# Symmetry group and irreducible representation

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.

• What about when $G(S)$ acts transitively on $S$ and $S$ intersects every subspace of $E$ nontrivially? – leibnewtz Dec 20 '18 at 21:44
• Or at least $G$ acts transitively on $S-\{0\}$ – leibnewtz Dec 20 '18 at 21:57
• @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? – Sebastien Palcoux Dec 21 '18 at 11:14
• 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. – Geoff Robinson Dec 21 '18 at 12:28
• @GeoffRobinson: so every irreducible finite Coxeter group has a faithful irreducible complex representation? – Sebastien Palcoux Dec 21 '18 at 13:37