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Let $(U,\omega),(V,\rho)$ be symplectic vector spaces. Call a relation $U \to V$ a (linear) Lagrangian relation (also Lagrangian correspondence) if it is a Lagrangian subspace of $\overline U \oplus V$, where $\overline U$ is the conjugate symplectic vector space $(U,-\omega)$.

These linear Lagrangian relations have the property that the composite (as relations) of two Lagrangian relations is again a Lagrangian relation. As any Lagrangian subspace of some space $U$ can be considered a Lagrangian relation $0 \to V$, this implies that Lagrangian relations map Lagrangian subspaces to Lagrangian subspaces. Does the converse hold?

That is, if a relation maps Lagrangian subspaces to Lagrangian subspaces, is it a linear Lagrangian relation?

EDIT: Nate Bottman provides an easy counterexample below. A refinement of this question to get closer to the intuition guiding it is to ask whether a relation that both preserves and reflects Lagrangian subspaces is Lagrangian. Any tips on this would be appreciated.

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Less precisely, in the category of symplectic vector spaces and linear Lagrangian relations, Lagrangian subspaces are thus the (monoidal unit-valued) points of each object. What can be said about Lagrangian relations as functions on this set of points, and is this a useful perspective? Is there any literature on this?

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(Cross-posted from math.stackexchange: https://math.stackexchange.com/questions/917808/if-a-linear-relation-maps-lagrangian-subspaces-to-lagrangian-subspaces-is-it.)

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  • $\begingroup$ Not sure on the etiquette here: someone let me know if I should start a new question instead of refining it. Will accept Nate's answer if no one provides a more helpful comment in the next few days. $\endgroup$
    – Brendan Fong
    Sep 19, 2014 at 3:45

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Counterexample: set $U$ to be anything, $V := \text{pt}$, and $\Lambda \subset \overline{U} \oplus \text{pt}$ to be a non-Lagrangian subspace.

(But maybe true with some hypotheses, e.g. $\Lambda$ induces an injection on Lagrangian subspaces? Hmm, a linear symplectic analogue of Orlov's theorem...)

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  • $\begingroup$ Thanks. I'm a little embarrassed to have missed that. I'll try to probe that to see whether anything more interesting counterexamples to my question can happen, but I think what I really want to get at is whether there are counterexamples when a Lagrangian relation AND its transpose/inverse/converse/whatever you want to call it both preserve Lagrangian subspaces. $\endgroup$
    – Brendan Fong
    Sep 19, 2014 at 3:35
  • $\begingroup$ I bet it's true under those hypotheses. Decompose $\Lambda$ as the direct sum of $A \oplus \{0\}$, $\text{graph}(\varphi)$, $\{0\} \oplus B$ for subspaces $A,B$ and $\varphi$ an isomorphism from a complement of $A$ to a complement of $B$. Then $A$ and $B$ must be isotropic. But I'm not sure how to show that $\text{graph}(\varphi)$ is isotropic. $\endgroup$
    – Nathaniel Bottman
    Sep 19, 2014 at 14:18
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    $\begingroup$ What about the graph of the map $v \mapsto 2v$ on any symplectic space $V$? Seems to identify Lagrangians and not be a Lagrangian itself... $\endgroup$
    – Max M
    Jan 14, 2015 at 15:31
  • $\begingroup$ Good point, Max. $\endgroup$
    – Nathaniel Bottman
    Jan 14, 2015 at 19:17

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