Finite dimensional irreducible representations of a semisimple Lie Algebra separate points of the universal enveloping algebra.

Let $$\mathfrak{g}$$ be a semisimple Lie Algebra, and $$U(\mathfrak g)$$ the universal enveloping algebra .

We know that for every representation $$\rho: \mathfrak g \to \mathfrak{gl}(V)$$, there exists a representation $$\tilde{\rho} : U(\mathfrak g) \to \mathfrak{gl}(V)$$, such that $$\rho = \tilde{\rho} \circ \iota$$, where $$\iota: \mathfrak g \to U(\mathfrak g)$$ is the natural inclusion. Besides that, using the standard notations, $$\tilde{\rho}(X_1 \cdot \ldots\cdot X_n) = \rho(X_1) \ldots \rho(X_n).$$

I'm very stuck in this problem

Question: Show that the finite dimensional irreducible representations of a semisimple Lie Algebra $$\mathfrak g$$ separate points of the universal algebra $$U(\mathfrak g)$$, i.e; if $$a \in U(\mathfrak g)$$ satisfies $$\tilde{\rho}(a) =0$$, for every irreducible representation $$\rho: \mathfrak g \to \mathfrak{gl}(V)$$, then $$a=0$$.

Can anyone help me?

• Where did this question arise? Nov 17, 2018 at 4:21
• I found this question on page 318 (chapter 11, problem 5) of the book "Algebras de Lie - Luiz A. B. San Martin" Nov 17, 2018 at 4:23

The following is an explicit argument building on the knowledge of the finite-dimensional irreducible representation of $${\mathfrak g}$$. At its heart is the non-degeneracy of the Shapovalov-form and the description of its determinant, but I tried to keep the exposition elementary.

Setup: Let $${\mathfrak g}={\mathfrak n}^-\oplus{\mathfrak h}\oplus{\mathfrak n}^+$$ be a triangular decomposition of $${\mathfrak g}$$ with respect to a Cartan subalgebra $${\mathfrak h}$$ of $${\mathfrak g}$$ and a choice of positive roots $$\Phi^+\subset{\mathfrak h}^{\ast}$$. Further, let $${\mathfrak b}:={\mathfrak h}\oplus{\mathfrak n}^+$$ be the associated Borel subalgebra. Finally, recall the PBW decomposition $${\mathscr U}{\mathfrak g}\cong{\mathscr U}{\mathfrak n}^-\otimes{\mathscr U}{\mathfrak h}\otimes{\mathscr U}{\mathfrak n}^+$$.

It is known (and not hard to show) that every finite-dimensional irreducible representation of $${\mathfrak g}$$ is uniquely of the form $$L(\lambda)=M(\lambda)/N(\lambda)$$, where $$\lambda\in{\mathfrak h}^{\ast}$$ is dominant integral, i.e. $$\lambda(h_\alpha)\in{\mathbb Z}_{\geq 0}$$ for all $$\alpha\in\Phi^+$$, and $$M(\lambda) := {\mathscr U}{\mathfrak g}\otimes_{{\mathscr U}{\mathfrak b}} {\mathbb C}_\lambda$$ for the $$1$$-dimensional $${\mathscr U}{\mathfrak b}$$-module $${\mathbb C}_\lambda$$ given by $${\mathfrak n}^+{\mathbb C}_\lambda = \{0\}$$ and $${\mathfrak h}$$ acting on $${\mathbb C}_\lambda$$ via $$\lambda$$.

It is important to get the idea of how $$M(\lambda)$$ and $$L(\lambda)$$ come about geometrically: The weight space diagram of $$M(\lambda)$$ is a downwards directed cone rooted in $$\lambda$$, while the one of $$L(\lambda)$$ is its largest symmetric subset with respect to the Weyl group action. See here, for example, where the dotted lines indicate the weight cone of $$M(\lambda)$$, and the solid area is where the weights of $$L(\lambda)$$ live.

