# Inscribed squares and relation avoiding paths.

Cole Hugelmeyer

###### Abstract.

We develop a connection between the inscribed square problem and the question
of understanding relation avoiding paths in a complex vector space. Our main
theorem is that a Jordan curve with no inscribed squares would have a
seemingly impossible structure which we call a square envelope. We will make
some conjectures about the nature of relation avoiding paths in vector spaces,
and show that these conjectures would imply the existence of inscribed squares
in Jordan curves with finitely many arbitrarily complicated singularities.

##  1\. Introduction

A Jordan curve is a continuous injective function
$\gamma:S^{1}\to\mathbb{R}^{2}$ which wraps counterclockwise around the region
it encloses. An inscribed square of $\gamma$ is a quadruple of distinct points
on $\text{im}(\gamma)$ which form a square in $\mathbb{R}^{2}$.

Figure 1. An inscribed square of a Jordan curve.

######  Conjecture 1 (Toeplitz 1911).

Every Jordan curve has an inscribed square.

Towards this end, we define a _bad Jordan curve_ to be a Jordan curve with no
inscribed squares. Our main result is that a bad Jordan curve must have a
structure which we call a _square envelope_. A square envelope is, loosely
speaking, a square moving with time, so that its first two corners are always
outside the Jordan curve, its second two corners are always inside the Jordan
curve, and the outside corners wrap completely around the Jordan curve as time
progresses.

To rigorously define a square envelope, we let
$T:\mathbb{R}^{2}\to\mathbb{R}^{2}$ be the linear map corresponding to 90
degree counterclockwise rotation. For a pair of points $a$ and $b$ in the
plane, we define $S_{1}(a,b)=a+T(b-a)$ and $S_{2}(a,b)=b+T(b-a)$. If $a\neq
b$, then the four points $a,b,S_{2}(a,b),S_{1}(a,b)$ form the vertices of a
square, labeled counterclockwise.

######  Definition 1.

Let $\gamma:S^{1}\to\mathbb{R}^{2}$ be a Jordan curve. A _square envelope_ of
$\gamma$ is a pair of continuous functions mapping from $\mathbb{R}$ to the
plane, $e_{1},e_{2}:\mathbb{R}\to\mathbb{R}^{2}$, such that all of the
following are true.

  * 1)

For all $x\in\mathbb{R}$, $e_{1}(x)$ and $e_{2}(x)$ are in the open exterior
of $\gamma$.

  * 2)

For all $x\in\mathbb{R}$, $S_{1}(e_{1}(x),e_{2}(x))$ and
$S_{2}(e_{1}(x),e_{2}(x))$ are in the open interior of $\gamma$.

  * 3)

$\lim_{x\to\infty}\|e_{2}(x)-e_{1}(x)\|=\lim_{x\to-\infty}\|e_{2}(x)-e_{1}(x)\|=0$.

  * 4)

Let $\ell_{\lambda}$ be the closed curve obtained by first following $e_{1}$
from $e_{1}(-\lambda)$ to $e_{1}(\lambda)$, then a straight line from
$e_{1}(\lambda)$ to $e_{2}(\lambda)$, then $e_{2}$ from $e_{2}(\lambda)$ to
$e_{2}(-\lambda)$, and finally a straight line from $e_{2}(-\lambda)$ back to
$e_{1}(-\lambda)$. If $p$ is any point inside $\gamma$, then there exists a
real number $\lambda_{0}>0$ such for all $\lambda>\lambda_{0}$, we have that
the winding number of $\ell_{\lambda}$ around $p$ is equal to one.

Figure 2. As square envelopes conjecturally do not exist, they are somewhat
difficult to draw. This is a rough approximation of what one might look like.
The square must somehow have its first two corners move around the outside of
the Jordan curve while the other two corners remain inside.

######  Theorem 1.

Every bad Jordan curve has a square envelope.

This theorem allows us to make a connection between the inscribed square
problem and the problem of understanding relation avoiding paths in a vector
space. The set of squares is the vector space, and the linear relation we
avoid is the first two corners of one square touching the second two corners
of another.

Let $V$ be a vector space over $\mathbb{C}$, and suppose
$R_{1},R_{2},...,R_{n}$ are linear relations, subspaces of the vector space
$V\times V$. We then let $R$ be the relation given by the union of sets
$\cup_{i=1}^{n}R_{i}$. We will assume that $R$ is symmetric, namely $aRb\iff
bRa$.

A _relation-avoiding path_ is a continuous function $p:[0,1]\to V$ such that
there does not exist any pair of times $t_{1},t_{2}$ in $[0,1]$ with
$p(t_{1})Rp(t_{2})$. We define a _relation-avoiding origin path_ to be a
continuous function $p:[0,\infty)\to V$ with $\lim_{t\to\infty}p(t)=0$, such
that there does not exist any pair of times $t_{1},t_{2}$ in $[0,\infty)$, for
which $p(t_{1})Rp(t_{2})$. The origin is not included as the endpoint of this
path because we always have $0R0$.

######  Definition 2.

We say two relation avoiding origin paths $p$ and $q$ are _weakly homotopic_
if there exists a continuous function $h:[0,1]\times[0,\infty)\to V$ such that

  * 1)

$\lim_{t\to\infty}h(s,t)=0$ for all $s$.

  * 2)

$h(0,t)=p(t)$ for all $t$.

  * 3)

$h(1,t)=q(t)$ for all $t$.

  * 4)

There does not exist any $(s,t)\in[0,1]\times[0,\infty)$ with
$h(s,t)\,R\,h(s,0)$.

We say $p$ and $q$ are _strongly homotopic_ if $h$ can be chosen so that
$h(s,-)$ is a relation avoiding origin path for all $s$.

We propose the following conjecture about relation avoiding origin paths.

######  Conjecture 2 (SC, Spiral Conjecture).

Every relation avoiding origin path $p$ is weakly homotopic to a relation
avoiding origin path $q$ for which there exist linearly independent vectors
$x_{1},...,x_{n}$ in $V$ and complex numbers $a_{1},...,a_{n}$ such that

$q(t)=\sum_{i=1}^{n}x_{i}\cdot e^{a_{i}t}$

for all $t$.

The following fact provides some evidence for this conjecture.

