most recent 30 from http://mathoverflow.net 2013-05-21T07:49:14Z http://mathoverflow.net/feeds/question/119918 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/119918/the-knot-whose-complement-is-the-hantzsche-wendt-manifold jf9rj4o 2013-01-26T06:42:54Z 2013-01-27T04:54:42Z

<p>Can someone identify the knot whose complement in $\mathbb{S}^{3}$ will produce the Hantzsche-Wendt manifold?</p> <p>Thanks</p>

http://mathoverflow.net/questions/119918/the-knot-whose-complement-is-the-hantzsche-wendt-manifold/119979#119979 Igor Rivin 2013-01-27T01:57:36Z 2013-01-27T01:57:36Z

<p>Much enlightenment (though not an explicit answer to the question) can be gleaned from Bruno Zimmermann's paper <a href="https://dl.dropbox.com/u/5188175/ZimmermanHW.pdf" rel="nofollow">"On the Hantzsche-Wendt manifold".</a></p>

http://mathoverflow.net/questions/119918/the-knot-whose-complement-is-the-hantzsche-wendt-manifold/119982#119982 Alex Suciu 2013-01-27T02:06:43Z 2013-01-27T04:54:42Z

<p>Supposing the question is, "Can the Hantsche-Wendt manifold be realized as a cyclic branched cover of a knot in the 3-sphere", then the answer is yes: it is a 3-fold cyclic branched cover along the figure-eight knot. Incidentally, it is also a 2-fold branched cover along the Borromean rings.</p> <p>On the other hand, if the question is, "Can the Hantsche-Wendt manifold $M$ be realized by performing Dehn surgery along a knot in the 3-sphere" (as Misha seems to suggest), then the answer is no, since $H_1(M,\mathbb{Z})=\mathbb{Z}_2 \oplus \mathbb{Z}_2$ is not a cyclic group. </p>