PTM-DMV Conference: Wild algebraic & geometric topology session

Posted on September 7, 2014 by Jeremy Brazas

I realize it has been a while since I have posted anything. I’m afraid teaching and research took up enough of my late summer schedule to keep me from finishing up a number of posts I had started. This is alright; I’ve been working hard trying to make my first experience with teaching topology at Georgia State a success.

This post is a reminder about the upcoming PTM-DMV Conference in Poznan, Poland: the joint meeting of the German Mathematical Society (Deutsche Mathematiker-Vereinigung) and Polish Mathematical Societies (Polskie Towarzystwo Matematyczne).

I will be giving a talk in the Session on Wild Algebraic & Geometric Topology.

Abstracts can be found on the session link above. Here is the schedule:

9/17/2014 (Wednesday)

  • 2:30 – 2:55 PM Wolfgang Herfort – Cotorsion and Homology
  • 3:00 – 3:25 PM Wolfram Hojka – Mapping the harmonic archipelago
  • 3:30 – 3:55 PM Isacc Goldbring – The fundamental group of a locally finite graph with ends: a hyperfinite approach
  • 4:00 – 4:25 PM Benoit Loridant – Fundamental group of Rauzy fractals
  • 4:30 – 5:00 PM Coffee Break
  • 5:00 – 5:25 PM Jean-Francois LaFont – One-dimensional geodesic spaces, Part I: Structure Theory
  • 5:30 – 5:55 PM David Constantine – One-dimensional geodesic spaces, Part II: Marked length rigidity

9/18/2014 (Thursday)

  • 2:30 – 2:55 PM Hanspeter Fischer – Word calculus in the fundamental group of the Menger Curve
  • 3:00 – 3:25 PM Katsuya Eda – Singular homology groups of one-dimensional Peano continua
  • 3:30 – 3:55 PM Janusz Przewocki – Milnor-Thurston homology of some wild topological spaces
  • 4:00 – 4:25 PM Thilo Kuessner – Measure homology and singular homology
  • 4:30 – 5:00 PM Coffee Break
  • 5:00 – 5:25 PM Jeremy Brazas – A characterization of the unique path lifting property for the whisker topology
  • 5:30 – 5:55 PM Ali Pakdaman – One point unions preserve having the categorical universal covering
  • 6:00 – 6:25 PM Alvaro Sanchez-Gonzalez – A shape topology for the universal path space

9/19/2014 (Friday)

  • 2:30 – 2:55 PM Matija Cencelj – Gropes and their fundamental groups
  • 3:00 – 3:25 PM Brendon LaBuz – Big free groups acting on \Lambda-trees
  • 3:30 – 3:55 PM Oleg Bogopolski – Generalized presentations of groups, in particular of Aut(F_{\omega})
  • 4:00 – 4:25 PM Andreas Zastrow – The obstruction to contractibility of snake cones and Alternating cones
  • 4:30 – 5:00 PM Coffee Break

There should be many great talks in this session on some pretty wild stuff!

My own talk will address the following theoretical barrier: to what extent can the structure of the fundamental group of a path-connected metric space be understood using generalized covering maps based on unique lifting of paths and homotopies of paths? Various generalizations have proven to be particularly useful for studying one-dimensional spaces like the Menger curve (and will be used in Hanspeter Fischer’s talk) and studying free topological groups.

I’ll use some categorical ideas to show that if a subgroup H\subset\pi_1(X) corresponds to any such covering-like map (with the most primitive unique lifting properties), then it can be understood using generalized coverings constructed with the so-called “whisker topology.” The whisker construction \widetilde{X}_H is not specialized, it appears in most textbooks that include the classification of covering spaces, however, it doesn’t not seem to be as well-known that it has much more general applications. The obstruction to whether or not these generalized coverings exists is precisely the unique path lifting property of the endpoint projection p_H:\widetilde{X}_H\to X. I’ll then characterize, for any arbitrary subgroup H, the existence of unique path lifting for this map in terms of a more practical sequential closure-like condition using test maps from the following one-dimensional planar Peano continuum \mathbb{D}.

bbd1

 

Here’s a sneak peak of my slides. Hope to see you there!

