\magnification\magstep1 \input picmac \input epsf \def\mpfig#1{\hbox{$\vcenter{\epsfbox{sand.#1}}$}} % This file uses PostScript illustrations sand.1 thru sand.5 made by MetaPost \font\rhfont=cmcsc8 \headline= {\ifnum\pageno>1 \rhfont The Electronic Journal of Combinatorics 1 (1994), \# A1 \hfil \tenrm\folio \else \hfil \fi} \footline={\hfil} % Reference numbers: \def\Ala{1} % Alizadeh 1991 \def\Alb{2} % Alizadeh 1993 \def\ALMSS{3} % Arora et al \def\Bol{4} % Bollobas \def\GLSell{5} % Groetschel et al 1981 \def\GLSrelax{6} % Groetschel et al 1986 \def\GLSbook{7} % Groetschel et al 1988 \def\House{8} % Householder \def\Ju{9} % Juhasz \def\KK{10} % Kashin \def\Ko{11} %Koniagin \def\Lov{12} % Lovasz 1979 \def\LL{13} % Lovasz 1986 \def\LLL{14} % Lovasz to appear \def\LS{15} % Lovasz, Schrijver \def\Ov{16} % Overton \def\Padberg{17} \def\Shannon{18} \let\SQRT=\sqrt \def\sqrt#1{\SQRT{\vphantom t#1}} \catcode`\@=\active \def@{\ifmmode\mskip1mu\else\char`\@ \fi} \def\0{^{\vphantom T}} \parskip3pt \parindent20pt \baselineskip14pt \def\rar{\rightarrow} \def\meno{\medskip\noindent} \def\adj{\mathrel-\joinrel\joinrel\mathrel-} %long minus \def\noadj{\hskip3pt\not\hskip-3pt\adj} %not long minus \def\ones{{1\mskip-4.5mu\rm l}} \def\bib{\par\noindent\hangindent 25pt} \def\disleft#1:#2:#3\par{\par\hangindent#1\noindent \hbox to #1{#2 \hfill \hskip .1em}\ignorespaces#3\par} \def\pfbox {\hbox{\hskip 3pt\lower1pt\vbox{\hrule \hbox to 7pt{\vrule height 7pt\hfill\vrule} \hrule}}\hskip3pt} \def\dddots{\mathinner{\mskip1mu\raise1pt\vbox{\kern7pt\hbox{.}}\mskip2mu \raise4pt\hbox{.}\mskip2mu\raise7pt\hbox{.}\mskip1mu}} \def\ddddots{\mathinner{\mskip1mu\raise7pt\vbox{\kern7pt\hbox{.}}\mskip2mu \raise4pt\hbox{.}\mskip2mu\raise1pt\hbox{.}\mskip1mu}} \pageno=0 \centerline{\bf The Sandwich Theorem} \bigskip \centerline{Donald E. Knuth} \bigskip\bigskip\bigskip\bigskip {\narrower\narrower\narrower\smallskip\noindent {\bf Abstract:}\enspace This report contains expository notes about a function $\vartheta(G)$ that is popularly known as the Lov\'asz number of a graph~$G$. There are many ways to define $\vartheta(G)$, and the surprising variety of different characterizations indicates in itself that $\vartheta(G)$ should be interesting. But the most interesting property of $\vartheta(G)$ is probably the fact that it can be computed efficiently, although it lies ``sandwiched'' between other classic graph numbers whose computation is NP-hard. I~have tried to make these notes self-contained so that they might serve as an elementary introduction to the growing literature on Lov\'asz's fascinating function. \smallskip} \vfill\eject \line{{\bf The Sandwich Theorem} \hfill DEK notes last revised 6 December 1993} \medskip \begingroup \advance\baselineskip 0pt minus1pt \halign{\hfil#\quad&\hbox to .9\hsize{# \dotfill\quad}&\hfil#\cr 0.&Preliminaries&2\cr 1.&Orthogonal labelings&4\cr 2.&Convex labelings&5\cr 3.&Monotonicity&6\cr 4.&The theta function&6\cr 5.&Alternative definitions of $\vartheta$&8\cr 6.&Characterization via eigenvalues&8\cr 7.&A complementary characterization&9\cr 8.&Elementary facts about cones&11\cr 9.&Definite proof of a semidefinite fact&12\cr 10.&Another characterization&13\cr 11.&The final link&14\cr 12.&The main theorem&14\cr 13.&The main converse&15\cr 14.&Another look at {\tt TH}&16\cr 15.&Zero weights&17\cr 16.&Nonzero weights&17\cr 17.&Simple examples&19\cr 18.&The direct sum of graphs&19\cr 19.&The direct cosum of graphs&20\cr 20.&A direct product of graphs&21\cr 21.&A direct coproduct of graphs&22\cr 22.&Odd cycles&24\cr 23.&Comments on the previous example&27\cr 24.&Regular graphs&27\cr 25.&Automorphisms&28\cr 26.&Consequence for eigenvalues&30\cr 27.&Further examples of symmetric graphs&30\cr 28.&A bound on $\vartheta$&31\cr 29.&Compatible matrices&32\cr 30.&Antiblockers&35\cr 31.&Perfect graphs&36\cr 32.&A characterization of perfection&37\cr 33.&Another definition of $\vartheta$&39\cr 34.&Facets of {\tt TH}&40\cr 35.&Orthogonal labelings in a perfect graph&42\cr 36.&The smallest non-perfect graph&43\cr 37.&Perplexing questions&45\cr} \endgroup \vfill\eject \centerline{\bf The Sandwich Theorem} \bigskip\noindent It is NP-complete to compute $\omega(G)$, the size of the largest clique in a graph~$G$, and it is NP-complete to compute $\chi(G)$, the minimum number of colors needed to color the vertices of~$G$. But Gr\"otschel, Lov\'asz, and Schrijver proved [\GLSell] that we can compute in polynomial time a real number that is ``sandwiched'' between these hard-to-compute integers: $$\omega(G)\leq\vartheta(\overline{G})\leq\chi(G)\,.\eqno(\ast)$$ Lov\'asz [\LL] called this a ``sandwich theorem.'' The book [\GLSbook] develops further facts about the function $\vartheta(G)$ and shows that it possesses many interesting properties. Therefore I~think it's worthwhile to study $\vartheta(G)$ closely, in hopes of getting acquainted with it and finding faster ways to compute~it. Caution: The function called $\vartheta(G)$ in [\LL] is called $\vartheta(\overline{G})$ in [\GLSbook] and~[\Lov]. I~am following the latter convention because it is more likely to be adopted by other researchers---[\GLSbook] is a classic book that contains complete proofs, while [\LL] is simply an extended abstract. In these notes I am mostly following [\GLSbook] and [\Lov] with minor simplifications and a few additions. I~mention several natural problems that I~was not able to solve immediately although I~expect (and fondly hope) that they will be resolved before I~get to writing this portion of my forthcoming book on Combinatorial Algorithms. I'm grateful to many people---especially to Martin Gr\"otschel and L\'aszl\'o Lov\'asz---for their comments on my first drafts of this material. These notes are in numbered sections, and there is at most one Lemma, Theorem, Corollary, or Example in each section. Thus, ``Lemma~2'' will mean ``the lemma in section~2''. \meno {\bf 0. Preliminaries.}\quad Let's begin slowly by defining some notational conventions and by stating some basic things that will be assumed without proof. All vectors in these notes will be regarded as column vectors, indexed either by the vertices of a graph or by integers. The notation $x\geq y$, when $x$ and~$y$ are vectors, will mean that $x_v\geq y_v$ for all~$v$. If $A$ is a matrix, $A_v$~will denote column~$v$, and $A_{uv}$ will be the element in row~$u$ of column~$v$. The zero vector and the zero matrix and zero itself will all be denoted by~0. We will use several properties of matrices and vectors of real numbers that are familiar to everyone who works with linear algebra but not to everyone who studies graph theory, so it seems wise to list them here: \medskip (i)\quad The {\it dot product\/} of (column) vectors~$a$ and~$b$ is $$a\cdot b=a^Tb\,;\eqno(0.1)$$ the vectors are {\it orthogonal\/} (also called perpendicular) if $a\cdot b=0$. The {\it length\/} of vector~$a$ is $$\Vert a\Vert=\sqrt{a\cdot a}\,.\eqno(0.2)$$ Cauchy's inequality asserts that $$a\cdot b\leq \Vert a\Vert\;\Vert b\Vert\,;\eqno(0.3)$$ equality holds iff $a$ is a scalar multiple of $b$ or $b=0$. Notice that if $A$ is any matrix we have $$(A^T\!A)_{uv}=\sum_{k=1}^n(A^T)_{uk}A_{kv}=\sum_{k=1}^nA_{ku}A_{kv}=A_u\cdot A_v\,;\eqno(0.4)$$ in other words, the elements of $A^T\!A$ represent all dot products of the columns of~$A$. \medskip (ii)\quad An {\it orthogonal matrix\/} is a square matrix~$Q$ such that $Q^TQ$ is the identity matrix~$I$. Thus, by (0.4), $Q$~is orthogonal iff its columns are unit vectors perpendicular to each other. The transpose of an orthogonal matrix is orthogonal, because the condition $Q^TQ=I$ implies that $Q^T$ is the inverse of~$Q$, hence $QQ^T=I$. \medskip (iii)\quad A given matrix $A$ is {\it symmetric\/} (i.e., $A=A^T$) iff it can be expressed in the form $$A=QDQ^T\eqno(0.5)$$ where $Q$ is orthogonal and $D$ is a diagonal matrix. Notice that (0.5) is equivalent to the matrix equation $$AQ=QD\,,\eqno(0.6)$$ which is equivalent to the equations $$AQ_v=Q_v\lambda_v$$ for all $v$, where $\lambda_v=D_{vv}$. Hence the diagonal elements of~$D$ are the eigenvalues of~$A$ and the columns of~$Q$ are the corresponding eigenvectors. Properties (i), (ii), and (iii) are proved in any textbook of linear algebra. We can get some practice using these concepts by giving a constructive proof of another well known fact: \proclaim Lemma. Given $k$ mutually perpendicular unit vectors, there is an orthogonal matrix having these vectors as the first $k$~columns. \noindent{\bf Proof.}\quad Suppose first that $k=1$ and that $x$ is a $d$-dimensional vector with $\Vert x\Vert=1$. If $x_1=1$ we have $x_2=\cdots =x_d=0$, so the orthogonal matrix $Q=I$ satisfies the desired condition. Otherwise we let $$y_1=\sqrt{(1-x_1)/2}\,,\qquad y_j=-x_j/(2y_1)\quad{\rm for}\ 13$. But $\overline{\chi}(G)$ is always an integer, and $\overline{\chi}(C_n) =\left\lceil{n\over 2}\right\rceil$ is greater than $\kappa(C_n)$ when $n$ is odd. Incidentally, these remarks establish the ``sandwich inequality'' $(\ast)$ stated in the introduction, because $$\alpha(G)\leq\vartheta(G)\leq\kappa(G)\leq\overline{\chi}(G)\eqno(4.9)$$ and $\omega(G)=\alpha(\overline{G})$, $\overline{\chi}(G)= \chi(\overline{G})$. \meno {\bf 5. Alternative definitions of $\vartheta$.}\quad Four additional functions $\vartheta_1$, $\vartheta_2$, $\vartheta_3$, $\vartheta_4$ are defined in~[\GLSbook], and they all turn out to be identical to~$\vartheta$. Thus, we can understand $\vartheta$ in many different ways; this may help us compute~it. We will show, following [\GLSbook], that if $w$ is any fixed nonnegative real labeling of~$G$, the inequalities $$\vartheta(G,w)\leq \vartheta_1(G,w)\leq\vartheta_2(G,w)\leq \vartheta_3(G,w)\leq\vartheta_4(G,w)\leq\vartheta(G,w)\eqno(5.1)$$ can be proved. Thus we will establish the theorem of~[\GLSbook], and all inequalities in our proofs will turn out to be equalities. We will introduce the alternative definitions~$\vartheta_k$ one at a time; any one of these definitions could have been taken as the starting point. First, $$\vartheta_1(G,w)=\min_a\;\max_v\bigl(w_v/c(a_v)\bigr)\,,\hbox{ over all orthogonal labelings $a$}.\eqno(5.2)$$ Here we regard $w_v/c(a_v)=0$ when $w_v=c(a_v)=0@$; but the max is~$\infty$ if some $w_v>0$ has $c(a_v)=0$. \proclaim Lemma. $\vartheta(G,w)\leq \vartheta_1(G,w)$. \noindent {\bf Proof.}\quad Suppose $x\in{\tt TH}(G)$ maximizes $w\cdot x$, and suppose $a$ is an orthogonal labeling that achieves the minimum value $\vartheta_1(G,w)$. Then $$\vartheta(G,w)=w\cdot x=\sum_vw_vx_v\leq \biggl(\max_v\, {w_v\over c(a_v)}\biggr)\,\sum_v c(a_v)x_v\leq \max_v\,{w_v\over c(a_v)}=\vartheta_1(G,w)\,.