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I am trying to figure out the formula for the quadratic Casimir of the adjoint representation of SO(N) but I can't find it anywhere.

I know that for the spinorial representation $C_2=N(N-1)/8$; for vectorial it is $C_2=N-1$. In the case of SU(N) the adjoint representation is $C_2=N$.

But for some reason, I just can't find the formula for the quadratic Casimir of the adjoint representation of SO(N). Can anybody please help me out?

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From your examples, I think the bilinear form on $\mathfrak{so}(N)$ you choose is $$B(X,Y)=\mathrm{Tr}(XY).$$ So we can choose a basis of $\mathfrak{so}(N)$ to be $X_{ij}=e_{ij}-e_{ji},1\leq i<j\leq N$, where $e_{ij}$ is a matrix whose entries are all zero except $(i,j)$-entry, which equals to $1$. The dual basis is $X^{ij}=-X_{ij}.$

We can check that for any $r<s$, $$\big(\sum_{i<j}\mathrm{ad}(X_{ij})\circ\mathrm{ad}(X^{ij})\big)(X_{rs})=2(N-2)X_{rs}.$$ This tells us the quadratic Casmir element of the adjoint representation is $2(N-2).$

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