######  Theorem 2.

The spiral conjecture is true when $V$ is one dimensional.

Finally, we will prove the following implication of Conjecture 2.

######  Theorem 3.

Assume the spiral conjecture. Then any Jordan curve $\gamma$ which is smooth
except at finitely many points has an inscribed square. The Jordan curve is
permitted to be arbitrarily complicated near these finitely many
singularities.

We propose that delving into the study of relation avoiding origin paths could
be a potential approach towards solving the inscribed square problem. At the
end of the paper, we will make more conjectures about relation avoiding origin
paths, which we hope might guide future research.

##  2\. Square Envelopes

If $\gamma$ is a bad Jordan curve, we will define an integer $b_{\gamma}$
which we call the bad wrapping number of $\gamma$. This integer is only well-
defined for bad Jordan curves. By counting inscribed squares in a generic
approximation of $\gamma$, we will prove that $b_{\gamma}$ is odd. Then, we
will use this fact to construct a square envelope of the Jordan curve.

Let $\gamma$ be a Jordan curve. We may choose a continuous function
$f:\mathbb{R}^{2}\to\mathbb{R}$ with the property that if $a$ is in the
interior of $\gamma$, then $f(a)<0$, if $a$ is on the image of $\gamma$, then
$f(a)=0$, and if $a$ is outside of $\gamma$, then $f(a)>0$. Using this
function, we can define a function $g_{f,\gamma}:S^{1}\times
S^{1}\to\mathbb{R}^{2}$, by the formula

$g_{f,\gamma}(x,y)=(f(S_{1}(\gamma(x),\gamma(y))),f(S_{2}(\gamma(x),\gamma(y)))).$

If $\gamma$ is a bad Jordan curve, then $g_{f,\gamma}(x,y)=(0,0)$ if and only
if $x=y$, because if we have $x\neq y$ and $g_{f,\gamma}(x,y)=(0,0)$, then
$\gamma(x),\gamma(y),S_{2}(\gamma(x),\gamma(y)),S_{1}(\gamma(x),\gamma(y)))$
form the vertices of an inscribed square.

Let $C=\\{(x,y)\in S^{1}\times S^{1}:x\neq y\\}$. Topologically, $C$ is an
open cylinder. By the above proposition, we see that if $\gamma$ is a bad
Jordan curve, then we can restrict the domain of $g_{f,\gamma}$ to $C$ to get
a map

$g_{f,\gamma}|_{C}:C\to\mathbb{R}^{2}\setminus\\{(0,0)\\}.$

We now fix homotopy equivalences between $C$,
$\mathbb{R}^{2}\setminus\\{(0,0)\\}$, and $S^{1}$. Our homotopy equivalence
$C\to S^{1}$ is to simply take the first coordinate of the ordered pair. Our
homotopy equivalence $\mathbb{R}^{2}\setminus\\{(0,0)\\}\to S^{1}$ is radial
projection onto the unit circle, which we orient counterclockwise.

Using these homotopy equivalences, we see that, $g_{f,\gamma}|_{C}$ induces a
map $S^{1}\to S^{1}$ up to homotopy, and therefore gives us an element of
$\pi_{1}(S^{1})\simeq\mathbb{Z}$. This integer is defined to be $b_{\gamma}$.

$b_{\gamma}$ does not depend on the choice of $f$, because given two choices
of $f$, one can be continuously transformed into the other while always
remaining a valid choice of $f$. This induces a homotopy between the resulting
maps $S^{1}\to S^{1}$, which implies that the value of $b_{\gamma}$ is
constant.

######  Lemma 1.

For any bad Jordan curve, $b_{\gamma}$ is odd.

The proof of this lemma will be given later in this section. The main idea
behind the proof is that each inscribed square of a generic smooth
approximation of $\gamma$ contributes parity to $b_{\gamma}$ in a way that
depends on the cyclic order of the vertices of the square on the Jordan curve.
The parity of inscribed squares with a given cyclic ordering is independent of
the Jordan curve, so we can simply calculate the parity of $b_{\gamma}$ by
summing the contributions from each square type. The result is odd parity.

Lemma 1 can be used to prove Theorem 1, which we will do at the end of this
section. The main idea is that because $b_{\gamma}\neq 0$, the map
$g_{f,\gamma}:C\to\mathbb{R}^{2}\setminus\\{(0,0)\\}$ wraps nontrivially
around the origin. This implies that there must be a path between the two ends
of the cylinder $C$ that maps into the lower left quadrant of the plane under
$g_{f,\gamma}$. This path is almost a square envelope, except that $e_{1}$ and
$e_{2}$ are on the Jordan curve rather than being inside its open exterior.
This can be remedied by simply pushing the paths off of the curve slightly.

For a manifold $M$, let $\tilde{C}_{4}(M)$ denote the manifold of cyclically
ordered quadruples of distinct points in $M$. In other words, if
$C_{4\;\text{dist}}(M)$ is the subset of $M^{4}$ consisting of ordered
quadruples of distinct points, then
$\tilde{C}_{4}(M)=C_{4\;\text{dist}}(M)/(\mathbb{Z}/4\mathbb{Z})$, where
$\mathbb{Z}/4\mathbb{Z}$ acts by cyclic permutation of the entries of the
4-tuple. Let $Sq$ denote the the submanifold of
$\tilde{C}_{4}(\mathbb{R}^{2})$ consisting of quadruples of points which form
squares labeled counterclockwise.

We call a Jordan curve $\gamma:S^{1}\to\mathbb{R}^{2}$ _generic_ if, within
$\tilde{C}_{4}(\mathbb{R}^{2})$, the subspace
$\tilde{C}_{4}(\text{im}(\gamma))$ intersects transversely with $Sq$.

Let $h:S^{1}\times[0,1]\to\mathbb{R}^{2}$ be a smooth homotopy between two
generic Jordan curves for which $h(\cdot,t)$ is a smooth embedding for all
$t\in[0,1]$. We say that $h$ is a _generic homotopy_ if, within
$\tilde{C}_{4}(\mathbb{R}^{2})\times[0,1]$, the subspace
$\\{(q,t):q\in\tilde{C}_{4}(\text{im}(h(\cdot,t))\\}$ intersects transversely
with the subspace $Sq\times[0,1]$.

