Fixed the clique-sum argument.

comparable-box-dimension.tex

@@ -100,63 +100,140 @@ or expressible in the first-order logic~\cite{logapx}.
 \section{Operations}
 
 Let us start with a simple lemma saying that addition of a vertex increases the comparable box dimension by at most one.
+In particular, this implies that $\cbdim(G)\le |V(G)$.
 \begin{lemma}\label{lemma-apex}
 For any graph $G$ and $v\in V(G)$, we have $\cbdim(G)\le \cbdim(G-v)+1$.
 \end{lemma}
 \begin{proof}
-Let $f$ be a comparable box representation of $G-v$ in $\mathbb{R}^d$, where $d=\cbdim(G-v)$.
+Let $f$ be a touching representation of $G-v$ by comparable boxes in $\mathbb{R}^d$, where $d=\cbdim(G-v)$.
 For each $u\in V(G)\setminus\{v\}$, let $h(u)=[0,1]\times f(u)$ if $uv\in E(G)$ and 
 $h(u)=[1/2,3/2]\times f(u)$ if $uv\not\in E(G)$.  Let $h(v)=[-1,0]\times [-M,M] \times \cdots \times [-M,M]$,
 where $M$ is chosen large enough so that $f(u)\subseteq [-M,M] \times \cdots \times [-M,M]$ for every $u\in V(G)\setminus\{v\}$.
-Then $h$ is a comparable box representation of $G$ in $\mathbb{R}^{d+1}$.
+Then $h$ is a touching representation of $G$ by comparable boxes in $\mathbb{R}^{d+1}$.
 \end{proof}
 
-Next, let us deal with clique-sums.  A \emph{clique-sum} of two graphs $G_1$ and $G_2$ is obtained from their disjoint union
-by identifying vertices of a clique in $G_1$ and a clique of the same size in $G_2$ and possibly
-deleting some of the edges of the resulting clique.  The main issue to overcome in obtaining a representation for a clique-sum
-is that the representations of $G_1$ and $G_2$ can be ``degenerate''. Consider e.g. the case that $G_1$ is represented
-by unit squares arrangef in a grid; in this case, there is no space to attach $G_2$ at the cliques formed by four squares intersecting
-in a single corner.  This can be avoided by increasing the dimension, but we need to be careful so that the dimension stays bounded
-even after, motivating the following definition.
+%We will need the fact that the chromatic number is at most exponential in the comparable box dimension;
+%this follows from~\cite{subconvex} and we include the argument to make the dependence clear.
+%\begin{lemma}\label{lemma-chrom}
+%If $G$ has a comparable box representation $f$ in $\mathbb{R}^d$, then $G$ is $3^d$-colorable.
+%\end{lemma}
+%\begin{proof}
+%We actually show that $G$ is $(3^d-1)$-degenerate.  Since every induced subgraph of $G$ also
+%has a comparable box representation in $\mathbb{R}^d$, it suffices to show that the minimum degree of $G$
+%is less than $3^d$.  Let $v$ be a vertex of $G$ such that $f(v)$ has the smallest volume.  For every neighbor $u$ of $v$,
+%there exists a translation $B_u$ of $f(v)$ such that $B_u\subseteq f(u)$ and $B_u$ touches $f(v)$.
+%Note that $f(v)\cup \bigcup_{u\in N(v)} B_u$ is a union of internally disjoint translations of $f(v)$ contained in
+%a box obtained from $f(v)$ by scaling it by a factor of three, and thus $1+|N(v)|\le 3^d$.
+%\end{proof}
+
+We need a bound on the clique number in terms of the comparable box dimension.
+For a box $B=I_1\times \cdots I_d$ and $i\in\{1,\ldots,d\}$, let $B[i]=I_i$.
+\begin{lemma}\label{lemma-cliq}
+If $G$ has a touching representation $f$ by comparable boxes in $\mathbb{R}^d$, then $\omega(G)\le 2^d$.
+\end{lemma}
+\begin{proof}
+...
+\end{proof}
 