Posted in Algebraic Topology, Conferences | Tagged , | Leave a comment

The Baer-Specker Group

Posted on July 2, 2014 by Jeremy Brazas

One of the infinite abelian groups that is important to infinite abelian group theory and which has shown up naturally in previous posts on infinitary fundamental groups is the Baer-Specker group, often just called the Specker group. This post isn’t all that topological, but I think it’s appropriate considering that infinite group theory is becoming increasingly important to this kind of infinite product-driven algebraic topology.

The Baer-Specker group is the countably infinite direct product of copies of the additive group of integers:

\displaystyle\prod_{n=1}^{\infty}\mathbb{Z}=\mathbb{Z}\times\mathbb{Z}\times...

The elements* of the Specker group are sequences (a_n) of integers a_n\in\mathbb{Z} and addition is component-wise (a_n)+(b_n)=(a_n+b_n).

We’ll write s_n=(0,0,..,0,1,0,0,...) for the sequence whose only non-zero component is a 1 in the n-th position. The countable set \{s_n|n\geq 1\} forms a basis for the free abelian subgroup \bigoplus_{n=1}^{\infty}\mathbb{Z} consisting of the sequences (a_1,a_2,...,a_n,0,0,...) that terminate in 0‘s.

But the Specker group \prod_{n=1}^{\infty}\mathbb{Z} is uncountable (as is any countably infinite product of sets with cardinality >2), which means the set \{e_n|n\geq 1\} cannot generate the whole group – equivalently \bigoplus_{n=1}^{\infty}\mathbb{Z} is proper.

So is \displaystyle\prod_{n=1}^{\infty}\mathbb{Z} a free abelian group or not?

On the face of it, the answer is not obvious. We answered the same question (with a negative answer) for the earring group \pi_1(\mathbb{E}) in a two-part post (Part I and Part II). This question about the Specker group is a bit easier since it is an abelian group with simpler combinatorial structure. In this post, I’ll walk through a nice, direct proof of the following theorem.

Theorem 1: \displaystyle\prod_{n=1}^{\infty}\mathbb{Z} is not a free group.

You’ll notice the argument is similar in spirit to the argument used to show \pi_1(\mathbb{E}) is not free: basically, we’ll assume the group is free and exploit the infinite structure to find an integer which has infinitely many divisors – something which obviously can’t happen.

*It is also common to view the Specker group as the group \mathbb{Z}^{\mathbb{N}} of functions f:\mathbb{N}=\{1,2,3,...\}\to\mathbb{Z} from the natural numbers to integers.

History

Apparently, Reinhold Baer [1] first proved Theorem 1 in 1937 (or at least stronger results that imply it). Ernst Specker [3] re-proved this result among a number of other original results about \prod_{n=1}^{\infty}\mathbb{Z}. So the foundational theory of \prod_{n=1}^{\infty}\mathbb{Z} is often attributed to Specker as well. The proof I’m using in this post is a nice one from the AMM [2].

Free abelian groups and divisibility

First, let’s review some terminology.

Recall that a basis for an abelian group A is a subset \{e_{\lambda}\mid\lambda\in\Lambda\}\subset A such that every element a\in A can be written uniquely as a linear combination a=\sum_{\lambda}n_{\lambda}e_{\lambda} where n_{\lambda}=0 for all but finitely many \lambda. An abelian group A is free if such a basis exists.

We will use the following elementary fact about free abelian groups.

Lemma 2: Suppose A is a free abelian group with basis \{e_{\lambda}\mid\lambda\in\Lambda\}, J\subset\Lambda is a subset, and H=\langle e_{\lambda}\mid\lambda\in J\rangle is the free subgroup generated with basis \{e_{\lambda}\mid\lambda\in J\}. The quotient A/H is a free abelian group for which the set of cosets \{e_{\lambda}+H \mid\lambda\in\Lambda\backslash J\} form a basis.

Also, we’ll deal with some divisibility issues.

If A is an abelian group, a\in A a non-zero element, then we say a natural number n divides a if

\displaystyle a=nx=\underbrace{x+x+...+x}_\text{n terms}

 for some x\in A. We call n a divisor of a and will be interested in cases when an element has finitely or infinitely many divisors. For example, every integer a\in \mathbb{Z} obviously has only finitely many divisors (using prime decomposition) but every non-zero rational number \frac{p}{q}\in \mathbb{Q} has infinitely many divisors since \frac{p}{q}=n\left(\frac{p}{nq}\right) for every n\geq 1.

