Denote $\Phi(n)$ as the root system of Lie algebra $\mathfrak{g}$ of type $A_{n}$. Call a disjoint union $\Phi(n) = \Phi_{1}\sqcup\dotsb\sqcup\Phi_{s}$ a decomposition of $\Phi$ if each $\Phi_{k}$ is a root subsystem of $\Phi$ such that $\Phi_{k} = \Phi(n)\cap\operatorname{span}_{\mathbb{R}}\Phi_{k}$.
Our question is to find the minimal number $s(n):=\inf \left\{s\mid\Phi = \Phi_{1}\sqcup\dotsb\sqcup\Phi_{s} \text{ is a decomposition of }\Phi(n)\right\}$ for all such decomposition of $A_{n}$ except $s=1$. (A related question on MO asked also by me is there.)
This question is from geometry when we study ADE singularity in hyperK"ahler ALE. The following is our partial result.
In terms of simple Lie algebra, recall the decomposition of Lie algebra $\mathfrak{g} := \bigoplus_{\alpha\in\Phi}\mathfrak{g}_{\alpha}$ and the decomposition of the positive nilpotent subalgebra $\mathfrak{n}^{+}:=\bigoplus_{\alpha\in\Phi^{+}}\mathfrak{g}_{\alpha}$. A decomposition $\Phi(n) = \Phi_{1}\sqcup\dotsb\sqcup\Phi_{s}$ gives $\Phi(n)^{+} = \Phi_{1}^{+}\sqcup\dotsb\sqcup\Phi_{s}^{+}$ for positive roots, and we obtain a direct sum (in the level of linear spaces): \begin{equation} \mathfrak{n}^{+} = \mathfrak{n}_{1}^{+} \oplus\cdots\oplus\mathfrak{n}_{s}^{+}\,,\,\mathfrak{n}_{k}^{+} := \bigoplus_{\alpha\in\Phi_{k}^{+}}\mathfrak{g}_{\alpha} \end{equation}
Example: For $A_{2}$, the positive roots corresponds to the upper triangular matrices as \begin{pmatrix} 0 &* &*\\ &0 &*\\ & &0 \end{pmatrix}
and a decomposition is given by \begin{equation} \begin{pmatrix} 0 &* &*\\ &0 &*\\ & &0 \end{pmatrix} = \begin{pmatrix} 0 &* &0\\ &0 &0\\ & &0 \end{pmatrix} \oplus \begin{pmatrix} 0 &0 &0\\ &0 &*\\ & &0 \end{pmatrix} \oplus \begin{pmatrix} 0 &0 &*\\ &0 &0\\ & &0 \end{pmatrix} \end{equation} where the direct sum is only for linear spaces. Thus, we can show for $A_{2}$, the minimal number $s(2) = 3$.
Example: For $A_{3}$, a decomposition is give by: \begin{equation} \begin{pmatrix} 0 &* &* &*\\ &0 &* &*\\ & &0 &*\\ & & &0 \end{pmatrix} = \begin{pmatrix} 0 &* &0 &0\\ &0 &0 &0\\ & &0 &*\\ & & &0 \end{pmatrix} \oplus \begin{pmatrix} 0 &0 &* &0\\ &0 &0 &*\\ & &0 &0\\ & & &0 \end{pmatrix} \oplus \begin{pmatrix} 0 &0 &0 &*\\ &0 &* &0\\ & &0 &0\\ & & &0 \end{pmatrix} \end{equation}. Moreover, we can show for $A_{3}$, $s(3)=3$.
Generally, we can construct examples as above for $A_{2k}$ and $A_{2k-1}$ with the number of root subsystems $s = k+2$. Thus, \begin{equation} s(n) \leq \lfloor \frac{n}{2} \rfloor + 2. \end{equation} and we calculate explicitly from $A_{2}$ to $A_{7}$, $s(n) = \lfloor \frac{n}{2} \rfloor + 2, n = 2,\cdots,7$. Hence, we conjecture that \begin{equation} s(n) = \lfloor \frac{n}{2} \rfloor + 2. \end{equation}
We want to find a proof of this in Lie theory.
