1
$\begingroup$

In grad school, I received some training in homological mirror symmetry and have begun learning about the classical Langlands program. I see that geometric Langlands at times explicitly mentions homological mirror symmetry, and now realize that much of what I saw in the mirror symmetry community at the very least has a large degree of overlap with geometric Langlands. I am wondering the extent to which one can make a precise statement here: Does the geometric Langlands program generalize homological mirror symmetry in any precise way?

Maybe there are dimensional restrictions (such as only correlating to 3d mirror symmetry) since the geometric Langlands program only studies vector bundles, etc. over a curve (i.e., of dimension 1)?

Any text references or articles would be greatly appreciated.

$\endgroup$
4
  • 5
    $\begingroup$ Katzarkov once told me that Langlands program is the first term (not zeroth) of the Taylor expansion of HMS (or maybe it was the other way around). To this day I still have no idea what he meant by this. $\endgroup$
    – M.G.
    Commented Aug 24 at 1:00
  • 2
    $\begingroup$ A slogan in the relative Langlands program: the Langlands program is part of the study of 4-dimensional (arithmetic, topological) quantum field theory. Then the Langlands correspondence is four-dimensional mirror symmetry. See web.ma.utexas.edu/users/vandyke/notes/langlands_sp21/… and math.ias.edu/~akshay/research/BZSVpaperV1.pdf for more details. $\endgroup$ Commented Aug 24 at 1:13
  • 2
    $\begingroup$ According to mathoverflow.net/questions/357653/… it seems to be a generalization of a special case. $\endgroup$
    – S. Carnahan
    Commented Aug 24 at 1:15
  • $\begingroup$ These are excellent! M.G., that is so interesting: I've heard of the derived category of an abelian category being the tangent space (i.e., first Taylor approximation) to that abelian category. And D^b_{Coh}(X) is slightly bigger than just K-theory or vector bundles, as considered in Bun from Langlands. coLaideronnette thank you very much for your references! These are exactly what I'm looking for. S. Carnahan thank you, I'll review this as well. $\endgroup$ Commented Aug 24 at 1:32

0

You must log in to answer this question.

Browse other questions tagged .