11institutetext:  Department of Mathematics, University of California at San
Diego
La Jolla, CA, 92093 USA
11email<EMAIL_ADDRESS>
#  Unavoidable Patterns in Complete Simple Topological Graphs††thanks:
Supported by NSF CAREER award DMS-1800746 and NSF award DMS-1952786.

Andrew Suk  Ji Zeng

###### Abstract

We show that every complete $n$-vertex simple topological graph contains a
topological subgraph on at least $(\log n)^{1/4-o(1)}$ vertices that is weakly
isomorphic to the complete convex geometric graph or the complete twisted
graph. This is the first improvement on the bound $\Omega(\log^{1/8}n)$
obtained in 2003 by Pach, Solymosi, and Tóth. We also show that every complete
$n$-vertex simple topological graph contains a plane path of length at least
$(\log n)^{1-o(1)}$.

###### Keywords:

Topological graph Unavoidable Patterns Plane path.

##  1 Introduction

A _topological graph_ is a graph drawn in the plane or, equivalently, on the
sphere, such that its vertices are represented by points and its edges are
represented by non-self-intersecting arcs connecting the corresponding points.
The arcs are not allowed to pass through vertices different from their
endpoints, and if two edges share an interior point, then they must properly
cross at that point in common. A topological graph is _simple_ if every pair
of its edges intersect at most once, either at a common endpoint or at a
proper crossing point. If the edges are drawn as straight-line segments, then
the graph is said to be _geometric_. If the vertices of a geometric graph are
in convex position, then it is called _convex_.

Simple topological graphs have been extensively studied [11, 13, 15, 17, 21],
and are sometimes referred to as _good drawings_ [1, 2], or simply as
_topological graphs_ [14]. In this paper, we are interested in finding large
unavoidable patterns in complete simple topological graphs. Two simple
topological graphs $G$ and $H$ are _isomorphic_ if there is a homeomorphism of
the sphere that transforms $G$ to $H$. We say that $G$ and $H$ are _weakly
isomorphic_ if there is an incidence preserving bijection between $G$ and $H$
such that two edges of $G$ cross if and only if the corresponding edges in $H$
cross as well. Clearly, any two complete convex geometric graphs on $m$
vertices are weakly isomorphic. Hence, let $C_{m}$ denote any complete convex
geometric graph with $m$ vertices.

By the famous Erdős-Szekeres convex polygon theorem [6] (see also [20]), every
complete $n$-vertex geometric graph contains a geometric subgraph on
$m=\Omega(\log n)$ vertices that is weakly isomorphic to $C_{m}$. (Note that
no three vertices in a complete geometric graph are collinear.) Interestingly,
the same is not true for simple topological graphs. The _complete twisted
graph_ $T_{m}$ is a complete simple topological graph on $m$ vertices with the
property that there is an ordering on the vertex set
$V(T_{m})=\\{v_{1},v_{2},\dots,v_{m}\\}$ such that edges $v_{i}v_{j}$ and
$v_{k}v_{\ell}$ cross if and only if $i<k<\ell<j$ or $k<i<j<\ell$. See Figure
1. It was first observed by Harborth and Mengerson [10] that $T_{m}$ does not
contain a topological subgraph that is weakly isomorphic to $C_{5}$. However,
in 2003, Pach, Solymosi, and Tóth [14] showed that it is impossible to avoid
both $C_{m}$ and $T_{m}$ in a sufficiently large complete simple topological
graph.

Figure 1: $C_{5}$ and $T_{5}$.

######  Theorem 1.1 (Pach-Solymosi-Tóth)

Every complete $n$-vertex simple topological graph contains a topological
subgraph on $m\geq\Omega(\log^{1/8}n)$ vertices that is weakly isomorphic to
$C_{m}$ or $T_{m}$.

The main result of this paper is the following improvement.

###### Theorem 1.2

Every complete $n$-vertex simple topological graph has a topological subgraph
on $m\geq(\log n)^{1/4-o(1)}$ vertices that is weakly isomorphic to $C_{m}$ or
$T_{m}$.

In the other direction, let us consider the following construction. Let
$V=\\{1,2,\dots,n\\}$ be $n$ vertices placed on the $x$-axis, and for each
pair $\\{i,j\\}\in V$, draw a half-circle connecting $i$ and $j$, with this
half-circle either in the upper or lower half of the plane uniformly randomly.
By applying the standard probabilistic method [3], one can show that there is
a complete $n$-vertex simple topological graph that does not contain a
topological subgraph on $m=\lceil 8\log n\rceil$ vertices that is weakly
isomorphic to $C_{m}$ or $T_{m}$. Another construction, observed by Scheucher
[18], is to take $n$ points in the plane with no $2\lceil\log n\rceil$ members
in convex position, and then draw straight-line segments between all pairs of
points.

It is not hard to see that both $C_{m}$ and $T_{m}$ contain a plane (i.e.
crossing-free) subgraph isomorphic to any given tree $T$ with at most $m$
vertices (see, e.g., [9]). Thus, as a corollary of Theorem 1.2, we obtain the
following.

###### Corollary 1

Every complete $n$-vertex simple topological graph contains a plane subgraph
isomorphic to any given tree $T$ with at most $(\log n)^{1/4-o(1)}$ vertices.

In the case when $T$ is a path, we improve this bound with the following
result, which is also recently obtained in [2] independently.

###### Theorem 1.3

Every complete $n$-vertex simple topological graph contains a plane path of
length at least $(\log n)^{1-o(1)}$.

In order to avoid confusion between topological and combinatorial edges, we
write $uv$ when referring to a topological edge in the plane, and write
$\\{u,v\\}$ when referring to an edge (pair) in a graph. Likewise, we write
$\\{u_{1},\dots,u_{k}\\}$ when referring to an edge ($k$-tuple) in a
$k$-uniform hypergraph. We systemically omit floors and ceilings whenever they
are not crucial for the sake of clarity in our presentation. All logarithms
are in base 2.

##  2 Monotone Paths and Online Ramsey Numbers

Before we prove Theorem 1.2, let us recall the following lemmas. Let $H$ be a
$k$-uniform hypergraph with vertex set $[n]=\\{1,2,\dots,n\\}$. We say that
$H$ contains a _monotone $k$-path_ of length $m$ if there are $m$ vertices
$v_{1}<v_{2}<\cdots<v_{m}$ such that $\\{v_{i},v_{i+1},\dots,v_{i+k-1}\\}\in
E(H)$ for $1\leq i\leq m-k+1$. We say that the edge set $E(H)$ is _transitive_
if for any $v_{1}<v_{2}<\cdots<v_{k+1}$ in $[n]$, the condition
$\\{v_{1},v_{2},\dots,v_{k}\\},\\{v_{2},v_{3},\dots,v_{k+1}\\}\in E(H)$
implies all $k$-element subsets of $\\{v_{1},\dots,v_{k+1}\\}$ are in $E(H)$.
We will need the following lemma due to Fox, Pach, Sudakov, and Suk.

