## Topologized Fundamental Groups: The Whisker Topology, Part 3

In Part 1 and Part 2, I gave detailed introductory exposition about the whisker topology on the fundamental group. In general, this topologized fundamental group $\pi_{1}^{wh}(X,x_0)$ is a left topological group and therefore a homogeneous space. Moreover, whenever this group is $T_1$, it’s also zero-dimensional. This bit about zero-dimensionality is a huge restriction which tells us the whisker topology is a very fine topology. Most often, we’re interested in metrizable spaces so it’s natural to ask what we can acheive in this situation.

The proof I’m going to give isn’t too hard but it does require working through several individual steps. If nothing else, this proof is basically the same as a certain proof that a universal covering space of a locally path-connected, metrizable space is metrizable.

## (Pseudo)metrizability of the whisker topology

Supposing that $X$ is metrizable, pick a metric $d$ that induces the topology of $X$. Let’s define a distance function $\rho$ on $\pi_{1}^{wh}(X,x_0)$. For loops $\alpha,\beta$ based at $x_0$, set

$\rho([\alpha],[\beta])=\inf\{diam(\epsilon)\mid [\epsilon]=[\alpha^{-}\cdot\beta]\}$

Now, it’s possible that $\rho([\alpha],[\beta])=0$ even when $[\alpha]\neq [\beta]$ so we should only expect that $\rho$ will be a pseudometric in general. Symmetry is pretty clear from the definition. The triangle inequality is maybe a little less clear but working it out is a nice exercise. Let’s do it. There are some different ways to write proofs like this involving infimums and supremums that use a variety of established/known facts but these are pretty efficient (though they are by contradiction).

Lemma: If $\alpha,\beta$ are loops in $X$ based at $x_0$ and $\gamma=\alpha\cdot\beta$, then $diam(\gamma)\leq diam(\alpha)+diam(\beta)$.

Proof. Recall the supremum definition of the diameter of a loop. We’ll use the general fact that $\sup(S)>a$ if and only if there exists $s\in S$ with $s>a$. If $diam(\gamma)> diam(\alpha)+diam(\beta)$, then there exists $a,b\in [0,1]$ with $d(\gamma(a),\gamma(b))>diam(\alpha)+diam(\beta)$. If $a,b\in [0,1/2]$, then $d(\gamma(a),\gamma(b))=d(\alpha(a/2),\alpha(b/2))>diam(\alpha)$, which is a contradiction. If $a,b\in [1/2,1]$, then $d(\gamma(a),\gamma(b))=d(\beta(2a-1),\alpha(2b-1))>diam(\beta)$. Without loss of generality, suppose $a\in[0,1/2]$ and $b\in [1/2,1]$. Then

$d(\alpha(2a),\beta(2b-1))= d(\gamma(a),\gamma(b))>diam(\alpha)+diam(\beta)\geq d(\alpha(2a),x_0)+d(x_0,\beta(2b-1))$

but this is a violation of the triangle inequality in $X$. $\square$

Lemma: Given $[\alpha],[\beta],[\gamma]\in\pi_{1}^{wh}(X,x_0)$, $\rho([\alpha],[\gamma])\leq \rho([\alpha],[\beta])+\rho([\beta],[\gamma])$.

Proof. We’ll play a similar game in this lemma using the analagous elementary fact for infimums:  $a>\inf(S)$ if and only if there exists $s\in S$ with $a>s$,

Suppose $\rho([\alpha],[\gamma])>\rho([\alpha],[\beta])+\rho([\beta],[\gamma])$. Since $\rho([\alpha],[\gamma])-\rho([\beta],[\gamma])>\rho([\alpha],[\beta])$, there exists a loop $\epsilon$ with $[\epsilon]=[\alpha^{-}\cdot\beta]$ and $\rho([\alpha],[\gamma])-\rho([\beta],[\gamma])>diam(\epsilon)$. Since $\rho([\alpha],[\gamma])-diam(\epsilon)>\rho([\beta],[\gamma])$, there exists a loop $\delta$ with $[\delta]=[\beta^{-}\cdot\gamma]$ and $\rho([\alpha],[\gamma])-diam(\epsilon)>diam(\delta)$. Now $[\epsilon\cdot\delta]=[\alpha^{-}\cdot \beta][\beta^{-}\cdot\gamma]=[\alpha^{-}\cdot\gamma]$ and $\rho([\alpha],[\gamma])>diam(\epsilon)+diam(\delta)$. We have $diam(\epsilon)+diam(\delta)\geq diam(\epsilon\cdot\delta)$ from the previous lemma so $\rho([\alpha],[\gamma])>diam(\epsilon\cdot\delta)$. Since $[\epsilon\cdot\delta]=[\alpha^{-}\cdot\gamma]$, this is a contradiction. $\square$.

