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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

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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

$\mathit{SH}^0(M)\cong\mathbb{K}[x]/(x^{m+1}),$

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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