Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $M \times S^{2n}$?

share|improve this question

3 Answers 3

up vote 10 down vote accepted

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.)

share|improve this answer
I thought I knew K\"unneth theorem but it was not true. Thank you. –  Hwang Feb 17 '11 at 7:06
This assumes M is compact, right? –  Eric O. Korman Feb 17 '11 at 13:12

Tom Church's answer is correct for $M$ compact. Here is a counter-example 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 $S \times \mathbb{R}^{2n+2}$.

share|improve this answer
What is $S$? Do you mean $S^{2n}$? –  Michael Albanese May 17 at 3:59

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 h-principle 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 $U=D^{n-1} \times [0,\infty) \to M$. 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 self-diffeomorphism $\psi$ 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 $m-1$-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.

share|improve this answer
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? –  Hwang Feb 18 '11 at 3:40
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. –  Hwang Feb 19 '11 at 5:49
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. –  Johannes Ebert Feb 19 '11 at 13:52
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? –  Hwang Feb 19 '11 at 15:02
$\psi$ is only an embedding, whose image does not contain the zero $p$. –  Johannes Ebert Feb 19 '11 at 18:31

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.