# Surgery diagram for the Seifert-Weber space

In every reference I see, the Seifert-Weber space is presented as an identification space (specifically identify opposite faces of a dodecahedron by a 3/10ths twist).

What I can't seem to find is a surgery diagram for this manifold. From its homology we know it doesn't arise as surgery on a knot. Its volume is approximately 11.1991. Checking with SnapPy there are a lot of candidate links in this range:

In [14]: len(filter(lambda x: x.num_cusps() >= 3,HTLinkExteriors[11.19:13.0]))
Out[14]: 1103
In [15]: len(filter(lambda x: x.num_cusps() >= 3,HTLinkExteriors[11.19:12.0]))
Out[15]: 365
In [16]: len(filter(lambda x: x.num_cusps() >= 3,HTLinkExteriors[11.19:12.5]))
Out[16]: 620


so my attempts at guessing have been unsuccessful.

One could calculate a surgery diagram by-hand from the Heegaard diagram that comes from the identification space and then try to simplify the result. This seems daunting to do by hand, but I'm also unaware of software for the task.

## Edit

I accepted Daniel's answer before working through the details. The process certainly gives the Seifert-Weber space a surgery on a manifold. The problem is that the branched-covers of $S^3$ over the trivial 2-component unlink are not themselves link complements, as in the case of a 1-component unlink, so the method in Rolfsen doesn't yield surgery on a link in $S^3$, but somewhere else. That somewhere else has non-separating 2-spheres in it, illustrated by this picture:

The cut disks are indicated by red and blue, the two light grey wireframe balls are the complements of the two knot components after cutting along the disks. As long as at least two of these gadgets are glued together light-to-dark, the yellow curve will pierce the solid grey 2-sphere only once.

• The Seifert-Weber space is a 5-fold cyclic branched cover of the Whitehead link complement. One may obtain it in Snappea (or Snappy) by performing (5,0) surgery on each component of the Whitehead, then taking cyclic covers of order 5 (there is only one that is a manifold). I'm not sure how to convert this to a surgery diagram though. – Ian Agol Jul 18 '13 at 22:28
• The volume would appear to be small enough so that you could drill curves out of the manifold (using SnapPy) and search for the resulting manifold in the HTLinkExterior census. Have you tried that? – Ryan Budney Jul 18 '13 at 23:33
• @IanAgol, there are two distinct isometry classes of 5-fold covers of the Whitehead link (Theorem 3 in this paper). – Edgar A. Bering IV Jul 26 '13 at 0:16
• @Edgar: yes, I was mistaken about the uniqueness. – Ian Agol Jul 26 '13 at 0:59

1. Untie the Whitehead link using $\pm 1$ framed surgeries along unknots ringing crossings (which induce crossing changes on the Whitehead link, while preserving $S^3$.) Actually you only need one surgery component, because a single crossing change unties the Whitehead link. You obtain a surgery presentation of the Whitehead link complement in the complement of a trivial 2-component unlink.
Edit: As the OP points out in the edit to his question, the above process gives a surgery presentation for the Seifert-Weber space, but unfortunately not in $S^3$.
There is another way to draw a surgery diagram (this time in $S^3$) for the Seifert-Weber space, starting again from the description as a 5-fold branched cover over the Whitehead link $L$. Namely, you can use the algorithm of Akbulut-Kirby (Branched covers of surfaces in 4-manifolds. Math. Ann. 252 (1979/80), no. 2, 111–131). This starts with a connected Seifert surface for $L$, where you want to see the surface as a disk with bands. In the case at hand, the initial picture will be a little complicated, with something like 12 surgery curves. I think that you could do some simplifications, though.
Since we are talking about a link, rather than a knot, there is some ambiguity about which branched cover is being discussed. (This same point is relevant to the responses above as well.) The simpler version, to which the algorithm applies directly, is the branched cover corresponding to a homomorphism $H_1(S^3 -L) \to Z/5$ that takes both meridians to a given generator $1 \in Z/5$. Up to homeomorphism, I think that the only other choice is to send one meridian to $1$ and the other to $2$; I'm sure that the Akbulut-Kirby algorithm could readily be adapted to this case, but haven't thought out exactly how.