Added the treewidth argument.

comparable-box-dimension.tex

@@ -138,12 +138,82 @@ In fact, we will prove the following stronger fact (TODO: Was this published som
 
 \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 $p(u)\preceq p(v)$, then $h(u)\sqsubseteq h(v)$.
+Then $G$ has a touching representation $h$ by axis-aligned hypercubes in $R^{t+1}$ such that
+for $u,v\in V(G)$, if $p(u)\prec p(v)$, then $h(u)\sqsubset h(v)$.
 Moreover, the representation can be chosen so that no two hypercubes have the same size.
 \end{lemma}
 \begin{proof}
-...
+Without loss of generality, we can assume that the root has a bag of size one
+and that for each $x\in V(T)$ with parent $y$, we have $|\beta(x)\setminus\beta(y)|=1$
+(if $\beta(x)\subseteq \beta(y)$, we can contract the edge $xy$; if $|\beta(x)\setminus\beta(y)|>1$,
+we can subdivide the edge $xy$, introducing $|\beta(x)\setminus\beta(y)|-1$ new vertices,
+and set their bags appropriately).  It is now natural to relabel the vertices of $G$
+so that $V(G)=V(T)$, by giving the unique vertex in $\beta(x)\setminus\beta(y)$
+the label $x$.  In particular, $p(x)=x$.  Furthermore, we can assume that $y\in\beta(x)$.
+Otherwise, if $y$ is not the root of $T$, we can replace the edge $xy$ by the edge from $x$
+to the parent of $y$.  If $y$ is the root of $T$, then the subtree rooted in $x$ induces a
+union of connected components in $G$, and we can process this subtree separately from the
+rest of the graph (being careful to only use hypercubes smaller than the one representing $y$
+and of different sizes from those used on the rest of the graph).
+
+Let us now greedily color $G$ by giving $x$ a color different from the colors of
+all other vertices in $\beta(x)$; such a coloring $\varphi$ uses only colors
+$\{1,\ldots,t+1\}$.
+
+Let $D=4\Delta(T)+1$.  Let $V(G)=V(T)=\{x_1,x_2,\ldots, x_n\}$, where
+for every $i<j$, $x_i$ and $x_j$ are either incomparable in $\prec$ or
+$x_j\prec x_i)$; in particular, $x_1$ is the root of $T$.  Let $\varepsilon=D^{-n-1}$.
+Let $s_i=D^{-i}$; we will represent $x_i$ by a hypercube $h(x_i)$ with edges of length $s_i$.
+Additionally, we will need to consider larger hypercubes around $h(x_i)$; let $h'(x_i)$
+be the hypercube with sides of length $2s_i$ and with $\min(h'(x_i)[j])=\min(h(x_i)[j])$
+for $j\in\{1,\ldots, t+1\}$, and $h''(x_i)$ the hypercube with sides of length $2s_i+\epsilon$
+and with $\min(h''(x_i)[j])=\min(h(x_i)[j])-\varepsilon$.  We will construct the representation $h$ so that the following
+invariant is satisfied:
+\begin{itemize}
+\item[(a)] For each $x,z\in V(T)$ such that $x\prec z$, we have $h'(x)\subset h''(z)$.
+\item[(b)] For each $y\in V(T)$ and distinct children $x$ and $z$ of $y$, we have $h''(x)\cap h''(z)=\emptyset$.
+\end{itemize}
+Note that this ensures that if $x$ and $z$ are vertices of $T$ and $h(x)\cap h(z)\neq\emptyset$, then $x\prec z$ or $z\prec x$.
+
+We now construct the representation $h$.  For the root $x_1$ of $T$, $h(r)$ is an arbitrary hypercube with sides
+of length $s_1$.  Assuming now we have already selected $h(y)$ for a vertex $y\in V(T)$, the hypercube $h(x_i)$ with sides of length $s_i$
+for a child $x_i$ of $y$ is chosen as follows. For $j\in\{1,\ldots, t+1\}$,
+\begin{itemize}
+\item[(i)] if $j=\varphi(w)$ for $w\in\beta(x_i)\setminus\{x_i\}$, we choose $h(x_i)[j]$ so that
+$\min(h(x_i)[j])=\max(h(w)[j])$ if $xw\in E(G)$ and so that $\min(h(x_i)[j])=\max(h(w)[j]) + \varepsilon$ otherwise.
+\item[(ii)] if $j$ is different from the colors of all vertices in $\beta(x_i)\setminus\{x_i\}$,
+then we choose $h(x_i)[j]$ so that $h''(x_i)[j]$ is a subset of the interior of $h(y)[j]$.  The interval $h''(x_i)[[j]$
+is furthermore chosen to be disjoint from $h''(x_m)[j]$ for any other child $x_m$ of $y$;
+this is always possible by the choice of $D$, $s_i$, and $s_m$.
+\end{itemize}
+Note that (ii) always applies for $j=\varphi(x_i)$ and this ensures that the invariant (b) holds.
+For the invariant (a), note that in the case (ii), we ensure $h''(x_i)[[j]\subseteq h(y)[j]$ and
+we have $h(y)[j]\subseteq h''(z)[j]$ by the invariant (a) for $y$ and $z$.  In the case (i),
+if $z\prec w$, then we have $w\in\beta(z)\setminus\{z\}$ and
+$\min(h(x_i)[j]),\min(h(z)[j])\in\{\max(h(w)[j]),\max(h(w)[j])+\varepsilon\}$.
+If $w\preceq z$, then note we choose $h'(x_i)[j]\subseteq h'(w)[j]$ by (i) and that
+we have $h'(w)[j]\subset h''(z)[j]$ by (a).  This verifies that the invariant (a)
+also holds at $x_i$.
+
+Consider now two adjacent vertices of $G$, say $x_i$ and $w$.  Note that any two
+adjacent vertices are comparable in $\prec$, and thus we can assume $x_i\prec w$
+and $w\in\beta(x_i)$.
+By (i), for $j=\varphi(w)$, the intervals $h(x_i)[j]$ and $h(w)[j]$
+intersect in a single point.  If $j\neq \varphi(w)$, then let $w_1$ be the child of $w$ on the path in $T$ from
+$w$ to $x_i$.  If no vertex in $\beta(w_1)\setminus\{w_1\}$ has color $j$, then by (ii), we have $h''(w_1)[j]\subset h(w)[j]$,
+Otherwise, $z\in \beta(w_1)\setminus\{w_1\}$ such that $\varphi(z)=j$; clearly, $w\prec z$ and $z\in\beta(w)\setminus\{w\}$.
+We have $\min(h(w_1)[j]),\min(h(w)[j])\in\{\max(h(z)[j]),\max(h(z)[j])+\varepsilon\}$ by (i).
+Hence, $\min(h(w)[j])-2\epsilon\le \min(h''(w_1)[j])\le \max(h''(w_1)[j])\le \max(h(w)[j])$.
+Since $h(x_i)[j]\subseteq h''(w_1)[j]$ by (a) and the length of the interval $h(x_i)[j]$ is greater than $2\varepsilon$,
+we conclude that $h(x_i)[j]\cap h(w)[j]\neq\emptyset$.  Therefore, the boxes $h(w)$ and $h(x_i)$ touch.
+
+Consider now two non-adjacent vertices of $G$, say $x_i$ and $w$.  As we noted before, if $x_i$ and $w$ are
+incomparable in $\prec$, then (a) and (b) implies that the boxes $h(x_i)$ and $h(w)$ are disjoint.
+Suppose now that say $x_i\prec w$, and let $j=\varphi(w)$.  If $w\in\beta(x_i)$, then $h(x_i)[j]$ and $h(w)[j]$ are disjoint by (i).
+Otherwise, let $y$ be the last vertex on the path from $w$ to $x_i$ in $T$ such that $w\in\beta(y)$ and let $z$ be the child of $y$
+on this path.  As we argued in the first paragraph, $y\neq w$, and by (i), the interior of $h(y)[j]$ is disjoint from $h(w)[j]$.
+By (ii), $h''(z)[j]$ is contained in the interior of $h(y)[j]$.  By (a), we conclude that $h(x_i)[j]\subseteq h''(z)[j]$,
+implying that the boxes $h(x_i)$ and $h(w)$ are disjoint.  Therefore, $h$ is a touching representation of $G$.
 \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
@@ -191,14 +261,14 @@ and thus $p(x)=x$.
 
