Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
It would help enormously if you reproduced at least the lemma, even better the proof, so that this question isn't only answerable by people with access to Humphreys. – Qiaochu Yuan Apr 29 '12 at 9:38
Ok ^^ I was hoping to avoid it as I am wriring this from a phone but you're right. – Olivier Bégassat Apr 29 '12 at 9:41
I included links and an image of the lemma and its proof. I hope that's what you intended. – t.b. Apr 29 '12 at 9:51
@t.b. That would be lovely, do I have to do something for these to appear in the question? – Olivier Bégassat Apr 29 '12 at 9:56
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. – rschwieb Apr 29 '12 at 14:13

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.

share|cite|improve this answer

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.

share|cite|improve this answer
This would make a good comment, but you'd really have to expand this more to be an answer. – rschwieb Sep 9 '13 at 14:49

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.