With the triangle inequality in hand, we have a pseudometric! Now we should check that it induces the topology $\pi_{1}^{wh}(X,x_0)$. In general, if $d$ is a (pseudo)metric on a set $S$, $O_d(s,r)=\{t\in S\mid d(t,s) will denote the open ball of radius $r$ about $s$.

Lemma: The pseudometric induces the topology of $\pi_{1}^{wh}(X,x_0)$.

Proof. Let $B([\alpha],U)$ be a basic neighborhood of $[\alpha]$ in $\pi_{1}^{wh}(X,x_0)$. Find $r>0$ such that $O_{d}(x_0,r)\subseteq U$ where $O_{d}(x_0,r)$ is the open $r$-ball about $x_0$. Suppose $[\beta]\in O_{\rho}([\alpha],r)$. Since $\rho([\alpha],[\beta]), there exists a loop $\gamma$ based at $x_0$ with $[\gamma]=[\alpha^{-}\cdot\beta]$ and $diam(\gamma). Since $\gamma$ has image in $O_{d}(x_0,r)$, it has image in $U$. This gives $[\beta]=[\alpha\cdot\epsilon]\in B([\alpha],U)$.

For the other direction, suppose $r>0$ and consider the neighborhood $O_{\rho}([\alpha],r)$. We’ll show that $B([\alpha],O_d(x_0,r/3))\subseteq O_{\rho}([\alpha],r)$. Suppose $[\beta]\in B([\alpha],O_d(x_0,r/3))$. Write $[\beta]=[\alpha\cdot\epsilon]$ for loop $\epsilon$ in $O_d(x_0,r/3))$. Then $[\epsilon]=[\alpha^{-}\cdot\beta]$ and $diam(\epsilon)\leq \frac{2r}{3}$. Thus $\rho([\alpha],[\beta])\leq \frac{2r}{3}, showing $[\beta]\in O_{\rho}([\alpha],r)$. $\square$

Theorem: If $X$ is metrizable, then $\pi_{1}^{wh}(X,x_0)$ is pseudometrizable.

Remark: This proof also goes through for the space $\widetilde{X}$ of all path-homotopy classes of paths starting at $x_0$ with the whisker topology of which $\pi_{1}^{wh}(X,x_0)$ is a subspace. Indeed, whenever $X$ is metrizable, $\widetilde{X}$ is pseudometrizable (See [5,Lemma 2.12]). Why I’m taking a second to do the proof is to emphasize that it works for ALL metrizable spaces, not just those for those having the usual assumptions of covering space theory.

But…can we get rid of the “pseudo?” In a pseudometric space it’s possible to have distinct points that have zero-distance from each other. Such a space is not even $T_0$.  This can certainly happen for $\pi_{1}^{wh}(X,x_0)$ and we saw an instance of it in an earlier post. But it’s also true that a pseudometrizable space is metrizable if and only if it’s Hausdorff and we characterized the Hausdorff property for $\pi_{1}^{wh}(X,x_0)$ in Part 2. Moreover, we showed that $\pi_{1}^{wh}(X,x_0)$ is zero-dimensional whenever it’s Hausdorff.

Let’s summarize. Taking several of our established results, we have the following.

Theorem: Let $X$ be a path-connected, metrizable, homotopically Hausdorff space and $x_0\in X$. Then $\pi_{1}^{wh}(X,x_0)$ is a homogeneous, zero-dimensional, metrizable space.

This applies in lots of cases, include all one-dimensional and planar spaces.