######  Lemma 1 ([7])

Let $n>k$, and let $H$ be a $k$-uniform hypergraph with vertex set $[n]$,
which contains a monotone path of length $n$, that is,
$\\{i,i+1,\dots,i+k-1\\}\in E(H)$ for all $1\leq i\leq n-k+1$. If $E(H)$ is
transitive, then $H$ is the complete k-uniform hypergraph on $[n]$.

Next, we need a lemma from Online Ramsey Theory. The _vertex online Ramsey
game_ is a game played by two players, _builder_ and _painter_. Let $t\geq 1$
and suppose vertices $v_{1},v_{2},\dots,v_{t-1}$ are present. At the beginning
of stage $t$, a new vertex $v_{t}$ is added. Then for each
$v_{i}\in\\{v_{1},\dots,v_{t-1}$}, builder decides (in any order) whether to
create the edge $\\{v_{i},v_{t}\\}$. If builder creates the edge, then painter
has to immediately color it red or blue. When builder decides not to create
any more edges, stage $t$ ends and stage $t+1$ begins by adding a new vertex.
Moreover, builder must create at least one edge at every stage except for the
first one. The _vertex online Ramsey number_ $r(m)$ is the minimum number of
edges builder has to create to guarantee a monochromatic monotone path of
length $m$ in a vertex online Ramsey game. Clearly, we have $r(m)\leq
O(m^{4})$, which is obtained by having builder create all possible edges at
each stage and applying Dilworth’s theorem [5] on the $m^{2}$ vertices. Fox,
Pach, Sudakov, and Suk proved the following.

######  Lemma 2 ([7])

We have $r(m)=(1+o(1))m^{2}\log_{2}m$.

##  3 Convex Geometric Graph versus Twisted Graph

In this section, we prove the following theorem, from which Theorem 1.2
quickly follows.

###### Theorem 3.1

Let $m_{1},m_{2},n$ be positive integers such that

$9(m_{1}m_{2})^{2}\log(m_{1})\log(m_{2})<\log n.$

Then every complete $n$-vertex simple topological graph contains a topological
subgraph that is weakly isomorphic to $C_{m_{1}}$ or $T_{m_{2}}$.

###### Proof

Let $G=(V,E)$ be a complete $n$-vertex simple topological graph. Notice that
the edges of $G$ divide the plane into several cells (regions), one of which
is unbounded. We can assume that there is a vertex $v_{0}\in V$ such that
$v_{0}$ lies on the boundary of the unbounded cell. Indeed, otherwise we can
project $G$ onto a sphere, then choose an arbitrary vertex $v_{0}$ and then
project $G$ back to the plane such that $v_{0}$ lies on the boundary of the
unbounded cell, moreover, the new drawing is isomorphic to the original one as
topological graphs.

Consider the topological edges emanating out from $v_{0}$, and label their
endpoints $v_{1},\dots,v_{n-1}$ in clockwise order. For convenience, we write
$v_{i}\prec v_{j}$ if $i<j$. Given subsets
$U,W\subset\\{v_{1},\dots,v_{n-1}\\}$, we write $U\prec W$ if $u\prec w$ for
all $u\in U$ and $w\in W$. Following the notation used in [14], we color the
triples of $\\{v_{1},\dots,v_{n-1}\\}$ as follows. For $v_{i}\prec v_{j}\prec
v_{k}$, let $\chi(v_{i},v_{j},v_{k})=xyz$, where $x,y,z\in\\{0,1\\}$ such that

  1. 1.

setting $x=1$ if edges $v_{j}v_{k}$ and $v_{0}v_{i}$ cross, and let $x=0$
otherwise;

  2. 2.

setting $y=1$ if edges $v_{i}v_{k}$ and $v_{0}v_{j}$ cross, and let $y=0$
otherwise;

  3. 3.

setting $z=1$ if edges $v_{i}v_{j}$ and $v_{0}v_{k}$ cross, and let $z=0$
otherwise.

Pach, Solymosi, and Tóth observed the following.

######  Observation 1 ([14])

The only colors that appear with respect to $\chi$ are $000$, $001$, $010$,
and $100$.

Figure 2: Configurations for 000, 010, 001, 100 respectively.

See Figure 2 for an illustration. We now make another observation.

###### Lemma 3

Colors $001$ and $100$ are transitive. That is, for $v_{i}\prec v_{j}\prec
v_{k}\prec v_{\ell}$,

  1. 1.

if $\chi(v_{i},v_{j},v_{k})=\chi(v_{j},v_{k},v_{\ell})=001$, then
$\chi(v_{i},v_{j},v_{\ell})=\chi(v_{i},v_{k},v_{\ell})=001$;

  2. 2.

if $\chi(v_{i},v_{j},v_{k})=\chi(v_{j},v_{k},v_{\ell})=100$, then
$\chi(v_{i},v_{j},v_{\ell})=\chi(v_{i},v_{k},v_{\ell})=100$.

###### Proof

Suppose $\chi(v_{i},v_{j},v_{k})=\chi(v_{j},v_{k},v_{\ell})=001$. Since edges
$v_{0}v_{\ell}$ and $v_{j}v_{k}$ cross, vertex $v_{\ell}$ must lie in the
closed region bounded by edges $v_{j}v_{k}$, $v_{i}v_{j}$, and $v_{0}v_{k}$.
See Figure 3. Hence, edge $v_{0}v_{\ell}$ crosses both $v_{i}v_{j}$ and
$v_{i}v_{k}$. Therefore, we have
$\chi(v_{i},v_{j},v_{\ell})=\chi(v_{i},v_{k},v_{\ell})=001$ as wanted. If
$\chi(v_{i},v_{j},v_{k})=\chi(v_{j},v_{k},v_{\ell})=100$, a similar argument
shows that we must have
$\chi(v_{i},v_{j},v_{\ell})=\chi(v_{i},v_{k},v_{\ell})=100$.∎

Figure 3: The closed region bounded by edges $v_{j}v_{k}$, $v_{i}v_{j}$, and
$v_{0}v_{k}$ in Lemma 3.

Based on the coloring $\chi$, we define a coloring $\phi$ of the pairs of
$\\{v_{1},v_{2},\dots,v_{n-1}\\}$ as follows. For $v_{i}\prec v_{j}$, let
$\phi(v_{i},v_{j})=(a,b)$ where $a$ is the length of the longest monotone
3-path ending at $\\{v_{i},v_{j}\\}$ in color $100$, and $b$ is the length of
the longest monotone 3-path ending at $\\{v_{i},v_{j}\\}$ in color $001$. We
can assume that $a,b<m_{2}$. Otherwise, by Lemma 3 and Lemma 1, we would have
a subset $U\subset V$ of size $m_{2}$ whose triples are all of the same color,
$100$ or $001$. And it is not hard to argue by induction that such a $U$
corresponds to a topological subgraph that is weakly isomorphic to $T_{m_{2}}$
as wanted.

