orders and length functions on finitely generated groups - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T07:01:45Z http://mathoverflow.net/feeds/question/42928 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/42928/orders-and-length-functions-on-finitely-generated-groups orders and length functions on finitely generated groups Mark Sapir 2010-10-20T19:42:20Z 2010-10-20T21:50:58Z <p>Let $G$ be a finitely generated group with the natural word length function ($|x|$ is the length of the shortest word in generators of $G$ representing $x$). We call a partial left invariant order $\le$ on $G$ <em>word order</em> if whenever $a\le b\le c$ we have $|b|\le C(|a|+|c|)$ for some constant $C$. Say, the standard order on $\mathbb Z$ is a word order. </p> <p><b> Question. </b> Is there a group $G$ with a left invariant linear order but without left invariant linear word order?</p> http://mathoverflow.net/questions/42928/orders-and-length-functions-on-finitely-generated-groups/42953#42953 Answer by Jesse Peterson for orders and length functions on finitely generated groups Jesse Peterson 2010-10-20T21:39:13Z 2010-10-20T21:39:13Z <p>If $G$ is a finitely generated infinite group and $\leq$ is a linear word order, then for each $a, c \in G$ there are only finitely many elements $b \in G$ such that $a \leq b \leq c$. From this it follows that $(G, \leq)$ is order isomorphic to $\mathbb Z$. If $\leq$ is also left invariant, then this isomorphism must be a group isomorphism as well.</p> http://mathoverflow.net/questions/42928/orders-and-length-functions-on-finitely-generated-groups/42954#42954 Answer by Sergei Ivanov for orders and length functions on finitely generated groups Sergei Ivanov 2010-10-20T21:40:37Z 2010-10-20T21:50:58Z <p>Let $G=\mathbb Z^2$. Every invariant linear order on $\mathbb Z^2$ is either induced from the standard order on $\mathbb R$ (or its inverse) by a linear map of the form $$(x,y)\mapsto x+\alpha y : \mathbb Z^2\to\mathbb R$$ where $\alpha\in\mathbb R\setminus\mathbb Q$, or is a composition of the lexicographic order $$(x,y)>(x',y') \quad\iff\quad x>x' \text{ or } (x=x' \text{ and } y>y')$$ with a bijective linear transformation $\mathbb Z^2\to\mathbb Z^2$.</p> <p>Indeed, the set of elements of $\mathbb Z^2$ that are greater or equal to zero is a semi-group $H$ such that <code>$H\cap(-H)=\{0\}$</code>. The closed convex hull of such $H$ must be a half plane. If this half-plane is bounded by an irrational line, we have the first type of a linear order. If this half-plane is bounded by a rational line, then we may assume that it is a coordinate line <code>$\{x=0\}$</code> (up to a linear change of coordinates in $\mathbb Z^2$), then the order is the lexicographic one (up to a change $y\mapsto -y$ depending on whether $(0,1)$ is "positive" or "negative" in this ordering).</p> <p>In both cases, there are infinitely many elements between $(0,0)$ and $(1,0)$, hence it is not a word order.</p>