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I’ve just uploaded to the arXiv my paper “An integration approach to the Toeplitz square peg problem“, submitted to Forum of Mathematics, Sigma. This paper resulted from my attempts recently to solve the Toeplitz square peg problem (also known as the inscribed square problem):

Conjecture 1 (Toeplitz square peg problem) Let ${\gamma}$ be a simple closed curve in the plane. Is it necessarily the case that ${\gamma}$ contains four vertices of a square?

See this recent survey of Matschke in the Notices of the AMS for the latest results on this problem.

The route I took to the results in this paper was somewhat convoluted. I was motivated to look at this problem after lecturing recently on the Jordan curve theorem in my class. The problem is superficially similar to the Jordan curve theorem in that the result is known (and rather easy to prove) if ${\gamma}$ is sufficiently regular (e.g. if it is a polygonal path), but seems to be significantly more difficult when the curve is merely assumed to be continuous. Roughly speaking, all the known positive results on the problem have proceeded using (in some form or another) tools from homology: note for instance that one can view the conjecture as asking whether the four-dimensional subset ${\gamma^4}$ of the eight-dimensional space ${({\bf R}^2)^4}$ necessarily intersects the four-dimensional space ${\mathtt{Squares} \subset ({\bf R}^2)^4}$ consisting of the quadruples ${(v_1,v_2,v_3,v_4)}$ traversing a square in (say) anti-clockwise order; this space is a four-dimensional linear subspace of ${({\bf R}^2)^4}$, with a two-dimensional subspace of “degenerate” squares ${(v,v,v,v)}$ removed. If one ignores this degenerate subspace, one can use intersection theory to conclude (under reasonable “transversality” hypotheses) that ${\gamma^4}$ intersects ${\mathtt{Squares}}$ an odd number of times (up to the cyclic symmetries of the square), which is basically how Conjecture 1 is proven in the regular case. Unfortunately, if one then takes a limit and considers what happens when ${\gamma}$ is just a continuous curve, the odd number of squares created by these homological arguments could conceivably all degenerate to points, thus blocking one from proving the conjecture in the general case.

Inspired by my previous work on finite time blowup for various PDEs, I first tried looking for a counterexample in the category of (locally) self-similar curves that are smooth (or piecewise linear) away from a single origin where it can oscillate infinitely often; this is basically the smoothest type of curve that was not already covered by previous results. By a rescaling and compactness argument, it is not difficult to see that such a counterexample would exist if there was a counterexample to the following periodic version of the conjecture:

Conjecture 2 (Periodic square peg problem) Let ${\gamma_1, \gamma_2}$ be two disjoint simple closed piecewise linear curves in the cylinder ${({\bf R}/{\bf Z}) \times {\bf R}}$ which have a winding number of one, that is to say they are homologous to the loop ${x \mapsto (x,0)}$ from ${{\bf R}/{\bf Z}}$ to ${({\bf R}/{\bf Z}) \times {\bf R}}$. Then the union of ${\gamma_1}$ and ${\gamma_2}$ contains the four vertices of a square.

In contrast to Conjecture 1, which is known for polygonal paths, Conjecture 2 is still open even under the hypothesis of polygonal paths; the homological arguments alluded to previously now show that the number of inscribed squares in the periodic setting is even rather than odd, which is not enough to conclude the conjecture. (This flipping of parity from odd to even due to an infinite amount of oscillation is reminiscent of the “Eilenberg-Mazur swindle“, discussed in this previous post.)

