7
$\begingroup$

Let $(X,\omega)$ be a symplectic manifold and $L\subset X$ be a Lagrangian subspace. Let $\mu_L:H_2(X,L;\mathbb{Z})\to \mathbb{Z}$ be the Maslov index homomorphism.

Usual hypothesis

Recall that $L$ is said to be monotone, if there exists $c>0$ such that the following identity holds for all $\beta\in \pi_2(X,L)$: $$c \mu_L(\beta)=\omega(\beta).$$ The minimal Maslov number is defined to be: $$\inf \{\mu_L(\beta) \ |\ \beta\in \pi_2(X,L), \ \omega(\beta)>0 \}.$$ Now given two monotone Lagrangians $L_0,L_1$ call assumption $A1$ $$(A1): \text{the minimal Maslov number of $L_0$ and $L_1$ are strictly greater than 2}$$ and assumption $A2$ $$(A2): \text{ $L_0$ is Hamiltonian isotopic to $L_1$}\phantom{aaaaaaaaaaaaaaaaaaaaaaaaaaaa}$$

In defining Lagrangian intersection Floer homology groups $HF(L_0,L_1)$ usually $A1$ or $A2$ is assumed.

  1. how are these assumption exploited in the construction of the homology? and what's the role of monotonicity?
  2. Why assuming the Lagrangians to be spin it is said to simplify things? How can we use the spin assumption?
$\endgroup$

1 Answer 1

9
$\begingroup$

When you try and prove that $d^2=0$ ($d$ being the Floer differential) you need to look at the boundary of the moduli space of index 2 J-holomorphic strips with one boundary on $L_0$, one on $L_1$. Certainly one component of the boundary will consist of "broken strips", i.e. pairs of index 1 strips with a common asymptote that contribute to $d^2$. If that were everything, we'd immediately conclude $d^2=0$. However, you could bubble off a disc at the side of the strip. The index of the disc and the strip have to add up to 2, but if you have e.g. Maslov 0 discs this could happen without affecting the index of the strip. Monotonicity ensures that zero index discs have zero area, so have to be constant and so don't bubble (they wouldn't eat up any energy/wouldn't be stable).

Maslov 1 discs only occur for nonorientable Lagrangians, so let's ignore those for now. Maslov 2 discs would leave index zero strips, which have to be constant (higher Maslov discs would give negative index strips, which don't exist generically, so don't contribute). This means that $d^2x$ ends up being a multiple of $x$, the multiple being the count of Maslov 2 discs through $x$ which bubble on $L_0$ minus the number of discs which bubble on $L_1$ (discs can bubble on either side of the strip).

If $L_0=L_1$ these numbers are the same, so $d^2=0$.

If the minimal Maslov index is bigger than 2 then these numbers are both zero, so again $d^2=0$.

In general, you either restrict to Lagrangians with the same number of Maslov 2 discs through each point, or you come to terms with the fact that $d^2$ is not zero and your Fukaya category is "curved".

The spin assumption allows you to orient your moduli space of J-holomorphic strips, and hence work with coefficients over fields of characteristic other than 2 (oriented moduli spaces tell you how to assign signs to your counts of strips). Edit: there are weaker assumptions (relatively pin) that suffice, but spin is (maybe?) easier to explain.

$\endgroup$
5
  • $\begingroup$ I have a couple questions: 1. How to deal with the Maslov 1 discs in general? 2. Why are index zero strips constant? $\endgroup$
    – warzasch
    Jan 21 at 17:18
  • 1
    $\begingroup$ 1. I wish I knew; they are problematic. 2. The moduli space of strips has an R-action by reparametrisations. Dividing by this cuts the virtual dimension down to -1... unless the strips are constant, in which case the action fixes the strips and doesn't affect the virtual dimension. $\endgroup$ Jan 22 at 18:09
  • $\begingroup$ Right. Could you also explain why sphere bubbles are absent in this setting? $\endgroup$
    – warzasch
    Jan 24 at 8:12
  • 1
    $\begingroup$ Monotonicity means that any sphere class with positive area has Chern number \geq 1 and hence Maslov number \geq 2 when considered as a relative class. If you compute the virtual dimension for the moduli space of strips in some relative class A which have bubbled off a sphere in a class B, it is therefore at least 2 less than the moduli space of strips in the class A+B. So if you are only interested in moduli spaces of dimension 0 or 1 then these bubble configurations don't appear generically. $\endgroup$ Jan 25 at 9:32
  • 1
    $\begingroup$ On a side note, I try not to use MathOverflow these days because it is hosted by StackExchange and StackExchange seems to be marketing itself as a factory for high-grade AI-food (which is surely better than low-grade, but it's not something I want to participate in any more). If you have further questions about Lagrangian Floer cohomology or symplectic geometry more generally, I'd be very happy to answer them by email! $\endgroup$ Jan 25 at 9:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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