Integral homology of lens space.

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Consider a $S^1$-fibration $L^n\to\mathbb CP^{n-1}$. Spectral sequence for this fibration gives an answer $$ H_0(L^n)=\mathbb Z,\,\,\, H_1(L^n)=H_3(L^n)=\ldots=H_{2n-3}=\mathbb Z_p,\,\,\, H_{2n-1}(L^n)=\mathbb Z. $$

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Updated on August 01, 2022

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  • Admin
    Admin over 1 year

    Let $p$ be an odd prime number. Regard the cyclic group $\pi$ of order $p$ as the group of $p$th roots of unity contained in $S^1$. Regard $S^{2n-1}$ as the unit sphere in $\mathbb{C}^n$, $n \ge 2$. Then $\pi \subset S^1$ acts freely on $S^{2n-1}$ via$$\zeta(z_1, \dots, z_n) = (\zeta z_1, \dots, \zeta z_n).$$Let $L^n = S^{2n-1}/\pi$ be the orbit space. The obvious quotient map $S^{2n-1} \to L^n$ is a universal covering. My question is, what is the integral homology of $L^n$, $n \ge2$?

    • Max
      Max about 8 years
      Can you compute the fundamental group of the lens space?
    • Admin
      Admin about 8 years
      Have you done the respective computation for $n=2$? You want to do the same thing you do for $\Bbb{RP}^n$: build it as an appropriate cell complex, after which the homology will be close to obvious. (Of course one should note that you're getting many different spaces, depending on the $z_i$.)