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Let $k$ be a commutative ring and $L$ a $k$-module. The tensor algebra $\otimes L$ is $\mathbb{Z}$-graded and $\mathbb{Z}_2$-graded (an element of $L^{\otimes n}$ has degree $n$ and $\mathbb{Z}_2$-degree $n\mod 2$); hence it is a superalgebra over $k$. This canonically induces a supercommutator $\left[\cdot,\cdot \right]_{\mathrm{s}}$ on $\otimes L$, which is simply given by


for any $U\in L^{\otimes n}$ and any $V\in L^{\otimes m}$.

Let $T:\otimes L\to \otimes L$ be the $k$-linear map which acts on pure tensors according to the formula

$$T\left(u_1\otimes u_2\otimes \ldots\otimes u_k\right) = \sum\limits_{i=1}^{k} \left(-1\right)^i u_i \otimes u_1 \otimes u_2 \otimes \ldots \otimes u_{i-1} \otimes u_{i+1} \otimes \ldots \otimes u_k$$

(this is clearly well-defined). It is easy to see that $L^{\otimes 0}\subseteq \operatorname{Ker} T$, that $\left[L, L\right]_{\mathrm{s}}\subseteq \operatorname{Ker} T$, and that

$$\operatorname{Ker} T\cdot \operatorname{Ker} T\subseteq \operatorname{Ker} T$$

(where multiplication is the multiplication in the tensor algebra $\otimes L$), so that $\operatorname{Ker} T$ is a subalgebra of $\otimes L$. (Thus, in particular, $\left[\operatorname{Ker} T,\operatorname{Ker} T\right]_{\mathrm{s}}\subseteq \operatorname{Ker} T$.) Also, $\left[L, \operatorname{Ker} T\right]_{\mathrm{s}}\subseteq \operatorname{Ker} T$.

Consequently, by induction, any nontrivial tree of supercommutator brackets decorated by elements of $L$ must evaluate to an element of $\operatorname{Ker} T$, and so must any tensor product of such trees (including empty products). I am wondering: do these generate (over $k$) all of $\operatorname{Ker} T$ or is there more? I am mostly interested in the case when the characteristic of $K$ is zero, but experimentation with Sage (see comments) suggests that the answer is characteristic-independent as long as $K$ does not have characteristic $2$. In characteristic $2$, one has to additionally consider the elements $x \otimes x \in \operatorname{Ker} T$ for all $x \in L$ (in all other characteristics, this follows from $\left[L, L\right]_{\mathrm{s}}\subseteq \operatorname{Ker} T$).

Here is a bit of motivation (I said this is a curiousity question, but in fact the original curiousity question was about bilinear forms):

Let $f:L\times L\to k$ be a bilinear form. We define a bilinear map $U:L\times\left( \otimes L\right)\to \otimes L$ by $U\left(u,v_1\otimes v_2\otimes ...\otimes v_k\right)=\sum\limits_{i=1}^k\left(-1\right)^{i-1}f\left(u,v_i\right)\cdot v_1\otimes v_2\otimes ...\otimes \hat{v_i}\otimes \ldots\otimes v_k$, where $\hat{v_i}$ means that the factor $v_i$ is omitted from the product. (Of course, we have just defined $U\left(u,V\right)$ for pure tensors $V$ only, but the rest is clear by linearity.)

This bilinear map $U$ is rather natural; people use to denote the similarly defined map $L\times\left( \wedge L\right)\to \wedge L$ (where all $\otimes$ signs have been replaced by $\wedge$ signs) as the "interior product" (the "exterior product" is just the wedge product) - best known in the context of differential forms.

Now I'm wondering what tensors $V\in \otimes L$ satisfy ($U\left(u,V\right)=0$ for every $u\in L$ and every bilinear form $f$). This is equivalent to $V\in\operatorname{Ker} T$.

EDIT: This seems closely related to Sections V and VI in Wilhelm Specht, Gesetze in Ringen I, Mathematische Zeitschrift (1950), Volume: 52, page 557-589. Essentially, Specht proves my conjecture for multilinear tensors (= tensors spanned by pure tensors whose tensorands are a permutation of the basis vectors). Apparently, Specht's motivation was understanding PI-algebras.

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You need to enclose complicated TeX in backticks, just as the sidebar "How to write math" instructs you to do. I've fixed it and made a couple of style/nomenclature changes, I hope it's all right. – Victor Protsak Aug 12 '10 at 7:46
Thanks a lot! I tried enclosing some signs in backticks, but I didn't think of enclosing just everything. – darij grinberg Aug 12 '10 at 9:34
Confirmed using Sage in the following settings: $k = \mathbb Q, L = k^2, \deg \leq 7$; $k = \mathbb Q, L = k^3, \deg \leq 5$; $k = \mathbb Q, L = k^4, \deg \leq 3$. Positive characteristic seems to behave in the same way! – darij grinberg Feb 19 at 3:16

1 Answer 1

up vote 0 down vote accepted

My conjecture was correct. This, and more, is now proven in The signed random-to-top operator on tensor space (draft) (not yet a paper, but probably soon to be posted in more or less this form on the arXiv).

(Some questions do remain, such as Todo 9.5.)

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