In connection to my previous question, A step in Victor Kac's book regarding the casimir element, let $\newcommand{\g}{\mathfrak{g}}$ $\g$ be a lie algebra with a root space decomposition and an invariant inner product $(,)$.
Let $\widetilde \rho =\sum_{\alpha>0} \alpha \times(\text{multiplicity } \alpha)$.
I am trying to show that for any root $\beta$ that $(\beta,\tilde \rho)=(\beta,\beta)$. Is this even true?
Motivation:
If true, this would imply that $ 2\sum_{\alpha >0} e^i_{-\alpha}e^i_{\alpha}+u_iu^i+\rho=\sum_{all\,\,roots} e^i_{-\alpha}e^i_\alpha-\tilde \rho +\rho$ commutes with any $x \in \g_\alpha$.
Notation: Here $\rho$ is any element of the cartan such that $\alpha(\rho)=(\alpha,\alpha)$, $e_\alpha^i$ is a basis of $\g_\alpha$ and $e_{-\alpha}^i$ is a dual basis with respect to $(,)$. $u_i$ is a basis of the cartan and $u^i$ is a dual basis of the cartan.
This is because $[\sum_{all\,\,roots} e^i_{-\alpha}e^i_\alpha-\tilde \rho +\rho,x]=[-\tilde \rho +\rho,x]=(-\alpha(\tilde \rho)+\alpha(\rho))x$=0
EDIT: I made a critical typo: I had written earlier that I was trying to show that $(\beta,\rho)=(\beta,\beta)$. This in fact holds by definition for any $\beta$. I am actually trying to show $(\beta,\tilde \rho)=(\beta,\beta)$ as I have now written.
