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?

  • 1
    $\begingroup$ 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? $\endgroup$
    – Ian Agol
    Feb 6, 2011 at 21:47
  • $\begingroup$ 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. $\endgroup$ Feb 6, 2011 at 22:09
  • $\begingroup$ @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 $f$). $\endgroup$
    – YCor
    Jun 14, 2018 at 10:43

2 Answers 2


I don't have a full answer to your question, but a hint. Any virtual knot group is the fundamental group of a knotted torus in 4-space. Constructing the torus from the virtual diagram is easy. The construction is due to Shin Satoh. Such a knotted torus has a Seifert solid --- the torus is the boundary of a 3-manifold in 4-space. There is also a meridonal class for the knotted torus. So I would imagine that the fundamental group of that knotted surface is an HNN extension. This is something that I should know, but don't know off the top of my head --- and I am heading out of the house in 5 minutes. If it is true, then it should not be hard to prove.

  • $\begingroup$ Thanks for the answer! What is the meridonal class for a knotted torus? Also, isn't Shin Satoh's construction highly non-canonical? (not that that would make a difference for my question; but it would make the answer less of an analogue to the regular knot case) $\endgroup$ Feb 7, 2011 at 15:42

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.


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