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Suppose $\Phi$ is an irreducible classical root system with simple roots $\Delta=\{\alpha_1,\ldots,\alpha_n\}$ and consider a subset $\Delta'=\{\alpha_{k_1},\ldots,\alpha_{k_m}\}\subseteq\Delta$. Given $c_1,\ldots,c_m\in\mathbb{N}_0$ not all zero such that $X=\{\sum_{j=1}^mc_j\alpha_{k_j}+\sum_{\alpha\in\Delta\backslash\Delta'}d_\alpha\alpha\in\Phi\colon d_\alpha\in\mathbb{N}_0\}$ is nonempty, say that an element $\beta\in X$ is maximal if $\beta+\alpha\notin X$ for any $\alpha\in\Delta\backslash\Delta'$, and minimal if $\beta-\alpha\notin X$ for any $\alpha\in\Delta\backslash\Delta'$. Clearly maximal and minimal elements exist, but for which root systems are they necessarily unique?

Note that $\sum_{j=1}^mc_j\alpha_{k_j}$ is not necessarily an element of $\Phi$.

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  • $\begingroup$ Given the classification and construction of irreducible root systems, I suppose one could work out (laboriously) an answer to the uniqueness question for each type. But your single tag 'root-systems' and the complete absence of motivation (including indications of what consequences an answer to your question might have) make the question look at first artificial. How does such a question arise? [By the way, I guess your symbol $\mathbb{N}_0$ means the set of non-negative integers. Some people just write $\mathbb{N}$ for this set.] $\endgroup$ Nov 18, 2017 at 19:29
  • $\begingroup$ @JimHumphreys Given a classical simple Lie algebra $\mathfrak{g}$ and a semisimple subalgebra $\mathfrak{a}\subseteq\mathfrak{g}$, I would like to know how $\mathfrak{g}$ decomposes into irreducible $\mathfrak{a}$-submodules with respect to the adjoint action. I have a partial description, provided the above claim holds. $\endgroup$ Nov 18, 2017 at 21:38
  • $\begingroup$ This subject of branching rules for simple Lie algebras has been looked at from many angles, so it may be worthwhile to check back on the literature involved over the years. (Meanwhile, I'm still uncertain about your notation involving the symbol $\mathbb{N}$. I also don't know what the "above claim" means; I don't see any claim. Please clarify.) $\endgroup$ Nov 22, 2017 at 15:53

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