# Is every virtual knot group an HNN extension?

A basic fact in knot theory is that a knot group $$\pi(K)$$ is an HNN extension of $$\pi(F)$$, the fundamental group of a Seifert surface complement. A nice discussion of this may be found in Chapter 11 of An introduction to the theory of groups by Rotman. This property means that a knot group is completely determined by the fundamental group of a Seifert surface complement, plus a choice of meridian. I wonder whether this is somehow the real reason that Seifert surfaces are important in knot theory. Morally, the fact that a knot group is an HNN extension means that all the information about a knot which you might care about is contained in any of its Seifert surfaces.
To remind you of the explicit group-theoretic statement, there is an isomorphism
$$\phi\colon\thinspace \frac{\pi(F)\ast \langle m\rangle}{\mathcal{N}}\longrightarrow \pi(K),$$
where $$\langle m\rangle$$ is the infinite cyclic group generated by the meridian, and $$\mathcal{N}$$ is the smallest normal subgroup of the free product $$\pi(F)\ast \langle m\rangle$$ containing the elements
$$m^{-1}\mu^+(z)m(\mu^{-}(z))^{-1}\qquad z\in \pi(F),$$
with $$\mu^{\pm}$$ denoting the pushoff maps.
There is a natural notion of a virtual knot group, by assigning a formal generator to each arc of a virtual knot diagram, and a Wirtinger relation to each real crossing (virtual crossings are ignored). Any Wirtinger presentation of deficiency $$0$$ or $$1$$ can be realized as a virtual knot group by Theorem 3 of a paper by Se-Goo Kim.

Question: Is every virtual knot group an HNN extension? (edit: over a finitely generated group?) Can the base group be described in terms of a group generated by a commutator at each real crossing? If not, is a virtual knot group "almost" an HNN extension in some useful sense?

I'm interested in this question because I wonder whether invariants coming from Seifert surfaces can be read off Gauss diagrams in any systematic way. Are Seifert surfaces an essential feature of knots, as opposed to virtual knots; or are they a non-essential luxury?

• It's an HNN extension in a trivial sense, since its abelianization is Z. Take the kernel of the homomorphism to Z, and take an extension of this by a meridian acting by conjugation. In general, the kernel will be infinitely generated (unless e.g. it is a fibered real knot), so this won't correspond to a Seifert surface. Maybe you want to rephrase your question for an HNN extension over finitely generated groups? Feb 6, 2011 at 21:47
• Thanks! Indeed, I want the base group to be something weakly analogous to the fundamental group of a Seifert surface complement in some sense, so I definitely want it finitely generated. Question edited. Feb 6, 2011 at 22:09
• @IanAgol to complement Ian's comment (and take into account the subsequent edit), it is a theorem of Bieri and Strebel (1978) that for every finitely presented group $G$, every surjective homomorphism $f:G\to\mathbf{Z}$ arises from some splitting of $G$ as an HNN extension over a finitely generated subgroup (of the kernel of $f$).
– YCor
Jun 14, 2018 at 10:43

According to a theorem of Kuperberg, a virtual knot corresponds canonically to an embedding of a knot in a thickened surface $$K\subset \Sigma_g\times [0,1]$$ of minimal genus $$g$$ (up to homeomorphism). There is therefore another natural fundamental group associated to the knot, namely the fundamental group of the knot complement $$\pi_1(\Sigma_g\times [0,1] - K)$$. This group certainly splits as an HNN extension (in many ways). The fundamental group of the virtual knot is obtained from this by killing the two peripheral subgroups corresponding to $$\Sigma_g \times \{0,1\}$$. One may think of this as the fundamental group $$\pi_1( S\Sigma -K)$$, where $$S\Sigma$$ is the suspension. If $$K$$ is homologically trivial in $$\Sigma\times [0,1]$$, then one could take an embedded minimal genus surface $$F \subset \Sigma \times [0,1]$$ spanning $$K$$, so $$\partial F=K$$. Unfortunately, though, there is no canonical choice of homology class for this surface. One has a geometric splitting of $$S\Sigma-K$$ along $$F$$, however $$F$$ might not be $$\pi_1$$-injective in this space since Dehn's lemma is not available.
If $$K$$ is not homologically trivial in $$\Sigma \times [0,1]$$, it is still homologically trivial in $$S\Sigma$$, so one could take a surface bounding it (which must intersect a singular point of $$S\Sigma$$). One could think of this as taking a minimal genus surface giving a cobordism between $$K$$ and an embedded curve in $$\Sigma \times \{0,1\}$$. Again, there is not a canonical homology class ($$H_2(S\Sigma)=\mathbb{Z}^{2g}$$) and the surface may not be $$\pi_1$$-injective (in fact, there are virtual knots where the longitude is trivial in the virtual knot fundamental group). Also, I don't think that linking numbers are well-defined (again, since $$H_2(S\Sigma)$$ is large), so it's not clear how to obtain an Alexander polynomial from such a surface.