## 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.

## 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.

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

## Topologized Fundamental Groups: The Whisker Topology, Part 1

This post follows the primer: How to “topologize” the fundamental group. I’m starting with the “whisker topology” because it comes from a very familiar construction. Also, with such a simple definition, a lot of the questions one could ask about it are pretty straightforward to answer. I found writing this to be a string of fun little exercises.

Despite the apparent simplicity, compact sets in the whisker topology are not always so easy to understand. To my own bewilderment, I’ve been needing to understand exactly such things is some recent projects on wild higher homotopy groups. So I’m also hoping these posts will help me clarify some of my own thoughts about it.

Let’s jump right into it. If you have seen covering space theory, you are likely to recognize the following construction.

## A construction from covering space theory

Let $X$ be a path-connected space with basepoint $x_0\in X$. If we were trying to construct a universal covering space for $X$ (whether it actually exists or not), we might try the following “standard” construction: Let $\widetilde{X}$ be the set of path-homotopy classes $[\alpha]$ of paths $\alpha:[0,1]\to X$ starting at $x_0$. We define a topology on $\widetilde{X}$ by determining a basis. A basic open neighborhood of $[\alpha]$ is of the form $N([\alpha],U)=\{[\alpha\cdot\delta]\mid \delta([0,1])\subseteq U\}$ where $U$ is an open neighborhood of $\alpha(1)$ in $X$. Here, $\alpha\cdot\delta$ denotes path concatenation. For later on, $\alpha^{-}$ will denote the reverse path of $\alpha$ and $c_{x_0}$ will be denote the constant path at $x_0$ so that $1=[c_{x_0}]$ is both the identity element of $\pi_1(X,x_0)$ and the basepoint of $\widetilde{X}$.

Visualizing a basic neighborhood $N([\alpha],U)$ of $[\alpha]$ in the whisker topology

Given a path $\alpha:([0,1],0)\to (X,x_0)$ and a neighborhood $U$ of $\alpha(1)$ in $X$, the neighborhood $N([\alpha],U)$ of $[\alpha]$ in $\widetilde{X}$ consists of only homotopy classes that differ from $[\alpha]$ by a “small” change or “whisker” at it’s end. This is why this topology is often referred to as the whisker topology. The directionality in the definition of the whisker topology is important! Nearby homotopy classes only differ by small differences on the right side.

This definition can be found in most textbooks that include covering space theory but it’s studied in detail in [5] and [6]. I’m not sure who first named it the “whisker” topology. In many settings, it does not need a name but since we are studying and comparing several different topologies, a name is appropriate for this discussion.

We define a function $p:\widetilde{X}\to X$ as the endpoint projection $p([\alpha])=\alpha(1)$. This function is certainly continuous since $p(N([\alpha],U))\subseteq U$ for any neighborhood $U$ and path $\alpha$ with $\alpha(1)\in U$.

If you recognize this construction, you might also recall the following theorem, which is often used to cover the existence part of the classifciation of covering maps: If $X$ is locally path connected and semilocally simply connected, then $p:\widetilde{X}\to X$ is a universal covering map.

The space $\widetilde{X}$ can be constructed for an arbitrary based space $(X,x_0)$. Even when $p:\widetilde{X}\to X$ does not end up being a universal covering map, this construction is often useful.

Since $\widetilde{X}$ is often a covering space, our minds quickly want to identify $\widetilde{X}$ with some familiar space. For example, if $X$ is a torus, we’d identify $\widetilde{X}$ with $\mathbb{R}^2$. But formally, $\widetilde{X}$ is still a set of homotopy classes that contains the fundamental group $\pi_1(X,x_0)$ as a genuine subset. Specifcially, the fundamental group is the basepoint fiber $p^{-1}(x_0)=\pi_1(X,x_0)$.

Therefore, we may give $\pi_1(X,x_0)$ the subspace topology inherited from $\widetilde{X}$. When we intersect the basic neighborhoods of $\widetilde{X}$ with $\pi_1(X,x_0)$, all of the small “whiskers” end up being loops.

Definition: The whisker topology on $\pi_1(X,x_0)$ is generated by basic open sets

$B([\alpha],U)=N([\alpha],U)\cap \pi_1(X,x_0)$

where $[\alpha]\in \pi_1(X,x_0)$ and $U$ is an open neighborhood of $x_0$ in $X$. We will write $\pi_{1}^{wh}(X,x_0)$ to denote the fundamental group equipped with the whisker topology.

