# Gap in Przytycki's computation of the skein module of links in a handlebody?

I am reading the paper [1], where the author proves that the skein module of links in a handlebody $F\times I$ has a free basis given by products $D_1 \cdots D_n$ where each $D_i$ is the closure of $v_i\in \pi_1(F)$, $v_i$ is the smallest length lexicographically smallest representative of its conjugacy class and $v_1\leq v_2\leq\cdots\leq v_n$. Besides several instances where the author makes a vague argument which presumably can be fixed using diamond lemma, I am stuck in the point where he proves that such products span the skein module.

The only place devoted to this issue is the second paragraph of "Proof of Theorem 2.11", p. 333. He says that, as an algebra both the skein module $\mathcal{S}(F\times I)$ and the symmetric algebra over the span of conjugacy classes of the fundamental group $S(R\hat{\pi}^\circ)$ are generated by the representatives of conjugacy classes of $\pi_1(F)$, hence the claim. But the map between these is not an algebra homomorphism, so this argument does not apply.

I don't know how to fix this gap. Naively one would think that induction on the number of crossings would work. But here is the problem. We need to be able to do two things: 1) switch the order of components, 2) change a representative of a component (i.e. conjugate the element $v_i\in\pi_1(F)$ to make it lexicographically smaller). Now the problem with these two operations is that they behave well with respect to different invariants. 1) works up to diagrams with smaller number of crossings, 2) works up to diagrams whose components have words of smaller lengths (see Lemma 1.7). Clearly 1) messes up the maximal length, indeed, usually the left-overs will have words whose lengths are sums of lengths of what we start with. On the other hand, 2) involves performing some isotopies. Even if the number of self-crossing of the component does not go up under such isotopies, the number of crossings with other components may increase. So we are chasing our own tail here.

UPDATE: First note that the question is about modules over the polynomial ring $\mathbb{Z}[v,v^{-1},h]$, not over the field of fractions. It turns out the question is not that easy. I did some experiments and it turns out the statement is actually false in almost all cases, except as stated by the author. For instance, if the surface is not planar, i.e. for the punctured torus it is false. If it is planar, i.e. a disk with several disks removed, and if the order of generators is different, or the choice of generators is different, it is false. The order of generators plays a role in saying what's lexicographically smaller. For instance, take a disk with $2$ disks removed. Suppose the generators of the fundamental group are $x$ going around one puncture, and $y$ going around both punctures. If the order on the generators is $x<y<y^{-1}<x^{-1}$, then the statement is false. It still seems to be true if the order of generators is as the author describes, namely $x<x^{-1}<y<y^{-1}$. Why?

Some explanations. We choose a base point $*$ on the boundary of $F$. For any element $\gamma\in \pi_1(F,*)$ we represent it by a path $\gamma:[0,1]\rightarrow F$ such that $\gamma(0)=\gamma(1)=*$ but $\gamma(t)\neq *$ for $0<t<1$. The graph of $\gamma$ is a path in $F\times [0,1]$ connecting $(*,0)$ to $(*,1)$. We close it up by adding the line $*\times [0,1]$. The resulting map from $S^1$ to $F\times [0,1]$ is a knot. More explicitly, if you view $S^1$ as $[0,1]/(0\sim 1)$ then $s:S^1\to F\times[0,1]$ is defined by $$s(t)=\begin{cases} (\gamma(2 t), 2t) & (t\leq \frac12),\\ (*, 2-2t) & (t\geq \frac12). \end{cases}$$ This is what I called closure of $\gamma\in \pi_1(F)$. For several elements $v_1,\ldots,v_n\in \pi_1(F)$ we stack the corresponding knot diagrams on top of each other and obtain what I denoted by $D_1\cdots D_n$.

Speaking about elements of $\pi_1(F,*)$ suppose we represent $\pi_1(F)$ as a free group on generators $g_1,\ldots,g_k$. Suppose we choose an order on the set $g_1,\ldots,g_k,g_1^{-1},\ldots,g_k^{-1}$. Then we define an order on $\pi_1(F)$ as follows. If we want to compare $v_1,v_2\in\pi_1(F)$ we write $v_1,v_2$ as a reduced word in letters $g_i$ and $g_i^{-1}$ and say that $v_1<v_2$ if the length of $v_1$ is strictly less than that of $v_2$ or if they have same length and the word of $v_1$ is lexicographically smaller than the word of $v_2$.

UPDATE: There is an interesting idea of inducting on the length of the knot with respect to a metric of negative curvature. Here is an example. Let $K$ be any knot in $F\times I$ represented by a geodesic. Suppose $K$ has several crossings. Take $K^n=K\cdot K \cdot \cdots\cdot K$, $n$ times. Choose an arc of $K$. Then over this arc there is $n$ arcs of $K^n$. Let us replace these $n$ arcs by some braid which lifts a cyclic permutation. We obtain a knot, denote $K_n'$, its diagram is again formed by geodesics in the limit. Now how do we commute $K_n'$ past $K_m'$?

