Morse Theory
Want to do Morse theory on the universal cover of a manifold. Have an infinite dimensional complex. Define $d \tilde x = \sum \tilde y$, where the sum is over lifts of gradient lines. If you fix one lift $\tilde x$ in each class, then you can reinterpret this as a complex generated by critical points but now over $R[ \pi_1(L) ]$ ($R$ is the base ring.)Floer theory
Look at image of $u(s,0)$ in differential to tell us how to interpret going between two different lifts of a given intersection point of $L, L'$. Problem: if $N_L =2$, $d^2 \ne 0$ in this case. Can bubble disks. The issue is then that(Thm, Oh) $$ d^2 x = \sum (disks) x$$, where we consider the count of Maslov 2 disks passing through $x$.
*Prop* (Oh): L monotone, $N_L = 2$, and $L, L'$ Hamiltonian isotopic. Then, the number passing through $x$ on $L$ and the number on $L'$ passing through $x$ have the same parity.
Therefore, at least $mod 2$, we have $d^2 = 0$, at least in classical Floer homology.
In this lifted picture on the universal cover, we have some trouble:
If $g \in \pi_1(L, x)$ has $g = [w( \partial D]$, then this disk on $L$ counts as $g$ and a similar disk on $L'$ counts as 1. Thus, $$d^2(\tilde x) = \sum_{g \in \pi_1(L) } (1+g) \#_2^g(x).$$
And where $\#_2^g(x)$ is the [mod 2] number of Maslov 2 disks whose boundaries are in homotopy class $g$.
In particular then, have well-defined if $N_L \ge 3$.
If not defined: Fix some $g$ so $\#_2^g = 1$. Now, take $\gamma \in \pi_1(L, x)$. We now construct some kind of cobordism to go from $g$ to $\gamma g \gamma^{-1}$. (Idea is to move the marked point in the evaluation map around the loop $\gamma$. It is important that we consider the based fundamental group and not the free one.)
Thus, the set $\{ \gamma g \gamma^{-1} \, | \, \gamma \in \pi_1(L, x) \}$ is finite. (Gromov compactness + rigidity)
Thus, $Z(g)$ is of finite index.
UPSHOT: defined/not-defined is a key tool in Mihai's work. This is a really cool variation of the usual pain we have to say something is well-defined.
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