A basic neighborhood $B([\alpha],U)$ of an element $[\alpha]$ in the fundamental group with the whisker topology consists of homotopy classes of the form $[\alpha\cdot\epsilon]$.

## What kind of thing is $\pi_{1}^{wh}(X,x_0)$?

It’s tempting to think that because the whisker topology comes from such a standard and simple construction that $\pi_{1}^{wh}(X,x_0)$ will be a topological group. However, it turns out that this very often not true. This is the functorial topology on $\pi_1$ that is, in a sense, the furthest from being a group topology.

Definition: Let $G$ be a group with a topology. Let $\lambda_{g}:G\to G$, $\lambda_{g}(a)=ga$ denote left translation by $g$. If $\lambda_{g}$ is continuous for all $g\in G$, then $G$ is a left topological group. Similarly, if the right translations $\rho_{g}(a)=ag$ are continuous for all $g$, then $G$ is a right topological group. If $G$ is both a left and right topological group and inversion is continuous, then $G$ is a quasitopological group.

If $G$ is a left topological group, then all of the left-translation maps $\lambda_{g}$ will be homeomorphisms since $\lambda_{g^{-1}}$ is the inverse of $\lambda_{g}$.

Proposition: $\pi_{1}^{wh}(X,x_0)$ is a left topological group.

Proof. Fix $[\beta]\in\pi_{1}^{wh}(X,x_0)$ and consider the left translation map $\lambda_{[\beta]}:\pi_{1}^{wh}(X,x_0)\to \pi_{1}^{wh}(X,x_0)$, $\lambda_{[\beta]}([\gamma])=[\beta][\gamma]$ is continuous. A straightforward check gives that $\lambda_{[\beta]}(B([\alpha],U))=B([\beta\cdot\alpha],U)$. This makes it clear that $\lambda_{[\beta]}$ is a homeomorphism. $\square$

Consequently, $\pi_{1}^{wh}(X,x_0)$ is a homogeneous space. But what about the other operations? For this, let’s explore an example.

Example: Let $\mathbb{E}=\bigcup_{n\geq 1}C_n$ be the usual earring space where $C_n\subseteq \mathbb{R}^2$ is the circle of radius $1/n$ centered at $(1/n,0)$. We take $x_0=(0,0)$ to be the wild point and and let $\ell_n:[0,1]\to \mathbb{E}$ denote a loop that traverses the $n$-th circle once counterclockwise. Let $U_n$ be a neighborhood of $x_0$ that contains $\bigcup_{k\geq n}C_k$ and meets $C_k$, $k in an open arc.

Then notice that $\{[\ell_n]\}_{n\geq 1}\to 1$ where $1$ denotes the identity element. Also, $\{[\ell_1\cdot\ell_n]\}_{n\geq 1}$ converges to $[\ell_1]$. However, $\{[\ell_n\cdot\ell_1]\}_{n\geq 1}$ does NOT converge to $[\ell_1]$ because $N([\ell_1],U_2)$ doesn’t contain any of the elements $[\ell_n\cdot\ell_1]$; it can only contain elements of the form $[\ell_1][\beta]$ where $\beta$ has image in $\bigcup_{k\geq 2}C_k$.

As a consequence, we see that $\pi_{1}^{wh}(\mathbb{E},x_0)$ is NOT a right topological group and certainly is not a topological group. Moreover, inversion is NOT always continuous. If inversion was continuous, $\{[\ell_1\cdot\ell_n]\}_{n\geq 1}\to [\ell_1]$ would give $\{[\ell_{n}]^{-1}[\ell_{1}]^{-1}\}_{n\geq 1}\to [\ell_{1}]^{-1}$ but this can’t occur because the sequence $\{[\ell_{n}]^{-1}[\ell_{1}]^{-1}\}_{n\geq 1}$ never enters the open set $B([\ell_{1}]^{-1},U_2)$.

## Functorality

I won’t say to much about the details here but $\pi_{1}^{wh}$ is a functor. Let $\mathbf{LTopGrp}$ be the category of left topological groups and continuous group homomorphisms.

