$\pi_1(X)\cong \mathbb Z_{p^n}$

Question

If $p$ is a prime. Can one construct a space $X$ such that $\pi_1(X)\cong \mathbb Z_{p^n}$, for any $n\in \mathbb N$?

This question doesn't make any sense. Fundamental groups are groups, finite fields are fields. You cannot have an isomorphism between a group and a field, because they live in different categories. What do you mean then? Also, how do you define $\mathbb{Z}_{p^n}$? — Martin Brandenburg, Sep 17 '13 at 23:14
It sounds like you expect there to be a complete list of topological spaces $X$ with $\pi_1(X) \cong \Bbb Z/p^n$. This simply won't be the case. Even for $\pi_1(X) \cong 0$ there's more spaces than you could possibly imagine. Are you instead looking for some examples? — Henry T. Horton, Sep 17 '13 at 23:16
First of all: http://en.m.wikipedia.org/wiki/Eilenberg–MacLane_space answers your edited question. — archipelago, Sep 17 '13 at 23:27
$[p^{n-1}]*[p]=[p^n]=[0]$, so for $n>1$ there are some zero divisors, hence its not a field. — archipelago, Sep 17 '13 at 23:28
It is a group under addition though. OP, since you used the term "finite field" at first, are you thinking about the additive group of ${\Bbb F}_{p^n}$ (which is ${\Bbb Z}_p^n$) or the cyclic group of order $p^n$, which is $\Bbb Z/p^n$? — anon, Sep 18 '13 at 00:26

score 2 · Answer 1 · answered Sep 17 '13 at 23:32

For any group $G$, you can construct a space $X$ such that $\pi_1(X) \cong G$.

In the present case, you can take the infinite-dimensional sphere over the complex numbers, i.e., the set of $(z_1, z_2, \ldots)$ with all but finitely many entries equal to $0$ and such that $\sum_i |z_i|^2 = 1$. This space is contractible. Then let $G$ be the set of $(p^n)^{\text{th}}$ roots of unity in the complex numbers, and define an action $G \times S^\infty \to S^\infty$ by $(\zeta; z_1, z_2, \ldots) \mapsto (\zeta z_1, \zeta z_2, \ldots)$. The orbit space $X = S^\infty/G$ gives a space you want. The covering space projection $S^\infty \to S^\infty/G$ in fact gives a universal bundle for this group $G$.

score 2 · Answer 2 · answered Sep 17 '13 at 23:34

For any group $G$ we can construct a topological space $X$ with $\pi_1(X) \cong G$.

In the case of $G = \Bbb Z/p^n$, we can just use lens spaces. If we consider the $3$-sphere $S^3$ as the set of unit vectors in $\Bbb C^4$, we can define a free $\Bbb Z/p^n$-action on $S^3$ by $$n \cdot (z_1, z_2) = (e^{2\pi n i/p} z_1, e^{2 \pi n i/p} z_2).$$ The quotient of $S^3$ by this action is a $3$-manifold $L(p^n; 1)$ with fundamental group $\Bbb Z/p^n$.

score 2 · Answer 3 · edited Apr 13 '17 at 12:20

The standard example would be the classifying space of $\mathbb{Z}_{k}$ which is homotopy equivalent to the infinite lens space $L(k,\infty)$ where $L(k,\infty)=S^{\infty}/{\sim}$. Here, $S^{\infty}$ is the infinite dimensional unit sphere in $\mathbb{C}^{\infty}$, and $\sim$ is the equivalence relation generated by the free group action of $\mathbb{Z}_k$ on $S^{\infty}$ given by scalar multiplication of the elements of $S^{\infty}\subset\mathbb{C}^{\infty}$ by the $k$th roots of unity.

That is, $\gamma\colon\mathbb{Z}_n\times S^{\infty}\rightarrow S^{\infty}$ is given by $$\gamma(n,(z_0,z_1,\ldots))=\exp\left(\frac{2ni\pi}{k}\right)(z_0,z_1,\ldots)$$ and for $z,z'\in S^{\infty}$, $$z\sim z'\iff \gamma(n,z)=z'$$ for some $n\in\mathbb{Z}_k$.

It's easy to see that $\pi_1(L(k,\infty))\cong\mathbb{Z}_k$ because $S^{\infty}$ is well known to be contractible and the quotient map on to the lens space is a universal covering space projection with finite fiber of cardinality $k$. We see that $\gamma(1,\cdot)$ generates the automorphism group of the fiber of the covering space and so the fundamental group is cylic of order $k$ hence $$\pi_1(L(k,\infty))\cong\mathbb{Z}_k.$$

Such a space is also called a $K(\mathbb{Z}_k,1)$ in reference to the usual notation used for an Eilenberg-MacLane Space.

$\pi_1(X)\cong \mathbb Z_{p^n}$

3 Answers3