By the transversality theory found in [3], we have the following facts about
generic Jordan curves.

  * 1)

Any Jordan curve can be approximated arbitrarily well in the $C^{0}$ topology
by generic Jordan curves.

  * 2)

Any two generic Jordan curves have a generic homotopy between them.

  * 3)

For a generic homotopy $h$, the intersection
$\\{(q,t):q\in\tilde{C}_{4}(\text{im}(h(\cdot,t))\\}\;\cap\;Sq\times[0,1]$ is
a compact 1-manifold with boundary. The boundary of this manifold consists of
the inscribed squares of the two generic Jordan curves at the beginning and
end of the homotopy.

Note that the manifold $\tilde{C}_{4}(S^{1})$ has three connected components.
This corresponds to three types of inscribed square.

######  Definition 3.

Let $\gamma$ be a Jordan curve. An inscribed square of $\gamma$ is called type
I if the counterclockwise order of points on the square is the same as that of
the Jordan curve. Such squares are also called gracing squares. A square is
called type II if when we label the vertices of the square as $1,2,3,4$ in
counterclockwise order, these vertices appear in the order $1,3,2,4$ when we
go counterclockwise along the Jordan curve. Finally, a square is called type
III if the counterclockwise order of the vertices is opposite to the
counterclockwise order of the Jordan curve.

Figure 3. Inscribed squares of type I, II, and III respectively.

######  Proposition 1.

If $\gamma$ is a generic Jordan curve, then of the inscribed squares of
$\gamma$, an odd number of them are type I, an even number are type II, and an
even number are type III.

###### Proof.

There are three connected components of $\tilde{C}_{4}(\text{im}(\gamma))$
corresponding to the three types of inscribed squares when we intersect with
$Sq$. This means that if $h$ is a generic homotopy between generic Jordan
curves, then the 1-manifold
$\\{(q,t):q\in\tilde{C}_{4}(\text{im}(h(\cdot,t))\\}\;\cap\;Sq\times[0,1]$ can
be separated into three distinct clopen pieces, one for each type of inscribed
square. This proves that the parity of each type of inscribed square is
invariant of which generic Jordan curve we choose. Finally we can compute the
parities by finding the inscribed squares of any generic Jordan curve. The
quintessential generic Jordan curve is a non-circular ellipse. This curve has
one gracing square, but no inscribed square of type II or III. ∎

Let $\gamma$ be a generic Jordan curve, and let $f$ be a smooth function
$\mathbb{R}^{2}\to\mathbb{R}$ which has $\gamma$ as its zero set, is negative
inside of $\gamma$, positive outside of $\gamma$, and has non-vanishing
gradient on $\gamma$. Since $\gamma$ is a smooth embedding, such an $f$ always
exists. Now, we will consider $g_{f,\gamma}:(S^{1})^{2}\to\mathbb{R}^{2}$, as
defined in the previous section.

We see that the zeros of $g_{f,\gamma}$ consist of the diagonal
$\Delta=\\{(a,a):a\in S^{1}\\}$, as well as four points for each inscribed
square in $\tilde{C}_{4}(\text{im}(\gamma))\cap Sq$, one point for each side
of the square. Furthermore, at these zeros corresponding to inscribed squares,
$g_{f,\gamma}$ has nonsingular derivative because $\gamma$ is generic and $f$
has non-vanishing gradient on $\text{im}(\gamma)$. Let
$X=\\{p\in(S^{1})^{2}:g_{f,\gamma}(p)\neq(0,0)\\}$. Let $\omega_{\gamma}$ be
the cohomology class in $H^{1}(X;\mathbb{Z}/2\mathbb{Z})$ given by pulling
back the nontrivial class of
$H^{1}(\mathbb{R}^{2}\setminus\\{(0,0)\\};\mathbb{Z}/2\mathbb{Z})$ under
$g_{f,\gamma}|_{X}$. A small loop around any of the zeroes corresponding to
sides of inscribed squares will evaluate nontrivially under $\omega$ because
$g_{f,\gamma}$ has nonsingular derivative at these zeroes. Furthermore, a loop
obtained by pushing the diagonal off to one side or the other will evaluate
trivially under $\omega_{\gamma}$ because it will correspond to a small square
sliding around the Jordan curve with two corners on $\gamma$ and the other two
corners off to one side of $\gamma$, so the corresponding loop in
$\mathbb{R}^{2}\setminus\\{(0,0)\\}$ will remain in just one quadrant of the
plane and therefore cannot wrap around the origin.

Putting this all together, we see that if $\ell$ is a simple closed curve in
$X$ which has the homotopy class of the diagonal in $S^{1}\times S^{1}$, then
$\omega_{\gamma}(\ell)$ is equal to the parity of the number of zeroes of
$g_{f,\gamma}$ in either connected component of
$(S^{1})^{2}\setminus(\Delta\cup\ell)$.

Let $\Delta^{\prime}$ be the anti-diagonal of $(S^{1})^{2}$, the loop
consisting of pairs $(a,b)$ where $a$ and $b$ are antipodes. Then
$(S^{1})^{2}\setminus(\Delta\cup\Delta^{\prime})$ has two connected
components, which we call $A$ and $B$. The component $A$ is the set of pairs
$(a,b)$ for which the angle from $a$ to $b$ is less than $\pi$ and the
component $B$ is the set of pairs for which the angle from $a$ to $b$ is
greater than $\pi$.

If $\gamma$ is a bad Jordan curve, and $\gamma^{\prime}$ is a sufficiently
$C^{0}$-close generic approximation, then the parity of $b_{\gamma}$ is equal
to $\omega_{\gamma^{\prime}}(\Delta^{\prime})$. We can therefore compute
$b_{\gamma}$ by counting the zeroes of $g_{f,\gamma^{\prime}}$ in $A$.

######  Proposition 2.

If $\gamma$ is a bad Jordan curve, and $\gamma^{\prime}$ is a sufficiently
$C^{0}$-close generic approximation, then each type I inscribed square of
$\gamma^{\prime}$ has exactly three of its corresponding zeroes of
$g_{f,\gamma^{\prime}}$ in $A$, each type II inscribed square has exactly two
of its corresponding zeroes in $A$, and each type III inscribed square has
exactly one of its corresponding zeroes in $A$.