Lemma 3: Every non-zero element of a free abelian group has only finitely many divisors.

Proof. Suppose A is a free abelian group with basis \{e_{\lambda}\mid\lambda\in\Lambda\} and a\in A is non-zero. Write a=\sum_{\lambda}n_{\lambda}e_{\lambda} uniquely using the basis. Fix some non-zero coefficient \lambda_0\in\Lambda such that n_{\lambda_0}\neq 0. Notice that if n divides a, then a=nx_n for some x_n=\sum_{\lambda}m_{\lambda}e_{\lambda}. But this means a is equal to both

nx_n=\sum_{\lambda}(nm_{\lambda})e_{\lambda} and \sum_{\lambda}n_{\lambda}e_{\lambda}.

By the uniqueness of the representation of a in the given basis n_{\lambda_0} is equal to nm_{\lambda} for some \lambda.

We have just argued that whenever n divides a, then n also divides the integer n_{\lambda_0}. Since integers only have finitely many divisors, a can only have finitely many divisors. \square

The Specker group is not free

Proof of Theorem 1. Suppose G=\prod_{n=1}^{\infty}\mathbb{Z} is a free abelian group with basis \{e_{\lambda}|\lambda\in\Lambda\}. Since G is uncountable, the indexing set \Lambda must also be uncountable.

Write each element s_n=(0,0,...,0,1,0,0,...) as a finite linear combination s_n=\sum_{\lambda}m_{\lambda,n}e_{\lambda} of basis elements. For fixed n, let B(n)=\{\lambda\in\Lambda\mid m_{\lambda,n}\neq 0\} so that B=\bigcup_{n=1}^{\infty}B(n) is the set of indices \lambda which has non-zero coefficient for some element s_n. Notice that B is countable since it is the countable union of finite sets B(n). This means the free abelian subgroup H=\langle e_{\lambda}\mid\lambda\in B\rangle generated by the basis \{e_{\lambda}\mid\lambda\in B\} is also countable.

Remark: It is worthwhile to point out that we have constructed H to be the smallest free subgroup (generated by our basis) containing all of the sequences s_n.

Here are a few important observations:

  1. \bigoplus_{n=1}^{\infty}\mathbb{Z}\subseteq H since s_n\in H for each n\geq 1.
  2. Since \Lambda is uncountable, the complement \Lambda\backslash B is also uncountable. Therefore the quotient group G/H is uncountable.
  3. By Lemma 2, G/H is a free abelian group with basis \{e_{\lambda}+H \mid\lambda\in\Lambda\backslash B\}.

Definition: Let’s say a sequence (a_n)\in G is multiplicative if for each n\geq 1, each quotient \displaystyle\frac{a_{n+1}}{a_n} of consecutive terms is an integer not equal to \pm 1 (automatically this means a_n\neq 0 for any n). Let M be the set of multiplicative sequences.

For example, (n!)=(1!,2!,3!,4!,...) and (2^n)=(2,4,8,16,...) are multiplicative. Let’s quickly convince ourselves that there are uncountably many multiplicative sequences.

Lemma 4: There are uncountably many multiplicative sequences in G.

Proof. Consider the set of sequences S=\{(b_n)\in G \mid |b_n|>1\text{ for each }n\}. Obviously this set is uncountable since, for instance, it contains the uncountable subset \prod_{n=1}^{\infty}\{2,3\}=\{(b_n)\in G\mid b_n\in\{2,3\}\}. Define a function f:S\to M by f(b_n)=(b_1,b_1b_2,b_1b_2b_3,...,b_1b_2\cdots b_k,...). Clearly f is an injection and since M has an uncountable subset, M must be uncountable. \square

Since M\subseteq G is uncountable and H\leq G is countable, there exists some multiplicative sequence a=(a_n)\in M such that a \notin H.

Recall that \bigoplus_{n=1}^{\infty}\mathbb{Z}\subseteq H and so we have

h_n=\sum_{k=1}^{n}a_ks_k=(a_1,a_2,....,a_n,0,0,...)\in H  for each  n\geq 1.

