Let $X=S^3\setminus B$ be the link complement of the Borromean rings.

    Borromean rings(source)

Then $G=\pi_1(X)$ has a presentation of the form $$ G = \langle \; a,b,c \mid [a,[b^{-1},c]],\; [b,[c^{-1},a]], \; [c,[a^{-1},b]]\;\rangle, $$ where any one of the relations is redundant.

Removing any component of the link results in a trivial link of two components. The effect on the fundamental group is to send one of the generators, say $c$, to $1$, resulting in a homomorphism $G\to F_2$ to the free group on $2$ generators. There results a split extension $$ 1\to K\to G\to F_2\to 1. $$

Is the kernel $K$ a free group?

  • $\begingroup$ could you specify the convention you use for the commutator ($aba^{-1}b^{-1}$ and $a^{-1}b^{-1}ab$ are both standard)? $\endgroup$
    – YCor
    Jun 6, 2013 at 12:48
  • $\begingroup$ @Yves: I tend to use $[a,b]=aba^{-1}b^{-1}$. $\endgroup$
    – Mark Grant
    Jun 6, 2013 at 12:59

2 Answers 2


Correction: Yes, it's a free group. The two component unlink complement may be regarded as a union of two handlebodies glued along a 4-punctured sphere (which surjects the fundamental group of both handlebodies), regarded as the bridge sphere in 2-bridge position, so that each handlebody group is $\pi_1$ isomorphic to the link complement group. The kernel of the 4-punctured sphere group to the unlink group is normally generated by a curve of intersection of the 4-punctured sphere with a splitting 2-sphere that separates the two link components.

The third component of the Borromean rings may be drawn as simple closed curve on this bridge sphere. The complement of this curve in the 4-punctured sphere consists of two 3-punctured spheres which $\pi_1$-inject into the handlebody group since the curve is disk-busting. When we take the universal cover of the 2-component unlink, the preimage of the 3-punctured spheres will be an infinite union of planes. The preimage of the two handlebodies have interiors homeomorphic to $\mathbb{R}^3$, but with boundary a copy of an infinite union of $\mathbb{R}^2$s. We then glue the two $\mathbb{R}^3$s together along the these planes in the boundaries (this is, double the handlebody cover along the planes in the boundary). Thus, by Van Kampen, the fundamental group is a graph of groups with two trivial vertex groups, and trivial edge groups, so it is an infinitely generated free group.

  • $\begingroup$ do you know if it's a finitely generated free group? $\endgroup$
    – YCor
    Jun 6, 2013 at 12:57
  • 1
    $\begingroup$ It is infinitely generated. $\endgroup$
    – Sam Nead
    Jun 6, 2013 at 14:13
  • 3
    $\begingroup$ Yves: There is a general fact of 3-dimensional topology, which you can find in Hempel's book: If $M$ is a compact 3-manifold, $H$ is a normal f.g. subgroup of $\pi_1(M)$ then either $H$ is virtually cyclic (in which case $M$ is Seifert) or $\pi(1(M)/H$ is virtually cyclic. $\endgroup$
    – Misha
    Jun 6, 2013 at 14:31
  • 1
    $\begingroup$ Ian: If I understand your picture correctly, $\pi_1$ of the 4-punctured sphere does not inject into $\pi_1$ of the 2-component link complement. Therefore the universal cover of the of the 2-component link complement does not contain a copy of the universal cover of the 4-punctured sphere (i.e. $\mathbb{R}^2$). Have I misunderstood something? $\endgroup$ Jun 6, 2013 at 17:03
  • 1
    $\begingroup$ @Kevin - The preimage of the bridge sphere is actually a planar surface of infinite topological type, with a slightly complicated boundary. (This is because longitudes die in $F_2$, so they lift to the cover. But meridians unwrap, giving paired of copies of $\mathbb{R}$ in the boundary.) But the infinite collection of lines coming from the third component cuts the planar surface into a collection of disks, so Ian is still ok. $\endgroup$
    – Sam Nead
    Jun 6, 2013 at 19:35

I believe that $K$ is an infinitely generated free group. First we find an infinite presentation $K = \langle X \mid R \rangle$, and then we argue that we can pick an infinite subset of $X$ that freely generates.

Presentation: Note that we can obtain the Borromean rings from the handlebody of genus two by drilling out the commutator curve and attaching a two-handle to the boundary along the usual separating curve. Since the attaching curve for the two-handle is trivial in the handlebody, we can take the universal cover of the handlebody and then attach infinitely many two-handles, and drill infinitely many lines. After drawing a few pictures, we can organize all of this information as follows.

Let $D^2$ be the Poincare disk model for the hyperbolic plane. Pick hyperbolic isometries $a$ and $b$ that have perpendicular axes and equal (very long) translation lengths. Thus $F_2 = \langle a,b \rangle$ is a free group of rank two. Let $T$ be the regular four-valent tree isometrically embedded in $H^2$, arising as the Cayley graph of $F_2$. Let $v_0$ be the vertex of $T$ corresponding to the identity of $F_2$. The universal cover of the handlebody above is obtained as a three-dimensional neighborhood of $T$, and $F_2$ is the deck group.

We define $X = \{ x_q \}$ where $q$ ranges over the connected components of $H^2 - T$. (These correspond to the drilled lines.) This gives the generating set for $K$ subject to the relations $R = \{ x_p X_q x_r X_s \}$ where the regions $p,q,r,s$ are arranged counter-clockwise about a vertex of $T$. (These correspond to the attached two-handles. Because we started with the commutator curve, the two-handles are attached to the neighborhood of $T$ via a "baseball curve" around each vertex.) Here we use the convention that $X_q = (x_q)^{-1}$. (Also, $p$ is always to the "north-east" of $v$ - we use translates of the picture about $v_0$ to set conventions.) For future use, let $w_v$ be the relation coming from vertex $v$.

Subset that freely generates: We recursively color the regions of $H^2 - T$ as follows. (1) All regions start white. (2) In the first step, we pick two regions $p, q$ adjacent to $v_0$. Color both of $p$ and $q$ black. (3) In general, pick a vertex $v$ having exactly two white regions $p, q$ adjacent to $v$. Note that $p$ and $q$ must be adjacent to each other. Color $p$ black and color $q$ grey.

The set of black regions gives a subset $X' \subset X$ of the generators. It is an exercise to prove inductively that $X'$ generates. Next, for a contradiction, suppose that $w$ is a word in the $X'$ that is trivial in $K$. Thus we can write $w$ as a product of conjugates of relations from $R$. Recall that relations are in bijection with the vertices of $T$. Pick one relation $w_v$ appearing in the product so that $v$ is as far as possible from $v_0$. Of the four regions adjacent to $v$, there are two regions $p, q$ that are combinatorially further than the others from $v_0$. Say $p$ and black and $q$ is grey. Then $x_q$ is not freely cancelled in the product of conjugates, a contradiction. QED


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