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Do the roots of a root space decomposition have a kernel? Since it is the duel space to the cartan subalgebra ,evaluation of the roots on a non-equal index cartan basis element should be zero.

Thanks

dylan7
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1 Answers1

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If I understand your question correctly, then yes, they do. And this has nothing to do with roots. For any linear function $f\in V^*$ the set $\{v\in V\mid <f,v>=0\}$ is a subspace of $V$ of codimension $1$ (if $f\neq 0$).

Sasha Patotski
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  • What I was basically asking, which might have not be clear, was that if the roots, which are functionals, have a kernel, like all other functionals, then one of the functionals evaluated on all the elements of the cartan algebra is not the same value regardless of the element of the cartan algebra, specifically when it is evaluated on a sub algebra of the cartan subalgebra of codimension 1. However in the definition of the root space decomposotion, a given root functiomal has the same value for all elements of the cartan algebra. Unless I am not understanding what roots are. – dylan7 Aug 05 '14 at 15:10
  • No, roots don't have the same value for all elements of the Cartan! By definition, an element $g\neq 0$ is in $\mathfrak{g}_\alpha$ iff $[h,g]=\alpha(h)g$. Consider example of $sl_2$. Then Cartan is $1$-dimensional, it's spanned by the diagonal matrices $diag(a,-a)$. Then if you take this $h=diag(a,-a)\in\mathfrak{h}$ and compute $[h,g]$ for $g=\begin{pmatrix}0 & 1\ 0 & 0\end{pmatrix}$, you will get $[h,g]=a\cdot g$. This means that there is a root $\alpha$ sending $h=diag(a,-a)$ to $a$. Sure this value is different for different $a$! – Sasha Patotski Aug 05 '14 at 16:12
  • So simultaneously diagonalizable doesn't mean that all the matrices in the cartan have the same eigenvalues? – dylan7 Aug 05 '14 at 16:41
  • Oh I realized my understanding of the definition of simultaneously diagonalizable was incorrect. Now I understand. Another question, so the basis of the cartan algebra are elements that when evaluated under the killing form with another element of the cartan algebra that is not another basis element is one of that cartan elements eigenvectors? Which is the same as a root being evaluated on that cartan element? The root associated with that cartan basis element. – dylan7 Aug 05 '14 at 17:43