Since a=(a_n) is multiplicative, the element

\displaystyle x_n=\left(0,0,...,0,\frac{a_{n+1}}{a_n},\frac{a_{n+2}}{a_n},\frac{a_{n+3}}{a_n},...\right),

where the first non-zero term is in the n+1-st position, is a well-defined element of the Specker group. Notice that

a_nx_n=(0,0,...,0,a_{n+1},a_{n+2},...)=a-h_n\in a+H.

So in the free abelian group G/H, we have a+H=a_n(x_n+H) for all n\in\mathbb{N} showing that a+H has infinitely many divisors a_1,a_2,a_3,... in G/H. But according to Lemma 3, this cannot happen in a free abelian group so we obtain a contradiction.

This concludes the proof that G=\prod_{n=1}^{\infty}\mathbb{Z} is not free. \square

How is all this related to the earring group?

In an earlier post, I wrote up a proof of the fact that \pi_1(\mathbb{E}) was not a free (non-abelian) group. The argument was similar the one above for the Specker group but was a bit more sophisticated. Actually, we proved a stronger fact:

Let g_n=[\ell_n]\in\pi_1(\mathbb{E}) be the homotopy class of the loop \ell_n which traverses the n-th circle of the earring space once in the counterclockwise direction.

Fact: The homomorphism \eta:Hom(\pi_1(\mathbb{E}),\mathbb{Z})\to\prod_{n=1}^{\infty}\mathbb{Z} to the Specker group given by \eta(f)=(f(g_n))=(f(g_1),f(g_2),f(g_3),...) is injective and has image \bigoplus_{n=1}^{\infty}\mathbb{Z}.

Neither identifying the image of \eta nor proving injectivity of \eta is easy business. See this post if you’re looking for the proof.

Viewing \mathbb{E} as the infinite wedge \bigvee_{n=1}^{\infty}S^1 of circles as a subspace of the infinite torus \prod_{n=1}^{\infty}S^1, the inclusion map \bigvee_{n=1}^{\infty}S^1\to\prod_{n=1}^{\infty}S^1 induces the surjective homomorphism \epsilon:\pi_1(\mathbb{E})\to\pi_1\left(\prod_{n=1}^{\infty}S^1\right)=\prod_{n=1}^{\infty}\mathbb{Z} to the Specker group. Another way to think about this map is that the n-th component of \epsilon([\alpha]) is the winding number of \alpha around the n-th circle. For example, \epsilon(g_n)=s_n and \epsilon([\ell_1\ell_2\ell_3...])=(1,1,1,...).

In a similar way, let’s define a homomorphism

\zeta:Hom\left(\prod_{n=1}^{\infty}\mathbb{Z},\mathbb{Z}\right)\to\prod_{n=1}^{\infty}\mathbb{Z}, by \zeta(f)=(f(s_n))=(f(s_1),f(s_2),...)

and prove the analogous theorem.

Theorem 5: The homomorphism \zeta:Hom\left(\prod_{n=1}^{\infty}\mathbb{Z},\mathbb{Z}\right)\to\prod_{n=1}^{\infty}\mathbb{Z} given by \eta(f)=(f(g_n))=(f(s_1),f(s_2),f(s_3),...) is injective and has image \bigoplus_{n=1}^{\infty}\mathbb{Z}. Consequently, Hom\left(\prod_{n=1}^{\infty}\mathbb{Z},\mathbb{Z}\right)\cong\bigoplus_{n=1}^{\infty}\mathbb{Z} is countable.

Proof. Since \epsilon(g_n)=s_n for each n\geq 1, the following diagram commutes:

speckerpost

Since \epsilon is surjective, Hom(\epsilon,\mathbb{Z}) is injective. Since both \eta and Hom(\epsilon,\mathbb{Z}) are injective, so is \zeta. Also Im(\zeta)\subseteq Im(\eta)=\bigoplus_{n=1}^{\infty}\mathbb{Z} (since the triangle commutes). For each n\geq 1, let f_n:\prod_{n=1}^{\infty}\mathbb{Z}\to\mathbb{Z}, f_n((a_n))=a_n be the projection onto the n-th coordinate. Then \zeta(f_n)=s_n showing that Im(\zeta)=\bigoplus_{n=1}^{\infty}\mathbb{Z}. \square

