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  "body": "\n## Example\n\nLet $p \\ge 2$ and $q$ be coprime integers, and let $\\zeta = e^{2\\pi i/p}$.\nView the $3$-sphere as $S^3 = \\set{(z_1, z_2) \\in \\C^2 : \\abs{z_1}^2 + \\abs{z_2}^2 = 1}$.\nThe map $g(z_1, z_2) = (\\zeta z_1, \\zeta^q z_2)$ generates an action of the cyclic group $\\Z/p$ on $S^3$.\nThe lens space $L(p, q)$ is the quotient $S^3 / (\\Z/p)$.\n\nThe action is free.\nIndeed, suppose $g^k$ fixes $(z_1, z_2)$ with $0 \u003c k \u003c p$.\nThen $\\zeta^k \\ne 1$ forces $z_1 = 0$, and $\\zeta^{kq} \\ne 1$ (because $q$ is prime to $p$) forces $z_2 = 0$, which is impossible on $S^3$.\nSo the quotient map is a [[covering map|3XZRRE165N]] and $L(p, q)$ is a closed orientable [[3-manifold|R20TB2KEQQ]].\nIts [[fundamental group|S1XK3HXHFX]] is $\\Z/p$, since $S^3$ is simply connected.\n\n## Remarks\n\nThe integral homology of $L(p, q)$ is $H_0 = \\Z$, $H_1 = \\Z/p$, $H_2 = 0$ and $H_3 = \\Z$.\nFor $p = 2$ one gets $L(2, 1)$, the [[real projective space|1DDYB6VVSR]] $\\mathbb{RP}^3$.\nA lens space is also the union of two solid tori glued along their boundary tori, so it has a [[Heegaard splitting|EDE9919B5R]] of genus $1$.\nThe spaces $L(p, q)$ and $L(p, q')$ are homeomorphic if and only if $q' \\equiv \\pm q^{\\pm 1} \\pmod p$.\nThey are homotopy equivalent if and only if $q q' \\equiv \\pm n^2 \\pmod p$ for some integer $n$.\nSo $L(7, 1)$ and $L(7, 2)$ are homotopy equivalent, because $2 \\equiv 3^2 \\pmod 7$, but they are not homeomorphic, because $2 \\not\\equiv \\pm 1 \\pmod 7$.\n\n## Explanations\n\n### L1 human\n\nA lens space is what one gets by folding the $3$-sphere onto itself by a rotation of order $p$ that moves every point.\nTwo lens spaces can have the same fundamental group and the same homology, and even be homotopy equivalent, while still being different manifolds.\nThis makes them the first test case for any new invariant of $3$-manifolds.\n",
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