## Topological wildness vs. algebraic wildness

What does “wildness” really refer to?

I’ll post about higher dimensional wildness soon but to avoid getting too general, I’m going to focus on a more specific question: what does it mean for a space to have a wild fundamental group?

If you surveyed mathematicians with this question most responses would likely say:

$X$ has a wild fundamental group if $X$ is not semilocally simply connected.

You might get a few answers of “not locally contractible,” which is a good answer to a more general question but that would be a deflection from our current focus on $\pi_1$. A space can easily fail to be locally contractible but still have tame 1-dimensional homotopy. The first answer is reasonable because we often study fundamental groups using covering space theory and, assuming locally path-connectivity, $X$ has a universal (i.e. simply connected) covering space if and only if $X$ is semilocally simply connected. But does failing to be semilocally simply connected really imply that the topology interacts with the algebra of $\pi_1$ in any kind of wild fashion?

The answer is: a lot of the time, but not always.

I’ll detail both parts of this answer but I think the not always part of the answer involves some fun topology. First, some quick and possibly skip-able review:

Definition: A space $X$ is semilocally simply connected at $x\in X$ if there exists an open neighborhood $U$ such that the inclusion $U\to X$ induces the trivial homomorphism $\pi_1(U,x)\to \pi_1(X,x)$ on fundamental groups. We say $X$ is semilocally simply connected if it is semilocally simply connected at all of its points.

Intuition: Despite the complicated name, the idea behind being “semilocally simply connected” is pretty straightforward. The triviality of the homomorphism $\pi_1(U,x)\to \pi_1(X,x)$ simply means that every loop in $U$ based at $x$ can be contracted in $X$ (but not necessarily in $U$). For instance, let $X\subset\mathbb{R}^3$ be a cone of height $1$ over a circle in the xy-plane. Then the open set $U=X\cap (\mathbb{R}^2\times [0,1/2))$ is a annular neighborhood of the base circle and is therefore homotopy equivalent to a circle. Any loop in $U$ is null homotopic in $X$ but a loop going around the base circle once is non-trivial in $U$ since you must use the top of the cone to contract it. In this case, $\pi_1(U,x)\to \pi_1(X,x)$ is trivial even though $\pi_1(U,x)$ is not.

Again, this property is a key assumption in classical covering space theory: If $X$ is path connected and locally path connected, then $X$ has a simply connected (i.e universal) covering space if and only if $X$ is semilocally simply connected.

Let’s analyze the situation where a space fails to have this property.

## Topological Wildness

Definition: The topological 1-wild set of $X$ is the subspace

$\mathbf{w}(X)=\{x\in X\mid X\text{ is not semilocally simply connected at }x\}$.

Hence, for locally path-connected $X$, the set $\mathbf{w}(X)$ is precisely the topological obstruction to the existence of a universal covering space. If $X$ isn’t locally path connected, remember that we have this handy dandy tool.

Examples:

• $\mathbf{w}(X)=\emptyset$ if and only if $X$ is semilocally simply connected. Included are all simply connected spaces, CW-complexes, polyhedra, and manifolds.
• If $\mathbb{H}$ is the Hawaiian earring with join point $b_0$, then $\mathbf{w}(\mathbb{H})=\{b_0\}$ is a singleton.
• The space pictured below has a shrinking sequence of Hawaiian earrings attached to the points $\frac{1}{2^{n-1}}\in [0,1]$. The wild set is $\{0,\dots,1/16,1/8,1/4,1/2,1\}$, which is not discrete.Notice that this space is wild at $0$ because every neighborhood of that point contains all but finitely many Hawaiian earrings.
• If $\mathbb{D}$ is the dyadic arc-space pictured below as the union of the base-arc $B=[0,1]\times \{0\}$ and countably many semi-circles, then $\mathbf{w}(\mathbb{D})=B$ is an interval and thus has uncountably many points.

The dyadic arc space

• It is possible that $\mathbf{w}(X)=X$! Examples include the Menger Curve, Sierpinski Triangle, and Sierpinski Carpet.

In all of the above examples, $\mathbf{w}(X)$ is closed in $X$; this is not a coincidence. Proving it’s true in general is a nice exercise to include in a course on fundamental groups.

Exercise: Prove that if $X$ is locally path-connected, then $\mathbf{w}(X)$ is closed in $X$.

In fact, even in the non-locally path-connected case, something can be said.

More general exercise: Show that if $f:Y\to X$ is a map from a locally path-connected space $Y$, then $f^{-1}(\mathbf{w}(X))$ is closed in $Y$.

