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For $SO(N)$ the quadratic Casimir for the spinorial representation is $N(N-1)/8$ and that of the vector representation is $N-1$, but what is the quadratic Casimir of the spin $2$ representation? or what is mathematically known as the symmetric traceless representation.

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Your representation is a Young tableau with two boxes in the first row and none in the other rows. So you just take $h = (2, 0, \dots, 0)$ and plug it into the formula $C_2 = \sum_{i = 1}^{\rfloor N/2 \lfloor} h_i^2 + (N-2i) h_i$.

You can prove this for $N = 2r$ and $N = 2r+1$ separately by considering the expression for the quadratic Casimir in the Cartan-Weyl basis. Namely $C_2 = \sum_{i = 1}^r H_i^2 + \sum_{\alpha \in \Delta} E_{\alpha} E_{-\alpha}$.

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