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I would like to figure out whether there is an irreducible complex (in the sense non-self-conjugate) representation of a group $SO(N)$, $N>2$, whose tensor square contains the fundamental representation.

$SO(N)$ only has complex representations for $N$ even, $N\geq 6$. When one allows projective representations, then the spinor representation of the algebra $so(6)=sl(4)$ does the job, but I am interested only in non-projective representations of the orthogonal group.

I think a higest-weight representation $\Gamma_{a_1,a_2,a_3}$ of $sl(4)$ gives a representation of $SO(6)$ if and only if $a_1+a_3$ is even, so that the total number of boxes in the corresponding Young tableau is $2(a_1+a_2)+a_1+a_3$, which is even. It follows from the Littlewood-Richardson rule that the tensor square of $\Gamma_{a_1,a_2,a_3}$ only contains irreps whose Young tableaux have $4k$ boxes, so can't contain the fundamental of $SO(6)$, i.e. $\Gamma_{0,1,0}$. Hence a possible example must have $N\geq 8$.

Any ideas?

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    $\begingroup$ You can try more numerical experiments with LiE (wwwmathlabo.univ-poitiers.fr/~maavl/LiE). You can write a few lines program that finds representations whose square contains the fundamental one and then you can check by yourself whether these are complex representations or not. $\endgroup$ Oct 16, 2013 at 9:19
  • $\begingroup$ What does exactly self-conjugate mean in terms of the highest weight? Does it mean that it is invariant under $-w_0$? $\endgroup$
    – Misha
    Oct 16, 2013 at 15:48

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