An inquisitive reader may wonder in what ways the space $\mathbf{w}(X)$ is a kind of homotopy invariant of $X$. Some partial answers are published, but I promised myself not to go down the rabbit hole of writing about this yet.

## What is algebraic wildness?

In particular, does $\mathbf{w}(X)\neq\emptyset$, imply that the fundamental group $\pi_1(X,x)$ is wild?

At first glance, it seems like it might. Surely, arbitrarily small non-trivial loops are going to result in complicated algebra, right? To find out, let’s clarify what kind of property “wildness” should be for $\pi_1$.

• It can’t be a purely group theoretic property since, by CW-approximation, every group is the fundamental group of some 2-dimensional CW-complex. I don’t care how complicated your group is or how hard the Whitehead conjecture is, the fundamental group of a CW-complex does not count as being wild.
• An algebraically “wild” fundamental group should admit at least one (possibly trivial) element that is represented by an infinite concatenation of non-contractible loops.
• It doesn’t just have to happen at the basepoint. Because of path-conjugation, wildness might occur at any point of $X$.

Therefore, since algebraic wildness is really about the natural presence of infinitary operations like infinite sums in calculus/analysis (and in contrast with binary, trinary, or other finitary operations that come for free in ordinary groups), we’ll use the more descriptive word “infinitary” instead of “wild.”

Definition: A fundamental group $\pi_1(X,x_0)$ is infinitary at $x\in X$ if there exists a loop $\alpha:[0,1]\to X$ based at $x$, and a closed set $\{0,1\}\subseteq C\subseteq \alpha^{-1}(x)$ with such that $[0,1]\backslash C$ has infinitely many components, and for each component $(a_n,b_n)$ of $[0,1]\backslash C$, the loop $\alpha_n=\alpha|_{[a_n,b_n]}$ is not null-homotopic. If $\pi_1(X,x_0)$ is not infinitary at $x$, then we say it is finitary at $x$. If $\pi_1(X,x_0)$ is finitary at all points of $x$, we say the group is finitary. If there is at least one point, where $\pi_1(X,x_0)$ is infinitary, we say the group is infinitary.

Two possible examples of path decompositions giving rise to an infinite product. The first example realizes an ordinary infinite product and the second case is a dense (or transfinite) product where $C$ is the middle third Cantor set.

Equivalent but more practical definition: A fundamental group $\pi_1(X,x_0)$ is infinitary at $x\in X$ if and only if there exists a map $f:(\mathbb{H},b_0)\to (X,x)$ from the Hawaiian earring such that $f$ restricted to each circle of $\mathbb{H}$ is not a null-homotopic loop.

Proof of Equivalence: If you start with the first definition, the loop $\alpha:[0,1]\to X$ satisfies $\alpha(C)=x$ and thus induces a map $f:[0,1]/C\to X$. However, because $[0,1]\backslash C$ has a countably infinite number of components, $[0,1]/C\cong \mathbb{H}$. Moreover, each loop $\alpha|_{[a,b]}$ induces the restriction of $f$ to a unique circle of $\mathbb{H}$. Conversely, if we start with a map $f:\mathbb{H}\to X$, let $x$ be the image of the wild point. If we take $\alpha_n$ to be the loop which is the image of $f$ on the $n$-th circle, consider the infinite concatenation $\alpha=\prod_{n=1}^{\infty}\alpha_n$ defined as $\alpha_n$ on $\left[\frac{n-1}{n},\frac{n}{n+1}\right]$, and $\alpha(1)=x$. The set $C=\{0,1/2,2/3,3/4,\dots,1\}$ now satisfies the first definition. $\square$

Definition: The algebraic 1-wild set of a space $X$ is the subspace

$\mathbf{aw}(X)=\{x\in X\mid \pi_1(X,x_0)\text{ is infinitary at }x\}$.

Exercise: Construct a locally path-connected space such that $\mathbf{aw}(X)$ is not sequentially closed.

Of course, topological 1-wildness (arbitrarily small non-contractible loops) and algebraic 1-wildness (a shrinking sequence of non-contractible loops) should feel closely related – they both involve the existence of arbitrarily small non-contractible loops at some point.

## A lot of the time, they’re the same

The next theorem shows where the two kinds of wildness agree.

Theorem: If $X$ is semilocally simply connected at $x$, then $\pi_1(X,x_0)$ is finitary at $x$. The converse is true if $X$ is first countable at $x$.

