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 the
simple 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 the
simple roots $\{\alpha_1,\dotsc, \alpha_{l}\}$ and $e_i, f_i$ ($1\le i\le
l$) 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 $E_i=1\otimes
e_i$ 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 it
to 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.
Best Answer
Let $\mathfrak{g}$ be a finite-dimensional simple Lie algebra with Dynkin diagram of type $X$. Let $\sigma$ be a diagram automorphism of $X$ of order $k$. Then the affine Dynkin diagram of type $X^{(k)}$ has a "distinguished node" so that after deleting it we are left with the Dynkin diagram of the fixed point subalgebra $\mathfrak{g}_0$. (This characterizes the node up to diagram automorphism.)
The labeling of the simple roots for affine Lie algebras used by Kac is the unique one so that $\alpha_0$ is the distinguished node of the Langlands dual Dynkin diagram. In other words, after reversing all the arrows in the affine Dynkin diagram, $\alpha_0$ labels the distinguished node of the resulting diagram.
It's worth noting that Kac's convention was already in the literature when Infinite Dimensional Lie Algebras was written. An earlier reference using it is
Kac, Victor G.; Peterson, Dale H., Infinite-dimensional Lie algebras, theta functions and modular forms, Adv. Math. 53, 125-264 (1984). ZBL0584.17007.
Here seems to be the main benefit of Kac's convention: both the article and the book use the fact that the primitive imaginary root $\delta$ differs from the highest root of the finite root system spanned by $\alpha_1,\ldots,\alpha_l$ by a multiple of $\alpha_0$. This is true with Kac's labeling but not with an "$\alpha_0=$ distinguished node" labeling. However, as you noticed, Kac's labeling complicates the usual construction of a twisted affine Lie algebra via an automorphism. Ultimately, the choice is yours to make.
Edit: Here is an example indicating the difference between the distinguished node and the one that Kac labels $\alpha_0$. The distinguished node in every diagram is indicated in red. The left and right sides are related by Langlands duality. Starting from a diagram on the left, to determine which node Kac would label $\alpha_0$, we look at which node is distinguished in the diagram on the right. The corresponding node on the left is $\alpha_0$.