# Heegaard genus of the hyperbolic dodecahedral space (is it 3 or 4?)

I have a question on the hyperbolic dodecahedral space, first described by C.Weber and H. Seifert in 1933 [Die beiden Dodekaederr\"aume, Math Z. 37 (1933), 237-253]. Is it known whether it admits a Heegaard splitting of genus 3?

We can see (even with elementary combinatorial tools, like discrete Morse theory) that it does admit a Heegaard splitting of genus 4; but I was wondering if this is best possible.

• I checked with SnapPy, and the smallest number of generators it can find seems to be 4. So it's possible that the rank is 4. – Ian Agol Oct 31 '13 at 22:04

## 1 Answer

The Whitehead link fibers with a twice-punctured genus 1 fiber. Since the Seifert-Webber Space is a cyclic 5-fold covering of the Whitehead link, the image of Whitehead link in this cover should again be a fibered link with a twice-punctured genus 1 fiber. So this will give a genus 3 Heegaard splitting.

I reckon this argument should work to show each of the two isometry classes of 5-fold cyclic covers (as mentioned in the comments to this question) have genus 3 Heegaard splittings.

Edit: Of course Agol's comment is right. I wasn't considering the other possible 5-fold cyclic branched covers.

So if I've got my head on straight about it, these other covers should be dual to a thrice-punctured genus 1 fiber of the link exterior. (Such a fiber is obtained from adding the twice-punctured torus Seifert surface with a suitably oriented once punctured torus bounded by one component and disjoint from the other.) Then in the cover, as well as in S^3, the two boundary components of the fiber on the same link component join together to make a non-orientable surface with $\chi = -3$ and one boundary component. The boundary of a neighborhood of this surface then gives a genus 4 Heegaard splitting. Presumably, this splitting is equivalent to the ones Bruno obtained.

• Ken, I think you have to be careful: I think the two 5-fold covers corresponding to the Seifert-Weber will not be dual to the genus 1 surface. If $\mathbb{Z}^2 = H_1(W)$, then there are 4 maps to $\mathbb{Z}/5$ (modulo automorphisms), given by $(1,0)\mapsto 1$, and $(0,1)\mapsto \pm 1, \pm 2$. By Zimmerman's criterion, the $(0,1)\mapsto \pm 1$ maps give homeomorphic covers, and these will be dual to the 2 genus 1 fibers. But I think the other two won't. – Ian Agol Nov 1 '13 at 22:27