Before we continue, let us give a rough outline of the rest of the proof. In
what follows, we will construct disjoint vertex subsets
$V^{a,b}\subset\\{v_{1},\dots,v_{n-1}\\}$, where $1<a,b<m_{2}$, such that
$\phi$ colors every pair in $V^{a,b}$ with color $(a,b)$. For each $V^{a,b}$,
we will play the vertex online Ramsey game by letting the builder create an
edge set $E^{a,b}$ and designing a painter’s strategy, which gives rise to a
coloring $\psi$ on $E^{a,b}$. We then apply Lemma 2 to show that if $n$ is
sufficiently large, some vertex set $V^{a,b}$ will contain a monochromatic
monotone 2-path of length $m_{1}$ with respect to $\psi$. Finally, we will
show that this monochromatic monotone 2-path will correspond to a topological
subgraph that is weakly isomorphic to $C_{m_{1}}$. The detailed argument
follows.

For integers $t\geq 0$ and $1<a,b<m_{2}$, we construct a vertex subset
$V^{a,b}_{t}\subset\\{v_{1},\dots,v_{n-1}\\}$, an edge set $E^{a,b}_{t}$ of
pairs in $V^{a,b}_{t}$, and a subset $S_{t}\subset\\{v_{1},\dots,v_{n-1}\\}$
such that the following holds.

  1. 1.

We have $\sum\limits_{1<a,b<m_{2}}|V^{a,b}_{t}|=t$.

  2. 2.

For all $1<a,b<m_{2}$, we have $V^{a,b}_{t}\prec S_{t}$.

  3. 3.

For $u_{1}\in V^{a,b}_{t}$, we have $\phi(u_{1},u_{2})=(a,b)$ for every
$u_{2}\in V^{a,b}_{t}\cup S_{t}$ with $u_{1}\prec u_{2}$.

  4. 4.

For each edge $\\{u_{1},u_{2}\\}\in E^{a,b}_{t}$, where $u_{1}\prec u_{2}$, we
have $\chi(u_{1},u_{2},u_{3})=\chi(u_{1},u_{2},u_{4})$ for all $u_{3},u_{4}\in
V^{a,b}_{t}$ such that $u_{1}\prec u_{2}\prec u_{3}\prec u_{4}$.

We start by setting $V^{a,b}_{0}=\emptyset$ for all $1<a,b<m_{2}$, and
$S_{0}=\\{v_{1},\dots,v_{n-1}\\}$. After stage $t$, we have $V^{a,b}_{t}$,
$E^{a,b}_{t}$, for $1<a,b<m_{2}$, and $S_{t}$ as described above.

At the beginning of stage $t+1$, let $w_{t+1}$ be the smallest element in
$S_{t}$ with respect to $\prec$. By the pigeonhole principle, there exists
integers $1<\alpha,\beta<m_{2}$ and a subset $S_{t,0}\subset
S_{t}\setminus\\{w_{t+1}\\}$ of size at least $(|S_{t}|-1)/m_{2}^{2}$, such
that $\phi(w_{t+1},u)=(\alpha,\beta)$ for all $u\in S_{t,0}$. Then we set
$V^{\alpha,\beta}_{t+1}:=V^{\alpha,\beta}_{t}\cup\\{w_{t+1}\\}$. For all
$1<a,b<m_{2}$ with $(a,b)\neq(\alpha,\beta)$, we set
$V^{a,b}_{t+1}:=V^{a,b}_{t}$ and $E^{a,b}_{t+1}:=E^{a,b}_{t}$.

###### Claim 1

For all $u\in V^{\alpha,\beta}_{t}$ and $v\in S_{t,0}$, we have
$\chi(u,w_{t+1},v)\in\\{000,010\\}$.

###### Proof

For the sake of contradiction, suppose $\chi(u,w_{t+1},v)=100$, where $u\in
V^{\alpha,\beta}_{t}$ and $v\in S_{t,0}$. Since
$\phi(u,w_{t+1})=(\alpha,\beta)$, the longest monotone 3-path in color 100
ending at $\\{u,w_{t+1}\\}$ has length $\alpha$. Hence, the longest monotone
3-path in color 100 ending at $\\{w_{t+1},v\\}$ has length at least
$\alpha+1$. This contradicts the fact that $\phi(w_{t+1},v)=(\alpha,\beta)$. A
similar argument follows if $\chi(u,w_{t+1},v)=001$.∎

Now that we have constructed $V^{\alpha,\beta}_{t+1}$ by adding $w_{t+1}$ to
$V^{\alpha,\beta}_{t}$, we play the vertex online Ramsey game so that builder
chooses and creates edges of the form $\\{u,w_{t+1}\\}$, where $u\in
V^{\alpha,\beta}_{t}$, according to his strategy. After each edge
$\\{u,w_{t+1}\\}$ is created, painter immediately colors it
$\psi(u,w_{t+1})\in\\{000,010\\}$ as follows. In painter’s strategy, after the
$j$-th edge $\\{u_{j},w_{t+1}\\}$ is created and colored, a set
$S_{t,j}\subset S_{t,0}$ will be constructed such that all triples
$\\{u_{j},w_{t+1},v\\}$ with $v\in S_{t,j}$ are colored by $\chi$ with the
same color in $\\{000,010\\}$. After the $(j+1)$-th edge
$\\{u_{j+1},w_{t+1}\\}$ is created, painter looks at all triples of the form
$\\{u_{j+1},w_{t+1},v\\}$ with $v\in S_{t,j}$. Since
$\chi(u_{j+1},w_{t+1},v)\in\\{000,010\\}$ by Claim 1, the pigeonhole principle
implies that there exists a subset $S_{t,j+1}\subset S_{t,j}$ with size at
least $|S_{t,j}|/2$ such that all triples $\\{u_{j+1},w_{t+1},v\\}$ with $v\in
S_{t,j+1}$ are colored by $\chi$ with the same color $xyz\in\\{000,010\\}$.
Then painter sets $\psi(u_{j+1},w_{t+1})=xyz$.

If builder decides to stop creating edges from $w_{t+1}$ to
$V^{\alpha,\beta}_{t}$ after $j$ edges are created and colored, the stage ends
and we set $S_{t+1}=S_{t,j}$, and we let $E^{\alpha,\beta}_{t+1}$ be the union
of $E^{\alpha,\beta}_{t}$ and all edges built during this stage. Let $e_{t+1}$
denote the total number of edges builder creates in stage $t+1$. Recall that
$e_{t+1}\geq 1$ unless $V^{\alpha,\beta}_{t}=\emptyset$. As long as
$|S_{t+1}|>0$, we continue this construction process by starting the next
stage. Clearly, $V^{a,b}_{t+1}$, $E^{a,b}_{t+1}$, for all $1<a,b<m_{2}$, and
$S_{t+1}$ have the four properties described above. We now make the following
claim.

