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When defining the Floer cohomology $HF(L_0,L_1)$ of 2 Lagrangians in a symplectic manifold $(M,\omega)$, one first has to choose some extra data such a 1-parameter family of almost complex structures $(J_t)$. Usually one requires that $J_t$ be compatible with $\omega$, ie that $g(u,v)=\omega(u,J_tv)$ defines a Riemannian metric.

However there is also the related notion of a tame $J$, one such that $\omega(u,Ju)>0$ for all nonzero $u$. My question is:

What goes wrong if we try to use tame but not necessarily compatible $J_t$ to define $HF(L_0,L_1)$?

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  • $\begingroup$ Tameness is somewhat an open condition which is easier to handle. On the other hand, $J$-holomorphic curves for compatible $J$ are minimal surfaces, which are nicer. But anyway, as Michael said in the following answer, as far as I know, there's no much difference between the two choices. $\endgroup$
    – Guangbo Xu
    Commented Nov 9, 2010 at 21:56

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As far as I know, nothing goes wrong if one uses tame instead of compatible almost complex structures (if anyone knows better please correct me). Either one of these conditions implies that the area of a holomorphic curve is bounded in terms of the integral of the symplectic form on it. This is what one needs to get Gromov compactness and to set up the Novikov ring. I don't think tameness or compatibility is needed elsewhere in the theory.

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    $\begingroup$ The linear analysis underlying a Floer theory typically involves (in the $L^2$ version) operators on $(d/ds)+A$ acting on maps $\mathbb{R}\to H$ for some Hilbert space $H$. Here $A$ should be a densely-defined symmetric operator on $H$. In Lagrangian Floer theory, this formulation arises when the inner product on $H$ is derived from the metric associated with a compatible $J$. In the tame setting one would presumably have to proceed differently. I'm not sure whether anyone has done this. $\endgroup$
    – Tim Perutz
    Commented Nov 9, 2010 at 22:58
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    $\begingroup$ Tim, what you say is correct, but can easily be worked around. For the linear analysis, you only need A to be asymptotically symmetric at the punctures, and it doesn't need to be with respect to the metric given by $\omega(\cdot, J\cdot)$. I can't find a reference right now, but I am certain I have seen the details worked out at least once in the tame case, essentially be deforming the metric near the intersections of the Lagrangians. (If you are willing to have tame except compatible in a neighbourhood of the Lagrangian intersection, then there is no need to do any work at all.) $\endgroup$
    – Sam Lisi
    Commented Nov 15, 2010 at 15:03

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