# Why are highest weight modules of integral highest weight $B$-equivariant?

Suppose $$G$$ is a connected semisimple algebraic group over $$\mathbb{C}$$, $$B \subset G$$ is a Borel subgroup, and $$T \subset B$$ is a maximal torus. Write $$\mathfrak g$$, $$\mathfrak b$$ and $$\mathfrak h$$ for the Lie algebras of $$G, B$$ and $$T$$ respectively.

Let $$\mathcal{U}$$ denote the universal enveloping algebra of $$\mathfrak g$$. Suppose $$M$$ is a finitely generated left $$\mathcal U$$-module that is $$\mathfrak h$$-semisimple, in the sense that we have a weight space decomposition $$M = \oplus_{\mu \in \mathfrak{h}^*} M_\mu$$ with $$\dim M_\mu < \infty$$. We say that $$M$$ is a highest weight module of weight $$\lambda \in \mathfrak{h}^*$$ if there exists a vector $$v^+ \in M_\lambda$$ such that $$\mathfrak b \cdot v^+ = 0$$ and $$\mathcal{U} \cdot v^+ = M$$. For example, $$M$$ might be the Verma module $$M(\lambda)$$, or its unique simple quotient $$L(\lambda)$$.

Now assume $$M$$ is a highest weight module of weight $$\lambda$$, where $$\lambda$$ is an integral weight (i.e. a $$\mathbb{Z}$$-linear combination of the fundamental dominant weights). The following fact is basic in representation theory, but I'm not satisfied I understand why it is true.

Why does the action of $$\mathfrak b$$ on $$M$$ integrate to an algebraic action on $$B$$ on $$M$$, in such a way that $$M$$ obtains the structure of a $$B$$-equivariant $$\mathfrak g$$-module?

The thing that I really want to understand is how to integrate the action (explicitly enough so the equivariance can be checked directly). Now there are results that say something along that lines of "Any locally finite representation of a semisimple Lie algebra can be integrated to the universal covering group" (see here for example), but of course, $$B$$ is not semisimple. So I suppose one needs to prove this result 'by hand'.

Note that every weight $$\mu$$ of $$M$$ is of the form $$\lambda - \sum^n_{j=1} i_j \alpha_j$$, where $$\alpha_1, \ldots, \alpha_n$$ are the simple roots and the $$i_j$$ are integers $$\geq 0$$. Therefore,

1. Every weight $$\mu: \mathfrak h \to \mathbb C$$ is integral, and hence integrates to an algebraic character $$e^\mu: T \to \mathbb{C}^\times$$.

2. The action of $$\mathfrak b$$ on $$M$$ is locally finite, since the finite dimensional space $$V_\mu := \oplus_{\nu \geq \mu} M_\nu$$ is $$\mathfrak b$$-stable.

I have a couple of thoughts on where to go from here. One could consider the subgroup $$B'$$ of $$\operatorname{Aut}(V_\mu)$$ generated by the automorphisms $$\exp y$$, for $$y \in \mathfrak b$$. To see that this makes sense, note that we have the Jordan decomposition $$y = h + x$$, where $$h \in \mathfrak h$$ and $$x \in \mathfrak n = [\mathfrak b, \mathfrak b]$$. Then $$\exp h$$ acts on $$M_\nu$$ as the scalar $$e^{\nu(h)}$$, while $$\exp x$$ is well-defined as an operator on $$V_\mu$$ since $$x$$ acts nilpotently on this space. One could then set $$\exp y := \exp h \circ \exp x$$ (although I'm wary of this since $$[h,x] \neq 0$$). Assuming that is correct, one still has to show that there is a homomorphism of algebraic groups $$B \to B'$$, and I'm not sure how to do this.

