Let $n > 1$. We define the infinite dimensional lens space $L$ as follows. Let $S^{\infty}$ be the unit sphere in the infinite dimensional complex vector space $C^{\infty}$, and let $\mathbb{Z}/n = \{\zeta^j | 0 \leq j \leq n-1\}$ where $\zeta$ is a primitive $n$th complex root of unity. Let $\mathbb{Z}/n$ act on $S^{\infty}$ by the formula
\begin{align*}
\zeta^j \cdot (z_1,z_2,…) = (\zeta^j z_1, \zeta^j z_2,…).
\end{align*}
I'm interested in computing all of the homotopy groups of $L$.
Here is what I have so far:
As $L$ is the orbit space of a free action of $\mathbb{Z}/n$ on $S^{\infty}$, we have the fibration sequence (or fiber sequence) $\mathbb{Z}/n \xrightarrow{i} S^{\infty} \xrightarrow{p} L$ with fiber $\mathbb{Z}/n$, total space $S^{\infty}$, base space $L$, fiber bundle (hence fibration) $p$ given by a covering space and inclusion $i$ given by a covering space action. Therefore, we obtain the long exact sequence of homotopy groups $\cdots \rightarrow \pi_k(\mathbb{Z}/n) \rightarrow \pi_k(S^{\infty}) \rightarrow \pi_k(L) \rightarrow \pi_{k-1}(\mathbb{Z}/n) \rightarrow \cdots \rightarrow \pi_0(S^{\infty}) \rightarrow 0$.
However, $S^{\infty}$ is a contractible space, so all of its homotopy groups are trivial. Further, $\mathbb{Z}/n$ is a discrete space, so all of its homotopy groups are trivial, as well. $L$ is also path-connected, so its $0$th homotopy group is trivial.
So, I'm getting that $\pi_n(L)$ is trivial for all $n \geq 0$. But this doesn't seem plausible — where am I going wrong?
Thanks!
Best Answer
You're almost right. We have
$$\dots \to \pi_k(S^{\infty}) \to \pi_k(L) \to \pi_{k-1}(\mathbb{Z}/n) \to \pi_{k-1}(S^{\infty}) \to \dots$$
so $\pi_k(L) \cong \pi_{k-1}(\mathbb{Z}/n)$ since $S^{\infty}$ is contractible. If $k > 1$, then $k - 1 > 0$ so $\pi_k(L) \cong \pi_{k-1}(\mathbb{Z}/n) = 0$. However, if $k = 1$, then we have $\pi_1(L) \cong \pi_0(\mathbb{Z}/n) = \mathbb{Z}/n$. Since $L$ is the continuous image of a path-connected space, namely $S^{\infty}$, it is path-connected so $\pi_0(L) = 0$.
Therefore $L$ has exactly one non-zero homotopy group which is $\pi_1(L) \cong \mathbb{Z}/n$. That is, $L$ is an Eilenberg-MacLane space, namely a $K(\mathbb{Z}/n, 1)$.