Compute the $\pi_{1} (S^{3} /G)$ and $H_{m} ( S^{3} / G)$ for the following $G$

Question

I am studying for a qualying exam and there is one exercise from the previous years exams which I don't know how to approach.

Let $n$ be a positive integer and $$ G = \{ g \in \Bbb C^{\times} | g^{n} =1 \} $$ the group of the $n$-roots of unity. The group $G$ acts by multiplication on the sphere $$ S^{3} = \{ a + bi +cj+dk \in \mathbb{H} \: | \: a^{2}+b^{2}+c^{2}+d^{2}=1 \} \subset \mathbb{H} = \{ a+bi+cj+dk | a,b,c,d \in \mathbb{R} \} $$ seen as subset of quaternions $\mathbb{H}$. Let $X = S^{3} / G $ be the quotient space. Compute the fundamental group $ \pi_{1} (X, x_{0})$ respect an arbitrary base point $x_{0} \in X$, and the homology groups $H_{m} ( X , \mathbb{Z} )$ for all $m$.

I have found this post:

Calculate the homology group of $S^3/G$, an Harvard qualifying exam problem with "unclear" solution

But I don't know what is Stiefel-Whitney class.

I don't know how to attack this problem, if my intuition with $S^{2}$ serves me on something, this space is a $n$ folded version of $S^{3}$, so we can probably expect groups with nonzero torsion. But I don't know how to procced. I have studied Hatcher's chapters 1 and 2, and Rotman's chapters 1-5, is there any way to solve this using that knowledge? Or should I learn additional material?

Answer 1

$G$ is finite group that acts freely on $S^3$. Because $S^3$ is simply connected, $\pi_1(S^3/G)\cong G\cong \Bbb Z/n$. (Since each element in $G$ represents a rotatation (homeomorphism) of a Hausdorff space, that $G$ acts freely on $S^3$ implies proper discontinuity, so you can apply proposition 1.40 in Hatcher, which you have studied.)

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Now, we compute the integral homology groups.

• We have $H_0(S^3/G)\cong \Bbb Z$ by connectedness and $H_1(S^3/G)\cong\pi_1(S^3/G)\cong \Bbb Z/n$ by abelianization.

We would like to use Poincare duality to obtain the information of other homology groups. To do so, we need to check the orientability first. Since the action is free, we know that $S^3/G$ is a spherical manifold, which is orientable.

• Orientability implies that $H_3(S^3/G)\cong \Bbb Z$.
• We know that $$H_2(S^3/G)\cong H^1(S^3/G)\cong\operatorname{Hom}(H_1(S^3/G),\Bbb Z)\oplus \operatorname{Ext}(H_0(S^3/G),\Bbb Z)\cong 0$$ by Poincare duality and Universal Coefficient Theorem.

In conclusion, $$H_k(S^3/G;\Bbb Z)\cong\begin{cases}\Bbb Z & k=0,3\\ \Bbb Z/n & k=1\\ 0 &\text{otherwise} \end{cases}$$

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To see that the action of $G$ on $S^3$ is free. We may identify $\Bbb H$ with $\Bbb R^4$ and elements in $G$ with matrices. Suppose $h\cdot x=x$ for some $h=\cos\theta+i\sin\theta\in G$ and $x=a+bi+cj+dk\in S^3\subset\Bbb H$ ($x$ will be identified with $\vec{x}=(a,b,c,d)\in\Bbb R^4$), then \begin{align} \begin{bmatrix} \cos\theta & -\sin\theta & 0 & 0\\ \sin\theta & \cos\theta & 0 & 0\\ 0 & 0 & \cos\theta & -\sin\theta\\ 0 & 0 & \sin\theta & \cos\theta \\ \end{bmatrix} \begin{bmatrix}a\\b\\c\\d\end{bmatrix} = \begin{bmatrix} a\cos\theta-b\sin\theta\\ a\sin\theta+b\cos\theta\\ c\cos\theta-d\sin\theta\\ c\sin\theta+d\cos\theta \end{bmatrix} =\begin{bmatrix}a\\b\\c\\d\end{bmatrix} \end{align} ($h$ is identified with the $4\times 4$ matrix) This shows that $\vec{x}=(a,b,c,d)$ is an eigenvector, and $1$ is the eigenvalue corresponding to $\vec{x}$. By direct computation, the characteristic polynomial of the matrix is $$((\cos\theta-\lambda)^2+\sin^2\theta)^2$$ Since $\lambda=1$ is a root, we have $$(\cos\theta-1)^2+\sin^2\theta=0$$ It's a sum of two squares, so we have $\sin\theta=0$ and $\cos\theta=1$, which proves that $h=1\in G$. Hence, the action is free.