0
$\begingroup$

Let $W$ be a Coxeter group with generators $s_1,s_2,s_3$, where $m(s_1,s_2)=3,m(s_1,s_3)=2$, and $m(s_2,s_3)=\infty$.

I understand that there's a surjective morphism $\varphi\colon W\to PGL(2,\mathbb{Z})$ sending $$ s_1\mapsto\begin{bmatrix} 0 & 1 \\ 1 & 0\end{bmatrix}, s_2\mapsto\begin{bmatrix} -1 & 1 \\ 0 & 1\end{bmatrix}, s_3\mapsto\begin{bmatrix} -1 & 0 \\ 0 & 1\end{bmatrix}. $$

Why is this map injective? I'm reading that if follows from the fact that we have the free product $PSL(2,\mathbb{Z})=\langle\varphi(s_1s_3)\rangle*\langle\varphi(s_1s_2)\rangle$, but this is not clear to me how this is used to prove injectivity.

This is an example in 5.1 of Reflection Groups and Coxeter Groups by James Humphreys.

$\endgroup$
  • $\begingroup$ For the hint, see also here $\endgroup$ – Dietrich Burde Jun 30 '16 at 8:11
  • $\begingroup$ @DietrichBurde Oops, thanks, I should have been more clear. I'm curious why $\varphi$ is injective, I knew that $PSL(2,\mathbb{Z})=C_2*C_3$. $\endgroup$ – Adelaide Dokras Jun 30 '16 at 8:16
  • $\begingroup$ Actually, we have $PGL(2,\mathbb{Z})=\langle \phi(s_1s_3)\rangle\ast_{\phi(s_1)}\langle \phi(s_1s_2)\rangle$, and we are done - see here $\endgroup$ – Dietrich Burde Jun 30 '16 at 8:26
  • 1
    $\begingroup$ The subgroup $H = \langle s_1s_3,s_1s_2 \rangle$ has index $2$ in $W$. You know that the restriction of $\phi$ to $H$ is an isomorphism and also $\phi(H)$ has index $2$ in $\phi(W) = {\rm PGL}(2,{\mathbb Z})$. It follows that $\phi$ is injective. $\endgroup$ – Derek Holt Jun 30 '16 at 8:26
  • $\begingroup$ @DerekHolt Thanks for the comment. Have I understood you correctly? If $w\in H\cap\ker\phi$, necessarily $w=1$ since $\phi|_H$ is an isomorphism. If $w\notin H$, then $W=H\sqcup wH$ since $H$ has index $2$, but we can't have $\phi(w)\in\phi(H)$, otherwise $\phi(wH)\subset\phi(H)$, contrary to $\phi$ being surjective, since $\phi(H)$ is proper in $PGL(2,\mathbb{Z})$. If $\phi(w)\notin\phi(H)$, $\phi(w)\neq 1$, so that $w\notin\ker\phi$? $\endgroup$ – Adelaide Dokras Jun 30 '16 at 8:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.