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How to show that a covering manifold of a symplectic manifold admits a symplectic structure? More precisely, let M be a $2n$-manifold and $(N, \omega)$ be a symplectic 2n-manifold. If there exists a covering $\pi: M\to N$, then $\pi^*\omega$ is a symplectic form on $M$. I don't have a clue how to prove this statement. Any help would be very appreciated.

Javi
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usr1988
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    What have you tried so far ? First, observe that $\pi$ is a local diffeomorphism, then try to express the pullback of the differential form $\omega$ on $M$ in order to check that $\pi^*\omega$ is closed and non-degenerate. – Bebop Jul 24 '18 at 10:54
  • Showing $d(\pi^\omega)=0$ is easy, but I do not know how to show $(\pi^\omega)^n \neq 0$, without explicitly specifying a local diffeomorphism. – usr1988 Jul 24 '18 at 12:51
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  • I think that I found an answer. Let $\omega_M=\pi^* \omega$ be a closed 2-form on M. Then we have $\int_M \omega_M^n=deg(\pi)\int_N \omega^n\neq 0$(assuming M is closed and orientable), since $\omega$ is non-degenerate on $N$. – usr1988 Jul 24 '18 at 13:04
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    Ok, so in general and assuming it is clear for you that $\pi$ is a local diffeomorphism, then $\forall p\in M, \forall X\in T_pM, \ (\pi^*\omega)p(X)=\omega{\pi(p)}(d\pi_{\pi(p)}(X))$. Can you check now that this 2-form is non degenerate ? – Bebop Jul 24 '18 at 13:44
  • Clearly, $d\pi_{\pi(p)}(X)\in T_{\pi(p)}N$ and $\omega_{\pi(p)}$ is non-degenerate on $T_{\pi(p)}N$, since $\omega$ is a symplectic form on N. Hence, $\pi^*\omega$ is also non-degenerate due to the equality you mentioned. – usr1988 Jul 24 '18 at 14:25
  • I made some mistakes in my last remark : $\forall p\in M, \forall X,Y\in T_pM, \ (\pi^\omega)p(X,Y)=\omega{\pi(p)}(d\pi_p(X),d\pi_p(Y))$. However, you got it, everything works because $\omega$ is a symplectic form on $N$ and also* because $d\pi_p$ is invertible. Do you see why ? – Bebop Jul 24 '18 at 15:07
  • Thanks a lot for your kind and quick answer. Assume that $d\pi_p$ is not invertible, which means its kernel is not trivial. Say $Z\in ker (d\pi_p)$, then for any nonzero $Y\in T_p M$, $(\pi^\omega)p(Y,Z)=\omega{\pi(p)}(d\pi_p(Y),0)=0$. This implies that $\pi^\omega$ is degenerate. What do you think? – usr1988 Jul 24 '18 at 15:17
  • Yes, that's something we don't want. More precisely, $d\pi_p$ being injective guaranties you that if $d\pi_p(X)=0$ then $X=0$ and its surjectivity tells you that $d\pi_p(Y)$ will reach any vector in $T_{\pi(p)}N$. – Bebop Jul 24 '18 at 16:10

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