This may be a standard problem in homotopy theory, but I don't know a good reference.

Let $\Sigma$ be a smooth, oriented surface, and let $X_1,X_2$ and $X_3$ be three smoothly embedded curves in $\Sigma$. Can I find smoothly embedded curves $Y_1,Y_2$ and $Y_3$ such that...

  • $Y_i$ is homotopic to $X_i$ through smooth embeddings,
  • each intersection in $Y_1\cup Y_2\cup Y_3$ is a transverse double intersection, and
  • the number of intersections between $Y_i$ and $Y_j$ is minimal among all pairs homotopic to $X_i$ and $X_j$ through smooth embeddings?

Roughly speaking, I have three curves, and I want to use homotopy to simultaneously minimize the number of intersections between each pair of curves.

  • $\begingroup$ you mean that ALL intersections should be double intersections? $\endgroup$ – Nikita Kalinin Dec 15 '11 at 19:41
  • $\begingroup$ I mean that all intersections should be between two local components, and no more. That is, the minimal number of local components for it to still be an intersection. Perhaps this is not called a double intersection? $\endgroup$ – Greg Muller Dec 15 '11 at 19:49

Firstly, "homotopic through smooth embeddings" is known as "isotopic", and there is a well known paper of David Epstein titled "Curves on 2-manifolds and isotopies" where he shows that homotopy is equivalent to isotopy in this setting.

Secondly, if you put a hyperbolic structure on your surface, then the curve shortening flow will isotope your curves into geodesics, which will automatically minimize intersection numbers (this can be seen by going to the hyperbolic plane as in the Klein model, and seeing what intersection numbers mean in terms of lifts and cross-ratios), and will also minimize self-intersection numbers, so will stay embedded if your curves are to begin with.

Thirdly, the geodesics, as above, might have triple intersections, but it is clear (by transversality, if you like) that a small local perturbation will remove the triple points (this would be less obvious if your curves $X_i$ were not embedded, but they are.)

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  • $\begingroup$ Thank you, Igor! Is "Closed geodesics on surfaces" (M. Freedman, J. Hass and P. Scott) the reference for this? I don't have access to that paper. $\endgroup$ – Greg Muller Dec 15 '11 at 21:43
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    $\begingroup$ I am not sure FHS is THE reference, though it probably has some of the stuff and is a beautiful paper. I will post a link. You can also look at my paper called "simple curves on surfaces" (there is an arxiv version) $\endgroup$ – Igor Rivin Dec 15 '11 at 22:04
  • $\begingroup$ You can find FHS here: dl.dropbox.com/u/5188175/FHS.pdf $\endgroup$ – Igor Rivin Dec 15 '11 at 22:07

EDIT: I've added a proof of the claim.

Suppose that $X_1$, $X_2$ and $X_3$ intersect transversely to begin with.

Claim. You can decrease the number of intersection between $X_1$ and $X_2$ if and only if there's an embedded disc $D$ on the surface, with $\partial D \subset X_1\cup X_2$, and such that the interior of $D$ is disjoint from $X_1\cup X_2$.

When you have a third curve $X_3$, there are two possibilities: either there's a smaller disc $D'\subset D$ that eliminates an intersection of $X_3$ and $X_1$ (or $X_2$), or, if you travel along $X_3$, you enter $D$ from the $X_1$-part of the boundary and you exit from the $X_2$-part (or vice-versa). You can proceed by eliminating all the first instances (by homotoping $X_3$ across $D'$ at each step, so that $X_1$ and $X_2$ don't move: these move only decrease the number of intersection points on $X_3$), until there are no more discs like $D'$ inside $D$. Now you can just ignore $X_3$, and you can move $X_1$ across $D$, thus decreasing the number of intersection points of $X_1$ and $X_2$ (and not increasing the other intersections): Jordan curve theorem is used in some form here, since you're basically using that $D\setminus X_3$ is a union of discs.

You can do this all the way to $Y_1$ and $Y_2$ that intersect transversely and minimise the number of intersections for $X_1$ and $X_2$. You also have a curve $Y_3$ that's homotopic to $X_3$, and has fewer intersections with $Y_1$ and $Y_2$ than $X_3$ had with $X_1$ and $X_2$. You can now proceed to isotope $Y_3$ across discs $D$ as before (using $Y_1$ first, and then $Y_2$): notice that you can't have any $D'$-case here, since $Y_1$ and $Y_2$ already minimise the number of intersections.

Proof (of the claim). Take an isotopy of $X_1$ that decreases the number of intersections with $X_2$, and perturb it to obtain a generic $F:C\to\Sigma$, where $C$ is $S^1\times I$. $F^{-1}(X_2)$ is a 1-dimensional submanifold in $C$, and since $F$ decreases the intersections, there's a properly embedded arc $\gamma$ with endpoints in $S^1\times\{0\}$. This bounds a disc $D_0$ together with an arc in $S^1\times \{0\}$, and you can choose $\gamma$ so that $D_0$ is minimal with respect to inclusion.

Now also $F^{-1}(X_1)$ is a 1-dimensional submanifold of $C$ (perhaps you need to be careful with $F$ at small times, but that's easy to achieve), and if you're careful for small times, it's just a bunch of closed curves. $F^{-1}(X_1)\cap D_0$ is a bunch of closed curves and properly embedded arcs. Closed curves in $D_0$ are not "essential", in the sense that you can replace $F$ with an isotopy that doesn't have any (these curves bound discs, and you can just quotient a compact neighbourhood of these discs and take the quotient map), so we'll suppose that there are just arcs.

In $D_0$ there's a smallest disc $D_1$ (but possibly more than one) that has a boundary component along $F^{-1}(X_1)$ and one along $F^{-1}(X_2)$. $D_1$ is the preimage under $F$ of a connected component $D$ of $\Sigma\setminus(X_1\cup X_2)$, and the boundary of $D$ is nullhomotopic in $D$ (using $F|_{D_1}$), therefore $D$ is a disc.

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  • $\begingroup$ Marco, I follow your argument, but how does one see the second paragraph? $\endgroup$ – Greg Muller Dec 15 '11 at 21:42
  • $\begingroup$ I had sort of taken it for granted (since it's very intuitive). I added a proof to the answer (it was too long to be a comment anyway), and I hope it does work. $\endgroup$ – Marco Golla Dec 16 '11 at 0:03

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