\ \pfbox$$ Incidentally, the fact that all inequalities are exact will imply later that every nonzero weight vector~$w$ has an orthogonal labeling~$a$ such that $$c(a_v)={w_v\over\vartheta(G,w)}\quad\hbox{for all $v$}.\eqno(5.3)$$ We will restate such consequences of (5.1) later, but it may be helpful to keep that future goal in mind. \meno {\bf 6. Characterization via eigenvalues.}\quad The second variant of~$\vartheta$ is rather different; this is the only one Lov\'asz chose to mention in~[\LL]. We say that $A$ is a {\it feasible matrix\/} for $G$ and $w$ if $A$ is indexed by vertices and $$\eqalignno{% &\hbox{$A$ is real and symmetric;}\cr \noalign{\smallskip} &\hbox{$A_{vv}=w_v$ for all $v\in V$;}\cr \noalign{\smallskip} &\hbox{$A_{uv}=\sqrt{w_uw_v}$ whenever $u\noadj v$ in $G$}&(6.1)\cr}$$ The other elements of $A$ are unconstrained (i.e., they can be anything between $-\infty$ and $+\infty$). If $A$ is any real, symmetric matrix, let $\Lambda(A)$ be its maximum eigenvalue. This is well defined because all eigenvalues of~$A$ are real. Suppose $A$ has eigenvalues $\{\lambda_1,\ldots,\lambda_n\}$; then $A=Q\,{\rm diag}(\lambda_1,\ldots,\lambda_n)@Q^T$ for some orthogonal~$Q$, and $\Vert Qx\Vert=\Vert x\Vert$ for all vectors~$x$, so there is a nice way to characterize $\Lambda(A)$: $$\Lambda(A)=\max\{\,x^T\!Ax\,\mid\,\Vert x\Vert =1\,\}\,.\eqno(6.2)$$ Notice that $\Lambda(A)$ might not be the largest eigenvalue in absolute value. We now let $$\vartheta_2(G,w)=\min\{\,\Lambda(A)\mid A\hbox{ is a feasible matrix for $G$ and $w$}\,\}\,.\eqno(6.3)$$ \proclaim Lemma. $\vartheta_1(G,w)\leq \vartheta_2(G,w)$. \noindent {\bf Proof.}\quad Note first that the trace ${\rm tr}\,A=\sum_v w_v\geq 0$ for any feasible matrix~$A$. The trace is also well-known to be the sum of the eigenvalues; this fact is an easy consequence of the identity $${\rm tr}\,XY=\sum_{j=1}^m\;\sum_{k=1}^nx_{jk}y_{kj}={\rm tr}\,YX \eqno(6.4)$$ valid for any matrices $X$ and $Y$ of respective sizes $m\times n$ and $n\times m$. In particular, $\vartheta_2(G,w)$ is always $\geq 0$, and it is $=0$ if and only if $w=0\ \bigl($when also $\vartheta_1(G,w)=0\bigr)$. So suppose $w\neq 0$ and let $A$ be a feasible matrix that attains the minimum value $\Lambda(A)=\vartheta_2(G,w)=\lambda >0$. Let $$B=\lambda I-A\,.\eqno(6.5)$$ The eigenvalues of $B$ are $\lambda$ minus the eigenvalues of $A$. $\bigl($For if $A=Q\,{\rm diag}(\lambda_1,\ldots,\lambda_n)@Q^T$ then $B=Q\,{\rm diag}(\lambda-\lambda_1,\ldots,\lambda-\lambda_n)@Q^T.\bigr)$ Thus they are all nonnegative; such a matrix~$B$ is called {\it positive semidefinite}. By (0.5) we can write $$B=X^TX\,,\qquad\hbox{i.e.,}\quad B_{uv}=x_u\cdot x_v\,,\eqno(6.6)$$ when $X={\rm diag}(\sqrt{\lambda-\lambda_1},\ldots,\sqrt{\lambda-\lambda_n}\,)@Q^T$. Let $a_v=(\sqrt{w_v}\,,x_{1v},\ldots,x_{rv})^T$. Then $c(a_v)=w_v/\,\Vert a_v\Vert^2=w_v/(w_v+x^2_{1v}+\cdots +x^2_{rv})$ and $x^2_{1v}+\cdots +x^2_{rv}=B_{vv}=\lambda-w_v$, hence $c(a_v)=w_v/\lambda$. Also if $u\noadj v$ we have $a_u\cdot a_v= \sqrt{w_uw_v}+x_u\cdot x_v = \sqrt{w_uw_v}+B_{uv}=\sqrt{w_uw_v}-A_{uv}=0$. Therefore $a$ is an orthogonal labeling and $\max_v\,w_v/c(a_v)=\lambda\geq \vartheta_1(G,w)$.\ \pfbox \meno {\bf 7. A complementary characterization.}\quad Still another variation is based on orthogonal labelings of the complementary graph~$\overline{G}$. In this case we let $b$ be an orthogonal labeling of~$\overline{G}$, normalized so that $\sum_v\Vert b_v\Vert^2=1$, and we let $$\eqalignno{\vartheta_3(G,w)=\max\left\{\,\sum_{u,v}\right. (\sqrt{w_u}\,b_u)&\cdot (\sqrt{w_v}\,b_v)\biggm| \cr \noalign{\smallskip} &\left.\phantom{\sum_{u,v}}\hskip-2em b\hbox{ is a normalized orthogonal labeling of $\overline{G}$}\,\right\}.&(7.1)\cr}$$ A normalized orthogonal labeling $b$ is equivalent to an $n\times n$ symmetric positive semidefinite matrix~$B$, where $B_{uv}=b_u\cdot b_v$ is zero when $u\adj v$, and where ${\rm tr}\,B=1$. \proclaim Lemma. $\vartheta_2(G,w)\leq\vartheta_3(G,w)$. This lemma is the ``heart'' of the proof that all $\vartheta$s are equivalent, according to~[\GLSbook]. It relies on a fact about positive semidefinite matrices that we will prove in \S9. \meno \proclaim Fact. If $A$ is a symmetric matrix such that $A\cdot B\geq 0$ for all symmetric positive semi\-definite~$B$ with $B_{uv}=0$ for $u\adj v$, then $A=X+Y$ where $X$ is symmetric positive semidefinite and $Y$ is symmetric and $Y_{vv}=0$ for all~$v$ and $Y_{uv}=0$ for $u\noadj v$. Here $C\cdot B$ stands for the dot product of matrices, i.e., the sum $\sum_{u,v}C_{uv}B_{uv}$, which can also be written ${\rm tr}\,C^TB$. The stated fact is a duality principle for quadratic programming. Assuming the Fact, let $W$ be the matrix with $W_{uv}=\sqrt{w_uw_v}$, and let $\vartheta_3=\vartheta_3(G,w)$. By definition (7.1), if $b$ is any nonzero orthogonal labeling of~$\overline{G}$ (not necessarily normalized), we have $$\sum_{u,v}(\sqrt{w_u}\,b_u)\cdot(\sqrt{w_v}\,b_v)\leq\vartheta_3\sum_v \Vert b_v\Vert^2\,.\eqno(7.2)$$ In matrix terms this says $W\cdot B\leq(\vartheta_3 I)\cdot B$ for all symmetric positive semidefinite~$B$ with $B_{uv}=0$ for $u\adj v$. The Fact now tells us we can write $$\vartheta_3 I-W=X+Y\eqno(7.3)$$ where $X$ is symmetric positive semidefinite, $Y$~is symmetric and diagonally zero, and $Y_{uv}=0$ when $u\noadj v$. Therefore the matrix~$A$ defined by $$A=W+Y=\vartheta_3 I-X$$ is a feasible matrix for $G$, and $\Lambda(A)\leq\vartheta_3$. This completes the proof that $\vartheta_2(G,w)\leq \vartheta_3(G,w)$, because $\Lambda(A)$ is an upper bound on $\vartheta_2$ by definition of~$\vartheta_2$.\ \pfbox \meno {\bf 8. Elementary facts about cones.}\quad A {\it cone\/} in $N$-dimensional space is a set of vectors closed under addition and under multiplication by nonnegative scalars. (In particular, it is convex: If $c$ and~$c'$ are in cone~$C$ and $0b$. \meno {\bf Proof.}\quad Let $d$ be the greatest lower bound of $\Vert x-c\Vert^2$ for all $c\in C$. Then there's a sequence of vectors~$c_k$ with $\Vert x-c_k\Vert^2 0$. Now let $y=x-c_{\infty}$ and $b=c_{\infty}\cdot y$. Clearly $x\cdot y=y\cdot y+b>b$. And if $c$ is any element of~$C$ and $\epsilon$ is any small positive number, the vector $\epsilon c+(1-\epsilon)c_{\infty}$ is in~$C$, hence $\bigl\Vert x-\bigl(\epsilon c+(1-\epsilon)c_{\infty}\bigr)\bigr\Vert^2\geq d$. But $$\eqalign{\bigl\Vert x-\bigl(\epsilon c+(1-\epsilon)c_{\infty}\bigr)\bigr\Vert^2-d &=\Vert x-c_{\infty}-\epsilon(c-c_{\infty})\Vert^2-d\cr \noalign{\smallskip} &=\null-2\epsilon y\cdot(c-c_{\infty})+\epsilon^2\,\Vert c-c_{\infty}\Vert^2\cr}$$ can be nonnegative for all small $\epsilon$ only if $y\cdot (c-c_{\infty})\leq 0$, i.e., $c\cdot y\leq b$.\ \pfbox \meno If $A$ is any set of vectors, let $A^{\ast}=\{\,b\mid a\cdot b\geq 0\hbox{ for all $a\in A$}\,\}$. The following facts are immediate: \meno {\bf F2.}\quad $A\subseteq A'$ implies $A^{\ast}\supseteq A'^{\ast}$. \meno {\bf F3.}\quad $A\subseteq A^{\ast\ast}$. \meno {\bf F4.}\quad $A^{\ast}$ is a closed cone. \medskip From F1 we also get a result which, in the special case that $C=\{\,Ax\mid x\ge0\,\}$ for a matrix~$A$, is called Farkas's Lemma: \meno {\bf F5.}\quad If $C$ is a closed cone, $C=C^{\ast\ast}$. \meno {\bf Proof.}\quad Suppose $x\in C^{\ast\ast}$ and $x\notin C$, and let $(y,b)$ be a separating hyperplane as in~F1. Then $(y,0)$ is also a separating hyperplane; for we have $x\cdot y>b\geq 0$ because $0\in C$, and we cannot have $c\cdot y>0$ for $c\in C$ because $(\lambda c)\cdot y$ would then be unbounded. But then $c\cdot(-y)\geq 0$ for all $c\in C$, so $-y\in C^{\ast}$; hence $x\cdot(-y)\geq 0$, a~contradiction. \ \pfbox \medskip If $A$ and $B$ are sets of vectors, we define $A+B=\{\,a+b\mid a\in A\ {\rm and}\ b\in B\,\}$. \meno {\bf F6.}\quad If $C$ and $C'$ are closed cones, $(C\cap C')^{\ast}=C^{\ast}+C'^{\ast}$. \meno {\bf Proof.}\quad If $A$ and $B$ are arbitrary sets we have $A^{\ast}+B^{\ast}\subseteq (A\cap B)^{\ast}$, for if $x\in A^{\ast}+B^{\ast}$ and $y\in A\cap B$ then $x\cdot y=a\cdot y+b\cdot y\geq 0$. If $A$ and~$B$ are arbitrary sets including~0 then $(A+B)^{\ast}\subseteq A^{\ast}\cap B^{\ast}$ by~F2, because $A+B\supseteq A$ and $A+B\supseteq B$. Thus for arbitrary~$A$ and~$B$ we have $(A^{\ast}+B^{\ast})^{\ast}\subseteq A^{\ast\ast}\cap B^{\ast\ast}$, hence $$(A^{\ast}+B^{\ast})^{\ast\ast}\supseteq (A^{\ast\ast}\cap B^{\ast\ast})^{\ast}\,.$$ Now let $A$ and $B$ be closed cones; apply F5 to get $A^{\ast}+B^{\ast}\supseteq(A\cap B)^{\ast}$.\ \pfbox \meno {\bf F7.}\quad If $C$ and $C'$ are closed cones, $(C+C')^{\ast}=C^{\ast}\cap C'^{\ast}$. (I~don't need this but I~might as well state it.) {\bf Proof.} F6~says $(C^{\ast}\cap C'^{\ast})^{\ast}=C^{\ast\ast}+C'^{\ast\ast}$; apply F5 and $\ast$ again. \ \pfbox \meno {\bf F8.}\quad Let $S$ be any set of indices and let $A_S=\{\,a\mid a_s=0\hbox{ for all }s\in S\,\}$, and let $\overline{S}$ be all the indices not in~$S$. Then $$A_S^{\ast}=A_{\overline{S}}\,.$$ \meno {\bf Proof.}\quad If $b_s=0$ for all $s\notin S$ and $a_s=0$ for all $s\in S$, obviously $a\cdot b=0@$; so $A_{\overline{S}}\subseteq A_S^{\ast}$. If $b_s\neq 0$ for some $s\notin S$ and $a_t=0$ for all $t\neq s$ and $a_s=-b_s$ then $a\in A_S$ and $a\cdot b <0@$; so $b\notin A_S^{\ast}$, hence $A_{\overline{S}}\supseteq A_S^{\ast}$.\ \pfbox \meno {\bf 9. Definite proof of a semidefinite fact.}\quad Now we are almost ready to prove the result needed in the proof of Lemma~7. Let $D$ be the set of real symmetric positive semidefinite matrices (called ``spuds'' henceforth for brevity), considered as vectors in $N$-dimensional space where $N={1\over 2}(n+1)n$. We use the inner product $A\cdot B={\rm tr}\,A^TB$; this is justified if we divide off-diagonal elements by~$\sqrt{2}$. For example, if $n=3$ the correspondence between 6-dimensional vectors and $3\times 3$ symmetric matrices~is $$(a,b,c,d,e,f)\ \leftrightarrow\ \pmatrix{a&d/\sqrt{2}&e/\sqrt{2}\cr \noalign{\smallskip} d/\sqrt{2}&b&f/\sqrt{2}\cr \noalign{\smallskip} e/\sqrt{2}&f/\sqrt{2}&c\cr}$$ preserving sum, scalar product, and dot product. Clearly $D$ is a closed cone. \meno {\bf F9.}\quad $D^{\ast}=D$. \meno {\bf Proof.}\quad If $A$ and $B$ are spuds then $A=X^TX$ and $B=Y^TY$ and $A\cdot B={\rm tr}\,X^TX\,Y^TY={\rm tr}\,XY^TYX^T=(YX^T)\cdot(YX^T)\geq 0@$; hence $D\subseteq D^{\ast}$. (In fact, this argument shows that $A\cdot B=0$ iff $AB=0$, for any spuds $A$ and~$B$, since $A=A^T$.) %\def\ddddots If $A$ is symmetric but has a negative eigenvalue $\lambda$ we can write $$A=Q\,{\rm diag}\,(\lambda,\lambda_2,\ldots,\lambda_n)\, %\pmatrix{\lambda\cr &\lambda_2\cr &&\ddddots\cr &&&\lambda_n\cr} Q^T$$ for some orthogonal matrix~$Q$. Let $B=Q\,{\rm diag}\,(1,0,\ldots,0)Q^T$; then $B$ is a spud, and $$A\cdot B={\rm tr}\;A^TB={\rm tr}\;Q\,{\rm diag}\,(\lambda,0,\ldots,0)\, Q^T=\lambda <0\,.