# Linearization of the Floer equation

In Floer Homology we want to prove that the Moduli spaces $$\mathcal{M}(x^{-},x^{+})$$ are finite dimensional manifolds. This is done by expressing them as the zero set of a Fredholm map. First one proves that the set $$\mathcal{P}^{1,p}(x^{-},x^{+})$$ of maps satisfying exponential decay conditions and the boundary conditions $$\lim_{s\rightarrow \pm \infty}u(s,\cdot)=x^{\pm}$$ is a Banach manifold . And then we have that $$\mathcal{M}({x^{-},x^{+}})$$ is the zero set of the map $$\mathcal{F}^{H,J}_u(Y):=\Phi_u(Y)^{-1}\mathcal{F}(\exp_u Y)$$ where $$Y$$ is a vector field along $$u$$ and $$\mathcal{F}$$ is the Floer equation.

I have been wondering for a while why we use this map $$\mathcal{F}^{H,J}_u$$. We will have that it's linearization will be a nice Fredholm operator since we can use Unitary trivializations of the bundles $$u^*(TM)$$ to show that $$d(\mathcal{F}_{u}^{H,J})_0$$ is conjugated to an operator of the form

$$D_u: W^{1,p}(\mathbb{R}\times S^1,\mathbb{R}^{2n})\rightarrow L^p(\mathbb{R}\times S^1,\mathbb{R}^{2n})$$ $$Y\mapsto \partial_sY+J_0\partial_t Y+S(s,t)Y$$

and these operators are very well understood and in particular we can compute it's Fredholm index. Is there any other reason other than this to why we use this map $$F^{H,J}_u$$?

I am asking this because if we want extra conditions on our Moduli spaces of maps $$\mathcal{M}(x^{-},x^{+})$$, which we call this space $$\mathcal{M}^k$$, such that we can prove that $$\mathcal{P}^{1,p,k}$$ is a banach manifold then I am not sure we can use the map $$\mathcal{F}^{H,J}_u$$ to prove that $$\mathcal{M}^k$$ is a banach manifold since this might not be the zero section of $$\mathcal{F}^{H,J}_u$$ due to the fact that $$\exp_u Y$$ might not need to verify the new boundary condition. For example the boundary condition could be a Lagrangian boundary conditions an in Lagrangian Floer Homology.

I am not sure if in this case we need to pick a metric $$\exp$$ so that all of this works or if there is another thing going on .

Any insight is appreciated, thanks in advance.

$$\mathcal{F}^{H,J}_u$$ is exactly the Floer operator (what you have called $$\mathcal{F}$$ in your post) locally near the cylinder $$u$$ in $$\mathcal{P}^{1,p}(x^-,x^+)$$, after we use the underlying metric & parallel transport to identify this local neighbourhood of $$u$$ with a neighbourhood of zero in the tangent space $$T_u \mathcal{P}^{1,p}(x^-,x^+) \simeq W^{1,p}(\mathbb{R} \times S^1; \mathbb{R}^{2n})$$ (where this latter identification is provided by recognizing that $$T_u \mathcal{P}^{1,p}(x^-,x^+) = W^{1,p}(u^*TM)$$ and then identifying $$W^{1,p}(u^*TM)$$ with $$W^{1,p}(\mathbb{R} \times S^1; \mathbb{R}^{2n})$$ via a trivialization of $$TM$$ along $$u$$).
A little more explicitly, the map $$exp_u: T_u \mathcal{P}^{1,p}(x^-,x^+) \rightarrow \mathcal{P}^{1,p}(x^-,x^+)$$ restricts to a diffeomorphism from some neighbourhood of $$0 \in T_u \mathcal{P}^{1,p}(x^-,x^+)$$ onto a neighbourhood of $$u \in \mathcal{P}^{1,p}(x^-,x^+)$$. So $$\mathcal{F}(exp_u(Y))$$ provides the value of the Floer operator at $$v=exp_u(Y)$$ a cylinder near $$u$$, and this resides in $$T_v \mathcal{P}^{1,p}(x^-,x^+)$$, but we would like to work with a fixed codomain for any $$Y \in T_u \mathcal{P}^{1,p}(x^-,x^+) \simeq W^{1,p}(\mathbb{R} \times S^1; \mathbb{R}^{2n})$$, so we use parallel transport by an appropriate connection to transport everything back to live in $$T_u \mathcal{P}^{1,p}(x^-,x^+)$$. This gives us $$\mathcal{F}^{H,J}_u=\Phi_u(Y)^{-1}\mathcal{F}(exp_u(Y))$$.
So somehow the answer as to why we look at $$\mathcal{F}^{H,J}_u$$ is because we're interested in $$\mathcal{F}$$, and $$\mathcal{F}^{H,J}_u$$ simply is $$\mathcal{F}$$ locally near $$u$$, once we've made the appropriate identifications so that we can view it as a non-linear operator between Banach spaces.
As for incorporating Lagrangian boundary conditions, this is done at the level of the mapping space that you're studying (so the conditions are imposed when you define the analogue of $$\mathcal{P}(x^-,x^+)$$, say $$\mathcal{P}^{1,p,k}$$ and then you check directly that the extra boundary conditions still give you that the ambient space $$\mathcal{P}^{1,p,k}$$ that you're working in is a Banach manifold just as you normally directly check that the exponential decay conditions and boundary conditions that you impose in defining $$\mathcal{P}^{1,p}(x^-,x^+)$$ give you a Banach manifold), but once that's done the abstract picture I described above is essentially the same.