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I'm working on the Poincaré Homology Sphere $P_3$ and would like to compute it's Homology $H_1$ and fundamental group. I would like to identify it's fundamental group with the binary icosahedral group $I^*$ in view of the representation of $P_3$ as the quotient space $S^3/I^*$. Also, using the isomorphism of the abelianized fundamental group $\pi_1(P_3)/[\pi_1(P_3),\pi_1(P_3)]$ onto $H_1$ I would like to conclude that $H_1$ is trivial. Anybody who can help me with this?

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I don't have a complete answer for you, but I can say a bit.

It's a general theorem that if we have a covering space $p:\tilde{X}\rightarrow X$ with $\tilde{X}$ simply-connected and $X$ path-connected and locally path-connected, then the group of deck transformations $G$ is isomorphic to $\pi_1(X)$. (For example, see Hatcher p.71 prop. 1.39.) All the hypotheses hold for this covering $S^3\rightarrow S^3/I^*=P_3$, so this shows that $I^*=\pi_1(P_3)$.

The rest is pure group theory. First look at the icosahedral rotational group, call it $I$. $I$ is a simple group of order 60. The commutator subgroup is always normal, so $[I,I]$ is either trivial or all of $I$. $[I,I]$ is not trivial because $I$ is not abelian, so $I=[I,I]$.

Now the center $Z$ of $I^*$ is a two-element group, and $I^*/Z = I$. Clearly $[I^*,I^*]$ maps onto $[I,I]$. So $I^*/[I^*,I^*]$ is either trivial or has two elements.

The correct answer is that $[I^*,I^*]=I^*$ (wikipedia), but I'm afraid I don't know a proof. Showing that $-1\in I^*$ is a product of commutators would do the trick.

A friend referred me to Weibel's book An Introduction to Homological Algebra. Section 6.9, pp.198-199, "Universal Central Extensions", proves (Example 6.9.1 plus Lemma 6.9.2) that $I^*$ is perfect, i.e., equal to its commutator subgroup. The key fact is that $I^*$ is a universal central extension of $I$. For this fact, Weibel refers us to Suzuki, Group Theory I. I don't have that handy.

However, playing with my dodecahedral paperweight suggests a proof strategy. Let $\tilde{a}$ and $\tilde{b}$ be the two generators of $I^*$ in the presentation $\langle \tilde{a},\tilde{b} | \tilde{a}^5 = \tilde{b}^3 = (\tilde{a}\tilde{b})^2\rangle$. Their images in the icosahedral group are $a$, a rotation of $72^\circ$, and $b$, a rotation of $120^\circ$ (and $ab$ is a rotation of $180^\circ$). In $I$, $a^5=b^3=(ab)^2=1$. Since 1 in $I$ has two preimages in $I^*$, namely $\pm 1$, we must have $\tilde{a}^5 = \tilde{b}^3 = (\tilde{a}\tilde{b})^2 = -1$. (If it equalled 1, then we'd get $I$ instead of $I^*$.)

Now $[a,b]$ is a rotation of $72^\circ$ in $I$ ("proof by paperweight"). So it's conjugate to $a$ in I, and thus $[a,b]^5=1$. That suggests that $[\tilde{a},\tilde{b}]^5$ should equal $-1$ in $I^*$. It should be possible to verify this by direct computation with $2\times 2$ matrices in $SU(2)$.

Hatcher's Example 2.38, "An Acyclic Space", p.142, is also interesting to look at.

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  • $\begingroup$ Thank you so much. This was very helpful! $\endgroup$ – Rune Christiansen Apr 17 '14 at 16:22
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The key is that since the action of $I^*$ on $S^3$ is free, the fundamental group (via the usual theory of covering spaces) should be easy enough to compute. As for homology, this is a purely algebraic problem. What can you say about the commutator subgroup of $I^*$? (In this case, I would look up the notion of a 'perfect' group)

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  • $\begingroup$ Thank you for your respond. I work with covering spaces for the first time, so there might be obvious things that are yet hard to understand for me. I know that the automorphism group of $S^3$ is isomorphic to the fundamental group of the quotient space $S^3/I^*$, so I guess I just need to convince myself that the automorphism group is equal to $I^*$. $\endgroup$ – Rune Christiansen Apr 17 '14 at 14:35
  • $\begingroup$ The automorphism group of $S^3$ is not $I^*$, but $I^*$ acts freely on $S^3$. The automorphism group of the covering, however, is $I^*$ as you say. Since $S^3$ is simply connected, a lot of the theorems about covering spaces are a lot simpler. $\endgroup$ – Simon Rose Apr 17 '14 at 14:48
  • $\begingroup$ Yes of the covering, that's what I meant, sorry. I will try to figure that out. For the homology part, I know that I would like to show that $I^*$ is perfect, i.e. equal to its commutator subgroup, but I find it very hard to compute since quaternion multiplication is non commutative which makes it very hard to express the elements of $I^*$ as a general word in powers of the two generators. But maybe I overlook something. $\endgroup$ – Rune Christiansen Apr 17 '14 at 15:46

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