$$ So $A$ is not in $D^{\ast}$; this proves $D\supseteq D^{\ast}$.\ \pfbox \medskip Let $E$ be the set of all real symmetric matrices such that $E_{uv}=0$ when $u\adj v$ in a graph~$G$; let $F$ be the set of all real symmetric matrices such that $F_{uv}=0$ when $u=v$ or $u\noadj v$. The Fact stated in Section~7 is now equivalent in our new notation to \meno {\bf Fact.}\qquad$(D\cap E)^{\ast}\subseteq D+F$. But we know that $$\vcenter{\halign{$#$\hfil\ &$#$\hfil\qquad&#\hfil\cr (D\cap E)^{\ast}&=D^{\ast}+E^{\ast}&by F6\cr \noalign{\smallskip} &=D+F&by F9 and F8.\ \pfbox\cr}}$$ \meno {\bf 10. Another characterization.}\quad Remember $\vartheta$, $\vartheta_1$, $\vartheta_2$, and $\vartheta_3$? We are now going to introduce yet another function $$\vartheta_4(G,w)=\max\left\{\,\sum_v c(b_v)w_v\biggm| b\hbox{ is an orthogonal labeling of }\overline{G}\,\right\}\,.\eqno(10.1)$$ \proclaim Lemma. $\vartheta_3(G,w)\leq \vartheta_4(G,w)$. \noindent {\bf Proof.}\quad Suppose $b$ is a normalized orthogonal labeling of~$\overline{G}$ that achieves the maximum~$\vartheta_3$; and suppose the vectors of this labeling have dimension~$d$. Let $$x_k=\sum_vb_{kv}\sqrt{w_v}\,,\qquad{\rm for}\ 1\leq k\leq d\,;\eqno(10.2)$$ then $$\vartheta_3(G,w)=\sum_{u,v}\sqrt{w_u}\,b_u\cdot b_v\,\sqrt{w_v} =\sum_{u,v,k}\sqrt{w_uw_v}\,b_{ku}b_{kv}=\sum_kx_k^2\,.$$ Let $Q$ be an orthogonal $d\times d$ matrix whose first row is $(x_1/\sqrt{\vartheta_3},\ldots,x_d/\sqrt{\vartheta_3})^T$, and let $b'_v=Qb_v$. Then $b'_u\cdot b'_v=b_u^TQ^TQb_v=b_u^Tb_v=b_u\cdot b_v$, so $b'$ is a normalized orthogonal labeling of~$\overline{G}$. Also $$\eqalignno{x'_k=\sum_vb'_{kv}\sqrt{w_v} &=\sum_{v,j}Q_{kj}b_{jv}\sqrt{w_v}\cr \noalign{\smallskip} &=\sum_jQ_{kj}x_j=\cases{\sqrt{\vartheta_3}\,,&$k=1$;\cr \noalign{\smallskip} 0\,,&$k>1$.\cr}&(10.3)\cr}$$ Hence by Cauchy's inequality $$\eqalignno{\vartheta_3(G,w)=\biggl(\sum_vb'_{1v}\sqrt{w_v}\biggr)^2 &\leq\biggl(\sum_v\Vert b'_v\Vert^2\biggr)\biggl( \sum_{\scriptstyle v\atop \scriptstyle b'_v\neq 0}\;{b'^2_{1v}\over\Vert b'_v\Vert^2}\,w_v\biggr)\cr \noalign{\smallskip} &=\sum_v c(b'_v)w_v\leq\vartheta_4(G,w)&(10.4)\cr}$$ because $\sum_v\Vert b'_v\Vert^2=\sum_v\Vert b_v\Vert^2=1$. \ \pfbox \meno {\bf 11. The final link.}\quad Now we can close the loop: \proclaim Lemma. $\vartheta_4(G,w)\leq \vartheta(G,w)$. \noindent {\bf Proof.}\quad If $b$ is an orthogonal labeling of $\overline{G}$ that achieves the maximum~$\vartheta_4$, we will show that the real labeling~$x$ defined by $x_v=c(b_v)$ is in ${\tt TH}(G)$. Therefore $\vartheta_4(G,w)=w\cdot x$ is $\leq \vartheta(G,w)$. We will prove that if $a$ is any orthogonal labeling of $G$, and if $b$ is any orthogonal labeling of~$\overline{G}$, then $$\sum_v c(a_v)@c(b_v)\leq 1\,.\eqno(11.1)$$ Suppose $a$ is a labeling of dimension $d$ and $b$ is of dimension~$d'$. Then consider the $d\times d'$ matrices $$A_v=a_v\0b_v^T\eqno(11.2)$$ as elements of a vector space of dimension~$dd'$. If $u\neq v$ we have $$A_u\cdot A_v={\rm tr}\;A_u^TA_v\0={\rm tr}\;b_u\0a_u^Ta_v\0b_v^T={\rm tr}\; a_u^Ta_v\0b_v^Tb_u\0=0\,,\eqno(11.3)$$ because $a_u^Ta\0_v=0$ when $u\noadj v$ and $b_v^Tb\0_u=0$ when $u\adj v$. If $u=v$ we have $$A_v\cdot A_v=\Vert a_v\Vert^2\,\Vert b_v\Vert^2\,.$$ The upper left corner element of $A_v$ is $a_{1v}b_{1v}$, hence the ``cost'' of~$A_v$ is $(a_{1v}b_{1v})^2/\,\Vert A_v\Vert^2=c(a_v)@c(b_v)$. This, with (11.3), proves (11.1). (See the proof of Lemma~1.)\ \pfbox \meno {\bf 12. The main theorem.}\quad Lemmas 5, 6, 7, 10, and 11 establish the five inequalities claimed in (5.1); hence all five variants of~$\vartheta$ are the same function of~$G$ and~$w$. Moreover, all the inequalities in those five proofs are equalities $\bigl($with the exception of (11.1)$\bigr)$. We can summarize the results as follows. \proclaim Theorem. For all graphs $G$ and any nonnegative real labeling $w$ of~$G$ we have $$\vartheta(G,w)=\vartheta_1(G,w)= \vartheta_2(G,w)=\vartheta_3(G,w)=\vartheta_4(G,w)\,.\eqno(12.1)$$ Moreover, if $w\neq 0$, there exist orthogonal labelings $a$ and $b$ of~$G$ and $\overline{G}$, respectively, such that $$\eqalignno{c(a_v)&=w_v/\vartheta\,;&(12.2)\cr \noalign{\smallskip} \sum c(a_v)@c(b_v)&=1\,.&(12.3)\cr}$$ \noindent {\bf Proof.}\quad Relation (12.1) is, of course, (5.1); and (12.2) is (5.3). The desired labeling~$b$ is what we called~$b'$ in the proof of Lemma~10. The fact that the application of Cauchy's inequality in (10.4) is actually an equality, $$\vartheta=\biggl(\sum_vb_{1v}\sqrt{w_v}\,\biggr)^2 =\biggl(\sum_v\,\Vert b_v\Vert^2\biggr)\biggl(\, \sum_{\scriptstyle v\atop\scriptstyle b_v\neq 0}\; {b^2_{1v}\over\Vert b_v\Vert^2}\, w_v\biggr)\,,\eqno(12.4)$$ tells us that the vectors whose dot product has been squared are proportional: There is a number~$t$ such that $$\Vert b_v\Vert =t\,{b_{1v}\sqrt{w_v}\over\Vert b_v\Vert}\,,\quad {\rm if}\ b_v\neq 0\,;\qquad\Vert b_v\Vert =0\quad{\rm iff} \quad b_{1v}\sqrt{w_v}=0\,.\eqno(12.5)$$ The labeling in the proof of Lemma 10 also satisfies $$\sum_v\,\Vert b_v\Vert^2=1\,;\eqno(12.6)$$ hence $t=\pm 1/\sqrt{\vartheta}\,$. We can now show $$c(b_v)=\Vert b_v\Vert^2\,\vartheta/w_v\,,\quad{\rm when}\ w_v\neq 0\,.\eqno(12.7)$$ This relation is obvious if $\Vert b_v\Vert=0@$; otherwise we have $$c(b_v)={b^2_{1v}\over\Vert b_v\Vert^2}={\Vert b_v\Vert^2\over t^2w_v}$$ by (12.5). Summing the product of (12.2) and (12.7) over $v$ gives (12.3).\ \pfbox \meno {\bf 13. The main converse.}\quad The nice thing about Theorem 12 is that conditions (12.2) and (12.3) also provide a {\it certificate\/} that a given value~$\vartheta$ is the minimum or maximum stated in the definitions of $\vartheta$, $\vartheta_1$, $\vartheta_2$, $\vartheta_3$, and $\vartheta_4$. \proclaim Theorem. If $a$ is an orthogonal labeling of $G$ and $b$ is an orthogonal labeling of\/ $\overline{G}$ such that relations (12.2) and (12.3) hold for some~$\vartheta$ and~$w$, then $\vartheta$ is the value of $\vartheta(G,w)$. \noindent {\bf Proof.}\quad Plugging (12.2) into (12.3) gives $\sum w_vc(b_v)=\vartheta$, hence $\vartheta\leq\vartheta_4(G,w)$ by definition of~$\vartheta_4$. Also, $$\max_v\;{w_v\over c(a_v)}=\vartheta\,,$$ hence $\vartheta\geq \vartheta_1(G,w)$ by definition of~$\vartheta_1$. \ \pfbox \meno {\bf 14. Another look at {\tt TH}.}\quad We originally defined $\vartheta(G,w)$ in (4.1) in terms of the convex set~{\tt TH} defined in section~2: $$\vartheta(G,w)=\max\{\,w\cdot x\mid x\in {\tt TH}(G)\,\}\,,\qquad{\rm when}\ w\geq 0\,.\eqno(14.1)$$ We can also go the other way, defining {\tt TH} in terms of~$\vartheta$: $${\tt TH}(G)=\{\,x\geq 0\mid w\cdot x\leq\vartheta(G,w)\hbox{ for all }w\geq 0\,\}\,.\eqno(14.2)$$ Every $x\in {\tt TH}(G)$ belongs to the right-hand set, by (14.1). Conversely, if $x$ belongs to the right-hand set and if $a$ is any orthogonal labeling of~$G$, not entirely zero, let $w_v=c(a_v)$, so that $w\cdot x=\sum_v c(a_v)x_v$. Then $$\vartheta_1(G,w)\leq\max_v\bigl(w_v/c(a_v)\bigr)=1$$ by definition (5.2), so we know by Lemma 5 that $\sum c(a_v)x_v\leq 1$. This proves that $x$ belongs to ${\tt TH}(G)$. Theorem 12 tells us even more. \proclaim Lemma. ${\tt TH}(G)=\{\,x\geq 0\mid\vartheta(\overline{G},x)\leq 1\,\}$. \noindent {\bf Proof.}\quad By definition (10.1), $$\vartheta_4(\overline{G},w)=\max\left\{\,\sum_v c(a_v)w_v\mid a\hbox{ is an orthogonal labeling of }G\,\right\}\,.\eqno(14.3)$$ Thus $x\in {\tt TH}(G)$ iff $\vartheta_4(\overline{G},x)\leq 1$, when $x\geq 0$.\ \pfbox \proclaim Theorem. ${\tt TH}(G)=\{\,x\mid x_v=c(b_v)\hbox{ for some orthogonal labeling $b$ of $\overline{G}$}\,\}$. \noindent {\bf Proof.}\quad We already proved in (11.1) that the right side is contained in the left. Let $x\in {\tt TH}(G)$ and let $\vartheta=\vartheta(\overline{G},x)$. By the lemma, $\vartheta\leq 1$. Therefore, by (12.2), there is an orthogonal labeling~$b$ of~$\overline{G}$ such that $c(b_v)=x_v/\vartheta\geq x_v$ for all~$v$. It is easy to reduce the cost of any vector in an orthogonal labeling to any desired value, simply by increasing the dimension and giving this vector an appropriate nonzero value in the new component while all other vectors remain zero there. The dot products are unchanged, so the new labeling is still orthogonal. Repeating this construction for each~$v$ produces a labeling with $c(b_v)=x_v$. \ \pfbox \medskip This theorem makes the definition of $\vartheta_4$ in (10.1) identical to the definition of~$\vartheta$ in~(4.1). \meno {\bf 15. Zero weights.}\quad Our next result shows that when a weight is zero, the corresponding vertex might as well be absent from the graph. \proclaim Lemma. Let $U$ be a subset of the vertices~$V$ of a graph~$G$, and let $G'=G\vert U$ be the graph induced by~$U$ (i.e., the graph on vertices~$U$ with $u\adj v$ in~$G'$ iff $u\adj v$ in~$G$). Then if $w$ and $w'$ are nonnegative labelings of~$G$ and~$G'$ such that $$w_v=w'_v\qquad\hbox{when}\quad v\in U\,,\qquad w_v=0\qquad\hbox{when}\quad v\notin U\,,\eqno(15.1)$$ we have $$\vartheta(G,w)=\vartheta(G',w')\,.\eqno(15.2)$$ \noindent {\bf Proof.}\quad Let $a$ and $b$ satisfy (12.2) and (12.3) for $G$ and~$w$. Then $c(a_v)=0$ for $v\notin U$, so $a\vert U$ and $b\vert U$ satisfy (12.2) and (12.3) for~$G'$ and~$w'$. (Here $a\vert U$ means the vectors~$a_v$ for $v\in U$.) By Theorem~13, they determine the same~$\vartheta$.\ \pfbox \meno {\bf 16. Nonzero weights.}\quad We can also get some insight into the significance of nonzero weights by ``splitting'' vertices instead of removing them. \proclaim Lemma. Let $v$ be a vertex of $G$ and let $G'$ be a graph obtained from~$G$ by adding a new vertex~$v'$ and new edges $$u\adj v'\quad{\rm iff}\quad u\adj v\,.\eqno(16.1)$$ Let $w$ and $w'$ be nonnegative labelings of~$G$ and~$G'$ such that $$\eqalignno{w_u&=w'_u\,,\qquad\hbox{when }u\neq v\,;&(16.2)\cr \noalign{\smallskip} w_v&=w'_v+w'_{v'}\,.&(16.3)\cr}$$ Then $$\vartheta(G,w)=\vartheta(G',w')\,.\eqno(16.4)$$ \noindent {\bf Proof.}\quad By Theorem 12 there are labelings $a$ and $b$ of~$G$ and~$\overline{G}$ satisfying (12.2) and (12.3). We can modify them to obtain labelings $a'$ and~$b'$ of~$G'$ and~$\overline{G'}$ as follows, with the vectors of~$a'$ having one more component than the vectors of~$a$: $$a'_u={a_u\choose 0}\,,\qquad b'_u=b_u\,,\qquad\hbox{when }u\neq v\,;\eqno(16.5)$$ \vskip-10pt $$a'_v={a_v\choose\alpha}\,,\qquad a'_{v'}={a_v\choose -\beta}\,,\qquad \alpha=\sqrt{{w'_{v'}\over w'_v}}\; \Vert a_v\Vert\,,\qquad \beta=\sqrt{{w'_v\over w'_{v'}}}\;\Vert a_v\Vert\,;\eqno(16.6)$$ \vskip-10pt $$b'_v=b'_{v'}=b_v\,.