Let $L\subset T^\ast S^n$ be a properly embedded Lagrangian submanifold homeomorphic to $\mathbb{R}^n$ with respect to the canonical symplectic structure on $T^\ast S^n$, and suppose $L$ intersects the zero section $S^n$ at exactly one point. Is it necessarily true that $L$ is Hamiltonian isotopic to a cotangent fiber?

  • $\begingroup$ You probably had this in mind when asking, but one should probably add some assumptions about the isotopy being well-behaved near infinity. Otherwise you can just scale $L$ outwards in the fiber direction (this stays Lagrangian) and it will approach a cotangent fiber. $\endgroup$ Jul 22, 2014 at 7:55
  • $\begingroup$ @Pardon Yes, you're correct. $\endgroup$
    – Acky
    Jul 22, 2014 at 8:08
  • $\begingroup$ This is not exactly your question, but may be of interest. If we assume our Lagrangian L to be conical at infinity and to agree with a cotangent fiber, then one can prove that in the wrapped Fukaya category, L is isomorphic to a cotangent fiber (shifted by some integer which depends on grading data). This is similar to what happens in recent attacks on the nearby Lagrangian problem. So, at first glance, it seems plausible to ask your question with, say, compactly supported Hamiltonian isotopies. $\endgroup$ Jul 24, 2014 at 2:33
  • $\begingroup$ @Pomerleano How to show that $L$ is isomorphic to a cotangent fiber? If $L$ is isomorphic to a cotangent fiber, then $L$ has to generate the wrapped Fukaya category. Is this true only for cotangent bundle of spheres or for all cotangent bundles of a closed manifold? $\endgroup$
    – Acky
    Jul 26, 2014 at 7:40
  • $\begingroup$ @ Acky It follows from Koszul duality. Basically, assume M is simply connected. Then we have a fully faithful functor from perfect $C_*(\Omega M)$ modules to homology finite modules over $C^*(M)$. Geometrically that functor can be thought of as taking Hom from the zero section to your given object. $\endgroup$ Aug 3, 2014 at 1:28

1 Answer 1


Let $\tau \in \mathrm{Symp}^c(T^*S^n)$ denote the generalised Dehn twist. Seidel has proven using Floer homology that $\tau^k(T_x^*S^n)$, $k \neq 0$, never is compactly supported Hamiltonian isotopic to the fibre $T_x^*S^n$. In some cases (e.g. $n=2$) there are no classical obstructions to this.

You could ask the question under the additional assumption that the Lagrangian is contained in an arbitrarily small neighbourhood of the fibre $T^*_xS^n$; in this situation the answer to the problem is (to my understanding) known only in the cases $n=1,2$, where the case $n=2$ is due to Eliashberg-Polterovich.


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