Theorem 5 implies Theorem 1: The fact that Hom\left(\prod_{n=1}^{\infty}\mathbb{Z},\mathbb{Z}\right)\cong\bigoplus_{n=1}^{\infty}\mathbb{Z} is countable gives another proof that the Specker group is not free since if it the Specker group is free with (necessarily uncountable) basis B, then the group

Hom\left(\prod_{n=1}^{\infty}\mathbb{Z},\mathbb{Z}\right)=Hom\left(\bigoplus_{b\in B}\mathbb{Z},\mathbb{Z}\right)\cong Set(B,\mathbb{Z})

of all functions B\to\mathbb{Z} is uncountable.

This viewpoint using the earring group should not be considered an easier proof of Theorem 1 since the hard work is still present on the non-abelian side. On the other hand, exploring the relationship between the two “non-freeness” results – namely that the non-abelian version implies the abelian one – is worth looking at.

References.

[1] R. Baer, Abelian groups without elements of finite order, Duke Math. J. 3 (1937) 68-122.

[2] S. Schroer, Baer’s Result: The infinite product of the integers has no basis, The American Mathematical Monthly 115, No. 7 (2008), 660-663.

[3] E. Specker, Additive Gruppen von Folgen ganzer Zahlen, Port. Math. 9 (1949) 131-140.

I found Specker’s paper freely available here but disappointingly, couldn’t get access to Baer’s paper without using my University library.
Posted in Baer-Specker group, earring group, earring space, Free abelian groups, Free groups, Group homomorphisms, Infinite Group Theory | Tagged , , , , , , , | 8 Comments

The Griffiths Twin Cone

Posted on June 28, 2014 by Jeremy Brazas

This post is about an important wild space which, in many ways, is similar to the harmonic archipelago space that I posted about a few weeks ago. The Griffiths twin cone (or Griffiths space) was first studied by H.B. Griffiths in the 1950’s [1]. Griffiths showed that despite being the union of two contractible subspaces, this beast has a non-trivial – in fact, uncountable – fundamental group.

Constructing the Griffiths twin cone

For any space X, the cone over X is the space \displaystyle CX=\frac{X\times [0,1]}{X\times \{1\}}. The image of X\times \{1\} is the vertex v of the cone and the (homeomorphic) image of X\times\{0\} is the base of the cone. Every cone CX is contractible (to the vertex point of the cone) and consequently has a trivial fundamental group.

Now suppose X has basepoint x and \ast is the image of (x,0) in the base of the cone CX. Now join two copies of the cone together at a single point to get the “wedge” space TC(X,x)=(CX,\ast)\vee (CX,\ast). Let’s call this space the twin cone over X. We’ve joined two contractible spaces together so this twin cone must be contractible right? Well… not exactly. If we had formed the wedge (CX,v)\vee (CX,v) by adjoining two cones at their vertices, then yes, we’d get back a contractible space. But the cone does not necessarily contract onto \ast if X is “wild” at x. The Griffiths twin cone is an example of a non-contractible twin cone.

Definition: Let \mathbb{E}\subset\mathbb{R}^2 be the usual earring space with basepoint x=(0,0). The Griffiths twin cone, denoted \mathbb{G} is the twin cone TC(\mathbb{E},x)=(C\mathbb{E},\ast)\vee(C\mathbb{E},\ast) over the earring space.

In other words, we construct \mathbb{G} by taking two copies of the cone of the earring space and pasting them together at the wild points on the base.

Griffiths twin cone

Griffiths twin cone

Notice that the bases of the adjoined cones form the one-point union of two copies of the  earring space (\mathbb{E},x)\vee(\mathbb{E},x) which is clearly homeomorphic to \mathbb{E} itself. This means we can construct this space in a slightly different way that will help clarify the relationship between the fundamental groups of \mathbb{E} and \mathbb{G}.

Alternative construction: For each integer n\geq 1, let C_n\subset\mathbb{R}^2 be the circle of radius \frac{1}{n} centered at \left(\frac{1}{n},0\right). Now view these circles and their union \mathbb{E}=\bigcup_{n\geq 1}C_n in the xy-plane of \mathbb{R}^3.

