In Hamiltonian dynamics and symplectic geometry a twisted cotangent bundle is the cotangent space $T^*N$ of a closed (compact without boundary) $n$-manifold $N$ equipped with a twisted symplectic structure: $T^*N$ carries the canonical symplectic structure $\omega=d\lambda$, where $\lambda$ is the Liouville 1-form. One can "twist" $\omega$ by adding a closed two-form $\sigma$ on $N$ as follows: $$ \omega_{\sigma}:=\omega + \pi^*\sigma. $$ Here $\pi:T^*N \to N$ denotes the footpoint map. It is easy to check that $\omega_{\sigma}$ is symplectic. Twisted cotangent bundles play an important role in Hamiltonian dynamics, but I am here interested in their symplectic topology. Many classical questions in symplectic topology concern the closed Lagrangian submanifolds of $(T^*N,\omega)$. But what about closed Lagrangian submanifolds in $(T^*N,\omega_{\sigma})$? Does anyone know a non-trivial example (meaning $\sigma$ is not exact) where $(T^*N,\omega_{\sigma})$ contains closed Lagrangian submanifolds with "good properties" (say weakly exact, monotone etc.)? Are any general statements known? Any examples, ideas, references or proofs will be highly appreciated!

It is easy to find non-compact Lagrangians in $(T^*N, \omega_{\sigma})$: If $X\subset N$ is a submanifold such that $\sigma|_{X}=0$ then its conormal space $$ \nu^*(X):=\{ p\in T^*N\ |\ p|_{TX}\equiv 0 \}\subset T^*N $$ is a non-compact Lagrangian submanifold. My questions therefore concerns closed Lagrangian submanifolds! It is easy to find closed Lagrangians when $\sigma$ is exact. Hence, my interest is really in the case when $\sigma$ is not exact.

Thanks in advance!

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    $\begingroup$ I would start with considering the graphs of one forms. For $\sigma=0$ the graph of a one-form is Lagrangian iff it is closed. If $\sigma$ is exact, you can see that the condition that the graph is lagrangian is exactly the condition that the one-form is a primitive of $\sigma$. These Lagrangians are of course all diffeomorphic to the base. $\endgroup$
    – Thomas Rot
    Nov 24, 2017 at 12:58
  • $\begingroup$ @ThomasRot Thanks. I agree that as long as $\sigma$ is exact it is quite easy to construct closed Lagrangians. One way to look at this is that $\omega_{\sigma}$ is an exact perturbation of $\omega$, so it should be symplectomorphic to it by Moser's argument. The question I am interested in is the case when $\sigma$ is not exact. In this case it doesn't seem so easy to me. $\endgroup$
    – MBIS
    Nov 24, 2017 at 16:14
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    $\begingroup$ After turning on the magnetic potential, it should still be true that a cotangent fiber generates the wrapped Fukaya category $\mathcal{W}(T^\ast N,\omega_\sigma)$, see the introduction part of the following paper: arxiv.org/pdf/1201.5880.pdf. Because of this, the classification of monotone Lagrangians is equivalent to the problem of classifying proper $A_\infty$-modules over the endomorphism algebra of a cotangent fiber. $\endgroup$
    – YHBKJ
    Nov 24, 2017 at 21:34
  • $\begingroup$ @YHBKJ thanks! This sounds interesting. I will definitely have a look. $\endgroup$
    – MBIS
    Nov 28, 2017 at 9:23
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    $\begingroup$ @MadsR.Bisgaard It's only a prediction, which tells you which kind of monotone Lagrangians are not allowed, but does not say anything about existence existence. Moreover, it's usually not easy to classify $A_\infty$-modules if the endomorphism algebra of a cotangent fiber is not formal and trivially graded. $\endgroup$
    – YHBKJ
    Nov 28, 2017 at 13:05

1 Answer 1


Let's begin by pointing out the following: you will not find monotone examples for the simple reason that a nontrivial such deformations creates a class of nonzero symplectic area, while the Chern class is always vanishing. The best you could hope for is Calabi-Yau, and such examples indeed exist. However, I know of no examples where a closed Lagrangian has a nonvanishing Floer homology.

Now a general observation: In the case when $T^*N \setminus N$ has no second cohomology with $\mathbb{R}$-coefficients, e.g. if $N=S^2,$ then for small closed forms $\sigma$ one can use Moser's trick to show that any compact Lagrangian submanifold of $T^*N \setminus N$ is preserved (up to smooth isotopy) after turning on a sufficiently small magnetic potential.

A more concrete example: taking $\sigma$ to be the area form on $S^2,$ we obtain the total space of the line bundle $\mathcal{O}(-2)$ on $\mathbb{C}P^1$ with its standard Kähler form. (The first reference coming to my mind is 2.4A in [Y. Eliashberg and L. Polterovich; Unknottedness of Lagrangian surfaces in symplectic 4-manifolds] but maybe there is something more to the point). Unlike $T^*S^2$, the latter symplectic manifold is an open toric Calabi-Yau manifold. Unfortunately, according to Theorem 5 in [Ritter; Floer theory for negative line bundles via Gromov-Witten invariants], its symplectic homology vanishes: this twisted cotangent bundle therefore contains no Lagrangians with interesting Floer homology. See [Ritter-Smith; The monotone wrapped Fukaya category and the open-closed string map] where a closed-open map is constructed in this setting.

A side note: if you compactify a subset of the total space of $\mathcal{O}(-2)$ to the Hirzebruch surface $F_2(\alpha)$ as studied in [Fukaya-Ohta-Ono-Oh; Toric degeneration and non-displaceable Lagrangian tori in $S^2\times S^2$] then Fukaya category actually becomes nontrivial.

  • $\begingroup$ Thanks a lot! This comes very close to the answer I was hoping for and it is definitely very helpful. I have one question concerning your example: Why does it follow from Ritter's result that the twisted $T^*S^2$ contains no Lagrangian $L$ with non-trivial Floer homology? I suppose the argument is that $HF(L)$ is a module over symplectic homology of the ambient symplectic homology. I know that this is true for the ambient quantum homology, but I have never seen this being discussed for SH. If this is the idea, do you have a reference? $\endgroup$
    – MBIS
    Dec 11, 2017 at 7:33
  • $\begingroup$ That is exactly the point! I added a reference. $\endgroup$
    – Nikolaki
    Dec 11, 2017 at 8:04
  • $\begingroup$ Great! Thanks a lot. This is definitely useful to me! $\endgroup$
    – MBIS
    Dec 11, 2017 at 11:57

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