Bourbaki

Exercise 3, § 2

Lie Groups and Lie Algebras, V

Bourbaki marks this one as harder than the rest.

Let V be a complex Hilbert space with basis e1,…,ele_1, \ldots, e_l. For 1≤i≤l1 \leq i \leq l, let sis_i be a unitary pseudo-reflection, with vector eie_i, such that si(ei)=cieis_i(e_i) = c_i e_i, with ci≠1c_i \neq 1; an element of V is invariant under sis_i if and only if it is orthogonal to eie_i. Let W be the subgroup of GL(V)\mathbf{GL}(V) generated by the sis_i.

a) Let ii be an integer ≥1\geq 1. Show that every element of ⋀iV\bigwedge^i V invariant under W is zero. (Argue by induction on ll; if V′V' is the subspace of V generated by e1,…,el−1e_1, \ldots, e_{l-1}, and if ee is a non-zero vector orthogonal to V′V', any element of ⋀iV\bigwedge^i V can be written in the form a+(b∧e)a + (b \wedge e), with a∈⋀iV′a \in \bigwedge^i V' and b∈⋀i−1V′b \in \bigwedge^{i-1} V'; if a+(b∧e)a + (b \wedge e) is invariant under W, aa and bb are invariant under the subgroup W′W' generated by s1,…,sl−1s_1, \ldots, s_{l-1}, and the induction hypothesis applies.)

b) Assume that W is finite. Show, by using a), that, for any endomorphism A of V,

∑w∈Wdet⁡(A−w)=Card⁡(W)⋅det⁡(A) \sum_{w \in W} \det(A - w) = \operatorname{Card}(W) \cdot \det(A)
∑w∈Wdet⁡(1−Aw)=Card⁡(W). \sum_{w \in W} \det(1 - Aw) = \operatorname{Card}(W).

Deduce that, for any A∈End⁡(V)A \in \operatorname{End}(V), there exists w∈Ww \in W such that AwAw has no non-zero fixed point.

c) Let Γ\Gamma be the graph whose set of vertices is {1,l}\{1, l\}, the arrows being the sets {i,j}\{i, j\} such that eie_i and eje_j are not orthogonal. Show that VV is a simple WW-module if and only if Γ\Gamma is connected and non-empty.

d) Assume that VV is a simple WW-module. Show that the WW-modules ⋀iV\bigwedge^i V (0≤i≤l0 \leq i \leq l) are simple. (Show that there exists an integer jj such that the graph Γ−{j}\Gamma - \{j\} is connected. Argue by induction on ll, and apply the induction hypothesis to the subspace V′V' generated by the ei,i≠je_i, i \neq j.) Show that these modules are pairwise non-isomorphic6^6.

Solution

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