The Griffiths Twin Cone

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]. he 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 $\mathbb{H}\subset\mathbb{R}^2$ be the usual Hawaiian earring space with basepoint $x=(0,0)$. The Griffiths twin cone, denoted $\mathbb{G}$ is the twin cone $TC(\mathbb{H},x)=(C\mathbb{H},\ast)\vee(C\mathbb{H},\ast)$ over the Hawaiian earring.

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

Griffiths twin cone

Notice that the bases of the adjoined cones form the one-point union of two copies of the Hawaiian earring $(\mathbb{H},x)\vee(\mathbb{H},x)$ which is clearly homeomorphic to $\mathbb{H}$ itself. This means we can construct this space in a slightly different way that will help clarify the relationship between the fundamental groups of $\mathbb{H}$ 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 $\mathbb{H}=\bigcup_{n\geq 1}C_n$ 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

Notice the intersection of $\mathbb{G}$ and the xy-plane is now the standard construction of the Hawaiian earring. From now on, we’ll identify $\mathbb{H}$ 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 $A_{S}=\bigcup_{n\in S}A_n\cong C\mathbb{H}$ is homeomorphic to the cone on the Hawaiian earring 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 $\pi_1(\mathbb{H})$ 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 $\displaystyle\mathbb{H}_{odd}=\bigcup_{n\text{ even}}C_n$ be the odd circle of the Hawaiian earring and $\displaystyle\mathbb{H}_{even}=\bigcup_{n\text{ even}}C_n$ be the even circles. Both of these subspaces are still homeomorphic to $\mathbb{H}$. By including these as subspaces of $\mathbb{H}$, we get two important subgroups of $\pi_1(\mathbb{H})$. Let

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

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

Theorem 1: The inclusion $i:\mathbb{H}\to\mathbb{G}$ induces a surjection $\phi:\pi_1(\mathbb{H})\to\pi_1(\mathbb{G})$ 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, $V_{odd}=\{(x,y,z)\in\mathbb{G}|z>1/3\}$ and $V_{even}=\{(x,y,z)\in\mathbb{G}|z<-1/3\}$.

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,
• $U\cong\mathbb{H}\times(-2/3,2/3)$ deformation retracts onto $\mathbb{H}$. Therefore, we can identify $\pi_1(U)=\pi_1(\mathbb{H})$,
• $V_{odd}\cap U\cong\mathbb{H}_{odd}\times (1/3,2/3)$ is path connected and deformation retracts onto $\mathbb{H}_{odd}$. This means the inclusion $V_{odd}\cap U\to U$ induces the canonical homomorphism $k_{odd}:\pi_1(\mathbb{H}_{odd})\to\pi_1(\mathbb{H})$.
• $V_{even}\cap U\cong\mathbb{H}_{even}\times (-1/3,-2/3)$ is path connected and deformation retracts onto $\mathbb{H}_{even}$. This means the inclusion $V_{even}\cap U\to U$ induces the canonical homomorphism $k_{even}:\pi_1(\mathbb{H}_{even})\to\pi_1(\mathbb{H})$.

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 $\pi_1(X)=\pi_1(V_{odd}\cup U)\cong\pi_1(\mathbb{H})/K_1$ where $K_1$, i.e. the conjugate closure of the image $G_{odd}=Im(\pi_1(\mathbb{H}_{odd})\to\pi_1(\mathbb{H}))$ 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 $\pi_1(\mathbb{H}_{even})\to\pi_1(X)$ induced by the inclusion $V_{even}\cap U=V_{even}\cap X\to X$.

Altogether, we see that both inclusions

$i:\mathbb{H}\to X\to\mathbb{G}$

induce a surjection of fundamental groups

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

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 $\mathbb{H}_{odd}$ or $\mathbb{H}_{even}$ 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 $a:\pi_1(\mathbb{H})\to\pi_1(X)$ and $b:\pi_1(X)\to\pi_1(\mathbb{H})$ 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 $Im(\pi_1(\mathbb{H}_{even})\to\pi_1(X))=a(G_{even})$. 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 $\mathbb{H}_{\geq m}=\bigcup_{n\geq m}C_n$ of the Hawaiian earring 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 $\beta:[0,1]\to\mathbb{H}_{\geq m}$ for every $m\geq 1$.

Proof. Fix $m\geq 1$. According to Lemma 1, $\alpha$ is homotopic to a based loop $\gamma:[0,1]\to\mathbb{H}$. According to our study of the Hawaiian 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 $\mathbb{H}_{\geq m}$ or $\bigcup_{1\leq k. But we observed above that any loop in $\mathbb{G}$ which has image in $\bigcup_{1\leq k is homotopically trivial. Therefore $\gamma$ is homotopic in $\mathbb{G}$ to a loop in $\mathbb{H}_{\geq m}$. $\square$

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

I’ve already written a few posts about different properties of the Hawaiian earring group $\pi_1(\mathbb{H})$. 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 $\pi_1(\mathbb{H})\to\mathbb{Z}$ 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 $f\circ\phi:\pi_1(\mathbb{H})\to\mathbb{Z}$ 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 $\pi_1(\mathbb{H})$ by the conjugate closure of the countable subset $\{[\ell_n]\in\pi_1(\mathbb{H})|n\geq 1\}\subset G_{odd}\cup G_{even}$, however, this quotient is “too big” since, to get $\pi_1(\mathbb{G})$, we must also kill the uncountable subgroups $G_{odd}=Im(\pi_1(\mathbb{H}_{odd})\to\pi_1(\mathbb{H}))$ and $G_{even}=Im(\pi_1(\mathbb{H}_{even})\to\pi_1(\mathbb{H}))$. 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 $f:\pi_1(\mathbb{H})\to G$ 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.