  • Let v_1=(0,0,1) and v_0=(0,0,-1).
  • If n is odd, let A_n be the union of all line segments from C_n to v_1.
  • If n is even, let A_n be the union of all line segments from C_n to v_0.

In short, A_n is the cone of C_n in \mathbb{R}^3 with vertex v_{n\text{ mod}2}. The Griffiths twin cone is the union \mathbb{G}=\bigcup_{n\geq 1}A_n with basepoint x_0=(0,0,0).

Alternative construction of the Griffiths twin cone

Alternative construction of the Griffiths twin cone

Notice the intersection of \mathbb{G} and the xy-plane is now the standard construction of the earring. From now on, we’ll identify \mathbb{E} as a subspace of \mathbb{G} in this way.

Here are a few quick observations we can go ahead and make about the fundamental group \pi_1(\mathbb{G}):

  • Every subspace A_n is contractible since it is the cone of a circle (and thus homeomorphic to the unit disk). Consequently, if we let \ell_n:[0,1]\to C_n\subset\mathbb{G} be the circle which traverses C_n counterclockwise in the xy-plane, then \ell_n is homotopic (in \mathbb{G}) to the constant loop and [\ell_n] is the identity element 1\in\pi_1(\mathbb{G}).
  • The previous point implies that if F\subset\{1,2,...\} is any finite set of integers and A_F=\bigcup_{n\in F}A_n, then [\alpha]=1 for any loop \alpha:[0,1]\to\mathbb{G} with image in A_F.
  • Suppose a given (finite or infinite) subset S\subset \{1,2,...\} contains all even or all odd integers. Then A_{S}=\bigcup_{n\in S}A_n\cong C\mathbb{E} is homeomorphic to the cone on the earring space and is therefore contractible. This means we have [\alpha]=1 for ANY based loop \alpha:[0,1]\to\mathbb{G} with image in A_S, even if it has image in infinitely many distinct cones.
  • What is less clear is what happens to the infinite concatenation \alpha=\ell_1\ell_2\ell_3\ell_4\cdots defined as \ell_n on \left[\frac{n-1}{n},\frac{n}{n+1}\right] and \alpha(1)=x_0. This loop winds around infinitely many even circles and infinitely many odd circles in an alternating fashion. Obviously, we can contract any finite number of these loops, showing that \alpha is homotopic to \ell_n\ell_{n+1}\ell_{n+1}\cdots for any n\geq 1. But even though we can deform \alpha arbitrarily close to the basepoint, it seems unlikely that this loop is homotopically trivial since we’d have to contract it all the way up (and down) to the two vertices infinitely many times.

The fundamental group of the Griffiths twin cone

To understand the relationship between \pi_1(\mathbb{E}) and \pi_1(\mathbb{G}) in elementary terms, we’ll use the following specialized case of the van Kampen theorem – one of the most useful computational results for fundamental groups.

van-Kampen Theorem (special case): Suppose X is the union of two path-connected open sets U_1,U_2 such that U_1\cap U_2 is path connected and contains the basepoint x_0. If U_1 is simply connected, then \pi_1(X)\cong \pi_1(U_2)/N where N is the conjugate closure of the image Im(\pi_1(U_1\cap U_2)\to \pi_1(U_2)) of the homomorphism induced by inclusion.

Let \displaystyle\mathbb{E}_{odd}=\bigcup_{n\text{ even}}C_n be the odd circle of the earring space and \displaystyle\mathbb{E}_{even}=\bigcup_{n\text{ even}}C_n be the even circles. Both of these subspaces are still homeomorphic to \mathbb{E}. By including these as subspaces of \mathbb{E}, we get two important subgroups of \pi_1(\mathbb{E}). Let

G_{odd}=Im(k_{odd}:\pi_1(\mathbb{E}_{odd})\to\pi_1(\mathbb{E})) and G_{even}=Im(k_{even}:\pi_1(\mathbb{E}_{even})\to\pi_1(\mathbb{E}))

be the respective images of the group monomorphisms induced by inclusion.

