There is a surprisingly simple, hundred-years-old conjecture in geometry, called the square peg problem.
Conjecture 1 (Toeplitz, 1911). Every Jordan curve in \( \mathbb{R}^2 \) has an inscribed square, i.e., a square that has vertices on the curve.
In a sense, this problem is in the spirit of the Borsuk–Ulam theorem; given an arbitrary map, we want to find something satisfying some given condition. But the situation is not that simple, and I would like to talk about why this problem is hard, unlike Borsuk–Ulam.
Let us first make some observations. Let \( \gamma : S^1 \rightarrow \mathbb{R}^2 \) be an embedding. We construct its four-fold product \( \Gamma : (S^1)^4 = T^4 \rightarrow (\mathbb{R}^2)^4 = \mathbb{R}^8 \). The space of squares in \( (\mathbb{R}^2)^4 = \mathbb{R}^8 \) is a \( 4 \)-dimensional subspace. (Given any two vertices, the square is uniquely determined.) Our goal is to check if the image of \( T^4 \) under \( \Gamma \) intersects the "squares space" \( \Sigma \cong \mathbb{R}^4 \subseteq \mathbb{R}^8 \).
But not so fast. The image \( \Gamma(T^4) \) trivially intersects \( \Sigma \), because for any point \( p \in \mathbb{R}^2 \) we have \( (p, p, p, p) \in \Sigma \). We need to exclude the degenerate squares somehow. This is something we do have to take care of. The problem then can be precisely formulated as: does the image \( \Gamma(T^4 \setminus \Delta) \) intersect \( \Sigma \), where \( \Delta \) is the diagonal?
Another small remark is that a dimension analysis shows that the number of squares is finite in most cases. The codimension of \( \Sigma \subseteq \mathbb{R}^8 \) is \( 4 \), which is the dimension of \( T^4 \). So we expect a finite number of solutions.
There is a very nice survey article by Benjamin Matschke. There is also a java applet on his website that you can play with.
1. The bordism argument#
I will now try to convince you that Conjecture 1 is intuitively true.
Theorem 2 (Schnirelman, 1944). It \( \gamma : S^1 \rightarrow \mathbb{R}^2 \) is a \( C^2 \) curve, then \( \gamma \) has an inscribed square.
Proof. The proof uses a very nice bordism argument. The idea is to deform the curve \( \gamma \) to another curve \( \gamma^\prime \). During the deformation, the set of squares will move around, and under nice transversality conditions, will form a \( 1 \)-dimensional submanifold of \( I \times (S^1)^4 \). The conclusion is that the number of squares in \( \gamma \) has the same parity as the number of squares in \( \gamma^\prime \).
Now observe that an ellipse has a unique inscribed square. It follows that \( \gamma \) has and odd number of inscribed squares and hence there is at least one square. ▨
So for a generic smooth curve, there is an inscribed square and moreover the number of squares is odd. Note that the smoothness condition is there to make sure there is a smooth-enough bordism between the ellipse and \( \gamma \).
In this direction (of making smoothness conditions) Stromquist proved a beautiful theorem:
Theorem 3 (Stromquist, 1989). If \( \gamma : S^1 \rightarrow \mathbb{R}^2 \) is a locally monotone embedding, then \( \gamma \) inscribes a square.
Here, \( \gamma \) being locally monotone means that for each \( t \in S^1 \), there exists a neighborhood \( U \ni t \) and a linear map \( f : \mathbb{R}^2 \rightarrow \mathbb{R} \) such that \( f \circ \gamma \) is monotone on \( U \). So graphs of functions rotated are locally monotone, while the Koch snowflake is not. \( C^2 \) curves are clearly locally monotone.
2. Smooth approximation#
Once we know this, a tempting way to approach to the original problem is through smooth approximation. Let \( \gamma : S^1 \rightarrow \mathbb{R}^2 \) be just a continuous embedding. Given \( \epsilon > 0 \), we can approximate it with a smooth curve \( \alpha _ \epsilon : S^1 \rightarrow \mathbb{R}^2 \) so that \( \lvert \gamma(t) - \alpha _ \epsilon(t) \rvert < \epsilon \) for all \( t \). Then there exist \( t _ {\epsilon 1}, t _ {\epsilon 2}, t _ {\epsilon 3}, t _ {\epsilon 4} \in S^1 \) such that \( \alpha _ \epsilon(t _ {\epsilon k}) \)s form a square. Since \( (S^1)^4 \) is sequentially compact, there exists a subsequence of \( (t _ {\epsilon k}) \in (S^1)^4 \) that converges to some quadruple \( (\tau _ k) \). Then \( \gamma(\tau _ k) \) form a square.
