I am reading about Kac–Moody algebras. I have a question about the construction of twisted affine algebras.
Let $\sigma$ be a graph automorphism of the simple Lie algebra $\mathfrak{g}$.
When $\mathfrak{g}=(A_{2l},2)$, $(A_{2l-1},2)$, $(D_{l+1},2)$, $(E_{6},2)$ or $(D_4,3)$, the fixed point $\mathfrak{g}_0\subseteq \mathfrak{g}$ of $\sigma$ is also a simple algebra and has the type $B_l$, $C_l$, $B_l$, $F_4$ or $G_2$ respectively. $2,2,2,2,3$ mean the corresponding the oder of $\sigma$ as a automorphism.
But when $\mathfrak{g}=A_{2l}$, the simple roots of $\mathfrak{g}_0$ are indexed by $\{0,1,\dotsc ,l-1\}$.
When $\mathfrak{g}=A_{2l-1}$, $D_{l+1}$, $E_6$, $D_4$, the simple roots of $\mathfrak{g}_0$ are indexed by $\{1,\dotsc ,l\}$, $\{1, \dotsc, l\}$, $\{1,2,3,4\}$, $\{1,2\}$ respectively.
When constructing twisted affine algebras, we need to add an extra point. For $A_{2l}$, the extra point is indexed by $l$. For other types, the extra point is indexed by $0$.
I'm wondering what internal reason this is for such a difference between $A_{2l}$ and other types?
New explanations of the problem
It comes from Kac's "Infinite dimensional Lie algebras" and a lecture from Introduction to Kac–Moody Lie algebras for twisted affine algebras.
As we know we have affine types $A^{(2)}_{2l}$, $A^{(2)}_{2l-1}$, $D^{(2)}_{l+1}$, $E^{(2)}_{6}$ and $D^{(4)}_{3}$, we want to construct the corresponding Kac–Moody algebra.
In Kac's book and the above lecture, we should firstly use the graph automorphism to get the corresponding fixed-point subalgebra $\mathfrak{g}_0$.
For example, when $\mathfrak{g}=A_{2l}$, the fixed-point subalgebra$\mathfrak{g}_0$ is the simple Lie algebra $B_l$ and we have thesimple roots $\{\alpha_0,\dotsc, \alpha_{l-1}\}$, generators $e_i, f_i$ ($0\le i\le l-1$) and Cartan subalgebra $\mathfrak{h}_0$.
When $\mathfrak{g}=A_{2l-1}$, the fixed-point subalgebra$\mathfrak{g}_0$ is the simple Lie algebra $C_l$ and we have thesimple roots $\{\alpha_1,\dotsc, \alpha_{l}\}$ and $e_i, f_i$ ($1\le i\lel$) and Cartan subalgebra $\mathfrak{h}_0$.
Let $\hat{\mathcal{L}}\left( \mathfrak{g} \right) =\left( \mathbb{C} \left[ t,t^{-1} \right] \otimes \mathfrak{g} \right) \oplus \mathbb{C} c\oplus \mathbb{C} d$ with Lie bracket $$\left[ t^a\otimes x+\lambda c+\mu d,t^b\otimes y+\lambda ^{\prime}c+\mu \prime d \right] =\left[ x,y \right] \otimes t^{a+b}+\mu y\otimes bt^b-\mu ^{\prime}x\otimes at^a+a\delta _{a,-b}\left( x,y \right) c.$$
When constructing the $A^{(2)}_{2l}$ and $A^{(2)}_{2l-1}$, $D^{(2)}_{l+1}$, $E^{(2)}_{6}$, $D^{(3)}_{4}$ in Kac's book and the above lectures, they put $A^{(2)}_{2l}$ in one case and put $A^{(2)}_{2l-1}$, $D^{(2)}_{l+1}$, $E^{(2)}_{6}$ and $D^{(3)}_{4}$ in the other case.
We consider the fixed points of $\hat{\mathcal{L}}\left( \mathfrak{g}\right) $ under the extended automorphism $\sigma$ by the formula$\sigma \left( t^a\otimes x+\lambda c+\mu d \right) =\frac{1}{\xi ^a}t^a\otimes \sigma \left( x \right) +\lambda c+\mu d$ where $\xi=e^{\frac{i2\pi}{m}}$ and $m$ is the order of $\sigma$.
For example, when $\mathfrak{g}=A_{2l-1}$, we can put $E_i=1\otimese_i$ and $f_i=1\otimes f_i(1\le i\le l)$ and $E_0=t\otimes e_0$,$F_0=t^{-1}\otimes f_0$ where $e_0\in E_{\theta}$ and $f_0 \in> E_{{-\theta}}$ ($\theta$ is the highest root of $C_l$).
When $\mathfrak{g}=A_{2l}$, we can put $E_i=1\otimes e_i$ and$f_i=1\otimes f_i(0\le i\le l-1)$ and $E_l=t\otimes e_l$,$F_l=t^{-1}\otimes f_l$ where $e_l\in E_{\theta}$ and $f_l\in> E_{{-\theta}}$ ($\theta$ is the highest root of $B_l$).
If we simply look at the Dynkin diagram, we have the picture for $A^{(2)}_{2l-1}$:
For $A^{(2)}_{2l}$, we have 
Now I wonder why there exist such a difference in the construction and the order of labelling between $A^{(2)}_{2l}$ and $A^{(2)}_{2l-1}$, $D^{(2)}_{l+1}$, $E^{(2)}_{6}$, $D^{(3)}_{4}$ ?
I think it's just a matter of numerical order, and we can renumber itto make $A^{(2)}_{2l}$ and $A^{(2)}_{2l-1}$, $D^{(2)}_{l+1}$,$E^{(2)}_{6}$, $D^{(3)}_{4}$ coincide. Is it true?
Any help and references are greatly appreciated.