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 !

  • 2
    $\begingroup$ Extending scalars to the algebraic closure will produce a subalgebra of the extended algebra of the same dimension and also commutative, so you may just as well assume that $A$ is a matrix algebra. $\endgroup$ – Mariano Suárez-Álvarez Aug 19 '13 at 15:55
  • $\begingroup$ And for matrix algebras, Schur's theorem gives the best possible dimension. $\endgroup$ – Dietrich Burde Aug 19 '13 at 19:23
  • $\begingroup$ $n$ shouldn't be replaced by $r$? I unsuccessfully tried to edit the post because of the error 'Edits must be at least 6 characters;' $\endgroup$ – Name Aug 20 '13 at 7:00

Sorry, I cannot add comment yet, so I just post my comment as an answer.

Extending scalars shows that the dimension is at most $[(rd)^2/4]+1$, but it does not prove that a commutative $k$-subalgebra with this dimension does exist.

For example, assume that $r=1$, i.e $A=D$ is division of degree $d$ . Then the maximal dimension of a commutative $k$-subalgebra is $d$ (because such as subalgebra is a subfield), but extending scalars gives you the bound $[d^2/4]+1$.

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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.

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