Let $K$ be a finite extension of $\mathbb{Q}_p$ and $G$ be a split connected reductive algebraic group over $K$ with Borel $B$. We have the associated Lie algebras $\mathfrak{g}=$Lie$(G)$ and $\mathfrak{b}=$Lie$(B)$.

Let $M$ be a $U(\mathfrak{g})$-module with $N \subset M$ a finite dimensional $K$-module, which is $B$-invariant and generates $M$ as a $U(\mathfrak{g})$-module.

I read that $M$ is then locally $\mathfrak{b}$-finite, i.e $U(\mathfrak{b}) \cdot m \subset M$ is finite dimensional for all $m \in M$, but I have trouble to see this. As $U(\mathfrak{b})$ seems so big for me, I cannot think of a finite basis for $U(\mathfrak{b}) \cdot m \subset M$ by knowing only $N$.

  • $\begingroup$ I am familiar only with complex numbers, but there the result is easy, because being $B$-invariant means that it is a space of highest weight vectors and then your $M$ is covered by Verma modules. $\endgroup$ – Vít Tuček May 29 at 14:51
  • $\begingroup$ Thanks for your fast answer but I cannot follow you. I'm still learning all these things. By space of highest weight vectors you mean $N$ has a basis of highest weight vectors? Why? And why is $M$ covered by Verma Modules? $\endgroup$ – CJS May 29 at 21:21

The module $U(\mathfrak{g})\otimes_{U(\mathfrak{b})} N$ is locally $U(\mathfrak{b})$-finite and there is a surjective $U(\mathfrak{g})$-homomorphism $\varphi\colon U(\mathfrak{g})\otimes_{U(\mathfrak{b})} N \to M$ given by $u \otimes n \mapsto u\cdot n.$ It is easy to see that every weight space of $M$ has only finitely-many preimages. It follows that $M$ cannot contain infinite-dimensional $U(\mathfrak{b})$-module. In the case $N$ is one-dimensional, the module I just constructed is called Verma module. In case of $N$ not being completely reducible, one would have to do some gymnastics with short exact sequences.

But perhaps for a beginner it's easier to just attack the problem directly. Assume first for simplicity that $m = u \cdot n$ for some $u \in \mathfrak{g}.$ Pick $X \in \mathfrak{b}.$ Then $$ X \cdot m = X\cdot u \cdot n = Xu \cdot n = [X,u]\cdot n + uX\cdot n = u'\cdot n + u \cdot n', $$ where $u'$ is some other element of $\mathfrak{g}$ and $n'$ is some other element of $N.$ Now iterate for general $u\in U(\mathfrak{g})$ and repeat for all possible $X \in \mathfrak{b}$. You will see that you pick up at most $\dim \bigotimes^k \mathfrak{g} \otimes N$ possible elements. They might not be all linearly independent, but they surely span $\mathfrak{b} \cdot m.$

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  • $\begingroup$ Thanks for this detailed answer. Could we argue also in the following way: $U(\mathfrak{g}) \otimes_{U(\mathfrak{b})} N$ lies in Category $\mathcal{O}$, which is closed under submodules and quotients. By the map you have given $M$ is a quotient of $U(\mathfrak{g}) \otimes_{U(\mathfrak{b})} N$ and therefore lies in category $\mathcal{O}$ too. Hence has to be locally $U(\mathfrak{b})$-finite. $\endgroup$ – CJS Jun 1 at 20:04
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    $\begingroup$ Yes. If you already know that $\mathcal{O}$ is closed under quotients then this is a good argument. $\endgroup$ – Vít Tuček Jun 1 at 21:40

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