Again, we are forgetting about degenerate squares. There is a fair chance that \( \tau _ 1 = \tau _ 2 = \tau _ 3 = \tau _ 4 \). One possible way to avoid this is to prove that a generic curve \( \gamma \) has an inscribed square of side length at least something, i.e., the radius of the largest disc contained inside \( \gamma \). But giving such a condition on the side-length makes the problem much harder, even for smooth curves.
However, if we can somehow topologically separate small squares from large squares, then approximation can work.
Theorem 4 (Matschke, 2011). Consider the annulus \( A = \{ v \in \mathbb{R}^2 : 1 \le \lVert x \rVert \le 1 + \sqrt{2}\} \). Then a curve \( \gamma : S^1 \rightarrow A \) that is non-zero in \( \pi _ 1(A) \) inscribes a square of side length at least \( \sqrt{2} \).
The squares in \( A \) fall into either the "big" category or the "small" category. If one can prove that a generic curve in \( A \) has an odd number of "big" squares, then the previous argument of taking the limit of squares does work.
3. A smaller problem#
In this section, we look at an analogous but simpler problem. The original problem asked if the intersection of \( \gamma(S^1)^4 \subseteq \mathbb{R}^8 \) and \( \Sigma \cong \mathbb{R}^4 \subseteq \mathbb{R}^8 \) is nontrivial. Why don’t we try \( 3 \) instead of \( 4 \)? For some \( 3 \)-dimensional subspace \( \Sigma^\prime \), we may ask if the intersection of \( \Sigma^\prime \) and \( \gamma(S^1)^3 \) is nontrivial. For the problem to make sense, we should better require \( \Sigma^\prime \) to contain the \( 2 \)-dimensional space \( \{ (p, p, p) : p \in \mathbb{R}^2 \} \). Then up to a linear transformation, the question we should ask is clear.
Does every Jordan curve \( \gamma : S^1 \rightarrow \mathbb{R}^2 \) inscribe an equilateral triangle with one side parallel to a given axis?
The short answer is no. If \( \gamma \) looks like a long and thin rhombus that is perpendicular to the given axis, then there is no such inscribed triangle. But there is a long answer.
This is an example of a question whose answer changes according to the smoothness condition on \( \gamma \). If \( \gamma \) is smooth, then the answer is yes. Intuitively, you can argue as follows. Suppose we want to find a equilateral triangle with one side parallel to the \( x \)-axis. Consider the family of segments with endpoints on \( \gamma \) and parallel to the \( x \)-axis. At each segment, imagine a equilateral erected to one side. Near the point where the \( y \)-coordinate is minimal, the third vertex will be in the interior of \( \gamma \). On the other hand, near the point where the \( y \)-coordinate is maximal, the third vertex will be on the exterior of \( \gamma \). By the intermediate value theorem, there exists a moment when the third vertex is on \( \gamma \).
In the non-smooth case, the claim that the third vertex lies in the interior/exterior of \( \gamma \) goes terribly wrong.
There is a more sophisticated proof of the fact in the smooth case. Consider the smallest equilateral triangle \( PQR \), the side \( PQ \) parallel to the \( x \) axis, that contains \( \gamma \) in its interior and boundary. Let \( A, B, C \) be the points on the curve that touches the sides \( QR, RP, PQ \). Then an application of homology theory shows that there exist points \( D, E, F \in \gamma \) with \( D \) between \( B \) and \( C \), \( E \) between \( C \) and \( A \), \( F \) between \( A \) and \( B \) such that \( DEF \) is equilateral and \( DE \) is parallel to the \( x \)-axis. It is clear from the order of the points that \( DEF \) is non-degenerate.
What goes wrong here if \( \gamma \) is not smooth? If \( \gamma \) is a general curve, the points \( A, B, C \) might not be distinct. This allows the points \( D, E, F \) to degenerate into a single point.