# Regular or h-regular CW-complex structure for the Poincaré homology sphere

I am looking for a regular (the characteristic maps of the cells are homeomorphisms) or h-regular (the characteristic maps of the cells are homotopy equivalences) CW-complex structure for the Poincaré homology sphere. I would like to find a more economic one than the triangulation having f-vector: [16, 106, 180, 90]. I would like to minimize the number of cells.

Thanks in advance and any idea or potential reference is welcome!

• You can construct the Poincare homology sphere from a dodecahedron by identifying opposite sides by a minimal clockwise turn. (see e.g. en.wikipedia.org/wiki/Homology_sphere). If I remmber correctly this identification identifies each vertex with three other vertices (so that the CW-structure has 20/4=5 zero-cells), each edge with two other edges (so that we have 30/3=10 one-cells) and each face with another face (so that we have 6 two-cells). And of course we just have one three-cell. Dec 19, 2020 at 14:36
• @HenrikRüping but...I believe that the CW structure you are providing it is not regular, not even h-regular since if it were so, then the space would have trivial homology in dimension 3, which can not be the case. Is my point clear or am I wrong? Dec 20, 2020 at 11:47
• Indeed. The attachin map of the 3-cell hits every two cell twice. I think this can be fixed by putting one more vertex in the middle of the dodecahedron, e.g by thinking of the full dodecahedron as the cone over its boundary. Dec 20, 2020 at 13:10

Henrik Rüping's suggestion (in the comments) decomposes the Poincaré homology three-sphere as 12 pentagonal pyramids. The resulting face vector is

[0 + 12, 6 + 30, 10 + 20, 5 + 1] = [12, 36, 30, 6]

for a total of 84 cells. You can reduce the number of three-cells, at the cost of increasing the number of cells overall, as follows. In Rüping construction, join the pyramids in pairs (along their pentagonal face) to get six bi-pyramids. This has an face vector of

[6, 30, 30, 6]

but is no longer regular -- every bi-pyramid meets the central vertex twice. We can fix this by "blowing up" the central vertex to obtain a small dodecahedron. This truncates the apex and nadir of each bi-pyramid. Now the face vector is

[6 + 1, 30 + 12, 30 + 30, 6 + 20] = [7, 42, 60, 26]

with sum 135. At least the number of three-cells is smaller...

Recently, I (with R. Chirivi and M. Spreafico) have explicitely given a decomposition of the Poincare homology sphere $$\mathbb{S}^3/\mathcal{I}$$ (with $$\mathcal{I}$$ the binary icosahedral group of order 120). For this see the Theorem 4.3.2 of https://arxiv.org/abs/2006.14417. In fact, we construct a regular $$\mathcal{I}$$-equivariant cellular structure on $$\mathbb{S}^3$$ using orbit polytopes. It has f-vector [1,5,5,1], which could be minimal. Moreover, the homology chain complex associated to it is quite simple, the first and third differentials are zero and the middle one is a sparse invertible circulant matrix. Hence and unfortunately, this decomposition of $$\mathbb{S}^3/\mathcal{I}$$ is not regular. Hope this still helps a bit though.

• Thank you very much for your answer. However, I don't fully understand this yet... Let's see, in a regular CW-complex decomposition, the coefficients of the boundary matrix must be +1 or -1 (working with integral homology). Am I right? So, the f-vector [1,5,5,1] should not be possible for the POincaré Homology Sphere since it would not give homology in dimension 3? Does it make sense what I am saying? Maybe, the f-vector could be [5,25,25,5]? But I would not know how to describe the boundary maps now...Thanks in advance for your time anyway!! Dec 21, 2020 at 9:42
• You're almost right ! For a regular CW-decomposition and with integral coefficients, the homology chain complex must only involve differentials with entries $\pm1$ or 0. As I have mentioned, the differentials in our complex do respect this condition. In particular, the third differential is zero and has source $\mathbb{Z}$, so $H_3=\mathbb{Z}$ as expected. The middle differential $d_2 : \mathbb{Z}^5 \to \mathbb{Z}^5$ is invertible, so $H_1=H_2=0$ and the first differential $d_1 : \mathbb{Z}^5\to\mathbb{Z}$ is zero, so $H_0=\mathrm{coker}(d_1)=\mathbb{Z}$. Dec 21, 2020 at 14:52
• But how do you get a matrix with only zeros if the attaching map of the three cell is a homeomorphism onto its image ? Dec 21, 2020 at 19:00
• Oops ! Mea culpa ! Indeed, the decomposition of $\mathbb{S}^3/\mathcal{I}$ is not regular indeed, but the $\mathcal{I}$-equivariant one we found on the sphere is. More zeros appear when applying the functor $-\otimes_{\mathbb{Z}[\mathcal{I}]}\mathbb{Z}$ to the equivariant chain complex to obtain the non-equivariant one on the orbit space. Thank you very much @HenrikRüping for your correction ! Dec 22, 2020 at 3:38