ConvexSubgroup 2007-05-10 10:38:57 2007-05-10 10:38:57 Definition convex subset \usepackage{amssymb,amscd} \usepackage{amsmath} \usepackage{amsfonts} \usepackage{mathrsfs} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions \usepackage{amsthm} % making logically defined graphics \usepackage{xypic} \usepackage{pst-plot} \usepackage{psfrag} % define commands here \newtheorem{prop}{Proposition} \newtheorem{thm}{Theorem} \newtheorem{ex}{Example} \newcommand{\real}{\mathbb{R}} \newcommand{\pdiff}[2]{\frac{\partial #1}{\partial #2}} \newcommand{\mpdiff}[3]{\frac{\partial^#1 #2}{\partial #3^#1}} We begin this article with something more general. Let $P$ be a poset. A subset $A\subseteq P$ is said to be \emph{convex} if for any $a,b\in A$ with $a\le b$, the poset interval $[a,b]\subseteq A$ also. In other words, $c\in A$ for any $c\in P$ such that $a\le c$ and $c\le b$. Examples of convex subsets are intervals themselves, antichains, whose intervals are singletons, and the empty set. One encounters convex sets most often in the study of partially ordered groups. A \emph{convex subgroup} $H$ of a po-group $G$ is a subgroup of $G$ that is a convex subset of the poset $G$ at the same time. Since $e\in H$, we have that $[e,a]\subseteq H$ for any $e\le a\in H$. Conversely, if a subgroup $H$ satisfies the property that $[e,a]\subseteq H$ whenever $a\in H$, then $H$ is a convex subgroup: if $a,b\in H$, then $a^{-1}b\in H$, so that $[e,a^{-1}b]\subseteq H$, which implies that $[a,b]=a[e,a^{-1}b]\subseteq H$ as well. For example, let $G=\mathbb{R}^2$ be the po-group under the usual Cartesian ordering. $G$ and $0$ are both convex, but these are trivial examples. Let us see what other convex subgroups $H$ there are. Suppose $P=(a,b)\in H$ with $(a,b)\ne (0,0)=O$. We divide this into several cases: \begin{enumerate} \item $ab>0$. If $a>0$, then $b>0$ ($P$ in the first quadrant), so that $O\le P$, which means $[O,P]\subseteq H$. If $a<0$, then $b<0$ ($P$ in the third quandrant), so that $O\le -P$. In either case, $H$ contains a rectangle ($[O,P]$ or $[O,-P]$) that generates $G$, so $H=G$. \item One of $a$ or $b$ is $0$. Suppose $a=0$ for now. Then either $0<b$ so that $[O,P]\subseteq H$ or $b<0$ so that $[O,-P]\subseteq H$. In either case, $H$ contains a line segment on the $y$-axis. But this line segment generates the $y$-axis. So $y$-axis $\subseteq H$. If $H$ is a subgroup of the $y$-axis, then $H$=$y$-axis. Otherwise, another point $Q=(c,d)\in H$ not on the $y$-axis. We have the following subcases: \begin{enumerate} \item If $cd>0$, then $H=G$ as in the previous case. \item If $cd< 0$, say $d<0$ (or $0<c$), then for some positive integer $n$, $0<d+nb$, so that $O\le Q+nP$, and $H=G$ as well. On the other hand, if $c<0$ (or $0<d$), then $-Q$ returns us to the previous argument and $H=G$ again. \item If $d=0$ (so $c\ne 0$), then either $O\le P+Q$ (when $0<c$) or $O\le P-Q$ (when $c<0$), so that $H=G$ once more. \end{enumerate} A similar set of arguments shows that if $H$ contains a segment of the $x$-axis, then either $H$ is the $x$-axis or $H=G$. In conclusion, in the case when $ab=0$, $H$ is either one of the two axes, or the entire group. \item $ab<0$. It is enough to assume that $0<a$ and $b<0$ (that $P$ lies in the fourth quadrant), for if $P$ lies in the second quadrant, $-P$ lies in the fourth. Since $O,P\in H$, $H$ could be a subgroup of the line group $L$ containing $O$ and $P$. No two points on $L$ are comparable, for if $(r,s)<(t,u)$ on $L$, then the slope of $L$ is positive $$0<\frac{u-s}{t-r},$$ a contradiction. So $L$, and hence $H$, is an antichaine. This means that $H$ is convex. Suppose now $H$ contains a point $Q=(c,d)$ not on $L$. We again break this down into subcases: \begin{enumerate} \item $Q$ is in the first or third quandrant. Then $H=G$ as in the very first case above. \item $Q$ is on either of the axes. Then $H=G$ also, as in case 2(b) above. \item $Q$ is in the second or fourth quadrant. It is enough to assume that $Q$ is in the same quadrant as $P$ (fourth). So we have $0<c$ and $d<0$. Since $L$ passes through $P$ and not $Q$, we have that $$\frac{a}{c}\ne \frac{b}{d}.$$ Let $0<r=a/c$ and $0<s=b/d$ and assume $r<s$. Then there is a rational number $m/n$ (with $0<m,n$) such that $$r<\frac{m}{n}<s.$$ This means that $na< mc$ and $nb<md$, or $nP<mQ$. But $nP,mQ\in H$, so is $R=mQ-nP\in H$, which is in the first quadrant. This implies that $H=G$ too. \end{enumerate} In summary, if $H$ contains a point in the second or fourth quadrant, then either $H$ is a subgroup of a line with slope $<0$, or $H=G$. \end{enumerate} The three main cases above exhaust all convex subgroups of $\mathbb{R}^2$ under the Cartesian ordering. If the Euclidean plane is equipped with the lexicographic ordering, then the story is quite different, but simpler. If $H$ is non-trivial, say $P=(a,b)\in H$, $P\ne O$. If $0<a$, then $(c,d)\le (a,b)$ for any $c< a$ regardless of $d$. Choose $Q=(c,d)$ to be in the first quadrant. Then $[O,Q]\subseteq H$, so that $H=G$. If $a<0$, then $-P$ takes us back to the previous argument. If $a=0$, then either $[O,P]$ (when $0<b$), or $[O,-P]$ (when $b<0$) is a positive interval on the $y$-axis. This implies that $H$ is at least the $y$-axis. If $H$ contains no other points, then $H=y$-axis. In summary, the po-group $\mathbb{R}^2$ with lexicographic order has the $y$-axis as the only non-trivial proper convex subgroup. \begin{thebibliography}{8} \bibitem{gb} G. Birkhoff {\em Lattice Theory}, 3rd Edition, AMS Volume XXV, (1967). \end{thebibliography}