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I am reading a paper "Stable equivalence of knots on surfaces and virtual knot cobordism." Following are the definitions from the paper of section 2.

  • Let ${\cal D}$ be the set of all pairs $(F, D)$ such that $F$ is a compact oriented surface and $D$ is a link diagram on $F$. By $(F_1, D_1) \overset{e}{\sim} (F_2, D_2)$ we mean that there exists a compact oriented surface $F_3$ and orientation-preserving embeddings $f_1: F_1 \to F_3$, $f_2:F_2 \to F_3$ such that $f_1(D_1)$ and $f_2(D_2)$ are related by Reidemeister moves on $F_3$.

  • Two elements $(F, D)$ and $(F', D')$ of ${\cal D}$ are stably Reidemeister equivalent, denoted by $(F, D) {\sim} (F', D')$, if there exists a sequence $(F, D) = (F_1, D_1) \overset{e}{\sim} (F_2, D_2) \overset{e}{\sim} \cdots \overset{e}{\sim} (F_n, D_n) = (F', D')$.

  • Let ${\cal C}$ be the set of all pairs $(F, C)$ such that $F$ is a compact oriented surface and $C$ is generic closed curves on $F$. By $(F_1, C_1) \overset{e}{\sim} (F_2, C_2)$ we mean that there exists a compact oriented surface $F_3$ and orientation-preserving embeddings $f_1: F_1 \to F_3$, $f_2:F_2 \to F_3$ such that $f_1(C_1)$ and $f_2(C_2)$ are homotopic in $F_3$.

  • Two elements $(F, C)$ and $(F', C')$ of ${\cal C}$ are stably equivalent, denoted by $(F, C) {\sim} (F', C')$, if there exists a sequence $(F, C) = (F_1, C_1) \overset{e}{\sim} (F_2, C_2) \overset{e}{\sim} \cdots \overset{e}{\sim} (F_n, C_n) = (F', D')$.

Clearly, we have a map $\pi: {\cal D} \rightarrow {\cal C}$ sending a knot diagram to its underlying immersed curve, i.e, forget the over and under crossing information.

Then it is written that we have induced map $\pi_{\sim}: {\cal D}/ \sim \rightarrow {\cal C}/ \sim.$

I am not able to understand why $\pi_{\sim}$ is a well-defined map. The reason given in the paper is that homotopy of curves is generated by the projection of Reidemeister moves.

I understand that the projection of first and second Reidemeister moves give a homotopy between curves, but how the projection of third Reidemeister move gives a homotopy. My doubt is because when we slide the vertical curve (in the third Reidemeister move ) from left to right, we will have a singularity which is not a transverse double point.

Can someone help in resolving this doubt?

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I think the trickiness here lies in the fact that Carter-Kamada-Saito are defining particular representatives first (knot diagrams on surfaces, stable equivalence moves, and generic immersed curves) since, at least at the point the paper was written, there was not really an underlying geometric object that the equivalence classes corresponded to.

In Kuperberg's "What is a virtual link?", he showed that the equivalence classes from Carter-Kamada-Saito can be described as being links in thickened closed oriented surfaces up to a kind of stabilization move from JSJ theory: whenever there is a properly embedded annulus $A\subset \Sigma\times[0,1]$ with $\partial A$ lying on both $\Sigma\times 0$ and $\Sigma\times 1$ (a vertical annulus) such that $A$ does not intersect the link, then we can destabilize by cutting along $A$ and attaching a $D^2\times [0,1]$ to each side of the cut.

If we're happy with manifolds with corners, then we can extend this to links in thickened compact oriented surfaces, and the above kind of stabilization can be thought of in the following way: if $f:\Sigma_1\to\Sigma_2$ is an orientation-preserving embedding of compact oriented surfaces and $L\subset \Sigma_1\times[0,1]$ is a link, then $(f\times\operatorname{id})(L)\subset\Sigma_2\times[0,1]$ is a stably equivalent link.

If we start from Kuperberg's point of view, of a link $L$ embedded in a thickened surface $\Sigma\times[0,1]$, then we can use standard arguments to show there exists a knot diagram $D$ on $\Sigma$ and that any two diagrams of $L$ (as a plain old link) are related by sequences of the three Reidemeister moves. Adding in stable equivalence is then allowing moves where you alter the surface in the region away from the diagram.

A flattened version of links in a thickened surface is immersed curves in a surface. Let $\mathcal{L}(\Sigma)$ denote the set of all links in $\Sigma\times[0,1]$, and let $\mathcal{C}(\Sigma)$ denote the set of all (multi-)curves in $\Sigma$. There is a map $\mathcal{L}(\Sigma)\to\mathcal{C}(\Sigma)$ given by projection. This map descends to $\mathcal{L}(\Sigma)/{\sim}\to\mathcal{C}(\Sigma)/{\sim}$, where $\mathcal{L}(\Sigma)/{\sim}$ is isotopy classes of links and $\mathcal{C}(\Sigma)/{\sim}$ is homotopy classes of curves, since every isotopy up in $\Sigma\times[0,1]$ when projected gives a homotopy in $\Sigma$.

It is a fact that every class of $\mathcal{C}(\Sigma)/{\sim}$ has a representative curve that is a generic immersion. One way to see this is to use the fact a closed curve $f:[0,1]\to\Sigma$ is homotopic to a smooth one where $f^{-1}(f(0))=\{0,1\}$, then take the path $f':[0,1]\to\Sigma\times[0,1]$ defined by $f'(t)=(f(t),t)$ and close it up into a knot which projects to $f([0,1])$. That knot diagrams exist is the fact that there is an isotopy of this knot such that its projection is a generic immersion, and since this isotopy projects to a homotopy, this shows the original curve is homotopic to a generic immersion. (A more straightforward way is to use some transversality argument to show almost every perturbation of a smooth representative will be generic.)

Each "Reidemeister move" of a generic immersion corresponds to an isotopy up in $\Sigma\times[0,1]$ whose projection is a homotopy of that curve. For each move, there is exactly one point in time where the homotopy passes through curves that are either not generic or not an immersion. Reidemeister I has a moment along the homotopy where the curve fails to be an immersion since it develops a cusp. Reidemeister II has a moment where the curve fails to be generic since the curve will intersect itself non-transversely. Reidemeister III has a moment where it fails to be generic since the curve will intersect itself in a triple point.

In summary. The equivalence relation on $\mathcal{C}$ (as in the paper) is that the generically immersed curves are stably equivalent up to homotopy. The homotopy itself does not need to pass through generic immersions. This is similar to how Reidemeister moves represent isotopies between link diagrams that pass through non-diagrams (in the sometimes-seen interpretation that a "link diagram" is a link in $S^2\times[0,1]$ whose projection to $S^2$ is a generic immersion).

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  • $\begingroup$ @Miller: This is a clear explanation. It raises some other doubts in my mind but also clears them simultaneously. Thank you for your time. $\endgroup$ Aug 5, 2019 at 5:11

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