Theorem 1: The inclusion i:\mathbb{E}\to\mathbb{G} induces a surjection \phi:\pi_1(\mathbb{E})\to\pi_1(\mathbb{G}) of fundamental groups. Moreover, \ker\phi is the conjugate closure of G_{odd}\cup G_{even}.

Proof. Define an open cover of \mathbb{G} as follows: Let U=\{(x,y,z)\in\mathbb{G}|-2/3<z<2/3\}, V_{odd}=\{(x,y,z)\in\mathbb{G}|z>1/3\} and V_{even}=\{(x,y,z)\in\mathbb{G}|z<-1/3\}.

Open cover of the Griffiths twin cone

Let’s make a few elementary observations about these open sets and their fundamental groups:

  • \mathbb{G}=V_{odd}\cup U\cup V_{even},
  • U, V_{odd}, and V_{even} are path connected,
  • V_{odd} and V_{even} are simply connected since each is a cone with deleted base and therefore contractible,
  • U\cong\mathbb{E}\times(-2/3,2/3) deformation retracts onto \mathbb{E}. Therefore, we can identify \pi_1(U)=\pi_1(\mathbb{E}),
  • V_{odd}\cap U\cong\mathbb{E}_{odd}\times (1/3,2/3) is path connected and deformation retracts onto \mathbb{E}_{odd}. This means the inclusion V_{odd}\cap U\to U induces the canonical homomorphism k_{odd}:\pi_1(\mathbb{E}_{odd})\to\pi_1(\mathbb{E}).
  • V_{even}\cap U\cong\mathbb{E}_{even}\times (-1/3,-2/3) is path connected and deformation retracts onto \mathbb{E}_{even}. This means the inclusion V_{even}\cap U\to U induces the canonical homomorphism k_{even}:\pi_1(\mathbb{E}_{even})\to\pi_1(\mathbb{E}).

Now, we’re ready to apply the special case of van-Kampen Theorem – first to X=V_{odd}\cup U and then to \mathbb{G}=X\cup V_{even}.

First application: Since V_{odd} is simply connected, the van-Kampen Theorem implies that \pi_1(X)=\pi_1(V_{odd}\cup U)\cong\pi_1(\mathbb{E})/K_1 where K_1, i.e. the conjugate closure of the image G_{odd}=Im(\pi_1(\mathbb{E}_{odd})\to\pi_1(\mathbb{E})) of the homomorphism on fundamental groups induced by the inclusion V_{odd}\cap U\to U.

Second application: Notice V_{even}\cap X=V_{even}\cap U. Since V_{even} is simply connected, the van-Kampen Theorem implies that \pi_1(\mathbb{G})=\pi_1(V_{even}\cup X)\cong\pi_1(X)/K_2 where K_2 is the conjugate closure of the image of the homomorphism \pi_1(\mathbb{E}_{even})\to\pi_1(X) induced by the inclusion V_{even}\cap U=V_{even}\cap X\to X.

Altogether, we see that both inclusions

i:\mathbb{E}\to X\to\mathbb{G}

induce a surjection of fundamental groups

\phi:\pi_1(\mathbb{E})\to\pi_1(X)\to\pi_1(\mathbb{G})

This proves the first statement of the theorem – that \phi is surjective.

Let N be the conjugate closure of G_{odd}\cup G_{even}. We need to show N=\ker\phi. By our observations above, any loop with image entirely in \mathbb{E}_{odd} or \mathbb{E}_{even} is null-homotopic in \mathbb{G}. Since \phi is induced by inclusion, \phi(G_{odd})=\phi(G_{even})=1 is trivial. This means that N\subseteq\ker\phi.

For the other inclusion, let a:\pi_1(\mathbb{E})\to\pi_1(X) and b:\pi_1(X)\to\pi_1(\mathbb{E}) be the surjections induced by inclusion that compose to give \phi. By our two applications of the van-Kampen theorem, \ker(a)=K_1 and \ker(b)=K_2. Thus \ker\phi=a^{-1}(K_2). Recall that K_2 is the conjugate closure of Im(\pi_1(\mathbb{E}_{even})\to\pi_1(X))=a(G_{even}). Therefore if g\in\ker\phi, then a(g) is of the form

