I = dP_{i_1} \ldots dP_{i_r}.
Show that the ωI\omega_I are linearly independent over SS, and that they belong to the subalgebra EGE^G consisting of the elements of EE invariant under GG. Deduce that, for all ω∈EG\omega \in E^G, there exist a,cI∈SGa, c_I \in S^G, with a≠0a \neq 0, such that
aω=∑IcIωI. a \omega = \sum_I c_I \omega_I.

c) Show that every element of EGE^G contained in S⊗⋀lVS \otimes \bigwedge^l V is of the form c.dP1…dPlc.dP_1 \ldots dP_l, with c∈SGc \in S^G. (Apply Prop. 5.)

d) Show that the ωI\omega_I form a basis of the SGS^G-module EGE^G. (With the notation in b), multiply both sides of the relation aω=∑IcIωIa \omega = \sum_I c_I \omega_I by an element ωJ\omega_J, and apply c); deduce that, if II is the complement of JJ, aa divides cIc_I; hence9^9 the fact that the ωI\omega_I generate EGE^G.)

e) Let SnS_n (resp. LmL_m) be the homogeneous component of SS (resp. LL) of degree nn (resp. mm). Put

En,m=Sn⊗Lm,En,mC=EG∩En,m;an,m=dim⁡En,mC, E_{n,m} = S_n \otimes L_m, \quad E^C_{n,m} = E^G \cap E_{n,m}; \quad a_{n,m} = \dim E^C_{n,m},
a(X,Y)=∑n,m≥0an,mXnYm. a(X, Y) = \sum_{n,m \geq 0} a_{n,m} X^n Y^m.
By using d), prove the formula
a(X,Y)=∏i=1l1+Y.Xpi−11−Xpi; a(X, Y) = \prod_{i=1}^l \frac{1 + Y.X^{p_i-1}}{1 - X^{p_i}};
where pi=deg⁡(Pi)p_i = \deg(P_i).

(9) For more details, see L. SOLOMON, Invariants of finite reflection groups, Nagoya Math. Journal, v. XXII (1963), p. 57-64.

f) If g∈Gg \in G, let Trn,m(g)\mathrm{Tr}_{n,m}(g) be the trace of the automorphism of En,mE_{n,m} defined by gg; put

Tr(X,Y)(g)=∑n,m≥0Trn,m(g)XnYm. \mathrm{Tr}(X, Y)(g) = \sum_{n,m \geq 0} \mathrm{Tr}_{n,m}(g) X^n Y^m.
Show that
Tr(X,Y)(g)=det⁡(1+Yg)det⁡(1−Xg). \mathrm{Tr}(X, Y)(g) = \frac{\det(1 + Yg)}{\det(1 - Xg)}.
g) Let q=Card(G)q = \mathrm{Card}(G). Show that
1q∑g∈GTr(X,Y)(g)=a(X,Y), \frac{1}{q} \sum_{g \in G} \mathrm{Tr}(X, Y)(g) = a(X, Y),
i.e.
1q∑g∈Gdet⁡(1+Yg)det⁡(1−Xg)=∏i=1l1+Y.Xpi−11−Xpi. \frac{1}{q} \sum_{g \in G} \frac{\det(1 + Yg)}{\det(1 - Xg)} = \prod_{i=1}^l \frac{1 + Y.X^{p_i-1}}{1 - X^{p_i}}.
(Use the following result: if GG acts on a finite dimensional vector space EE, the dimension of the space of elements of EE invariant under GG is equal to 1q∑g∈GTrE(g)\frac{1}{q} \sum_{g \in G} \mathrm{Tr}_E(g).) What does this formula give for Y=0Y = 0? h) For any integer p≥0p \geq 0, let HpH_p be the set of elements g∈Gg \in G that have 1 as an eigenvalue with multiplicity pp. Let hp=Card(Hp)h_p = \mathrm{Card}(H_p). Prove the formula
∑p=0lhpTp=∏i=1l(pi−1+T). \sum_{p=0}^l h_p T^p = \prod_{i=1}^l (p_i - 1 + T).
(In the formula in g) above, replace YY by −1+T(1−X)-1 + T(1 - X) and then put X=1X = 1 in the result. If g∈Hpg \in H_p, the term Tr(X,Y)(g)\mathrm{Tr}(X, Y)(g) becomes TpT^p.)

Solution

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