Generic choice of non-degenerate Hamiltonians $H$ in Floer theory When developing floer theory for an Hamiltonian $H:M\times S^{1}\rightarrow \mathbb{R}$ we will want $H$ to satisfy a non-degenerancy condition, that is, for every $x\in \mathcal{P}(H)$, a periodic solution of the hamiltonian system, $1$ is not an eigenvalue of $d\phi^{1}_{H}(x(0))\in GL(T_{x(0)}T^*M)$. From all the places I have been reading about Floer homology, it's said that this assumption holds for a generic choice of $H$. However I would like to see the exact statement regarding this result, is it that we have a dense set of such functions in the $C^{\infty}-$topology ? Does anyone know a reference for this ?
Any insight is appreciated, thanks in advance.
 A: You can find a statement (and proof) of such a theorem in Hofer-Salamon's Floer homology and Novikov rings, where it appears as Theorem $3.1$. They require also that no holomorphic spheres with first Chern number $\leq 1$ lie on the periodic orbit (which becomes necessary when one deals with bubbling off in the Floer-theoretic setting), but the first part of the proof pertains just to the non-degeneracy condition.
The genericity is in the sense that there is a set $\mathcal{U} \subset C_{\epsilon}^\infty(S^1 \times M)$ of the second category in the sense of Baire such that every Hamiltonian $H \in \mathcal{U}$ is non-degenerate in your sense. Here $\epsilon=(\epsilon_k)_{k=0}^{\infty}$ is a sequence of positive numbers which decrease to $0$ sufficiently rapidly, and $C_{\epsilon}^\infty(S^1 \times M)$ denotes the Banach space of smooth functions $H : S^1 \times M \rightarrow \mathbb{R}$ such that $\sum_{k=0}^\infty \epsilon_k \| H \|_{C_k} < \infty$. If the $\epsilon_k$ are chosen to decrease sufficiently rapidly, then this space is dense in $L^2(S^1 \times M)$ (see Lemma $5.1$ in Floer's The unregularized gradient flow of the symplectic action, or the proof of Proposition $8.3.1$ in Audin-Damien's book on Floer homology), and so since $\mathcal{U}$ is in particular dense in $C_{\epsilon}^\infty(S^1 \times M)$ and since $C^\infty(S^1 \times M) \subset L^2(S^1 \times M)$, $\mathcal{U}$ is also dense in $C^\infty(S^1 \times M)$.
