I'm looking for examples of Riemannian manifolds M of dimension $\geq 2$ such that the isometry group $Isom(M)$ contains as subgroup a finite Coxeter group $G$ such that $Tor(Z(G))$, the torsion group of the center of $G$, contains an element of order $\geq 3$.

  • 1
    $\begingroup$ The center of a finite Coxeter group never contains an element of order $\ge 3$ (m-hikari.com/imf/imf-2013/29-32-2013/…). $\endgroup$ – Qiaochu Yuan May 19 '15 at 0:34
  • 2
    $\begingroup$ @Qiaochu: I'm surprised you refer to a paper from 2013 for a fact which was certainly well known before (see e.g. the reference that Max gave in his answer). I'm even more surprised that such papers are getting published... $\endgroup$ – Tom De Medts May 19 '15 at 12:05
  • $\begingroup$ No wonder I was having a hard time finding such an example :O) Many thanks! $\endgroup$ – Ferran V. May 19 '15 at 14:02
  • $\begingroup$ @Tom: it was just the first reference I could find. Googling is faster than looking things up in Humphreys... $\endgroup$ – Qiaochu Yuan May 19 '15 at 16:51
  • $\begingroup$ What about when $G$ is a finite group given by the quotient of a Coxeter group? Still the center never contains elements of order $\geq 3$? $\endgroup$ – Ferran V. May 19 '15 at 17:16

Such a thing cannot exist, as the center of a Coxeter group $(W,S)$ is always an elementary abelian 2-groups.

Indeed, it is easy to reduce to the irreducible case, that is: If $S=S_1\cup \ldots S_k$ is a decomposition of $S$ into irreducible components, then $W= \langle S \rangle \cong \langle S_1 \rangle \times \cdots \times \langle S_k \rangle$, hence $Z(W) = Z(\langle S \rangle) \cong Z(\langle S_1 \rangle) \times \cdots \times Z(\langle S_k \rangle)$.

Now, for an irreducible Coxeter group $(W,S)$, if $W$ is infinite then its center is always trivial, and if $W$ is finite, it can be trivial or of order 2 (see e.g. exercise 1 in Section 6.3 of "Reflection Groups and Coxeter Groups" by James E. Humphreys. (There are perhaps better, and certainly older references for this, but I'll leave it to others to dig those up).

| cite | improve this answer | |
  • $\begingroup$ No wonder I was having a hard time finding such an example :O) Many thanks Max! $\endgroup$ – Ferran V. May 19 '15 at 14:02

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.