Let $g$ be a semi simple Lie algebra, $\rho$ be the half sum of all positive roots. I want to show that
The Verma module $M(-\rho)$ is irreducible
I have no idea how to approache this. Any hint or reference would be helpful.
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- $\begingroup$ @DavidHill sorry, it should be $-\rho$. I edited the text. $\endgroup$User X– User X2019-09-19 18:47:02 +00:00Commented Sep 19, 2019 at 18:47
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2 Answers
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1 Any submodule of a Verma is a highest weight module, so it coincides with the image of another Verma module (EDIT: not correct!). If $M(\lambda) \to M(-\rho)$ is a non-trivial homomorphism, by the Harish-Chandra theorem on infinitesimal characters, we must have $\lambda + \rho = w (-\rho + \rho) =0$, for some $w$ in the Weyl group. Thus $M(-\rho) \to M(-\rho)$ is the only possibility for a non-trivial homomorphism. But it cannot have proper image, because $M(-\rho)$ has a unique vector (up to scalar) of weight $-\rho$. Hence $M(-\rho)$ is irreducible.
Note that a very similar argument proves that any $M(\lambda)$ with $\lambda$ antidominant is irreducible.
EDIT: Sorry, I was not careful. The first sentence is not correct in general, so the whole argument breaks. Here is an alternative, which works for any antidominant $M(\lambda)$: Any simple composition factor $L(\mu)$ of $M(\lambda)$ must satisfy:
- $\mu = w(\lambda + \rho) -\rho$ for some $w \in W$, by Harish-Chandra
- $\mu \leq \lambda$, by Poincare-Birkhoff-Witt and definition of Verma.
If $\lambda$ is antidominant, the above two conditions can only be satisfied if $\mu = \lambda$. (In your case $\lambda = \rho$ this follows immediately from 1.)
So $M(\lambda)$ can only contain $L(\lambda)$ as a composition factor. But it can appear only once, since the dimension of the $\lambda$-weight space in $M(\lambda)$ is 1. Hence $M(\lambda)=L(\lambda)$.
- $\begingroup$ That is very nice! Thank you. $\endgroup$User X– User X2019-09-24 13:31:58 +00:00Commented Sep 24, 2019 at 13:31
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1 There is a paper by Shapovalov called "On a Bilinear Form on the Universal Enveloping Algebra of a Complex Semisimple Lie Algebra" in which the author considers an invariant bilinear form on $U(\mathfrak{n}^-)$ (and the Verma module $M(\lambda-\rho)$). The radical of this form coincides with the unique maximal left ideal in $M(\lambda-\rho)$. In particular, the determinant of the form is nonzero precisely when $M(\lambda-\rho)$ is irreducible.
The author computes the determinant of the form on each weight space of $U(\mathfrak{n}^-)$ which, when evaluated at $\lambda-\rho$ is the determinant of the form on each weight space of $M(\lambda-\rho)$. The formula for the determinant on the $\eta$ weight space of $U(\mathfrak{n}^-)$ (up to rescaling by a nonzero constant) is $$ \prod_{\alpha\in \Delta^+}\prod_{n>0}(h_\alpha+\rho(h_\alpha)-n)^{P(\eta-n\alpha)} $$ where $P$ denotes the Kostant partition function. This means that the value of the determinant on the $-\eta-\rho$ weight space of $M(0-\rho)$ is (up to rescaling by a nonzero constant) $$ \prod_{\alpha\in \Delta^+}\prod_{n>0}(-n)^{P(\eta-n\alpha)}\neq 0. $$ Therfore, the module is irreducible.
- $\begingroup$ Thank you for your detailed answer! Although I suppose this is a little too over-killed. $\endgroup$User X– User X2019-09-19 20:09:02 +00:00Commented Sep 19, 2019 at 20:09