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

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

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    $\begingroup$ 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, 2015 at 2:07
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    $\begingroup$ @MikeMiller: What if we take a symplectic $(4n+2)$-manifold whose boundary doesn't support an orientation-reversing diffeomorphism? $\endgroup$
    – Kyle
    May 29, 2015 at 17:04
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    $\begingroup$ 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, 2015 at 17:16
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    $\begingroup$ @tessellation: I think the OP is asking whether all compact symplectic manifolds have such a diffeomorphism. $\endgroup$
    – Kyle
    May 30, 2015 at 0:15
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    $\begingroup$ 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, 2015 at 16:36

1 Answer 1


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.

  • $\begingroup$ +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 $\endgroup$ Jun 2, 2015 at 9:19

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