###### Proof.

Let $\gamma_{1},\gamma_{2},\gamma_{3},...$ be a sequence of generic Jordan
curves that limit to our bad Jordan curve $\gamma$ in the $C^{0}$ topology.
Let $s_{n}$ be the side length of the largest inscribed square of
$\gamma_{n}$. We know that $\lim_{n\to\infty}s_{n}=0$ because otherwise there
would be a convergent subsequence of squares that limit to an inscribed square
of our bad Jordan curve. We can give $S^{1}$ a metric by identifying it with
the unit circle in $\mathbb{R}^{2}$ and using euclidean distance. Using this
identification, we let $M$ be the set of all pairs $(a,b)\in(S^{1})^{2}$ which
have $\|a-b\|\geq 1$. Let $N\in\mathbb{Z}$ be sufficiently large that both
$\sup_{x\in S^{1}}\|\gamma_{N}(x)-\gamma(x)\|$ and $s_{N}$ are smaller than
the quantity $\varepsilon=\frac{1}{4}\cdot\min_{(a,b)\in
M}\|\gamma(a)-\gamma(b)\|$. We claim that $\gamma^{\prime}=\gamma_{N}$ is an
approximation for which the proposition holds.

Let $a,b,c,d$ be the vertices of an inscribed square of $\gamma_{N}$, labeled
counterclockwise. Let $Q$ be the set of four points in $S^{1}$ which map onto
$\\{a,b,c,d\\}$ under $\gamma_{N}$. We claim that the diameter of $Q$ is less
than one. To prove this, suppose there were elements $x$ and $y$ in $Q$ with
$\|x-y\|\geq 1$. Then $(x,y)\in M$, so $\|\gamma(x)-\gamma(y)\|\geq
4\varepsilon$, so $\|\gamma_{N}(x)-\gamma_{N}(y)\|\geq 2\varepsilon$. However,
$2\varepsilon>\sqrt{2}\cdot s_{N}$ and $\gamma_{N}(x)$ and $\gamma_{N}(y)$ are
vertices of a square with side length at most $s_{N}$, so this is impossible.

Since the diameter of $Q$ is less than $1$, all four points of $Q$ must lie
within some interval $I$ of angular length $\pi/3$ radians. Orient $I$
counterclockwise, and let $w,x,y,z$ be the four elements of $Q$ in the order
they appear along $I$. We know that $(w,x),(w,y),(w,z),(x,y),(x,z),$ and
$(y,z)$ are in $A$, and all the other ordered pairs are in $B$.

We know that $\gamma_{N}$ maps $\\{w,x,y,z\\}$ onto $\\{a,b,c,d\\}$, but it
might not preserve the ordering. Without loss of generality, we can assume
$\gamma_{N}(w)=a$ because we can permute the labels $a,b,c,d$ cyclically.
Then, we have six possibilities to check for the six permutations of the
remaining three letters. We will now check all of the possibilities.

  * 1)

If $(w,x,y,z)\mapsto(a,b,c,d)$, the square is type I and the corresponding
zeroes in $A$ are $(w,x),(x,y),(y,z)$.

  * 2)

If $(w,x,y,z)\mapsto(a,c,b,d)$, the square is type II and the corresponding
zeroes in $A$ are $(w,y),(x,z)$.

  * 3)

If $(w,x,y,z)\mapsto(a,b,d,c)$, the square is type II and the corresponding
zeroes in $A$ are $(w,x),(x,z)$.

  * 4)

If $(w,x,y,z)\mapsto(a,d,c,b)$, the square is type III and the only
corresponding zero in $A$ is $(w,z)$.

  * 5)

If $(w,x,y,z)\mapsto(a,d,b,c)$, the square is type II and the corresponding
zeroes in $A$ are $(w,y),(y,z)$.

  * 6)

If $(w,x,y,z)\mapsto(a,c,d,b)$, the square is type II and the corresponding
zeroes in $A$ are $(w,z),(x,y)$.

This confirms the proposition. The type I squares have three corresponding
zeroes in $A$, the type II squares have two corresponding zeroes in $A$, and
the type III squares have only one corresponding zero in $A$.

∎

We can now prove Lemma 1 and Theorem 1.

###### Proof of Lemma 1..

Let $\gamma$ be a bad Jordan curve, let $\gamma^{\prime}$ be a sufficiently
$C^{0}$-close generic approximation, and let $f$ be a function with
$\gamma^{\prime}$ as its zero set as above. We now count zeroes of
$g_{f,\gamma^{\prime}}$ in $A$, the parity of which is the parity of
$b_{\gamma}$. We count the zeroes by those corresponding to the squares of
each type. We have an odd number of threes, an even number of twos, and an
even number of ones. This totals to an odd number.

###### Proof of Theorem 1..

We wish to prove that every bad Jordan curve has a square envelope. We have
the function $g_{f,\gamma}|_{C}:C\to\mathbb{R}^{2}\setminus\\{(0,0)\\}$, which
we know to be homotopically nontrivial. Let
$\pi:\mathbb{R}^{2}\setminus\\{(0,0)\\}\to S^{1}$ be radial projection onto
the unit circle, and let $r=\pi\circ g_{f,\gamma}|_{C}$. We now take
$r^{\prime}$ to be a smooth approximation of $r$ with
$\|r(x)-r^{\prime}(x)\|<1/4$ for all $x\in C$. Let $u\in S^{1}$ be a regular
value for $r^{\prime}$ within distance $1/4$ of the point
$(-1/\sqrt{2},-1/\sqrt{2})$. Then $r^{-1}(u)$ is a 1-dimensional manifold,
$L$, which is mapped under $r$ into a circle of radius $1/2$ around
$(-1/\sqrt{2},-1/\sqrt{2})$, which is entirely in the lower left quadrant of
the plane. Therefore, $g_{f,\gamma}|_{C}$ maps $L$ into the lower left
quadrant of the plane. This implies that the square corresponding to a point
of $L$ has its first two corners on $\gamma$, and the other two in the open
interior of $\gamma$. Furthermore, since $b_{\gamma}$ is odd, we see that $L$
must have an odd number of connected components which are homeomorphic to
$\mathbb{R}$ with the two ends limiting to the two boundaries of $C$.
Parameterizing such a connected component, we have functions
$e_{1},e_{2}:\mathbb{R}\to\mathbb{R}^{2}$ with
$\text{im}(e_{1})\cup\text{im}(e_{2})\subseteq\text{im}(\gamma)$, and
$\lim_{x\to\pm\infty}\|e_{1}(x)-e_{2}(x)\|=0$, and with the property that
$S_{1}(e_{1}(x),e_{2}(x))$ and $S_{2}(e_{1}(x),e_{2}(x))$ are always in the
open interior of $\gamma$. Since the interior of gamma is open, we may choose
a continuous function $\varepsilon:\mathbb{R}\to\mathbb{R}_{>0}$ with the
property that, for all $x$, if $\|a-e_{1}(x)\|$ and $\|b-e_{2}(x)\|$ are less
than $\varepsilon(x)$, then $S_{1}(a,b)$ and $S_{2}(a,b)$ are in the open
interior of $\gamma$ as well. Thus, if we choose a continuous deformation of
$(e_{1},e_{2})$ in which we push $e_{1}(x)$ and $e_{2}(x)$ off of the Jordan
curve without exceeding a distance of $\varepsilon(x)$, we obtain a square
envelope for $\gamma$.

