A subset of a symplectic manifold is called **strongly non-displaceable** if it cannot be displaced by symplectomorphisms. A meridian in a $2$-torus is displaceable by a symplectomorphism, but not by a Hamiltonian diffeomorphism. I am reading "Rigid subsets of symplectic manifolds" by Entov and Polterovich and trying to understand Theorem 1.11 saying that the special fiber of the moment map is strongly non-displaceable. For an explicit example $S^2$ with moment map as the height function $H\colon S^2 \rightarrow [-1,1]$, the special fiber is the equator $X:= H^{-1}(0)$.


They actually prove that $X$ is **superheavy**, which means that $\zeta(K) \leq \sup\_X K$ for all smooth function $K$ on $S^2$. Here symplectic quasi-state $\zeta$ is defined as $\zeta(K):= \lim\_{r \rightarrow \infty}\frac{c(rK)}{r}$, where $c$ is the spectral invariant defined using filtered Floer homology. More explicitly,
$$c(K):= \inf \{\alpha \mid PSS([S^2]) \in HF\_\*^{\alpha}(K)\}$$
where $PSS\colon QH\_\*(S^2) \rightarrow HF\_\*(S^2)$ is the PSS-isomorphism from quantum homology to Floer homology by Piunikhin-Salamon-Schwarz. The upper index $\alpha$ indicates that elements are coming from elements whose value of the action functional is less than $\alpha$.


According to their Proposition 4.3, if a function $K$ on $S^2$ is zero on a superheavy set, then $\zeta(K) = 0$. Because I am a beginner, I tried to compute this directly for $K=H$, but I got $\zeta(H)=1$. I must be wrong somewhere but I don't know where. The following is my computation.


I follow the sign convention of Entov and Polterovich. Take a non-integer $\epsilon>0$ so that the only orbits of period $1$ of Hamiltonian $K= \epsilon H$ are fixed points $N$ and $S$. Critical points of the action functional
$$\mathcal{A}\_K(\gamma, u)= \int\_0^1 K(\gamma(t))dt - \int u^\*\omega$$
are either $(N,u)$ or $(S,u)$, where $u\colon S^2 \rightarrow S^2$ is any map. If we compute the Conley-Zehnder index, for an integer $k$ with $k-1 < \epsilon < k$, we have
$$\mu\_{CZ}(N, u)= 1+k+4m, \quad \mu\_{CZ}(S, u) = 1-k+4m$$
where $m$ is an integer representing the degree of $u$. So the differential is trivial and elements of $HF\_\*(K)$ are generated by $(N,u)$ and $(S,u)$. The image of the fundamental class by $PSS$ should be the north pole $(N, pt)$ and $\mathcal{A}\_K(N,pt)= \epsilon$. We get
$$c(\epsilon H) = \epsilon.$$
Taking $\epsilon \rightarrow \infty$, we have $\zeta(H) = 1$.


Where am I doing wrong?


**[Added]** the Conley-Zehnder index should be
$$\mu\_{CZ}(N, u)= 2k+4m, \quad \mu\_{CZ}(S, u) = 2-2k+4m,$$
so odd degree elements don't appear.



Computation of symplectic quasi-state
I don't think your computations of CZ indices are correct -- in particular, I am surprised you have a factor of $4$ showing up.


The key mistake, and the one that explains the key reason this theorem works, is that the image of the PSS map will be of a fixed grading (depending on your convention). The grading of $(N, pt)$ changes as $\epsilon \to \infty$ (i.e. as $k \to \infty$). To compensate for this, the image of the PSS map must be this North pole with a different capping. If the CZ indices are correct, this will give you an action that looks like $\epsilon - k$, which stays bounded, and thus when you divide this by $\epsilon$, the quantity goes to $0$ as desired.


