Let $\alpha \in \Omega^1(M)$ be a closed $1$-form on a closed Riemannian manifold $(M,g)$. Denote by $X$ the corresponding dual vector field. Consider a flow line $$\gamma \colon \mathbb{R} \to M$$ connecting the two zeros $x^+,x^-$ of $\alpha$, i.e. $\gamma$ is an integral curve of $-X$ such that $$\lim_{t\to +\infty}\gamma(t)=x^+, \; \lim_{t \to -\infty}\gamma(t)=x^-.$$
Let $\mathcal{M}(x^-,x^+)$ denote the set of all such integral curves connecting the two zeros $x^-,x^+$ of $\alpha$. If the form $\alpha$ is Morse-Smale then we know that $\mathcal{M}(x^-,x^+)$ is a smooth submanifold of $M$. For a general smooth path $\gamma_0 \colon \mathbb{R} \to M$ one can define the energy $l^2$ of $\gamma_0$ as follows:
$$l^2(\gamma_0):=\int_{-\infty}^{\infty} \vert \dot{\gamma}_0(s) \vert^2ds,$$ where $g=\langle \cdot,\cdot \rangle$ and $\vert \cdot \vert$ is the corresponding induced norm.
Now if we restrict to the flow lines $\gamma \in \mathcal{M}(x^-,x^+)$ we get the following nice expression: $$l^2(\gamma)=-\int_{\gamma} \alpha.$$
My $\textbf{question}$: Is the functional $l^2$ locally constant on $\textbf{any}$ $\mathcal{M}(x^-,x^+)$ or just whenever we have the index relation $$ind(x^-)-ind(x^+)=1?$$ In the latter case I was able to write down a proof using the fact that $\mathcal{M}(x^-,x^+)$ modulo the obvious $\mathbb{R}$-action is discrete if the above index relation holds.