5
$\begingroup$

For my project in bivariant K-theory for locally convex algebras, I'm looking how to call a particular notion of groups, too simple to be never considered elsewhere.

Let $G$ b a group with length $|\cdot|$, $H$ its subgroup, and $\bar G$ be a set of representatives of the cosets $G/H$ in $G$. We assume that all the elements in $\bar G$ have the minimal length among their cosets. Set $\rho\colon G \to \bar G$ to be the obvious map i.e. $gH = \rho(g)H$ for all $g\in G$.

Let now $g\in G$, $\bar g \in \bar G$, and $h=\rho(g\bar g)^{-1}g\bar g\in H$. The requirement is:

The length $|h|$ depends only on $|g|$, and the dependence is at most polynomial.

There is to be a name for this notion, or at least it had appear in some paper, I just need to know who am I to refer to.

Improve this question
$\endgroup$
4
  • 3
    $\begingroup$ Why do you use the term 'Quasi-orthogonal'? Do orthogonal groups viewed as subgroups of general linear groups satisfy a stronger version of this property? If so, what is it? $\endgroup$
    – Vincent
    Commented Oct 20, 2015 at 7:01
  • $\begingroup$ Well, as to the term, this is the best one I can come up with, out of purely geometric picturing of the group as some kind of "orthogonal sum" of the coset set and a subgroup. As to your second question - sorry, never thought of it. My closest examples are the groups with finite subgroups, central subgroups of finite index, and several more, like $\mathbb{Z}/2\mathbb{Z}$ as a subgroup of the fundamental groups of n-cusp space. $\endgroup$
    – Kolya Ivankov
    Commented Oct 20, 2015 at 7:24
  • $\begingroup$ I haven't seen it before so I cannot answer your question, but I still find it interesting. Here is another question. What about the weaker following condition: "Let $g \in G$ and let $h' = \rho(g)^{-1}g \in H$. The weaker requirement now is: The length $|h'|$ depends only on $|g|$, and the dependence is at most polynomial."? This would sound more natural to me. Do you know if this weak condition has a name? Or does it follow directly from the definition of length and hence doesn't need a name? $\endgroup$
    – Vincent
    Commented Oct 20, 2015 at 12:27
  • $\begingroup$ Well, I think Your condition is stronger, and actually, after reconsidering, it was the one I came up with some time ago and them forgotten - so thank You. I also think that this is satisfied if the group is hyperbolic, and I am also not sure if it's not just a trivial consequence from the definition of the length. I haven't been working with the properties of groups which were less trivial than just compactness or commutativity, therefore the question. $\endgroup$
    – Kolya Ivankov
    Commented Oct 20, 2015 at 13:11

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .