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Is there a relation between number of roots of a finite dimensional semi-simple Lie algebra L and dimension of the maximal toral sub-algebra H(Cartan sub-algebra) of L?

Thanks!

sumit
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2 Answers2

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There is the relation $$ \dim (H)+\mid \Phi\mid =\dim (L), $$ where $\mid \Phi\mid$ denotes the cardinality of the root system of $L$. For example, with $L$ of type $A_n$ we have $\dim (H)=n$, $\mid \Phi\mid=n^2+n$ and $\dim(L)=n^2+2n$.

Dietrich Burde
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  • Thanks.Yeah, I know about the root space decomposition. And, using that only , I concluded that 2 dim(H) <= no. of roots. Now, I think for each type of semisimple Lie algebra like sl_n(C) or so_n(C), there would be relations. But, I was thinking if we can say something without using the classification of complex semi-simple Lie algebras. – sumit Apr 20 '15 at 06:35
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With notations as in the question, 2 dim(H) <= no. of roots. This is true because roots span H*(Dual space of H) and roots always occur in pairs.(I mean x is a root iff -x is a root.)

I do not hope there is any other relation, in general. But, may be, if we consider some special classes of L, we may have some better results!

sumit
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