###### Claim 2

For $t\geq 1$, we have

$|S_{t}|\geq\frac{n-1}{m_{2}^{2t}\cdot
2^{\sum_{i=2}^{t}e_{i}}}-\sum\limits_{i=2}^{t}\frac{1}{m_{2}^{2(t+1-i)}\cdot
2^{\sum_{j=i}^{t}e_{j}}}.$

###### Proof

We proceed by induction on $t$. For the base case $t=1$, there’s no edge for
the builder to build in the first stage, so
$|S_{1}|=|S_{0,0}|\geq(n-1)/m_{2}^{2}$ as desired. For the inductive step,
assume the statement holds for $t\geq 1$. When we start stage $t+1$ and
introduce vertex $w_{t+1}$, the set $S_{t}$ shrinks to $S_{t,0}$ whose size is
guaranteed to be at least $(|S_{t}|-1)/m_{2}^{2}$, and each time builder
creates an edge from $w_{t+1}$ to $V^{\alpha,\beta}_{t}$, our set decreases by
a factor of two. Since builder creates $e_{t+1}$ edges during stage $t+1$, we
have

$\displaystyle|S_{t+1}|$
$\displaystyle\geq\frac{|S_{t}|-1}{m_{2}^{2}2^{e_{t+1}}}\geq\frac{n-1}{m_{2}^{2(t+1)}\cdot
2^{\sum_{i=2}^{t+1}e_{i}}}-\sum\limits_{i=2}^{t}\frac{1}{m_{2}^{2((t+1)+1-i)}\cdot
2^{\sum_{j=i}^{t+1}e_{j}}}-\frac{1}{m_{2}^{2}2^{e_{t+1}}}$
$\displaystyle=\frac{n-1}{m_{2}^{2(t+1)}\cdot
2^{\sum_{i=2}^{t+1}e_{i}}}-\sum\limits_{i=2}^{t+1}\frac{1}{m_{2}^{2((t+1)+1-i)}\cdot
2^{\sum_{j=i}^{t+1}e_{j}}},$

which is what we want.∎

After $t$ stages, builder has created a total of $\sum_{i=1}^{t}e_{i}$ edges,
such that each edge has color $000$ or $010$ with respect to $\psi$. If there
is no monochromatic 2-path of length $m_{1}$ with respect to $\psi$ on any
$(V^{a,b}_{t},E^{a,b}_{t})$, this implies that

$\sum_{i=1}^{t}e_{i}<m_{2}^{2}r(m_{1})\leq 2(m_{1}m_{2})^{2}\log m_{1}.$

Also, since $e_{i}\geq 1$ for all but $m_{2}^{2}$ many indices $1\leq i\leq
t$, we have

$t\leq m_{2}^{2}+\sum_{i=1}^{t}e_{i}<3(m_{1}m_{2})^{2}\log m_{1}.$

Since we assumed

$n>2^{9(m_{1}m_{2})^{2}\log(m_{1})\log(m_{2})},$

we have

$\displaystyle|S_{t}|$ $\displaystyle\geq\frac{n-1}{m_{2}^{2t}\cdot
2^{\sum_{i=2}^{t}e_{i}}}-\sum\limits_{i=2}^{t}\frac{1}{m_{2}^{2(t+1-i)}\cdot
2^{\sum_{j=i}^{t}e_{j}}}$
$\displaystyle\geq\frac{n-1}{2^{8(m_{1}m_{2})^{2}\log(m_{1})\log(m_{2})}}-\sum\limits_{i=2}^{t}\frac{1}{2^{t-i+1}}>1.$

Hence, we can continue to the next stage and introduce vertex $w_{t+1}$.
Therefore, when this process stops, say at stage $s$, we must have a
monochromatic monotone 2-path of length $m_{1}$ with respect to $\psi$ on some
$(V^{a,b}_{s},E^{a,b}_{s})$.

Now let $W^{*}=\\{w_{1}^{*},\dots,w_{m_{1}}^{*}\\}$, where
$w_{1}^{*}\prec\cdots\prec w_{m_{1}}^{*}$, be the vertex set that induces a
monochromatic monotone 2-path of length $m_{1}$ with respect to $\psi$ on
$(V^{a,b}_{s},E^{a,b}_{s})$. Since $\phi$ colors every pair in $W^{*}$ with
the color $(a,b)$, by following the proof of Claim 1, we have
$\chi(w_{i}^{*},w_{j}^{*},w_{k}^{*})\in\\{000,010\\}$ for every $i<j<k$.
Hence, the following argument due to Pach, Solymosi, and Tóth [14] shows that
$W^{*}$ induces a topological subgraph that is weakly isomorphic to
$C_{m_{1}}$. For the sake of completeness, we include the proof.

###### Claim 3

Let $W^{*}=\\{w^{*}_{1},\dots,w^{*}_{m_{1}}\\}$ be as described above. Then
$W^{*}$ induces a topological subgraph that is weakly isomorphic to
$C_{m_{1}}$.

###### Proof

Suppose $\psi(w^{*}_{i},w^{*}_{i+1})=000$ for all $i$. It suffices to show
that every triple in $W^{*}$ has color 000 with respect to $\chi$. For the
sake of contradiction, suppose we have $w^{*}_{i}\prec w^{*}_{j}\prec
w_{k}^{*}$ such that $\chi(w^{*}_{i},w^{*}_{j},w^{*}_{k})=010$, and let us
assume that $j-i$ is minimized among all such examples. Since
$\\{w_{i}^{*},w_{i+1}^{*}\\}\in E^{a,b}_{s}$, we have
$\chi(w^{*}_{i},w^{*}_{i+1},w^{*}_{k})=\psi(w^{*}_{i},w^{*}_{i+1})=000$. This
implies that $j>i+1$ and the edge $w^{*}_{i+1}w^{*}_{k}$ crosses
$v_{0}w^{*}_{j}$ (see Figure 4), which contradicts the minimality condition. A
similar argument follows if $\psi(w^{*}_{i},w^{*}_{i+1})=010$ for all $i$.∎

This completes the proof of Theorem 3.1∎

Figure 4: A figure illustrating Claim 3.

##  4 Plane Path

In this section, we prove Theorem 1.3. We will need the following lemma, which
was observed by Fulek and Ruiz-Vargas in [8].

###### Lemma 4

If a complete simple topological graph $G$ contains a topological subgraph
that is isomorphic to a plane $K_{2,m^{2}}$, then $G$ contains a plane path of
length $\Omega(m)$.

Let us briefly explain how to establish this lemma, as it is not explicitly
stated in [8]. In [22], Tóth proved that every $n$-vertex geometric graph with
more than $2^{9}k^{2}n$ edges contains $k$ pairwise disjoint edges. His proof
easily generalizes to simple topological graphs whose edges are drawn as
$x$-monotone curves, and, in fact, shows the existence of a plane path of
length $2k$.

Given a plane topological subgraph $K_{2,m^{2}}$ inside a complete simple
topological graph $G$, Fulek and Ruiz-Vargas [8] showed that there exists a
topological subgraph $G^{\prime}\subset G$, with $m^{2}$ vertices and
$\Omega(m^{4})$ edges, that is weakly-isomorphic to an $x$-monotone simple
topological graph $G^{\prime\prime}$. Hence, we can conclude Lemma 4 by
applying Tóth’s result stated above with $k=\Omega(m)$.

