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A complex Lie group may have several real forms. Are there any duality/trinity... between them? Maybe a trivial question to ask, is $SL(3,\mathbb{C})$ a real form of $SL(3,\mathbb{C})\times SL(3,\mathbb{C})$ ?

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  • $\begingroup$ The only thing that I know that relates different real forms is the Cayley transform. I myself would welcome a good reference for that. $\endgroup$ Oct 10, 2014 at 10:41
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    $\begingroup$ With the usual definitions, the answer to your trivial question is yes if you regard $SL(3,\mathbb{C})$ as the underlying real Lie group. $\endgroup$ Oct 10, 2014 at 11:09

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I think not. For a compact connected real Lie group $G$, the real forms of the complexification of $G$ are classified by the set of elements of order 2 in a maximal torus of $G$ modulo the action of the Weyl group. This set appears to have no obvious symmetries. (The quoted result can be found, for example, in Serre's Galois Cohomology).

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I'm not sure that this is what you had in mind, but there is one non-trivial correspondence between different real forms of the same complex semi-simple Lie algebra. Their complex representation theory is identical. This is because an irreducible complex representation of a real form always extends to a complex linear representation of the complexification, which contains the other real forms.

On the other hand, the real representation theory of different real forms can be quite different. In particular their irreducible real representations do not even need to have the same dimension. The minimal example is a good illustration: The complex representations of $\mathfrak{su}(2)$ and $\mathfrak{sl}(2,\mathbb{R})$ are identical, but their real irreps do not even have the same dimension. $\mathfrak{sl}(2,\mathbb{R})$ has a nontrivial irreducible real 2d rep (the tautological one), but the smallest nontrivial real irreducible rep of $\mathfrak{su}(2)$ is of dim 3 (the adjoint rep).

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