Return to Answer

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 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.

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.

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.

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 forces 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 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.

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.

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 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.

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.

This is almost never true in general. 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 forces 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 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.

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.

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 forces 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 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.

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.