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


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As far as I know, the HW manifold is closed (it is the flat 3-manifold that Seifert-fibers over the projective plane). – Bruno Martelli Jan 26 '13 at 6:56
What is the definition of this manifold? There's not much that's googlable. – Ryan Budney Jan 26 '13 at 7:47
The HW manifold is the orbit space $\mathbb{R}^{3} / <x , y , z |xy^{2}x^{-1}y^{2} = y x^{2} y^{-1} x^{2} = 1 , z = xy > $ where $X = diag[1 , -1 , 1] , Y = diag[-1, 1 , -1 ] $ and $Z = diag[-1,-1,1]$ and $x = (\frac{1}{2} e_{1} , X), y = ( \frac{1}{2} (e_{2} - e_{3} ) , Y ) $ and $z=(\frac{1}{2}(e_{1} - e_{2} + e_{3} ) , Z)$, this group is a subgroup of the group of affine motions of $3$-space. – jf9rj4o Jan 26 '13 at 13:04
@jf9: Does "produce" mean Dehn suegery? Otherwise, Bruno answered your question. – Misha Jan 26 '13 at 18:43

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.

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.

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As shown in the reference I gave. – Igor Rivin Jan 27 '13 at 2:08
Sorry, I saw your post after I finished writing my answer. And yes, that's the right reference. – Alex Suciu Jan 27 '13 at 2:51

Much enlightenment (though not an explicit answer to the question) can be gleaned from Bruno Zimmermann's paper "On the Hantzsche-Wendt manifold".

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