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Let $E:=\{x\in \mathbb{R}^{l+1}:x_1+x_2+\cdots + x_{l+1}=0\}$ and let $\Phi\subseteq E$ denote its root system of type $A_l$ given the basis $\Delta=\{e_i-e_{i+1}, 1\leq i \leq l\}$ and with $\{e_i\}$ denoting the standard orthonormal basis of $\mathbb{R}^{l+1}$.

Let further $\Delta'\subseteq \Delta$ and consider $\Phi'=\text{span}_\mathbb{Z}(\Delta')\cap \Phi$ and $E'=\text{span}_\mathbb{R}(\Delta')\cap E$.

I want to determine the $\textit{type}$ of the root system $\Phi'\subseteq E'$.

Am I correct in assuming that one could consider the cut-out of the Cartan matrix of $\Phi$ corresponding to the Cartan matrix of $\Phi'$ in order to conclude that the Dynkin diagram of $\Phi'$ is given as a sum $\bigoplus A_{k}$ such that $\sum k= \dim(E')$ (and that all such choices of $k$ are possible)?

Thank you

Arctic Char
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Deracless
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    You are correct that the "standard parabolic" sub-root-systems (those systems of the form $\phi$ intersected with a subspace spanned by some of the simple roots) are all a direct sum of smaller $A_k$. However not all such sums appear: for example $A_2$ has only $A_1$ as a strict parabolic subsystem, you need to go up to $A_3$ before $A_1 \times A_1$ appears. – Joppy Jan 19 '22 at 10:46

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So if we take a subset of the simple roots $\Delta' \subset \Delta$ this corresponds to taking a subset of nodes of the Dynkin diagram. This gives a subdiagram where the nodes are this subset and the edges are only those edges connecting two nodes in this subset (i.e. no dangling edges). This diagram is then exactly the diagram of $\Phi'$. It is not too hard to prove this so I will leave that as an exercise. NB: we are only allowed to make these subdiagrams by removing nodes and the edges attached to them. We cannot just remove edges.

Thus by removing nodes from $A_l$ we can see that all the possible subdiagrams are of the form $A_{l_1}\times \cdots \times A_{l_k}$ for some $k$. Note that not all possible $l_i$'s will work and indeed we do not even have $l_1 + \cdots + l_k = l$. This should make sense because $A_l$ corresponds to $\mathfrak{sl}_{l+1}$ and what we are doing is dividing $\mathbb{C}^{l+1}$ into a direct sum of subspaces (They don't have to span the whole of $\mathbb{C}^{l+1}$) so we have $(l_1 + 1) + \cdots \cdots + (l_k + 1) = l_1 + \cdots + l_k + k \leq l +1$.

So some low dimensional examples. $A_2$ has only 1 proper, non-trivial subdiagram and that is of the form $A_1$. This corresponds to finding a copy of $\mathbb{C}^{2}$ inside $\mathbb{C}^{3}$. $A_3$ has three: $A_2$, $A_1 \times A_1$ and $A_1$ where we have removed one of the outer nodes, the middle node and two of the nodes respectively. These correspond to finding $\mathbb{C}^{3}\leq \mathbb{C}^{4}$, $\mathbb{C}^{2}\oplus\mathbb{C}^{2} = \mathbb{C}^{4}$ and $\mathbb{C}^{2}\leq \mathbb{C}^{4}$.

Callum
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  • Thank you very much! – Deracless Jan 19 '22 at 20:32
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    As a side point this is not the only way to make a root subsystem. Another good way is to take the extended Dynkin diagram (i.e. add a node corresponding to the lowest root) and then start removing nodes. This doesn't produce anything new for $A_l$ but for the others we can get some things you might expect like $D_n \times D_n$ as a subsystem of $D_{2n}$ (i.e. splitting $\mathbb{C}^{4n}$ into $ \mathbb{C}^{2n} \oplus \mathbb{C}^{2n}$). – Callum Jan 20 '22 at 10:48
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    +1 and +1 for that comment. I encountered a lot of root subsystems I had not expected when answering https://math.stackexchange.com/q/3864501/96384. Callum, I would actually appreciate it if you check my answer there. – Torsten Schoeneberg Jan 23 '22 at 20:27
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    @TorstenSchoeneberg I think you may be more of an expert here than me. I think adding in the lowest short root to create new extended diagrams is an inspired way to find more root subsystems. I have checked against the classification in Oshima's note (https://arxiv.org/abs/math/0611904) and this seems to provide a comprehensive list although I can't see an easy proof as to why. In particular I am impressed that it finds the long and short copies of $D_4$ inside $F_4$ separately. – Callum Jan 26 '22 at 11:37
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    Ah, nice: That paper has exactly those Dynkin diagrams extended by highest short root in Proposition 9.3. – Torsten Schoeneberg Jan 27 '22 at 18:41