# Absolutely irreducible finite reflection/rotation groups

Suppose that $$G$$ is a finite irreducible reflection group with irreducible orthogonal representation $$\rho: G\rightarrow \mathrm{O}(d)$$, and let $$\rho^+: G^+\rightarrow \mathrm{SO}(d)$$ be its restriction to the rotation subgroup $$G^+$$ of $$G$$.

Question: For what $$G$$ (respectively, $$G^+$$) is $$\rho$$ (respectively, $$\rho^+$$) absolutely irreducible, i.e., remains irreducible when viewed as a complex representation?

For instance, this is certainly the case when $$G=S_d$$ is the symmetric group and $$\rho$$ is the standard $$(d-1)$$-dimensional representation (and would seem to be true for $$\rho^+$$, in which case $$G^+=A_d$$ is the alternating group, at least when $$d\geq 3$$). For the standard 2-dimensional real representation of Dihedral groups this is also true, but false when one restricts to the index 2 cyclic subgroups. I am especially interested in the case when $$d=3$$ and $$G$$ (respectively, $$G^+$$) are the reflection (rotation) groups of the octahedron and icosahedron. ,

• The representation $\rho$ is always absolutely irreducible, see for instance W. G. Dwyer , C. W. Wilkerson, Centers and Coxeter elements, Lemma 2.10. You should check their proof, it is likely that $\rho_+$ is always absolutely irreducible for non-dihedral groups. – Moishe Kohan Jan 6 at 17:16
• Dwyer and Wilkerson papers are always interesting. Link to the article referenced by @MoisheKohan: Dwyer and Wilkerson - Centers and Coxeter elements. – LSpice Jan 6 at 17:22
• Thanks! Their proof seems to rely on the product decomposition of essential reflection groups over any field, so it is not immediately obvious (to me anyway!) that it will carry over to rotation groups as well – Bob Jan 6 at 19:27

For $$\rho$$, take a non-trivial complex (absolutely) irreducible constituent $$\chi$$ of the character $$\theta$$ afforded by $$\rho.$$ Then for some reflection $$t \in G$$, we have $$\chi(t) = \chi(1)-2,$$ and $${\rm Res}^{G}_{\langle t \rangle }(\chi)$$ contains the non-trivial linear character $$\lambda$$ of $$\langle t \rangle$$ with multiplicity $$1$$ ( and the trivial character with multiplicity $$\chi(1)-1$$).

Hence $$\chi$$ occurs with multiplicity one in the (character of) the representation $${\rm Ind}_{\langle t \rangle }^{G}(\lambda)$$, which is a representation explicitly realised over the real field $$\mathbb{R}.$$

By the general theory of the Schur index, $$\chi$$ has Schur index $$1$$, and may be realised over the field of its character. But note that $$\chi$$ and $$\overline{\chi}$$ occur with equal multiplicity in the real valued character $$\theta.$$ If $$\overline{\chi} \neq \chi$$, then the above reflection $$t$$ has the eigenvalue $$-1$$ with multiplicity $$2$$ or more in the representation $$\rho$$, a contradiction. Hence $$\chi$$ is real valued, and since its Schur index is one, $$\chi$$ is realizable over $$\mathbb{R}.$$ Since $$\rho$$ is irreducible as a real representation, we have $$\chi = \theta.$$

Now let $$H= G^{+}$$ be the rotation (normal) subgroup of index $$2$$ in $$G$$. Since $$\theta = \chi$$ is absolutely irreducible, Clifford's theorem tells us that $${\rm Res}^{G}_{H}(\theta)$$ is either (absolutely) irreducible, or the sum of two distinct (absolutely) irreducible characters/ Note that by consideration of the character inner product, the possibility that $${\rm Res}^{G}_{H}(\chi)$$ is twice a (complex) irreducible character is excluded.

If $${\rm Res}^{G}_{H}(\theta)$$ is irreducible as a complex character, then $$\rho^{+}$$ is an absolutely irreducible representation. If the restriction is not (absolutely) irreducible, then there are distinct complex irreducible character $$\alpha, \beta$$ of $$H = G^{+}$$ such that $${\rm Res}^{G}_{H}(\theta)$$ = $$\alpha + \beta$$ and $$\alpha(1) = \beta(1).$$

In the latter case, the character inner product tells us that $$\theta$$ vanishes identically outside $$H = G^{+}.$$ In that case, each reflection of $$G$$ has the eigenvalue $$1$$ with mutiplicity $$\alpha(1)$$ in the representation $$\rho^{+}.$$ Since $$G$$ is a reflection group, it follows that $$\alpha(1) = \beta(1) = 1.$$

Irreducibility of $$\theta$$ tells us that $$\alpha$$ is not real-valued, (otherwise $$G^{+}$$ has order $$2$$ and $$G$$ is Abelian). Hence $$\beta = \overline{\alpha}$$ and $$G^{+}$$ is cyclic. Then $$G$$ is dihedral.

So the answer is that (as noted in comments) the representation $$\rho$$ is always absolutely irreducible. Also, $$\rho^{+}$$ is absolutely irreducible unless $$\rho(1) = 2$$ and $$G$$ is a dihedral group with at least six elements (the latter condition can be omitted if you do not consider the Klein $$4$$-group as a dihedral group).

Yes. As has been observed by several others, the representation $$\rho: G \rightarrow O(V)$$ ($$V={\mathbb R}^d$$) is absolutely irreducible. The proof is elementary. Suppose $$0\neq W\subset V_{\mathbb C}$$ is a $$G$$ invariant subspace. The group $$G$$ is generated by reflections $$r$$. If all these reflections act trivially on $$W$$, then $$W$$ consists of $$G$$ invariants, which is a subspace defined over $${\mathbb R}$$ and is hence $$0$$ by irreducibility of $$V$$. Therefore, some reflection $$r\in G$$ does not act trivially on $$W$$.

But this means that the image $$(r-1)(W)$$ is non-zero. Since $$r$$ is a reflection, this image is one dimensional and consists of complex multiples of a real vector $$w$$ generating $$(r-1)V \simeq \mathbb R$$. Hence $$W$$ contains the $$G$$ module generated by $$w$$; by the $$\mathbb R$$- irreducibility of $$V$$, this means that $$W\supset V$$, and hence $$W=V_{\mathbb C}$$.

• I did not look at the second part of the question about $G^+$ but that has been answered completely by Geoff Robinson – Venkataramana Jan 7 at 12:25
• This is a really nice argument for the reflection groups – Bob Jan 7 at 15:05
• @Bob : Thank you. – Venkataramana Jan 9 at 8:38

The classification of the irreducible complex reflection groups (see for example here) contains the complexifications of all real reflection groups:

$$I_1,I_2(n),A_d,B_d,D_d,H_3,H_4,F_4,E_6,E_7\text{ and }E_8.$$

They are therefore absolutely irreducible.

The classes $$A_n,B_n$$ and $$D_n$$ are contained among the $$G(m,p,n)$$. See also "Unitary Reflection Groups" by Lehrer and Tyalor, Example 2.11.