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The Poincaré homology sphere $X$ is constructed by identifying opposite faces of a dodecahedron by a (clockwise) twist of 36 degree.

Many books say its fundamental group $\pi_1(X)$ is the binary icosahedron group, my question is how to prove this. In Ratcliffe's hyperbolic manifold book, there is a theorem of Poincaré about computing $\pi_1$ of such spaces, following this method one can check $\pi_1(X)$ has a presentation of $6$ generators, $a, b, c, d, e, f$ and relations $a=bd=ce=df=eb=fc$ and $f=be$, how to show this is isomorphic to the 120-order icosahedron group? Is there any easy way to see this? Thanks!

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  • $\begingroup$ This is an easy exercise I used to give to undergraduate students of algebraic topology. The quotient space inherits a cell structure with only one vertex from the dodecahedron. The fundamental group can therefore be easily computed from the cell decomposition in the usual way. $\endgroup$
    – Fernando Muro
    Commented Mar 20, 2012 at 22:33
  • $\begingroup$ Well, I don't remember now whether the inherited cell structure has exactly one vertex, or maybe more, but this is not really important for the computation. $\endgroup$
    – Fernando Muro
    Commented Mar 20, 2012 at 22:52
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    $\begingroup$ Fernando: I think the question is not "how does one compute a presentation of $\pi_1(X)$", but why the presentation is a presentation of the binary icosahedral group. In particular, why it finite (and of order 120). I don't know any simple way to see this. $\endgroup$
    – Paul
    Commented Mar 20, 2012 at 22:53
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    $\begingroup$ Have you looked at en.wikipedia.org/wiki/Binary_icosahedral_group and tried to map $a,\ldots,f$ to some unit quaternions? $\endgroup$
    – Mark Grant
    Commented Mar 20, 2012 at 23:51
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    $\begingroup$ I believe that there is the following approach: In $SO(3)$ there is a group $G$ of order $60$, symmetries of the dodecahedron. Thinking of $SO(3)$ as a Riemannian manifold, the twelve elements of $G$ closest to the identity are the $36$ degree rotations in $G$. The elements of $SO(3)$ that are closer to the identity than to any other element of $G$ form a domain shaped like a dodecahedron, and the manifold $SO(3)$ can be assembled by gluing together the sixty translates of this by the elements of $G$. The $3$-sphere, double cover of $SO(3)$, is made of $120$ of these. $\endgroup$
    – Tom Goodwillie
    Commented Mar 21, 2012 at 3:28

4 Answers 4

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You can read the book by Seifert and Threlfall "A textbook of topology", pages 223-225: They start by writing down a presentation of $\pi_1$ of the Poincare homology sphere (by reading off generators and relators from the identification of faces of the spherical dodecahedron). Then they describe how to transform this presentation to the one of the icosahedral group. No computers are required, just some patience with their somewhat archaic style.

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The presentation for $\pi_1(X)$ that you write down simplifies to $\langle b, e \mid beb=eb^2e, ebe=be^2b\rangle$. As is well-known, the binary icosahedral group $I^*$ is isomorphic to the group of unimodular, $2\times 2$ matrices over $\mathbb{Z}_5$. An explicit isomorphism $\pi_1(M) \to {\rm SL}(2,5)$ is given by $$ b \mapsto \begin{pmatrix} 3 & 1 \\\\ -1 & 0 \end{pmatrix}, \quad e \mapsto \begin{pmatrix} -1 & -1 \\\\ 0 & -1 \end{pmatrix}. $$

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  • $\begingroup$ @ Alex: What is the simplest way to see that the map you wrote is injective? You are right that the fact that this presents $I^*$ is "well-known", but the arguments I know involve some geometry, eg Goodwillie's comment. $\endgroup$
    – Paul
    Commented Mar 21, 2012 at 22:39
  • $\begingroup$ Well, I have to confess, I found this isomorphism with the help of GAP, which assures me this map (which can be checked "by hand" is a homomorphism) is indeed a bijection. I would need more motivation to verify this by hand, but surely it can be done. Now, as to why $I^*$ is isomorphic to ${\rm SL}(2,5)$. One could use the same kind argument: start with a known presentation of $I^*$, say, $\langle x, y \mid x^3=y^5=(xy)^2 \rangle$, write down a map to ${\rm SL}(2,5)$, and check it's an isomorphism. Or, one could use a more geometric approach (symmetries of the icosahedron, quaternions, etc). $\endgroup$
    – Alex Suciu
    Commented Mar 22, 2012 at 1:07
  • $\begingroup$ Or, one could cite <en.wikipedia.org/wiki/Binary_icosahedral_group> or <groupprops.subwiki.org/wiki/Special_linear_group:SL(2,5)> (if url's would work nicely in this markup language, that is...) $\endgroup$
    – Alex Suciu
    Commented Mar 22, 2012 at 1:26
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A quite direct argument is to solve exercise 4.4.17 in page 252 of Thurston's book 3-D Geom & Topo - the published one in 1997; in this exercise Thurston considered the action of the icosahedral group on the unit circle bundle $UTS^2=RP^3$; this is a free action, and the point is to describe a fundamental domain and show that the pattern of gluing of its boundary is exactly the same as in the Poincare dodecahedral space. When doing this exercise, one may want to notice two things: 1. On $S^2$, each of the $5$ angles of a spherical pentagon face of a dodecahedron is $120$ degrees, in contrast to $108$ degrees for a pentagon on plane - this is useful when you try to create a local coordinate system on the circle bundle $UTS^2$; 2. The fundamental domain Thurston created here does not look like a dodecahedron in the metric sense - it look like a cylinder over a pentagon base; however other fundamental domains will mark a line, with quite a slope, on each of the five side faces and therefore each side face splits into two "pentagons"... With these in mind, exercise 4.4.17 should be very pleasant to work out.

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Here we take a 3D pentagon instead of a "spherical pentagon" - does it correspond to Thurston's idea of the fundamental domain? http://quantumcinema.uni-ak.ac.at/site/research/current-topics/

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