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This is a question pertaining to Humphrey's Introduction to Lie Algebras and Representation Theory

Is there an explanation of the lemma in §4.3-Cartan's Criterion? I understand the proof given there but I fail to understand how anybody could have ever devised it or had the guts to prove such a strange statement...

Lemma: Let $k$ be an algebraically closed field of characteristic $0$. Let $V$ be a finite dimensional vector space over $k$, and $A\subset B\subset \mathrm{End}(V)$ two subspaces. Let $M$ be the set of endomorphisms $x$ of $V$ such that $[x,B]\subset A$. Suppose $x\in M$ is such that $\forall y\in M, \mathrm{Tr}(xy)=0$. Then, $x$ is nilpotent.

The proof uses the diagonalisable$+$nilpotent decomposition, and goes on to show that all eigenvalues of $x$ are $=0$ by showing that the $\mathbb{Q}$ subspace of $k$ they generate has only the $0$ linear functional.

Added: (t.b.) here's the page from Google books for those without access:

Lemma from Humphreys

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    I partially understand what you mean, I know for instance that the Jacobson radical is nilpotent when the ring is artinian. But I don't think $M$ is generally a Lie algebra for one and don't expect to have $B\subset M$ generally. Even though there have been no answers to this question, I feel a lot less bewildered than before. I knew the content of the proof, now I understand it but remain impressed ^^ thanks for your help :)2012-04-29

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This lemma troubled me until I saw proof of Cartan's criterion for complex case in an appendix of Fulton & Harris book.

Denote $x_s = d(\lambda_1, \ldots, \lambda_n)$ the semisimple part of $x$, and $\bar{x}_s = d(\bar{\lambda}_1, \ldots, \bar{\lambda}_n)$, the matrix you get when you ''complex-conjugate'' $x_s$. Lemma is then replaced with the following simple observation: $\operatorname{Tr}(x \bar{x}_s) = |\lambda_1|^2 + \ldots + |\lambda_n|^2 = 0$ implies $x$ nilpotent.

For completeness, proof of Cartan's criterion using this fact:

Let $V$ be a finite-dimensional complex vector space, $L$ Lie subalgebra of $\mathfrak{gl} V$.

Let's assume that $\operatorname{Tr}(xy) = 0$, for all $x \in [LL], y \in L$. If we show that every $x \in [LL]$ is nilpotent, it will follow (by Engel's theorem) that $[LL]$ is nilpotent and that will imply that $L$ is solvable. We'll do this using the aforementioned simple observation. Write $x = \sum_i [y_i, z_i]$. $\operatorname{Tr}(x\bar{x}_s) = \sum_i \operatorname{Tr}([y_i, z_i]\bar{x}_s) = \sum_i \operatorname{Tr}(y_i, [z_i, \bar{x}_s])$ Now, we can use Lagrange's interpolation to write $\operatorname{ad}(\bar{x}_s)$ as a polynomial in $\operatorname{ad}(x_s)$ without constant term, so it follows that $[z_i, \bar{x}_s]$ is an element of $[L, L]$. Now, our assumption gives us $\operatorname{Tr}(x \bar{x}_s) = 0$.

As you can see, more general proof follows this one closely. Using the nice properties of complex conjugation, we don't have to check $\operatorname{Tr}(xy) = 0$ for all $y \in M$ to get $x$ nilpotent, we just have to check that for one element in $M$, namely $\bar{x}_s$. I can imagine Cartan proving first the complex case, then generalizing the proof to the one you read in Humphreys.

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This doesn't entirely answer your question but the key ingredients are (1) the rationals are nice in that their squares are non-negative (2) you can get from general field elements to rationals using a linear functional f (3) getting a handle on x by way of the eigenvalues of s.

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    This would make a good comment, but you'd really have to expand this more to be an answer.2013-09-09