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Let $(M,\omega)$ be a compact $2n$ dimensional symplectic manifold and $T$ be a compact smooth $(n-1)$ dimensional manifold.

Let $CT$ be the cone over $T$, i.e. $CT = T\times [0,1] / \sim $ where $\sim$ is an equivalence relation $(x,1)\sim(y,1)$. Clearly $CT$ is a smooth manifold away from the point $[(y,1)]$.

A conical Lagrangian $C \subset M$ over $T$ is a homeomorphism $\phi:CT\to M$ whose image is $C$ and which is a smooth Lagrangian embedding away from the conical singular point.

Is there a softer version of Weinstein Theorem which says that if $C’$ is a conical Lagrangian over $T$ in a different symplectic manifold $M’$ , there are neighborhoods of $C$ and $C’$ which are symplectomorphic?

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    $\begingroup$ Let me call the manifold that is being coned as $A$ instead of $T$ for notational ease later, so $C = C(A)$. Let's say $U$ is a neighborhood of $C$ in $M$. This contains a standard Darboux ball $B_\epsilon$ around the cone point, scoop it out, note that $\partial B_\epsilon \cap C$ is a Legendrian embedded copy of $A$ in the sphere $\partial B_\epsilon$. The rest $(A \times [\epsilon, 1], A \times \epsilon) \hookrightarrow (M \setminus B_\epsilon, \partial B_\epsilon)$ is a Lagrangian w/ Legendrian boundary in a symplectic manifold w/ contact concave boundary, so there Weinstein goes through.. $\endgroup$
    – Balarka Sen
    Commented Dec 7, 2022 at 2:59
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    $\begingroup$ Given this, it seems $U \setminus B$ is $T^*(A \times [\epsilon, 1])$ and thus $U = B \cup_{J^1(A)} T^*(A \times [\epsilon, 1])$ where the subbundle $J^1 A = T^*(A \times [\epsilon, 1])|_{A \times \epsilon}$ is glued to the contact Weinstein nbhd of $A = C \cap \partial B$ in $\partial B$. From this, it at least follows that any two such neighborhoods are diffeomorphic. $\endgroup$
    – Balarka Sen
    Commented Dec 7, 2022 at 3:03
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    $\begingroup$ Namely, this is because $T^*(A \times [\epsilon, 1]) \cong J^1 A \times [\epsilon, 1]$ and so $U = B \cup_{J^1(A)} J^1(A) \times [\epsilon, 1]$. This is a special case of a more general fact: for any submanifold $N \subset M$ and tubular nbhd $\nu(N)$, $M \cup_{\nu(N)} \nu(N) \times I$ is diffeomorphic to $M$ ("melt $\nu(N) \times I$ back in $M$ using the radial coordinate of $\nu(N)$"). Given this, it seems to me that this should be true symplectically as well, where we do the melting trick with the radial coordinate given by the Liouville vector field. $\endgroup$
    – Balarka Sen
    Commented Dec 7, 2022 at 3:12
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    $\begingroup$ Corrections: (1) I meant $W \cup_{\nu(N)} \nu(N) \times I$ is diffeomorphic to $W$, where $\partial W = M$. (2) Reeb, not Liouville vector field, although these are symplectically paired. Note also that this argument says a neighborhood of any conical Lagrangian is symplectically just a ball. $\endgroup$
    – Balarka Sen
    Commented Dec 7, 2022 at 3:47
  • $\begingroup$ Ah, this seems very nice! I feel at this stage there should be some Moser-esque argument to change to symplectomorphism. $\endgroup$
    – Soham
    Commented Dec 7, 2022 at 17:39

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The answer is 'no', even in the simplest nontrivial case $n=2$, if you assume that the symplectomorphism is to take $C$ to $C'$. (If you don't assume that the symplectomorphism is supposed to take $C$ to $C'$, then I think it's 'yes'.)

In the case $n=2$, you can take $T\simeq S^1$, so $CT$ is topologically a disk $\Delta = \{z\in\mathbb{C}\,|\,|z|<1\}$. Let $M$ be a complex algebraic surface with holomorphically trivial canonical bundle, say $M = \mathbb{C}^2/\Lambda$ where $\Lambda\subset\mathbb{C}^2$ is a suitable lattice, and let your symplectic structure be given by the real part of the nonvanishing holomorphic $2$-form. Then any injective holomorphic map $\phi:\Delta\to M$ whose differential only vanishes at $z=0$ will give you a conical Lagrangian $\phi(\Delta)\subset M$, but the various different types of singularities of $\phi$ at $z=0$ will give you conical Lagrangians that are not even topologically equivalent near $\phi(0)$ because their links will be topologically distinct. For example, in local holomorphic coordinates $(w_1,w_2)$ near a point of $M$, you could have $\phi(z) = (z^p,z^q)$ where $p>q>1$ are relatively prime integers, and all of these will be distinct topologically in a neighborhood of $w_1=w_2=0$.

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  • $\begingroup$ It also fails for n = 1 with this assumption. Take two configurations of four lines through the origin in (R^2, dx ^ dy) with different cross ratios. I'd hoped the weaker version is what OP meant by "is there a softer version..." $\endgroup$
    – Balarka Sen
    Commented Dec 7, 2022 at 13:25
  • $\begingroup$ Indeed my intention was to have $C$ go to $C'$, thanks for the example! I guess for such a local result, the assumption should incorporate the link information. Say for example $C$ and $C'$ have the same Legendrian isotopy class of the link near the conical point, would it be possible to map $C$ to $C'$ ? $\endgroup$
    – Soham
    Commented Dec 7, 2022 at 17:41
  • $\begingroup$ @Soham: Well, that's a good question. Certainly, the Legendrian isotopy class of the link is an invariant, but it is not clear to me that there is anything beyond that. In some sense, you are asking whether the 'natural' radial vector field on $CT$ can be extended to a conformally symplectic vector field on a neighborhood of the singular point, or something like that. I'll think about whether that can be made precise. In any case, Balarka Sen's cautionary counterexample for $n=1$ would have to be dealt with, maybe by requiring that $C$ be connected. $\endgroup$
    – Robert Bryant
    Commented Dec 7, 2022 at 18:31

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