Take a closed manifold $\mathcal{L}$ and endow its cotangent bundle $T^*\mathcal{L}$ by the standard symplectic form $\omega = -d\lambda$, $\lambda$ being the Liouville form.

I was wondering if it was possible to embedd a neighbourhood of the zero section $0_{T^*\mathcal{L}}$ into another compact symplectic manifold. Or in other words, can $\mathcal{L}$ be a Lagrangian submanifold of a closed symplectic manifold.


1 Answer 1


Yes, any smooth manifold can be embedded as a Lagrangian in a closed symplectic manifold. In fact, Lisca and Matic proved that any Stein domain (like $T^*L$) embeds into a smooth projective variety with ample canonical bundle, so you can do better than just any old closed symplectic manifold. Here's a citation:

  • Lisca, P., Matić, G. Tight contact structures and Seiberg-Witten invariants. Invent. Math. 129 (1997), no.3, 509–525.

Going a little further, your question is in fact equivalent to asking whether the unit cotangent bundle admits a "symplectic cap", which can then be glued to $T^*L$ to give a closed symplectic manifold. (The not-as-obvious implication is given by the Weinstein tubular neighborhood theorem.) In fact, it is now known more generally that any contact manifold admits a symplectic cap. This was proved in three dimensions by Etnyre and Honda and in arbitrary dimensions independently by Lazarev as well as by Conway and Etnyre.

  • Etnyre, J., Honda, K. On symplectic cobordisms. Math. Ann. 323 (2002), no.1, 31–39.
  • Lazarev, O. Maximal contact and symplectic structures. J. Topol. 13 (2020), no.3, 1058–1083.
  • Conway, J., Etnyre, J. Contact surgery and symplectic caps. Bull. Lond. Math. Soc. 52 (2020), no.2, 379–394.
  • $\begingroup$ Thank you very much for being so exhaustive. I think I have some reading to catch up on as I don't see directly why domains like the unit cotangent bundle $T^*L$ are indeed a Stein domain. Your other statements seem neat ! $\endgroup$
    – S.C
    Jan 27 at 11:18
  • $\begingroup$ From the symplectic viewpoint, $T^*L$ is easily seen to be a Weinstein domain. Then one can look at the book of Cieliebak and Eliashberg "From Stein to Weinstein and Back" to turn this into a Stein domain. $\endgroup$
    – KSackel
    Jan 29 at 21:42
  • $\begingroup$ That is exactly what I needed. Thanks a lot ! $\endgroup$
    – S.C
    Jan 31 at 18:53

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