1

Let $M$ be a smooth manifold. Recall that a 2-form $\omega$ on $M$ is called nondegenerate if for each $p\in M$ and $v\in T_pM-\{0\}$, there exists a $w\in T_pM$ with $\omega(v,w)\neq 0$. By linear algebra it can be seen that existence of a nondegenerate form implies that $ \dim M=n$ is even. Is it true that the following converse holds?: if $M$ is even-dimensional then there is a nondegenerate $2$-form on $M$.

First I thought that this can be shown by an elementary partition-of-unity argument, but I realized that sum of nondegenerate forms need not be nondegenerate..

user302934
  • 1,738
  • 1
  • 9
  • 27

1 Answers1

2

The comment proves only that there exist manifolds without closed nondegenerate 2-forms. Dropping the closed condition results in a structure called an "almost symplectic structure".

Not every manifold admits an almost symplectic structure. In fact, if you admit an almost symplectic structure you (non-obviously) also admit an almost complex structure, and vice versa. It is a well-known result (of Bott? please forgive my historical ignorance) that the only spheres admitting an almost complex structure are $S^2$ and $S^6$, so $S^4$ gives a counterexample to your question.

In fact you can read in the book of Gompf and Stipsicz a complete characterization of which simply connected closed 4-manifolds admit almost complex structures. $\Bbb{CP}^2 \# \Bbb{CP}^2$, for instance, does not.

user940357
  • 36
  • The argument is obstruction-theoretic: The existence of a nondegenerate 2-form on an $n$-manifold is equivalent to a reduction of the (principal) frame bundle to an $Sp(n)$-bundle. The group $U(n)$ is maximal compact in $Sp(n)$, hence, a reduction to an $Sp(n)$-bundle is equivalent to a reduction to a $U(n)$-bundle. The existence of a latter reduction is equivalent to the existence of an almost complex structure. – Moishe Kohan Jun 13 '21 at 18:50