Tensor product and dual of $k$-vector spaces over a $k$-algebra: what is the dimension?

Here is the setting: let $$k$$ be a field, $$V$$ a finite dimensional $$k$$-vector space with $$k$$-dual $$V^*$$, $$E$$ a $$k$$-subalgebra of $$\mathrm{End}_k(V)$$. $$V$$ has thus a canonical structure of faithful left $$E$$-module. The object that we are interested in is

$$V \otimes_E V^* = V \otimes_k V^* / \langle \{f(v) \otimes \eta - v \otimes f^*(\eta) \mid f \in E, v \in V, \eta \in V^*\} \rangle,$$

where $$f^*: \eta \mapsto \eta \circ f$$ is the canonical faithful right action of $$E$$ on $$V^*$$. If $$V$$ is free as an $$E$$-module, then so is $$V^*$$, and they have the same $$E$$-dimension, so that

$$\dim_k(V \otimes_E V^*) = \dim_E(V)^2 \cdot \dim_k(E) = \frac{\dim_k(V)^2}{\dim_k(E)}.$$

If $$V$$ is not $$E$$-free, then the central expression is not defined, but the first and the last one are. Which leads to the question:

Question 1: is it always true that $$\dim_k(V \otimes_E V^*) =\frac{\dim_k(V)^2}{\dim_k(E)}$$? If not, does this at least give an upper or a lower bound?

Remark: More in general, one could ask whether, given two finite-dimensional $$E$$-modules (with $$E$$ a finite-dimensional $$k$$-algebra), we always have $$\dim_k(V \otimes_E W) =\frac{\dim_k(V) \cdot \dim_k(W)}{\dim_k(E)}.$$ This, however, is false, see for instance here for an example. However, maybe the fact that in my case the action of $$E$$ is faithful and $$W=V^*$$ are somehow of help...

So far for the tensor product. My second question, which is probably closely related, concerns the dual:

Question 2: Consider the dual of $$V$$ as an $$E$$-module, i.e., $$V^\circ:= \mathrm{Hom}_E(V,E)$$. What is its relationship with $$V^*$$?

• One observation: if $E$ contains the identity endomorphism, then $V^\circ \subset V^*$. Jul 30, 2021 at 17:19
• @57Jimmy: You can't expect equality as there is no reason $\frac{\dim_k(V)^2}{\dim_k(E)}$ is an integer. For example, $V$ can be $\mathbb{k}^2$ and $E$ can be the algebra of upper triangular matrices (identified with endomorphisms after choosing some basis) which is $3$ dimensional. Jul 30, 2021 at 18:36
• @BenGrossmann $E$ does contain the identity (I should have specified). But why do we have an inclusion? I see that choosing any element of $E^*$ gives a map $V^\circ \to V^*$, by composition. But why should it be injective? Jul 30, 2021 at 21:07
• @BenGrossmann Actually, isn't the last example in the answer by Eric Wofsey a counterexample to your comment? Jul 30, 2021 at 23:06

(The definition of the tensor product that you have used is a bit nonstandard: normally, when you write $$V\otimes_E V^*$$, that means $$V$$ is a right $$E$$-module and $$V^*$$ is a left $$E$$-module. So I will instead consider $$V^*\otimes_E V$$; this is essentially the same as what you have defined but is the standard way of writing it.)

Recall that $$V^*\otimes_k V$$ can be identified with $$\operatorname{End}_k(V)$$ by mapping $$\eta\otimes v$$ to the endomorphism $$w\mapsto \eta(w)v$$. The two actions of $$E\subseteq\operatorname{End}_k(V)$$ on $$V^*\otimes_k V\cong \operatorname{End}_k(V)$$ then just correspond to composition on each side. In other words, $$V^*\otimes_E V$$ is the quotient of $$\operatorname{End}_k(V)$$ by the span of the commutators $$ax-xa$$ where $$a\in \operatorname{End}_k(V)$$ and $$x\in E$$.

Let us now consider some examples. First, let us take $$V=k^n$$ and $$E\subseteq \operatorname{End}_k(V)=M_n(k)$$ to consist of the upper triangular matrices. Write $$e_{ij}$$ for the matrix with $$ij$$ entry $$1$$ and other entries $$0$$, so $$E$$ is generated by those $$e_{ij}$$ with $$i\leq j$$. We can see that identifying $$ae_{ij}$$ with $$e_{ij}a$$ for all matrices $$a$$ will kill $$e_{kj}$$ and $$e_{ik}$$ whenever $$k\neq i,j$$ (take $$a=e_{ki}$$ or $$a=e_{jk}$$) and will identify $$e_{jj}$$ with $$e_{ii}$$ (take $$a=e_{ji}$$). So, modding out all the commutators $$ax-xa$$ where $$x\in E$$ kills all $$e_{ij}$$ for $$i\neq j$$ and identifies together all $$e_{ii}$$, and the resulting quotient $$V^*\otimes_E V$$ has dimension $$1$$. Explicitly, $$V^*\otimes_E V\cong k$$ by the map taking a matrix to its trace. For $$n>1$$, this dimension is less than $$\frac{\dim_k(V)^2}{\dim_k(E)}=\frac{n^2}{\binom{n+1}{2}}$$.

Now let use take $$E=k\times k$$. Write $$M=k\times 0$$ and $$N=0\times k$$ (considered as $$E$$-modules in the obvious way) and let $$V=M^m\oplus N^n$$ for some $$m$$ and $$n$$. As long as $$m,n>0$$, $$E$$ acts faithfully on $$V$$. Also, it is easy to see that $$V^*\cong V$$ as an $$E$$-module (this makes sense because $$E$$ is commutative so we don't care about which side the action is on). Also, $$M\otimes_E M\cong M$$, $$N\otimes_E N\cong N$$, and $$M\otimes_E N=0$$. So, $$V^*\otimes_E V\cong V\otimes_E V\cong M^{m^2}\oplus N^{n^2}$$. We thus have $$\dim_k V^*\otimes_E V=m^2+n^2$$ whereas $$\frac{\dim_k(V)^2}{\dim_k(E)}=\frac{(m+n)^2}{2}$$. As long as $$m\neq n$$, we have $$m^2+n^2>\frac{(m+n)^2}{2}$$ (since their difference is $$\frac{(m-n)^2}{2}$$).

So, $$\frac{\dim_k(V)^2}{\dim_k(E)}$$ is neither an upper bound nor a lower bound for $$\dim_k(V^*\otimes_E V)$$ in general.

As for your second question, I don't think there's any nice relationship between $$V^*$$ and $$V^\circ$$ in general. For instance, consider the following example. Let $$E=k[x,y]/(x^2,xy,y^2)$$ and let $$V=E\oplus E/(x,y)$$. Note that $$\operatorname{Hom}_E(E/(x,y),E)$$ is actually two-dimensional over $$k$$ even though $$E/(x,y)$$ is one-dimensional: you can map the generator of $$E/(x,y)$$ to any linear combination of $$x$$ and $$y$$ in $$E$$. So, $$V^\circ$$ is $$5$$-dimensional while $$V$$ is $$4$$-dimensional.

• @57Jimmy: Yeah, sorry, I'll delete my comment. Aug 1, 2021 at 8:30