# Influence of symplectic invariants of the complement on being superheavy

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.

 Rigid subsets of symplectic manifolds, Entov and Polterovich

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 $$K^{c}$$.)