\eqno(16.7)$$ (We can assume by Lemma 15 that $w'_v$ and $w'_{v'}$ are nonzero.) All orthogonality relations are preserved; and since $v\noadj v'$ in~$G'$, we also need to verify $$a'_v\cdot a'_{v'}=\Vert a_v\Vert^2-\alpha\beta =0\,.$$ We have $$c(a'_v)={c(a_v)\;\Vert a_v\Vert^2\over \Vert a_v\Vert^2+\alpha^2} ={c(a_v)\over 1+w'_{v'}/w'_v}={c(a_v)w'_v\over w_v}={w'_v\over\vartheta}\,,$$ and similarly $c(a'_{v'})=w'_{v'}/\vartheta$; thus (12.2) and (12.3) are satisfied by~$a'$ and~$b'$ for~$G'$ and~$w'$. \ \pfbox \medskip Notice that if all the weights are integers we can apply this lemma repeatedly to establish that $$\vartheta(G,w)=\vartheta(G')\,,\eqno(16.8)$$ where $G'$ is obtained from~$G$ by replacing each vertex~$v$ by a cluster of~$w_v$ mutually nonadjacent vertices that are adjacent to each of~$v$'s neighbors. $\bigl($Recall that $\vartheta(G')=\vartheta(G',\ones)$, by definition (4.2).$\bigr)$ In particular, if $G$ is the trivial graph~$K_2$ and if we assign the weights $M$ and~$N$, we have $\vartheta\bigl(K_2,(M, N)^T\bigr)=\vartheta(K_{M,N})$ where $K_{M,N}$ denotes the complete bipartite graph on~$M$ and~$N$ vertices. A similar operation called ``duplicating'' a vertex has a similarly simple effect: \proclaim Corollary. Let $G'$ be constructed from $G$ as in the lemma but with an additional edge between~$v$ and~$v'$. Then $\vartheta(G,w)=\vartheta(G',w')$ if $w'$ is defined by (16.2) and $$w_v=\max(w'_v,w'_{v'})\,.\eqno(16.9)$$ \noindent {\bf Proof.}\quad We may assume that $w_v=w'_v$ and $w'_{v'}\neq 0$. Most of the construction (16.5)--(16.7) can be used again, but we set $\alpha=0$ and $b'_{v'}=0$ and $$\beta=\sqrt{{w_v-w'_{v'}\over w'_{v'}}}\;\Vert a_v\Vert\,.$$ Once again the necessary and sufficient conditions are readily verified. \ \pfbox \medskip If the corollary is applied repeatedly, it tells us that $\vartheta(G)$ is unchanged when we replace the vertices of~$G$ by cliques. \meno {\bf 17. Simple examples.}\quad We observed in section 4 that $\vartheta(G,w)$ always is at least $$\vartheta_{\min}=\vartheta(K_n,w)=\max\{w_1,\ldots,w_n\}\eqno(17.1)$$ and at most $$\vartheta_{\max}=(\overline{K}_n,w)=w_1+\cdots +w_n\,.\eqno(17.2)$$ What are the corresponding orthogonal labelings? For $K_n$ the vectors of $a$ have no orthogonal constraints, while the vectors of~$b$ must satisfy $b_u\cdot b_v=0$ for all $u\neq v$. We can let $a$ be the two-dimensional labeling $$a_v={\sqrt{w_v}\choose \sqrt{\vartheta-w_v}}\,,\qquad \vartheta=\vartheta_{\min}\eqno(17.3)$$ so that $\Vert a_v\Vert^2=\vartheta$ and $c(a_v)=w_v/\vartheta$ as desired; and $b$ can be one-dimensional, $$b_v=\cases{(1)\,,&if $v=v_{\max}$\cr \noalign{\smallskip} (0)\,,&if $v\neq v_{\max}$\cr}\eqno(17.4)$$ where $v_{\max}$ is any particular vertex that maximizes~$w_v$. Clearly $$\sum_v c(a_v)@c(b_v)={c(a_{v_{\max}})\over \vartheta}= {w_{v_{\max}}\over\vartheta}=1\,.$$ For $\overline{K}_n$ the vectors of $a$ must be mutually orthogonal while the vectors of~$b$ are unrestricted. We can let the vectors~$a$ be the columns of any orthogonal matrix whose top row contains the element $$\sqrt{w_v/\vartheta}\,,\qquad\vartheta=\vartheta_{\max}\eqno(17.5)$$ in column $v$. Then $\Vert a_v\Vert^2=1$ and $c(a_v)=w_v/\vartheta$. Once again a one-dimensional labeling suffices for~$b$; we can let $b_v=(1)$ for all~$v$. \meno {\bf 18. The direct sum of graphs.}\quad Let $G=G'+G''$ be the graph on vertices $$V=V'\cup V''\eqno(18.1)$$ where the vertex sets~$V'$ and $V''$ of~$G'$ and~$G''$ are disjoint, and where $u\adj v$ in~$G$ if and only if $u,v\in V'$ and $u\adj v$ in~$G'$, or $u,v\in V''$ and $u\adj v$ in~$G''$. In this case $$\vartheta(G,w)=\vartheta(G',w')+\vartheta(G'',w'')\,,\eqno(18.2)$$ where $w'$ and $w''$ are the sublabelings of~$w$ on vertices of~$V'$ and~$V''$. We can prove (18.2) by constructing orthogonal labelings $(a,b)$ satisfying (12.2) and (12.3). Suppose $a'$ is an orthogonal labeling of $G'$ such that $$\Vert a'_v\Vert^2=\vartheta'\qquad a'_{1v}=\sqrt{w'_v}\,,\eqno(18.3)$$ and suppose $a''$ is a similar orthogonal labeling of~$G''$. If $a'$ has dimension~$d'$ and~$a''$ has dimension~$d''$, we construct a new labeling~$a$ of dimension $d=d'+d''$ as follows, where $j'$ runs from~2 to~$d'$ and $j''$ runs from~2 to~$d''$: $$\vcenter{\halign{\hfil$#\;$&$#$\hfil\qquad &\hfil$#\;$&$#$\hfil\cr &\hbox{ if $v\in V'$}&&\hbox{ if $v\in V''$}\cr \noalign{\medskip} a_{1v}&=\sqrt{w_v'}=a'_{1v}\,,&a_{1v}&=\sqrt{w''_v}=a''_{1v}\,,\cr \noalign{\smallskip} a_{j'v}&=\sqrt{\vartheta/\vartheta'}\,a'_{j'v}\,,&a_{j'v}&=0\,,\cr \noalign{\smallskip} a_{(d'+1)v}&=\sqrt{\vartheta''w'_v/\vartheta'}\,, &a_{(d'+1)v}&=-\sqrt{\vartheta'w''_v/\vartheta''}\,,\cr \noalign{\smallskip} a_{(d'+j'')v}&=0\,,&a_{(d'+j'')v} &=\sqrt{\vartheta/\vartheta''}\,a''_{j''v}\,.\cr}}\eqno(18.4)$$ Now if $u,v\in V'$ we have $$a_u\cdot a_v=\sqrt{w'_uw'_v}+{\vartheta\over\vartheta'}\,\bigl(a'_u\cdot a'_v-\sqrt{w'_uw'_v}\,\bigr)+{\vartheta''\over\vartheta'}\,\sqrt{w'_uw'_v}= {\vartheta\over\vartheta'}\,a'_u\cdot a'_v\,;\eqno(18.5)$$ thus $a_u\cdot a_v=0$ when $a'_u\cdot a'_v=0$, and $$\Vert a_v\Vert^2={\vartheta\over\vartheta'}\,\Vert a'_v\Vert^2=\vartheta\,.\eqno(18.6)$$ It follows that $c(a_v)=w_v/\vartheta$ as desired. A similar derivation holds for $u,v\in V''$. And if $u\in V'$, $v\in V''$, then $$a_u\cdot a_v=\sqrt{w'_uw''_v}-\sqrt{w'_uw''_v}=0\,.\eqno(18.7)$$ The orthogonal labeling $b$ of $\overline{G'+G''}$ is much simpler; we just let $b_v=b'_v$ for $v\in V'$ and $b_v=b''_v$ for $v\in V''$. Then (12.2) and (12.3) are clearly preserved. This proves (18.2). There is a close relation between the construction (18.4) and the construction (16.6), suggesting that we might be able to define another operation on graphs that generalizes both the splitting and direct sum operation. \meno {\bf 19. The direct cosum of graphs.}\quad If $G'$ and $G''$ are graphs on disjoint vertex sets~$V'$ and~$V''$ as in section~18, we can also define $$G=G'\mathbin{\overline+} G''\ \Longleftrightarrow \overline{G}=\overline{G'}+ \overline{G''}\,.\eqno(19.1)$$ This means $u\adj v$ in $G$ if and only if either $u\adj v$ in~$G'$ or $u\adj v$ in~$G''$ or $u$ and~$v$ belong to opposite vertex sets. In this case $$\vartheta(G,w)=\max\bigl(\vartheta(G',w'),\; \vartheta(G'',w'')\bigr)\eqno(19.2)$$ and again there is an easy way to construct $(a,b)$ from $(a',b')$ and $(a'',b'')$ to prove (19.2). Assume ``without lots of generality'' that $$\vartheta(G',w')\geq\vartheta(G'',w'')\eqno(19.3)$$ and suppose again that we have (18.3) and its counterpart for~$a''$. Then we can define $$\vcenter{\halign{\hfil$#\;$&$#$\hfil\qquad &\hfil$#\;$&$#$\hfil\cr &\hbox{ if $v\in V'$}&&\hbox{ if $v\in V''$}\cr \noalign{\medskip} a_{1v}&=\sqrt{w_{v'}}=a'_{1v}\,,&a_{1v}&=\sqrt{w_{v''}}=a''_{1v}\,,\cr \noalign{\smallskip} a_{j'v}&=a'_{j'v}\,,&a_{j'v}&=0\,,\cr \noalign{\smallskip} a_{(d'+1)v}&=0\,, &a_{(d'+1)v}&=\sqrt{(\vartheta'-\vartheta'')w''_v/\vartheta''}\,,\cr \noalign{\smallskip} a_{(d'+j'')v}&=0\,,&a_{(d'+j'')v}&=\sqrt{\vartheta'/\vartheta''}\,a''_{j''v} \,.\cr}}\eqno(19.4)$$ Now $a_v$ is essentially unchanged when $v\in V'$; and when $u,v\in V''$ we have $$a_u\cdot a_v=\sqrt{w''_uw''_v}+\left(\,{\vartheta'\over\vartheta''}-1\right)\, \sqrt{w''_uw''_v}+{\vartheta'\over\vartheta''}\, \bigl(a''_u\cdot a''_v-\sqrt{w''_uw''_v}\,\bigr)={\vartheta'\over\vartheta''} \, a''_u\cdot a''_v\,.\eqno(19.5)$$ Again we retain the necessary orthogonality, and we have $c(a_v)=w_v/\vartheta$ for all~$v$. For the $b$'s, we let $b_v=b'_v$ when $v\in V'$ and $b_v=0$ when $v\in V''$. \meno {\bf 20. A direct product of graphs.}\quad Now let $G'$ and $G''$ be graphs on vertices~$V'$ and~$V''$ and let $V$ be the $n=n'n''$ ordered pairs $$V=V'\times V''\,.\eqno(20.1)$$ We define the `strong product', $$G=G'\;\ast\;G''\eqno(20.2)$$ on $V$ by the rule $$\eqalignno{&(u',u'')\adj (v',v'')\quad{\rm or}\quad (u',u'')=(v',v'')\quad {\rm in} \ G\cr \noalign{\smallskip} &\quad\Longleftrightarrow\ (u'\adj v'\ {\rm or}\ u'=v'\ {\rm in}\ G')\quad {\rm and}\quad (u''\adj v''\ {\rm or}\ u''=v''\ {\rm in}\ G'')\,.&(20.3)\cr}$$ In this case we have, for example, $K_{n'}\;\ast\; K_{n''}=K_{n'n''}$ and $\overline{K}_{n'}\;\ast\;\overline{K}_{n''}=\overline{K}_{n'n''}$. More generally, if $G'$ is regular of degree $r'$ and $G''$ is regular of degree~$r''$, then $G'\;\ast\;G''$ is regular of degree $(r'+1)(r''+1)-1=r'r''+r'+r''$. I don't know the value of $\vartheta(G,w)$ for arbitrary~$w$, but I do know it in the special case $$w_{(v',v'')}=w'_{v'}w''_{v''}\,.\eqno(20.4)$$ \proclaim Lemma. If $G$ and $w$ are given by (20.2) and (20.4), then $$\vartheta(G,w)=\vartheta(G',w')\,\vartheta(G'',w'')\,.\eqno(20.5)$$ \noindent {\bf Proof.}\quad [\Lov]\enspace Given orthogonal labelings $(a',b')$ and $(a'',b'')$ of $G'$ and~$G''$, we let $a$ be the Hadamard product $$a_{(j',j'')(v',v'')}=a'_{j'v'}a''_{j''v''}\,,\qquad 1\leq j'\leq d'\,,\quad 1\leq j''\leq d''\,,\eqno(20.6)$$ where $d'$ and $d''$ are the respective dimensions of the vectors in $a'$ and~$a''$. Then $$\eqalignno{a_{(u',u'')}\cdot a_{(v',v'')} &=\sum_{j',j''}a'_{j'u'}a''_{j''u''}a'_{j'v'}a''_{j''v''}\cr \noalign{\smallskip} &=(a'_{u'}\cdot a'_{v'})(a''_{u''}\cdot a''_{v''})\,.&(20.7)\cr}$$ Thus $\Vert a_{(v',v'')}\Vert^2=\Vert a'_{v'}\Vert^2\,\Vert a''_{v''}\Vert^2$ and $$c(a_{(v'v'')})=c(a'_{v'})@c(a''_{v''})\,.\eqno(20.8)$$ The same construction is used for $b$ in terms of $b'$ and $b''$. All necessary orthogonalities are preserved, because we have $$\eqalign{% &(u',u'')\adj (v',v'')\hbox{ and }(u',u'')\neq (v',v'')\hbox{ in }G\cr \noalign{\vskip3pt} &\qquad \Rightarrow (u'\adj v'\hbox{ and }u'\neq v'\hbox{ in }G') \hbox{ or }(u''\adj v''\hbox{ and }u''\neq v''\hbox{ in }G'')\cr \noalign{\vskip3pt} &\qquad \Rightarrow b_{(u',u'')}\cdot b_{(v',v'')}=0\,;\cr \noalign{\smallskip} &(u',u'')\noadj (v',v'')\hbox{ and }(u',u'')\neq (v',v'')\hbox{ in }G\cr \noalign{\vskip3pt} &\qquad \Rightarrow (u'\noadj v'\hbox{ and }u'\neq v'\hbox{ in }G') \hbox{ or }(u''\noadj v''\hbox{ and }u''\neq v''\hbox{ in }G'')\cr \noalign{\vskip3pt} &\qquad \Rightarrow a_{(u',u'')}\cdot a_{(v',v'')}=0\,.\cr}$$ (In fact one of these relations is $\Leftrightarrow$, but we need only $\Rightarrow$ to make (20.7) zero when it needs to be zero.) Therefore $a$ and~$b$ are orthogonal labelings of~$G$ that satisfy (12.2) and (12.3). \pfbox \meno {\bf 21. A direct coproduct of graphs.}\quad Guess what? We also define $$G=G'\;\overline{\ast}\;G''\ \Longleftrightarrow\ \overline{G}= \overline{G'}\;\ast\;\overline{G''}\,.