######  Proof (of Theorem 1.3)

First, we keep the following notations from the proof of Theorem 1.2. Let
$G=(V,E)$ be a complete $n$-vertex simple topological graph. We can assume
that there is a vertex $v_{0}\in V$ such that $v_{0}$ lies on the boundary of
the unbounded cell. We label the other vertices by $v_{1},\dots,v_{n-1}$ such
that the edges $v_{0}v_{i}$, for $1\leq i<n$, emanate out from $v_{0}$ in
clockwise order. We write $v_{i}\prec v_{j}$ if $i<j$, and color every triple
$v_{i}\prec v_{j}\prec v_{k}$ by
$\chi(v_{i},v_{j},v_{k})\in\\{000,010,100,001\\}$.

For each $v_{i}$, we arrange the vertices $\\{v_{i+1},\dots,v_{n-1}\\}$ into a
sequence $\theta(v_{i})=(v_{j_{1}},\dots,v_{j_{n-1-i}})$ such that the
topological edges
$v_{i}v_{0},v_{i}v_{j_{1}},v_{i}v_{j_{2}},\dots,v_{i}v_{j_{n-1-i}}$ emanate
out from $v_{i}$ in counterclockwise order. See Figure 5. We call a sequence
of vertices $S=(v_{i_{1}},\dots,v_{i_{k}})$ increasing (or decreasing) if
$v_{i_{1}}\prec v_{i_{2}}\prec\dots\prec v_{i_{k}}$ (or $v_{i_{1}}\succ
v_{i_{2}}\succ\dots\succ v_{i_{k}}$).

Figure 5: An example with $\theta(v_{1})=(v_{4},v_{3},v_{2},v_{5})$.

###### Lemma 5

If there exists a vertex $u$ such that $\theta(u)$ contains an increasing
subsequence $(u_{1},\dots,u_{m^{2}})$, then the edges $v_{0}u_{i}$ and
$uu_{i}$, for all $1\leq i\leq m^{2}$, form a plane subgraph $K_{2,m^{2}}$.

###### Proof

It suffices to show $v_{0}u_{i}$ and $uu_{j}$ do not cross each other for
every $1\leq i,j\leq k$. When $i=j$, this follows from $G$ being simple. When
$j>i$, by the increasing assumption, the edges $uv_{0}$, $uu_{i}$, and
$uu_{j}$ emanate out from $u$ in counterclockwise order. Observe that this
condition forces $u_{j}$ to be outside the region $\Delta_{v_{0}uu_{i}}$
bounded by the topological edges $v_{0}u$, $uu_{i}$, and $u_{i}v_{0}$. Then
the Jordan arc $uu_{j}$ starting at $u$, initially outside
$\Delta_{v_{0}uu_{i}}$, cannot enter $\Delta_{v_{0}uu_{i}}$ then leave again
to end at $u_{j}$. In particular, $uu_{j}$ doesn’t cross $v_{0}u_{i}$. See
Figure 6 for an illustration. A similar argument follows if $j<i$. ∎

Figure 6: An increasing subsequence of $\theta(u)$ induce a plane
$K_{2,m^{2}}$.

We set $m=\left\lfloor\frac{\log n}{2\log\log n}\right\rfloor$ and prove that
$G$ contains a plane path of length $\Omega(m)$. We can assume $m>1$,
otherwise there’s nothing to prove. If some sequence $\theta(v_{i})$ contains
an increasing subsequence of length $m^{2}$, then by Lemma 5 and Lemma 4, we
are done. Therefore, we assume that $\theta(v_{i})$ doesn’t contain an
increasing subsequence of length $m^{2}$ for every $i$.

For integer $t\geq 1$, we inductively construct subsets
$U_{t},S_{t}\subset\\{v_{1},\dots,v_{n-1}\\}$ with
$U_{t}=\\{u_{1},\dots,u_{t}\\}$, where $u_{1}\prec\dots\prec u_{t}$, and
$U_{t}\prec S_{t}$. Initially we set $U_{1}=\\{u_{1}:=v_{1}\\}$ and
$S_{1}=\\{v_{2},\dots,v_{n-1}\\}$. Suppose for some $t$, we have already
constructed $U_{t}$ and $S_{t}$. If $|S_{t}|\leq m^{2}$, we stop this
construction process, otherwise we continue to construct $U_{t+1}$ and
$S_{t+1}$ as follows: Let $\theta^{\prime}$ be the subsequence of
$\theta(u_{t})$ that contains exactly those vertices in $S_{t}$. Note that the
length of $\theta^{\prime}$ equals to $|S_{t}|$. According to our assumption,
the length of the longest increasing subsequence in $\theta^{\prime}$ is less
than $m^{2}$. Hence, by Dilworth’s theorem [5], $\theta^{\prime}$ contains a
decreasing subsequence of length at least $|S_{t}|/m^{2}$. Let
$S^{\prime}_{t+1}$ be the set of vertices that appear in this decreasing
subsequence of $\theta^{\prime}$. Next, we take $u_{t+1}$ to be the smallest
element of $S^{\prime}_{t+1}$ with respect to $\prec$ and let
$U_{t+1}:=U_{t}\cup\\{u_{t+1}\\}$. Consider the region
$\Delta_{v_{0}u_{t}u_{t+1}}$ bounded by the topological edges $v_{0}u_{t}$,
$u_{t}u_{t+1}$, and $u_{t+1}v_{0}$. Each vertex in
$S^{\prime}_{t+1}\setminus\\{u_{t+1}\\}$ is either inside or outside
$\Delta_{v_{0}u_{t}u_{t+1}}$. So, by the pigeonhole principle, there exists a
subset $S_{t+1}\subset S^{\prime}_{t+1}\setminus\\{u_{t+1}\\}$ with
$|S_{t+1}|\geq|S^{\prime}_{t+1}\setminus\\{u_{t+1}\\}|/2$ such that the whole
set $S_{t+1}$ is either inside or outside $\Delta_{v_{0}u_{t}u_{t+1}}$.
Clearly, we have $U_{t+1}\prec S_{t+1}$ and

$|S_{t+1}|\geq\frac{|S_{t}|/m^{2}-1}{2}\geq\frac{|S_{t}|}{(2m)^{2}}.$

Using the inequality above and the fact that $|S_{1}|=n-2$, we can inductively
prove $|S_{t}|\geq\frac{n}{(2m)^{2t}}$. When $t=m-1$, this gives us

$|S_{m-1}|\geq\frac{n}{(2m)^{2(m-1)}}>\frac{n}{(2m)^{\log n/\log\log
n-2}}>m^{2}\cdot\frac{n}{(\log n)^{\log n/\log\log n}}\geq m^{2}.$

Hence, the construction process ends at a certain $t$ larger than $m-1$, and
we will always construct $U_{m}=\\{u_{1},\dots,u_{m}\\}$.

Now we show that $u_{i}u_{i+1}$, for $1\leq i<m$, form a plane path. Our
argument is based on the following two claims.

###### Claim 4

For any vertices $u_{i}\prec u_{i+1}\prec u_{j}\prec u_{k}$, we have $u_{j}$
and $u_{k}$ either both inside or both outside the region
$\Delta_{v_{0}u_{i}u_{i+1}}$.

