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    Homotopy classification of lens spaces

    E-HAT-4.2-29

    Area
    Topology
    Topics
    Higher Homotopy Groups
    Status
    draft

    Finish the homotopy classification of lens spaces begun in Exercise 2 of §3.E by showing that two lens spaces \(L_m(\ell_1, \ldots, \ell_n)\) and \(L_m(\ell_1', \ldots, \ell_n')\) are homotopy equivalent if \(\ell_1 \cdots \ell_n \equiv \pm k^n \ell_1' \cdots \ell_n' \pmod{m}\) for some integer \(k\).

    Solution

    Put \[ P=\ell_1\cdots\ell_n, \qquad P'=\ell_1'\cdots\ell_n'. \] Assume \[ P\equiv \pm k^nP'\pmod m. \] Since \(P\) and \(P'\) are units modulo \(m\), so is \(k\). Replacing the chosen generator of the \(\mathbb Z_m\)-action on the target by its \(k\)-th power replaces every \(\ell_j'\) by \(k\ell_j'\) without changing the lens space. Thus we may reduce to the case \[ P\equiv\pm P'\pmod m. \]

    Choose nonzero integers \(k_j\) satisfying \[ k_j\ell_j\equiv\ell_j'\pmod m. \] Exercise 2 of §3.E constructs an equivariant map of universal covers and hence a map \[ f:L_m(\ell_1,\dots,\ell_n)\to L_m(\ell_1',\dots,\ell_n') \] that induces an isomorphism on \(\pi_1\) and has degree \[ d_0=k_1\cdots k_n. \] Modulo \(m\), \[ d_0P\equiv P', \] so the reduced congruence gives \[ d_0\equiv\pm1\pmod m. \] Choose \(d\in\mathbb Z\) with \[ d_0+dm=\pm1. \]

    Modify \(f\) inside a small ball in the source as follows. Pinch that ball to add an \(S^{2n-1}\)-summand, map this new sphere to \(S^{2n-1}\) with degree \(d\), then compose with the \(m\)-sheeted covering \[ S^{2n-1}\to L_m(\ell_1',\dots,\ell_n'). \] The resulting map \(g\) agrees with \(f\) on the \(1\)-skeleton, so it induces the same isomorphism on \(\pi_1\), while its degree is \[ \deg g=d_0+dm=\pm1. \]

    Lift \(g\) to universal covers. Since the covering maps on both sides have degree \(m\), the lift \[ \widetilde g:S^{2n-1}\to S^{2n-1} \] has degree \(\pm1\), hence is a homotopy equivalence. Therefore \(g\) induces isomorphisms on all higher homotopy groups as well as on \(\pi_1\). Lens spaces are CW complexes, so Whitehead’s theorem gives \[ \boxed{L_m(\ell_1,\dots,\ell_n)\simeq L_m(\ell_1',\dots,\ell_n').} \] This proves the converse and completes the classification.