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Let $\mathfrak{g}$ be a semisimple Lie algebra and $I$ its vertices of Dynkin diagram. The weight $\rho$ is defined by $\rho = \sum_{i \in I} \omega_i = \frac{1}{2} \sum_{\alpha \in \Phi^+} \alpha$, where $\Phi^+$ is the set of all positive roots, $\omega_i$ are fundamental weights.

There is a corresponding co-weight $\rho^{\vee}$, see for example the paper page 16, line 6. By definition, $\rho^{\vee} = \frac{1}{2} \sum_{\alpha \in \Phi^+} \alpha^{\vee}$.

Do we have $\rho^{\vee} = \sum_{i \in I} \omega_i^{\vee}$ ($\omega_i^{\vee}$ are fundamental co-weights)?

In general, in a root system $\Phi$, suppose that $\beta \in \Phi$. Is $\beta^{\vee} \in \Phi^{\vee}$ defined by $(\beta, \beta^{\vee})=2$? I think that in general, $(\beta_1 + \beta_2)^{\vee} \neq \beta_1^{\vee} + \beta_2^{\vee}$. Am I correct?

Thank you very much.

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    $\begingroup$ Yes. The $\alpha^\vee$'s also define a root system. $\endgroup$ – Ben Webster May 20 '18 at 18:48
  • $\begingroup$ In addition to @BenWebster's remark, which is the crucial point, it may be worth remarking that, while $(\beta, \beta^\vee) = 2$, this is not a definition; for rank $> 1$, there are many other elements with the same pairing. $\endgroup$ – LSpice Sep 7 '18 at 16:16

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