View \(S^3\) as the set of unit vectors in \(\RR^4\), and consider the equivalence relation on them induced by \(u\sim v\) if \(A\cdot u=v\), where \[A=\begin{pmatrix}0&1&0&0\\-1&0&0&0\\0&0&0&1\\0&0&-1&0\end{pmatrix}.\] Compute the fundamental group of the quotient space \(S^3/\sim\).
Solution
Let \(G=\langle A\rangle\le O(4)\), acting on \(S^3\). The equivalence relation generated by \(u\sim Au\) has the \(G\)-orbits as its classes, so \(S^3/\sim\) is the orbit space \(S^3/G\).
1. \(G\) is cyclic of order \(4\).
\(A\) is block diagonal with two copies of \(J=\begin{pmatrix}0&1\\-1&0\end{pmatrix}\), and \(J^2=-I_2\), so \(A^2=-I\) and \(A^4=I\). Since \(A^2\neq I\), the order of \(A\) is \(4\).
2. \(G\) acts freely on \(S^3\).
The nonidentity elements of \(G\) are \(A\), \(A^2=-I\), and \(A^3\). If \(Av=v\), then \(-v=A^2v=v\), so \(v=0\notin S^3\). The element \(-I\) has no fixed point on \(S^3\). If \(A^3v=v\), then applying \(A\) gives \(v=Av\), which is the first case.
3. The quotient map \(p\colon S^3\to S^3/G\) is a covering map with deck transformation group \(G\).
A finite group acting freely by homeomorphisms on a Hausdorff space acts properly discontinuously, so by step 2 the orbit map is a covering map. For the orbit map of a properly discontinuous action of \(G\) on a connected space, the deck transformation group is \(G\).
4. Q.E.D.
\(S^3\) is simply connected, so by step 3 the covering \(p\) is universal, and the fundamental group of its base is isomorphic to its deck transformation group. By step 1, \[\pi_1(S^3/\!\sim)\cong\pi_1(S^3/G)\cong G\cong\boxed{\ZZ/4\ZZ}.\]