##  3\. Relation avoiding paths

In this section, we will prove that if the spiral conjecture is true, then the
inscribed square conjecture holds for Jordan curves which are smooth except at
a finite number of arbitrarily complicated singularities.

However, we will first present a proof of Theorem 2, that the spiral
conjecture is indeed true in the one-dimensional case.

###### Proof of Theorem 2..

Identifying $V$ with $\mathbb{C}$, our relation avoiding origin paths are
paths to the origin in the complex plane $p:[0,\infty)\to\mathbb{C}$.
Furthermore, we know that $p(t)$ is never zero, because $(0,0)$ is in every
linear relation. Thus, without loss of generality, we can ignore any of the
$R_{i}$ corresponding to $\\{0\\}\times\mathbb{C}$ or
$\mathbb{C}\times\\{0\\}$. We can therefore rewrite the relation $R$ as

$xRy\iff\exists
i\in\\{1,...,n\\}\;\;\;x=\alpha_{i}y\;\;\text{or}\;\;y=\alpha_{i}x$

where $\alpha_{1},...,\alpha_{n}$ are nonzero complex numbers with norm at
most one. In particular, our path $p$ is relation avoiding if and only if it
is disjoint from the paths $\alpha_{1}p,\alpha_{2}p,...,\alpha_{n}p$. Without
loss of generality, we can assume that $p(0)=1$, and that $|p(t)|\leq 1$ for
all $t$. The reason we can do this is that we can always homotope $p$ to such
a path by first contracting it within its image so that the starting point is
at the point with maximal norm, and then rescaling within $\mathbb{C}$ so that
the starting point becomes $1$. We can also assume without loss of generality
that $p$ follows the path of a logarithmic spiral inside some small
neighborhood around 1, because making this happen only requires a small
perturbation. Working with these assumptions, we fix a branch of the
logarithm, and let $\ell_{1},...,\ell_{n}$ be paths in upper-half plane such
that $e^{2\pi i\ell_{i}(t)}=\alpha_{i}p(t)$ with $\ell_{i}(0)=\frac{1}{2\pi
i}\ln(\alpha_{i})$, and similarly let $\ell$ be a lift of $p$ with
$\ell(0)=0$. The path $\ell$ divides the upper-half plane, and we define $U$
to be the set of all points of $\mathbb{H}\setminus\text{im}(\ell)$ from which
a path to $-\infty$ has even intersection parity with $\ell$. We then define
integers $k_{1},...,k_{n}$, where $k_{i}$ is the largest integer such that
$\ell_{i}(0)+k_{i}\in U$.

We see that $\ell$ is disjoint from $\ell_{i}+k$ for any index $i$ and integer
$k$, and this gives us paths disjoint from $\ell$, going up to $i\infty$, from
every point of the form $\ell(0)+k$. Therefore, the homotopy type of $\ell$
can be completely determined by knowing which points of the form
$\ell_{i}(0)+k$ are in $U$. Thus, to prove the one-dimensional spiral
conjecture, it suffices to prove that there exists a $\theta\in(0,\pi)$ so
that for all $i\in\\{1,...,n\\}$, we have the inequalities
$\arg(\ell_{i}(0)+k_{i})\geq\theta\geq\arg(\ell_{i}(0)+k_{i}+1)$. This
$\theta$ would then denote the angle of a straight line in the upper half
plane that exponentiates to the desired logarithmic spiral. To prove that such
a $\theta$ exists, it suffices to prove that for all $i$ and $j$ we have
$\arg(\ell_{j}(0)+k_{j}+1)\leq\arg(\ell_{i}(0)+k_{i})$.

We define a _split pair_ to be a pair of points in the upper half-plane
$(p,q)$, such that $p+\ell$ and $q+\ell$ are both disjoint from $\ell$, with
$p\in U$ and $q\not\in U$. For instance,
$(\ell_{i}(0)+k_{i},\ell_{j}(0)+k_{j}+1)$ is always a split pair.

Figure 4. A depiction of a split pair, disjoint translates of $\ell$, one on
each side.

Given a split pair $(p,q)$, we can construct a new split pair as follows. If
$\text{im}(p)\geq\text{im}(q)$, then $(p-q,q)$ is the new split pair. If
$\text{im}(q)>\text{im}(p)$, then $(p,q-p)$ is the new split pair. The split
pair obtained by applying this transformation is called the derived split
pair. We claim that the derived split pair is always another split pair.

Figure 5. One way to understand derived split pairs is that one draws a
parallelogram spanning the origin and the two points, then moves the uppermost
point to the new corner.

To prove this claim, we consider the case that $\text{im}(p)\geq\text{im}(q)$.
We have that $\ell+q$ is entirely on the right side of $\ell$, and $\ell+p$ is
entirely on the left side of $\ell$. Furthermore, since all of $\ell+p$ has
greater imaginary coordinate than $\text{im}(q)$, we therefore have that
$p\in\\{a\in U:\text{im}(a)\geq\text{im}(q)\\}\subseteq U+q$. This tells us
that $p-q$ is in $U$, so $(p-q,q)$ is a valid split pair. The argument for the
other case is similar.

