Topologized Fundamental Groups: The Whisker Topology, Part 2

This post is the second on the “whisker topology” on fundamental groups, in a larger sequence of posts on topologized fundamental groups. If you haven’t seen it, you’ll probably want to start with the first post to get the basic definitions and terminology. The topology on $\pi_{1}^{wh}(X,x_0)$ is generated by the sets $B([\alpha],U)=\{[\alpha\cdot\epsilon]\in \pi_1(X,x_0)\mid \epsilon([0,1])\subseteq U\}$ where $U$ is an open neighborhood of $x_0$ .

One thing I kind of glossed over earlier is the fact that changing the basepoint can change the whisker topology. To convince you of this, let’s take a look at the following question, which I left as an exercise in the previous post.

When is the Whisker Topology discrete?

In the first post, we saw that $\pi_{1}^{wh}(X,x_0)$ need not be a topological group but that it is still a homogeneous space. Consequently, if any one-point set is open, including the trivial subgroup, then the space will be discrete. Based on this observation, we know that $\pi_{1}^{wh}(X,x_0)$ will be discrete if and only if some basic neighborhood $B(1,U)$ of the identity contains only the identity element, i.e. $B(1,U)=\{1\}$ . This will happen if and only if $[\alpha]=1$ for any loop $\alpha$ in $U$ , that is, if any loop in $U$ based at $x_0$ is null-homotopic in $X$ . This is precisely the definition of being semilocally simply connected at $x_0$ !

Here, we’re using the “based” version of this property: $X$ is semilocally simply connected at $x\in X$ if there exists an open neighborhood of $U$ such that the homomorphism $\pi_1(U,x)\to \pi_1(X,x)$ induced by inclusion is trivial.

Proposition: $\pi_{1}^{wh}(X,x_0)$ is discrete if and only if $X$ is semilocally simply connected at $x_0$ .

So the whisker topology only detects wildness only the basepoint. Algebraically, if $x_1\in X$ , then there is a basepoint-change (group) isomorphism $\pi_1(X,x_0)\cong \pi_1(X,x_1)$ . However, the topologized groups $\pi_{1}^{wh}(X,x_0)$ and $\pi_{1}^{wh}(X,x_0)$ will not always be homeomorphic!

Example: If $y_0$ is a point in the earring space $\mathbb{E}$ other than the wild point, then $\pi_{1}^{wh}(\mathbb{E},y_0)$ will be an uncountable discrete group. But if $x_0$ is the wild point, we saw in the last post that $\pi_{1}^{wh}(\mathbb{E},y_0)$ is not discrete. So these two fundamental groups (with the same space but different basepoints) are isomorphic as groups but not homeomorphic as spaces.

That changing the basepoint can change the whisker topology emphasizes, once again, that this topology really only tells us about the topology at the basepoint.

How Connected is the Whisker Topology?

One of the interesting things about the whisker topology is the following lemma, which tells us that for a fixed neighborhood $U$ of $x_0$ , the basic neighborhoods $B([\alpha],U)$ partition $\pi_{1}^{wh}(X,x_0)$ . Recall that $\widetilde{X}$ is the set of homotopy classes of paths starting at $x_0$ and $N([\alpha],U)$ denotes the corresponding basic neighborhoods in $\widetilde{X}$ .

Neighborhood Lemma: In $\widetilde{X}$ , if $[\beta]\in N([\alpha],U)$ , then $N([\alpha],U)=N([\beta],U)$ .

Proof. If $[\beta]\in N([\alpha],U)$ , write $[\beta]=[\alpha\cdot\epsilon]$ for $\epsilon$ with image in $U$ . Then $[\alpha]=[\beta\cdot\epsilon^{-}]$ . With $\epsilon$ fixed, we prove both subset inclusions. Given $[\gamma]\in N([\alpha],U)$ , write $[\gamma]=[\alpha\cdot\delta]$ for $\delta$ with image in $U$ . Thus $[\gamma]=[\beta\cdot(\epsilon^{-}\cdot\delta)]$ where $\epsilon^{-}\cdot\delta$ has image in $U$ (see the illustration below). Thus $[\gamma]\in N([\beta],U)$ . This proves $N([\alpha],U)\subseteq N([\beta],U)$ . The other inclusion is similar. $\square$

Proof of the Neighborhood Lemma in $\widetilde{X}$ .

Corollary: Given an open neighborhood $U$ of $x_0$ in $X$ , two sets $B([\alpha],U)$ and $B([\beta],U)$ in $\pi_{1}^{wh}(X,x_0)$ are either equal or disjoint.

