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Let $\Gamma_{(\lambda_1, \dots, \lambda_{n})}$ denote an irreducible $SO(2n)$-module with highest weight $(\lambda_1, \dots, \lambda_n)$ and let more specifically $X = \Gamma_{(2\lambda, \dots, 0)}$ and $Y = \Gamma_{(2\lambda_1, \dots, 2\lambda_n)}$, where at least one of the $\lambda_j$ with $j>1$ is not equal to $0$. Furthermore, let $W \leq X$ be a (irreducible) $U(n)$-submodule of $X$, and $\tilde{W} \leq Y$ a (irreducible) $U(n)$-submodule of $Y$. My Questions:

1) Is it true that for any $X,Y,W,\tilde{W}$ choosen in such a fashion \begin{align*} W \ncong \tilde{W}? \end{align*}

2) Is there a branching rule for $U(n)$-submodules of $SO(2n)$-modules? (This obviously relates to Question 1)

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  • $\begingroup$ @AmirSagiv Maybe it is worth mentioning that some users do not agree to adding MathJax/LaTeX to the titles which are perfectly readable without it. (As far as I can say, there is no clear consensus about this.) See this discussion on meta or this message in chat. (Regardless of this issue, I appreciate efforts to improve posts by editing them, be it for typos, grammar, LaTeX or anything else.) $\endgroup$
    – Martin Sleziak
    Commented May 25, 2016 at 5:50
  • $\begingroup$ I'll read and comment in Meta. Thanks for the heads up. In the post themselves it is a consensus, right? $\endgroup$
    – Amir Sagiv
    Commented May 25, 2016 at 11:48

1 Answer 1

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you can find branching rule ${\rm SO}(2n) \to U(n)$ in the book of Knapp, Lie groups beyond an introduction, and also in some paper of his, and in the book of Zelobenko, compact Lie groups and their rep's, also try the book of Tom Dieck.

best
jorge

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  • $\begingroup$ I cannot find such a rule in the book by Knapp. Could you give a more precise reference? $\endgroup$
    – Felix
    Commented May 4, 2016 at 9:01
  • $\begingroup$ I am sorrry, I had an old version of Knapp's book not containing the section about branching rules! Indeed in the new version one finds Kostant's branching law which applies to this situation! From calculations in dimension 4 it is then not to hard to see that the answer to 1) is negative! $\endgroup$
    – Felix
    Commented May 23, 2016 at 9:48

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