We say a split pair is _good_ if $\arg(p)>\arg(q)$. We say a split pair is
_bad_ otherwise. We see that the derived split pair of a bad split pair is
bad, and the derived split pair of a good split pair is good. To complete the
proof of the theorem, it suffices to prove that all split pairs are good, so
we will consider what occurs under repeated derivation of a bad split pair.
First of all, note that if $\text{im}(p)/\text{im}(q)$ is a rational number,
then we will eventually have a split pair with a real number as one of the
terms, and we can see that any split pair containing a real number is good.
Thus, we have shown that $\text{im}(p)/\text{im}(q)$ is irrational for any bad
split pair. Now, we will proceed by contradiction to prove that there are no
bad split pairs. Suppose that $(p,q)$ is a bad split pair. We will treat the
cases of $\arg(p)<\arg(q)$ and $\arg(p)=\arg(q)$ separately.

Suppose that $(p,q)$ is a bad split pair with $\arg(p)<\arg(q)$. Then, the
$i$-th derived split pair is of the form $(a_{i}p-b_{i}q,c_{i}q-d_{i}p)$, for
nonnegative integers $a_{i},b_{i},c_{i},d_{i}$ such that

$\lim_{i\to\infty}a_{i}/b_{i}=\lim_{i\to\infty}d_{i}/c_{i}=\text{im}(q)/\text{im}(p)$

This tells us that while the imaginary parts of the terms of the derived split
pairs approach zero, the real parts approach $+\infty$ and $-\infty$
respectively. However, this would force the second term to eventually be in
$U$, which gives us a contradiction.

Now, suppose $\arg(p)=\arg(q)$. In this case, the terms of the derived split
pairs stay on a fixed straight line, and they approach zero. However, earlier
in the proof, we assumed without loss of generality that $p$ followed the path
of a logarithmic spiral in some small neighborhood around 1, which means
$\ell$ is a straight line in some small neighborhood around zero. This means
that there are no bad split pairs in this neighborhood, so we have a
contradiction. This completes the proof.

∎

Before we prove Theorem 3, we need to develop some notation surrounding square
envelopes, and prove a couple lemmas.

######  Definition 4.

Let $\gamma:S^{1}\to\mathbb{R}^{2}$ be a Jordan curve, and let $(e_{1},e_{2})$
be a square envelope. We write $\sigma_{p}^{*}(e_{1},e_{2},\gamma)$, where
$p\in\\{1,2\\}$ and $*\in\\{+,-\\}$, to denote a sign in $\\{+1,-1\\}$
determined as follows. For $\lambda\in\mathbb{R}$, let $P_{\lambda}$ be the
path that starts at $e_{1}(\lambda)$, then goes along $e_{1}$ to $e_{1}(0)$,
then goes in a straight line from $e_{1}(0)$ to $e_{2}(0)$, then follows
$e_{2}$ to $e_{2}(\lambda)$, then goes in a straight line back to the starting
point $e_{1}(\lambda)$. We then let $n(\lambda)$ be the wrapping number of
$P_{\lambda}$ around $S_{p}(e_{1}(0),e_{2}(0))$. Finally, we set
$\sigma_{p}^{*}(e_{1},e_{2},\gamma)=\lim_{\lambda\to*\infty}(-1)^{n(\lambda)}$.

It is worth observing that these parities are determined by how the Jordan
curve winds between the vertices of a small square near $t=\pm\infty$. For
curves with finitely many singularities, they are related to each other by the
following lemma.

######  Lemma 2.

If $\gamma$ is a Jordan curve which is smooth except at finitely many points,
then the following facts are true for any square envelope $(e_{1},e_{2})$ of
$\gamma$.

  * 1)

$\sigma_{1}^{+}(e_{1},e_{2},\gamma)=\sigma_{2}^{+}(e_{1},e_{2},\gamma)$ and
$\sigma_{1}^{-}(e_{1},e_{2},\gamma)=\sigma_{2}^{-}(e_{1},e_{2},\gamma)$

  * 2)

$\sigma_{1}^{+}(e_{1},e_{2},\gamma)=-\sigma_{1}^{-}(e_{1},e_{2},\gamma)$

  * 3)

If $*\in\\{+,-\\}$ is such that $\sigma_{1}^{*}(e_{1},e_{2},\gamma)=-1$, then
the limits $\lim_{t\to*\infty}e_{1}(t)$ and $\lim_{t\to*\infty}e_{2}(t)$
exist, and are equal to each other.

###### Proof.

First of all, by property (4) in the definition of a square envelope, we
immediately have that
$\sigma_{1}^{+}(e_{1},e_{2},\gamma)=-\sigma_{1}^{-}(e_{1},e_{2},\gamma)$ and
$\sigma_{2}^{+}(e_{1},e_{2},\gamma)=-\sigma_{2}^{-}(e_{1},e_{2},\gamma)$
because the defining loops for the parities in question compose to give us a
loop that wraps around the Jordan curve exactly once.

Next, we claim that if $*\in\\{+,-\\}$, and the limit
$\lim_{t\to*\infty}e_{1}(t)$ does not exist, then we have
$\sigma_{1}^{*}(e_{1},e_{2},\gamma)=\sigma_{2}^{*}(e_{1},e_{2},\gamma)=1$. To
prove this claim, note that for the limit to fail to exist, there must be a
limit point of $e_{1}(t)$, $t\to*\infty$ at one of the points where $\gamma$
is smooth. Taking a sufficiently small square of the envelope, in the
$*\infty$ direction, near such a point, we see that the Jordan curve must
separate the vertices of the square in a way locally equivalent to how a
straight line could separate the vertices, and there is only one such way that
separates the $e_{1},e_{2}$ vertices from the other two vertices. Since all
squares sufficiently near $t=\pm\infty$ will have side length smaller than the
diameter of some disk inside of the Jordan curve, the paths that define
$\sigma_{1}^{*}(e_{1},e_{2},\gamma)$ and $\sigma_{2}^{*}(e_{1},e_{2},\gamma)$
cannot wrap around either of the two interior vertices of the square at $t=0$.
Therefore, we have
$\sigma_{1}^{*}(e_{1},e_{2},\gamma)=\sigma_{2}^{*}(e_{1},e_{2},\gamma)=1$.

