# 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.

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 $$C^\infty(S^1 \times M)$$.