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As the title asks, is the restricted root system of a simple real Lie group irreducible? I believe this is true but I need a reference to cite.

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    $\begingroup$ You should be more explicit about saying what you mean by "restricted root system". Also, is the Lie group assumed to be the group of real points of a simple algebraic group defined over $\mathbb{R}$? (And why do you believe the statement is true? What examples have you looked at?) $\endgroup$ Oct 2, 2015 at 18:19

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As my comment suggested, the question itself lacks detail and seems to be out of focus. For the somewhat intricate structure theory of a (real) simple or semisimple Lie group, there are standard textbook treatments: see for instance Chapter VI of Knapp's book Lie Groups Beyond an Introduction, PM 140, Birkhauser, 1996. Usually the structure theory can be best understood through the study of the corresponding real Lie algebra and its complexification, via root space decompositions.

Here the real roots are studied in detail by Knapp in VI.4, with special attention to restricted roots in the setting of the Iwasawa decomposition. But in the restricted root decomposition of a real simple Lie algebra, one often gets root spaces of dimension higher than 1. So this is not a "root system" in the classical sense as developed for complex semisimple Lie algebras. In particular, it's unclear what it would mean for the set of restricted roots to be "irreducible". The examples worked out by Knapp show the range of possibilities, and he develops the complete classification of real semisimple Lie algebras (evolving from the original work of E. Cartan). All of this requires considerable detail (and notation), since the groups and Lie algebras in question vary a lot from compact to split types.

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I think that Springer's Linear Algebraic Groups (2nd edition), no. 15.5.6 gives a general argument why the answer is "yes" for absolutely simple groups (in a much more general setup and if we agree on terminology, which should be clarified as per Jim Humphreys' comment). Of course, the tables (like in Knapp's book cited by Jim Humphreys, end of chapter VI) also show this case by case.

Now if you drop the "absolutely", you basically only have to add scalar restrictions to the consideration, and here the restricted root system is of the same type as the absolute one -- thus irreducible. The argument (on the Lie algebra level) is on page 366 of Knapp's book.

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