-For a box $B=I_1\times \cdot\times I_d$ and $i\in\{1,\ldots,d\}$ let $B[i]$ denote the interval $I_i$.
-Let $B_1$, \ldots, $B_k$ be pairwise touching boxes in $\mathbb{R}^d$.
-A box $B$ \emph{touches $B_1$, \ldots, $B_k$ generically} if there exist distinct $i_1,\ldots, i_k\in\{1,\ldots, d\}$ such that for
-$j=1,\ldots, k$,
-\begin{itemize}
-\item $B[i_j]\cap B_j[i_j]$ consists of a single point, and
-\item for every $i\in\{1,\ldots,d\}\setminus \{i_j\}$, $B[i]\cap B_j[i]$ is a non-empty interval of non-zero length.
-\end{itemize}
-A clique $K$ in a touching box representation $f$ of a graph $G$ in $\mathbb{R}^d$ is \emph{exposed} if there exists a box $B$
+A \emph{tree decomposition} of a graph $G$
+is a pair $(T,\beta)$, where $T$ is a rooted tree and $\beta:V(T)\to 2^{V(G)}$ assigns a \emph{bag} to each of its nodes,
 such that
 \begin{itemize}
-\item the interior of $B$ is disjoint from $f(v)$ for every $v\in V(G)$,
-\item $B$ touches the boxes $\{f(u):u\in K\}$ generically, and
-\item there exists $i\in\{1,\ldots,d\}$ such that $B[i]\subseteq f(u)[i]$ for every $u\in K$.
+\item for each $uv\in E(G)$, there exists $x\in V(T)$ such that $u,v\in\beta(x)$, and
+\item for each $v\in V(G)$, the set $\{x\in V(T):v\in\beta(x)\}$ is non-empty and induces a connected subtree of $T$.
 \end{itemize}
-The representation is \emph{exposed} if all cliques are exposed.
-
-\begin{lemma}\label{lemma-expose}
-Every graph $G$ has an exposed comparable box representation in $\mathbb{R}^{\chi(G)}$.
+For each vertex $v\in V(G)$, let $p(v)$ be the node $x\in V(T)$ such that $v\in \beta(x)$ nearest to the root of $T$.
+The \emph{adhesion} of the tree decomposition is the maximum of $|\beta(x)\cap\beta(y)|$ over distinct $x,y\in V(T)$,
+and its \emph{width} is the maximum of the sizes of the bags minus $1$.  The \emph{treewidth} of a graph is the minimum
+of the widths of its tree decompositions.  We will need to know that graphs of bounded treewidth have bounded dimension.
+In fact, we will prove the following stronger fact (TODO: Was this published somehere before?)
+
+\begin{lemma}\label{lemma-tw}
+Let $(T,\beta)$ be a tree decomposition of a graph $G$ of width $t$.
+Then $G$ has a touching representation $h$ by hypercubes in $R^{t+1}$ such that
+for $u,v\in V(G)$, if $p(u)\neq p(v)$ and $p(u)$ is an ancestor of $p(v)$ in $T$,
+then $\vol(h(u))>\vol(h(v))$.
 \end{lemma}
 \begin{proof}
 ...
 \end{proof}
 
