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Given just the root system of a Lie algebra. How can we tell if the Lie algebra will or will not admit a complex representation? (e.g. a representation in terms of complex $N\times N$ matrices which cannot be written in terms of real $N \times N$ matrices).

Is there some easy way to tell just from looking at the root system? Such as does the root system have to have a certain symmetry?

As an example from looking at $E_6$ and $E_8$ root systems, is it easy to tell that $E_6$ admits a complex representation (27D complex matrices) $E_8$ does not (fundamental representation is 248 real adjoint)?

See also this question.

zooby
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  • When you say "the root system of a Lie algebra", do you mean a (semisimple) real Lie algebra? (Because I think of complex Lie algebras first, and they obviously have complex representations.) But then, what do you mean by "the root system"? The one of the complexification? Because that one does not even determine the Lie algebra up to isomorphism. – Torsten Schoeneberg May 08 '18 at 06:36
  • Yes, I'm mainly thinking of simple compact Lie groups. As in the ones that can be used with Yang-Mills Theory. They have to have a complex representation (unless there are mirror bosons). Can we tell if they do just from the root system? – zooby May 08 '18 at 14:13
  • Every compact Lie group has a finite-dimensional faithful complex representation. This is a well-known corollary of the Peter-Weyl theorem. – Spenser May 08 '18 at 15:35
  • OK. Let's talk about fundamental representations then. $E_6$ has a fundamental 27 dimensional complex representation. The fundamental representation of $E_8$ is the 248 real dimensional adjoint representation. I'm sure you know what I mean even if I get some of the words wrong. Just because a group is a subgroup of GL(n,C) doesn't mean it's representation is complex. Since real numbers are a subset of complex numbers. – zooby May 08 '18 at 16:26
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    For mathematics people, I think the meaning of "complex representation" is a repn of of a real Lie algebra/group on a complex vector space which does not arise by tensoring with $\mathbb C$ a repn on a real vector space. – paul garrett May 08 '18 at 16:42
  • Your effort to clarify the question is appreciated, but I still have trouble understanding what exactly you want. In my answer I assumed you ask: "Which fundamental representations of a (simple) compact Lie group/algebra do not admit a real structure, i.e. do not come from just tensoring a real representation with $\Bbb C$?" – Torsten Schoeneberg May 11 '18 at 22:29
  • Cf. https://math.stackexchange.com/q/1408871/96384 for the unfortunate conflict of terminologies /although I think we got it figured out here). – Torsten Schoeneberg Jul 30 '19 at 22:04

1 Answers1

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I'm not at all an expert in the representation theory of compact Lie groups/algebras, but it seems to me that Bourbaki's Lie Groups and Algebras, chapter 9, §7 no.2 proposition 1, answers your question in principle. Bourbaki makes a distinction of complex representations of a compact group into three types: a) real, b) complex and c) quaternionic (often called "pseudoreal" especially in physics). "Real" basically means that it's just the complexification of a real representation. In an earlier comment, I assumed your question was: "Which fundamental representations are not of type a?", whereas after seeing Are the physics and math definitions of a complex representation equivalent?, it seems that according to physicists' terminology, your question is rather: "Which representations are of type b?".

In any case, Bourbaki says (I paraphrase):

An irreducible representation of highest weight $\lambda$ (for $\lambda$ dominant w.r.t. a chosen set of simple roots) is of type b) if and only if $$-w_0(\lambda) \neq \lambda$$ where $w_0$ is the longest element of the Weyl group (w.r.t. that set of simple roots).

Further, if it happens that $-w_0(\lambda) = \lambda$, then we are

in case a) if $\sum_{\alpha\in \Phi^+} \lambda(\check\alpha)$ is even;

in case c) if $\sum_{\alpha\in \Phi^+} \lambda(\check\alpha)$ is odd.

Here $\check\alpha$ means the dual root to $\alpha$, but with all lattices (weight, root, coroot) appropriately identified. In the simply-laced cases $ADE$, one can just insert $\lambda(\alpha)$ for $\lambda(\check\alpha)$.

The condition on $-w_0(\lambda)$ is maybe not immediately visible from the Dynkin diagram of the root system, but it's certainly known. Cf. https://math.stackexchange.com/a/59789/96384. In particular, for types $A_1, B_n, C_n, D_{2n}, E_7, E_8, F_4$ and $G_2$ it is clear that $w_0 = -id$ and hence you will certainly not be in case b) for any $\lambda$.

Now whether you are in a) or c) in those cases, or in a), b) or c) in the remaining cases, apparently depends on which weight $\lambda$ you're looking at. I admit I don't know enough to say anything about that parity distinction yet -- maybe a true expert can take it from here.

E.g. for $A_n$ with even $n$, at least all fundamental weights do not get sent to their exact negative, so the corresponding representations are in case b) and hence "genuinely complex"; whereas for the cases $A_{2n+1}, D_{2n+1}$ and $E_6$, some of their fundamental representations are in b), but others are not, and then one has to do some calculations to figure out if they are be in case a) or c). Luckily that has been done, for all irreps with -$w_0(\lambda) = \lambda$, see e.g. the table on p. 175 (178 in the pdf-file) of http://cds.cern.ch/record/134739/files/198109187.pdf.

Community
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Torsten Schoeneberg
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  • For type $A$ applying $-w_0$ (when the weight is written in the fundamental basis) reverses the order of the coefficients. So in this case it is easy to see which weights are self-dual. I seem to recall it being the identity for all the type $D$, but I did not actually check this – Tobias Kildetoft May 11 '18 at 17:10
  • @TobiasKildetoft: Bourbaki (tables at the end of ch. 6) and A. Geraschenko's answer which I linked to agree that in type $D_n$, the element $-w_0$ is the identity for even $n$, but flips the two "horns" of the Dynkin diagram for odd $n$. That's why I separated the cases $D_{2n}$ and $D_{2n+1}$. – Torsten Schoeneberg May 11 '18 at 19:25
  • Ahh, then I misremembered it. Thanks. – Tobias Kildetoft May 11 '18 at 20:45
  • @TobiasKildetoft: By the way, I came across this mnemonic to remember (after realising there is a parity distinction for $D_n$) which case is which: In case $D_4$, there is no distinct fork in the diagram (there are three), so $-w_0$ (which is always of order $\le 2$) cannot choose one of them, hence has to be the identity. Now 4 is even, so ... – Torsten Schoeneberg May 12 '18 at 22:34