a(\ell_{1}^{-1}k_1\ell_1\ell_{2}^{-1}k_2\ell_2...\ell_{n}^{-1}k_n\ell_n)=[h_{1}^{-1}a(k_1)h_{1}][h_{2}^{-1}a(k_2)h_{2}]...[h_{n}^{-1}a(k_n)h_{n}]

where k_j\in G_{even} and a(\ell_j)=h_j. Notice k=\ell_{1}^{-1}k_1\ell_1\ell_{2}^{-1}k_2\ell_2...\ell_{n}^{-1}k_n\ell_n lies in the conjugate closure of G_{even}. Thus gk^{-1}\in\ker(a)=K_1 which is the conjugate closure of G_{odd}. Thus g is an element of the coset K_1k\subset N. This concludes the proof that \ker\phi=N. \square

For each m\geq 1, we have a smaller copy \mathbb{E}_{\geq m}=\bigcup_{n\geq m}C_n of the earring space that we can view as a subspace of \mathbb{G}. The next Corollary basically says the we can continuously deform any loop as close as we want to the basepoint.

Corollary 2: Every based loop \alpha:[0,1]\to\mathbb{G} is homotopic in \mathbb{G} to a based loop \beta:[0,1]\to\mathbb{E}_{\geq m} for every m\geq 1.

Proof. Fix m\geq 1. According to Lemma 1, \alpha is homotopic to a based loop \gamma:[0,1]\to\mathbb{E}. According to our study of the earring group (in the original post and also in Lemma 5 of this post) \gamma is homotopic to a finite concatenation of loops each of which has image in either \mathbb{E}_{\geq m} or \bigcup_{1\leq k<m}C_k. But we observed above that any loop in \mathbb{G} which has image in \bigcup_{1\leq k<m}C_k is homotopically trivial. Therefore \gamma is homotopic in \mathbb{G} to a loop in \mathbb{E}_{\geq m}. \square

Another quick observation about \pi_1(\mathbb{G})

I’ve already written a few posts about different properties of the earring group \pi_1(\mathbb{E}). Let’s use one of these oldies-but-goodies to prove something interesting about \pi_1(\mathbb{G}).

Theorem 3: Hom(\pi_1(\mathbb{G}),\mathbb{Z})=0. Consequently, \pi_1(\mathbb{G}) cannot be a free group.

Proof. In Lemma 6 of this post, we decided that every homomorphism \pi_1(\mathbb{E})\to\mathbb{Z} that sends [\ell_n] to 0 for all n\geq 1, must be the trivial homomorphism. But notice that \phi([\ell_n])=1 for all n\geq 1. So if f:\pi_1(\mathbb{G})\to\mathbb{Z} is any homomorphism, then f\circ\phi:\pi_1(\mathbb{E})\to\mathbb{Z} must be trivial. Since \phi is surjective, f must be trivial.

Honestly, it can – for me – be tempting to think of \pi_1(\mathbb{G}) as the quotient of \pi_1(\mathbb{E}) by the conjugate closure of the countable subset \{[\ell_n]\in\pi_1(\mathbb{E})|n\geq 1\}\subset G_{odd}\cup G_{even}, however, this quotient is “too big” since, to get \pi_1(\mathbb{G}), we must also kill the uncountable subgroups G_{odd}=Im(\pi_1(\mathbb{E}_{odd})\to\pi_1(\mathbb{E})) and G_{even}=Im(\pi_1(\mathbb{E}_{even})\to\pi_1(\mathbb{E})). This, in my opinion, makes dealing with \pi_1(\mathbb{G}) a little more complicated than it is to deal with the Harmonic archipelago group \pi_1(\mathbb{HA}). For instance, if G is any non-trivial finite group, there are uncountably many homomorphisms f:\pi_1(\mathbb{E})\to G such that f([\ell_n])=1 for all n\geq 1, however it does not follow immediately – as it did for \pi_1(\mathbb{HA}) – that Hom(\pi_1(\mathbb{G}),G) is uncountable (though I have heard by word of mouth that this is true).

References.

[1] H.B. Griffiths, The fundamental group of two spaces with a common point, Quart. J. Math. Oxford (2) 5 (1954) 175-190.

[2] K. Eda, A locally simply connected space and fundamental groups of one point unions of cones, Proc. Amerc. Math. Soc. 116 no. 1 (1992) 239-249.

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