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Let $\mathfrak{g}$ be a complex semisimple Lie algebra and $M$ a finite-dimensional representation of $\mathfrak{g}$ (or more generally an element of the category $\mathcal{O}$).

Let $\mathfrak{h}$ be a Cartan subalgebra, $\Phi^+$ a system of positive root and consider the partial ordering on $\mathfrak{h}^*$ induced by $\Phi^+$. Let $\Psi(M)\subseteq\mathfrak{h}^*$ be the set of weights of $M$. Here are two standard definitions:

$(1)$ A highest weight is a maximal element of $\Psi(M)$ with respect to this partial ordering. That is, $\lambda$ is a highest weight if for all $\mu\in\Psi(M)$, either $\mu$ doesn't compare to $\lambda$, or $\mu\leq\lambda$.

$(2)$ $\lambda\in\Psi(M)$ is a highest weight if there exists a nonzero $v\in M_\lambda$ such that $e_\alpha\cdot v=0$ for all $\alpha\in\Phi^+$ where $e_\alpha$ is a root vector of $\alpha$.

I can easily see that $(1)\Rightarrow(2)$. Is the converse also true?

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Choose $\lambda$ such that both $\lambda$ and $\lambda-\alpha$ are dominant. Then $L(\lambda)\oplus L(\lambda-\alpha)$ is a counter example.

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No. Consider e.g. the Verma module $M(0)$ for $\mathfrak{sl}_2$ with highest weight $0$. This has quotient the $1$-dimensional irreducible representation, and the kernel is the Verma module $M(-2)$, a submodule of $M(0)$ generated by a highest weight vector of weight $-2$. Since $-2 < 0$ in the ordering you are considering, this gives you a counterexample (and also shows that in Verma modules for integral weights, you should expect the implication (2) $\implies$ (1) to fail almost all the time).

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    $\begingroup$ The hypothesis is that the module is finite dimensional. $\endgroup$ – David Hill Apr 26 '17 at 1:46
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    $\begingroup$ @DavidHill ... or more generally in category O. $\endgroup$ – Stephen Apr 26 '17 at 2:13
  • $\begingroup$ As you were. . . $\endgroup$ – David Hill Apr 26 '17 at 3:18

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