\eqno(21.1)$$ This graph tends to be ``richer'' than $G'\;\ast\;G''$; we have $$\eqalignno{&(u',u'')\adj (v',v'')\hbox{ and }(u',u'')\neq (v',v'') \hbox{ in }G\cr &\qquad \Longleftrightarrow (u'\adj v'\hbox{ and }u'\neq v' \hbox{ in }G')\hbox{ or }(u''\adj v''\hbox{ and }u''\neq v'' \hbox{ in }G'')\,.&(21.2)\cr}$$ Now, for instance, if $G'$ is regular of degree~$r'$ and $G''$ is regular of degree~$r''$, then $$G'\;\bar{\ast}\;G''\hbox{ is regular of degree } n'n''-(n'-r')(n''-r'')=r'n''+r''n'-r'r''\,.$$ (This is always $\geq r'r''+r'+r''$, because $r'(n''-1-r'')+r''(n'-1-r')\geq 0$.) Indeed, $G'\;\overline{\ast}\;G''\supseteq G'\;\ast\;G''$ for all graphs~$G'$ and~$G''$. The Hadamard product construction used in section~20 can be applied word-for-word to prove that $$\vartheta(G,w)=\vartheta(G',w')\,\vartheta(G'',w'')\eqno(21.3)$$ when $G$ satisfies (21.1) and $w$ has the special factored form (20.4). It follows that many graphs have identical $\vartheta$'s: \proclaim Corollary. If $G'\,\ast\,G''\subseteq G\subseteq G'\;\overline{\ast}\; G''$ and $w$ satisfies (20.4), then (21.3) holds. \noindent {\bf Proof.}\quad This is just the monotonicity relation (4.3). The reason it works is that we have $a_{(u',v')}\cdot a_{(u'',v'')}=b_{(u',v')}\cdot b_{(u'',v'')}=0$ for all pairs of vertices $(u',u'')$ and $(v',v'')$ whose adjacency differs in $G'\,\ast\,G''$ and $G'\;\overline{\ast}\;G''$. \ \pfbox \medskip Some small examples will help clarify the results of the past few sections. Let $P_3$ be the path of length~2 on 3~vertices, $\bullet\!$---$\!\bullet\!$---$\!\bullet$, and consider the four graphs we get by taking its strong product and coproduct with~$\overline{K}_2$ and~$K_2$: $$\leftline{\qquad$\overline K_2\,\ast\,P_3 = \mpfig1\qquad\qquad \vartheta=\max(u+w,v)+\max(x+z,y)$}$$ $\bigl($Since $P_3$ may be regarded as $\overline{K}_2 \mathbin{\overline +} K_1$ and $\overline{K}_2$ is $K_1+K_2$, this graph is $$\bigl((K_1+K_1) \mathbin{\overline +} K_1\bigr)+\bigl((K_1+K_1) \mathbin{\overline +}K_1\bigr)$$ and the formula for $\vartheta$ follows from (18.2) and (19.2).$\bigr)$ $$\leftline{\qquad$\overline K_2\,\mathbin{\overline\ast}\,P_3 = \mpfig2 \qquad\qquad \vartheta=\max(u+w+x+z,v+y)$}$$ (This graph is $\overline{K}_2\mathbin{\overline +}\overline{K}_4$; we could also obtain it by applying Lemma~16 three times to~$P_3$.) $$\leftline{\qquad$K_2\,\ast\,P_3 = \mpfig3\qquad\qquad \vartheta=\max\bigl(\max(u,x)+\max(w,z),\max(v,y)\bigr)$}$$ $$\leftline{\qquad$K_2\,\mathbin{\overline\ast}\,P_3 = \mpfig4\qquad\qquad \vartheta=\max\bigl(\max(u+w,x+z),\max(v,y)\bigr)$}$$ If the weights satisfy $u=\lambda x$, $v=\lambda y$, $w=\lambda z$ for some parameter~$\lambda$, the first two formulas for~$\vartheta$ both reduce to $(1+\lambda)\max(u+w,v)$, in agreement with (20.5) and (21.3). Similarly, the last two formulas for~$\vartheta$ reduce to $\max(1,\lambda)\max(u+w,v)$ in such a case. \meno {\bf 22. Odd cycles.}\quad Now let $G=C_n$ be the graph with vertices $0,1,\ldots,n-1$ and $$u\adj v\ \Longleftrightarrow u-v\equiv \pm 1\pmod{n}\,,\eqno(22.1)$$ where $n$ is an {\it odd\/} number. A~general formula for $\vartheta(C_n,w)$ appears to be very difficult; but we can compute $\vartheta(C_n)$ without too much labor when all weights are~1, because of the cyclic symmetry. It is easier to construct orthogonal labelings of $\overline{C}_n$ than of~$C_n$, so we begin with that. Given a vertex~$v$, $0\leq v7$ because it has more edges than automorphisms. Also, $\overline{C}_7$ has no automorphism that takes $0\adj 2$ into $0\adj 3$. \proclaim Lemma. If $G$ is edge-symmetric and regular, equality holds in Lemma~24. \noindent {\bf Proof.}\quad Say that $A$ is an optimum feasible matrix for~$G$ if it is a feasible matrix with $$\Lambda(A)=\vartheta(G)$$ as in section~6. We can prove that optimum feasible matrices form a convex set, as follows. First, $tA+(1-t)B$ is clearly feasible when $A$ and~$B$ are feasible. Second, $$\Lambda\bigl(tA+(1-t)B\bigr)\leq t\Lambda(A)+(1-t)\Lambda(B)\,,\qquad 0\leq t\leq 1\eqno(25.2)$$ holds for all symmetric matrices $A$ and~$B$, by (6.2); this follows because there is a unit vector~$x$ such that $\Lambda\bigl(tA+(1-t)B\bigr)=x^T\bigl(tA+(1-t)B\bigr)x=tx^T\!Ax+(1-t)x^TBx \leq t\Lambda(A)+(1-t)\Lambda(B)$. Third, if $A$ and~$B$ are optimum feasible matrices, the right side of (25.2) is $\vartheta(G)$ while the left side is $\geq\vartheta(G)$ by (6.3). Therefore equality holds. If $A$ is an optimum feasible matrix for~$G$, so is $p(A)$, the matrix obtained by permuting rows and columns by an automorphism~$p$. (I~mean $p(A)_{uv}=A_{p(u)p(v)}$.) Therefore the average, $\bar{A}$, over all~$p$ is also an optimal feasible matrix. Since $p(\bar{A})=\bar{A}$ for all automorphisms~$p$, and since $G$ is edge-symmetric, $\bar{A}$~has the form $J+xB$ where $B$ is the adjacency matrix of~$G$. The bound in Lemma~24 is therefore tight.\ \pfbox (Note: If $p$ is a permutation, let $P_{uv}=1$ if $u=p(v)$, otherwise~0. Then $(P^T\!AP)_{uv}=\sum P^T_{uj}A_{jk}P_{kv}=A_{p(u)p(v)}$, so $p(A)=P^T\!AP$.) The argument in this proof shows that the set of all optimum feasible matrices~$A$ for~$G$ has a common eigenvector~$x$ such that $Ax=\vartheta(G)x$. The argument also shows that, if $G$ has an edge automorphism taking $u\adj v$ into $u'\adj v'$, we can assume without loss of generality that $A_{uv}=A_{u'v'}$ in an optimum feasible matrix. This simplifies the computation of $\Lambda(A)$, and justifies our restriction to circulant matrices (22.14) in the case of cyclic graphs. \proclaim Theorem. If $G$ is vertex-symmetric, $\vartheta(G)@\vartheta(\,\overline{G}\,)=n$. \noindent {\bf Proof.}\quad Say that $b$ is an optimum normalized labeling of $\overline{G}$ if it is a normalized orthogonal labeling of~$\overline{G}$ achieving equality in (7.1) when all weights are~1: $$\vartheta=\sum_{u,v}b_u\cdot b_v\,,\qquad \sum_v\Vert b_v\Vert^2=1\,,\qquad b_u\cdot b_v=0\hbox{ when }u\adj v\,.\eqno(25.3)$$ Let $B$ be the corresponding spud; i.e., $B_{uv}=b_u\cdot b_v$ and $\vartheta=\sum_{u,v}B_{uv}$. Then $p(B)$ is also equivalent to an optimum normalized labeling, whenever $p$ is an automorphism; and such matrices~$B$ form a convex set, so we can assume as in the lemma that $B=p(B)$ for all automorphisms~$p$. Since $G$ is vertex-symmetric, we must have $B_{vv}=1/n$ for all vertices~$v$. Thus there is an optimum normalized labeling~$b$ with $\Vert b_v\Vert^2=1/n$, and the arguments of Lemma~10 and Theorem~12 establish the existence of such a~$b$ with $$c(b_v)=\vartheta(G)/n\eqno(25.4)$$ for all $v$. But $b$ is an orthogonal labeling of~$\overline{G}$, hence $$\vartheta_1(\,\overline{G},\ones)\leq n/\vartheta(G)$$ by the definition (5.2) of $\vartheta_1$. Thus $\vartheta(\,\overline{G}\,) \vartheta(G)\leq n$; we have already proved the reverse inequality in Lemma~23. \ \pfbox \meno {\bf 26. Consequence for eigenvalues.}\quad A curious corollary of the results just proved is the following fact about eigenvalues. \proclaim Corollary. If the graphs $G$ and~$\overline{G}$ are vertex-symmetric and edge-symmetric, and if the adjacency matrix of~$G$ has eigenvalues $$\lambda_1\geq \lambda_2\geq \cdots\geq \lambda_n\,,\eqno(26.1)$$ then $$(\lambda_1-\lambda_n)(n-\lambda_1+\lambda_2) =-\lambda_n(\lambda_2+1)n\,.\eqno(26.2)$$ \noindent {\bf Proof.}\quad By Lemma 25 and Theorem 25, $${n\Lambda(-B)\over\Lambda(B)+\Lambda(-B)}\ {n\Lambda(-\overline{B}\,)\over \Lambda(\,\overline{B}\,) +\Lambda(-\overline{B}\,)} =n\,,\eqno(26.3)$$ where $B$ and $\overline{B}$ are the adjacency matrices of $G$ and~$\overline{G}$, and where we interpret 0/0 as~1. We have $$\overline{B}=J-I-B\,.\eqno(26.4)$$ If the eigenvalues of $B$ are given by (26.1), the eigenvalues of~$\overline{B}$ are therefore $$n-1-\lambda_1\geq -1-\lambda_n\geq\cdots\geq -1-\lambda_2\,.\eqno(26.5)$$ (We use the fact that $G$ is regular of degree~$\lambda_1$.) Formula (26.2) follows if we plug the values $\Lambda(B)=\lambda_1$, $\Lambda(-B)=-\lambda_n$, $\Lambda(\,\overline{B}\,)=n-1-\lambda_1$, $\Lambda(-\overline{B}\,)=1+\lambda_2$ into (26.3).\ \pfbox \meno {\bf 27. Further examples of symmetric graphs.}\quad Consider the graph $P(m,t,q)$ whose vertices are all ${m\choose t}$ subsets of cardinality~$t$ of some given set~$S$ of cardinality~$m$, where $$u\adj v\ \Longleftrightarrow\ \vert u\cap v\vert=q\,.\eqno(27.1)$$ We want $0\leq q1$. Then the matrix~$A$ with $A_{uv}=a_u\cdot a_v$ is easily seen to be $\lambda$-compatible with $G$ and $w$. Conversely, if such a matrix $A$ exists, there are $n+1$ vectors $a_0$, \dots,~$a_n$ such that $A_{uv}=a_u\cdot a_v$; in particular, $\Vert a_{0}\Vert^2=\lambda$. Let $Q$ be an orthogonal matrix with first row $a_0^T\!/\sqrt\lambda$, and define $a'_v=Qa_v$ for all~$v$. Then $a'_{10}=\sqrt\lambda$ and $a'_{j0}=0$ for all $j>1$. Also $a'_u\cdot a'_v= a_u\cdot a_v=A_{uv}$ for all $u$ and~$v$. Hence $\sqrt\lambda a'_{1v}= a'_0\cdot a'_v=A_{0v}=w_v$ and $\Vert a'_v\Vert^2=a'_v\cdot a'_v=A_{vv}=w_v$, for all $v\in G$, proving that $c(a'_v)=w_v/\lambda$. Finally $a'$ is an orthogonal labeling, since $a'_u\cdot a'_v=A_{uv}=0$ whenever $u\noadj v$.\quad\pfbox \proclaim Corollary. $x\in{\tt TH}(G)$ iff there exists a matrix $1$-compatible with $\overline G$ and~$x$. \noindent {\bf Proof.}\quad Set $\lambda=1$ in the lemma and apply Theorem 14.\quad\pfbox \medskip The corollary and definition (4.1) tell us that $\vartheta(G,w)$ is $\max(w_1x_1+\cdots+w_nx_n)$ over all $x$ that appear in matrices that are $1$-compatible for $\overline G$ and $x$. Theorem~12 tells us that $\vartheta(G,w)$ is also the minimum~$\lambda$ such that there exists a $\lambda$-compatible matrix for $G$ and~$w$. The ``certificate'' property of Theorem~13 has an even stronger formulation in matrix terms: \proclaim Theorem. Given a nonnegative weight vector $w=(w_1,\ldots,w_n)^T$, let $A$ be $\lambda$-compatible with $G$ and~$w$, where $\lambda$ is as small as possible, and let $B$ be $1$-compatible with $\overline G$ and~$x$, where $w_1x_1+\cdots+w_nx_n$ is as large as possible. Then $$A@@DB=0\,,\eqno(29.1)$$ where $D$ is the diagonal matrix with $D_{00}=-1$ and $D_{vv}=+1$ for all $v\ne0$. Conversely, if $A$ is $\lambda$-compatible with $G$ and~$w$ and if $B$ is $1$-compatible with $\overline G$ and~$x$, then (29.1) implies that $\lambda=w_1x_1+\cdots+w_nx_n=\vartheta(G,w)$. \noindent {\bf Proof.