Let's consider the point separation for elements of $${\mathscr U}{\mathfrak n}^-$$ first. For those, their action on $$M(\lambda)$$ is very simple: As a $${\mathscr U}{\mathfrak n}^-$$-module, $$M(\lambda)\cong {\mathscr U}{\mathfrak n}^-$$ with $$1\otimes 1\mapsto 1$$ because $${\mathscr U}{\mathfrak g}\cong{\mathscr U}{\mathfrak n}^-\otimes{\mathscr U}{\mathfrak h}\otimes{\mathscr U}{\mathfrak n}^+$$ by PBW. So no non-zero element of $${\mathscr U}{\mathfrak n}^-$$ acts trivially on $$M(\lambda)$$, because it doesn't kill the highest weight vector $$1\otimes 1$$. The idea is now to make $$\lambda\gg 0$$ large enough, for any fixed element of $${\mathscr U}{\mathfrak n}^-\setminus\{0\}$$, so that this argument can be carried over to $$L(\lambda)$$, showing that the element under consideration doesn't annihilate the highest weight vector. Intuitively, this should be possible, because the larger $$\lambda$$ gets, the further 'away from' $$\lambda$$ does the submodule $$N(\lambda)$$ start that is annihilated from $$M(\lambda)$$ when passing to $$L(\lambda)$$.

Starting to be precise, you have the following:

Proposition: For any simple root $$\alpha\in\Delta$$, there is a unique embedding $$M(s_\alpha\cdot\lambda)\subset M(\lambda)$$, and $$L(\lambda)=M(\lambda)/\sum_{\alpha\in\Delta} M(s_\alpha\cdot\lambda).$$

NB: Pursuing this further, you get the BGG resolution of $$L(\lambda)$$ in terms of $$M(w\cdot \lambda)$$, with $$w\in W$$ in the $$l(w)$$-th syzygy.

Corollary: If $$\mu\preceq\lambda$$ (i.e. $$\lambda-\mu\in{\mathbb Z}_+\Phi^+$$, so $$\mu$$ is in the cone below $$\lambda$$) but $$\lambda - \mu = \sum_{\alpha\in\Delta} c_\alpha \alpha$$ with $$c_\alpha < \lambda(h_\alpha)$$ for all $$\alpha\in\Delta$$, then the projection $$M(\lambda)_\mu\twoheadrightarrow L(\lambda)_\mu$$ is an isomorphism.

In other words, it is only in the union of the 'shifted' cones rooted at $$s_\alpha\cdot\lambda$$ that $$L(\lambda)$$ starts looking different from $$M(\lambda)$$. This should be somewhat intuitive.

From that we get separation of points as follows:

Corollary: Let $$\theta = \sum_{\alpha\in\Delta} c_\alpha \alpha$$ with $$c_\alpha\in{\mathbb Z}^+$$, and suppose $$y\in{\mathscr U}{\mathfrak n}^-_{-\theta}$$; that is, $$x$$ is a sum of products $$y_{\alpha_{i_1}}\cdots y_{\alpha_{i_k}}$$ such that $$\theta = \sum_i \alpha_{i_j}$$. Then for any $$\lambda\in{\mathfrak h}^{\ast}$$ with $$\lambda(h_\alpha)\in{\mathbb Z}^{> c_\alpha}$$ for all $$\alpha\in\Delta$$, $$y.v_\lambda\neq 0$$ for the highest weight vector $$v_\lambda$$ of $$L(\lambda)$$. In particular, $$xy$$ doesn't act trivially on $$L(\lambda)$$.

Proof: If $$\tilde{v}_\lambda$$ denotes the highest weight vector of $$M(\lambda)$$, then by the previous proposition we have $$y.\tilde{v}_\lambda\in M(\lambda)\setminus N(\lambda)$$. In particular, $$x$$ acts nontrivially on the image $$v_\lambda$$ of $$\tilde{v}_\lambda$$ in $$L(\lambda)$$.

Corollary: Let $$\theta$$, $$y\in{\mathscr U}{\mathfrak n}^-_{-\theta}$$ and $$\lambda$$ be as before. Then there exists some $$x\in{\mathscr U}{\mathfrak n}^+_{\theta}$$ such that $$(xy)_0(\lambda)\neq 0$$, where $$(xy)_0\in {\mathscr U}{\mathfrak h}\cong {\mathscr P}({\mathfrak h}^{\ast})$$ is the projection of $$xy\in{\mathscr U}{\mathfrak g}_{0}$$ onto $${\mathscr U}{\mathfrak h}\subset {\mathscr U}{\mathfrak g}_{0}$$ with respect to the PBW decomposition.

