In a Liouville manifold $M$ having a Liouville subdomain $i: N \hookrightarrow M$, there is the so-called Viterbo restriction map in symplectic cohomology $$SH^*(i): SH^*(M)\rightarrow SH^*(N).$$ In particular, given an exact Lagrangian $i_L: L\hookrightarrow M$, its Weinstein neighborhood $T^*L$ is a Liouville subdomain of $M,$ hence there is a map $$SH^*(i_L):SH^*(M)\rightarrow SH^*(T^*L).$$ Is there a statement that says that, under certain conditions on $L$, the map $SH^*(i_L)$ is surjective? The examples which I have in mind are when $M$ is Weinstein and $L$ is a component of its Liouville skeleton.

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    $\begingroup$ This is already false for the inclusion of a circle in a punctured genus 1 surface. What kind of condition do you have in mind? $\endgroup$ Jan 12, 2020 at 19:55
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    $\begingroup$ Hmm, that is true -- maybe I should be a bit more restrictive, then. I had in mind a setup of holomorphic symplectic manifold $M$ with a Lagrangian skeleton $L$ which is a holomorphic projective variety. Its irreducible components $L_i$ are exact holomorphic Lagrangians, if smooth. In particular, the ambient M could be even hyperkahler, e.g. ADE plumbings of $T^*S^2$. Here the skeleta are ADE trees of Lagrangian $S^2$-spheres, and I think (though I did not prove) that this surjectivity statement is true for those spheres. $\endgroup$
    – Filip
    Jan 12, 2020 at 22:34
  • $\begingroup$ ++ The fact that makes me think this surjectivity to be true in these ADE plumbings examples is that the rank of $SH^i(M)$ is exactly the sum of the ranks of $SH^i(T^*S^2)$, summing by all components of skeleton, for all $i\in \mathbb{Z}$, hence one may think that Viterbo restrictions are sending different subspaces of $SH^*(M)$ isomorphically to $SH^*$ of Weinstein neighborhoods of different spheres. $\endgroup$
    – Filip
    Jan 13, 2020 at 13:10

1 Answer 1


This is almost never true in general, although it's obviously true for boundary connected sums. For example, it is usually the case that Weinstein handle attachment will kill (non-trivial) invertible elements in $\mathit{SH}^0(T^\ast L)$, so the corresponding Viterbo restriction map can never be surjective. The counterexample proposed by Mohammed belongs to such a case. More generally, I think $L$ should never be a torus, because otherwise there are a lot of non-trivial units in $\mathit{SH}^0(T^\ast T^n)$ to be killed by attaching handles.

This would force you to look at Weinstein manifolds $M$ which do not contain exact Lagrangian tori, which is the case, for example, when $M$ admits a dilation in the sense of Seidel-Solomon. More generally, one could consider smooth affine varieties $M$ with log Kodaira dimension $-\infty$, which should be manifolds whose first Gutt-Hutchings capacities are finite. For Milnor fibers, this means that the configuration of vanishing cycles is very "sparse", in other words, it's "close" to a boundary connected sum of $D^\ast S^n$'s.

For the case of 4-dimensional $A_m$ Milnor fibers, explicit calculations of Hochschild cohomologies imply that


so it really makes sense to expect that the Viterbo restriction map is actually surjective. Similar computations can be done in the $D_m$ and $E_m$ case under the assumption that $\mathrm{char}(\mathbb{K})=0$.

Also a simple observation is that $1\pm x\in\mathit{SH}^0(M)^\times$ doesn't get killed under obvious handle attachments in these examples, so I think probably what you should look at is under which circumstances non-trivial units in $\mathit{SH}^0(M)$ are preserved under handle attachments. Unfortunately the description of the product structure on $\mathit{SH}^\ast(M)$ under Legendrian surgery is quite involved, so it's difficult to extract a simple and explicit condition. I'll update this answer once I have any new ideas.


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