For $n > 1$, $2n$dimensional sphere $S^{2n}$ does not admit symplectic structures. Then how about the product with a manifold? Are there any results about the symplectic structures on $M \times S^{2n}$?
From the Künneth theorem you can check that there is no class $\omega\in H^2(M^{2d}\times S^{2n};\mathbb{R})$ such that $\omega^{d+n}\neq 0$. (This is an excellent thing for you to work out for yourself.) Since a symplectic form is closed and nondegenerate, this shows that no symplectic structure on $M\times S^{2n}$ can exist for $n>1$ if $M$ is compact.
(Thanks to Eric and David for pointing out that I was assuming $M$ to be compact; this is necessary so that nondegeneracy implies that the top power of $\omega$ represents a nonzero multiple of the fundamental class. In his answer David Speyer gives a nice counterexample if $M$ is not required to be compact.)

$\begingroup$ I thought I knew K\"unneth theorem but it was not true. Thank you. $\endgroup$ – Hwang Feb 17 '11 at 7:06

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Tom Church's answer is correct for $M$ compact. Here is a counterexample for $M$ not compact: The cotangent bundle $T^* S$ has the tautological symplectic structure: At a point of $T^* S$ lying over $x \in S$, the tangent space to $T^* S$ is naturally isomorphic to $T_x S \oplus (T_x S)^*$. Take the natural pairing between the first facts and the second factor and skew symmetrize it to get a symplectic structure on $T^* S$.
Now, look at $T^* S \times \mathbb{R}^2$ with the standard symplectic structure on $\mathbb{R}^2$. So this is a symplectic manifold.
But the cotangent bundle to $S$ is stably trivial: $T^* S \times \mathbb{R} \cong S \times \mathbb{R}^{2n+1}$. So $T^* S \times \mathbb{R}^2 \cong S \times \mathbb{R}^{2n+2}$ and there is a symplectic structure on $S \times \mathbb{R}^{2n+2}$.
If $M$ is symplectic, noncompact and connected, then there is a symplectic structure on $M \times S^{2n}$, for each $n$. The proof can be given with the hprinciple for symplectic structures. It says the following: let $M$ be a connected noncompact (''open'' in the sequel) manifold, $a \in H^2 (M; \mathbb{R})$ and $J$ an almost complex structure on $M$. Then there exists a symplectic structure $\omega$ on $M$, such that the cohomology class of $\omega$ is $a$ and such that there is a compatible almost complex structure $I$ with $I$ homotopic to $J$. Therefore, any open almost complex manifold has a symplectic structure.
Now I claim: $M$ open and almost complex, then $M \times S^{2n}$ is almost complex (and of course open).
Step 1: $M$ has a vector field without zeroes. To see this, take a vector field $X$ with isolated zeroes and let $p$ be a zero. Choose an ''escape path'', i.e. an embedding $u:[0,\infty] \to M$ with $u$ proper and $u(1)=p$. Moreover, $u$ should avoid the other zeroes. Pick a tubular neighborhood of $u$. The result is that you extend $u$ to a proper embedding $U=D^{n1} \times [0,\infty) \to M$. Let $\phi_t:U \to U$, $t \in [0,1]$ be an isotopy of embeddings $\phi_1=id$ and $\phi_0([0,\infty)) \subset [0,1/2]$. This istopy should be constant near $t=0,1$ and near $x=0$. Then define $\psi:U \to U$ by the formula $\psi(v,x):= (v, \phi_{v^2} (x))$. This is an embedding $U \to U$ whose image does not contain $0$. Extend to a selfdiffeomorphism $\psi$ of $M$. Then $\psi^* X$ is a vector field without the zero $p$.
This decomposes the tangent bundle into $TM = \mathbb{R} \oplus V$.
EDIT: A vector field without zeroes can also be found using obstruction theory: if $M$ is open of dimension $m$, then $M$ is homotopy equivalent to an $m1$dimensional CW complex.
Step 2: Use the trivial factor to show that $T (M \times S^{2n}) \cong TM \times \mathbb{R}^{2n}$. Therefore, the tangent bundle of $M \times S^{2n}$ has a complex structure and $M \times S^{2n}$ is an almost complex manifold.

$\begingroup$ I couldn't understand Step 1. It seems that there is a nowhere vanishing vector field on an open manifold, but I have no idea how to prove it. Could you explain what it means "moving zero outside the manifold", or what reference I should read? $\endgroup$ – Hwang Feb 18 '11 at 3:40

$\begingroup$ I am sorry I am confused. What do you mean by $\psi^{*}X$? Is it a pushforward of a vector field $X$ by a diffeomorphism $\psi^{1}$? But then the resulting vector field still has zeroes. $\endgroup$ – Hwang Feb 19 '11 at 5:49

$\begingroup$ But $\psi^X$ has one less zero than $X$. If $X$ had only finitely many zeroes, this would give an inductive argument. If there were infinitely many (isolated) zeroes, a slightly more careful argument is needed. I would choose paths to all of these zeroes at the same time. If $dim (M) \geq 3$, I can pick all these paths to be disjoint and perform the pushing away of the zeroes simultaneously at one time. In dimension $2$, this does not work, but an open surface is always parallelizable. $\endgroup$ – Johannes Ebert Feb 19 '11 at 13:52

$\begingroup$ A diffeomorphism $\psi$ on $M$ induces an isomorphism on the tangent space at each point. So two vector fields $X$ and $(\psi^{1})_{*} X$ have the same number of zeroes on $M$. What am I thinking wrong? $\endgroup$ – Hwang Feb 19 '11 at 15:02

$\begingroup$ $\psi$ is only an embedding, whose image does not contain the zero $p$. $\endgroup$ – Johannes Ebert Feb 19 '11 at 18:31