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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.

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  • $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ @GeoffRobinson: so every irreducible finite Coxeter group has a faithful irreducible complex representation? $\endgroup$ – Sebastien Palcoux Dec 21 '18 at 13:37

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