## Fundamental groups of topological groups.

Let $G$ be a topological group, and $\pi_1(G,e)$ its fundamental group at the identity. If $G$ is the trivial group then $G \cong \pi_1(G,e)$ as abstract groups. My question is:

If $G$ is a non-trivial topological group can $G \cong \pi_1(G,e)$ as abstract groups?

About all I know now is that $G$ would have to be abelian.

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 This seems very unlikely, and citing the trivial group as an example is not really the evidence to start from. I can only image it would require some non-T_0 topology on G and/or uncountable pi_1. – David Roberts Oct 21 2010 at 5:35 One thing to consider would be to prove there is no natural isomorphism between the underlying group functor U:TopGrp --> Grp and the fundamental group functor pi_1:TopGrp --> Grp (even perhaps when restricted to a big subcategory of TopGrp). This doesn't prove there are no such topological groups, but it rules out any isomorphisms that exist due to any systematic reason. – David Roberts Oct 21 2010 at 5:39 I was only meaning to eliminate the trivial group from the answers, but now that I look at the question again I see how it looks. Thanks. – Chris Oct 21 2010 at 13:43 @David,Chris:Well,I personally think beginning with the simplest possible case and working from there is always a good idea. – Andrew L Oct 21 2010 at 17:37 Do you have any motivation for this question? (It's okay if not!) – Nate Eldredge Oct 23 2010 at 0:36
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Here is an example: a product of infinitely many $\mathbb{RP}^\infty$'s.

The crucial thing thing to see is that $\mathbb{RP}^\infty$ (or, easier to see, its universal cover $S^\infty$) has a group structure whose underlying group is a vector space of dimension $2^{\aleph_0}$. This is not hard: the total space $S^\infty$ of the universal $\mathbb{Z}_2$-bundle is obtained by applying a composite of functors to the group structure $\mathbb{Z}_2$ in the category of sets:

$$Set \stackrel{K}{\to} Cat \stackrel{nerve}{\to} Set^{\Delta^{op}} \stackrel{R}{\to} Top$$

($Top$ here is the category of compactly generated Hausdorff spaces and continuous maps). Here $K$ is the right adjoint to the "underlying set of objects" functor; it takes a set to the category whose objects are the elements of the set and there is exactly one morphism between any two objects. The functor $R$ is of course geometric realization.

Each of these functors is product-preserving, and since the concept of group can be formulated in any category with finite products, a product-preserving functor will map a group object in the domain category to one in the codomain category. Even more: the concept of a $\mathbb{Z}_2$-vector space makes sense in any category with finite products since we merely need to add the equation $\forall_x x^2 = 1$ to the axioms for groups, which can be expressed by a simple commutative diagram.

Thus $S^\infty$ is an internal vector space over $\mathbb{Z}_2$ in $Top$. Its underlying group (in $Set$) is clearly a vector space of dimension $2^{\aleph_0}$. We make take this vector space to be the countable product $\mathbb{Z}_2^{\mathbb{N}}$. Modding out by $\mathbb{Z}_2$ (modding out by a 1-dimensional subspace), the space $\mathbb{RP}^\infty$ is also, as an abstract group, isomorphic to this. And so is a countably infinite product $(\mathbb{RP}^\infty)^{\mathbb{N}}$ of copies of $\mathbb{RP}^\infty$.

Finally, the functor $\pi_1$ is product-preserving, and so

$$\pi_1((\mathbb{RP}^\infty)^{\mathbb{N}}) \cong \mathbb{Z}_{2}^{\mathbb{N}}$$

and we are done.

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You mean $x^2=0$ as an axiom for vector spaces in categories. – Mariano Suárez-Alvarez Oct 21 2010 at 13:42
Well, if you like, but the way I said it is fine. The '1' is traditional notation for the group identity. It becomes the 0 element of a vector space structure if we assume the axiom $x^2 = 1$. – Todd Trimble Oct 21 2010 at 13:55
I know what you're trying to say, but is S^\infty actually given by applying the functor you give to Z_2?? I would prefer to say it is some model of EZ_2... But otherwise this is nice argument to show that BZ_2 can be chosen as a Z_2-module plus some topology. – David Roberts Oct 21 2010 at 23:54
Whuh, you don't believe me? :-) The answer anyhow is *yes*; it's another way of describing the standard Milgram construction of the classifying bundle, viz., the realization of the nerve applied to the groupoid map $K\mathbb{Z}_2 \to B\mathbb{Z}_2$ (that '$B$' being our nLab notation for the 1-object category attached to the group), and that's homeomorphic to the projection $S^\infty \to \mathbb{RP}^\infty$. Incidentally, the same Lawvere theory line of argument I'm using here shows that $S^\infty$ is a topological Boolean algebra for which negation is the antipodal map, and the (cont.) – Todd Trimble Oct 22 2010 at 1:34
identification of a point with its antipode = negation is the same as identifying a point $x$ with $1+x$ in the topological Boolean ring, which is the same as modding out by the subspace $\{0, 1\}$. This is one way of making precise an argument behind one of the claims in my answer. – Todd Trimble Oct 22 2010 at 1:38

First, $\pi_1(G;e)=\pi_1(G_e)$, where $G_e$ is the connected component of $e$, therefore we may assume that $G$ is connected. If $G$ is a finite dimensional compact connected manifold, then it is an $n$-torus (because it is abelian), and its fundamental group is discrete (${\mathbb Z}^n$). In this case, $\pi_1(G)$ is very different from $G$. I suspect that in general $\pi_1$ of an abelian group is either trivial or non-compact. It is hard to see how it could not be discrete (if it is discrete, then $G$ has to be trivial). If $\pi_1(G)$ is not discrete, there are arbitrarily small loops that are not homotopic to a point; thus $G$ is full of holes! Definitely a strange beast.

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If G is Hausdorff, and your guess about the compactness is true, then it would have to be not locally compact, because connected locally compact Hausdorff abelian topological groups are isomorphic to R^n \times K for K compact. – David Roberts Oct 21 2010 at 6:39
So the first question is: how is the $\pi_1$ of an Abelian topological group? Is it torsion-free, or maybe even free? If so, how is the $\pi_1$ of a torsion-free Abelian TG (resp, free Abelian TG)? Is it trivial? – Pietro Majer Oct 21 2010 at 8:01
There are quite a lot of familiar properties that a nontrivial group $G$ with $\pi_1 (G)\cong G$ cannot have. To give an example, $G$ cannot be locally simply connected and satisfy the second axiom of countability - because then the fundamental group is countable (exercise). I think groups with that property are either trivial or at least as weird as Todd Trimble's example. – Johannes Ebert Oct 21 2010 at 16:33

This question occurred as Advanced Problem 5889 in the Amer. Math. Monthly 80 (1973), no. 1, 82.  It was listed as still unsolved five years later, in vol. 85, no. 10, p. 834, of the Monthly; however, my recollection is that it mysteriously vanished from the Monthly's "unsolved" list the next time this got updated, but without a solution having appeared in the interim.

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Cool! The business about $S^\infty$ I gave above was my own response to a problem Paul Halmos mentions in his autobiography: does there exist a nontrivial connected Hausdorff topological group of exponent 2? (This problem is one of fifteen on a take-home final he once gave.) Ever since I found this solution, I've enjoyed trotting it out on odd occasions, and I'm happy that MO provided me with an opportunity to do so again! – Todd Trimble Oct 23 2010 at 11:39
I think you ought to submit your solution to the Monthly. The more so in light of this interesting history you mention. – Greg Marks Oct 25 2010 at 18:31