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I am trying to solve the following exercise:
Let $(M,\omega)$ be a symplectic manifold and $L$ a compact Lagrangian submanifold such that $H^{1}(L)=0$. Let $\{L_{t}\}_{t\in(-1,1)}$ be a smooth family of Lagrangian submanifolds of $(M,\omega)$ with $L_{0}=L$. Does there exist $\epsilon>0$ such that for all $t\in(-\epsilon,\epsilon)$: $L\cap L_{t}\neq\emptyset$?

So far, all counterexamples I could find have nontrivial $H^{1}$, like for instance a family of circles in $T^{*}S^{1}\cong S^{1}\times\mathbb{R}$. This makes me believe that the statement is true. Any suggestions how to prove it?

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    $\begingroup$ Do you know the Lagrangian tubular neighborhood theorem? $\endgroup$
    – Jack Lee
    Apr 17, 2016 at 16:28
  • $\begingroup$ I think so. The theorem that states that there exist neighborhoods $U$ of $L$ in $M$, and $V$ of $L$ in $T^{*}L$ and a symplectomorphism $\phi: U\rightarrow V$ that is identity on $L$? $\endgroup$
    – studiosus
    Apr 17, 2016 at 16:52
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    $\begingroup$ Yes, that's true. Now think of a family of Lagrangians $L_t$ in the cotangent buundle, $L_0$ the zero section. For small $t$, these are Lagrangian sections. Can you prove that assertion, and use this to do the exercise? $\endgroup$
    – user98602
    Apr 17, 2016 at 17:10
  • $\begingroup$ Thank you. I don't quite see why the $L_{t}$ are sections, but I believe I can solve the exercise assuming that: As $L_{t}$ is a lagrangian section, it must be the graph of a closed 1-form on L. Since $H^{1}(L)=0$, this is even an exact form, so $L_{t}=graph(df_{t})$. As $L$ is compact, $f_{t}$ has to reach extremal value on $L$, hence $df_{t}=0$ at some point. So $L_{t}$ intersects the zero section, which is what we had to show. $\endgroup$
    – studiosus
    Apr 17, 2016 at 18:03
  • $\begingroup$ @Almagast Good work. I wrote out the details of why $L_t$ is a section below. $\endgroup$
    – user98602
    Apr 17, 2016 at 18:15

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By the symplectic tubular neighborhood theorem, there is a neighborhood $U$ of $L_0$, symplectomorphic to $T^*L_0$ by a symplectomorphism that takes $L_0$ to the zero section of $T^*L_0$. For sufficiently small $t$, $L_t \subset U$ (this is nothing more than the fact that $L_t$ varies continuously: the map $f_t: L \times I \to M$ has $f_t^{-1}(U)$ open, and hence by the so-called tube lemma, and the compactness of $L$, there is an open set of $t$ including zero such that $L_t \subset U$.

A similar argument will prove that, for small $t$, $L_t$ is a section of the cotangent bundle: for consider the projection $\pi \circ f_t : L \times I \to L_0$. For $t=0$, this is a diffeomorphism (the identity). By continuity of the derivative, $f_t$ is also a submersion for small $t$ (and hence a diffeomorphism, since it's homotopic to the identity). That this projection is a diffeomorphism means we can identify $L_t$ as the image of a section of the cotangent bundle; call this section $\alpha_t$.

Now Lagrangian sections of the cotangent bundle can be identified with closed forms. Now we invoke $H^1(L) = 0$; $\alpha_t = dg_t$ for some $g_t$. (One could take that to be continuous in $t$, but that doesn't actually matter unless you're interested in how the intersection points move around.) Now $g_t$ has a maximum somewhere ($x$, say) on $L_0$, and hence at that point $0 = dg_t(x) = \alpha_t(x)$. Hence $L_t \cap L_0$ is nonempty: it contains $x$.

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  • $\begingroup$ Thank you for your detailed answer. There is one last thing: why is a submersion that is homotopic to the identity a diffeomorphism? $\endgroup$
    – studiosus
    Apr 17, 2016 at 21:11
  • $\begingroup$ @Almagast It's a covering map, by compactness of the base. Nontrivial covering maps do not induce the identity map on the fundamental group, whereas our map does. This is just one approach; another says that for small $t$, there's still only one point in the fiber, and hence the covering map is a diffeomorphism (as opposed to some n-to-1 map). $\endgroup$
    – user98602
    Apr 17, 2016 at 21:13

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