Claim 4 is obviously guaranteed by the construction process of $U_{m}$.

###### Claim 5

For any vertices $u_{i}\prec u_{j}\prec u_{k}$, the topological edges
$v_{0}u_{i}$ and $u_{j}u_{k}$ do not cross each other.

###### Proof

Consider the region $\Delta_{v_{0}u_{i}u_{j}}$ bounded by the topological
edges $v_{0}u_{i}$, $u_{i}u_{j}$, and $u_{j}v_{0}$, then $u_{k}$ is either
inside or outside $\Delta_{v_{0}u_{i}u_{j}}$. If $u_{k}$ is inside
$\Delta_{v_{0}u_{i}u_{j}}$, then $v_{0}u_{k}$ must cross $u_{i}u_{j}$. By
Observation 1, we have $\chi(u_{i},u_{j},u_{k})=001$, which implies
$v_{0}u_{i}$ and $u_{j}u_{k}$ do not cross. See the third configuration in
Figure 2.

Suppose $u_{k}$ is outside $\Delta_{v_{0}u_{i}u_{j}}$. By the construction
process of $U_{m}$, the edges $u_{i}v_{0}$, $u_{i}u_{k}$ and $u_{i}u_{j}$ must
emanate from $u_{i}$ in counterclockwise order, this implies that $u_{i}u_{k}$
crosses $v_{0}u_{j}$. Then, by Observation 1, $\chi(u_{i},u_{j},u_{k})=010$
and $u_{j}u_{k}$ doesn’t cross $v_{0}u_{i}$. See the second configuration in
Figure 2. ∎

Figure 7: For $u_{i}u_{i+1}$ and $u_{j}u_{j+1}$ with $i+1<j$ to cross each
other, either $u_{j}$ and $u_{j+1}$ are not both inside or both outside
$\Delta_{v_{0}u_{i}u_{i+1}}$ (left graph), or the topological edge
$u_{j}u_{j+1}$ crosses one edge in $\\{v_{0}u_{i},v_{0}u_{i+1}\\}$ (right
graph).

Finally, we argue that the edges $u_{i}u_{i+1}$ and $u_{j}u_{j+1}$ do not
cross for any $i<j$. When $j=i+1$, this follows from $G$ being simple. When
$j>i+1$, by Claim 4, the vertices $u_{j}$ and $u_{j+1}$ are either both inside
or both outside the region $\Delta_{v_{0}u_{i}u_{i+1}}$. So, the edge
$u_{j}u_{j+1}$ crosses the boundary of $\Delta_{v_{0}u_{i}u_{i+1}}$ an even
number of times. On the other hand, by Claim 5, $u_{j}u_{j+1}$ doesn’t cross
$v_{0}u_{i}$ or $v_{0}u_{i+1}$. So $u_{j}u_{j+1}$ doesn’t cross
$u_{i}u_{i+1}$. See Figure 7 for an illustration. This concludes the proof of
Theorem 1.3. ∎

##  5 Concluding Remarks

Answering a question of Pach and Tóth [15], Suk showed that every complete
$n$-vertex simple topological graph contains $\Omega(n^{1/3})$ pairwise
disjoint edges [19] (see also [8]). This bound was later improved to
$n^{1/2-o(1)}$ by Ruiz-Vargas in [16]. Hence, for plane paths, we conjecture a
similar bound should hold.

###### Conjecture 1

There is an absolute constant $\varepsilon>0$, such that every complete
$n$-vertex simple topological graph contains a plane path of length
$n^{\varepsilon}$.

Let $h=h(n)$ be the smallest integer such that every complete $n$-vertex
simple topological graph contains an edge crossing at most $h$ other edges. A
construction due to Valtr (see page 398 in [4]) shows that
$h(n)\geq\Omega(n^{3/2})$. In the other direction, Kynčl and Valtr [12] used
an asymmetric version of Theorem 1.1 to show that $h(n)=O(n^{2}/\log^{1/4}n)$.
By using Theorem 3.1 instead, their arguments show that $h(n)\leq n^{2}/(\log
n)^{1/2-o(1)}$. We conjecture the following.

###### Conjecture 2

There is an absolute constant $\varepsilon>0$ such that $h(n)\leq
n^{2-\varepsilon}$.

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frequency and relative intensity of several vibrational modes of BH2
complexes. These results are expected to provide guidance for future
experiments toward a better identification of these centers.

Finally, regarding the LeTID of boron-doped Si solar cells, and in the light
of our results, we recall the sequence of reactions proposed by de Guzman _et
al._ [59] for the degradation and recovery steps under dark annealing
conditions,

$\begin{array}[]{rcl}\textrm{Latent}&&\hskip
11.38092pt\textrm{H}_{2}+2\textrm{B}^{-}+2\textrm{h}^{+}\\\ &&\hskip
39.83368pt\bigg{\downarrow}\hskip
5.69046pt(1)\>\>\tau,\rho\>\textrm{degradation}\\\ \textrm{Degraded}&&\hskip
14.22636pt\textrm{B}\textrm{H}_{2}^{+}(\textrm{D})+\textrm{B}^{-}\\\ &&\hskip
39.83368pt\bigg{\downarrow}\hskip
5.69046pt(2)\>\>\tau\>\textrm{regeneration}\\\ \textrm{Passivated}&&\hskip
31.2982pt2\textrm{BH}\\\ &&\hskip 39.83368pt\bigg{\downarrow}\hskip
5.69046pt(3)\>\>\rho\>\textrm{recovery}\\\
\textrm{Recovered}&&\textrm{H}_{2}(\textrm{sink})+2\textrm{B}^{-}+2\textrm{h}^{+}\end{array}$

The above reaction sequence adds a _Degraded_ state to the three-state-model
of Voronkov and Falster [41] (see Eq. 1), which has been used to explain
changes in resistivity of hydrogenated boron doped Si crystals during
heat/illumination treatments. We refer to the three states of the original
model as _Latent_ , _Passivated_ and _Recovered_. LeTID initiates from a
Latent state, where H2 molecules linger in the as-fired/quenched devices.
Lifetime ($\tau$) degradation occurs during step (1) when heat is provided,
and H2 is able to escape from trapping sites, most notably O impurities, and
react with B to form BH2. We note that already at this stage, for each
BH${}_{2}^{+}$ complex that is formed, two free-holes are subtracted. This
means that the resistivity ($\rho$) change and lifetime degradation are
concurrent, and any measured resistivity increase rate, also reflects the
lifetime degradation rate. The calculated activation energy for step (1) is
1.1 eV and corresponds to the dissociation of H2 next to B. This quantity
compares well with the LeTID activation energy of 1.08 eV from the study of
Vargas $et\leavevmode\nobreak\ al.$ [35].

The recovery of lifetime and resistivity occur separately. Lifetime
regeneration involves the annealing of BH${}_{2}^{+}(\textrm{D})$ and
formation of BH pairs in step (2) According to our calculations, the
activation energy for
$\textrm{BH}_{2}^{+}(\textrm{D})\rightarrow\textrm{BH}+\textrm{H}^{+}$ is 0.8
eV and the proton should be captured by another boron atom. Importantly, step
(2) does not involve a change in the free-hole concentration, but rather
attaining a passivated state with a weak recombination activity.

