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.

  • $\begingroup$ How do you compute the Conley-Zehnder index concretely? Do you have a worked out example? $\endgroup$
    – sab
    Jan 14 at 10:20
  • 1
    $\begingroup$ @sab: If I remember correctly, I consulted a paper by Alexandru Oancea when I posted this question. See Section 3 and references therein. $\endgroup$
    – Hwang
    Jan 17 at 0:38

1 Answer 1


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.

  • $\begingroup$ Thanks a lot. I didn't see that the image of 1 by PSS should change as $\epsilon$ varies. For CZ indices, the computation seems to be OK for me. 4 appears because it is twice the first Chern number of $S^2$, which is the difference for different cappings. $\endgroup$
    – Hwang
    Mar 17, 2014 at 0:38
  • $\begingroup$ Oh I am sorry, you are right. My computations of CZ indices are wrong. I should have put $2k-1$ instead of $k$. $\endgroup$
    – Hwang
    Mar 17, 2014 at 7:22
  • $\begingroup$ In actual fact, we were both wrong with our CZ indices originally :-) I'm glad this got you on the right track despite that. $\endgroup$
    – Sam Lisi
    Mar 17, 2014 at 13:12

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