= Solution
The <complexification of a Lie algebra> gives $\mathfrak{so}(2n,\mathbb C)$ of type <Dn root system> and $\mathfrak{so}(2n+1,\mathbb C)$ of type <Bn root system>. For $D_n$, the <simple roots> $\alpha_i=e_i-e_{i+1}$ for $1\le i<n$ and $\alpha_n=e_{n-1}+e_n$ give a chain $1,\ldots,n-2$ with two terminal nodes $n-1,n$ joined to $n-2$. All edges are single. For $B_n$, use $\alpha_i=e_i-e_{i+1}$ for $i<n$ and $\alpha_n=e_n$. The <Bn Dynkin diagram> is a chain with a double final edge and an arrow pointing to the short node $n$.
\Image[/past-exam-of-the-groups-and-representations-course-of-the-university-of-oxford/2025-orthogonal-dynkin-diagrams.png]
{title=Dynkin diagrams of the odd and even orthogonal Lie algebras, with the final arrow pointing toward the short root}
The usual family diagrams apply to $D_n$ for $n\ge4$ and $B_n$ for $n\ge2$. The small cases are $D_3\cong A_3$, $D_2\cong A_1\oplus A_1$, $B_1\cong A_1$; $\mathfrak{so}(2)$ is one-dimensional and Abelian.
To fix <Casimir normalization for orthogonal vector representations>, use Hermitian <Lie algebra generators> $t_{ij}=i(E_{ij}-E_{ji})/\sqrt2$, so the defining <representation> has <Dynkin index> $c(\mathbf N)=1$. With long <roots> of squared length two, the <quadratic Casimir operator> has <Casimir eigenvalue>
$$
C_2(\lambda)=\frac12(\lambda,\lambda+2\rho)=\frac12\lambda^TG(\lambda+2\mathbf1),
$$
where $\rho$ is the <Weyl vector> and $\lambda$ in the final expression is the column of <Dynkin labels>. The defining <fundamental representation> has <highest weight> $(1,0,\ldots,0)$ in the stated family conventions.
For $D_n$, the row supplied in the source has sum $n-1$ and first entry one, so
$$
C_2(\mathbf{2n})=\frac12\{1+2(n-1)\}=\frac{2n-1}{2}.
$$
For $B_n$, the supplied row has sum $n-\tfrac12$, giving
$$
C_2(\mathbf{2n+1})=\frac12\{1+2(n-\tfrac12)\}=n.
$$
Taking the <trace> of $\sum_a t_at_a=C_2(R)I_R$ gives the relation between the <Dynkin index> and <Casimir eigenvalue>, $c(R)\dim\mathfrak g=C_2(R)\dim R$. Therefore
$$
\boxed{C_2(\mathbf N)=\frac{N-1}{2},\qquad c(\mathbf N)=1.}
$$
There is also a direct check: each coordinate appears in $N-1$ planes and $t_{ij}^2=(E_{ii}+E_{jj})/2$, so $\sum_{i<j}t_{ij}^2=(N-1)I/2$.
If the <Casimir operator> is instead defined by $C(\lambda)=(\lambda,\lambda+2\rho)$, every <Lie algebra generator> is larger by $\sqrt2$ and the corresponding answers are $C(\mathbf N)=N-1$ and $c(\mathbf N)=2$. These are the same <Lie algebra> in a different normalization. At the exceptional small ranks the family <Dynkin labels> of the vector change, but the direct <matrix> computation remains valid.
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