Are there interesting legendrians at infinity in the cotangent space of bundles?

A question from Laura Schaposnik:

If you have a 3-manifold $M$ with boundary in a surface $S$, you get a holomorphic Lagrangian $Loc(M) \subset Loc(S)$.  Translate over to the Hitchin system, you have a usual Lagrangian inside $T^* Bun_G(S)$.  Now forget the Lagrangian and just remember its asymptotics, $\Lambda$.  What’s the meaning of $Sh_{\Lambda}(Bun_g(S))$?

I think the usual yoga tells you it’s something like $D^b(Loc(M))$.  Is that true?  And, if so, can we actually describe this $\Lambda$ in a reasonable way and use this fact…

3 comments

  1. Harold Williams says:

    Why is $Loc(M)$ still Lagrangian, not to mention exact, after hyperkahler rotation? I could imagine a story at least for why it’s Lagrangian for the $\mathbb{C}^*$ in the twistor sphere but the poles seem much more subtle. Seems like a good point that $Sh_\Lambda(Bun_g(S))$ looks like an important object though given that it’s pretty interesting even when $\Lambda$ is empty…

  2. andyneitzke says:

    $Loc(M)$ is Lagrangian for an $S^1$ worth of complex structures. This is just because it is a holomorphic Lagrangian with respect to one of the complex structures — call that one $J$ say — which means the holomorphic symplectic form $\Omega_J$ vanishes on $Loc(M)$. Since $\Omega_J = \omega_K + i \omega_I$ (using $\omega$ for the three real symplectic forms ie Kahler forms), any real linear combination of $\omega_I$ and $\omega_K$ vanishes; in particular the combination $(\cos \theta) \omega_I + (\sin \theta) \omega_K$ vanishes, and this says $Loc(M)$ is Lagrangian for an $S^1$ of complex structures, including the structures $\pm I$ corresponding to the poles of the twistor sphere, ie the “Higgs bundle” structures.

Leave a Reply

Your email address will not be published. Required fields are marked *