Influence of symplectic invariants of the complement on being superheavy

Question

Let $(M,\omega)$ be a symplectic manifold.

I'm trying to show that a compact subset $K\subset M$ is $1$-superheavy [1, Definition 1.3] where $1=PD([M])$ is the unit in $QH^0(M)$.

My question is: How do symplectic invariants of $K^c=M\setminus K$ influence this property? I'm thinking about invariants like $c_1(K^c)$, $QH^*(K^c)$ and $SH^*(K^c)$.

An interesting special case would be where $c_1(K^c)=0$ or $SH^*(K^c)=0$.

I have the feeling that the answer goes through properties of symplectic quasi-states (something like: as $SH^*(K^c)=0$ the quasi-state is only affected by the behaviour on $K$), but can't see how.

[1] Rigid subsets of symplectic manifolds, Entov and Polterovich

Answer 1

Two things I am aware of:

1. If $K^{c}$ is a finite disjoint union of stably-displaceable sets. Then $K$ is a stable stem. (This is explained in 1.2 in the paper you cited Entov and Polterovich - Rigid subsets of symplectic manifolds.) And a stable stem is super-heavy. (Theorem 1.8 in the same paper.)

2. If $M$ is Calabi–Yau or negatively monotone, then Ishikawa proved that the complement of disjoint open balls is super-heavy. (See Theorem 1.1 of Spectral invariants of distance functions for the general statement.)

Remark: these two results both support your feeling that there is a relation between $SH^{*}(K^{c})=0$ and $K$ being super-heavy, in a rough sense. (I am not sure what version of symplectic cohomology you used here for $K^{c}$.)