Let $(M, \omega)$ be a connected closed symplectic manifold of dimension $2n$.


Assume there exist smooth covering maps $\phi\_k:M\to M$ such that $\phi\_k^\* \omega=\sqrt[n]{k}\omega$ for all $k\geq 1$. Is $n=1$ then?



Smooth covers rescaling the symplectic form
It **is** the case that $n \in \{0,1\}$. (I include $n=0$ because $0$-manifolds with $\omega = 0$ are symplectic.)


**Two-sentence summary:** We know what integral elements of cohomology are, and pullback acts integrally. This allows us to think of the coefficients of $\omega$ as living in some fixed abelian subgroup of $\mathbb{R}$ of finite rank, which yields a contradiction.


**Details:** Consider the action of the pullback of these maps on cohomology, $$\phi\_k^\*~ \colon H^2(M;\mathbb{Z})/\mathrm{Tors} \rightarrow H^2(M;\mathbb{Z})/\mathrm{Tors},$$ which upon tensoring with $\mathbb{R}$ yields the pull-back map $$\phi\_k^\* \otimes \mathbb{R}~\colon H^2(M;\mathbb{R}) \rightarrow H^2(M;\mathbb{R}).$$ (Typically one just denotes this as $\phi\_k^\*$, but we make explicit that it comes from an integral map.) You are asking that for each $k$, $\omega$ is an eigenvector of $\phi\_k^\* \otimes \mathbb{R}$ with eigenvalue $\sqrt[n]{k}$.


For simplicity, pick a $\mathbb{Z}$-basis $(e\_1,\ldots,e\_r)$ for $H^2(M;\mathbb{Z})/\mathrm{Tors}$ (where $r = \beta\_2$ is the 2nd Betti number), which therefore also determines a basis for $H^2(M;\mathbb{R})$. We may write $$\omega = \sum\_{i=1}^{r} c\_i e\_i,$$ where WLOG $c\_1 \neq 0$. The elements $c\_i$ generate a subgroup $G \leq \mathbb{R}$ (which is a free abelian group of finite rank at most $r$), and since the maps $\phi\_k^\*$ act integrally on the $e\_i$, we have that the coefficients of $(\phi\_k^\* \otimes \mathbb{R})(\omega)$ with respect to the chosen basis lie again in $G$. But also $$(\phi\_k^\* \otimes \mathbb{R})(\omega) = \sqrt[n]{k} \cdot \omega = \sum\_{i=1}^{r} (\sqrt[n]{k} \cdot c\_i)e\_i.$$ In particular, looking at the first coefficient, we have $\sqrt[n]{k} \cdot c\_1 \in G$ for each $k$. Dividing by $c\_1$, it follows that $\sqrt[n]{k}$ lies in the subgroup $\frac{1}{c\_1}G$ for every $k$. But when $n \geq 2$, the elements $\sqrt[n]{k}$ generate an infinite rank subgroup of $\mathbb{R}$, so it is impossible for all of these elements to lie in the finite rank group $\frac{1}{c\_1}G$. Thus $n \in \{0,1\}$.


