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Let $G$ be a semi-simple compact Lie group. Let $V$ be a real vector space and let:

$\rho : G \to Aut_{\mathbb{R}}(V)$

be an irreducible real representation of $G$ on $V$. We say that $\rho$ is a real representation of complex type if and only if there exist a $J\in Aut_{\mathbb{R}}(V)$ satisfying:

$J^{2} = - Id\, , \qquad \rho\circ J = J \circ \rho$.

My question is the following: Is there any sort of classification of real irreducible representations of complex type for $G$ compact and semi-simple? I am particularly interested in the case in which $\rho$ is the Adjoint representation of $G$ on its real Lie algebra $\mathfrak{g}$.

Thanks.

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  • $\begingroup$ I think that a simple compact Lie group $G$ is the group of real points of an absolutely simple anisotropic algebraic $\mathbb{R}$-group $\bf G$. It follows that the adjoint representation of $G$ is absolutely irreducible, and therefore, it cannot be of complex type. $\endgroup$ Feb 18, 2017 at 20:05
  • $\begingroup$ @MikhailBorovoi. Thanks for your comment, that is in fact what I was expecting. Can you please give me a reference to verify your statement? $\endgroup$
    – Bilateral
    Feb 18, 2017 at 20:17
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    $\begingroup$ $R_{\mathbb{C}/\mathbb{R}}\bf G$ is Weil's restriction of scalars. This means that you regard a $d$-dimensional complex group as a $2d$-dimensional real group. $\endgroup$ Feb 18, 2017 at 20:46
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    $\begingroup$ Concerning complex algebraic groups see for example the book by Onishchik and Vinberg. $\endgroup$ Feb 18, 2017 at 20:55
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    $\begingroup$ As a simple example, consider the 3-dimensional simple complex group $\mathbf{G}=\mathrm{SL}(2,\mathbb{C})$, which you can regard as a 6-dimensional simple real group $G$. The adjoint representation of $G$ is clearly of complex type. Of course $G$ is not compact. $\endgroup$ Feb 18, 2017 at 21:17

2 Answers 2

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Irreducible real representations of complex type of a compact group correspond to irreducible complex representations that do not admit an invariant bilinear form. Irreducible real representations of quaternionic type correspond to irreducible complex representations that admit an alternating invariant bilinear form. See Serre, Linear representations of finite groups, Section 13.2, Prop. 38.

Concerning invariant bilinear forms on irreducible complex representations of simple groups, see Onishchik and Vinberg, Lie groups and algebraic groups, Table 3 (page 297).

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  • $\begingroup$ The OP is using the terminology "complex type" with a meaning different from the usual one (he only asks that $\mathbf{C}$ occur inside the endomorphism algebra over $\mathbf{R}$, not that it coincides; i.e., quaternionic cases are also being called "complex type" by him). $\endgroup$
    – nfdc23
    Feb 18, 2017 at 20:42
  • $\begingroup$ @nfdc23: Yes, I have noticed this, but he can learn the standard terminology form Serre's book or any other book. $\endgroup$ Feb 18, 2017 at 20:49
  • $\begingroup$ Thanks for the comments. I was aware of the terminology, but I wanted to simplify the question so I just defined the complext type to mean both. I will edit it accordingly. $\endgroup$
    – Bilateral
    Feb 18, 2017 at 20:52
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Let $\pi$ be a complex representation of the compact connected Lie group $G$ (no need for semisimplicity here) on a (finite-dimensional) vector space $V$. We say that $\pi$ is of real type if it comes from a representation of $G$ on a real vector space by extension of scalars, and we say that $\pi$ is of quaternionic type if it comes from a representation of G on a quaternionic vector space by restriction of scalars. If $\pi$ is neither of real type nor of quaternionic type, we say that $\pi$ is of complex type.

Let $\rho$ be a real irreducible representation of $G$ on a real vector space $W$. By Schur’s lemma, the centralizer of $\rho(G)$ in $\mathrm{End}(W)$ is an associative real division algebra, thus, by Frobenius’ theorem, isomorphic to one of (a) $\mathbb R$, (b) $\mathbb H$ or (c) $\mathbb C$.

The relation between real and complex representations is, respectively, that: (a) the complexification $\rho^c$ is irreducible (and we say that $\rho$ is absolutely irreducible) and $\rho^c = \pi$ is a representation of real type; (b) the complexification $\rho^c$ is reducible and $\rho^c=\pi\oplus\pi$ where $\pi$ is an irreducible representation of quaternionic type; (c) the complexification $\rho^c$ is reducible and $\rho^c=\pi\oplus\pi^*$ where $\pi$ is an irreducible representation of complex type and $\pi^*$ is not equivalent to $\pi$ (where $\pi^*$ denotes the induced representation on $V^*=\bar V$).

Also $\rho $ is a real form of $\pi$ in the first case ($\rho^c=\pi$), but $\rho$ is $\pi$ viewed as a real representation in the other two cases ($\rho=\pi^r$).

Regarding the adjoint representation of a compact simple Lie group, it is always absolutely irreducible. Simplicity of the group is equivalent to irreducibility of the representation. Even in the semisimple case, admiting and invariant complex structure would mean that the Lie group is a complex Lie group viewed as real, not possible due to the compactness.

Cartan's theory of real representations of semisimple Lie algebras is masterfully presented in modern form in

ESI Lectures in Mathematics and Physics

Arkady L. Onishchik (Yaroslavl State University, Russia)

Lectures on Real Semisimple Lie Algebras and Their Representations ISBN print 978-3-03719-002-9, ISBN online 978-3-03719-502-4 DOI 10.4171/002 February 2004, 95 pages, softcover, 17 x 24 cm. 24.00 Euro

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  • $\begingroup$ Claudio Gorodski can you explain why complexification of the adjoint representation of a compact simple Lie group/algebra is irreducble? I understand that simplicity implies irreducibility of the adjoint representation but I don't see why the complexification is also irreducible which is needed for absolute irreducibility. $\endgroup$ May 4, 2017 at 2:05
  • $\begingroup$ @AdamLeibniz The complexification would be reducible if and only if there is an ad -invariant complex structure on the (real) Lie algebra of the group. This would mean that the Lie group is a complex Lie group viewed as real. However this is impossible due to compactness of the group. $\endgroup$ May 4, 2017 at 16:26

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