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While the fundamental group $\pi_1$ preserves products, it is not true in general that an inverse limit of simply connected topological spaces is simply connected. I would like to know if similar things can happen with topological groups.

Let $G=\varprojlim_{n}(G_n,p_{n+1,n}:G_{n+1}\to G_n)$ be the inverse limit of an inverse sequence of 1-connected topological groups $G_n$ and continuous homomorphisms. If $e$ is the identity element of $G$, must $\pi_1(G,e)$ be trivial? What if each $G_n$ is a Lie group?

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  • $\begingroup$ I do not know the answer, but why not take a sequence simply connected topological spaces $X_i$ with non-simply connected inverse limit, then for each $X_i$ take the "free Abelian topological group" $G(X_i)$ generated by $X_i$ (I think Markov had a construction like that). These groups should be simply connected and form an inverse limit which would not be simply connected (I am guessing). $\endgroup$
    – user6976
    Commented Jul 2, 2017 at 21:12
  • $\begingroup$ @MarkSapir Thanks. Your suggestion about the most general situation is likely correct using the Graev (based) free abelian or non-abelian topological groups. Actually, I had considered this but the details elude me on a few points. The literature on free topological groups is vast but contains only partial results on their homotopy theory. I am hoping there is a simpler example or a more obvious answer for Lie groups. $\endgroup$
    – Jeremy Brazas
    Commented Jul 2, 2017 at 21:50
  • $\begingroup$ You of course know Markov's construction. Perhaps it is better than Graev's construction in a sense that the homotopy groups are easier computed? $\endgroup$
    – user6976
    Commented Jul 2, 2017 at 23:35
  • $\begingroup$ @JeremyBrazas, if the tower is a tower of fibrations then there is a Milnor $\lim^1$-sequence $1 \to \lim^1 \pi_2(G_n, e) \to \pi_1(\lim G_n, e) \to \lim \pi_1(G_n,e) \to 1$, and so for example if the sequence of second homotopy groups is a sequence $\dots \to \Bbb Z \to \Bbb Z \to \Bbb Z$ where all the maps are multiplication by $p$ the $\lim^1$-term is nonzero. To construct such a sequence of groups you can do something similar to what Mark Sapir suggests: take the associated tower of $K(\Bbb Z,3)$'s and apply a Kan loop group construction to get a tower of $K(\Bbb Z,2)$'s which are groups. $\endgroup$
    – Tyler Lawson
    Commented Oct 14, 2017 at 23:37
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    $\begingroup$ In the definition of "simply connected", some authors ask that the (pointed) space is path-connected and others not. This gives two interpretations of the question. $\endgroup$
    – YCor
    Commented Oct 15, 2017 at 7:16

1 Answer 1

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It seems that the inverse limit of simply connected Lie groups is simply connected. The argument uses the well-known fact that the second homotopy group of any Lie group is trivial (see e.g. Homotopy groups of Lie groups).

Applying the long exact sequence of homotopy groups to the bonding homomorphism $p_{n+1,n}:G_{n+1}\to G_n$ between the simply connected topological groups, we conclude that its kernel $K_{n+1}=p_{n+1,n}^{-1}(e)$ has trivial homotopy group $\pi_1(K_{n+1})=\pi_2(G_n)=0$. Using this fact, and also the fact that a locally trivial bundle over a contractible space is trivial, for any loop $\gamma:\partial\mathbb D\to G$ defined on the boundary of the unit disk $\mathbb D$ we can inductively construct a sequence of maps $\bar\gamma_n:\mathbb D\to G_n$ such that $p_{n+1,n}\circ \bar\gamma_{n+1}=\bar\gamma_n$ and $\bar\gamma_n|\partial\mathbb D=p_{\infty,n}\circ\gamma$ for all $n$. Here $p_{\infty,n}:G\to G_n$ denotes the limit projection. Then the maps $\bar \gamma_n$ determine a map $\bar \gamma:\mathbb D\to G$ extending the map $\gamma$ and witnessing that the topological group $G$ is simply connected.

I hope that this argument is correct (this is a question to specialists in Algebraic Topology).

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  • $\begingroup$ Something here is not so clear to me. The bonding maps are indeed fibrations onto their image. If each bonding map is surjective, then one can apply Prop. 4.67 of Hatcher to prove the affirmative answer (also using $\pi_2=0$). What contractible space are you taking a trivial bundle over? I believe that a simply connected Lie group need not be contractible. $\endgroup$
    – Jeremy Brazas
    Commented Oct 14, 2017 at 23:40
  • $\begingroup$ @Jeremy-Brazas I had in mind the contractibility of the disk $\mathbb D$. If the map $\bar\gamma_n:\mathbb D\to G_n$ is given, then the locally trivial bundle $p_{n+1,n}:G_{n+1}\to G_n$ induces a locally trivial bundle over $\mathbb D$ with fiber $\mathbb K_{n+1}$. The contractibility of $\mathbb D$ guarantees that this locally trivial bundle is trivial. So, we can lift the map $\bar \gamma_n$ to the map $\bar\gamma_{n+1}$. $\endgroup$
    – Taras Banakh
    Commented Oct 15, 2017 at 4:55
  • $\begingroup$ I see what you mean now. Yes, this inductive pullback construction works. The only thing left to mention is that I think the LES applies to the fiber bundle $p_{n+1,n}:G_{n+1}\to Im(p_{n+1,n})$, right? Everything else goes through the same. $\endgroup$
    – Jeremy Brazas
    Commented Oct 15, 2017 at 13:56

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