Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $2<n$ and $(2,n)=1$.

If $T$ is the standard torus obtained by identifying images of opposite edges of unit square properly, $p\colon \mathbb{R}^2\rightarrow T$ is the quotient map, then $K_{2,n}$, the image under $p$ of the line in $\mathbb{R}^2$, passing through origin, with slope $2/n$ is a torus knot, and $\pi_1(\mathbb{R}^3\backslash K_{2,n}) \cong \mathbb{Z}/2 *\mathbb{Z}/n$..

As we vary $n$ in odd integers, we obtain topological spaces $\mathbb{R}^3\backslash K_{2,n}$ with fundamental group $\mathbb{Z}/2*\mathbb{Z}/n$.

Can we do modifications in this example to get topological space with fundamental group $\mathbb{Z}/2*\mathbb{Z}$, and in general $\mathbb{Z}/m *\mathbb{Z}$? ?

I was taking a line through origin with irrational slope ($<1$), but its image under $p$ need not be a closed curve.

share|improve this question
    
I'm not too familiar with this stuff, but it looks as if the image of a line with rational slope $p/q$ is just the $(p,q)$-torus knot, which Wikipedia and Mathworld say have presentation $$\langle x,y|x^p=y^q\rangle.$$ –  anon Sep 16 '11 at 8:41
add comment

1 Answer

up vote 2 down vote accepted

Edited to fix three-dimensional limit:

Knot groups are torsion free. (This follows from the fact that knot complements are $K(\pi,1)$'s. I don't know a more direct proof.) The groups you describe are the result of killing the center of the torus knot group. This gives the orbifold fundamental group of the base orbifold. So, for example

$$\pi_1(S^3 - K_{p,q}) = \langle x, y \mid x^p = y^q \rangle$$

and the center is generated by $x^p$. By the way, $x$ and $y$ have geometric meaning. They are the cores of the two solid torus components of $S^3 - T$. So the torus knot group is one of the early examples of a free product with amalgamation: $\newcommand{\ZZ}{\mathbb Z}$ $\ZZ \ast_\ZZ \ZZ$.

Ok, if we kill the center (corresponding to crushing the circle fibers of the given Seifert fibered space) then we get the group

$$\langle x, y \mid x^p = y^q = 1 \rangle.$$

This is the free product $\ZZ/p \ast \ZZ/q$ that you are interested in. As mentioned above, this is the orbifold fundamental group of the orbifold $S^2(p,q,\infty) = D^2(p,q)$: that is, a sphere with three orbifold points, of orders $p$, $q$ and infinity or equivalently a disk with two points of orders $p$ and $q$. Now you want to take a limit as $q$ goes to $\infty$.

There is a "limit" of knots: namely $\lim_{q \to \infty} K_{p,q} = L_{2,2p}$ the $(2,2p)$ torus link. We have that

$$\pi_1(S^3 - L_{2,2p}) = \langle x, z \mid zx^p = x^pz \rangle.$$

As expected, $x^p$ generates the center. Killing $x^p$ gives the group

$$\langle x, z \mid x^p = 1 \rangle \cong \ZZ/p \ast \ZZ$$

as desired. We can also take the base orbifolds of this sequence of Seifert fibered spaces. In the limit we get $S^2(p,\infty, \infty)$ or equivalently an annulus with one cone point of order $p$. Again, this has orbifold fundamental group $\ZZ/p \ast \ZZ$. A reference for much of this material is Peter Scott's beautiful paper on the eight Thurston geometries. Taking geometric limits of knots is typically done in the hyperbolic setting - an example in Thurston's book is that the $(2,q)$ twist knots limit to the Whitehead link as $q$ tends to infinity.

We can avoid orbifold fundamental groups by instead taking connect sums of lens spaces: for example, the connect sum of $\mathbb{RP}^3$ and $S^2 \times S^1$ has fundamental group $\ZZ/2 \ast \ZZ$. For that matter, we can get the same fundamental group by taking the one point union of $\mathbb{RP}^2$ and a circle, $S^1$. However the above answer is directly inspired by your question.

share|improve this answer
    
I always naïvely thought that a one-relator group $\langle s_1,\ldots,s_n\mid r\rangle$ is torsion-free if and only if $r$ is not a proper power in the free group $F(s_1,\ldots,s_n)$ (what else could create torsion?). Am I missing something? –  t.b. Sep 16 '11 at 10:46
    
Knot groups need not be one-relator groups. But your proof is more direct for torus knot groups? –  Sam Nead Sep 16 '11 at 11:09
    
@Theo Buehler: This is true, and was first proved by Lyndon (I think) in his paper on the cohomology of one-relator groups. –  user641 Sep 16 '11 at 19:46
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.