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


1 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 $K^{c}$.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.