Exercise: If $f:(X,x_0)\to (Y,y_0)$ is a based map, prove that the induced homomorphism $f_{\#}:\pi_{1}^{wh}(X,x_0)\to \pi_{1}^{wh}(Y,y_0)$, $f_{\#}([\alpha]=[f\circ\alpha]$ is continuous.

Once the exercise is done, the fact that $\pi_{1}$ is a a functor, direclty implies that $\pi_{1}^{wh}:\mathbf{Top_{\ast}}\to \mathbf{LTopGrp}$ is a functor.

In fact, since $f_{\#}=g_{\#}$ whenever $f,g:(X,x_0)\to (Y,y_0)$ are homotopic rel. basepoint, $\pi_{1}^{wh}$ induces a functor $\pi_{1}^{wh}:\mathbf{HTop_{\ast}}\to \mathbf{LTopGrp}$ on the pointed-homotopy category (objects are based spaces and morphisms are basepoint-preserving homotopy classes of based maps).

## When CAN we expect continuous operations?

Some of us might be a little unhappy that so many operations in $\pi_{1}^{wh}(X,x_0)$ are going to be discontinuous so often. You might think, well…what if we allowed for whiskers to appear on both ends instead of just on the right. Here, I mean define $B([\alpha],U)=\{[\epsilon\cdot\alpha\cdot\delta]\mid \epsilon([0,1])\cup \delta([0,1])\subseteq U\}$ instead. It’s not a bad idea to consider this for a moment but remember that $\pi_{1}^{wh}(X,x_0)$ is a left topological group. Right multiplication is often discontinuous precisely because we have the whiskers on the right. So if we allow for whiskers on the left and right, then left multiplication won’t be continuous either.

It may seem like this could not possibly be useful until we recall that this topology came from $\widetilde{X}$. So the whisker topology of $\pi_{1}^{wh}(X,x_0)$ is really meant to be more like the vertex set some kind of graph or tree that branches off in infinite and topologically non-trivial ways (maybe think of something that you could consider a non-discrete vertex set in an $\mathbb{R}$-tree if that idea is familiar to you).

However, if you’re really hung up on the whole “fails to be a topological group” thing, we can work to understand this failure better through the following theorem. I’ve never seen this theorem in the literature before but it wasn’t that hard to figure out or prove. If anyone know of a reference, please let me know so I can provide some attribution.

Theorem: Let $G=\pi_{1}^{wh}(X,x_0)$. Then the following are equivalent:

1. Group inversion $in:G\to G$, $in(g)=g^{-1}$ is continuous,
2. $G$ is a topological group,
3. For every $g\in G$, conjugation $c_{g}:G\to G$, $c_g(h)=ghg^{-1}$ is continuous.

Proof. The hardest direction is 1. $\Rightarrow$ 2 so we’ll prove that first. Suppose group inversion $in:G\to G$, $in(g)=g^{-1}$ is continuous. To show that $G$ is a topological group, it’s enough to show that multiplication $([\alpha],[\beta])\mapsto [\alpha\cdot\beta]$ is continuous. Let $U$ be a neighborhood of $x_0$ in $X$ so that $B([\alpha\cdot\beta],U)$ is a basic open neighborhood of $[\alpha\cdot\beta]$. Since inversion is continuous, specifically at $[\beta^{-}]\in G$, we may find an open neighborhood $V$ of $x_0$ such that $V\subseteq U$ and $in(B([\beta^{-}],V))\subseteq B([\beta],U)$. The second inclusion means that whenever $\delta$ is a loop in $V$ based at $x_0$, we have $[\delta^{-}][\beta]=[\beta][\gamma]$ for some loop $\gamma$ in $U$.

We will show that group multiplication maps $B([\alpha],V)\times B([\beta],V)$ into $B([\alpha\cdot\beta],U)$. Let $[\alpha\cdot\delta]\in B([\alpha],V)$ and $[\beta\cdot\epsilon]\in B([\beta],V)$ for loops $\delta,\epsilon$ in $V$. Since $\delta^{-}$ has image in $V$, we have $[\delta][\beta]=[\beta][\gamma]$ for some loop $\gamma$ in $U$. Thus

$[\alpha\cdot\delta][\beta\cdot\epsilon]=[\alpha][\delta][\beta][\epsilon]=[\alpha][\beta][\gamma][\epsilon]=[\alpha\cdot\beta][\gamma\cdot\epsilon]$

where $\gamma\cdot\epsilon$ has image in $U$. Thus the product $[\alpha\cdot\delta][\beta\cdot\epsilon]$ lies in $B([\alpha\cdot\beta],U)$, completing the proof of the first direction.