 We are now ready to deal with the clique-sums.
 
-\begin{theorem}
+\begin{theorem}\label{thm-cs}
 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 $\mathbb{R}^{a+1}$ such that
+$T_\beta$ has a touching representation $h$ by axis-aligned hypercubes in $\mathbb{R}^{a+1}$ such that
 $h(x)\sqsubseteq h(y)$ whenever $x\preceq y$.
 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 $G_x$ of $x$ by comparable boxes in $\mathbb{R}^d$,
@@ -233,6 +303,8 @@ Moreover, $x\prec y$, implying that $E_i(u)\cap E_i(v)\neq\emptyset$ for $i=0,\l
 $xy\in E(T_\beta)$.  Hence, again we have $f(u)\cap f(v)\neq\emptyset$.
 \end{proof}
 
+Note that in Theorem~\ref{thm-cs}, we only get a representation of a supergraph
+of $G$.  
 
 %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.
@@ -249,7 +321,7 @@ $xy\in E(T_\beta)$.  Hence, again we have $f(u)\cap f(v)\neq\emptyset$.
 %\end{proof}
 
 
-\section{Exploiting the product structure}
+\section{The product structure and minor-closed classes}
 
 \subsection*{Acknowledgments}
 This research was carried out at the workshop on Geometric Graphs and Hypergraphs organized by Yelena Yuditsky and Torsten Ueckerdt