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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 $V^*$?

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  • $\begingroup$ One observation: if $E$ contains the identity endomorphism, then $V^\circ \subset V^*$. $\endgroup$ Jul 30, 2021 at 17:19
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    $\begingroup$ @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. $\endgroup$
    – levap
    Jul 30, 2021 at 18:36
  • $\begingroup$ @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? $\endgroup$
    – 57Jimmy
    Jul 30, 2021 at 21:07
  • $\begingroup$ @BenGrossmann Actually, isn't the last example in the answer by Eric Wofsey a counterexample to your comment? $\endgroup$
    – 57Jimmy
    Jul 30, 2021 at 23:06

1 Answer 1

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(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.

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  • $\begingroup$ @57Jimmy: Yeah, sorry, I'll delete my comment. $\endgroup$
    – levap
    Aug 1, 2021 at 8:30

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