-Note that the chromatic number of $G$ is at most exponential in the comparable box dimension;
-this follows from~\cite{subconvex} and we include the argument to make the dependence clear.
-\begin{lemma}\label{lemma-chrom}
-If $G$ has a comparable box representation $f$ in $\mathbb{R}^d$, then $G$ is $3^d$-colorable.
+Next, let us deal with clique-sums.  A \emph{clique-sum} of two graphs $G_1$ and $G_2$ is obtained from their disjoint union
+by identifying vertices of a clique in $G_1$ and a clique of the same size in $G_2$ and possibly
+deleting some of the edges of the resulting clique.  The main issue to overcome in obtaining a representation for a clique-sum
+is that the representations of $G_1$ and $G_2$ can be ``degenerate''. Consider e.g. the case that $G_1$ is represented
+by unit squares arrangef in a grid; in this case, there is no space to attach $G_2$ at the cliques formed by four squares intersecting
+in a single corner.  This can be avoided by increasing the dimension, but we need to be careful so that the dimension stays bounded
+even after arbitrary number of clique-sums.
+
+It will be convenient to work in the setting of tree decompositions.
+Consider a tree decompostion $(T,\beta)$ of a graphs $G$.
+For each $x\in V(T)$, the \emph{torso} $G_x$ of $x$ is the graph obtained from $G[\beta(x)]$ by adding a clique on $\beta(x)\cap\beta(y)$
+for each $y\in V(T)$.  For a class of graphs $\GG$, we say that the tree decomposition is \emph{over $\GG$} if all the torsos belong to $\GG$.
+We use the following well-known fact.
+\begin{observation}
+A graph $G$ is obtained from graphs in a class $\GG$ by repeated clique-sums if and only if $G$ has a tree decomposition over $\GG$.
+\end{observation}
+For each note $x\in V(T)$, let $\pi(x)=\{x\}\cup \{p(v):v\in\beta(x)\}$.  Let $T_\beta$ be the graph with vertex set $V(T)$ such that
+$xy\in E(T_\beta)$ if and only if $x\in\pi(y)$ or $y\in\pi(x)$.
+\begin{lemma}\label{lemma-legraf}
+If $(T,\beta)$ is a tree decompositon of $G$ of adhesion $a$, then $(T,\pi)$ is a tree decomposition of $T_\beta$ of width at most $a$.
 \end{lemma}
 \begin{proof}
-We actually show that $G$ is $(3^d-1)$-degenerate.  Since every induced subgraph of $G$ also
-has a comparable box representation in $\mathbb{R}^d$, it suffices to show that the minimum degree of $G$
-is less than $3^d$.  Let $v$ be a vertex of $G$ such that $f(v)$ has the smallest volume.  For every neighbor $u$ of $v$,
-there exists a translation $B_u$ of $f(v)$ such that $B_u\subseteq f(u)$ and $B_u$ touches $f(v)$.
-Note that $f(v)\cup \bigcup_{u\in N(v)} B_u$ is a union of internally disjoint translations of $f(v)$ contained in
-a box obtained from $f(v)$ by scaling it by a factor of three, and thus $1+|N(v)|\le 3^d$.
+For each edge $xy\in E(T_\beta)$, we have $x,y\in \pi(x)$ or $x,y\in \pi(y)$ by definition.
+Moreover, for each $x\in V(T_\beta)$, we have
+$$\{y:x\in\pi(y)\}=\{x\}\cup \bigcup_{v\in V(G):p(v)=x} \{y:v\in\beta(y)\},$$
+and all the sets on the right-hand size induce connected subtrees containing $x$,
+implying that $\{y:x\in\pi(y)\}$ also induces a connected subtree containing $x$.
+Hence, $(T,\pi)$ is a tree decomposition of $T_\beta$.
+
+Consider a node $x\in V(T)$.  Note that for each $v\in \beta(x)$, the vertex $p(v)$ is an ancestor of $x$ in $T$.
