# Viterbo restriction map surjective on Weinstein neighbourhood

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.

• 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? Jan 12, 2020 at 19:55
• 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. Jan 12, 2020 at 22:34
• ++ 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. Jan 13, 2020 at 13:10

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.