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.

wildness

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 $X$ . wildness2

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 $X_{\omega}$ .

hoopspace

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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Topological wildness vs. algebraic wildness

Topological Wildness

What is algebraic wildness?

A lot of the time, they’re the same

but not always

2 Responses to Topological wildness vs. algebraic wildness

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Topological wildness vs. algebraic wildness

Topological Wildness

What is algebraic wildness?

A lot of the time, they’re the same

but not always

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