**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](http://mit.edu/~darij/www/algebra/algebra.html#chevalley). 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?)



Invariant space of lifted Chevalley automorphisms of the tensor algebra
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 $S$.


