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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?

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    $\begingroup$ I don't know the answer, but among finite groups there are many cases in which this happens. For example, every representation of a finite symmetric group is defined over the rationals, so in particular it is real. $\endgroup$
    – Angelo
    Commented Nov 27, 2010 at 9:23
  • $\begingroup$ See also this question, which is concerned with finite groups: mathoverflow.net/questions/42646/… $\endgroup$
    – Alex B.
    Commented Nov 27, 2010 at 9:28
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    $\begingroup$ @John: for finite groups, aren't you just asking for examples where the character table is real?? This is not at all uncommon. Dihedral group order 8, quaternion group order 8, all symmetric groups, product of groups for which the character table is real... $\endgroup$
    – Kevin Buzzard
    Commented Nov 27, 2010 at 12:28
  • $\begingroup$ @Kevin: Yes, that's what I'm asking for. I'd known all irreps of symmetric groups were defined over the rationals but somehow hadn't applied that knowledge to this puzzle (I think this is known as "stupidity"). Torsten's answer hits the nail on the head and I'm actually pleased to know there are so many examples. $\endgroup$
    – John Baez
    Commented Nov 27, 2010 at 16:24

3 Answers 3

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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 $G_2$.

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    $\begingroup$ $w_0 = -1$ if and only if the center of the ( simply connected semisimple) Lie group is isomorphic to $(\mathbb Z/2)^k$ for some $k\ge 0$. $\endgroup$
    – André Henriques
    Commented Nov 27, 2010 at 14:02
  • $\begingroup$ Otherwise, the difference between $w_0$ and $-1$ is measured by the outer automorphism of $G$ that acts as $g\mapsto -g$ for $g\in Z(G)$. $\endgroup$
    – André Henriques
    Commented Nov 27, 2010 at 14:17
  • $\begingroup$ Thanks, Torsten and André. I guess for type $D$ I can just think about the spinor reps (which I actually know and love) and figure out when they're real or quaternionic; all the other fundamental reps are real, since they're exterior powers of the tautologous rep of $SO(n)$ on $\mathbb{R}^n$. $\endgroup$
    – John Baez
    Commented Nov 27, 2010 at 16:28
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    $\begingroup$ Torsten wrote: "All representations are real-valued precisely when..." I think it might be a bit clearer to say "All characters are real-valued precisely when..." $\endgroup$
    – John Baez
    Commented Nov 28, 2010 at 4:52
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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.

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    $\begingroup$ I believe that the usage of real/complex/quaternionic both in John's question and in Noah's answer are only superficially different. In fact, the relation between the two is explained, for instance, in §II.6 of the book Representation theory of compact Lie groups by Bröcker and tom Dieck. For example, let $U$ be an irreducible real representation. If $U$ is real (resp. quaternionic) in the sense of this answer (and Noah's), if and only if its complexification is of real (resp. quaternionic) type. $\endgroup$
    – José Figueroa-O'Farrill
    Commented Nov 27, 2010 at 18:26
  • $\begingroup$ The section in Bröcker and tom Dieck that you point to contained the definition I had in mind (and expressed poorly) -- this seems like the natural definition to me. Now that you mention it, I see that their/my definition agrees with John's in the very special case where the group is compact and the representation is irreducible. (If we allow, say $V \oplus V^*$, then you can always endow that with a nondegenerate symmetric or skew-symmetric bilinear form.) So in the strict sense of the question, I stand corrected, the two definitions agree. $\endgroup$
    – Skip
    Commented Nov 28, 2010 at 2:50
  • $\begingroup$ Yes, sorry, perhaps I should have said explicitly in my comment that this is of course only true in the context of unitary irreps. $\endgroup$
    – José Figueroa-O'Farrill
    Commented Nov 28, 2010 at 17:00
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    $\begingroup$ For future readers of this list of comments, here are some counterexamples to the equivalence if the hypotheses are relaxed. If $G$ is not compact, the spin reps of Spin(5,3) are complex in my sense -- meaning not defined over $\mathbb{R}$ -- but have an invariant symmetric bilinear form. If the representation is not irreducible: take $G$ compact and $V$ an irreducible representation that is complex in my sense, so $V^*$ is the complex conjugate of $V$ and is not $V$. Then $V \oplus V^*$ is both real and quaternionic in Baez's sense but is only real in my sense ("no quaternionic structure"). $\endgroup$
    – Skip
    Commented Nov 29, 2010 at 16:03
  • $\begingroup$ @Skip: I know it's almost ten years since you wrote this, but I just came across it and wonder if these contradictory definitions are the clue to explaining the contradiction between the accepted answer and my answer to math.stackexchange.com/q/310678/96384. Would you mind having a look at those, as well as my recent answer to math.stackexchange.com/q/3738143/96384 which talks about when certain spin representations are complex/quaternionic/real? $\endgroup$
    – Torsten Schoeneberg
    Commented Jul 7, 2020 at 23:38
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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.

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  • $\begingroup$ $D_{even}$ has $2$-fold symmetry, but it goes unused, one might say. $\endgroup$
    – Allen Knutson
    Commented Nov 28, 2011 at 14:27

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