Proof. Suppose $X$ is semilocally simply connected at $x$ and $f:(\mathbb{H},b_0)\to (X,x)$ is a map. We must show that $f$ applied to at least one circle of $\mathbb{H}$ is null-homotopic. Find an open neighborhood $U$ of $x$ such that every loop in $U$ based at $x$ is null-homotopic in $X$. Write $\mathbb{H}=\bigcup_{n\geq 1}C_n$ as the usual union of circles $C_n$ of radius $1/n$. By the continuity of $f$, there is a neighborhood $V$ of $b_0$ such that $f(V)\subseteq U$. Given the topology of $\mathbb{H}$, we may find an $N$ such that $C_n\subseteq V$ for all $n\geq N$. Therefore, $f$ maps $C_N$ to a loop in $U$, which must be null-homotopic in $X$. This completes the first direction.

For the converse, suppose $X$ is first countable at $x$ and that $X$ is not semilocally simply connected at $x$. Let $U_1\supseteq U_2\supseteq U_3\supseteq \cdots$ be a neighborhood base at $x$. By assumption, for each $n\geq 1$, there is a loop $\alpha_n:[0,1]\to U_n$ based at $x$, which is not null-homotopic in $X$.

Define a function $f:\mathbb{H}\to X$ to be $\alpha_n$ on $C_n$. To check the continuity of $f$, we really only need to check the continuity of $f$ at $b_0$: If $U$ is a neighborhood of $x$, find $N$ with $U_N\subseteq U$. Then it is clear that $f\left(\bigcup_{n\geq N}C_n\right)\subseteq U_N\subseteq U$. For $n\in \{1,2,\dots N-1\}$ find a neighborhood $V_n$ of $b_0$ in $C_n$ such that $\alpha_n(V_n)\subseteq U$. Now $V=\bigcup_{n\geq N}C_n\cup\bigcup_{1\leq n\leq N-1}V_n$ is a neighborhood of $b_0$ such that $f(V)\subseteq U$. Since $f$ is continuous an non-trivial on each circle, $\pi_1(X,x_0)$ is infinitary. $\square$

Corollary: For any space $X$, we have $\mathbf{aw}(X)\subseteq \mathbf{w}(X)$ with equality if $X$ is first countable.

So, for first countable spaces, the two notions of wildness are the same. Of course, even with some non-first countable spaces like very large CW-complexes, we still have equivalence.

This all suggests that to find a difference between topological and algebraic wildness, we need to dig into difference between nets and sequences of loops.

## but not always

General Topology permits all sorts of phenomenon, including interesting spaces that are not first countable. To conclude this post, I’m going to describe a space that isn’t semilocally simply connected (and doesn’t have a universal covering space) but whose fundamental group is not wild, i.e. is finitary.

Let $\omega_1$ be the first uncountable ordinal and $\omega_1+1=\omega_1\cup\{\omega_1\}$ be it’s successor, the first compact uncountable ordinal. Here, $\omega_1$ denote the maximal point of $\omega_1+1$ and we take it to be the basepoint. The key topological fact we’ll need to remember is that there does not exist any sequence in $[0,\omega_1)$ that converges to $\omega_1$. Let

$X=\Sigma (\omega_1+1)=\frac{(\omega_1+1)\times [0,1]}{(\omega_1+1)\times \{0,1\}\cup\{\omega_1\}\times[0,1]}$

be the reduced suspension with canonical basepoint $x_0$. For each countable ordinal $\alpha<\omega_1$, the image of $\{\alpha\}\times [0,1]$ will result in a unique circle $X_{\alpha}$ in $X$. These circles will all be joined at $x_0$. However, the topology of this space is not the one you’d give to a wedge of CW-complexes. To help visualize this monster, consider the first convergent sequence $\{n\}\to\omega$ in $\omega_1+1$. This corresponds to the subspace $\Sigma(\{1,2,\dots,\omega\}\cup\{\omega_1\})$ of $X$ illustrated below as the sequence of circles $\{X_n\}$ converging to the limit circle $X_{\omega}$.

From this subspace, imagine building $X$ inductively by creating larger and larger bouquet’s of circles parameterized by countable ordinals. In the limit, the circles will “converge” to $x_0$ in the sense that if $U$ is an open neighborhood of $x_0$, then there exists a $\beta<\omega_1$ such that $\bigcup_{\alpha\geq \beta}X_{\alpha}\subseteq U$.

