9
$\begingroup$

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

$$\left[U,V\right]_{\mathrm{s}}=UV-\left(-1\right)^{nm}VU$$

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.

$\endgroup$
1
  • $\begingroup$ 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! $\endgroup$ Feb 19, 2015 at 3:16

1 Answer 1

3
$\begingroup$

My conjecture was correct. This, and more, is now proven in The signed random-to-top operator on tensor space (draft) (aka arXiv preprint arXiv:1505.01201).

(Some questions do remain, such as those in §9.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.