I therefore tried to construct counterexamples to Conjecture 2. I began perturbatively, looking at curves ${\gamma_1, \gamma_2}$ that were small perturbations of constant functions. After some initial Taylor expansion, I was blocked from forming such a counterexample because an inspection of the leading Taylor coefficients required one to construct a continuous periodic function of mean zero that never vanished, which of course was impossible by the intermediate value theorem. I kept expanding to higher and higher order to try to evade this obstruction (this, incidentally, was when I discovered this cute application of Lagrange reversion) but no matter how high an accuracy I went (I think I ended up expanding to sixth order in a perturbative parameter ${\varepsilon}$ before figuring out what was going on!), this obstruction kept resurfacing again and again. I eventually figured out that this obstruction was being caused by a “conserved integral of motion” for both Conjecture 2 and Conjecture 1, which can in fact be used to largely rule out perturbative constructions. This yielded a new positive result for both conjectures:

Theorem 3

• (i) Conjecture 1 holds when ${\gamma}$ is the union ${\{ (t,f(t)): t \in [t_0,t_1]\} \cup \{ (t,g(t)): t \in [t_0,t_1]\}}$ of the graphs of two Lipschitz functions ${f,g: [t_0,t_1] \rightarrow {\bf R}}$ of Lipschitz constant less than one that agree at the endpoints.
• (ii) Conjecture 2 holds when ${\gamma_1, \gamma_2}$ are graphs of Lipschitz functions ${f: {\bf R}/{\bf Z} \rightarrow {\bf R}, g: {\bf R}/{\bf Z} \rightarrow {\bf R}}$ of Lipschitz constant less than one.

We sketch the proof of Theorem 3(i) as follows (the proof of Theorem 3(ii) is very similar). Let ${\gamma_1: [t_0, t_1] \rightarrow {\bf R}}$ be the curve ${\gamma_1(t) := (t,f(t))}$, thus ${\gamma_1}$ traverses one of the two graphs that comprise ${\gamma}$. For each time ${t \in [t_0,t_1]}$, there is a unique square with first vertex ${\gamma_1(t)}$ (and the other three vertices, traversed in anticlockwise order, denoted ${\gamma_2(t), \gamma_3(t), \gamma_4(t)}$) such that ${\gamma_2(t)}$ also lies in the graph of ${f}$ and ${\gamma_4(t)}$ also lies in the graph of ${g}$ (actually for technical reasons we have to extend ${f,g}$ by constants to all of ${{\bf R}}$ in order for this claim to be true). To see this, we simply rotate the graph of ${g}$ clockwise by ${\frac{\pi}{2}}$ around ${\gamma_1(t)}$, where (by the Lipschitz hypotheses) it must hit the graph of ${f}$ in a unique point, which is ${\gamma_2(t)}$, and which then determines the other two vertices ${\gamma_3(t), \gamma_4(t)}$ of the square. The curve ${\gamma_3(t)}$ has the same starting and ending point as the graph of ${f}$ or ${g}$; using the Lipschitz hypothesis one can show this graph is simple. If the curve ever hits the graph of ${g}$ other than at the endpoints, we have created an inscribed square, so we may assume for contradiction that ${\gamma_3(t)}$ avoids the graph of ${g}$, and hence by the Jordan curve theorem the two curves enclose some non-empty bounded open region ${\Omega}$.

Now for the conserved integral of motion. If we integrate the ${1}$-form ${y\ dx}$ on each of the four curves ${\gamma_1, \gamma_2, \gamma_3, \gamma_4}$, we obtain the identity

$\displaystyle \int_{\gamma_1} y\ dx - \int_{\gamma_2} y\ dx + \int_{\gamma_3} y\ dx - \int_{\gamma_4} y\ dx = 0.$

This identity can be established by the following calculation: one can parameterise

$\displaystyle \gamma_1(t) = (x(t), y(t))$

$\displaystyle \gamma_2(t) = (x(t)+a(t), y(t)+b(t))$

$\displaystyle \gamma_3(t) = (x(t)+a(t)-b(t), y(t)+a(t)+b(t))$

$\displaystyle \gamma_4(t) = (x(t)-b(t), y(t)+a(t))$

for some Lipschitz functions ${x,y,a,b: [t_0,t_1] \rightarrow {\bf R}}$; thus for instance ${\int_{\gamma_1} y\ dx = \int_{t_0}^{t_1} y(t)\ dx(t)}$. Inserting these parameterisations and doing some canceling, one can write the above integral as

$\displaystyle \int_{t_0}^{t_1} d \frac{a(t)^2-b(t)^2}{2}$

which vanishes because ${a(t), b(t)}$ (which represent the sidelengths of the squares determined by ${\gamma_1(t), \gamma_2(t), \gamma_3(t), \gamma_4(t)}$ vanish at the endpoints ${t=t_0,t_1}$.

