# Checking axiom of Category $\mathcal{O}$

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 $$N$$.

• 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. – Vít Tuček May 29 at 14:51
• 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? – 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 $$\mathfrak{b} \cdot m.$$
• 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. – CJS Jun 1 at 20:04
• Yes. If you already know that $\mathcal{O}$ is closed under quotients then this is a good argument. – Vít Tuček Jun 1 at 21:40