Let $k$ be field. Let $A$, $B$ be $k$-algebras, and let ${}_AM_B$ be a dualizable bimodule.
Pre-Question (too naive): Is the algebra of $A$-$B$-bilinear endomorphisms of $M$ necessarily finite dimensional?
Answer: No. Take $A$ some infinite dimensional commutative algebra, and $M={}_AA_A$. Then $End({}_AA_A)=A$ is not finite dimensional.
Question: Assume that $A$ and $B$ have finite dimensional centers. Is it then true that the algebra of $A$-$B$-bilinear endomorphisms of $M$ has to be finite dimensional?
Special case for which I know the answer to be positive:
If $k=\mathbb C$ or $\mathbb R$, and if we're in a C*-algebra context, then I know how to prove that $End({}_AM_B)$ is finite dimensional. But my proof relies on certain inequalities, and it does not generalize.
Definitions:
A bimodule ${}_AM_B$ is called left dualizable if there is an other bimodule ${}_BN_A$ (the left dual) and maps $r:{}_AA_A\to {}_AM\otimes_BN_A$ and $s:{}_BN\otimes_AM_B\to {}_BB_B$ such that $(1\otimes s)\circ(r\otimes 1) = 1_M$ and $(s\otimes 1)\circ(1\otimes r) = 1_N$.
There's a similar definition of right dualizability. I'm guessing that right dualizability is not equivalent to left dualizability, but I don't know a concrete example that illustrates the difference between these two notions.
In my question above, I've just used the term "dualizable", which you
might either should interpret to be "left dualizable" or as
"both left and right dualizable".
There's also the notion of fully dualizable, which means that the left dual has its own left dual, which should in turn have its own left dual etc., and similarly for right duals. Once again, I'm a bit vague as to whether all these infinitely many conditions are really needed, or whether they are implied by finitely many of them.
PS: I hope that I didn't mix my left and my right.
