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Introduction. Let $k$ be a field of characteristic $0$, and let $n\in\mathbb N$. Let $V=k^n$. The group $\mathrm{GL}_n\left(k\right)=\mathrm{GL} V$ acts on $\mathrm{End} V$ by conjugation, and thus also on the space of $s$-multilinear forms $\left(\mathrm{End} V\right)^s\to k$ for each $s\in\mathbb N$.

For every $p\in\mathbb N$ and every $p$-multilinear form $f:\left(\mathrm{End} V\right)^p \to k$, we define the antisymmetrization of $f$ to be the $p$-multilinear form

$g:\left(\mathrm{End} V\right)^p \to k,$

$\left(A_1,A_2,...,A_p\right) \mapsto \dfrac{1}{p!}\sum\limits_{\sigma\in S_p}\left(-1\right)^{\sigma} f\left(A_{\sigma\left(1\right)},A_{\sigma\left(2\right)},...,A_{\sigma\left(p\right)}\right)$.

This $g$ is an antisymmetric $p$-multilinear form.

For any integers $p\geq 0$ and $q\geq 0$, any antisymmetric $p$-multilinear form $\alpha:\left(\mathrm{End} V\right)^p\to k$ and any antisymmetric $q$-multilinear form $\beta:\left(\mathrm{End} V\right)^q\to k$, we can define an antisymmetric $p+q$-multilinear form $\alpha\wedge\beta:\left(\mathrm{End} V\right)^{p+q}\to k$ as the antisymmetrization of the form

$\left(\mathrm{End} V\right)^{p+q}\to k,$

$\left(A_1,A_2,...,A_p,B_1,B_2,...,B_q\right)\mapsto \alpha\left(A_1,A_2,...,A_p\right)\beta\left(B_1,B_2,...,B_q\right)$.

(We could also define it using shuffle products, but that's not important in characteristic $0$.)

For any $p\in\mathbb N$, let $\Omega_p : \left(\mathrm{End} V\right)^p \to k$ be the antisymmetrization of the form

$\left(\mathrm{End} V\right)^p \to k,$

$\left(A_1,A_2,...,A_p\right)\mapsto \mathrm{Tr}\left(A_1A_2...A_p\right)$.

Then, it is known that the ring of antisymmetric $\mathrm{GL}_n\left(k\right)$-invariant multilinear forms on $\mathrm{End}V$ (with multiplication being given by $\wedge$) is generated by the $\Omega_p$ for $p\in\mathbb N$ (this follows from the First Fundamental Theorem for $\mathrm{GL}_n\left(k\right)$, which actually gives all multilinear invariants rather than just the antisymmetric ones). It is also easy to see that $\Omega_p=0$ for all even $p$, and the Amitsur-Levitzki theorem yields that $\Omega_p=0$ for all $p\geq 2n$.

Thus, the family $\left(\Omega_{p_1}\wedge\Omega_{p_2}\wedge ...\wedge\Omega_{p_r}\right)$ (indexed by all strictly increasing sequences $\left(p_1,p_2,...,p_r\right)$ of odd positive integers smaller than $2n$) generates the vector space of all antisymmetric $\mathrm{GL}_n\left(k\right)$-invariant multilinear forms on $\mathrm{End}V$.

Question. How to prove that this family is a basis of this space?

Context. This is quoted as a consequence of (not further specified) invariant theory in Pierre Cartier's A primer of Hopf algebras, page 9, §2.1. I am suspecting Cartier wants to involve some kind of Second Fundamental Theorem, but I don't know it well enough. Maybe there is a slick proof in the same vein as one shows that Amitsur-Levitzki does not hold in smaller degrees than $2n$ ?