}\quad Assume that $A$ is $\lambda$-compatible with $G$ and $w$, and $B$ is $1$-compatible with $\overline G$ and~$x$. Let $B'=DBD$, so that $B'$ is a spud with $B'_{00}=1$, $B'_{0v}=B'_{v0}=-x_v$, and $B'_{uv}=B_{uv}$ when $u$ and~$v$ are nonzero. Then the dot product $A\cdot B'$ is $$\lambda-w_1x_1-\cdots-w_nx_n-w_1x_1-\cdots-w_nx_n+w_1x_1+\cdots+w_nx_n =\lambda-(w_1x_1+\cdots+w_nx_n)\,,$$ because $A_{uv}B_{uv}=0$ when $u$ and $v$ are vertices of $G$. We showed in the proof of F9 in section~9 that the dot product of spuds is nonnegative; in fact, that proof implies that the dot product is zero if and only if the ordinary matrix product is zero. So $\lambda=w_1x_1+\cdots+w_nx_n=\vartheta(G,w)$ iff $AB'=0$, and this is equivalent to~(29.1).\quad\pfbox \medskip Equation (29.1) gives us further information about the orthogonal labelings $a$ and~$b$ that appear in Theorems 12 and~13. Normalize those labelings so that $\Vert a\Vert^2=w_v$ and $\Vert b\Vert^2=x_v$. Then we have $$\openup 1\jot \eqalignno{\sum_{t\in G}w_t\,(b_t\cdot b_v)&=\vartheta\,x_v\,,&(29.2)\cr \sum_{t\in G}x_t\,(a_t\cdot a_v)&=w_v\,,&(29.3)\cr \sum_{t\in G}(a_t\cdot a_u)@(b_t\cdot b_v)&=w_ux_v\,,&(29.4)\cr}$$ for all vertices $u$ and $v$ of $G$. (Indeed, (29.2) and (29.4) are respectively equivalent to $(AB')_{0v}=0$ and $(AB')_{vv}=0$; (29.3) is equivalent to $(B'A)_{0v}=0$.) Notice that if $\widehat A$ and $\widehat B$ are the $n\times n$ spuds; obtained by deleting row~0 and column~0 from optimum matrices $A$ and~$B$, these equations are equivalent to $$\widehat Bw=\vartheta x\,,\qquad \widehat Ax=w\,,\qquad \widehat A\widehat B=wx^T\,.\eqno(29.5)$$ Equation (29.1) is equivalent to (29.5) together with the condition $w_1x_1+\cdots+w_nx_n=\vartheta$. Since $AB'=0$ iff $B'A=0$ when $A$ and $B'$ are symmetric matrices, the optimum matrices $A$ and~$B'$ commute. This implies that they have common eigenvectors: There is an orthogonal matrix~$Q$ such that $$A=Q\,{\rm diag}\,(\lambda_0,\ldots,\lambda_n)\,Q^T\,,\qquad\qquad B'=Q\,{\rm diag}\,(\mu_0,\ldots,\mu_n)\,Q^T\,.\eqno(29.6)$$ Moreover, the product is zero, so $$\lambda_0\mu_0=\cdots=\lambda_n\mu_n=0\,.\eqno(29.7)$$ The number of zero eigenvalues $\lambda_k$ is $n+1-d$, where $d$ is the smallest dimension for which there is an orthogonal labeling~$a$ with $A_{uv}=a_u\cdot a_v$. A similar statement holds for $B'$, since the eigenvalues of $B$ and $B'$ are the same; $y$ is an eigenvector for $B$ iff $Dy$ is an eigenvector for $B'$. In the case $G=C_n$, studied in section~22, we constructed an orthogonal labeling (22.3) with only three dimensions, so all but~3 of the eigenvalues~$\mu_k$ were zero. When all the weights~$w_v$ are nonzero and $\vartheta(G)$ is large, Theorem~28 implies that a large number of $\lambda_k$ must be nonzero, hence a large number of $\mu_k$ must be zero. The ``optimum feasible matrices'' $A$ studied in section 6 are related to the matrices $\widehat A$ of (29.5) by the formula $$\vartheta \widehat A=ww^T-\vartheta\,{\rm diag}\,(w_1,\ldots,w_n) -{\rm diag}\,(\sqrt{w_1},\ldots,\sqrt{w_n}\,)\,A\, {\rm diag}\,(\sqrt{w_1},\ldots,\sqrt{w_n}\,)\,,\eqno(29.8)$$ because of the construction following (6.6). If the largest eigenvalue $\Lambda(A)=\vartheta$ of~$A$ occurs with multiplicity~$r$, the rank of $\vartheta I-A$ will be $n-r$, hence $\widehat A$ will have rank $n-r$ or $n-r+1$, and the number of zero eigenvalues $\lambda_k$ in (29.6) will be $r+1$ or $r$. \meno {\bf 30. Antiblockers.}\quad The convex sets {\tt STAB}, {\tt TH}, and {\tt QSTAB} defined in section~2 have many special properties. For example, they are always nonempty, closed, convex, and nonnegative; they also satisfy the condition $$0\leq y\leq x \quad{\rm and}\quad x\in X\ \Rightarrow\ y\in X\,.\eqno(30.1)$$ A set $X$ of vectors satisfying all five of these properties is called a {\it convex corner}. If $X$ is any set of nonnegative vectors we define its {\it antiblocker\/} by the condition $${\rm abl}\,X=\{\,y\geq 0\mid x\cdot y\leq 1\hbox{ for all }x\in X\,\}\,.\eqno(30.2)$$ Clearly abl$\,X$ is a convex corner, and ${\rm abl}\,X\supseteq{\rm abl}\,X'$ when $X\subseteq X'$. \proclaim Lemma. If $X$ is a convex corner we have ${\rm abl}\,{\rm abl}\,X=X$. \noindent {\bf Proof.}\quad (Compare with the proof of F5 in section~8.) The relation $X\subseteq {\rm abl}\,{\rm abl}\,X$ is obvious by definition (30.2), so the lemma can fail only if there is some $z\in {\rm abl}\,{\rm abl}\,X$ with $z\notin X$. Then there is a hyperplane separating~$z$ from~$X$, by~F1; i.e., there is a vector~$y$ and a number~$b$ such that $x\cdot y\leq b$ for all $x\in X$ but $z\cdot y>b$. Let $y'$ be the same as~$y$ but with all negative components changed to zero. Then $(y',b)$ is also a separating hyperplane. [{\it Proof:\/} If $x\in X$, let $x'$ be the same as~$x$ but with all components changed to zero where $y$ has a negative entry; then $x'\in X$, and $x\cdot y'=x'\cdot y\leq b$. Furthermore $z\cdot y'\geq z\cdot y>b$.] If $b=0$, we have $\lambda y'\in {\rm abl}\,X$ for all $\lambda >0@$; this contradicts $z\cdot\lambda y'\leq 1$. We cannot have $b<0$, since $0\in X$. Hence $b>0$, and the vector $y'/b\in {\rm abl}\,X$. But then $z\cdot (y'/b)$ must be $\leq 1$, a~contradiction. \ \pfbox \proclaim Corollary. If $G$ is any graph we have $${\tt STAB}(\overline{G})={\rm abl}\;{\tt QSTAB}(G)\,,\eqno(30.3)$$ \vskip-20pt $${\tt TH}(\overline{G})={\rm abl}\;{\tt TH}(G)\,,\eqno(30.4)$$ \vskip-20pt $${\tt QSTAB}(\overline{G})={\rm abl}\;{\tt STAB}(G)\,.\eqno(30.5)$$ \noindent {\bf Proof.}\quad First we show that $${\rm abl}\,X=\hbox{abl convex hull $X$}.\eqno(30.6)$$ The left side surely contains the right. And any element $y\in{\rm abl}\,X$ will satisfy $$(\alpha_1x^{(1)}+\cdots +\alpha_kx^{(k)})\cdot y\leq 1$$ when the $\alpha$'s are nonnegative scalars summing to~1 and the $x^{(j)}$ are in~$X$. This proves (30.6), because the convex hull of~$X$ is the set of all such $\alpha_1x^{(1)}+\cdots +\alpha_kx^{(k)}$. Now (30.6) implies (30.5), because the definitions in section~2 say that $$\eqalign{{\tt QSTAB}(\overline{G})&={\rm abl}\,\{\,x\mid x\hbox{ is a clique labeling of $\overline{G}$}\,\}\cr \noalign{\smallskip} &={\rm abl}\,\{\,x\mid x\hbox{ is a stable labeling of $G$}\,\}\,,\cr \noalign{\smallskip} {\tt STAB}(G)&=\hbox{convex hull}\,\{\,x\mid x\hbox{ is a stable labeling of $G$}\,\}\,.\cr}$$ And (30.5) is equivalent to (30.3) by the lemma, because ${\tt STAB}(G)$ is a convex corner. (We must prove (30.1), and it suffices to do this when $y$ equals~$x$ in all but one component; and in fact by convexity we may assume that $y$ is~0 in that component; and then we can easily prove~it, because any subset of a stable set is stable.) Finally, (30.4) is equivalent to Theorem~14, because ${\tt TH}(G)={\rm abl}\,\{\,x\mid x_v=c(a_v)$ for some orthogonal labeling of~$G\,\}$. \ \pfbox \medskip The sets {\tt STAB} and {\tt QSTAB} are polytopes, i.e., they are bounded and can be defined by a finite number of inequalities. But the antiblocker concept applies also to sets with curved boundaries. For example, let $$X=\{\,x\geq 0\bigm| \,\Vert x\Vert\leq 1\,\}\eqno(30.7)$$ be the intersection of the unit ball and the nonnegative orthant. Cauchy's inequality implies that $x\cdot y\leq 1$ whenever $\Vert x\Vert \leq 1$ and $\Vert y\Vert\leq 1$, hence $X\subseteq{\rm abl}\,X$. And if $y\in{\rm abl}\,X$ we have $y\in X$, since $y\neq 0$ implies $\Vert y\Vert =y\cdot(y/\,\Vert y\Vert)\leq 1$. Therefore $X={\rm abl}\,X$. In fact, the set $X$ in (30.7) is the only set that equals its own antiblocker. If $Y={\rm abl}\,Y$ and $y\in Y$ we have $y\cdot y\leq 1$, hence $Y\subseteq X$; this implies ${\rm abl}\,Y\supseteq X$. \meno {\bf 31. Perfect graphs.}\quad Let $\omega(G)$ be the size of a largest clique in~$G$. The graph~$G$ is called {\it perfect\/} if every induced subgraph~$G'$ of~$G$ can be colored with $\omega(G')$ colors. (See section~15 for the notion of induced subgraph. This definition of perfection was introduced by Claude Berge in 1961.) Let $G^+$ be $G$ with vertex~$v$ duplicated, as described in section~16. This means we add a new vertex~$v'$ with the same neighbors as~$v$ and with $v\adj v'$. \proclaim Lemma. If $G$ is perfect, so is $G^+$. \noindent {\bf Proof.}\quad Any induced subgraph of $G^+$ that is not $G^+$ itself is either an induced subgraph of~$G$ (if it omits $v$ or~$v'$ or both), or has the form $G'^+$ for some induced subgraph~$G'$ of~$G$ (if it retains $v$ and~$v'$). Therefore it suffices to prove that $G^+$ can be colored with $\omega(G^+)$ colors. Color $G$ with $\omega(G)$ colors and suppose $v$ is red. Let $G'$ be the subgraph induced from~$G$ by leaving out all red vertices except~$v$. Recolor $G'$ with $\omega(G')$ colors, and assign a new color to the set $G^+\backslash G'$, which is stable in~$G^+$. This colors $G^+$ with $\omega(G')+1$ colors, hence $\omega(G^+)\leq\omega(G')+1$. We complete the proof by showing that $\omega(G^+)=\omega(G')+1$. Let $Q$ be a clique of size $\omega(G')$ in~$G'$. {\bf Case 1.} $v\in Q$. Then $Q\cup\{v'\}$ is a clique of~$G^+$. {\bf Case 2.} $v\notin Q$. Then $Q$ contains no red element. \noindent In both cases we can conclude that $\omega(G^+)\geq \omega(G')+1$. \ \pfbox \proclaim Theorem. If $G$ is perfect, ${\tt STAB}(G)={\tt QSTAB}(G)$. \noindent {\bf Proof.}\quad It suffices to prove that every $x\in{\tt QSTAB}(G)$ with {\it rational\/} coordinates is a member of ${\tt STAB}(G)$, because ${\tt STAB}(G)$ is a closed set. Suppose $x\in{\tt QSTAB}(G)$ and $qx$ has integer coordinates. Let $G^+$ be the graph obtained from~$G$ by repeatedly duplicating vertices until each original vertex~$v$ of~$G$ has been replaced by a clique of size~$qx_v$. Call the vertices of that clique the {\it clones\/} of~$v$. By definition of ${\tt QSTAB}(G)$, if $Q$ is any clique of~$G$ we have $$\sum_{v\in Q}x_v\leq 1\,.$$ Every clique $Q'$ of $G^+$ is contained in a clique of size $\sum_{v\in Q}qx_v$ for some clique~$Q$ of~$G$. (Including all clones of each element yields this possibly larger clique.) Thus $\omega(G^+)\leq q$, and the lemma tells us that $G^+$ can be colored with $q$~colors because $G^+$ is perfect. For each color $k$, where $1\leq k\leq q$, let $x^{(k)}_v=1$ if some clone of~$v$ is colored~$k$, otherwise $x_v^{(k)}=0$. Then $x^{(k)}$ is a stable labeling. Hence $${1\over q}\ \sum_{k=1}^qx^{(k)}\in{\tt STAB}(G)\,.$$ But every vertex of $G^+$ is colored, so $\sum_{k=1}^q x_v^{(k)}=qx_v$ for all~$v$, so $q^{-1}\sum_{k=1}^qx^{(k)}=x$.\ \pfbox \meno {\bf 32. A characterization of perfection.}\quad The converse of Theorem 31 is also true; but before we prove it we need another fact about convex polyhedra. \proclaim Lemma. Suppose $P$ is the set $\{\,x\geq 0\mid x\cdot z\leq 1\hbox{ for all }z\in Z\,\}={\rm abl}\,Z$ for some finite set~$Z$ and suppose $y\in {\rm abl}\,P$, i.e., $y$ is a nonnegative vector such that $x\cdot y\leq 1$ for all $x\in P$. Then the set $$\postdisplaypenalty=10000 Q=\{\,x\in P\mid x\cdot y=1\,\}\eqno(32.1)$$ is contained in the set $\{\,x\mid x\cdot z=1\,\}$ for some $z\in Z$ (unless $Q$ and~$Z$ are both empty). \noindent {\bf Proof.