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Let the highest weight of a $SO(2n+1)$ representation be given as $(m_1,m_2,...,m_n)$ ($m_1\geq m_2 \geq .. \geq m_n \geq 0$) and the highest weight of a $SO(2n)$ representation be $(s_1,s_2,...,s_n)$ ($s_1\geq s_2 \geq .. \geq \vert s_n \vert$).

The branching rule for the first to contain the second is given as, $m_1 \geq s_1 \geq m_2 \geq s_2 \geq m_3 \geq s_3 \geq ... \geq m_n \geq \vert s_n \vert$

Now if for $SO(2n)$ I am interested in the representation, $(s,0,0,0..,0)$ then the branching rule would be that $m_1 \geq s \geq m_2 \geq 0$ and all the $m$s and $s$s are zero. (...for $n=1$ one would use $\vert m_2 \vert$ in the above)

  • But there seems to be a condition of saying that the $SO(2n)$ representation is contained "maximally" in the required $SO(2n+1)$ representation - which sets $m_2 = s$ (or $\vert m_2\vert = s$ for $n=1$)

    I would like to understand what this condition is! (..hopefully there is some simple way of seeing this!...)

It seems that one can think of the $(s,0,..,0)$ representation as being given by symmetric traceless transverse tensors of rank $s$ and then this maximality condition is tantamount to the "transverse" condition...but I don't really understand this

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If $n=1$, there is no any $m_2$ coefficient. You should delete the phrases between "(" and ")". – emiliocba Sep 25 '13 at 11:33
What reference did you use to get your above formulas? – Turion Nov 14 '13 at 15:54

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