7
$\begingroup$

The three-dimensional lens spaces $L(p;q)$ are quotients of $S^3$ by $\mathbb{Z}/p$-actions. More precisely, let$p$ and $q$ be coprime integers and consider $S^3$ as the unit sphere in $\mathbb C^2$. Then the $\mathbb{Z}/p$-action on $S^3$ generated by the homeomorphism :$$(z_1,z_2) \mapsto (e^{2\pi i/p} \cdot z_1, e^{2\pi i q/p}\cdot z_2)$$ is free. The resulting quotient space is called the '''lens space''' $L(p;q)$.

This can be generalized to higher dimensions as follows: Let $p,q_1,\ldots,q_n$ be integers such that the q_i are coprime to $p$ and consider $S^{2n-1}$ as the unit sphere in $\mathbb C^n$. The lens space $$L(p;q_1,\ldots q_n)$$ is the quotient of $S^{2n-1}$ by the free $\mathbb Z/p$-action generated by : $$(z_1,\ldots,z_n) \mapsto (e^{2\pi iq_1/p} \cdot z_1,\ldots, e^{2\pi i q_n/p}\cdot z_n).$$

my inquiries:

  1. In Wikipedia it says: In three dimensions we have $L(p;q)=L(p;1,q).$ Why is that? Isnt that one is in 3 real and the other in 5 real dimensions?

  2. How to show: The fundamental group of all the lens spaces $L(p;q_1,\ldots, q_n)$ is $\mathbb Z/p\mathbb Z$ independent of the $q_i$?

  3. $L ( 5 ; 1 )$ and $L ( 5 ; 2 )$ were not homeomorphic even though they have isomorphic fundamental groups and the same homology, though they do not have the same homotopy type. What are the homotopy groups of $L ( 5 ; 1 )$ and $L ( 5 ; 2 )$, respectively? And their homology groups?

$\endgroup$
6
$\begingroup$

1) The latter notation corresponds to $n = 2$; the equivalence is just a matter of unwinding the definition.

2) Any such lens space $L$ is just a quotient of $S^{2n-1}$ by a nice action of $\mathbb{Z}_m$. Since $S^{2n-1}$ is simply-connected, $\pi_1 L= \mathbb{Z}_m$.

3) Similarly, $\pi_i(L) = \pi_i(S^{2n-1})$ for $i > 1$. The latter group is very complicated above dimension $2n-1$, but at least the lower $\pi_i(L)$ are clear. As for homology, the most straightforward way is to work with cellular homology; the computation is done in Hatcher. (There are other approaches that require more machinery but less computation. Normally I'm totally in favor of that, but lens spaces are one of the canonical spaces for which homology can be worked out by hand, and just about any algebraic topology book will go through all the details.) The punchline is that in the interesting range $p = 1, \dots, 2n-2$, the (integral) homology $H_p(L)$ is $\mathbb{Z}_m$ for odd $p$ and $0$ for even $p$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.