5
$\begingroup$

Let $A$ be an algebra over $\mathbb{C}$. Let $M$ be a left $A$-module, let $N$ be a right $A$-module and consider the tensor product $N \otimes_A M$, which is a complex vector space.

Q1: Can this tensor product be the trivial vector spacefor non-zero $M$, $N$? If yes, what are examples where this happens?

Q2: Are there criteria on $M$ and $N$ which ensure that the tensor product is non-zero?

Edit: Thanks for the answers below. It seems that Q2 is too broad to answer in general. However, I am mostly interested in the case of an infinite-dimensional Clifford algebra; in other words, $A$ is central simple.

$\endgroup$
  • $\begingroup$ Thanks, edited. $\endgroup$ – Matthias Ludewig May 28 '19 at 4:06
  • 3
    $\begingroup$ In the commutative case you can take two modules with disjoint support (maybe I need f.p. here just to be safe). $\endgroup$ – Qiaochu Yuan May 28 '19 at 4:12
4
$\begingroup$

I give an answer in case the modules are finite dimensional (not 100% sure whether the algebra also has to be finite dimensional but I think it is not needed). The base field can be arbitrary and does not need to be $\mathbb{C}$. Let the algebra $A$ be over a field $K$ and $D=Hom_K(-,K)$. Then we have $N \otimes_A M \cong D Hom_A(N,D(M))$.

This makes it very easy to at least calculate the vector space dimension of $N \otimes_A M$ and give alot of examples for Q1: Just take $M$ and $N$ simple such that $N$ is not isomorphic to $D(M)$ and by Schur's lemma $N \otimes_A M \cong D Hom_A(N,D(M))=0$.

Example: Let $A$ be a finite dimensional algebra with $n$ fixed idempotent $e_1,...,e_n$ and simple right modules $S_1,...,S_n$ and simple left modules $G_1,...,G_n$ (corresponding to the idempotents ). Then $D(G_i) \cong S_i$ and thus $S_j \otimes_A G_i \neq 0$ if and only if $i=j$.

At least when the algeba is a finite dimensional quiver algebra deciding when $D Hom_A(N,D(M))=0$ can be reduced to linear algebra and answers Q2.

| cite | improve this answer | |
$\endgroup$
4
$\begingroup$

This is intuitively dual to the problem of finding left $A$-modules $M$ and $V$, neither having zero action, such that ${}_A{\rm hom}(M,V)=0$. Then a natural thing to do is to take two inequivalent (fin dim) complex irreps of some finite group $G$ (Schur orthogonality then tells us the hom-space between two inequivalent irreps of $G$ is zero).

You then recover examples for Q1 by taking $V=N^*$. The ease with which such examples are found makes me pessimistic about answering Q2 unless you can narrow down the class of modules or algebras you are interested in.

We can come up with very simple examples based on choosing the easiest possible $G$, namely ${\bf Z}/2{\bf Z}$, which has two inequivalent 1-dim modules given by the two characters of this group.

This translates into the following setup. $A$ can be identified with ${\bf C}\oplus {\bf C}$ (Fourier transform of ${\bf C}G$ for $G$ being the group above) and then there are two characters $\phi_1, \phi_2: A \to {\bf C}$ defined by $\phi_j((a_1,a_2)) = a_j$ ($j=1,2$). The corresponding 1-dimensional modules $V_1$ and $V_2$ satisfy $V_1\otimes_A V_2 =0$, because $$ 1 \otimes 1 = 1\cdot x\otimes 1 - 1 \otimes x\cdot 1 \quad\hbox{if $x=(1,0)$} $$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you for your answer; I now have a better grasp on what's going on. In my case, I am interested in a very special situation though, see my edit above: The case of an infinite-dimensional Clifford algebra, which is central simple. $\endgroup$ – Matthias Ludewig May 28 '19 at 4:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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