## symplectic 4-manifolds with free circle action

Hi. I have a question.

Let $(M,\omega)$ be a closed symplectic 4-manifold equipped with a free circle action which preserves $\omega$ (symplectic circle action).

My question is , is there an example of $M$ which is not homeomorphic to $S^1 \times N$? ($N$ is a closed oriented 3-manifold)

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I'm guessing you want $M$ to be compact? – Eric O. Korman Mar 9 2011 at 1:59
Yes. I fixed my question. Thank you. – YCho Mar 9 2011 at 2:07
This sort of question is addressed here (although no explicit example is given):front.math.ucdavis.edu/1102.0821 – Agol Mar 9 2011 at 7:10

Here's an example, using a construction of Fernandez, Gray and Morgan (1991):

Take a closed surface $S$ with area form $\omega$, let $\phi$ be an area-preserving diffeomorphism, and $p\colon S_\phi \to S^1$ its mapping torus. This carries a closed 2-form $\omega_\phi$ induced by $\omega$, and a closed 1-form $p^\ast dt$. Take a class $e\in H^2(S_\phi;\mathbb{Z})$ which restricts trivially to $H^2(S;\mathbb{Z})$. Then $e$ has a de Rham representative of form $p^\ast dt\wedge a$, where $a$ is a closed 1-form. Take $L\to S_\phi$ be a hermitian line bundle with a connection form $i\eta$ of curvature $(-2\pi i) p^\ast dt\wedge a$, and let $\pi\colon M\to S_\phi$ be the unit circle bundle in $L$. Then the 4-manifold $M$ carries the $S^1$-invariant symplectic form $\Omega:= \pi^* \omega_\phi + \pi^*p^\ast dt\wedge \eta$.

Let's take $S$ to have genus $>1$ and $\phi$ to be a Dehn twist along a non-separating curve. The Wang exact sequence identifies $\ker (H^2(S_\phi)\to H^2(S))$ with the cokernel of $(1-\phi^\ast)$ acting on $H^1(S)$. In this case, the cokernel is $\mathbb{Z}$, and we take $e$ to be the generator. The fibration by $S^1$-orbits is non-trivial, but we need to check that the resulting $M$ is not homeomorphic to $S^1\times N^3$ in some weirder fashion.

Well, $\pi_1(M)$ is a non-trivial central $\mathbb{Z}$-extension of $\pi_1(S_\phi)$, and the latter is a semidirect product of $\pi_1(S)$ with $\mathbb{Z}$, where $1\in \mathbb{Z}$ acts on $\pi_1(S)$ by $\phi^{-1}$. If I'm not mistaken, $\pi_1(S_\phi)$ has trivial centre. Hence the centre of $\pi_1(M)$ is $\mathbb{Z}$, the subgroup generated by the $S^1$-fibre $\gamma$. If $M = S^1\times N$ then $\pi_1(M)$ is a trivial central $\mathbb{Z}$-extension of $\pi_1(N)$. But the central $\mathbb{Z}$-subgroup defined by this product splitting must be generated by a multiple of $\gamma$, and so the extension it defines is not trivial after all.

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 Yes, $\pi_1(S_\phi)$ has trivial center, since it is a graph manifold (not a Seifert-fibered space). – Agol Mar 9 2011 at 16:58 This is a nice answer! – J.C. Ottem Mar 9 2011 at 19:26 @Agol: Thanks for helping with this point. @J.C.O.: glad you like it! – Tim Perutz Mar 10 2011 at 3:20