Question. Let $k$ be a field of characteristic $0$. Let $L$ be a $k$-vector space. Consider the subspace $S$ of $L\otimes L\otimes L\otimes L$ spanned by all tensors of the form

$\left[a,\left\lbrace b,\left[c,d\right]\right\rbrace \right]$ for $a,b,c,d\in L$,

where $\left[u,v\right]$ denotes $u\otimes v-v\otimes u$, and where $\left\lbrace u,v\right\rbrace$ denotes $u\otimes v+v\otimes u$.

Let $u$, $v$, $w$, $t$ be four vectors in $L$. Does the tensor $\left[u\otimes v,\left\lbrace x, y\right\rbrace\right]$ lie in $S$ ?

Motivation. This all started with me wondering why the Clifford algebra of a quadratic space is isomorphic to its exterior algebra, as a vector space. This is a kind of "baby PBW theorem" (with a bilinear form instead of a Lie bracket, which indeed makes things easier). Many books give various proofs of this fact, some actually being wrong, the other being (at least) hard to understand. This made me search for an own proof, and I found a purely computational one. While "computational" really means lots of computation, it has its hidden advantages: It works not just for a vector space with a quadratic form, but for any $k$-module over a commutative ring $k$ with unity, with any bilinear (not necessarily symmetric) form. (No, quadratic forms that don't come from bilinear forms are not supported (yet).) The so-called Chevalley map, which is a $k$-module isomorphism $\wedge L\to \mathrm{Cl}\left(L,f\right)$ (here, $L$ is our $k$-module, $f$ is the bilinear form, and $\mathrm{Cl}\left(L,f\right)$ is the corresponding Clifford algebra) turns out to be the projection of a $k$-module automorphism $\alpha^f : \otimes L\to \otimes L$ of the tensor algebra $\otimes L$, which has a nice combinatorial/inductive definition and surprising properties (for example, $\alpha^f \circ \alpha^g = \alpha^{f+g}$ for any two bilinear forms $f$ and $g$).

Now I have been wondering which kind of tensors in $\otimes L$ are invariant under $\alpha^f$ for all symmetric bilinear forms $f$. (Without the "symmetric" this is yet another question.) Tensors of the form

$a$ for $a\in L$;

$\left[a,b\right]$ for $a,b\in L$;

$\left\lbrace a,\left[b,c\right]\right\rbrace$ for $a,b,c\in L$;

$\left[a,\left\lbrace b,\left[c,d\right]\right\rbrace \right]$ for $a,b,c,d\in L$;

etc. (alternately commutator and anticommutator, with the innermost one being a commutator)

are examples of such tensors, and I am wondering whether they span all of them. If that's the case, they should span $\left[u\otimes v,\left\lbrace x, y\right\rbrace\right]$.

I would already be happy to know the answer over fields of characteristic $0$.

Remark. The invariant theory tag is due to the fact that $f\mapsto \alpha^f$ is an algebraic action of the additive group of all symmetric binary forms (or just binary forms) on the tensor algebra of $L$, and we are looking for its invariants. But I do not know whether there are any results from invariant theory that can be applied here; it's an action of a commutative algebraic group.

Feel free to remove the cyclic-homology and free-lie-algebras tags. I chose them because cyclic homology and free Lie algebras involve (at least superficially) similar algebraic expressions (like iterated commutators), and because I feel that people who do computations with cyclic homology and free Lie algebras probably know how to efficiently check properties of tensor products (maybe there is a good CAS for that?)


Starting with the tensor $[e_1,\{e_2,[e_3,e_4]\}]$, I calculated $[e_{\sigma(1)},\{e_{\sigma(2)},[e_{\sigma(3)},e_{\sigma(4)}]\}]$ for $\sigma$ running over a set of representatives of the cosets of (34) (since interchanging the last two inputs only changes the sign of the original tensor). Encoding the 12 resulting 24-vectors as a $12\times 24$ matrix of $0$'s and $\pm 1's$, I calculated that this matrix had rank 10. Augmenting this matrix with the vector corresponding to $[e_1\otimes e_2,\{e_3,e_4\}]$, the resulting $13\times 24$ matrix has rank 11, so unless I overlooked something, it appears that tensors of the form $[u\otimes v,\{x,y\}]$ are not in $S$.

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  • $\begingroup$ Thank you very much! Just wondering, what CAS did you use? $\endgroup$ – darij grinberg Mar 24 '11 at 16:20
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    $\begingroup$ Actually I don't have access to Maple or Mathematica or any other CAS right now, so I worked out the vectors by hand (I seem to have too much free time currently). Once I had the vectors, I ran the matrix rank computations on an online matrix calculator. $\endgroup$ – ARupinski Mar 24 '11 at 16:45
  • $\begingroup$ Just wow. $\endgroup$ – darij grinberg Mar 24 '11 at 17:00

Why doesn't ARupinski's answer get more votes?

Anyway I think I've just found a simpler answer to my question, and I am writing it here in case someone might have another use for the trick.

Let $\left(e_1,e_2,...,e_n\right)$ be a basis of the $k$-vector space $L$ (we assume $k$ to be a field of characteristic $0$). Then, $\left(e_a\otimes e_b\otimes e_c\otimes e_d\right)_{1\leq a,b,c,d\leq n}$ is a basis of the $k$-vector space $L\otimes L\otimes L\otimes L$. For every $1\leq i\neq j\leq n$, let $\epsilon_{i,j}:L\otimes L\otimes L\otimes L\to k$ be the linear map which sends $e_a\otimes e_b\otimes e_c\otimes e_d$ to:

$1$ if both $i$ and $j$ appear in the list $\left(a,b,c,d\right)$, and the first $i$ appears before the first $j$;

$0$ otherwise.

Now $\left[a, \left\lbrace b, \left[c,d\right]\right\rbrace \right]$ lies in the kernel of $\epsilon_{i,j}$ for all $a,b,c,d\in L$ and for all $i\neq j$ (to prove this, consider the case of basis vectors). But $\left[u\otimes v,\left\lbrace x,y\right\rbrace\right]$ doesn't (at least not for $u=e_1$, $v=e_2$, $w=e_3$, $t=e_4$, $i=1$, $j=3$). So the latter tensor cannot lie in the span of the former.

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  • $\begingroup$ Actually my first thought on seeing this problem was to find some linear form such that the elements of $S$ were in the kernel of this form and then check whether $[u\otimes v,\{w,x\}]$ was also in the kernel, but I didn't see how to find the form without computer help, so I just went with the brute force method instead. I really like this solution. As for my answer not getting more votes, it really doesn't bother me too much. $\endgroup$ – ARupinski Mar 25 '11 at 15:59

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