Using this conserved integral of motion, one can show that

$\displaystyle \int_{\gamma_3} y\ dx = \int_{t_0}^{t_1} g(t)\ dt$

which by Stokes’ theorem then implies that the bounded open region ${\Omega}$ mentioned previously has zero area, which is absurd.

This argument hinged on the curve ${\gamma_3}$ being simple, so that the Jordan curve theorem could apply. Once one left the perturbative regime of curves of small Lipschitz constant, it became possible for ${\gamma_3}$ to be self-crossing, but nevertheless there still seemed to be some sort of integral obstruction. I eventually isolated the problem in the form of a strengthened version of Conjecture 2:

Conjecture 4 (Area formulation of square peg problem) Let ${\gamma_1, \gamma_2, \gamma_3, \gamma_4: {\bf R}/{\bf Z} \rightarrow ({\bf R}/{\bf Z}) \times {\bf R}}$ be simple closed piecewise linear curves of winding number ${1}$ obeying the area identity

$\displaystyle \int_{\gamma_1} y\ dx - \int_{\gamma_2} y\ dx + \int_{\gamma_3} y\ dx - \int_{\gamma_4} y\ dx = 0$

(note the ${1}$-form ${y\ dx}$ is still well defined on the cylinder ${({\bf R}/{\bf Z}) \times {\bf R}}$; note also that the curves ${\gamma_1,\gamma_2,\gamma_3,\gamma_4}$ are allowed to cross each other.) Then there exists a (possibly degenerate) square with vertices (traversed in anticlockwise order) lying on ${\gamma_1, \gamma_2, \gamma_3, \gamma_4}$ respectively.

It is not difficult to see that Conjecture 4 implies Conjecture 2. Actually I believe that the converse implication is at least morally true, in that any counterexample to Conjecture 4 can be eventually transformed to a counterexample to Conjecture 2 and Conjecture 1. The conserved integral of motion argument can establish Conjecture 4 in many cases, for instance if ${\gamma_2,\gamma_4}$ are graphs of functions of Lipschitz constant less than one.

Conjecture 4 has a model special case, when one of the ${\gamma_i}$ is assumed to just be a horizontal loop. In this case, the problem collapses to that of producing an intersection between two three-dimensional subsets of a six-dimensional space, rather than to four-dimensional subsets of an eight-dimensional space. More precisely, some elementary transformations reveal that this special case of Conjecture 4 can be formulated in the following fashion in which the geometric notion of a square is replaced by the additive notion of a triple of real numbers summing to zero:

Conjecture 5 (Special case of area formulation) Let ${\gamma_1, \gamma_2, \gamma_3: {\bf R}/{\bf Z} \rightarrow ({\bf R}/{\bf Z}) \times {\bf R}}$ be simple closed piecewise linear curves of winding number ${1}$ obeying the area identity

$\displaystyle \int_{\gamma_1} y\ dx + \int_{\gamma_2} y\ dx + \int_{\gamma_3} y\ dx = 0.$

Then there exist ${x \in {\bf R}/{\bf Z}}$ and ${y_1,y_2,y_3 \in {\bf R}}$ with ${y_1+y_2+y_3=0}$ such that ${(x,y_i) \in \gamma_i}$ for ${i=1,2,3}$.