To prove all three parts of the lemma, all that remains is to eliminate the
possibility that the limits $\lim_{t\to*\infty}e_{1}(t)$ and
$\lim_{t\to*\infty}e_{2}(t)$ exist, but $\sigma_{1}^{*}(e_{1},e_{2},\gamma)$
and $\sigma_{2}^{*}(e_{1},e_{2},\gamma)$ are not equal to one another. The
reason this is impossible is that for this to be the case, the paths
$e_{1}(t),e_{2}(t),S_{1}(e_{1}(t),e_{2}(t)),$ and $S_{2}(e_{1}(t),e_{2}(t))$
would need to all approach some point in the plane, none of them intersecting
each other, in such a way that as one encircles the point counterclockwise,
the paths appear in a cyclic order that alternates between the sets
$\\{e_{1}(t),e_{2}(t)\\}$ and
$\\{S_{1}(e_{1}(t),e_{2}(t)),S_{2}(e_{1}(t),e_{2}(t))\\}$. This would then
contradict the fact that the simple closed curve $\gamma$ separates the paths
in $\\{e_{1}(t),e_{2}(t)\\}$ from those in
$\\{S_{1}(e_{1}(t),e_{2}(t)),S_{2}(e_{1}(t),e_{2}(t))\\}$. ∎

If $\gamma$ is a Jordan curve which is smooth except at finitely many points,
and $(e_{1},e_{2})$ is a square envelope, we say $(e_{1},e_{2})$ is positively
oriented if $\sigma_{1}^{+}(e_{1},e_{2},\gamma)=-1$ and negatively oriented if
$\sigma_{1}^{+}(e_{1},e_{2},\gamma)=+1$. Note that if $(e_{1}(t),e_{2}(t))$ is
a negatively oriented square envelope, we can obtain a positively oriented one
by taking $(e_{1}(-t),e_{2}(-t))$.

We need one more lemma before we can prove Theorem 3.

######  Lemma 3.

Let $\alpha_{1},\alpha_{2},\beta_{1},\beta_{2}$ be complex numbers such that
the ratio $\beta_{1}/\beta_{2}$ is a positive real number. Let $t_{0}$ and $p$
be arbitrary real numbers with $p\neq 0$, and let $\lambda,r_{1},r_{2}$ be
real numbers in $[0,1)$. If there exist real numbers $t_{1}$ and $t_{2}$ in
$[0,\infty)$ such that

$e^{\alpha_{1}+\beta_{1}p(t_{1}-t_{0})}\left(1+\lambda
r_{1}e^{ip(t_{1}-t_{0})}\right)=e^{\alpha_{2}+\beta_{2}p(t_{2}-t_{0})}\left(1+\lambda
r_{2}e^{ip(t_{2}-t_{0})}\right)$

then there also exist $t_{1}^{\prime}$ and $t_{2}^{\prime}$ in $[0,\infty)$
such that

$e^{\alpha_{1}+\beta_{1}p(t_{1}^{\prime}-t_{0})}\left(1+r_{1}e^{ip(t_{1}^{\prime}-t_{0})}\right)=e^{\alpha_{2}+\beta_{2}p(t_{2}^{\prime}-t_{0})}\left(1+r_{2}e^{ip(t_{2}^{\prime}-t_{0})}\right)$

###### Proof.

Making the substitutions $\alpha_{1}-\alpha_{2}=\alpha$, $p(t_{1}-t_{0})=t$,
and $p(t_{2}-t_{0})=(\beta_{1}/\beta_{2})t+s$, the equation

$e^{\alpha_{1}+\beta_{1}p(t_{1}-t_{0})}\left(1+\lambda
r_{1}e^{ip(t_{1}-t_{0})}\right)=e^{\alpha_{2}+\beta_{2}p(t_{2}-t_{0})}\left(1+\lambda
r_{2}e^{ip(t_{2}-t_{0})}\right)$

rearranges to

$e^{\alpha-\beta_{2}s}=\frac{1+\lambda r_{1}e^{it}}{1+\lambda
r_{2}e^{i(\frac{\beta_{1}}{\beta_{2}}t+s)}}$

Rearranging, and setting $a=r_{1}$, $b(s)=-r_{2}e^{\alpha+(i-\beta_{2})s}$,
and $c(s)=e^{\alpha+\beta_{2}s}-1$, we have

$c(s)=\lambda\cdot\left(a\cdot e^{it}+b(s)\cdot
e^{i\left(\frac{\beta_{1}}{\beta_{2}}t\right)}\right)$

Also, let

$T(s)=\\{t:\exists t_{1}\geq 0,\exists t_{2}\geq
0,\;\;\;p(t_{2}-t_{0})=(\beta_{1}/\beta_{2})t+s,\;\;\;p(t_{1}-t_{0})=t\\}$

be the allowable values of $t$ for fixed $s$, when $t_{1}$ and $t_{2}$ are
nonnegative. This will always be an interval of the form $(-\infty,z]$ or
$[z,\infty)$ for some real number $z$. Now, let $H(s)$ denote the unique
minimal simply connected set containing the set

$\\{a\cdot e^{it}+b(s)\cdot e^{i(\beta_{1}/\beta_{2})t}:t\in T(s)\\}$

Then, we see that there exists a pair of nonnegative real numbers $t_{1}$ and
$t_{2}$ such that

$e^{\alpha_{1}+\beta_{1}p(t_{1}-t_{0})}\left(1+\lambda
r_{1}e^{ip(t_{1}-t_{0})}\right)=e^{\alpha_{2}+\beta_{2}p(t_{2}-t_{0})}\left(1+\lambda
r_{2}e^{ip(t_{2}-t_{0})}\right)$

if and only if there exists a real number $s$ such that $c(s)\in\lambda H(s)$.
Since $\beta_{1}/\beta_{2}$ is positive, when $\beta_{1}/\beta_{2}$ is
rational, $H(s)$ is the region enclosed by the outer boundary of an
epitrochoid curve. When $\beta_{1}/\beta_{2}$ is irrational, $H(s)$ it is a
disk missing some subset of its boundary. In particular, $H(s)$ is always a
radial set, meaning that for any $x\in H(s)$, we have $\lambda x\in H(s)$ for
all $\lambda\in[0,1]$. Therefore, increasing the parameter $\lambda$ preserves
the existence of solutions to our original equation. Setting $\lambda=1$ gives
us the desired result.

