Classical Lie Groups
This article is one of Lie Group & Representation contents.
$A_N$: Special Unitary group: $SU(N+1)$
Unitary transformation of complex vector with determinant 1.
$$U^\dagger U=1$$
Root became $n(n-1)$ permutation of $(1,-1,0,\cdots,0)$.
Simple root became $(1,-1,0,\cdots,0,0),\ (0,1,-1,\cdots,0,0),\cdots (0,0,0,\cdots,1,-1).$
Rank is $n-1$ and Cartan Matrix became $$\begin{pmatrix}2&-1&0&\cdots & 0&0\\ -1&2&-1&\cdots& 0&0\\ 0&-1&2&\cdots & 0&0\\ \vdots& \vdots& \vdots& \ddots& \vdots& \vdots \\ 0& 0& 0& \cdots& 2& -1\\ 0& 0& 0& \cdots& -1 & 2\end{pmatrix}$$
Adjoint representation is traceless tensor product of fundamental and antifundamental representation, $T^i_j$ (traceless), in Dynkin Coefficient, $[1,0,\cdots,0,1]$
$B_N$: Odd Special Orthogonal Group: $SO(2N+1)$
Orthogonal transformation of real vector with determinant 1.
$$U^T U=I_{2N+1}$$
For Lie group, $U=\exp (Y)$, $$-Y^T=Y$$
When $Y=\begin{pmatrix} Y_1&Y_2\\ Y_3&Y_4\end{pmatrix}$, $$Y_4=-Y_1^T,\ Y_2=-Y_2^T,\ Y_3=-Y_3^T$$
Let's set maximal torus(Cartan Subalgebra) on $Y_1$.
Root become $2N^2+N$ permutation of two $(\pm 1,0,\cdots,0)$.
Simple root is $(1,-1,0,\cdots,0),\ (0,1,-1,\cdots,0),\ (0,\cdots,1,-1),\ (0,\cdots, 0,1)$
Rank is $n-1$ and Cartan Matrix became $$\begin{pmatrix}2&-1&0&\cdots & 0&0\\ -1&2&-1&\cdots& 0&0\\ 0&-1&2&\cdots & 0&0\\ \vdots& \vdots& \vdots& \ddots& \vdots& \vdots \\ 0& 0& 0& \cdots& 2& -2\\ 0& 0& 0& \cdots& -1 & 2\end{pmatrix}$$
Adjoint representation is antisymmetric tensor product of two fundamental representation since $-Y^T=Y$. It write as $T^{[ij]}$, in Dynkin Coefficient, $[0,1,\cdots,0]$
$C_N$: Symplectic Group: $Sp(2N)$
$$U^\dagger U=I_{2N}$$
$$U^T \epsilon U = \epsilon$$
when $$\epsilon=\begin{pmatrix}0& I_N\\ -I_N&0\end{pmatrix}$$
For Lie algebra, $U=\exp (Y)$, $$Y^T=\epsilon Y\epsilon$$
When $Y=\begin{pmatrix} Y_1&Y_2\\ Y_3&Y_4\end{pmatrix}$, $$Y_4=-Y_1^T,\ Y_2=Y_2^T,\ Y_3=Y_3^T$$
Let's set maximal torus(Cartan Subalgebra) on $Y_1$.
Root become $2N^2+N$ permutation of two $(\pm 1,0,\cdots,0)$.
Simple root is $(1,-1,0,\cdots,0),\ (0,1,-1,\cdots,0),\ (0,\cdots,1,-1),\ (0,\cdots, 0,2)$
Rank is $n-1$ and Cartan Matrix became $$\begin{pmatrix}2&-1&0&\cdots & 0&0\\ -1&2&-1&\cdots& 0&0\\ 0&-1&2&\cdots & 0&0\\ \vdots& \vdots& \vdots& \ddots& \vdots& \vdots \\ 0& 0& 0& \cdots& 2 &-1\\ 0& 0& 0& \cdots& -2 & 2\end{pmatrix}$$
Adjoint representation is symmetric tensor product of two fundamental representation, $T^{[ij]}$, in Dynkin Coefficient, $[0,1,\cdots,0]$
$D_N$: Even Special Orthogonal Group: $SO(2N)$
Orthogonal transformation of real vector with determinant 1.
$$U^T U=I_{2N}$$
For Lie group, $U=\exp (Y)$, $$-Y^T=Y$$
When $Y=\begin{pmatrix}b^1&c^1& c^2 \\ d^1 &Y_1&Y_2\\ d^2& Y_3&Y_4\end{pmatrix}$, $$Y_4=-Y_1^T,\ Y_2=Y_2^T,\ Y_3=Y_3^T, b^1=0, c^1=-d^{2T}, c^2=-d^{2T}$$
Let's set maximal torus(Cartan Subalgebra) on $Y_1$.
Root become $2N^2-N$ permutation of two $(\pm 1,0,\cdots,0)$.
Simple root is $(1,-1,0,\cdots,0),\ (0,1,-1,\cdots,0),\ (0,\cdots,1,-1),\ (0,\cdots, 1,1)$
Rank is $n-1$ and Cartan Matrix became $$\begin{pmatrix}2&-1&0&\cdots & 0&0&0\\ -1&2&-1&\cdots& 0&0&0\\ 0&-1&2&\cdots & 0&0&0\\ \vdots& \vdots& \vdots& \ddots& \vdots& \vdots&\vdots \\ 0& 0& 0& \cdots& 2 &-1& -1\\ 0& 0& 0& \cdots& -1 & 2& 0 \\ 0& 0& 0& \cdots& -1& 0& 2\end{pmatrix}$$
Adjoint representation is antisymmetric tensor product of two fundamental representation since $-Y^T=Y$. It write as $T^{[ij]}$, in Dynkin Coefficient, $[0,1,\cdots,0]$
Reference
Matthew Foster - Highest-weight representations
Matthew Foster - SU(n), SO(n), and Sp(2n) Lie groups