Proof. If $B([\alpha],U)$ and $B([\beta],U)$ are not disjoint for loops $\alpha,\beta$ based at $x_0$ , then we have $[\alpha\cdot\delta]=[\beta\cdot\epsilon]$ for loops $\delta,\epsilon$ with image in $U$ . Then $[\alpha]=[\beta\cdot(\epsilon\cdot\delta^{-})]$ where $epsilon\cdot\delta^{-}$ has image in $U$ and so $[\alpha]\in B([\beta],U)$ . Recalling that $B([\alpha],U)=\pi_1(X,x_0)\cap N([\alpha],U)$ , it follows from the Neighborhood Lemma that $B([\alpha],U)=B([\beta],U)$ . $\square$

This corollary implies that all basic open sets in $\pi_{1}^{wh}(X,x_0)$ are always clopen, that is, the small inductive dimension of $\pi_{1}^{wh}(X,x_0)$ is zero.

Theorem: Whenever $\pi_{1}^{wh}(X,x_0)$ is Hausdorff, it is zero-dimensional (in the sense of small inductive dimension).

Another way to see what’s going on here is to notice that the neighborhoods $B(1,U)$ of the identity element $1$ are all clopen. Therefore, if $\mathscr{U}(x_0)$ is the set of open neighborhoods of $x_0$ , then $\bigcap_{U\in\mathscr{U}(x_0)}B(1,U)$ is equal to the closure of the trivial subgroup $\overline{\{1\}}$ .

When is the Whisker Topology Hausdorff?

At this point, we’ve seen that whenever $\pi_{1}^{wh}(X,x_0)$ is Hausdorff, is pretty highly structured – zero dimensional homogeneous spaces are only so complicated. So what would it take for $\pi_{1}^{wh}(X,x_0)$ to be Hausdorff?

Topological group theory tells us that a topological group is completely regular (Tychonoff) if and only if the trivial subgroup is closed. In fact, we have close to the same thing for the whisker topology even though it’s often far from being a topological group.

Theorem: The following are equivalent.

The trivial subgroup is closed in $\pi_{1}^{wh}(X,x_0)$ ,
$\pi_{1}^{wh}(X,x_0)$ is Hausdorff,
$\pi_{1}^{wh}(X,x_0)$ is $T_3$ ,

Proof. 3. $\Rightarrow$ 2. $\Rightarrow$ 1. is clear. Also, every zero-dimensional Hausdorff space is $T_3$ . So it suffices to show 1. $\Rightarrow$ 2. Suppose the trivial subgroup $\{1\}$ is closed and $[\alpha]\neq [\beta]$ in $\pi_{1}^{wh}(X,x_0)$ . Then $[\beta^{-}\cdot\alpha]\neq 1$ and so there exists an open neighborhood $U$ of $x_0$ such that $1\notin B([\beta^{-}\cdot\alpha],U)$ . Suppose, to obtain a contradiction, that $B([\alpha],U)$ and $B([\beta],U)$ are disjoint. The Neighborhood Lemma gives $[\alpha]\in B([\beta],U)$ . Thus $[\alpha]=[\beta\cdot\delta]$ for loop $\delta$ in $U$ . Then $[\beta^{-}\cdot\alpha]=[\delta]$ , which means $[\beta^{-}\cdot\alpha]\in B(1,U)$ . The Neighborhood Lemma then implies $1\in B([\beta^{-}\cdot\alpha],U)$ ; a contradiction. $\square$

While this theorem tells us that we gain a lot of ground by just knowing the trivial subgroup is closed, understanding when this happens is a different story. It really depends on the space $X$ , particularly how homotopy classes interact with the local topology at $x_0$ .

There’s an idea that I wrote about previously called the “Homotopically Hausdorff” property. I’ll remind you of that here.

Definition: A space $X$ is Homotopically Hausdorff at $x\in X$ if for every non-trivial element $1\neq [\alpha]\in \pi_1(X,x)$ , there exists an open neighborhood such that no loop in $U$ based at $x$ is path-homotopic to $\alpha$ .

This definition is stated a little differently than in my earlier post but, as noted in the remark afterward, the two are equivalent. I’d say this definition is conceptually simpler but the other one is easier to apply. I would imagine that the following theorem is apparently the original justification for the name “homotopically Hausdorff” but the origin of the definition is surprisingly difficult to track down. I first encountered this proof in [2]. It’s a pretty straightforward proof that would make a nice exercise. It’s utility is just that it gives a different characterization of the homotopically Hausdorff property, namely, one in terms of a functorial topology on $\pi_1$ .

Theorem: The group $\pi_{1}^{wh}(X,x_0)$ is Hausdorff if and only if $X$ is homotopically Hausdorff at $x_0$ .