This conjecture is easy to establish if one of the curves, say ${\gamma_3}$, is the graph ${\{ (t,f(t)): t \in {\bf R}/{\bf Z}\}}$ of some piecewise linear function ${f: {\bf R}/{\bf Z} \rightarrow {\bf R}}$, since in that case the curve ${\gamma_1}$ and the curve ${\tilde \gamma_2 := \{ (x, -y-f(x)): (x,y) \in \gamma_2 \}}$ enclose the same area in the sense that ${\int_{\gamma_1} y\ dx = \int_{\tilde \gamma_2} y\ dx}$, and hence must intersect by the Jordan curve theorem (otherwise they would enclose a non-zero amount of area between them), giving the claim. But when none of the ${\gamma_1,\gamma_2,\gamma_3}$ are graphs, the situation becomes combinatorially more complicated.

Using some elementary homological arguments (e.g. breaking up closed ${1}$-cycles into closed paths) and working with a generic horizontal slice of the curves, I was able to show that Conjecture 5 was equivalent to a one-dimensional problem that was largely combinatorial in nature, revolving around the sign patterns of various triple sums ${y_{1,a} + y_{2,b} + y_{3,c}}$ with ${y_{1,a}, y_{2,b}, y_{3,c}}$ drawn from various finite sets of reals.

Conjecture 6 (Combinatorial form) Let ${k_1,k_2,k_3}$ be odd natural numbers, and for each ${i=1,2,3}$, let ${y_{i,1},\dots,y_{i,k_i}}$ be distinct real numbers; we adopt the convention that ${y_{i,0}=y_{i,k_i+1}=-\infty}$. Assume the following axioms:

• (i) For any ${1 \leq p \leq k_1, 1 \leq q \leq k_2, 1 \leq r \leq k_3}$, the sums ${y_{1,p} + y_{2,q} + y_{3,r}}$ are non-zero.
• (ii) (Non-crossing) For any ${i=1,2,3}$ and ${0 \leq p < q \leq k_i}$ with the same parity, the pairs ${\{ y_{i,p}, y_{i,p+1}\}}$ and ${\{y_{i,q}, y_{i,q+1}\}}$ are non-crossing in the sense that

$\displaystyle \sum_{a \in \{p,p+1\}} \sum_{b \in \{q,q+1\}} (-1)^{a+b} \mathrm{sgn}( y_{i,a} - y_{i,b} ) = 0.$

• (iii) (Non-crossing sums) For any ${0 \leq p \leq k_1}$, ${0 \leq q \leq k_2}$, ${0 \leq r \leq k_3}$ of the same parity, one has

$\displaystyle \sum_{a \in \{p,p+1\}} \sum_{b \in \{q,q+1\}} \sum_{c \in \{r,r+1\}} (-1)^{a+b+c} \mathrm{sgn}( y_{1,a} + y_{2,b} + y_{3,c} ) = 0.$

Then one has

$\displaystyle \sum_{i=1}^3 \sum_{p=1}^{k_i} (-1)^{p-1} y_{i,p} < 0.$

Roughly speaking, Conjecture 6 and Conjecture 5 are connected by constructing curves ${\gamma_i}$ to connect ${(0, y_{i,p})}$ to ${(0,y_{i,p+1})}$ for ${0 \leq p \leq k+1}$ by various paths, which either lie to the right of the ${y}$ axis (when ${p}$ is odd) or to the left of the ${y}$ axis (when ${p}$ is even). The axiom (ii) is asserting that the numbers ${-\infty, y_{i,1},\dots,y_{i,k_i}}$ are ordered according to the permutation of a meander (formed by gluing together two non-crossing perfect matchings).

Using various ad hoc arguments involving “winding numbers”, it is possible to prove this conjecture in many cases (e.g. if one of the ${k_i}$ is at most ${3}$), to the extent that I have now become confident that this conjecture is true (and have now come full circle from trying to disprove Conjecture 1 to now believing that this conjecture holds also). But it seems that there is some non-trivial combinatorial argument to be made if one is to prove this conjecture; purely homological arguments seem to partially resolve the problem, but are not sufficient by themselves.

While I was not able to resolve the square peg problem, I think these results do provide a roadmap to attacking it, first by focusing on the combinatorial conjecture in Conjecture 6 (or its equivalent form in Conjecture 5), then after that is resolved moving on to Conjecture 4, and then finally to Conjecture 1.