Sorry, Not Sorry: The space $X$ is compact and it’s locally path connected at $x_0$, but it’s not locally path-connected at all of its points. I take no responsibility for this. Countable limit ordinals like $\omega$ are to blame. Fortunately, we’re interested in the topology around $x_0$ so there is no harm in sweeping this under the rug. If you’re unhappy about it, remember that you can take the locally path-connected coreflection without any loss of homotopy/homology group data. The downside to doing so is that the coreflection is not compact. So it goes.

Observation 1: $X$ is not semilocally simply connected. In particular, $\mathbf{w}(X)=\{x_0\}$.

Proof. Since $X$ is a quotient space of $(\omega_1+1)\times[0,1]$ and $[0,1]$ is compact, for every neighborhood $U$ of $x_0$, there is a countable ordinal $\beta$ such that $\bigcup_{\alpha\geq \beta}X_{\alpha}\subseteq U$. Also, $X$ retracts onto each circle $X_{\alpha}$. Therefore, the loop traversing $X_{\beta}$ is contained in $U$ and is not contractible in $X$. Since ordinals are totally path disconnected, every path component of $X\backslash \{x_0\}\cong \omega_1\times (0,1)$ is an open interval. Therefore, $X$ is semilocally simply connected at every point in $X\backslash \{x_0\}$. $\square$

Observation 2: $\pi_1(X,x_0)$ is finitary, i.e. $\mathbf{aw}(X)=\emptyset$.

Proof. Since $\mathbf{aw}(X)\subseteq \mathbf{w}(X)=\{x_0\}$, if $\mathbf{aw}(X)$ contains a point, it must be $x_0$. Consider any map $f:(\mathbb{H},b_0)\to (X,x_0)$ and suppose, to obtain a contradiction, that $f|_{C_n}$ is not null-homotopic for each circle $C_n$, $n\geq 1$ of $\mathbb{H}$. In order for this to happen, it must be the case that, for each $n$, there exists some arc of the circle $C_n$ that maps onto some circle $X_{\alpha_n}$.

In particular, we may find $t_n\in C_n\backslash \{b_0\}$ such that $f(t_n)$ is the image of the point $(\alpha_n,1/2)$ in the circle $X_{\alpha_n}$ (this is the point on $X_{\alpha_n}$ furthest from/antipodal to $b_0$). Since $\{t_n\}\to b_0$ in $\mathbb{H}$, the continuity of $f$ tells us that the sequence $\{ (\alpha_n,1/2)\}$ converges to $x_0$ in $X$.

Maybe you already see the problem. Remember the one thing I said we’d need to use about $\omega_1+1$? There is no sequence of countable ordinals converging to $\omega_1$. But if $(\beta,\omega_1]=\{\alpha\mid \beta<\alpha\leq \omega_1\}$ is an arbitrary neighborhood of $\omega_1$ in $\omega_1+1$, then we may take $V$ to be the open neighborhood of $x_0$, which is the image of $(\beta,\omega_1]\times [0,1]\cup (\omega_1+1)\times [0,1/3)\cup(2/3,1]$ in $X$. The only way for the sequence $\{ (\alpha_n,1/2)\}$ to eventually be in $V$ is for the sequence $\{\alpha_n\}$ of countable ordinals to eventually be inside $(\beta,\omega_1]$. This implies $\{\alpha_n\}\to \omega_1$; a contradiction of the topology of $\omega_1$. Since $x_0$ is not in $\mathbf{aw}(X)$, this set must be empty. $\square$

Having made Observations 1 and 2, we conclude that even though $X$ fails to have a simply connected covering space, it’s fundamental group is completely tame. In fact, from a fundamental group perspective, it’s basically the same as an ordinary wedge of circles $\bigvee_{\alpha\in\omega_1}X_{\alpha}$, which is a CW-complex! Intuitively, the neighborhood base at $x_0$ is so large (more precisely, the “tightness” is so large) that shrinking sequences of loops don’t actually converge to the constant loop at $x_0$ and this sequential convergence is exactly what is required to have an infinitary fundamental group.

If you put together the ideas behind the proofs of Observations 1 and 2, it’s not too hard to prove the following more detailed results:

Theorem: The continuous identity function $\bigvee_{\alpha\in\omega_1}X_{\alpha}\to X$ from the ordinary (i.e. CW) wedge of circles to $X$ is a weak homotopy equivalence.

Corollary: $\pi_1(X,x_0)$ is canonically isomorphic to the free group $F(\omega_1)$ on uncountably many generators.

Here’s a fun exercise I’ll leave you with.

Challenge Exercise: Construct a locally path-connected space $X$ for which $\mathbf{aw}(X)$ is not closed.

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