On the other hand, one could try to define the action directly, in the following way. Suppose $$U$$ is the unipotent radical of $$B$$, so that $$B = TU$$ with $$T \cap U = \{ 1 \}$$; then every element $$b \in B$$ has a Jordan decomposition $$b = tu$$ with $$t \in T, u \in U$$. Now it's clear that $$t$$ should act on the weight space $$M_\mu$$ via the character $$e^\mu$$. But how should $$u$$ act? I was thinking that if $$G$$ were a matrix group and $$x \in \mathfrak n = \operatorname{Lie} U$$ were such that $$u = \exp x$$ (matrix exponential), then one could let $$u$$ act as $$\exp x$$ on $$V_\mu$$. But I doubt this is reasonable, since one may have $$x^2 = 0$$ as a matrix while $$x^2$$ does not act as zero on $$M$$; consider the case where $$\mathfrak g = \mathfrak{sl}_2 (\mathbb{C})$$, $$x = \big( \begin{smallmatrix} 0 & 1 \\ 0 & 0 \end{smallmatrix} \big)$$, and $$M$$ is basically any highest weight module.

A third possibility would be to take a more analytic approach. By Lie's correspondence, the map of the underlying real Lie algebras $$\mathfrak b \to \mathfrak{gl} (V_\mu)$$ integrates to a map of the underlying real Lie groups $$B \to \mathrm{GL}(V_\mu)$$. Perhaps one can then complexify this map, and use the integrality assumption to show that $$V_\mu = \oplus_{\nu \geq \mu} M_\nu$$ is a weight space decomposition for $$T$$. Could this then imply that the action of $$B$$ is algebraic?

Anyway, as is probably clear, I don't know what the right approach is. I would really appreciate a proof outline or reference. :)

UPDATE

As Callum pointed out in his answer, the fact that $$\exp: \mathfrak n \to U$$ is an isomorphism of varieties can be used to integrate the action of $$\mathfrak n$$ to $$U$$. Unfortunately, my proposal to integrate the actions of $$\mathfrak h$$ and $$\mathfrak n$$ separately, and then combine them to an action of $$B$$, doesn't make a lot of sense. After all, the action of $$t$$ must commute with the action of $$u$$, but the two actions I defined won't commute. Similarly, it doesn't make sense to set $$\exp (y) = \exp(h) \circ \exp(x)$$, since as I noted $$[h,x] \neq 0$$; of course, we want $$\exp(y) = \exp(h+x)$$!

I'm not sure that this answers the full question for certain but this was a bit too long for a comment. In your second attempt you talk about matrix groups and exponentials a bit too loosely.

First things first, we don't need a matrix group for $$\exp$$ to be a diffeomorphism from $$\mathfrak{n}$$ to $$U$$. This happens automatically because $$\mathfrak{n}$$ contains only nilpotent elements.

Secondly, writing an element of $$\mathfrak{g}$$ as a specific matrix (i.e. as a operator on a specific representation) and then using it to act on a different representation leads to immediate problems. In particular, $$x^2$$ is not a concept that stretches between reps. This should be clear just thinking about $$\mathfrak{sl}_2$$ reps.

The element represented by $$\begin{pmatrix}0&1\\0&0\end{pmatrix}$$ on $$\mathbb{C}^2$$ could be represented by $$\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix}$$ on $$\mathbb{C}^3$$. In the first case its square is $$0$$ but in the second it would be $$\begin{pmatrix}0&0&1\\0&0&0\\0&0&0\end{pmatrix}$$.

So $$u$$ can act as $$\exp x$$ but this should not be thought of as the matrix exponential unless the representation is precisely the one you are claiming $$G$$ is a matrix group for.

Instead if $$\pi:\mathfrak{g} \to \mathfrak{gl}(V)$$ is a Lie algebra representation. Then we can define (on a neighbourhood of the identity for which $$\exp$$ is invertible) a group representation $$\Pi(\exp x) = \exp \pi(x) \in GL(V)$$. The latter here can be thought of as the usual power series/matrix exponential but note how it depends on the representation chosen. Away from the neighbourhood this can lead to $$\Pi$$ having multiple values but it is well defined for $$\exp$$ invertible. In particular, it is well defined for $$\mathfrak{n}$$ and $$U$$.

• Thanks, that's a helpful critique. Your last paragraph doesn't quite answer the question, but only because I realised my proposal to integrate the actions of $\mathfrak h$ and $\mathfrak n$ separately doesn't make any sense. :) Jun 15, 2022 at 1:41