Math 2.0

Manifolds and cell complexes Three-manifolds Example

Lens space

Example

Let p≥2p \ge 2 and qq be coprime integers, and let ζ=e2πi/p\zeta = e^{2\pi i/p}. View the 33-sphere as S3={(z1,z2)∈ℂ2:∣z1∣2+∣z2∣2=1}S^3 = \set{(z_1, z_2) \in \C^2 : \abs{z_1}^2 + \abs{z_2}^2 = 1}. The map g(z1,z2)=(ζz1,ζqz2)g(z_1, z_2) = (\zeta z_1, \zeta^q z_2) generates an action of the cyclic group ℤ/p\Z/p on S3S^3. The lens space L(p,q)L(p, q) is the quotient S3/(ℤ/p)S^3 / (\Z/p).

The action is free. Indeed, suppose gkg^k fixes (z1,z2)(z_1, z_2) with 0<k<p0 < k < p. Then ζk≠1\zeta^k \ne 1 forces z1=0z_1 = 0, and ζkq≠1\zeta^{kq} \ne 1 (because qq is prime to pp) forces z2=0z_2 = 0, which is impossible on S3S^3. So the quotient map is a covering map and L(p,q)L(p, q) is a closed orientable 3-manifold. Its fundamental group is ℤ/p\Z/p, since S3S^3 is simply connected.

Remarks

The integral homology of L(p,q)L(p, q) is H0=ℤH_0 = \Z, H1=ℤ/pH_1 = \Z/p, H2=0H_2 = 0 and H3=ℤH_3 = \Z. For p=2p = 2 one gets L(2,1)L(2, 1), the real projective space ℝℙ3\mathbb{RP}^3. A lens space is also the union of two solid tori glued along their boundary tori, so it has a Heegaard splitting of genus 11. The spaces L(p,q)L(p, q) and L(p,q′)L(p, q') are homeomorphic if and only if q′≡±q±1(modp)q' \equiv \pm q^{\pm 1} \pmod p. They are homotopy equivalent if and only if qq′≡±n2(modp)q q' \equiv \pm n^2 \pmod p for some integer nn. So L(7,1)L(7, 1) and L(7,2)L(7, 2) are homotopy equivalent, because 2≡32(mod7)2 \equiv 3^2 \pmod 7, but they are not homeomorphic, because 2≢±1(mod7)2 \not\equiv \pm 1 \pmod 7.

Further reading

  1. Web https://en.wikipedia.org/wiki/Lens_space
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MSC57K30, 57M10
Instance of3-manifold
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Changed2 November 2026

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