The implication 2. $\Rightarrow$ 3. is clear so it is suffices to prove 3. $\Rightarrow$ 1. Suppose for every $g\in G$, that conjugation by $g$ is continuous. Let $[\beta]\in G$. We will check that inversion is continuous at $[\beta]$. Let $U$ be an open neighborhood of $x_0$ in $X$ so that $B([\beta^{-}],U)$ is an open neighborhood of $in([\beta])=[\beta^{-}]$. Since conjugation $c_{[\beta]}$ is continuous at the identity element, we may find a neighborhood $V$ of $x_0$ such that $[\beta]B(1,V)[\beta^{-}]\subseteq B(1,U)$. In other words, if $\delta$ is any loop in $V$ based at $x_0$, then there is a loop $\epsilon$ in $U$ such that $[\beta][\delta][\beta^{-}]=[\epsilon]$. Now it suffices to show that $in(B([\beta],V))\subseteq B([\beta^{-}],U)$. Let $\delta$ be a loop in $V$ based at $x_0$. Since $\delta^{-}$ is also in $V$, we may find a loop $\epsilon$ in $U$ such that $[\beta][\delta^{-}][\beta^{-}]=[\epsilon]$. Then $in([\beta\cdot\delta])=[\delta^{-}\cdot\beta^{-}]=[\beta^{-}][\epsilon]\in B([\beta^{-}],U)$. $\square$

The whisker topology becomes a group topology if and only if for every $\beta$ and neighborhood $U$, there exists a neighborhood $V$ such that for any loop $\epsilon$ in $V$, the conjugate $\beta\cdot\epsilon\cdot\beta^{-}$ is homotopic to some loop in $U$.

What I like about this theorem is that the proof really uses the whisker topology (showing how it depends on the topology of $X$ at $x_0$) and can’t just be proven by analyzing compositions of operations. I made sure to include Condition 3. in the theorem because even if we know $\pi_{1}^{wh}(X,x_0)$ is abelian, it’s not immediately clear that inversion is continuous. However, we do know that these conjugation maps in abelian groups are always the identity!

Corollary: If $\pi_1(X,x_0)$ is abelian, then $\pi_{1}^{wh}(X,x_0)$ is a topological group.

But just being non-commutative doesn’t guarantee that $\pi_{1}^{wh}(X,x_0)$ will fail to be a topological group.

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

Example: If $X=\bigvee_{a\in A}S^1$ is an ordinary wedge of circles with the weak topology, then $\pi_{1}(X,x_0)$ is a free group but $\pi_{1}^{wh}(X,x_0)$ will be discrete and all discrete groups are topological groups.

So, being a topological group here is not just about purely algebraic things like commutativity or conjugation. What’s really going on? If you read the proof closely you can see that for Condition 3. you really only need to use that the conjugation maps are continuous at the identity element. This idea is exactly what is described in the above illustration. So for $\pi_{1}^{wh}(X,x_0)$ to be a topological group, all elements $g$, no matter how “large,” have to conjugate small loops to small loops. Or maybe think about the negation this way…If there is some $g\neq 1$ and also arbitrarily small $a\neq 1$ with $gag^{-1}$ remaining “large,” then you’re in trouble.

Example: The harmonic archipelago gives an example where $\pi_{1}^{wh}(X,x_0)$ is non-commutative but is a topological group because it is indiscrete

So now here’s a question. I haven’t thought too deeply about it so I don’t exactly know how hard it is. If you know of an answer, feel free to share.

Question: Is there a space $X$ where $\pi_{1}^{wh}(X,x_0)$ is a non-discrete, Hausdorff, and non-commutative topological group?

In the next post, I’ll discuss more about separation axioms where we’ll run into the familiar term “homotopically Hausdorff” and see that when this happens the group $\pi_{1}^{wh}(X,x_0)$ is always zero dimensional!

## 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.

[6] Z. Virk, A. Zastrow, The comparison of topologies related to various concepts of
generalized covering spaces, Topology Appl. 170 (2014) 52–62.