Can the fundamental group $\pi_1(X)$ of topological space $X$ be non-abelian? or be continuous?

Question

We are familiar with the homotopy group of spheres $S^n$: https://en.wikipedia.org/wiki/Homotopy_groups_of_spheres. There we learn that $\pi_d(S^n)$ must be abelian and discrete. They are direct sum of $\mathbb{Z}$ and $\mathbb{Z}/p$ for some positive integer $p\geq 2$.

But there are general homotopy group of topological spaces $X$.

For $\pi_d(X)$, $d \geq 2$, am I correct to say that $\pi_d(X)$ must be an abelian group and a discrete finite group?
For the fundamental group $\pi_1(X)$:

Is it possible that $\pi_1(X)$ can be non-abelian?
Is it possible that $\pi_1(X)$ can be continuous, not just discrete?

For example, we can construct space like the wedge sum $S^1 \vee S^1 \vee S^m \vee S^n \vee \dots$?

And the space you stated at the end has non-abelian $\pi_1$. — Arctic Char, Aug 02 '21 at 16:12
continuous group like $S^1 \cong U(1)$ is a continuous abelian group — Марина Marina S, Aug 02 '21 at 16:12
Ok..... it is better to give the definitions, instead of throwing out two examples. — Arctic Char, Aug 02 '21 at 16:13
You're confusing something. The fundamental group (a priori) does not come with a topology (that's why Arctic Char was asking for a definition of discrete groups). — Qi Zhu, Aug 02 '21 at 16:15
Yes, higher homotopy groups are abelian. This is by the famous Eckmann-Hilton argument. — Qi Zhu, Aug 02 '21 at 16:15
Here is a thread discussing how to endow a topology on the fundamental group: https://mathoverflow.net/questions/26680. — Qi Zhu, Aug 02 '21 at 16:19
Post related to the comment above: every group is a fundamental group. It seems that settles your question, no matter what the definition of discrete and continuous are. — Arctic Char, Aug 02 '21 at 16:29
The fundamental group of a bouquet of $n$ circles is the free group of rank $n$. — Arturo Magidin, Aug 02 '21 at 16:31
many thanks - I wanted to vote up but I do not have enough reputation to vote up (at least 15 or 50 reputation for me to do something...) thanks again! — Марина Marina S, Aug 02 '21 at 16:41

score 2 · Accepted Answer · answered Aug 02 '21 at 16:34

For any group $G$ there is actually a CW complex $X$ with $\pi_1(X)=G,$ so yes the fundamental group can be nonabelian. The construction is pretty simple and goes like this: Let $\langle \{g_\alpha\}_{\alpha\in I}\;|\;\{r_\beta\}_{\beta\in J}\rangle$ be a presentation for $G,$ where the $g_\alpha$ are generators and the $r_\beta$ are words in these generators which give the relations for $G.$ Then we let $X$ be the CW complex given by taking one circle for each generator $g_\alpha$ and wedging them all together. Then we attach 2-cells to this wedge of circles according to the relations. I think there's a theorem in Hatcher's book somewhere in Chapter 1 which proves that this gives the fundamental group you want.

For example, if $G=\mathbb Z\times\mathbb Z=\langle a,b\;|\;aba^{-1}b^{-1}\rangle$ then we start with a wedge of two circles, one corresponding to $a$ and the other to $b$ and we attach a disk by attaching one quarter of the boundary along $a,$ then the next quarter along $b,$ then the next quarter along $a$ but this time in reverse, and the final quarter along $b$ but again in reverse. After some thought and some drawings you should convince yourself that this is just a torus.

The trouble with your second question is that there is usually no topology on the fundamental group. Qi Zhu linked a nice discussion on MathOverflow discussing this in the comments. But if you just want a fundamental group isomorphic to $S^1$ or something as groups then the construction above will give you one.

Can the fundamental group $\pi_1(X)$ of topological space $X$ be non-abelian? or be continuous?

1 Answers1