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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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    It does look a bit strange. A possibility that comes to mind is that may be it was first proved in a context (Lie groups?), where $F$ contains the reals. After all, the last step works for an $\mathbf{R}$-linear mapping $f$.2012-04-29
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    @Jyrki Lahtonen Wether or not the reals are involved, this seems like a crazy proof strategy to me. Hoever, I would really enjoy a proof of this which uses real or complex analysis. If such an analytical proof exists I could understand the motivation behind this proof.2012-04-29
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    My best attempt at reverse engineering a proof of this in the complex case: Let $r$ be a polynomial without a constant term that maps all the eigenvalues $a_i, i=1,2,\ldots,n,$ to their complex conjugates. Let $y=r(s)$. Then $y$ is a polynomial without a constant term in $x$, hence $y\in M$. Furthermore, $y$ acts as scalar multiplication by $a_i^*$ in the generalized eigenspace of $x$ corresponding to $\lambda=a_i$. Therefore $tr(yx)=\sum_i|a_i|^2=0$. The claim follows.2012-04-29
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    It's strange how you can at once follow every step of a proof and still not understand it ^^ but this thread is helping me get there.2012-04-29
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    I'm not very handy with Lie algebra, but a few things jumped out at me as kind of familiar from associative algebra. $M$ is the largest Lie subring in which $B$ is a Lie ideal. Tr(-,-) is acting like a bilinear form, and this is saying the elements of its kernel are nilpotent. One expects this in associative algebras because the kernel elements are in the Jacobson radical, which is nilpotent in an Artinian algebra. Thanks for your patience with my ramblings.2012-04-29
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    I forgot in my previous comment that we are looking at ad-action. Sorry about that.2012-04-29
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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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