Fredholm property of linearization of Floer map

Question

I am reading Audin and Damian's book "Morse theory and Floer homology". In Proposition 8.1.4 which reveals the transversality property of moduli space of solutions of Floer equation, the authors define an operator $$ \begin{aligned} \begin{aligned}\Gamma:W^{1,p}(\mathbf{R}\times S^1;\mathbf{R}^{2n})\times C_\varepsilon^\infty(H_0)\end{aligned}& \longrightarrow L^p(\mathbf{R}\times S^1;\mathbf{R}^{2n}) \\ (Y,h)& \longmapsto(dF^H)_u(Y)+\operatorname{grad}_uh. \end{aligned} $$ Here $C_\varepsilon^\infty(H_0)$ is the space of perturbations of Hamiltonian function $H $, F is the Floer map $$\begin{aligned}&F:C^\infty(\mathbf{R}\times S^1;W)\longrightarrow C^\infty(\mathbf{R}\times S^1;TW)\\&u\longmapsto\frac{\partial u}{\partial s}+J\frac{\partial u}{\partial t}+\mathrm{grad}_u(H_t).\end{aligned} $$

My question is the exercise 44 of this book. Why is the kernel of $\Gamma$ is not finite-dimensional? I know that the difference between finite-dimensional space and infinite-dimensional space is that unit ball in infinite-dimensional space is not compact. But I don't know how to show it in the exercise. Any hint is welcome!

Answer 1

They show in Section 8.7.c that $(dF^H)_u$ is a Fredholm operator.

What remains is simple linear algebra (even continuity is irrelevant). Suppose $A: V \times W \to U$ is a linear map, with $B = A|_V$ a Fredholm map and $C = A|_W$ an arbitrary linear map. Then $\ker(A)$ is infinite-dimensional if and only if $W$ is infinite-dimensional.

For consider the projection map $p: \ker(A) \to W$ sending $(v,w)$ to $w$. An element of the domain has $A(v,w) = 0$, so $Bv = -Cw$. Thus $\ker(p) = \ker(B) \times \{0\}$.

Suppose $C$ has finite-dimensional kernel, then $\text{im}(p) = C^{-1}(\text{im}(B))$ has finite codimension in $W$. Because $W$ is infinite-dimensional, it follows that $\ker(A)$ is infinite-dimensional.

Now suppose $\ker(C)$ is infinite-dimensional. Then $0 \times \ker(C) \subset \ker(A)$, so $\ker(A)$ is infinite-dimensional.

On the other hand, if $W$ was finite-dimensional, it is easy to see $A$ remains Fredholm.