Following Refs. [41, 45], one may speculate that the recovery of resistivity
upon prolonged dark annealings, involves the dissociation of BH pairs and
subsequent trapping of hydrogen at some kind of sink. This would explain the
full recovery of the devices and reactivation of the B dopants as described by
step (3). The calculated dissociation barrier of BH (1.2 eV) is close to the
measured counterpart (1.28 eV [60]) and also close to the activation energy
for the recovery of resistivity (1.1-1.3 eV [35, 44, 49]).

##  Appendix A Configurational entropy for
$\textrm{B}^{-}+\textrm{H}^{+}\rightarrow\textrm{BH}$

For the evaluation of configurational entropy, we have to estimate the total
number of equivalent microstates in the reactants and products sides of the
reaction of interest, $W_{\textrm{R}}$ and $W_{\textrm{P}}$ , respectively.
This is only tractable upon consideration of a few assumptions.

Firstly, we consider a crystalline sample made of $n_{\textrm{Si}}$ silicon
sites, containing $n_{\textrm{B}}$ substitutional boron acceptors and
$n_{\textrm{H}}$ hydrogen atoms subject to $n_{\textrm{Si}}\gg
n_{\textrm{B}}\gg n_{\textrm{H}}$;

Next, because hydrogen is a fast diffuser and boron is strongly _anchored_ to
the lattice, at low temperatures all hydrogen is inferred to be trapped by
boron (in the form of BH pairs), so that the number of pairs is also
$n_{\textrm{H}}$. The remaining boron is electrically activated;

At high temperatures, all pairs dissociate into uncorrelated B- and H+
elements. The number of free carriers is conserved along the process and no
change of electronic entropy is considered;

Finally, at low temperatures, BH pairing is only considered to occur upon
formation of first neighboring complexes. Any other configurations are, for
the sake of the calculation of configurational entropy, equivalent to isolated
B- and H+ species.

The relevant reaction is therefore,

$n_{\textrm{B}}\textrm{B}^{-}+n_{\textrm{H}}\textrm{H}{}^{+}\rightarrow(n_{\textrm{B}}-n_{\textrm{H}})\textrm{B}{}^{-}+n_{\textrm{H}}\textrm{BH}.$
(15)

When all BH are dissociated, the number of microstates for the reactants side
is found by first distributing the $n_{\textrm{B}}$ acceptors (there are
$n_{\textrm{Si}}!/(n_{\textrm{Si}}-n_{\textrm{B}})!n_{\textrm{B}}!$ ways to do
that), and for each boron-related microstate we distribute the
$n_{\textrm{H}}$ hydrogen atoms among the available
$2n_{\textrm{Si}}-4n_{\textrm{B}}$ bond center sites not adjacent to boron.
The total number of ways to perform the two operations is,

$W_{\textrm{R}}=\frac{n_{\textrm{Si}}!}{(n_{\textrm{Si}}-n_{\textrm{B}})!n_{\textrm{B}}!}\times\frac{(2n_{\textrm{Si}}-4n_{\textrm{B}})!}{(2n_{\textrm{Si}}-4n_{\textrm{B}}-n_{\textrm{H}})!n_{\textrm{H}}!}.$
(16)

On the other hand, when all H is trapped by boron, the total number of
possible microstates for the products side is given by the number of ways to
distribute the $n_{\textrm{B}}$ boron atoms times the number of ways to
distribute $n_{\textrm{H}}$ hydrogen atoms on $4n_{\textrm{B}}$ bond centered
sites next to boron,

$W_{\textrm{P}}=\frac{n_{\textrm{Si}}!}{(n_{\textrm{Si}}-n_{\textrm{B}})!n_{\textrm{B}}!}\times\frac{(4n_{\textrm{B}})!}{(4n_{\textrm{B}}-n_{\textrm{H}})!n_{\textrm{H}}!}.$
(17)

Using Stirling’s approximation, we find that the total entropy change is
linear with $n_{\textrm{H}}$ , and we may derive the change of configurational
entropy per BH pair as,

$\Delta
S_{\textrm{conf}}=\frac{k_{\textrm{B}}}{n_{\textrm{H}}}\ln\left(\frac{W_{\textrm{P}}}{W_{\textrm{R}}}\right)=k_{\textrm{B}}\ln(2c_{\textrm{B}}),$
(18)

where $k_{\textrm{B}}$ is the Boltzmann constant and
$c_{\textrm{B}}=n_{\textrm{B}}/n_{\textrm{Si}}$ is the fractional
concentration of boron. If we were to consider up to second neighboring bonds
to distinguish between pairing formation/dissociation, the factor of 2 next to
$c_{\textrm{B}}$ would have to be replaced by 8, thus having a minute impact
on the final result.

##  Appendix B Configurational entropy for
$\textrm{B}^{-}+\tfrac{1}{2}\textrm{H}_{2}+\textrm{h}^{+}\rightarrow\textrm{BH}$

Now we implicitly assume that multi-trapping of H at B is negligible, so that
the number of resulting pairs is also $n_{\textrm{H}}$, and that after the
reaction, the remaining $(n_{\textrm{B}}-n_{\textrm{H}})$ boron ions are
ionized. Again, taking $n_{\textrm{Si}}\gg n_{\textrm{B}}\gg n_{\textrm{H}}$,
the relevant reaction is,

$n_{\textrm{B}}(\textrm{B}^{-}+\textrm{h}^{+})+\frac{n_{\textrm{H}}}{2}\textrm{H}_{2}\rightarrow(n_{\textrm{B}}-n_{\textrm{H}})(\textrm{B}^{-}+\textrm{h}^{+})+n_{\textrm{H}}\textrm{BH},$
(19)

which highlights the fact that each hydrogen molecule is capable of
subtracting two holes from the sample. Since the number of free holes is not
conserved across reaction 19, the effect of electronic entropy should be
assessed as well. In this appendix we evaluate the configurational term.

For the reactants side, we assume that the $n_{\textrm{B}}$ boron ions and the
$n_{\textrm{H}}/2$ molecules are perfectly uncorrelated, and they can occupy
any of the $n_{\textrm{Si}}$ substitutional and $n_{\textrm{Si}}$ tetrahedral
interstitial sites, respectively. The number of ways to distribute such
populations is

$W_{\textrm{R}}=\frac{n_{\textrm{Si}}!}{(n_{\textrm{Si}}-n_{\textrm{B}})!n_{\textrm{B}}!}\times\frac{2^{n_{\textrm{H}}/2}n_{\textrm{Si}}!}{(n_{\textrm{Si}}-n_{\textrm{H}}/2)!(n_{\textrm{H}}/2)!},$
(20)

where the factor of $2^{n_{\textrm{H}}/2}$ accounts for the two possibilities
that we have to form each molecule.

