## Yet more inscribed squares: a word from Moscow, Idaho

Somehow while writing the first two posts I managed to miss this clear and thorough write-up about the inscribed square problem by Mark Nielsen, a math professor at the University of Idaho. From his page one learns that indeed every polygon has an inscribed square. In fact, every curve which is not somehow horrifically fractal does have an inscribed square — this is a theorem of Walter Stromquist from 1989. Nielsen also includes many related theorems, references, exercises, and a few toothsome unsolved problems at the bottom of the page. Go look!

## 4 thoughts on “Yet more inscribed squares: a word from Moscow, Idaho”

1. Richard says:

That first link is messed up.

At the Thurston birthday conference, R. Schwartz mentioned this problem during the problem session. Thurston’s response was that it was “curious.” :) He also suggested limit sets of quasifuchsian groups as a special case. More generally, you could ask the question for any quasicircle (image of a round circle under a quasiconformal map).

I guess I should look at Stromquist to see what “nice enough” means. Thanks for the update.

2. JSE says:

Arg. Link fixed. Hope you were able to find the page from Nielsen’s homepage.

Can you say more what you mean by “limit sets of quasifucshian groups”? I think Stromquist’s theorem appiles whenever each point has a neighborhood in which the curve looks like a line segment (but I didn’t look at the paper.) Does a quasicircle have this property?

3. Richard says:

One way to define a quasifuchsian group is to say that you take a (torsion free) fuchsian group and conjugate it by an equivariant quasiconformal mapping of the Riemann sphere. In this case the limit set is just the image of the invariant circle for the fuchsian group.

(Recall that a $K$–qc map is a homeomorphism $f$ with locally square integrable derivatives (in the sense of distributions) that satisfy
$|f_{\bar z}| \leq k f_z$ where $k = (K-1)/(K+1)$. Unfortunately, this definition is never very satisfactory to the uninitiated. “In the sense of distributions” just means that the derivatives look like real derivatives when you intergrate against compactly supported smooth functions, so just pretend the derivatives make sense and are what you think they are.)

More generally, the limit set of a Kleinian group $G$ (a discrete subgroup of $PSL(2,C)$) is the smallest nonempty closed $G$–invariant set in the Riemann sphere. The usual definition of a quasifuchsian group is that its limit set (with this definition) is a Jordan circle and the group preserves the complementary domains setwise. (It’s a theorem of Bers that the two definitions I’ve given agree for qf groups.)

These circles (qf limit sets) won’t be differentiable anywhere. There are quasicircles with this property, though (take a qc map whose support doesn’t contain part of the circle, and then a point outside the support will still be differentiable).

Maybe this special case is in fact too special to yield intuition about the general case. Here are some ideas:

A qc map can be factored into qc maps whose conformal distortion is very small (first observed by Bers?), and maybe one can use this fact to obtain a sequence of approximating curves with squares of uniformly bounded diameter.

If that doesn’t work out by itself (this is all just brainstorming), McMullen showed, in his “Iteration on Teichmuller space” paper, that a $K$–quasicircle (image of round circle by K–qc map) has bounded “turning” depending only on $K$. You can think of this as some sort of deviation from being convex. If $K$ is very close to one, then you’d think that you should be able to produce a square, since you’re pretty close to convexity. Maybe you could also bound the diameter of the square in terms of $K$, too. Then, using Bers’ observation about factorization, maybe this would yield a proof.

That’s all pretty speculative, I guess. (And long. :))

4. Richard says:

Oops, there should be absolute value bars around $f_z$ in the definition of qc.

Note that $f$ is conformal iff $k = 0$.