# Coxeter Groups and representations of $+$ type

Background:

A Coxeter group is generated by reflections, which all have the real eigenvalues $$\pm 1$$, so intuitively it makes sense that all the representations would be real. Certainly every Coxeter group must have one faithful real valued irrep, since every finite Coxeter group is just a Euclidean reflection group.

Question:

I have noticed that all the irreps of a finite Coxeter group are of $$+$$ type (I have not checked every finite Coxeter group but this seems to be the pattern) (by $$+$$ type I mean they have Frobenius-Schur indicator $$+1$$). I would imagine this fact (if true) is well know. Does anyone know an explanation/proof/reference for this?

More generally, does anyone know any necessary or sufficient (or necessary and sufficient!) conditions for a finite group to have all $$+$$ type irreps?

• What does "$+$ type" mean? This term is impossible to Google for... Oct 1, 2022 at 16:01
• Oops I didn't realize that term was nonstandard. I just mean Frobenius Schur indicator +1. I think I first saw terms like " $-$ type" and "$+$ type" used here math.stackexchange.com/questions/4536063/… anyway I'll edit the question to clarify Oct 1, 2022 at 16:19

Some stuff I scrounged up. First, an easier result: an element $$g$$ of a finite group $$G$$ is real if it is conjugate to $$g^{-1}$$, or equivalently if the character of $$g$$ is always real in every representation. So the following conditions are equivalent:

1. Every element of $$G$$ is conjugate to its inverse.
2. Every character of $$G$$ is real.
3. Every finite-dimensional representation of $$G$$ is either real or quaternionic.

Such groups are called ambivalent. This is a necessary condition for all representations being real; the symmetric groups obviously satisfy it, but I'm not sure if all finite Coxeter groups do (edit: they do, see the comments). However, it turns out to be satisfied by all Weyl groups.

The stronger condition that every representation is real is called being totally orthogonal; apparently it's known (see the link) that any such group must be generated by involutions. It doesn't seem like necessary and sufficient conditions are known in general.

Finally, the Weyl groups satisfy an even stronger condition which implies total orthogonality, that every representation is realizable over $$\mathbb{Q}$$: this is claimed in Humphreys in Section 8.10 and the citation is to Benard's On the Schur indices of characters of the exceptional Weyl groups. Such a group must in particular satisfy the property that every character is integer-valued and this condition is called being rational.

• great answer. Interesting to learn that the condition I describe is known as a group being being "totally orthogonal." And the theorem 3.3 you point out from your reference , that every totally orthogonal group is generated by involutions, is very interesting. Just a small typo, I think the page you link says that dihedral groups are ambivalent groupprops.subwiki.org/wiki/Dihedral_groups_are_ambivalent. So I guess that still leaves open the question of whether all finite Coxeter groups are totally orthogonal. Oct 1, 2022 at 19:10
• Ah, sorry, I made a silly error re: the dihedral groups. I'm not sure what happens for the non-crystallographic finite Coxeter groups but presumably one could just go through them all case-by-case? Oct 1, 2022 at 20:13
• The groups $H_3$ and $H_4$ have all characters of $+$ type. For dihedral groups, the FS indicator is (up to scalar) the sum of $\chi(g^2)$. Since $D_{2n}$ is more than half elements with $g^2=1$, this contributes more than $|G|/2\cdot \chi(1)$ to the sum. The rest has modulus strictly less than $|G|/2\cdot \chi(1)$ as $|\chi(g)|\leq \chi(1)$ and traingle inequality. Thus the FS indicator is positive. Oct 1, 2022 at 20:24
• Ok great with the Benard result on Weyl groups together with the fact for dihedral groups and $H_3$ and $H_4$ I guess that covers all the cases and we can conclude that every finite Coxeter group is totally orthogonal, that's super! Oct 1, 2022 at 21:56
• Also interesting to note that although all totally orthogonal groups are generated by involutions there are lots and lots of groups which are generated by involutions and are not totally orthogonal. For example mathoverflow.net/questions/26297/… states that all finite simple groups other than $PSU(3,3)$ are generated by three involutions. So $GL(3,2)$ of order $168$ for example would be a group which is generated by involutions but is not totally orthogonal. Oct 1, 2022 at 22:33