6
$\begingroup$

Let $S$ be a bounded geometric shape in the Euclidean space $E=\mathbb{R^n}$. Assume that the origin of $E$ is a fixed point of every element of the symmetry group $G(S)$ of $S$, and assume that $G(S) \le O(n)$.

Let $V:= E \otimes_{\mathbb{R}} \mathbb{C}$ be the complexification of $E$. It is a faithful complex representation of $G(S)$.

Question: Under which conditions on $S$, the representation $V$ is irreducible?

Remark: Here are two cases where $V$ is not irreducible:

  • $G(S) = \{ 1 \}$ and $n>1$,
  • the vector space generated by $S$, denoted $ \langle S \rangle$, is a strict subspace of $E$.

For people thinking my question too broad, here are more specific questions.

Let's assume that $n>1$, $G(S) \neq \{ 1 \}$ and $ \langle S \rangle = E$.

Question 1: Is $V$ irreducible if $S$ is a regular polytope?
Question 2: If so, can we extend to semiregular polytope?
Question 3: If so, what is your better extension?

Remark: The symmetry group of any regular polytope is an irreducible finite Coxeter group, and every such group is of this form, except those of type $D_n$, $E_6$, $E_7$, and $E_8$ which are symmetry groups of semiregular polytopes.

$\endgroup$
  • $\begingroup$ What about when $G(S)$ acts transitively on $S$ and $S$ intersects every subspace of $E$ nontrivially? $\endgroup$ – leibnewtz Dec 20 '18 at 21:44
  • $\begingroup$ Or at least $G$ acts transitively on $S-\{0\}$ $\endgroup$ – leibnewtz Dec 20 '18 at 21:57
  • $\begingroup$ @leibnewtz: Good sufficient condition (assuming the complexification keeps the irreducibility in this case). For example, $O(n)$ acts transitively on the unit sphere $\mathbb{S}$ of $E$, and $\mathbb{S}$ intersects every subspace of $E$ nontrivially. Now if $G(S)$ is finite whereas $S$ is not, then the action cannot be transitive; this happens for example if $S$ is a regular polytope; now, by definition, it is a polytope whose symmetry group acts transitively on its flags. Is it sufficient to answer Question 1 positively? $\endgroup$ – Sebastien Palcoux Dec 21 '18 at 11:14
  • 2
    $\begingroup$ If $G$ is a finite subgroup of ${\rm GL}(n,\mathbb{C})$ generated by reflections, then $G$ is a direct product of irreducible such subgroups generated by reflections. Hence if a finite Coxter group $G$ does not have a faithful irreducible complex representation, then $G$ is a direct product of smaller reflection groups. $\endgroup$ – Geoff Robinson Dec 21 '18 at 12:28
  • 1
    $\begingroup$ @GeoffRobinson: so every irreducible finite Coxeter group has a faithful irreducible complex representation? $\endgroup$ – Sebastien Palcoux Dec 21 '18 at 13:37

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.