symplectic 4-manifolds with free circle action - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T05:59:28Z http://mathoverflow.net/feeds/question/57894 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/57894/symplectic-4-manifolds-with-free-circle-action symplectic 4-manifolds with free circle action YCho 2011-03-09T01:06:41Z 2011-03-09T16:13:09Z <p>Hi. I have a question. </p> <p>Let $(M,\omega)$ be a closed symplectic 4-manifold equipped with a free circle action which preserves $\omega$ (symplectic circle action). </p> <p>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)</p> <p>Thank you in advance. </p> http://mathoverflow.net/questions/57894/symplectic-4-manifolds-with-free-circle-action/57968#57968 Answer by Tim Perutz for symplectic 4-manifolds with free circle action Tim Perutz 2011-03-09T16:13:09Z 2011-03-09T16:13:09Z <p>Here's an example, using a construction of Fernandez, Gray and Morgan (1991): </p> <p>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 $\Omega:= \pi^* \omega_\phi + \pi^*p^\ast dt\wedge \eta$.</p> <p>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.</p> <p>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.</p>