Skip to main content

All Questions

4 votes
0 answers
233 views

How to find generators of translation subgroup of an abstract reflection (coxeter) group

I have an infinite reflection group https://en.wikipedia.org/wiki/Coxeter_group Take for example the affine groups $[4,4],[4,3,4],[4,3,3,4]$... I'd like to get an explicit expression for generators ...
unknown's user avatar
  • 1,010
7 votes
1 answer
399 views

On groups with presentations $ \langle a,b,c\mid a^2=b^2=c^2=(ab)^p=(bc)^q=(ca)^r=(abc)^s=1\rangle $...

$$ \langle a,b,c\mid a^2=b^2=c^2=(ab)^p=(bc)^q=(ca)^r=1\rangle =\Delta(p,q,r) $$ This is a presentation of a triangle group $\Delta(p,q,r)$, a special kind of Coxeter group. EDIT In fact, these are ...
draks ...'s user avatar
  • 18.6k
5 votes
0 answers
317 views

Mapping $\Delta(2,2,2)\mapsto \Delta(4,4,2)$... [closed]

Looking at the images below, you recognize that the adajency matrix of the graph $A_G$ splits up into three different color submatrices, with $A_G=A_r+A_b+A_d$ (where $d$ is dark, damn...). It's ...
draks ...'s user avatar
  • 18.6k
4 votes
1 answer
499 views

Coxeter presentation of Hyperoctahedral group $(\mathbb{Z}/2\mathbb{Z})^n \rtimes S_n$.

I know that the hyperoctahedral group $(\mathbb{Z}/2\mathbb{Z})^n \rtimes S_n$ has the presentation $$\langle s_{\text{1}},\ldots,s_n\mid s_{\text{1}}^{\text{2}}=s_i^2=1, (s_1s_2)^4=(s_is_{i+1})^3=(...
budi's user avatar
  • 1,810
3 votes
0 answers
273 views

About the order of Coxeter groups

I have a question about Coxeter groups with $3$ generators: Suppose, as a group, $G$ is generated by $a,b$ and $c$, with the relations $a^2 =b^2 =c^2 =1$, $(ab)^m = (bc)^n = (ca)^p =1$ where $2 \...
sunkist's user avatar
  • 1,155