I didn't understand step b) of this proof and would be happy if someone could help me with this.

Let dimV be finite. Let L be a semisimple Lie algebra, $\ L_\alpha $ a weight space. Let $\ \Delta $= {$\ \alpha_1 ,...,\alpha_l $} be a base, H the maximal toral subalgebra of L.

a) Then the Borel subalgebra $\ B(\Delta)=H\bigoplus\ L_\alpha$ has a common eigenvector (killed by all $\ L_\alpha, \alpha \succ 0$, thanks to Lie's Theorem.

b) This eigenvector has to lie in any $L_\lambda$ and it has to satisfy $L_\alpha v^+ \subseteq V_\lambda \bigcap V_{\alpha+\lambda}=0$ for all $\alpha \succ0$, hence it has to be a maximal vector. On the other hand it is obvious that every maximal vector is common eigenvector of B.


1 Answer 1


[Edit: So glad that the system jumped this question. I had dropped the ball here. Trying to fix...]

Ok. Let $v$ be a common eigenvector for $B(\Delta)$. It is a weight vector, because it is an eigenvector for all of $H$. So $v\in V_\lambda$ for some weight $\lambda$.

From the general theory we know that for any positive root $\alpha\succ0$ we have $L_\alpha V_\lambda\subseteq V_{\lambda+\alpha}.$ Because $V_{\lambda+\alpha}\cap V_{\lambda}=\{0\}$, the vector $v$ can be an eigenvector of $x_\alpha\in L_\alpha$ only, if the eigenvalue is zero. Therefore $v$ is a maximal vector (the definition of a maximal vector is a weight vector annihilated by all the positive root operators).

OTOH, if $v$ is a maximal vector, it is an eigenvector of all of $H$ by virtue of being a weight vector. It is also an eigenvector belonging to eigenvalue zero of all the positive root operators. Therefore it is a common eigenvector of all of $B(\Delta)$.


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