16
$\begingroup$

Let $(M,\omega)$ be a compact symplectic manifold.

Is there always a diffeomorphism $f$ on M with $f^{*}\omega =-\omega$?

$\endgroup$
  • 6
    $\begingroup$ I would really like the answer to be no, but I can't find a counterexample. One obvious thing to try is to find a $(4n+2)$-manifold that supports no orientation-reversing diffeomorphisms. This is not possible in dimension 6(!!!) and the first example I can find is in dimension 14, which supports no symplectic structure for cohomological reasons. Interesting question! $\endgroup$ – user98602 May 25 '15 at 2:07
  • 2
    $\begingroup$ @MikeMiller: What if we take a symplectic $(4n+2)$-manifold whose boundary doesn't support an orientation-reversing diffeomorphism? $\endgroup$ – Kyle May 29 '15 at 17:04
  • 1
    $\begingroup$ I was hoping to find a symplectic filling of a $(4n+1)$-contact manifold that doesn't have an orientation-reversing diffeomorphism. Certain lens spaces and products of lens spaces make for nice classes of chiral odd-dimensional manifolds, but I don't know anything about contact structures on them (and symplectic fillings thereof). $\endgroup$ – Kyle May 29 '15 at 17:16
  • 2
    $\begingroup$ @tessellation: I think the OP is asking whether all compact symplectic manifolds have such a diffeomorphism. $\endgroup$ – Kyle May 30 '15 at 0:15
  • 2
    $\begingroup$ I'm not sure if the above comment is coherent. I'll rephrase. I think the best approach is to find a symplectic 4-manifold $(W,\omega)$, where $\omega$ restricts to the boundary as a non-exact form, such that $\partial W$ has no diffeomorphisms that reverse $H^2$ at all. If we want to do this for closed manifoilds, we would want to find a closed $W$ which has no $H^2$-reversing diffeomorphisms; this seems harder if there aren't nice simply-connected examples. $\endgroup$ – user98602 May 30 '15 at 16:36
14
$\begingroup$

Serious thanks to squirrel for the idea to think about things with boundary and the tipoff that the 2-form below is not exact, and to Paul Plummer for helping me find the asymmetric hyperbolic manifold.

The answer to your question for manifolds with boundary is no. Here's a (not very explicit!) construction.

1) Kojima proved the existence of closed oriented hyperbolic 3-manifolds with given finite isometry group $G$ in his paper "Isometry transformations of hyperbolic 3-manifolds". In particular, we may take $G$ to be the trivial group. Because hyperbolic manifolds are $K(\pi,1)$s, homotopy classes of maps $Y \to Y$ are classified by the homomorphism they induce on the fundamental group. Mostow rigidity and the assumption that $\text{Isom}(Y) = 1$ implies that $\text{Out}(\pi_1(Y)) = 1$, and hence every automorphism of $\pi_1(Y)$ is inner, and hence induces the identity map on $H_1(Y) = \pi_1(Y)^{ab}$. The naturality of Poincare duality implies that every orientation-preserving self-homeomorphism $Y \to Y$ induces the identity on $H^2(Y)$. In Kojima's construction, we may take $b_1(Y) > 0$. Because $Y$ is hyperbolic, $Y$ must be irreducible.

So henceforth let $Y$ be an irreducible closed oriented 3-manifold with $b_1(Y) > 0$ and such that all orientation-preserving self-homeomorphisms of $Y$ induce the trivial map on $H^2(Y)$.

2) Gabai proved that if $Y$ is a closed, oriented, irreducible 3-manifold with nonzero Betti number, then $Y$ admits an orientable taut foliation (with a closed leaf). (See "Foliations and the topology of 3-manifolds"; this is a corollary of his theorem 5.5 and our assumption that $b_1(Y) > 0$.) So our $Y$ above supports a taut foliation. One equivalent definition to the claim that a (co-orientable) 2-plane distribution $\xi$ gives a taut foliation is that there is a transverse vector field $X$ to $\xi$ that preserves some volume form $\Omega$. This is similar to the second equivalent definition here. This is immediately seen to be equivalent to the claim that there is a closed 2-form $\omega$ on $Y$ such that $\omega \big|_{\xi}$ is nowhere zero. In particular, if we orient $\xi$, we can take $\omega$ to be positive on $\xi$. Now we see that $\omega$ is not exact: for restrict it to one of the closed leaves; because it's positive on its tangent bundle, it must be a volume form for the closed leaf, and hence cannot be exact.

So let $\xi$ be such a 2-plane distribution on $Y$ and $\omega$ such a closed 2-form.

3) This construction is from Eliashberg and Thurston's monograph "Confoliations". Let $\alpha$ be a 1-form such that $\text{ker}(\alpha) = \xi$. Consider $Y \times I$ and the 2-form $\omega_\varepsilon = p^* \omega + \varepsilon d(t p^*\alpha)$ on it. For $\varepsilon$ small enough, this is non-degenerate. $\xi \times \{0,1\}$ has a canonical orientation as a 2-plane distribution on the boundary of a symplectic manifold, and we see that $\omega_\varepsilon$ is positive on $\xi \times \{0,1\}$ for $\varepsilon$ small enough. Pick a contact structure $\xi'$ on $Y$ that is $C^0$ close to $\xi$; then the above discussion applies to this. This is called a symplectic semi-filling of the contact manifold $(Y,\xi')$.

4) Because $(Y \times I, \omega_\varepsilon)$ is a symplectic filling of $(Y, \xi')$, this article by Eliashberg shows that we can cap off the right side in such a way that the symplectic form $\omega_\varepsilon$ extends to a symplectic form, say $\omega'$, on a larger manifold $M$ whose boundary is $Y$. We have constructed a symplectic manifold $M$ whose boundary is $Y$. This is our desired symplectic manifold.

5) Now suppose there was a diffeomorphism $f: M \to M$ such that $f^*\omega' = -\omega'$. This would be orientation preserving, because $\omega \wedge \omega$ is a volume form of $M$. Then $f$ restricts to an orientation-preserving diffeomorphism of its boundary (because it sends outward normal vectors to outward normal vectors). By hypothesis $\omega'$ restricts to the 2-form $\omega$ on $Y$, our assumption that $f: M \to M$ has $f^* \omega' = -\omega'$ means that $f\big|_Y : Y \to Y$ has $f\big|_Y^* \omega = -\omega$. Recall from 1) that all orientation-preserving homeomorphisms of $Y$ induce the identity on $H^2(M)$. Because $\omega$ is not exact, this contradicts the existence of $f$! So there is no such $f$ on our manifold $M$.

Note that in the construction of $M$, there was a certain symmetry: we could easily have constructed $[-1,1] \times Y$ and capped off both ends via Eliashberg's result. One might hope that such a double would also carry a symplectic form that doesn't support an $\omega$-reversing diffeomorphism; perhaps chasing through the proofs and doing everything here a bit more constructively would allow one to check that it doesn't.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ +1 for your deep answer. I need to more times to understand the details. Befor that your answer appear, I presented the question in MO, too mathoverflow.net/questions/208175/…. thanks for your help $\endgroup$ – Ali Taghavi Jun 2 '15 at 9:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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