# Symplectic form in each class of cohomology

Is there a topological problem? Can we find in each class $c$ of cohomology $H^2 (M,Z)$ of a manifold $M$ of dimension $2n$, such that $c^n=1$, a representative which is a symplectic form? Where is the obstruction if any?

## 1 Answer

If $M$ is a closed manifold of dimension $2n$ endowed with a symplectic form $\omega$, $\omega^n$ is a volume form.

Consider $T^4$ which is the quotient of $R^4$ by the group $G$ generated by the translations $t_{e_i}$ of direction $e_i, i=1,2,3,4$, $\omega' =dx_1\wedge dx_2$ is a closed two form invariant by $G$ thus defines a $2$-closed form $\omega$ on $T^4$, $\omega^2=0$, so it does not define a symplectic form.

If I understand well the change in your question, you want to know wether we can find in each class $[\omega]$ of $H^2(M,Z)$ such that $[\omega]^n\neq 0$ a symplectic structure. Consider $M$ the connected sum of two copies of $CP^2$ the complex projective plane, it is not endowed with a structure of a symplectic manifold, and there exists $[\omega]\in H^2(M,Z)$ such that $[\omega]\wedge[\omega]\neq 0$.

• But perhaps that in the cohomology class of $\omega$, there is a representative that is a symplectic form? – Antoine Balan Sep 8 '16 at 21:59
• The wedge product of forms descend to the cohomology, math.stackexchange.com/questions/1060176/… – Tsemo Aristide Sep 8 '16 at 22:04
• But the volum form can be zero in the cohomology of $M$, no? – Antoine Balan Sep 8 '16 at 22:07
• No if the manifold is closed without boundary. – Tsemo Aristide Sep 8 '16 at 22:07
• So I choice cohomology classes $c$ such that $c^n=1$. Are there symplectic forms? – Antoine Balan Sep 8 '16 at 22:13