TITLE: Dimension of commutative subalgebras of a central simple algebra
QUESTION [8 upvotes]: let $k$ be a field, and let $A$ be a central simple $k$-algebra over $k$.
What is the maximal dimension of a commutative $k$-subalgebra of $A$?
If $A=M_r(D)$, where $D$ is a central division $k$-algebra of degree $d$, then we may find a commutative subalgebra of degree $([n^2/4]+1)d$, by choosing a commutative subalgebra $E$ of $M_r(k)$ of dimension $[n^2/4]+1$ and a maximal subfield $L$ of $D$, and forming the tensor product $E\otimes_k L$.
But is it the best possible dimension ?
Thanks in advance !

REPLY [2 votes]: This is a funny question, which has arisen as a critical component now and then. Yes, if we insisted that the subalgebra be semi-simple, then Schur's lemma gives the result, as in a comment. Perhaps some readers overlooked the possibility of getting $[n^2/4]+1$ (for the matrix case) by taking scalars together with upper-right-corner blocks, with everything else $0$.
The best reasonably-provable result I know was proven by D. Kazhdan as a part of J./I. Bernstein's proof of the admissibility of supercuspidal repns of p-adic reductive groups, and (for the matrix=split case) says that the dimension for a subalgebra with $\ell$ non-scalar generators is bounded by $(n^2)^{1-2^{-\ell}}$. 
Since it's not obvious how to exceed $n^2/4$, this bound seems needlessly weak, but we can suspect that Kazhdan would have given a sharper bound if a simple one really held, at least as a meta-argument for no sharper bound holding. I've not tried to construct an algebra exceeding the $n^2/4$, although I have tried (and failed) to improve that bound. The argument is surprisingly non-trivial. A very sketchy version appears in Bernstein's original paper (in J. Fun. An. and Applications), a somewhat fuller version (revised after some comments from D. Renard) in my essay http://www.math.umn.edu/~garrett/m/v/proving_admissibility.pdf, and as an appendix in David Renard's relatively recent book on repn theory of p-adic groups.