Alternating multilinear invariants of GL(n) on End (k^n) - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T23:47:18Z http://mathoverflow.net/feeds/question/90178 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/90178/alternating-multilinear-invariants-of-gln-on-end-kn Alternating multilinear invariants of GL(n) on End (k^n) darij grinberg 2012-03-04T03:20:20Z 2012-03-05T09:04:00Z <p><strong>Introduction.</strong> 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 $s\in\mathbb N$.</p> <p>For every $p\in\mathbb N$ and every $p$-multilinear form $f:\left(\mathrm{End} V\right)^p \to k$, we define the <em>antisymmetrization</em> of $f$ to be the $p$-multilinear form</p> <p>$g:\left(\mathrm{End} V\right)^p \to k,$</p> <p>$\left(A_1,A_2,...,A_p\right) \mapsto \dfrac{1}{p!}\sum\limits_{\sigma\in S_p}\left(-1\right)^{\sigma} f\left(A_{\sigma\left(1\right)},A_{\sigma\left(2\right)},...,A_{\sigma\left(p\right)}\right)$.</p> <p>This $g$ is an antisymmetric $p$-multilinear form.</p> <p>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</p> <p>$\left(\mathrm{End} V\right)^{p+q}\to k,$</p> <p>$\left(A_1,A_2,...,A_p,B_1,B_2,...,B_q\right)\mapsto \alpha\left(A_1,A_2,...,A_p\right)\beta\left(B_1,B_2,...,B_q\right)$.</p> <p>(We could also define it using shuffle products, but that's not important in characteristic $0$.)</p> <p>For any $p\in\mathbb N$, let $\Omega_p : \left(\mathrm{End} V\right)^p \to k$ be the antisymmetrization of the form</p> <p>$\left(\mathrm{End} V\right)^p \to k,$</p> <p>$\left(A_1,A_2,...,A_p\right)\mapsto \mathrm{Tr}\left(A_1A_2...A_p\right)$.</p> <p>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 <a href="http://en.wikipedia.org/wiki/Amitsur%25E2%2580%2593Levitzki_theorem" rel="nofollow">Amitsur-Levitzki theorem</a> yields that $\Omega_p=0$ for all $p\geq 2n$.</p> <p>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 $\mathrm{End}V$.</p> <p><strong>Question.</strong> How to prove that this family is a basis of this space?</p> <p><strong>Context.</strong> This is quoted as a consequence of (not further specified) invariant theory in <a href="http://www.math.osu.edu/~kerler.2/VIGRE/InvResPres-Sp07/Cartier-IHES.pdf" rel="nofollow">Pierre Cartier's <em>A primer of Hopf algebras</em></a>, 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 $2n$ ?</p> http://mathoverflow.net/questions/90178/alternating-multilinear-invariants-of-gln-on-end-kn/90212#90212 Answer by Vladimir Dotsenko for Alternating multilinear invariants of GL(n) on End (k^n) Vladimir Dotsenko 2012-03-04T17:02:57Z 2012-03-04T17:02:57Z <p>One possible proof goes as follows, I think: </p> <p>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 $U_{n-1}\hookrightarrow U_n\twoheadrightarrow S^{2n-1}$). </p> <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/90178/alternating-multilinear-invariants-of-gln-on-end-kn/90262#90262 Answer by Vladimir Dotsenko for Alternating multilinear invariants of GL(n) on End (k^n) Vladimir Dotsenko 2012-03-05T08:54:22Z 2012-03-05T09:04:00Z <p>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): </p> <p>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). </p> <p>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.</p> <p>Of course, this also uses cohomology, but in a much less heavy way :-)</p>