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I'm looking for the precise definition of "Lagrangian skeleton", as I'm eventually going to give a talk on this topic. As I asked a professor in my university about references on Lagrangian skeleta, he listed the following:

As I googled "Lagrangian skeleton", the only useful information I've found is this post of MO Has anything precise been written about the Fukaya category and Lagrangian skeletons? in which Ben Webster said the following:

As I understand it, a Lagrangian skeleton is a union of Lagrangian submanifolds which a symplectic manifold retracts to.

Two references above never used the term "Lagrangian skeleton", but I have often seen "skeleton of a Morse function", which is the union of all stable manifolds. Is this what Lagrangian skeleton is? Or is Webster's definition the right/standard one? In fact, I haven't seen Ben's definition in the papers. If the latter is the case, which concept in the above papers correspond to Lagrangian skeleton? I'm 100% confident that it is in the papers, especially the latter one, which is, according to the professor, the canonical text on the topic.

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  • $\begingroup$ a symplectic manifold surrounds $\endgroup$
    – JMP
    Commented Sep 28, 2016 at 3:27
  • $\begingroup$ A simplest concrete example is given by the union of zero sections in a plumbing of cotangent bundles, or the union of vanishing cycles in a Milnor fiber of any simple singularity. $\endgroup$
    – YHBKJ
    Commented Oct 30, 2016 at 0:03

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Indeed the Cieliebak-Eliashberg reference is canonical. I summarize some things here:

The notion of skeleton in symplectic geometry is generally used in the exact setting, i.e. the symplectic form is $\omega = d\lambda$. Note that in this case the symplectic manifold $M$ must be noncompact.

Having fixed the primitive $\lambda$, one forms the Liouville vector field $Z$ by demanding $\omega(Z, \cdot) = \lambda$. In this case one says that the skeleton is the locus in $M$ which does not escape every compact set under the flow of $Z$. One generally restricts further to the ``Liouville'' setting where there is a convex contact hypersurface $V$ whose flow under $Z$ exhausts the non-compactness of the symplectic manifold; in this case the flow gives a retraction from $M$ to the skeleton.

The relation to the Morse theoretic notion is that one often asks moreover for a ``Weinstein structure'', i.e. a Morse function, convex with respect to some compatible almost-complex structure, with respect to which $Z$ is gradient-like. In this case one can show (easily) that the skeleton is necessarily isotropic.

One reason for some of the confusion regarding the definitions in this area is the surprisingly large number of open (to my knowledge) questions about really basic things. Here are some examples. In the first two I take $M$ to be a Liouville manifold, i.e. I have fixed a primitive for the symplectic form and the manifold is asymptotically a cylinder on a contact manifold.

(1) Suppose the skeleton is Lagrangian. Does there exist a Weinstein function with respect to which the Liouville flow is gradient like?

(2) Suppose given a 1-parameter family $L_t$ of Lagrangian subvarieties of M, with $L_0$ the skeleton determined by $\lambda$. When does there exist a family $\lambda_t$ of 1-forms for which these are the skeleta?

(3) Now let $(M, \omega)$ be any symplectic manifold, and let $L \subset M$ be a singular Lagrangian. When is it true that $L$ admits a neighborhood $W$ such that $(W, \omega)$ admits a Weinstein structure for which L is the skeleton?

[Mathematical pun: an answer to the 3rd question above would be a ``Weinstein neighborhood theorem'']

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  • $\begingroup$ After reading your post, everything made sense. I realized my silly misunderstanding. I was thinking that Lagrangian skeleton was something totally new rather than just a skeleton which is Lagrangian. $\endgroup$ Commented Sep 28, 2016 at 6:10
  • $\begingroup$ In your first question, why can't the skeleton be isotropic? $\endgroup$
    – YHBKJ
    Commented Nov 17, 2016 at 0:26
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    $\begingroup$ Sure, it can, but anyway if you can find a function for an isotropic skeleton, then by perturbing you can find one for a Lagrangian skeleton. $\endgroup$ Commented Nov 17, 2016 at 4:05
  • $\begingroup$ @VivekShende Hi, do you know of any references for the claim that one can get a Lagrangian skeleton by perturbing the Lyapunov function? $\endgroup$
    – Paul
    Commented Apr 1, 2022 at 4:27

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