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Can someone answer the following question: Is there any classification of maximal proper Zariski-closed real subgroups of $SO(p,q)$ which are not parabolic, and satisfy the following conditions:

  1. they have rank at least the rank of $SO(p) \times SO(q)$ .

  2. they should act irreducibly on the natural representation of $SO(p,q)$, i.e. on $\mathbb{R}^{p,q}$.

Thanks.

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    $\begingroup$ The references given in this answer classify all maximal subgroups; perhaps extracting the ones you want is not too hard? $\endgroup$ Commented Sep 20, 2014 at 16:46
  • $\begingroup$ Forgive me asking this: I tried to get the papers you suggested from my library, but I could not get them. Do you have access to them, and in case if they are allowed to be shared, would you please share them with me?. $\endgroup$
    – Vanya
    Commented Sep 20, 2014 at 22:42
  • $\begingroup$ Sorry, I don't have copies at hand; have you tried interlibrary loan? $\endgroup$ Commented Sep 21, 2014 at 0:36
  • $\begingroup$ @Francois: The paper by Taufik contains classification of irreducible maximal subalgebras. In that paper, he mentions that the reducible ones are given in mathnet.ru/php/… $\endgroup$
    – Vanya
    Commented Sep 28, 2014 at 3:38
  • $\begingroup$ I'm afraid Mušic's paper hasn't been translated. OTOH, you should be able to find translations of Dynkin's, in "Selected papers of E. B. Dynkin with commentary". $\endgroup$ Commented Sep 28, 2014 at 5:30

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A Russian mathematician whose name is Komrakov, has listed the maximal subgroups of SO(p,q), if you do not find the articles, i have the pdf fil's, best regards jorge

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  • $\begingroup$ Note that the key paper by Komrakov is one of the references mentioned by Francois, though it lacks full details. The basic source is the older classification work by Dynkin. $\endgroup$ Commented Sep 26, 2014 at 18:09
  • $\begingroup$ Thanks to Francois, I found the references. The paper by Komrakov contains final results at the Lie algebra level. If you have the papers that have proofs, please send it to me. $\endgroup$
    – Vanya
    Commented Sep 27, 2014 at 17:25

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