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So that I understand the question, the point is that $\Omega_p$ is an odd-degree form if $p$ is odd (and already $0$ if $p$ is even), and so there cannot be any terms of form $\Omega_p \wedge \Omega_p = -\Omega_p \wedge\Omega_p$. So it follows that your sequences are a spanning set, and the only thing missing is the independence? Put another way, the question is why the ring of invariants is a polynomial algebra (on odd variables) and not a quotient thereof? –  Theo Johnson-Freyd Mar 4 '12 at 12:33
Yes. It is the linear independence that I don't understand. –  darij grinberg Mar 4 '12 at 15:08
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2 Answers

up vote 1 down vote accepted

Another proof of linear independence (which I should have typed in the first place - but there are too many spectral sequences on my mind these days):

To prove linear independence of the forms you are considering, it is enough to prove that $\Omega_1\wedge\Omega_3\wedge\ldots\wedge\Omega_{2n-1}\ne0$ (if there is a linear combination equal to zero, you can always multiply it by something so that one of the [lowest degree] terms in that combination becomes $\Omega_1\wedge\Omega_3\wedge\ldots\wedge\Omega_{2n-1}$, and all others disappear).

Now, let us note that in the case of a finite-dimensional algebra homology is dual to the cohomology, so it is sufficient to prove that the top homology of $gl_n$ is non-zero. But it is very easy to check that $\wedge_{i,j=1}^ne_{ij}$ is closed in the Chevalley--Eilenberg complex, and there is nothing that may kill that homology class, so we are done.

Of course, this also uses cohomology, but in a much less heavy way :-)

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Okay. I should have actually learnt some Lie algebra cohomology before asking this question, as I could expect that algebraizing Lie group cohomology will lead into Lie algebra cohomology. At the moment I don't know what is the relation between cohomology, invariants and homology that makes you "identify" $\Omega_1\wedge\Omega_3\wedge ...\wedge\Omega_{2n-1}$ with $\bigwedge_{i,j=1}^n e_{ij}$. But this being a failure of mine rather than of your proof, I accept it. –  darij grinberg Mar 5 '12 at 15:49
Just a quick line that might give you some pointers. First: for a compact Lie group $G$ with the Lie algebra $g$, its de Rham complex is homotopy equivalent to the subcomplex of left-invariant forms, which is the Chevalley--Eilenberg complex computing the Lie algebra cohomology of $g$, and is homotopy equivalent to the subcomplex of bi-invariant forms, which has zero differential and consists of invariants in the exterior algebra of $g^*$ which you want to study. Second: the dual of the Chevalley--Eilenberg complex is a chain complex computing the homology of the Lie algebra $g$ –  Vladimir Dotsenko Mar 5 '12 at 16:15
(continued)... and the wedge product of all basis elements of $g$ lives in that homological complex. So the last step is a little bit sneaky - instead of proving that some component in a graded vector space is one-dimensional, we prove that the appropriate component in the dual vector space is one-dimensional. –  Vladimir Dotsenko Mar 5 '12 at 16:18
Thanks! I am just seeing that (more or less) your argument without the geometric detour is given in §8.11 of Procesi's "Lie Groups - An Approach through Invariants and Representations" (as Proof of the Theorem on page 454). –  darij grinberg Mar 5 '12 at 19:25
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One possible proof goes as follows, I think:

First, let us replace $End(V)$ and $GL(V)$ by the Lie algebra $u_n$ and the Lie group $U_n$ for the moment. In that case, invariant skew-symmetric forms give you precisely the Lie algebra cohomology of $u_n$, and the de Rham cohomology of $U_n$ (it's about averaging over group and stuff, explained in Weyl's book on invariants of classical groups), and those are given by the free skew-symmetric algebra on those generators (which can be proved, for instance, by considering the Leray spectral sequence of the fibration $U_{n-1}\hookrightarrow U_n\twoheadrightarrow S^{2n-1}$).

Second, there is no difference between $gl_n$ and $u_n$ over complex numbers, so the "unitary trick" of a sort shows that your question has the same answer.

It remains to remark that some of the above works better over one of our favourite fields, like $\mathbb{R}$, but the cohomology of the Chevalley-Eilenberg complex we are interested in is defined over $\mathbb{Q}$, so in the end your result is valid in char 0 always.

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Okay. This is exactly the reverse of Cartier's argument, as he uses the invariant-theoretical statement to prove the cohomology of $U_n$ rather than the other way round. (He is doing this for the sake of historical authenticity, I assume - there were no Leray spectral sequences at the time when the cohomology of Lie groups was first computed.) But +1 for a nice application of geometry in algebra. –  darij grinberg Mar 4 '12 at 17:52
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