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

Question

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?

Answer 1

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 $\theta$.