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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!

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    $\begingroup$ 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. $\endgroup$ Dec 19, 2020 at 14:36
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    $\begingroup$ @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? $\endgroup$
    – D1811994
    Dec 20, 2020 at 11:47
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    $\begingroup$ 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. $\endgroup$ Dec 20, 2020 at 13:10

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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...

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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.

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  • $\begingroup$ 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!! $\endgroup$
    – D1811994
    Dec 21, 2020 at 9:42
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    $\begingroup$ 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 $H_0=\mathrm{coker}(d_1)=\mathbb{Z}$. $\endgroup$ Dec 21, 2020 at 14:52
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    $\begingroup$ But how do you get a matrix with only zeros if the attaching map of the three cell is a homeomorphism onto its image ? $\endgroup$ Dec 21, 2020 at 19:00
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    $\begingroup$ 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 ! $\endgroup$ Dec 22, 2020 at 3:38

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