Comparison between Hamiltonian Floer cohomology and Lagrangian Floer cohomology of the diagonal

Question

Let X be a compact symplectic manifold with a form $\omega$. And $X \times X$ is equipped with the symplectic form $(\omega,-\omega)$. The diagonal $\Delta:X \mapsto X \times X$ is a Lagrangian submanifold. So, in this question, Hochschild (co)homology of Fukaya categories and (quantum) (co)homology. Tim Perutz says "PSS is a canonical ring isomorphism from QH∗(X) to the Hamiltonian Floer cohomology of X, and the latter can be compared straightforwardly to the Lagrangian Floer cohomology of the diagonal." I have no doubt that this second assertion is straightforward, since I have consulted a couple of references and no one spells this out. But, I don't quite see it. I believe what I am missing is the relationship between holomorphic strips in $X\times X$ and holomorphic cylinders in X. (Edit: I would also like to understand the comparison of the product structures too)

Edit: here is a rough geometric idea which might have something to do with the truth. I want to assume that my Hamiltonian is time independent and that all orbits are actually fixed points. Given a map of a strip into $X\times X$ the two projections give us two strips into X. The idea is to glue the strips together to form a cylinder which is a map into X. Of course, this doesn't take into account issues of compactifications and so on... Anyways, if someone would be happy to spell it out I would appreciate it.

Answer 1

Let $f:X\to X$ be a (Hamiltonian) symplectomorphism. The claim is that the fixed point Floer homology of $f$ agrees with the Lagrangian intersection Floer homology of the graph of $f$ with the diagonal in $X\times X$. I think the argument is that if we choose almost complex structures for the two Floer theories in the following way then the two chain complexes are isomorphic. Or at least, the corresponding holomorphic curves agree. (I'm ignoring issues of transversality, which coefficient ring to use, etc.)

The Floer homology of $f$ is the homology of a chain complex which is generated by fixed points of $f$ and whose differential counts maps $u:{\mathbb R}\times {\mathbb R}\to X$ such that $u(s,t+1)=f(u(s,t))$ and $(\partial_s + J_t\partial_t)u=0$ where $J_t$ is a family of $\omega$-compatible almost complex structures on $X$ parametrized by $t\in{\mathbb R}$ such that $J_{t+1}=f_*\circ J_t\circ (f^{-1})_*$.

The Lagrangian Floer homology of the graph and the diagonal is the homology of a chain complex which is generated by intersection points and whose differential counts maps $v=(v_-,v_+):{\mathbb R}\times[0,1]\to X\times X$ such that $v_-(s,0)=v_+(s,0)$, $v_+(s,1)=f(v_-(s,1))$, and $(\partial_s + \tilde{J}_t\partial_t)v=0$ where $\tilde{J}_t$ is a family of $-\omega\oplus\omega$-compatible almost complex structures on $X\times X$ parametrized by $t\in[0,1]$.

To relate these, we first observe that there is an obvious bijection between the generators of the chain complexes. To get the holomorphic curves to match up, first choose the family $J_t$, then define

$\tilde{J}_t = (-J_{\frac{1-t}{2}})\oplus J_{\frac{1+t}{2}}$.

Now given a holomorphic cylinder $u$ as above, you can cut it into two halves to get a holomorphic strip $v$:

$v(s,t)=(u(\frac{s}{2},\frac{1-t}{2}),u(\frac{s}{2},\frac{1+t}{2})).$

Conversely, given a holomorphic strip $v$, one can glue together its two components $v_-$ and $v_+$ to get a holomorphic cylinder $u$ where $u(s,t)=v_-(2s,1-2t)$ for $t\in[0,1/2]$ and $u(s,t)=v_+(2s,2t-1)$ for $t\in[1/2,1]$. Because of the boundary conditions this is at least $C^1$ where we glue the pieces together, and so by elliptic regularity it is actually smooth.

Answer 2

Michael's answer describes the sort of thing I had in mind when I brusquely dismissed this as "straightforward" in the earlier, big-picture-style answer that Daniel quoted. (Michael, thanks for explaining.)

This isomorphism intertwines the product structures: think of cutting a pair of pants into two identical pieces.

I'd like to add a cautionary note. Before you can compare Hamiltonian Floer homology to the Lagrangian Floer self-homology of the diagonal, you have to define both groups. The difficulties in doing so are not identical. In the Lagrangian case, disc-bubbling potentially obstructs the construction.

For particularly simple manifolds - monotone ones, where $c_1$ is a positive multiple of $[\omega]$ when evaluated on spherical homology classes - both groups have reasonably straightforward definitions. This is a setting where the isomorphism really works.

For Calabi-Yau manifolds, the definition of Hamiltonian Floer homology is still not too complicated (it involves understanding transversality of evaluation on moduli spaces of somewhere-injective holomorphic spheres). But Lagrangian Floer homology of a Maslov-index-zero Lagrangian in a Calabi-Yau is hard to construct. So far as I know, one generally has to work rationally and to compute obstructions in the sense of Fukaya-Oh-Ohta-Ono. Presumably, in the case of the diagonal these obstructions vanish.

So there is a definite reason for preferring the Hamiltonian Floer homology model when trying to formulate things precisely: it bypasses the obstructions, and (for Calabi-Yau manifolds) the virtual transversality theory.