Figure 6. The region enclosed by an epitrochoid curve is always a radial set.

∎

Now, we can prove Theorem 3.

###### Proof of Theorem 3..

We begin by considering the system of linear relations on $\mathbb{C}^{2}$
given by letting $(x_{1},y_{1})R(x_{2},y_{2})$ when

$\\{x_{1},y_{1},x_{2},y_{2}\\}\cap\\{x_{1}+i(y_{1}-x_{1}),y_{1}+i(y_{1}-x_{1}),x_{2}+i(y_{2}-x_{2}),y_{2}+i(y_{2}-x_{2})\\}\neq\varnothing$

Now, take a positively oriented square envelope $(e_{1},e_{2})$. Due to Lemma
2, restricting the domain to $[0,\infty)$ immediately gives us a relation
avoiding origin path
$(e_{1}(t),e_{2}(t))-\lim_{t\to\infty}(e_{1}(t),e_{2}(t))$ for $R$. We are
assuming the spiral conjecture, so we may mow take $h:[0,1]\times[0,\infty)$
satisfying the conditions of the spiral conjecture. This gives us complex
numbers $a_{1},a_{2}$, and vectors $x_{1},x_{2}\in\mathbb{C}^{2}$, such that
$(f_{1}(t),f_{2}(t))=f(t)=x_{1}e^{a_{1}t}+x_{2}e^{a_{2}t}$ is a relation
avoiding origin path for $R$ which has the property that the path in
$\mathbb{C}$ starting at $0$, following $f_{2}$ to $f_{2}(0)$, going in a
straight line from $f_{2}(0)$ to $f_{1}(0)$, then following $f_{1}$ back to
$0$ has nontrivial wrapping number parity around the points
$f_{1}(0)+i(f_{2}(0)-f_{1}(0))$ and $f_{2}(0)+i(f_{2}(0)-f_{1}(0))$. Without
loss of generality, we can assume that $a_{1}$ and $a_{2}$ have the same real
part, because otherwise one term will dominate at large $t$, reducing us to
the case of a pure logarithmic spiral, which we will cover later. Similarly,
we may assume $a_{1}\neq a_{2}$. Now, to prove that no such
$a_{1},a_{2},x_{1},x_{2}$ exist, we see that for the path to be relation
avoiding, we must be able to find some fixed complex number $\beta$, and
nonzero real number $p$, such that all four paths
$f_{1}(t),f_{2}(t),f_{1}(t)+i(f_{2}(t)-f_{1}(t))$, and
$f_{2}(t)+i(f_{2}(t)-f_{1}(t))$ are of the form

$e^{\alpha+q\beta p(t-t_{0})}\left(1+re^{ip(t-t_{0})}\right)$

with $\alpha\in\mathbb{C},t_{0}\in\mathbb{R},q\in(0,\infty),$ and $r\in[0,1)$.
Therefore, Lemma 3 allows us to continuously scale down all of the $r$
parameters to zero while staying a relation avoiding origin path.

All that remains is to show that no pure logarithmic spiral $f(t)=xe^{at}$ can
be a relation avoiding origin path with the stated wrapping number parity
condition. To prove this, note that the region swept out by a line segment
between $f_{1}$ and $f_{2}$ would have the same area as the region swept out
by the line segment between $f_{1}+i(f_{2}-f_{1})$ and $f_{2}+i(f_{2}-f_{1})$.
However, our wrapping number condition implies that the former region would
need to strictly contain the latter, which would contradict them having equal
areas. ∎

To wrap things up, we make a couple more conjectures about relation avoiding
origin paths.

######  Conjecture 3 (SSC, Strong Spiral Conjecture).

Every relation avoiding origin path $p$ is strongly homotopic to a relation
avoiding origin path $q$ for which there exist linearly independent vectors
$x_{1},...,x_{n}$ in $V$ and complex numbers $a_{1},...,a_{n}$ such that

$q(t)=\sum_{i=1}^{n}x_{i}\cdot e^{a_{i}t}$

for all $t$.

######  Conjecture 4 (MSC, Monotonic Spiral Conjecture).

For any continuous function $p:[0,\infty)\to V\setminus\\{0\\}$ with
$\lim_{t\to\infty}p(t)=0$, there exists a continuous function
$h:[0,1]\times[0,\infty)\to V$ with the following properties.

  * 1)

$\lim_{t\to\infty}h(s,t)=0$ for all $s$.

  * 2)

$h(0,t)=p(t)$ for all $t$.

  * 3)

There exist linearly independent vectors $x_{1},...,x_{n}$ in $V$ and complex
numbers $a_{1},...,a_{n}$ such that

$h(1,t)=\sum_{i=1}^{n}x_{i}\cdot e^{a_{i}t}$

for all $t$.

  * 4)

If $0\leq s_{1}\leq s_{2}\leq 1$, then $B(s_{2})\subseteq B(s_{1})$, where

$B(s):=\bigcup_{(t_{1},t_{2})\in[0,\infty)^{2}}\\{f\in(V\times
V)^{\prime}:f(h(s,t_{1}),h(s,t_{2}))=0\\}$

It is not too difficult to prove the implications $MSC\implies SSC\implies
SC$. The strong spiral conjecture clearly implies the spiral conjecture, but
it might not be so easy to see why the monotonic spiral conjecture implies the
strong spiral conjecture. The reason is that a small perturbation of the
homotopy can be made to avoid any relations with codimension greater than one
in $V\times V$, and the homotopy for the monotonic spiral conjecture will
always increase the set of codimension one relations that it avoids.

Morally speaking, the monotonic spiral conjecture should be true in the one-
dimensional case because curve shortening flow in a logarithmic metric should
have the desired property. This argument should at least cover the case of
smooth paths which are periodic deviations from a logarithmic spiral, but the
analysis is more complicated for arbitrary continuous paths. It would be
useful if there was a way to generalize such an argument to higher dimensions,
though it is not at all obvious how to achieve this. Regardless, we hope that
a deep enough understanding of relation avoiding paths will lead to new
progress on the inscribed square problem.

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