Which groups have only real and quaternionic irreducible representations? - MathOverflow

Which groups have only real and quaternionic irreducible representations?

2010-11-27T08:55:35Z

Consider a continuous irreducible representation of a compact Lie group on a finite-dimensional complex Hilbert space. There are three mutually exclusive options:

1) it's not isomorphic to its dual (in which case we call it 'complex')

2) it has a nondegenerate symmetric bilinear form (in which case we call it 'real')

3) it has a nondegenerate antisymmetric bilinear form (in which case we call it 'quaternionic')

It's 'real' in this sense iff it's the complexification of a representation on a real vector space, and it's 'quaternionic' in this sense iff it's the underlying complex representation of a representation on a quaternionic vector space.

Offhand, I know just four compact Lie groups whose continuous irreducible representations on complex vector spaces are all either real or quaternionic in the above sense:

1) the group Z/2

2) the trivial group

3) the group SU(2)

4) the group SO(3)

Note that I'm so desperate for examples that I'm including 0-dimensional compact Lie groups, i.e. finite groups!

1) is the group of unit-norm real numbers, 2) is a group covered by that, 3) is the group of unit-norm quaternions, and 4) is a group covered by that. This probably explains why these are all the examples I know. For 1), 2) and 4), all the continuous irreducible representations are in fact real.

What are all the examples?

Answer by Torsten Ekedahl for Which groups have only real and quaternionic irreducible representations?

2010-11-27T11:34:03Z

An irreducible representation is real or quaternionic precisely when its character is real-valued. By the Peter-Weyl theorem all characters are real-valued precisely when every element in the group is conjugate to its inverse. When the group is connected a more precise answer is as follows: The Weyl group (in its tautological representation) must contain multiplication by $-1$ and this is true precisely when all indecomposable root system factors have that property. I don't remember off hand which indecomposable root systems have this property but it is of course well known (type A is out, type B/C is in, type D depends on the parity of the rank).

Addendum: Found the relevant places in Bourbaki. All characters are real-valued precisely when the element he calls $w_0$ is $-1$ (Ch. VIII,Prop. 7.5.11) and one can also read off if a given representation is real or quaternionic (loc. cit. Prop 12). From the tables in Chapter 6 one gets that $w_0=-1$ precisely for $A_1$, B/C, D for even rank, $E_7$, $E_8$, $F_4$ and $G_2$.

Answer by Skip for Which groups have only real and quaternionic irreducible representations?

2010-11-27T16:35:25Z

Torsten answered this question perfectly for the definition of real/complex/quaternionic in John's original question. But this usage of real/complex/quaternionic is foreign to my experience. Specifically, if you look at an irreducible real representation of a group, then its endomorphism ring is (by Schur and Frobenius) R, C, or H. And this seems to give a natural meaning of the terms "real", "complex" and "quaternionic" for irreps. This definition does not agree with John's, as you can see by considering the spin reps of Spin(7,1).

My definition is also what you find in Noah Snyder's answer here and in Wikipedia's definition of quaternionic representation.

Answer by Faisal for Which groups have only real and quaternionic irreducible representations?

2010-11-27T16:37:49Z

This was a comment on Torsten's answer, but it got too long.

Suppose $G$ is connected and semisimple. Fixing a choice $\Phi^+$ of positive roots for $G$, we can describe $w_0$ as the unique element of the Weyl group of $G$ that takes $\Phi^+$ to the negative roots $\Phi^- = -\Phi^+$. Now, $-w_0$ is an involution of the Dynkin diagram of $G$. This involution is trivial when the components of the Dynkin diagram lack two-fold symmetry, and this happens precisely for components of type $A_1$, $B_n$, $C_n$, $D_{2n}$, $E_7$, $E_8$, $F_4$ and $G_2$, in which case $-w_0=1$. For type $A_n$ ($n>1$), the involution is given by $\alpha_i \leftrightarrow \alpha_{n-i+1}$, for $D_n$ it's given by $\alpha_i \leftrightarrow \alpha_{i-1}$, and for $E_6$ it's given by $\alpha_1 \leftrightarrow \alpha_6$ and $\alpha_2 \leftrightarrow \alpha_5$.

Now if $V$ is an irrep of highest weight $\lambda$, then $V^\ast$ has highest weight $-w_0\lambda$. So $V \cong V^\ast$ whenever $-w_0=1$, and the above discussion tells us when this happens.

Side note: There's a closely related MO question, which was asked not too long ago, whose answers might be helpful.