Bourbaki

Exercise 15, § 6

Lie Groups and Lie Algebras, I, 104

Bourbaki marks this one as harder than the rest.

(a) Let X,YX, Y be two matrices of gl(n,K)\mathfrak{gl}(n, K). Show that, if [X,Y]=X[X, Y] = X, XX is nilpotent. (For every polynomial f∈K[T]f \in K[T], observe that

[f(X),Y]=Xf′(X). [f(X), Y] = Xf'(X).
(b) Let gg be a Lie algebra and h,xh, x two elements of gg such that [h,x]=x[h, x] = x and there exists z∈gz \in g with [x,z]=h[x, z] = h. Show that there exists y∈gy \in g such that [x,y]=h[x, y] = h and [h,y]=−y[h, y] = -y. (Observe first that [z,h]+z[z, h] + z belongs to the centralizer nn of xx in gg. Then show that nn is stable under ad h\mathrm{ad}\ h and that the restriction to nn of I−ad hI - \mathrm{ad}\ h is bijective; to do this, note that ad x\mathrm{ad}\ x is nilpotent, using (a), and prove that, if gig_i is the image of gg under (ad x)i(\mathrm{ad}\ x)^i, I−ad hI - \mathrm{ad}\ h gives on taking quotients a bijection of (n∩gi−1)/(n∩gi)(n \cap g_{i-1})/(n \cap g_i); for this use the relation
[ad z,(ad x)k]=−k(ad x)k−1(ad h+k−12I). [\mathrm{ad}\ z, (\mathrm{ad}\ x)^k] = -k(\mathrm{ad}\ x)^{k-1}(\mathrm{ad}\ h + \frac{k-1}{2} I).
(c) Let gg be a subalgebra of gl(n,K)\mathfrak{gl}(n, K); suppose that, for every nilpotent matrix X∈gX \in g, there exist two other matrices H,YH, Y in gg such that [H,X]=X, [H,Y]=−Y, [X,Y]=H[H, X] = X,\ [H, Y] = -Y,\ [X, Y] = H. Let bb be a subalgebra of gg such that there exists a subspace mm supplementary to hh in gg and such that [h,m]⊂m[h, m] \subset m. Show that hh has the same property as gg for all its nilpotent matrices (use (b)).

Solution

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