## Separability

Now I’m wondering about separability. Here, I’m still assuming $(X,d)$ is a homotopically Hausdorff metric space so that we’re in the situation of the previous theorem. Clearly $\pi_{1}^{wh}(X,x_0)$ is separable if it is countable. That’s not so interesting. What about in other cases? Honestly, I’m not too sure how to characterize this in general. Here’s how I think we can understand it. Let $U_n$ be the $1/n$-ball about $x_0$ in $X$. If $S=\{s_1,s_2,s_3,\dots\}$ was a dense countable subset of $\pi_{1}^{wh}(X,x_0)$, then for every homotopy class $g\in \pi_{1}^{wh}(X,x_0)$ and neighborhood $n\in\mathbb{N}$, we’d have $s_k\in B(g,U_n)$ for some $k=k(g,n)$. Another way to say this is that there is a sequence $s_{k(g,1)},s_{k(g,2)},s_{k(g,3)},\dots$ in $S$ where the products $gs_{k(g,1)}^{-1},gs_{k(g,3)}^{-1},gs_{k(g,3)}^{-1},\dots$ have representatives that get arbitrarily small. So we can take any element of the fundamental group and make it small by multiplying on the right by inverses of the elements of $S$.

In the earring group, a Cantor diagonalization argument should be possible to see it’s not separable. I don’t want to get hung up on this here and get caught up in examples, but it’s a curious question and maybe someone else wants to go down this rabbit hole. This could make a nice student project.

Question: If $X$ is a homotopically Hausdorff metric space, when is $\pi_{1}^{wh}(X,x_0)$ separable? What are some nice examples illustrating when it is and isn’t separable?

To finish the post, I’m going to go back to why I actually spent 3 posts writing about a $\pi_1$  topology I’ve never really cared much about.

## Why am I starting to care more about the whisker topology?

I’ll be honest… I used to think the whisker topology wasn’t too interesting or useful. Here, I’ll spend a little time trying to explain why I’ve changed my tune on this. I want to be clear that I’m not so much trying to publicize my own results as must as I want to publicize the fact that I don’t really understand a certain naturally occuring group as well as I’d like to.

Consider the one-point union $Y=\mathbb{E}_1\vee\mathbb{E}_n$ of the earring space and the n-dimensional earring as seen below for $n=2$.

The one-point union $\mathbb{E}_1\vee\mathbb{E}_2$.

The group $\pi_n(Y)$ is fairly complicated because $\pi_1(Y)=\pi_1(\mathbb{E}_1)$ is uncountable and infinitary. Unlike with $S^1\vee S^2$, the usual $\pi_1$-action is not enough to really understand $\pi_2$. Recently, I proved that $\pi_n(Y)$ canonically embeds into the group $\prod_{j\in\mathbb{N}}\oplus_{\pi_1(\mathbb{E}_1)}\mathbb{Z}$. I’d like to understand the image in this product of $\mathbb{Z}$‘s better.

We can think of elements of $G= \prod_{j\in\mathbb{N}}\oplus_{\pi_1(\mathbb{E}_1)}\mathbb{Z}$ as functions $g:\mathbb{N}\times \pi_1(\mathbb{E}_1)\to\mathbb{Z}$, which has finite support in the second variable: for each $j\in\mathbb{N}$, the set $\{[\alpha]\in\pi_1(\mathbb{E}_1)\mid g(j,[\alpha])\neq 0\}$ is finite.

The canonical homomorphism $\Psi:\pi_n(Y)\to G$ is not onto! Here’s the characterization of the image: an element $g\in G$ is in the image of $\Psi$ and therefore uniquely represents a homotopy class in $\pi_n(Y)$ if and only if $\{[\alpha]\in \pi_1(\mathbb{E}_1)\mid g(j,[\alpha])\neq 0\}$ is countable and has compact closure in $\pi_{1}^{wh}(\mathbb{E}_1)$. So elements of $\pi_n(Y)$ are classified by functions $g:\mathbb{N}\times\pi_1(\mathbb{E}_1)\to\mathbb{Z}$ with countable support and such that that $g$ is non-trivial on only a “bounded” set in the second variable (but only with the whisker topology!). Even though I have a geometric understanding of this group and feel this description is good enough for most any use I can imagine, there is topology embedded in the indexing sets. My instinct tells me that it’s built-in topological nature can’t be done away with but this idea is not really something I know how to formalize. Anyway, I never would have guessed the whisker topology would become relevant to wild higher homotopy theory.

