Bourbaki

Exercise 18, § 1

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

Lie Groups and Lie Algebras, VIII, 226

Bourbaki marks this one as harder than the rest.

Assume that k=R,Ck=\mathbf{R},\mathbf{C} or a field complete for a discrete valuation and with residue field of characteristic p≠0p\not= 0 (for example a finite extension of the pp-adic field Qp\mathbf{Q}_p).

a) Let n\mathfrak{n} be a nilpotent Lie algebra, N the Lie group obtained by giving n\mathfrak{n} the Hausdorff law (Chap. III, §9, no. 5), and ρ:n→gl(E)\rho :\mathfrak{n}\rightarrow \mathfrak{g}\mathfrak{l}(E) a linear representation of n\mathfrak{n} on a finite dimensional vector space E. Assume that ρ(x)\rho (x) is nilpotent for all x∈nx\in \mathfrak{n}, and put π(x)=\pi (x) = exp(ρ(x)\rho (x)). Show that π\pi is the only homomorphism of Lie groups φ:N→GL(E)\varphi : N\rightarrow \mathbf{G}\mathbf{L}(E) such that L(φ)=ρL(\varphi ) =\rho.

(When k=Rk=\mathbf{R} or C\mathbf{C}, use the fact that N is connected. When kk is ultrametric, show that the eigenvalues of φ(x),x∈N\varphi (x),x\in N, are equal to 1; for this, prove first of all that, if k′k' is a finite extension of kk, and if (λ1,...,λn,...)(\lambda_1, . . . , \lambda_n, . . .) is a sequence of elements of k′k' such that λn=λn+1p\lambda_n=\lambda^p_{n+1} for all nn and λ1=1\lambda_1= 1, then λn=1\lambda_n= 1 for all n.n.)

b) Let ρ:s→gl(E)\rho :\mathfrak{s}\rightarrow \mathfrak{g}\mathfrak{l}(E) be a finite dimensional linear representation of s\mathfrak{s}, and let π\pi be the homomorphism from SL(2,k)\mathbf{S}\mathbf{L}(2, k) to GL(E)\mathbf{G}\mathbf{L}(E) compatible with ρ\rho (no. 4). Show that π\pi is the unique homomorphism of Lie groups φ:SL(2,k)→\varphi :\mathbf{S}\mathbf{L}(2, k)\rightarrow GL(E)\mathbf{G}\mathbf{L}(E) such that L(φ)=ρL(\varphi ) =\rho. (Use aa) to prove that π\pi and ρ\rho coincide on the exp(nn), with nn nilpotent in s\mathfrak{s}, and remark that SL(2,k)\mathbf{S}\mathbf{L}(2, k) is generated by the exp(nn).).)

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