}\quad This lemma is ``geometrically obvious''---it says that every vertex, edge, etc., of a convex polyhedron is contained in some ``facet''---but we ought also to prove~it. The proof is by induction on~$\vert Z\vert$. If $Z$ is empty, the result holds because $P$ is the set of all nonnegative~$x$, hence $y$ must be~0 and $Q$ must be empty. Suppose $z$ is an element of~$Z$ that does not satisfy the condition; i.e., there is an element $x\in P$ with $x\cdot y=1$ and $x\cdot z\neq 1$. Then $x\cdot z<1$. Let $Z'=Z\setminus\{z\}$ and $P'={\rm abl}\,Z'$. It follows that $x'\cdot y\leq 1$ for all $x'\in P'$. For if $x'\cdot y>1$, a~convex combination $x''=\epsilon x+ (1-\epsilon)x'$ will lie in~$P$ for sufficiently small~$\epsilon$, but $x''\cdot y>1$. Therefore by induction, $Q'=\{\,x\in P'\mid x\cdot y=1\,\}$ is contained in $\{\,x\mid x\cdot z'=1\,\}$ for some $z'\in Z'$, unless $Q'$ is empty, when we can take $z'= z$. And $Q\subseteq Q'$, since $P\subseteq P'$. \pfbox \proclaim Theorem. $G$ is perfect if and only if ${\tt STAB}(G)={\tt QSTAB}(G)$. \noindent {\bf Proof.}\quad As in section 15, let $G\vert U$ be the graph induced from~$G$ by restriction to vertices~$U$. If $X$ is a set of vectors indexed by~$V$ and if $U\subseteq V$, let $X\vert U$ be the set of all vectors indexed by~$U$ that arise from the vectors of~$X$ when we suppress all components~$x_v$ with $v\notin U$. Then it is clear that $${\tt QSTAB}(G\vert U)={\tt QSTAB}(G)\vert U\,,\eqno(32.2)$$ because every $x\in{\tt QSTAB}(G\vert U)$ belongs to ${\tt QSTAB}(G)$ if we set $x_v=0$ for $v\notin U$, and every $x\in{\tt QSTAB}(G)$ satisfies $\sum_{v\in Q}x_v\leq 1$ for every clique $Q\subseteq U$. Also $${\tt STAB}(G\vert U)={\tt STAB}(G)\vert U\,,\eqno(32.3)$$ because every stable labeling of $G\vert U$ is a stable labeling of~$G$ if we extend it with zeros, and every stable labeling of~$G$ is stable for $G\vert U$ if we ignore components not in~$U$. Therefore ${\tt STAB}(G)={\tt QSTAB}(G)$ iff ${\tt STAB}(G')={\tt QSTAB}(G')$ for all induced graphs. By Theorem~31 we need only prove that ${\tt STAB}(G)={\tt QSTAB}(G)$ implies $G$ can be colored with $\omega(G)$ colors. Suppose ${\tt STAB}(G)={\tt QSTAB}(G)$. Then by Corollary~30, $${\tt STAB}(\overline{G})={\tt QSTAB}(\overline{G})\,.\eqno(32.4)$$ Let $P={\tt STAB}(\overline{G})$, and let $y=\ones/\omega(G)$. Then $x\cdot y\leq 1$ whenever $x$ is a clique labeling of~$G$, i.e., whenever $x$ is a stable labeling of~$\overline{G}$; so $x\cdot y\leq 1$ for all $x\in P$. Let $Z$ be the set of all stable labelings of~$G$, i.e., clique labelings of~$\overline{G}$. Then $P={\tt QSTAB}(\overline{G})={\rm abl}\,Z$ and $Z$ is nonempty. So the lemma applies and it tells us that the set~$Q$ defined in (32.1) is contained in $\{\,x\mid x\cdot z=1\,\}$ for some stable labeling~$z$ of~$G$. Therefore every maximum clique labeling~$x$ satisfies $x\cdot z=1$; i.e., every clique of size $\omega(G)$ intersects the stable set~$S$ corresponding to~$z$. So $\omega(G')=\omega(G)-1$, where $$G'=G\vert(V\setminus S)\,.\eqno(32.5)$$ By induction on $\vert V\vert$ we can color the vertices of~$G'$ with $\omega(G')$ colors, then we can use a new color for the vertices of~$S$; this colors~$G$ with $\omega(G)$ colors.\ \pfbox \medskip Lov\'asz states in [\LL] that he knows no polynomial time algorithm to test if $G$ is perfect; but he conjectures (``guesses'') that such an algorithm exists, because the results we are going to discuss next suggest that much more might be provable. \meno {\bf 33. Another definition of $\vartheta$.}\quad The following result generalizes Lemma~9.3.21 of~[\GLSbook]. \proclaim Lemma. Let $a$ and $b$ be orthogonal labelings of $G$ and~$\overline{G}$ that satisfy the conditions of Theorem~12, normalized so that $$\Vert a_v\Vert^2\Vert b_v\Vert^2=w_vc(b_v)\,,\qquad a_{1v}\geq 0\,,\qquad \hbox{and}\qquad b_{1v}\geq 0\,,\eqno(33.1)$$ for all $v$. Then $$\sum_va_{jv}b_{kv}=\cases{\sqrt{\vartheta(G,w)}\,,&\rm if $j=k=1$;\cr 0,&\rm otherwise.\cr}\eqno(33.2)$$ \noindent {\bf Proof.}\quad Let $a_0=(\sqrt\vartheta,0,\ldots,0)^T$ and $b_0=(-1,0,\ldots,0)^T$. Then the $(n+1)\times(n+1)$ matrices $A=a^Ta$ and $B=b^Tb$ are spuds, and $A\cdot B=0$. (In the special case $\Vert a_v\Vert^2=w_v$, $\Vert b_v\Vert^2=c(b_v)$, matrix~$B$ is what we called $B'$ in the proof of Theorem~29.) Therefore $0={\rm tr}\,A^TB={\rm tr}\,a^Tab^Tb={\rm tr}\,ba^Tab^T= (ab^T)\cdot(ab^T)$, and we have $ab^T=0$. In other words $$a_{j0}b_{k0}+\sum_va_{jv}b_{kv}=0$$ for all $j$ and $k$.\quad\pfbox \medskip We now can show that $\vartheta$ has yet another definition, in some ways nicer than the one we considered in section~6. (Someday I~should try to find a simpler way to derive all these facts.) Call the matrix~$B$ {\it dual feasible\/} for~$G$ and~$w$ if it is indexed by vertices and $$\eqalignno{&\hbox{$B$ is real and symmetric;}\cr &\hbox{$B_{vv}=w_v$ for all $v\in V$;}\cr &\hbox{$B_{uv}=0$ whenever $u\noadj v$ in $G$;}&(33.3)\cr}$$ and define $$\eqalignno{ \vartheta_6(G,w)=\max\{\,&\Lambda(B)\,\vert\cr &B\hbox{ is positive semidefinite and dual feasible for $G$ and $w$}\,\}\,.&(33.4)\cr}$$% $\bigl($Compare with the analogous definitions in (6.1) and (6.3).$\bigr)$ \proclaim Theorem. $\vartheta(G,w)=\vartheta_6(G,w)$. \noindent{\bf Proof.}\quad If $B$ is positive semidefinite and dual feasible, and if $\lambda$ is any eigenvalue of~$B$, we can write $B=QDQ^T$ where $Q$ is orthogonal and $D$ is diagonal, with $D_{11}=\lambda$. Let $b=\sqrt{D}\,Q^T$; then $b$ is an orthogonal labeling of~$\overline{G}$ with $\Vert b_v\Vert^2=w_v$ for all~$v$. Furthermore $c(b_v)=b^2_{1v}/w_v=\lambda\,q^2_{v1}/w_v$, where $(q_{11},\ldots,q_{n1})$ is the first column of~$Q$. Therefore $\sum_v c(b_v)w_v=\lambda\sum_v q^2_{v1}=\lambda$, and we have $\lambda \leq \vartheta_4(G,w)$ by (10.1). This proves that $\vartheta_6\leq \vartheta$. Conversely, let $a$ and $b$ be orthogonal labellings of~$G$ and~$\overline{G}$ that satisfy the conditions of Theorem~12. Normalize them so that $\Vert a_v\Vert^2=c(b_v)$ and $\Vert b_v\Vert^2=w_v$. Then $a^2_{1v}=c(a_v)c(b_v)= w_vc(b_v)/\vartheta=b^2_{1v}/\vartheta$. The lemma now implies that $(b_{11},\ldots,b_{1n})^T$ is an eigenvector of~$b^Tb$, with eigenvalue~$\vartheta$. This proves that $\vartheta\leq \vartheta_6$. \ \pfbox \proclaim Corollary. $\vartheta(G)=1+\max\{\,\Lambda(B)/\Lambda(-B) \,\vert\, B\hbox{ is dual feasible for $G$ and 0}\,\}$. \noindent{\bf Proof.}\quad If $B$ is dual feasible for $G$ and~0, its eigenvalues are $\lambda_1\geq \cdots\geq \lambda_n$ where $\lambda_1=\Lambda(B)$ and $\lambda_n=-\Lambda(-B)$. Then $B'=I+B/\Lambda(-B)$ has eigenvalues $1-\lambda_1/\lambda_n,\ldots,\break 1-\lambda_n/\lambda_n=0$. Consequently $B'$ is positive semidefinite and dual feasible for~$G$ and~$\ones$, and $1+\Lambda(B)/\Lambda(-B)=\Lambda(B')\leq \vartheta_6(G)$. Conversely, suppose $B'$ is positive semidefinite and dual feasible for~$G$ and~$\ones$, with $\Lambda(B')=\vartheta=\vartheta(G)$. Let $B=B'-I$. Then $B$ is dual feasible for~$G$ and~0, and $0\leq\Lambda(-B)\leq 1$ since the sum of the eigenvalues of~$B$ is $\mathop{\rm tr} B=0$. It follows that $\vartheta -1=\Lambda(B) \leq\Lambda(B)/\Lambda(-B)$. \ \pfbox \meno {\bf 34. Facets of {\tt TH}.}\quad We know that ${\tt TH}(G)$ is a convex corner set in $n$-dimensional space, so it is natural to ask whether it might have $(n-1)$-dimensional facets on its nontrivial boundary---for example, a~straight line segment in two dimensions, or a region of a plane in three dimensions. This means it has $n$~linearly independent vectors~$x^{(k)}$ such that $$\sum_vx_v^{(k)}c(a_v)=1\eqno(34.1)$$ for some orthogonal labeling $a$ of~$G$. \proclaim Theorem. If ${\tt TH}(G)$ contains linearly independent solutions $x^{(1)},\ldots,x^{(n)}$ of (34.1), then there is a maximal clique~$Q$ of~$G$ such that $$c(a_v)=\cases{1\,,&$v\in Q$;\cr 0\,,&$v\notin Q$.\cr}\eqno(34.2)$$ \noindent {\bf Proof.}\quad Theorem 14 tell us that every $x^{(k)}\in {\tt TH}(G)$ has $x_v^{(k)}=c(b_v^{(k)})$ for some orthogonal labeling of~$\overline{G}$. Set $w_v= c(a_v)$; then $\vartheta(G,w)=1$, by Theorem 13. We can normalize the labelings so that $\Vert a_v\Vert^2=a_{1v}=w_v$ and $\Vert b_v^{(k)}\Vert^2 =b_{1v}^{(k)}=x_v^{(k)}$. Hence, by Lemma~33, $$\sum_vx_v^{(k)}a_v=\pmatrix{1\cr 0\cr \vdots\cr 0\cr}=e_1\,.\eqno(34.3)$$ Let $$Q=\{\,v\mid a_{1v}\neq 0\,\}=\{\,v\mid c(a_v)\neq 0\,\}\eqno(34.4)$$ and suppose $Q$ has $m$ elements. Then (34.3) is equivalent to the matrix equation $$Ax^{(k)}=e_1\eqno(34.5)$$ where $A$ is a $d\times m$ matrix and $x^{(k)}$ has $m$ components $x_v^{(k)}$, one for each $v\in Q$. By hypothesis there are $m$ linearly independent solutions to (34.5), because there are $n$ linearly independent solutions to (34.3). But then there are $m-1$ linearly independent solutions to $Ax=0$, and it follows that $A$ has rank~1: Every row of~$A$ must be a multiple of the top row (which is nonzero). And then (34.5) tells us that all rows but the top row are zero. We have proved that $$c(a_v)\neq 0\ \Rightarrow \ c(a_v)=1\,.\eqno(34.6)$$ Therefore if $u$ and $v$ are elements of~$Q$ we have $a_u\cdot a_v\neq 0$, hence $u\adj v$; $Q$~is a clique. Moreover, $Q$ is maximal. For if $v\notin Q$ is adjacent to all elements of~$Q$, there is a $k$ such that $x_v^{(k)}>0$. But the characteristic labeling of $Q\cup\{v\}$ is an orthogonal labeling~$a'$ such that $\sum_ux_u^{(k)}c(a'_u)=1+x_v^{(h)}>1$, hence $x^{(k)}\notin {\tt TH}(G)$. \ \pfbox \medskip Conversely, it is easy to see that the characteristic labeling of any maximal clique~$Q$ does have $n$~linearly independent vectors satisfying (34.1), so it does define a facet. For each vertex~$u$ we let $x_u^{(u)}=1$, and $x_v^{(u)}=0$ for all $v\neq u$ except for one vertex $v\in Q$ with $v\noadj u$ (when $u\notin Q$). Then $x^{(u)}$ is a stable labeling so it is in ${\tt TH}(G)$. The point of the theorem is that a constraint $\sum_vx_vc(a_v)\leq 1$ of {\tt TH}$(G)$ that is not satisfied by all $x\in{\tt QSTAB}(G)$ cannot correspond to a facet of {\tt TH}$(G)$. \proclaim Corollary. $$\eqalign{{\tt TH}(G)\hbox{ is a polytope}\ &\Longleftrightarrow\ {\tt TH}(G)={\tt QSTAB}(G)\cr &\Longleftrightarrow\ {\tt TH}(G)={\tt STAB}(G) \ \Longleftrightarrow \ G \ \hbox{\rm is perfect}.\cr}$$ \noindent {\bf Proof.