As for the number of microstates in the products side ($W_{\textrm{P}}$) we
can use Eq. 17 from Appendix A. Hence, the configurational entropy change per
BH pair can be approximated to

$\Delta
S_{\textrm{conf}}=\frac{k_{\textrm{B}}}{n_{\textrm{H}}}\ln\left(\frac{W_{\textrm{P}}}{W_{\textrm{R}}}\right)=\frac{k_{\textrm{B}}}{2}\left[\ln\left(\frac{8c_{\textrm{B}}^{2}}{c_{\textrm{H}}}\right)+1\right],$
(21)

where $c_{\textrm{B}}=n_{\textrm{B}}/n_{\textrm{Si}}$ and
$c_{\textrm{H}}=n_{\textrm{H}}/n_{\textrm{Si}}$ are the fractional
concentrations of boron and hydrogen, respectively.

##  Appendix C Configurational entropy for
$\textrm{C}+\textrm{H}_{2}\rightarrow\textrm{CH}_{2}$

It is known that CH pairs are not very stable. They dissociate at $T\sim 100$
°C in darkness and even below 230 K upon illumination with above band gap
light [127]. Therefore, we assume that interactions between H2 and C upon
quenching from H2 exposure at high temperatures, essentially lead to formation
of CH2 complexes [110].

Again, taking $n_{\textrm{Si}}\gg n_{\textrm{C}}\gg n_{\textrm{H}}$, where
$n_{\textrm{C}}$ is the number of substitutional carbon impurities in the
sample, and the number of resulting CH2 complexes as $n_{\textrm{H}}/2$, the
relevant reaction is,

$n_{\textrm{C}}\textrm{C}+(n_{\textrm{H}}/2)\textrm{H}_{2}\rightarrow(n_{\textrm{C}}-n_{\textrm{H}}/2)\textrm{C}+(n_{\textrm{H}}/2)\textrm{CH}_{2},$
(22)

On the reactants side, we have
$n_{\textrm{Si}}!/(n_{\textrm{Si}}-n_{\textrm{C}})!n_{\textrm{C}}!$ ways to
distribute the C atoms while H2 molecules can occupy tetrahedral interstitial
sites. Considering also that there are two ways to form each molecule (see Eq.
20), the total number of microstates becomes,

$W_{\textrm{R}}=\frac{n_{\textrm{Si}}!}{(n_{\textrm{Si}}-n_{\textrm{C}})!n_{\textrm{C}}!}\times\frac{2^{n_{\textrm{H}}/2}n_{\textrm{Si}}!}{(n_{\textrm{Si}}-n_{\textrm{H}}/2)!(n_{\textrm{H}}/2)!}.$
(23)

Conversely, when all H2 is trapped at the C atoms, the total number of
possible microstates for the products side is now the number of ways to
distribute the $n_{\textrm{C}}$ carbon atoms times the number of ways to
combine them with the $n_{\textrm{H}}/2$ pairs of H atoms. Considering that
each CH2 complex has four-fold orientational degeneracy, and again the two
possibilities for the formation of each complex (due to the existence of two H
atoms), one finds,

$W_{\textrm{P}}=\frac{n_{\textrm{Si}}!}{(n_{\textrm{Si}}-n_{\textrm{C}})!n_{\textrm{C}}!}\times\frac{2^{n_{\textrm{H}}/2}(4n_{\textrm{C}})!}{(4n_{\textrm{B}}-n_{\textrm{H}}/2)!(n_{\textrm{H}}/2)!}.$
(24)

Following the same procedure as in Appendix A and B, we have the
configurational entropy change per H atom,

$\Delta
S_{\textrm{conf}}=\frac{k_{\textrm{B}}}{2}\ln\left(4c_{\textrm{C}}\right),$
(25)

where $c_{\textrm{C}}=n_{\textrm{C}}/n_{\textrm{Si}}$ is the fractional
concentrations carbon.

##  Appendix D Electronic free energy

For solid state reactions where there is a variation in the number of free
carriers, the contribution of electronic entropy to the free energy change can
be comparable to other terms. This is particularly relevant when we are
dealing with reactions that involve the thermal promotion of carriers, such as
$\textrm{D}^{0}\rightarrow\textrm{A}^{-}+\textrm{h}^{+}$, where D is a deep or
electronically inactive defect, while A stands for an arbitrary shallow
acceptor, and $\textrm{h}^{+}$ a free hole in thermal equilibrium with the
sample.

We restrict our analysis to the low doping regime at room temperature and up
to few hundred Kelvin (LeTID conditions), where virtually all shallow
acceptors (boron or BH2(A) complexes) are ionized. At these temperatures,
intrinsic excitations are negligible and that is reflected in the excellent
agreement between the calculated specific heat, obtained from vibrational
degrees of freedom only, and the measurements [86, 91]. For low doping
conditions, the effective density of valence band states largely exceeds the
free hole concentration ($N_{\textrm{v}}\gg p$) and we can apply Boltzmann
statistics. For instance, for the reaction

$\textrm{B}^{-}+\tfrac{1}{2}\textrm{H}_{2}+\textrm{h}^{+}\rightarrow\textrm{BH},$
(26)

the electronic free energy change is estimated from that of a free-hole gas
with density $\Delta p=fc_{\textrm{H}}[\textrm{Si}]$, where
$f=\exp(-E_{\textrm{h}}/k_{\textrm{B}}T)$ is the Boltzmann distribution
function, and $E_{\textrm{h}}$ the hole binding energy to the acceptor
($E_{\textrm{h}}=46$ meV for the boron acceptor [77]).

In the classical limit, and as proposed by Estreicher _et al._ [86], the
electronic free energy change due to thermal release of a hole from a shallow
state into the electronic thermal bath of the sample, is given by [128]

$\Delta F_{\textrm{elec,h}}=f(\mu-k_{\textrm{B}}T),$ (27)

where $\mu$ is the electronic chemical potential with respect to the valence
band top. Equation 27 finds the Helmholtz free energy by simply subtracting
$pV$ (which is $k_{\textrm{B}}T$ for a classical gas of non-interacting holes)
to the Gibbs free per hole ($\mu$).

The chemical potential is approximated to $\mu=k_{\textrm{B}}T\ln(\Delta
p/N_{\textrm{v}})$, where $N_{\textrm{v}}$ is obtained numerically in the
range $T=200\textrm{-}500$ K as $N_{\textrm{v}}=3.10\times
10^{19}(T/300)^{1.85}$ cm-3 [129]. For the temperatures of interest, i.e.,
$T\approx 300\textrm{-}500$ K, we have $f\approx 1$ and provided that $\Delta
p\ll p\ll N_{\textrm{v}}$, equation 27 allows us to estimate the electronic
free energy variation due to small increase of free holes, across a reaction
that involves the full ionization/passivation of shallow acceptors.

###### Acknowledgements.

We acknowledge the FCT through projects LA/P/0037/2020, UIDB/50025/2020,
UIDP/50025/2020 and 2021.09643.CPCA (Advanced Computing Project using the
Oblivion supercomputer). The work in the UK was funded by EPSRC via grant
EP/TO25131/1.

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