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Let $k$ denote the killing form on the Lie algebra $\mathfrak{gl}_n(\mathbb{C})$ and let $\mathfrak{h},\mathfrak{n}_+, \mathfrak{n}_-$ denote the subspaces of diagonal matrices, strictly upper triangular matrices, and strictly lower triangular matrices respectively. How can I show that $\mathfrak{h}$ and $\mathfrak{n}_+\oplus \mathfrak{n}_-$ are orthogonal with respect to the killing form $k$?

Christina
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  • Have you computed the Killing form or that is exactly the problem? – Phicar Nov 14 '20 at 17:31
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    So I computed what is the action of of $ad(x)\circ ad(y)$ on $z$ for $x\in \mathfrak{h}$, $y\in \mathfrak{n}+\oplus \mathfrak{n}-$, and $z\in \mathfrak{gl}_n$, but I am having trouble to calculate trace of just $ad(x)\circ ad(y)$. – Christina Nov 14 '20 at 18:08
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    Minor problem: since $\mathfrak{gl}_n$ is reductive, but not semi-simple, its Killing form is actually degenerate on its center, which is not what you want... Not hard to repair, ... – paul garrett Nov 16 '20 at 20:21

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It should not be necessary to compute $ad(x)\circ ad(y)$ explicitly. The point is that you can use a basis of $\mathfrak{gl}_n(\mathbb C)$ that consits of joint eigenvectors for $ad(x)$ for all elements $x\in\mathfrak h$. By bilinearity, it suffices to consider the case that $y$ is one of these basis elements cotained in $\mathfrak n_+$ or $\mathfrak n_-$. Then the action of $ad(y)$ maps any eigenvector for $ad(\mathfrak{h})$ to an eigenvector with different eigenvalue, while $ad(x)$ preserves each of the eigenspaces. This shows that in a matrix representation with respect to the above basis, the map $ad(x)\circ ad(y)$ has all entries on the main diagonal equal to zero, so in particular, the trace vanishes.

Andreas Cap
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