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In particular, I am interested in $A_4$ root system. Considering simpler cases of $A_2$ and $A_3$ my guess would be $(n+1)^2-(n+1)$ (where n is rank of the root system), but I'm not certain if it's true for higher ranks.

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  • $\begingroup$ In which way are you given the root system? Concretely as a subset of some Eucledian space - in this case you could of course just count - or abstractly in terms of its Dynkin diagram? $\endgroup$
    – Hanno
    Sep 13, 2015 at 9:28
  • $\begingroup$ Yes, in terms of Euclidean space. But Dynkin diagrams give only simple roots with angles. I can visualize and count all roots in 3D space (A3), but I struggle to do so in higher dimensions. $\endgroup$
    – Kosm
    Sep 13, 2015 at 9:45

1 Answer 1

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The Cartan decomposition of a simple Lie algebra $L$ of type $A_n$ implies that $$ |\Phi|=\dim L-\dim H=(n+1)^2-1-n=n^2+n $$ Here $\dim A_n=(n+1)^2-1$ and $\dim H$, the dimension of a Cartan subalgebra $H$, is equal to the rank of $A_n$, which is $n$. In fact, $H$ consists of diagonal matrices of size $n+1$, but with trace zero. So your result is correct.

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