## So what are the compact subsets?

That stuff about homotopy groups is why I was interested in understanding compact subsets of $\pi_{1}^{wh}(X,x_0)$. The theorem we worked out in these posts says that $\pi_{1}^{wh}(X,x_0)$ is often zero-dimensional and metrizable. In this case, a compact subset $K$ of $\pi_{1}^{wh}(X,x_0)$ will be a zero-dimensional compact metric space.

This leaves open the possibility $K$ could be homeomorphic to a Cantor set, which does seem reasonable. Here’s an example.

Example: $X=\prod_{n=1}^{\infty}\mathbb{RP}_2$ is the infinite direct product of copies of the projective plane, then $\pi_1(X,x_0)$ is abelian and therefore $\pi_{1}^{wh}(X,x_0)\cong \prod_{n=1}^{\infty}\mathbb{Z}/2\mathbb{Z}$ is a topological group by results in Post 1. But we can say more. We can identify $\pi_{1}^{wh}(\mathbb{RP}^2,x)$ with the discrete group $\mathbb{Z}/2\mathbb{Z}$ for any choice of basepoint. Now, a basic neighborhood of the basepoint in $\prod_{n=1}^{\infty}\mathbb{RP}_2$ can be taken to be of the form $V= \prod_{i=1}^{k}U\times \prod_{i > k}\mathbb{RP}_2$ where $U$ is a contractible neighborhood of the basepoint in $\mathbb{RP}_2$ .Therefore, if $\alpha=(\alpha_i)$ is a loop in $X$, then $B([\alpha],V)=\prod_{i=1}^{k}\{[\alpha_i]\}\times \prod_{i > k}\mathbb{Z}/2\mathbb{Z}$. This is the key idea needed to check that $\pi_{1}^{wh}(X,x_0)$ is canonically isomorphic to the topological group $\prod_{n=1}^{\infty}\mathbb{Z}/2\mathbb{Z}$ with the product topology. In particular, $X$ is a metric space for which $\pi_{1}^{wh}(X,x_0)$ is homeomorphic to a Cantor set. $\square$

So, yes, a compact subset $K$ of $\pi_{1}^{wh}(X,x_0)$ can be as complicated as a Cantor set. But descriptive set theory tells us it can’t be much worse. Every Polish space is the disjoint union of a countable scattered subspace and (possibly) a perfect set. In our situation, zero-dimensionality ensures that if there is a perfect subset, it must be a Cantor set.

Theorem: Let $X$ be a path-connected, metrizable, homotopically Hausdorff space and $x_0\in X$. If $K\subseteq \pi_{1}^{wh}(X,x_0)$ is a compact, then $K$ is either homeomorphic to a countable compact ordinal or the union of a countable scattered space and a Cantor set.

I don’t know that this solves the entirety of my lack of understanding of my representation of $\pi_n(Y)$ using the whisker topology but it does give me a way to handle compact sets in $\pi_{1}^{wh}(X,x_0)$ if I need to.

## References

The references here include several papers that involve the whisker topology from a group of Iranian researchers who have done a lot of research in this area.

[1] N. Jamali, B. Mashayekhy, H. Torabi, S.Z. Pashaei, M. Abdullahi Rashid, On
topologized fundamental groups with small loop transfer viewpoints, Acta Math. Vietnamica, 44 (2019) 711–722.

[2] M. Abdullahi Rashid, N. Jamali, B. Mashayekhy, S.Z. Pashaei, H. Torabi, On subgroup topologies on the fundamental group. Hacettepe Journal of Mathematics & Statistics 49 (2020), no. 3, 935 – 949.

[3] M. Abdullahi Rashid, B. Mashayekhy, H. Torabi, S.Z. Pashaei, On subgroups of
topologized fundamental groups and generalized coverings, Bull. Iranian Math. Soc.
43 (2017), no. 7, 2349–2370.

[4] 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). Note: easily found through a google search but I can’t get a link to work.

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

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