Absolutely irreducible finite reflection/rotation groups

Question

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. ,

Answer 1

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).

Answer 2

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 $W=V_{\mathbb C}$.

Answer 3

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.