Proof. The previous theorem tells us that it suffices to prove that the trivial subgroup is not closed if and only if $X$ is not homotopically Hausdorff at $x_0$ . Suppose $\{1\}$ is not closed. Then there exists $1\neq [\alpha]$ in the closure of $\{1\}$ . Now let $U$ be an open neighborhood of $x_0$ . Since $B([\alpha],U)$ is an open neighborhood of $[\beta]$ , we have $1\in B([\alpha],U)$ . Thus $1=[\alpha][\epsilon]$ for some loop $\epsilon$ in $U$ . It follows that $[\alpha]=[\epsilon^{-}]$ , i.e. $\alpha$ is path-homotopic to a loop in $U$ . This proves $X$ is not homotopically Hausdorff at $x_0$ .

For the converse, suppose $X$ is not homotopically Hausdorff at $x_0$ . Then there is a non-null-homotopic $\alpha$ based at $x_0$ , which is path-homotopic to a loop in every neighborhood of $x_0$ . We claim that $[\alpha]$ lies in the closure of $1$ . Let $U$ be a neighborhood of $x_0$ . Then we have $[\alpha]=[\epsilon]$ for some loop $\epsilon$ in $U$ . Equivalently, $1=[\alpha\cdot \epsilon^{-}]$ . Since $\epsilon^{-}$ has image in $U$ , this means $1\in B([\alpha],U)$ . Since $1$ lies in every neighborhood of $[\alpha]$ , we conclude that $[\alpha]$ lies in the closure of the trivial subgroup. $\square$

Example: There are lots of homotopically Hausdorff spaces. In my bestiary, I almost always list whether or not this property holds – given an example space, it’s usually easy to decicde whether this property is present or not. Generally, all one-dimensional Hausdorff spaces and subsets of surfaces are homotopically Hausdorff at all of their points and so whenever $X$ is one of these spaces $\pi_{1}^{wh}(X,x_0)$ will be Hausdorff. A fairly non-trivial example is the harmonic pants space. Of course, if you’ve got a locally “nice” space like a manifold or CW-complex, then $\pi_{1}^{wh}(X,x_0)$ is trivially Hausdorff because it is discrete. Examples of spaces, which are not homotopically Hausdorff include the harmonic archipelago and Griffiths twin cone. Actually, for both of these spaces, $\pi_{1}^{wh}(X,x_0)$ is an uncountable indiscrete group.

Example: For an example that is neither Hausdorff nor indiscrete, you can take $\mathbb{G}$ to be the Griffiths twin cone and set $X=\mathbb{G}\vee S^1$ where $x_0$ is the wedgepoint (see the figure below). Let $r: X\to S^1$ be the based retraction that collapses $\mathbb{G}$ to $x_0$ . We know $\pi_{1}^{wh}(S^1,x_0)$ is isomorphic to the discrete group $\mathbb{Z}$ and functorality tells us that $r_{\#}:\pi_{1}^{wh}(X,x_0)\to \pi_{1}^{wh}(S^1,x_0)$ is continuous. Therefore $N=\ker (r_{\#})$ is a proper, non-trivial open subgroup of $\pi_{1}^{wh}(X,x_0)$ . So $\pi_{1}^{wh}(X,x_0)$ is not discrete. However, we may view $\pi_1(\mathbb{G},x_0)$ naturally as a subgroup of $\pi_{1}^{wh}(X,x_0)$ . It’s not too hard to see that $\pi_1(\mathbb{G},x_0)$ is actually equal to the closure of the trivial subgroup in $\pi_{1}^{wh}(X,x_0)$ (but it’s not equal to $N$ !). In particular, $\pi_{1}^{wh}(X,x_0)$ is not Hausdorff.

The Griffiths twin cone with an extra circle attached.

Remark: At the start of Part 1, I mentioned that I was particularly interested in compact subsets of $\pi_{1}^{wh}(X,x_0)$ . At this point, we can say that if $X$ is homotopically Hausdorf,f then a compact subset $K\subseteq \pi_{1}^{wh}(X,x_0)$ will be zero-dimensional. This is helpful but it would be nice to know when $K$ is metrizable too. So in the third and final post on the whisker topology, I’ll dive into the metrizability of the group $\pi_{1}^{wh}(X,x_0)$ .

References

[1] M. Abdullahi Rashid, S.Z. Pashaei, B. Mashayekhy, H.Torabi, On the Whisker Topology on Fundamental Group. Conference Paper from 46th Annual Iranian Mathematics Conference 46 (2015). pdf available here.

[2] H. Fischer, A. Zastrow, Generalized universal covering spaces and the shape
group, Fund. Math. 197 (2007) 167-196.

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