I wonder if there is an example of rational homology sphere that is not a Seifert manifold. If there is, how can one construct such a rational homology sphere from a surgery of a knot in $S^3$?
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2 Answers
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By Thurston, all but finitely many $(p,q)$-surgeries on a hyperbolic knot in $S^3$ result in hyperbolic rational homology spheres for $p\neq 0$. In particular there are infinitely many integral homology spheres among them.
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If you glue opposite faces of a dedecahedron with a twist of $\frac\pi5$, you obtain the classical Poincaré sphere. It is an integral homology sphere, and a Seifert manifold, see BS' comment below.
But if you do the gluing with a twist of $\frac{3\pi}5$, then you obtain the Seifert-Weber manifold, which is a rational homology sphere with a hyperbolic structure, so definitely not a Seifert manifold. It has $H_1\cong(\mathbb Z/5)^3$, see Neil Hoffman's comment. I do not know what knot would give this manifold. But you can deduce many of its properties by drawing a picture of a dodecahedron with the appropriate corners, edges and so on identified.
answered Feb 13, 2016 at 10:04
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3$\begingroup$ Poincaré sphere is Seifert fibered over $S^2$ with three singular fibers of orders $2,3,5$ : simply look at orbits of $SO(2)$ on $SO(3)/A_5$. I wonder if Poincaré himself ever saw "his sphere" as $SO(3)/A_5$ (he constructed it as a Heegard diagram). $\endgroup$– BS.Commented Feb 13, 2016 at 13:31
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2$\begingroup$ In the case of a $3\pi/5$ twist, the manifold obtained is the so-call Seifert-Weber Manifold, which is a rational homology sphere, but not an integral homology sphere since the first homology is $(\mathbb{Z}/5\mathbb{Z})^3$. $\endgroup$ Commented Feb 17, 2016 at 23:00
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$\begingroup$ Since $H_1$ of the Seifert-Weber manifold is not cyclic, it can't be surgery on a knot; you'd need a link of at least 3 components to get $(Z/5Z)^3$. $\endgroup$ Commented Feb 18, 2016 at 14:56