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 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?

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?