+In particular, if $x$ is the root of $T$, then $\pi(x)=\{x\}$.  Otherwise, if $y$ is the parent of $x$ in $T$, then
+$p(v)=x$ for every $v\in \beta(x)\setminus\beta(y)$, and thus $|\pi(x)|\le |\beta(x)\cap \beta(y)|+1\le a+1$.
+Hence, the width of $(T,\pi)$ is at most $a$.
+\end{proof}
+
+We are now ready to deal with the clique-sums.
+
+\begin{theorem}
+If $G$ is obtained from graphs in a class $\GG$ by clique-sums, then there exists a graph $G'$ such that
+$G\subseteq G'$ and $\cbdim(G')\le (\cbdim(\GG)+1)(\omega(\GG)+1)$.
+\end{theorem}
+\begin{proof}
+Let $(T,\beta)$ be a tree decomposition of $G$ over $\GG$; the adhesion $a$ of $(T,\beta)$ is at most $\omega(\GG)$.
+By Lemma~\ref{lemma-legraf}, $(T,\pi)$ is a tree decomposition of $T_\beta$ of width at most $a$. By Lemma~\ref{lemma-tw},
+$T_\beta$ has a touching representation $h$ by hypercubes in $\mathbb{R}^{a+1}$. Moreover,
+letting $\prec$ be a linear ordering on $V(T)$ in a non-decreasing order according to the volume of the hypercubes assigned by $h$,
+we have that $x\prec y$ whenever $x$ is a descendant of $y$ in $T$.
+Since $T_\beta$ has treewidth at most $a$, it has a proper coloring $\varphi$ by colors $\{0,\ldots,a\}$.
+For every $x\in V(T)$, let $f_x$ be a touching representation of the torso of $x$ by comparable boxes in $\mathbb{R}^d$
+for $d=\cbdim(\GG)$.  We scale and translate the representations so that for every $x\in V(T)$ and $i\in\{0,\ldots,a\}$,
+there exists a box $E_i(x)$ such that
+\begin{itemize}
+\item whenever $x\prec y$, we have $E_i(x)\subseteq E_i(y)$ and a translation of $E_i(x)$ is a subset of every box of
+the representation $f_y$ whenever $x\prec y$,
+\item if $i=\varphi(x)$, then all boxes of $f_x$ are subsets of $E_i(x)$, and
+\item if $i\neq p(v)$ and $K=\{v\in\beta(x):\varphi(p(v))=i\}$, then letting $y=p(v)$ for $v\in K$ (and noting that this $y$
+is unique, since $\varphi$ is a proper coloring of $T_\beta$ and that $K$ is a clique in $G_y$), the box $E_i(x)$
+contains a point belonging to $\bigcap_{v\in K} f_y(v)$.
+\end{itemize}
+
+Let us now define $f(v)=h(p(v))\times E_0(v)\times \cdots\times E_a(v)$ for each $v\in V(G)$,
+where $E_i(v)=f_{p(v)}(v)$ if $i=\varphi(p(v))$ and $E_i(v)=E_i(p(v))$ otherwise.  We claim this gives a touching representation
+of a supergraph of $G$ by comparable boxes in $\mathbb{R}^{(d+1)(a+1)}$.  First, note that the boxes are indeed comparable;
+if $p(u)=p(v)$, then this is the case since $f_{p(v)}$ is a representation by comparable boxes, and if say
+$p(u)\prec p(v)$, then this is due to the scaling of $f_{p(u)}$.  Next, let us argue $f(u)$ and $f(v)$ have disjoint
+interiors.  If $p(u)=p(v)$, this is the case since $f_{p(v)}$ is a touching representation, and if $p(u)\neq p(v)$,
+then this is the case because $h$ is a touching representation.  Finally, suppose that $uv\in E(G)$.  Let $x$
+be the node of $T$ nearest to the root such that $u,v\in \beta(x)$.  Without loss of generality, $p(u)=x$.
+Let $y=p(v)$.  If $x=y$, then $f(u)\cap f(v)\neq\emptyset$, since $f_x$ is a touching representation of $G_x$.
+If $x\neq y$, then $y\in\pi(x)$ and $xy\in E(T_\beta)$, implying that $h(x)\cap h(y)\neq\emptyset$.
+Moreover, $x\prec y$, implying that $E_i(u)\cap E_i(v)\neq\emptyset$ for $i=0,\ldots,a$.
+Hence, again we have $f(u)\cap f(v)\neq\emptyset$.
 \end{proof}
 
+
+
 \section{Exploiting the product structure}
 
 \subsection*{Acknowledgments}