Consider \(S^3\) as the space of all pairs of complex numbers \((z_1, z_2)\) satisfying the equation \(\abs{z_1}^2 + \abs{z_2}^2 = 1\). Given relatively prime positive integers \(n\) and \(k\), define \(h: S^3 \to S^3\) by the equation
\[ h(z_1, z_2) = (z_1 e^{2\pi i/n}, z_2 e^{2\pi i k/n}). \]
Show that \(h\) generates a subgroup \(G\) of the homeomorphism group of \(S^3\) that is cyclic of order \(n\), and that only the identity element of \(G\) has a fixed point. The orbit space \(S^3/G\) is called the lens space \(L(n, k)\).
Show that if \(L(n, k)\) and \(L(n', k')\) are homeomorphic, then \(n = n'\). [It is a theorem that \(L(n, k)\) and \(L(n', k')\) are homeomorphic if and only if \(n = n'\) and either \(k \equiv k' \pmod{n}\) or \(kk' \equiv 1 \pmod{n}\). The proof is decidedly nontrivial.]
Show that \(L(n, k)\) is a compact 3-manifold.
Solution
- The map \[ h(z_1,z_2)=(e^{2\pi i/n}z_1,e^{2\pi ik/n}z_2) \] is a homeomorphism of \(S^3\), with inverse obtained by replacing the exponents by their negatives. Also \(h^n=1\).
Suppose \(h^j\) fixes \((z_1,z_2)\in S^3\), with \(0<j<n\). At least one coordinate is nonzero. If \(z_1\ne0\), then \[ e^{2\pi ij/n}=1, \] forcing \(n\mid j\), impossible. If \(z_1=0\), then \(z_2\ne0\) and \[ e^{2\pi i k j/n}=1. \] Since \(\gcd(k,n)=1\), again \(n\mid j\), impossible. Thus no nonidentity power has a fixed point. In particular \(h\) has exact order \(n\), so \[ G=\langle h\rangle\cong\mathbb Z/n. \]
The free finite action makes \[ S^3\longrightarrow L(n,k)=S^3/G \] a regular \(n\)-sheeted covering. Since \(S^3\) is simply connected, it is the universal cover, and its group of covering transformations is \(G\). Therefore \[ \pi_1(L(n,k))\cong G\cong\mathbb Z/n. \] A homeomorphism \(L(n,k)\cong L(n',k')\) induces an isomorphism of fundamental groups, so \[ \mathbb Z/n\cong\mathbb Z/n'. \] Finite cyclic groups are isomorphic only when their orders agree. Hence \(n=n'\).
The sphere \(S^3\) is compact. Its continuous image under the quotient map is therefore compact. Because the action is free and \(G\) is finite, for every \(x\in S^3\) one can choose a small open ball \(U\) such that \[ gU\cap U=\varnothing\qquad(g\ne1). \] The quotient map restricts to a homeomorphism from \(U\) onto its image. Thus every point of \(L(n,k)\) has a neighborhood homeomorphic to an open subset of \(S^3\), hence to an open subset of \(\mathbb R^3\). The quotient is Hausdorff and second countable (finite group quotients preserve these properties). Therefore \(L(n,k)\) is a compact 3-manifold.