[1] Przytycki, J. (1992). Skein module of links in a handlebody. Topology '90 (pp. 315-342).

• I would suggest to expand a bit the definitions... it is not clear to me what the $D_i$'s are. – Bruno Martelli Aug 28 '17 at 14:39
• @BrunoMartelli I tried to expand a bit. Does it look ok now? – Anton Mellit Aug 28 '17 at 15:59

Here's something that isn't a complete rigorous proof, but maybe it can be completed to one (unless I'm missing something). Let's fix a non-positively curved metric on $F$ (which rules out the sphere, but I'd guess you're ok with that). Extend this to $F \times [0,1]$ using the product metric (where $[0,1]$ has its standard Euclidean metric). Pick an isotopy representative $L_\alpha$ of each class $\alpha \in \hat \pi^\circ$ which is $\epsilon$ close to a geodesic. (I'm not sure exactly how $\epsilon$ needs to depend on $\alpha$, but I'll ignore this.) Then the claim is that there is a linear isomorphism $Sym(\{L_\alpha\}) \to Sk_q(F)$.

Now we can assign to a link the sum of the lengths of its components, and induct on this number. A couple properties:

1) Knots can be reordered using the skein relations, modulo terms of strictly lower length (by the triangle inequality, the resolution is strictly shorter).

2) Knots in different "layers" (in the $[0,1]$ direction) can be simplified separately since changing one knot doesn't change the lengths of the others.

3) Knots can be simplified to a unique geodesic using Reidemeister moves and skein relations, modulo terms of strictly lower length.

I'm being a little vague on the third point, but I think it follows from Theorem 1.10 in [Hass, Scott, "Shortening curves on surfaces," Topology 1994], which is stated slightly differently in Theorem 2.10 in [D. Thurston, "Positive basis for surface skein algebras" PNAS 2013]. (The paper [HS] doesn't deal with knots in a thickened surface, but with their projections, which means that some of the Reidemeister 2 moves that their algorithm does need to be done with the skein relation. But this is ok because the resolution term in the skein relation has strictly shorter length.)

I'm not sure, but my guess would be that translating this choice of ordering (the sum of the lengths) into a combinatorial description using the fundamental group might be not very nice.

• So let me try to "fix" Przytycki's statement. The ordering on the set of conjugacy classes remains arbitrary, for instance lexocographic ordering, but instead of choosing for each conjugacy class the closure of the lexicographically smallest representative I need to take the shortest geodesic representative. For instance, for the torus I would pick torus knots, right? – Anton Mellit Aug 29 '17 at 13:21
• That's what I had in mind - then the [HS] curve shortening flow would move each knot towards its isotopy representative using Reidemeister moves. Of course, it's this last part which is still vague... – Peter Samuelson Aug 29 '17 at 13:59
• I think the idea is useful, but still is not complete. See my update for a possible counter-example. – Anton Mellit Sep 4 '17 at 9:13

[This is a partial repair of the original incorrect argument. The proof is not complete, but perhaps it is helpful anyway.]

I have not read the Przytycki paper, but I think one can prove that such products span using standard techniques.

I assume the skein relation is the HOMFLYPT skein relation, with terms $L_+$, $L_-$ and $L_0$.

Let $S$ be the span of the products you describe. Assume inductively that diagrams with $k$ or fewer crossings lie $S$. The skein relation implies that $L_+ - L_-$ lies in $S$.

[Incorrect original argument: It thus suffices to show that some link related to our original $k+1$ crossing link by crossing changes lies in $S$. This is clearly the case.]

There are $2^{k+1}$ diagrams related to the original one by crossing changes (keeping the projection fixed). If we can show that one of those diagrams is in $S$, we are done.
Choose a base point on each component of the link. Choose a height function from the link to $\mathbb R$ such that outside of a small neighborhood of the base points, the height function on the $j$-th component increases monotonically from $j$ to $j + 0.9$. The height function determines a choice of crossing (over or under) at each double point of the projection.
At this point we have put the components in any order we like (modulo $S$), and the isotopy class of each component depends only on the homotopy class of the projection to $F$, relative to a neighborhood of the chosen base point for the component. If we could remove the "relative to a neighborhood of the base point", we would be done, but I don't see a way to do that at the moment.
• I was hoping that I explained in my question why this kind of proof doesn't work. It is not true that you can get any representative of a conjugacy class by flipping crossings. Take a disk with 2 discs removed, let $a$ be the loop going around one puncture counterclockwise, $b$ around the other one. Depending on how you draw, one of the knots corresponding to the words $ab$, $ba$ will have $0$ crossings after some isotopy, another will have $2$. This means you cannot get the second one from the first simply because there is no crossing to flip... – Anton Mellit Aug 8 '17 at 18:01