# Pullback of Morse form satisfies Palais Smale

Let $(\alpha,g)$ be a Morse-Smale pair on a closed smooth manifold $M$, i.e. $\alpha$ is a Morse form and $g$ a Riemannian metric on $M$ such that stable and unstable manifolds of the gradient vector field $X$ intersect transversally.

Let $\pi \colon \tilde{M} \to M$ be the associated cover to $\ker [\alpha]$. Denote by $\tilde{f} \colon \tilde{M} \to \mathbb{R}$ a primitive of the exact pullback $\pi^* \alpha$ and set $\tilde{g}=\pi^*g$.

It is clear that $(\tilde{f},\tilde{g})$ is a Morse-Smale pair on the cover $\tilde{M}$ and it is also clear that $\tilde{M}$ is non-compact whenever the Morse form $\alpha$ is non-exact. This leads to the natural question:

Does $\tilde{f} \colon \tilde{M} \to \mathbb{R}$ satisfy the Palais-Smale condition in the non-exact case?

For various reasons I'd like this to be true, but I'm not able to find a proof for this (neither by looking in the literature, nor by trying myself). Is anyone aware of some place in the literature that answers this (or vaguely related) question?

EDIT 1:

I think that I have a valid argument for the special case of $\mathrm{Deck}(\tilde{M})\cong \mathbb{Z}$, but the proof is slightly akward as it does not use all the hypothesis:

Pick a sequence $(\tilde{x}_n) \subseteq \tilde{M}$ that satisfies the PS-condition with respect to $(\tilde{f},\tilde{g})$. Define $x_n:=\pi(\tilde{x}_n)$ and pick a covergent subsequence $x_n \to x$ (recall that $M$ is closed). Now fix a preferred lift $\tilde{x}_0 \in \pi^{-1}(x)$ and let $U_x$ be a small neighborhood of $x$ such that $$\pi^{-1}(U_x) \cong \bigsqcup_{\tilde{x} \in \pi^{-1}(x)} U_{\tilde{x}}.$$ For $n$ large enough the $\tilde{x}_n$ will all lie in $\pi^{-1}(U_x)$. Denote by $A_n \in \mathrm{Deck}(\tilde{M})=\mathbb{Z}$ the unique decktransformation such that $$A_n \tilde{x}_n \in U_{\tilde{x}_0}.$$ Due to the choice of the subsequence $(x_n)$ we have $A_n \tilde{x}_n \to \tilde{x}_0$. Observe that the period homomorphism $$\Phi_{[\alpha]} \colon \pi_1(M) \to \mathbb{R}$$ induces an injective group homomorphism $$\chi \colon \mathrm{Deck}(\tilde{M})=\mathbb{Z} \to \mathbb{R},$$ satisfying the formular $$\chi(A_n)=\tilde{f}(\tilde{x}_n)-\tilde{f}(A_n\tilde{x}_n).$$ The PS-assumption and the convergence of $A_n \tilde{x}_n$ show that the RHS is bounded, thus $$\sup_n \vert \chi(A_n) \vert <+\infty.$$ This shows that there are only finitely many distinct $A_n$'s due to $\chi \colon \mathbb{Z} \to \mathbb{R}$ being an injective homomorphism. The subsequence $\tilde{x}_n$ is thus contained in the closure of finitely many neighborhoods $U_{\tilde{x}}$, which is a compact set by Hopf Rinow.

Note: In the general case one has $\mathrm{Deck}(\tilde{M}) \cong \mathbb{Z}^k$, but the proof above fails in the case $k \geq 2$ since $\chi(A_n)$ might be bounded with infinitely many distinct $A_n$'s.

Edit 2:

If we are allowed to swap $\alpha$ with any other cohomologous Morse-form, say $\beta$, then the answer is positive in the $n$-torus case $\Pi^n=S^1 \times \dots \times S^1$: Any $1$-form on $\Pi^n$ is cohomologous to some linear combination of angular forms $\beta=\sum_i a_i \, dx_i$. Since $\beta$ has no zeros it is automatically a Morse form. Furthermore the norm of the gradient of the corresponding $\tilde{f} \colon \tilde{\Pi^n} \to \mathbb{R}$ is constant and positive, i.e. $$\Vert \nabla \tilde{f} \Vert=\mathrm{const}>0.$$ Thus $\tilde{f}$ satisfies the PS-condition. However, this is pretty useless (for what I need at least) as the Novikov homology of the $n$-torus dies in the non-exact case...