Fundamental group of the Poincaré Homology Sphere

Question

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?

Answer 1

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.

Answer 2

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)

Answer 3

The key point that seems to be missing from the other answers is a proof that the binary icosahedral group is perfect. Here are two proofs; the first is indirect but convenient if one is comfortable using a bit of character theory and is anyway interested in the McKay correspondence, while the second is an explicit computation.

For the first proof: by the McKay correspondence (which one can work out explicitly for each of the binary polyhedral groups), the binary icosahedral group has only one $1$-dimensional representation and is therefore perfect. (By contrast, the binary octahedral group has two $1$-dimensional representations, and the binary tetrahedral group has three).

Now for the second proof, which is elementary and quite direct, if computation-intensive. The quaternions $\mathbf{H}$ are the $\mathbf{R}$-algebra generated by $i,j$ and $k$ subject to $$i^2=j^2=k^2=ijk=-1.$$ Thus $ij=k$ and $ki^{-1}=-j$, implying that$$iji^{-1}j^{-1}=ki^{-1}j^{-1}=-jj^{-1}=-1$$ is a commutator in any subgroup of $\mathbf{H}^\times$ containing $i$ and $j$.

The element $$\varpi=\frac{1}{2}(1+i+j+k)$$ in $\mathbf{H}^\times$ is of order $6$ (since its trace is $1$), and together with $i$ and $j$ generates the binary tetrahedral group $T$ (of which the group of quaternions is a normal subgroup, with conjugation by $\varpi$ inducing the automorphism cycling $i \mapsto j \mapsto k \mapsto i$).

The binary icosahedral group $I$ is then generated by $T$ and $$\sigma=\frac{1}{2}(\alpha+\alpha^{-1} i+j)$$ where $$\alpha=\frac{-1+\sqrt{5}}{2}.$$

Now the computation: the conjugation action of $\sigma$ on the space of pure quaternions $ai+bj+ck$, identified with $\mathbf{R}^3$, induces a rotation of order $5$ about the axis joining the points $\pm(\alpha^{-1}i+j)$. Moreover, the $T$-orbit of these points is of size twelve, and stable by $\sigma$, hence by $I$. The convex hull of these twelve points is the icosahedron, and the action of $I$ identifies the group of rigid motions of the icosahedron (which is $A_5$) with the quotient $I/\{\pm 1\}$. Thus $I/\{\pm 1\} \cong A_5$ is simple. As we have seen above, $-1$ is in the kernel of any one-dimensional representation of $I$, and hence any one dimensional representation of $I$ factors through a one-dimensional representation of $A_5$. But there is only one of these.