Here, we used that $${\mathscr U}{\mathfrak h}\cong {\mathfrak S}({\mathfrak h})\cong {\mathscr P}({\mathfrak h}^{\ast})$$ can be viewed as the algebra of polynomial functions on $${\mathfrak h}^{\ast}$$.

Proof: Since $$y.v_\lambda\neq 0$$ in $$L(\lambda)$$ and $$L(\lambda)$$ is simple, we have $$L(\lambda)={\mathscr U}{\mathfrak g}.y.v_\lambda={\mathscr U}{\mathfrak n}^-{\mathscr U}{\mathfrak b}.y.v_\lambda$$. In particular, there exists $$x\in {\mathscr U}{\mathfrak n}^+$$ such that $$(xy).v_\lambda\neq 0$$ in $$L(\lambda)_\lambda$$. For such $$x$$, we must have $$(xy)_0\neq 0$$ since the $$({\mathscr U}{\mathfrak g}){\mathfrak n}^+$$-component of $$xy$$ acts trivially on the highest weight vector $$v_\lambda$$. Finally, note that $$z\in{\mathscr U}{\mathfrak h}$$ acts on $$v_\lambda$$ by $$z(\lambda)$$.

In the previous corollary, the roles of $$x$$ and $$y$$ can be reversed:

Corollary: For $$\theta$$, $$\lambda$$ as before and $$x\in {\mathscr U}{\mathfrak n}^+_{\theta}$$, there exists an $$y\in {\mathscr U}{\mathfrak n}^-_{-\theta}$$ such that $$(xy)_0(\lambda)\neq 0$$.

Proof: Apply the corollary to $$\tau(y)\in {\mathscr U}{\mathfrak n}^-_{-\theta}$$, where $$\tau:{\mathscr U}{\mathfrak g}^{\text{opp}}\to{\mathscr U}{\mathfrak g}$$ is the auto-involution of $${\mathscr U}{\mathfrak g}$$ swapping $${\mathfrak n}^+$$ and $${\mathfrak n}^-$$.

Theorem (Separation of Points): For any $$z\in {\mathscr U}{\mathfrak g}\setminus\{0\}$$ there exists a finite-dimensional $$L(\lambda)$$ such that $$z.L(\lambda)\neq 0$$.

Proof: Assume $$z=\sum_\theta y_\theta h_\theta x_\theta$$ where $$x_\theta\in{\mathscr U}{\mathfrak n}^+_\theta$$ and $$y_\theta\in{\mathscr U}{\mathfrak n}^-$$, $$h_\theta\in{\mathscr U}{\mathfrak h}$$; in other words, you group PBW terms by the weight on the $${\mathfrak n}$$-side. Now, consider $$\theta$$ maximal w.r.t. the ordering $$\lambda\preceq\mu:\Leftrightarrow \mu-\lambda\in{\mathbb Z}^+\Phi^+$$ such that $$y_\theta h_\theta x_\theta$$ nonzero. Then, we know from our previous work that there's some $$\lambda\gg 0$$ such that for any $$\lambda^{\prime}$$ such $$\lambda\preceq\lambda^{\prime}$$ there exists some $$y^{\prime}_\theta\in{\mathscr U}{\mathfrak n}^-_{-\theta}$$ (depending on $$\lambda^{\prime}$$) such that $$(x_\theta y^{\prime}_\theta)_0(\lambda^{\prime})\neq 0$$. Picking $$\lambda^{\prime}$$ large enough, we may assume that also $$h_\theta(\lambda^{\prime})\neq 0$$; this is because the polynomial $$h_\theta\in {\mathscr P}{\mathfrak h}\cong{\mathscr P}({\mathfrak h}^{\ast})$$ cannot vanish on the shifted half-lattice $$\lambda + {\mathbb Z}^+\Phi^+$$. Putting everything together, in $$L(\lambda^{\prime})$$ we then have $$(y_\theta h_\theta x_\theta).(y^{\prime}_\theta v_{\lambda^{\prime}}) = h_\theta(\lambda^{\prime}) (x_\theta y^{\prime}_\theta)_0(\lambda^{\prime}) y_\theta v_{\lambda^{\prime}}\neq 0$$, where for the last step we potentially have to enlarge $$\lambda^{\prime}$$ again. What about the other summands in $$z$$? They all annihilate $$y^{\prime}_\theta v_{\lambda^{\prime}}$$ because of the maximality of $$\theta$$.