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I've recently become familiar with the notion of topological group and the theorem that the binary operation on topolgical group and the one on it's fundamental group are the same.

However I questioned whether all topological groups must be path connected, whether it makes actually sense to talk about loops on it generally? And the answer was no, there are topological groups that are not path connected.

My question is: is it that there is just a silent assumption in the theorem (the one stating the binary operations are the same), that the topological group is path connected? or am I missing some important point here?

edit: I mean why there is no assumption on path connectedness in the theorem itself?

Community
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Kristina
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You don't need to assume that a topological group is path connected. Normally in the definition of the fundamental group you have to pick a base point. However the fundamental group of a topological group does not depend on the choice of the base point. That is because for any $a, b\in G$ there is a homeomorphism $f:G\to G$ that maps $a\mapsto b$ namely $f(x)=a^{-1}x$. This homeomorphism induces isomorphism of $\pi(G, a)\simeq \pi(G, b)$. Actually the same is true for any space that has this property (i.e. the existance of such homeomorphism for any two points). So that's why the notion $\pi(G)$ makes sense.

Also it is convenient to always take $\pi(G, e)$ where $e$ is the neutral element of $G$. That's because the path connected component of $e$ is a closed normal subgroup of $G$. So from the other point of view you can always assume that $G$ is path connected.

freakish
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  • But how can I even construct a loop if it is not path connected? – Kristina Mar 16 '17 at 10:32
  • @Kristina Why do you assume you need path connectness to construct a loop? For example take $X=S^1\times\mathbb{Z}^2$ (two copies of a circle). This group/space is not path connected (it has two path components: $S^1\times{0}$ and $S^1\times{1}$). But you can normally construct loops. Its just that each loop has to be in one of the components. – freakish Mar 16 '17 at 10:32
  • See also https://math.stackexchange.com/questions/727999/g-is-topological-implies-pi-1g-e-is-abelian/729009#729009 – Ronnie Brown Dec 02 '18 at 10:19
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Suppose the topological group is connected. Is it necessarily path-connected?

I would guess so. But I'm not sure.

Certainly an infinite topological group that is not discrete (like the discrete integers) need not have any path component containing more than one point. For instance, the countable direct product of 2-element groups is a compact and totally disconnected abelian group that, as a space, is homeomorphic to the Cantor set.

But if the topological group has a connected component, I would guess it must be a path-component.

Edit: There exist compact metrizable topological groups that are connected but not path-connected: Solenoids are topological groups defined as the inverse limit of a series of maps S1 → S1 each of the form z |→ znk for various nk > 1. (The simplest solenoid has all nk = 2.) According to Wikipedia: "Solenoids are compact metrizable spaces that are connected, but not locally connected or path connected."

Dan Asimov
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