TITLE: Is the endomorphism algebra of a dualizable bimodule necessarily finite dimensional?
QUESTION [18 upvotes]: 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 
should interpret 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.

REPLY [4 votes]: This is more or less orthogonal to Kevin's answer; he puts a stronger restriction on $A$ and $B$, while I put a stronger restriction (perhaps too strong!) on ${_A}M_B$. I likewise apologize for the lack of pictures.
If we assume that

${_A}M_B$ has a 2-sided dual ${_B}N_A$, and
the duality pairings exhibit ${_B}N \otimes_A M_B$ as a retract of ${_B}B_B$,

then $\operatorname{End}({_A}M_B)$ is dualizable as a $Z(A)$-module, and hence, if $Z(A)$ is finite dimensional, it will follow that $\operatorname{End}({_A}M_B)$ is finite dimensional.
To prove this, we exhibit a duality pairing between ${_{Z(A)}}\operatorname{End}({_A}M_B)$ and $\operatorname{End}({_B}N_A)_{Z(A)}$.
The evaluation map $\epsilon: {_{Z(A)}}\operatorname{End}({_A}M_B) \otimes_k \operatorname{End}({_B}N_A)_{Z(A)} \to {_{Z(A)}}Z(A)_{Z(A)}$ is given by first mapping to $\operatorname{End}({_A}M \otimes_B N_A)$ and then pre- and post-composing by the appropriate unit and counit, respectively, from the duality pairings between $M$ and $N$ to get an endomorphism of ${_A}A_A$, i.e., an element of $Z(A)$.
The coevaluation map $\eta: k \to \operatorname{End}({_B}N_A) \otimes_{Z(A)} \operatorname{End}({_A}M_B)$ is given by identifying the codomain with $\operatorname{End}({_B}N \otimes_A M_B)$, which follows from condition 2, whence $\eta$ is just the unit of this algebra.
That $\epsilon$ and $\eta$ give a duality pairing follows from condition 2. (The nontrivial morphism in one of the triangle diagrams sends a map $f: {_A}M_B \to {_A}M_B$ to the identity on ${_A}M_B$ tensored with the "trace" of $f$ via the duality pairings. Condition 2 tells us that the identity string can absorb the trace bubble, and similarly for the other triangle diagram.)
Unfortunately, condition 2 (or its "mirror," which would show dualizability as a $Z(B)$-module) seems very strong, but it is crucial for the above proof.