}\quad If ${\tt TH}(G)$ is a polytope it is defined by facets as in the theorem, which are nothing more than the constraints of ${\tt QSTAB}(G)$; hence ${\tt TH}(G)={\tt QSTAB}(G)$. Also the antiblocker of a convex corner polytope is a polytope, so ${\tt TH}(\overline{G})$ is a polytope by (30.4); it must be equal to ${\tt QSTAB}(\overline{G})$. Taking antiblockers, we have ${\tt TH}(G)={\tt STAB}(G)$ by (30.3). The converses are easy since {\tt STAB} and {\tt QSTAB} are always polytopes. The connection to perfection is an immediate consequence of Theorem~32 and Lemma~2. \ \pfbox \medskip We cannot strengthen the corollary to say that $\vartheta(G)=\alpha(G)$ holds if and only if $\vartheta(G)=\kappa(G)$; the Petersen graph (section~27) is a counterexample. \meno {\bf 35. Orthogonal labelings in a perfect graph.}\quad A perfect graph has $$\vartheta(G,w)=\alpha(G,w)=\max\{x\cdot w\mid \hbox{$x$ is a stable labeling of $G$}\}\,,\eqno(35.1)$$ and Theorem 12 tells us there exist orthogonal labelings of $G$ and $\overline G$ such that (12.2) and (12.3) hold. But it isn't obvious what those labelings might be; the proof was not constructive. The problem is to find vectors $a_v$ such that $a_u\cdot a_v=0$ when $u\noadj v$ and such that (12.2) holds; then it is easy to satisfy (12.3) by simply letting $b$ be a stable labeling where the maximum occurs in~(35.1). The following general construction gives an orthogonal labeling (not necessarily optimum) in any graph: Let $g(Q)$ be a nonnegative number for every clique~$Q$, chosen so that $$\sum_{v\in Q} g(Q)=w_v\,,\qquad\hbox{for all $v$}\,.\eqno(35.2)$$ Furthermore, for each clique $Q$, let $$a_{Qv}=\cases{\sqrt{g(Q)},&if $v\in Q$;\cr 0,&otherwise.\cr}\eqno(35.3)$$ Then $$a_u\cdot a_v\;=\;\sum_{\{u,v\}\subseteq Q}g(Q)\,,$$ hence $a_u\cdot a_v=0$ when $u\noadj v$. If we also let $a_{Q0}=\sqrt{g(Q)}$ for all~$Q$, $a_{00}=0$, we find $$a_0\cdot a_v=a_v\cdot a_v=\sum_{v\in Q}g(Q)=w_v\,.$$ We have constructed a matrix $A$ that is $\lambda$-compatible with $G$ and~$w$, in the sense of section~29, where $$\lambda=a_0\cdot a_0=\sum_Q g(Q)\,.\eqno(35.4)$$ An orthogonal labeling with costs $c(a'_v)=w_v/\lambda$ can now be found as in the proof of Lemma~29. The duality theorem of linear programming tells us that the minimum of (35.4) subject to the constraints (35.2) is equal to the maximum value of $w\cdot x$ over all $x$ with $\sum_{v\in Q}x_v\le1$ for all~$Q$. When $x$ maximizes $w\cdot x$, we can assume that $x\ge0$, because a negative $x_v$ can be replaced by~0 without decreasing $w\cdot x$ or violating a constraint. (Every subset of a clique is a clique.) Thus, we are maximizing $w\cdot x$ over ${\tt QSTAB}(G)$; the construction in the previous paragraph allows us to reduce $\lambda$ as low as $\kappa(G,w)$. But $\kappa(G,w)=\vartheta(G,w)$ in a perfect graph, so this construction solves our problem, once we have computed $g(Q)$. The special case of a bipartite graph is especially interesting, because its cliques have only one or two vertices. Suppose all edges of~$G$ have the form $u\adj v$ where $u\in U$ and $v\in V$, and consider the network defined as follows: There is a special source vertex~$s$ connected to all $u\in U$ by a directed arc of capacity~$w_u$, and a special sink vertex~$t$ connected from all $v\in V$ by a directed arc of capacity~$w_v$. The edges $u\adj v$ of $G$ are also present, directed from~$u$ to~$v$ with infinite capacity. Any flow from $s$ to~$t$ in this network defines a suitable function~$g$, if we let $$\eqalign{g(\{u,v\})&=\hbox{the flow in $u\rar v$}\,,\cr g(\{u\})&=\hbox{$w_u$ minus the flow in $s\rar u$}\,,\cr g(\{v\})&=\hbox{$w_v$ minus the flow in $v\rar t$}\,,\cr}$$ for all $u\in U$ and $v\in V$. Let $S$ be a subset of $U\cup V$. If we cut the edges that connect~$s$ or~$t$ with vertices not in~$S$, we cut off all paths from $s$ to~$t$ if and only if $S$ is a stable set. The minimum cut (i.e., the minimum sum of capacities of cut edges) is equal to the maximum flow; and it is also equal to $$\sum_{u\in U}w_u+\sum_{v\in V}w_v-\max\{w\cdot x\mid \hbox{$x$ is a stable labeling}\} =\sum_{u\in U}w_u+\sum_{v\in V}w_v-\alpha(G,w)\,.$$ Thus the value of $\lambda=\sum_Qg(Q)$ is $\sum_{u\in U}w_u-\{\hbox{flow from $s$}\} +\sum_{v\in V}w_v-\{\hbox{flow to $t$}\} +\{\hbox{flow in $u\rar v$ arcs}\}=\alpha(G,w)= \vartheta(G,w)$ as desired. For general perfect graphs $G$, a solution to (35.4) with $\lambda =\vartheta(G,w)$ can be found in polynomial time as shown in equation (9.4.6) of [\GLSbook]. However, the methods described in [\GLSbook] are not efficient enough for practical calculation, even on small graphs. \meno {\bf 36. The smallest non-perfect graph.}\quad The cyclic graph $C_5$ is of particular interest because it is the smallest graph that isn't perfect, and the smallest case where the function $\vartheta(G,w)$ is not completely known. The discussion following Theorem 34 points out that ${\tt TH}(G)$ always has facets in common with ${\tt QSTAB}(G)$, when those facets belong also to ${\tt STAB}(G)$. It is not hard to see that ${\tt QSTAB}(C_5)$ has ten facets, defined by $x_j=0$ and $x_j+x_{j\bmod5}=1$ for $0\le j<5$; and ${\tt STAB}(C_5)$ has an additional facet defined by $x_0+x_1+x_2+x_3+x_4=2$. The weighted functions $\alpha$ and $\kappa$ of section~4 are evaluated by considering the vertices of {\tt STAB} and {\tt QSTAB}: $$\eqalignno{ \alpha(C_5,\{w_0,\ldots,w_4\}^T)&=\max(w_0+w_2,w_1+w_3, w_2+w_4,w_3+w_0,w_4+w_1)\,;&(36.1)\cr \kappa(C_5,\{w_0,\ldots,w_4\}^T)&=\max\bigl(\alpha(C_5,\{w_0,\ldots,w_4\}^T), (w_0+\cdots+w_4)/2\bigr)\,.&(36.2)\cr}$$ Where these functions agree, they tell us also the value of $\vartheta$. For example, let $f(x)=\vartheta(C_5,\{x,1,1,1,1,1\}^T)$. Relations (36.1) and (36.2) imply that $f(x)=x+1$ when $x\ge2$. Clearly $f(0)=2$, and section~22 tells us that $f(1)=\sqrt5$. Other values of $f(x)$ are not yet known. Equation (23.2) gives the lower bound $f(x)^2\ge x^2+4$. Incidentally, the $a$ vectors $$\openup1\jot \pmatrix{\sqrt x\cr 1\cr 0\cr 0\cr 0\cr}\qquad \pmatrix{\sqrt x\cr 1\cr \sqrt{x+1}\cr \sqrt{(x-2)(x+1)}\cr 0\cr}\qquad \pmatrix{1\cr -\sqrt x\cr 0\cr 0\cr 0\cr}\qquad \pmatrix{1\cr -\sqrt x\cr 0\cr 0\cr 0\cr}\qquad \pmatrix{\sqrt x\cr 1\cr -\sqrt{x+1}\cr 0\cr \sqrt{(x-2)(x+1)}\cr}$$ and $b=(1)\,(0)\,(0)\,(0)\,(0)$ establish $f(x)$ for $x\ge2$ in the fashion of Theorems 12 and~13. Let $\phi=(1+\sqrt5\,)/2$ be the golden ratio. The matrices $A$ and $B'$ of Theorem 29, when $G=C_5$ and $w=\ones$, are $$\openup2\jot \eqalign{A&=\pmatrix{\sqrt5&1&1&1&1&1\cr 1&1&\phi-1&0&0&\phi-1\cr 1&\phi-1&1&\phi-1&0&0\cr 1&0&\phi-1&1&\phi-1&0\cr 1&0&0&\phi-1&1&\phi-1\cr 1&\phi-1&0&0&\phi-1&1\cr}\,;\cr B'&={1\over\sqrt5}\pmatrix{\sqrt5&-1&-1&-1&-1&-1\cr -1&1&0&\phi-1&\phi-1&0\cr -1&0&1&0&\phi-1&\phi-1\cr -1&\phi-1&0&1&0&\phi-1\cr -1&\phi-1&\phi-1&0&1&0\cr -1&0&\phi-1&\phi-1&0&1\cr}\,.\cr}$$ They have the common eigenvectors $$\pmatrix{\sqrt5\,\cr 1\cr 1\cr 1\cr 1\cr 1\cr}\qquad \pmatrix{\sqrt5\,\cr -1\cr -1\cr -1\cr -1\cr -1\cr}\qquad \pmatrix{0\cr \phi\cr 1\cr -1\cr -\phi\cr 0\cr}\qquad \pmatrix{0\cr 0\cr \phi\cr 1\cr -1\cr -\phi\cr}\qquad \pmatrix{0\cr 1\cr -\phi\cr \phi\cr -1\cr 0\cr}\qquad \pmatrix{0\cr 0\cr 1\cr -\phi\cr \phi\cr -1\cr}\,,$$ with respective eigenvalues $(\lambda_0,\ldots,\lambda_5)= (2\sqrt5,0,\sqrt5/\phi,\sqrt5/\phi,0,0)$ and $(\mu_0,\ldots,\mu_5)=(0,2,0,0,1/\phi,1/\phi)$. (Cf.~(29.6) and (29.7).) \meno {\bf 37. Perplexing questions.}\quad The book [\GLSbook] explains how to compute $\vartheta(G,w)$ with given tolerance~$\epsilon$, in polynomial time using an ellipsoid method, but that method is too slow and numerically unstable to deal with graphs that have more than 10 or so vertices. Fortunately, however, new ``interior-point methods'' have been developed for this purpose, especially by Alizadeh [\Ala,\Alb], who has computed $\vartheta(G)$ when $G$ has hundreds of vertices and thousands of edges. He has also shown how to find large stable sets, as a byproduct of evaluating $\vartheta(G,w)$ when $w$ has integer coordinates. Calculations on somewhat smaller cyclically symmetric graphs have also been reported by Overton [\Ov]. Further computational experience with such programs should prove to be very interesting. Solutions to the following four concrete problems may also help shed light on the subject: {\bf P1.}\quad Describe ${\tt TH}(C_5)$ geometrically. This upright set is isomorphic to its own anti\-blocker. (Namely, if $(x_0,x_1,x_2,x_3,x_4)\in{\tt TH}(C_5)$, then so are its cyclic permutations $(x_1,x_2,x_3,x_4,x_0)$, etc., as well as the cyclic permutations of $(x_0,x_4,x_3,x_2,x_1)$; ${\tt TH}(\overline{C}_5)$ contains the cyclic permutations of $(x_0,x_2,x_4, x_1,x_3)$ and $(x_0,x_3,x_1,x_4,x_2)$.) Can the values $f(x)=\vartheta(C_5,\{x,1,1,1,1\}^T)$, discussed in section 36, be expressed in closed form when $00$ such that no polynomial algorithm can compute a number between $\alpha(G)$ and $n^{\epsilon}\alpha(G)$ for all $n$-vertex graphs~$G$, unless $P=NP$. Therefore it would be surprising if the answer to~P6 turns out to be that $\vartheta_-(G)$ is, say, $O(\log n)^2$ with probability $\rar 1$ for random~$G$. Still, this would not be inconsistent with [\ALMSS], because the graphs for which $\alpha(G)$ is hard to approximate might be decidedly nonrandom. Lov\'asz has called my attention to papers by Kashin and Konyagin [\KK, \Ko], which prove (in a very disguised form, related to (6.2) and Theorem~33) that if $G$ has no stable set with 3~elements we have $$\vartheta(G)\leq 2^{2/3}n^{1/3}\,;\eqno(37.5)$$ moreover, such graphs exist with $$\vartheta(G)=\Omega(n^{1/3}\!/\sqrt{\log n}\,)\,.\eqno(37.6)$$ Further study of methods like those in [\LS] promises to be exciting indeed. Lov\'asz has sketched yet another approach in~[\LLL]. \bigskip \centerline{\bf References} \bib [\Ala] Farid Alizadeh, ``A sublinear-time randomized parallel algorithm for the maximum clique problem in perfect graphs,'' {\sl ACM-SIAM Symposium on Discrete Algorithms\/ \bf 2} (1991), 188--194. \bib [\Alb] Farid Alizadeh, ``Interior point methods in semidefinite programming with applications to combinatorial optimization,'' preprint, 1993, available from {\tt alizadeh@icsi.\allowbreak berkeley.edu}. \bib [\ALMSS] Sanjeev Arora, Carsten Lund, Rajeev Motwani, Madhu Sudan, and Mario Szegedy, ``Proof verification and intractability of approximation problems,'' {\sl Proceedings of the 33rd IEEE Symposium on Foundations of Computer Science\/} (1992), 14--23. \bib [\Bol] B\'ela Bollob\'as, {\sl Random Graphs\/} (Academic Press, 1985). \bib [\GLSell] Martin Gr\"otschel, L. Lov\'asz, and A. 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