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The central character does not determine the primitive ideal

Statement refuted

Assume the Axiom of Choice. The assertion that a central character determines the primitive ideal over it is false. In $U ({s l}_{2} (C))$ , for every integer $n \geq 0$ the primitive ideals $I (n) = Ann L (n)$ and $J (n) = Ann M (- n - 2)$ are distinct, although $L (n)$ and the simple Verma module $M (- n - 2)$ have the same central character $χ_{n} = χ_{- n - 2}$ . Thus the fibre of the central-character map on primitive ideals has at least two points over every such $χ_{n}$ , and the assignment $λ \mapsto Ann L (λ)$ is not injective. In particular the fibre over the regular integral central character $χ_{0}$ already has two points.

Facts & Assumptions

Given: The Axiom of Choice, $g = {s l}_{2} (C)$ with Casimir $Ω$ , an integer $n \geq 0$ , the finite-dimensional simple module $L (n)$ , and the Verma module $M (- n - 2)$ .

[F1]

$L (n)$ is a finite-dimensional simple module, and $I (n) = {Ann}_{U (g)} L (n)$ is primitive; here $Ω : = e f + f e + \frac{1}{2} h^{2}$ is the normalized central Casimir, and the central-reduction lemma also gives $Z (U (g)) = C [Ω]$ and the eigenvalue $Ω \mapsto n (n + 2) / 2$ on $L (n)$ (Finite-dimensional simple modules are classified by dominant highest weights, Annihilators of simple highest-weight modules are primitive, The central reduction of U(sl2) is simple away from the finite-dimensional central characters).

[F2]

$M (- n - 2)$ is simple — its highest weight is antidominant, $⟨ λ + ρ, α^{\lor} ⟩ < 0$ for the positive root — so $J (n) = {Ann}_{U (g)} M (- n - 2)$ is primitive; $Ω$ acts on $M (- n - 2)$ by $(- n - 2) (- n) / 2 = n (n + 2) / 2$ , and every weight space of $M (- n - 2)$ is finite-dimensional with weights $(- n - 2) - 2 k$ , $k \in Z_{\geq 0}$ , so $M (- n - 2)$ is infinite-dimensional (Antidominant regular Verma modules are simple, Verma modules, The central reduction of U(sl2) is simple away from the finite-dimensional central characters, Weights of a Verma module lie below lambda, The Axiom of Choice).

[F3]

For ${s l}_{2}$ , equal values of the character on $Ω$ give equal central characters: $χ_{n} = χ_{- n - 2}$ by [F1], [F2] and Central characters are dot-Weyl orbits, because $0$ and $- 2$ (and generally $n$ and $- n - 2$ ) lie in the same dot orbit. [F1, F2]

[F4]

If $K$ is the annihilator of a module $M$ , then $U (g) / K$ embeds into ${End}_{C} (M)$ and acts faithfully on $M$ ; the image of a finite-dimensional vector space under a linear map is finite-dimensional, since the images of a finite basis span the image (The annihilator of a module over an enveloping algebra, Primitive ideals of an enveloping algebra).

[F5]

If the normalized Casimir value $c$ is not $m (m + 2) / 2$ for any integer $m \geq 0$ , every Verma module with that value is simple and its annihilator is the central ideal (The central reduction of U(sl2) is simple away from the finite-dimensional central characters). A simple Verma module equals its unique simple quotient $L (λ)$ (Annihilators of simple highest-weight modules are primitive).

Counterexample

technique · contradiction

1.1F1F2F3

By [F1] and [F2] both $I (n)$ and $J (n)$ are primitive ideals, and both modules have the same central character $χ_{n}$ by [F3].

1.2F1F5algebra

To verify the separate noninjectivity assertion, take the distinct highest weights $1 / 2$ and $- 5 / 2$ . Their normalized Casimir eigenvalue is $5 / 8$ , and their central characters agree because $Z (U (g)) = C [Ω]$ . This value is not a finite-dimensional Casimir value: $n (n + 2) / 2$ is $0$ at $n = 0$ and at least $3 / 2$ at every integer $n \geq 1$ . Therefore The central reduction of U(sl2) is simple away from the finite-dimensional central characters makes both Verma modules simple and gives the same central ideal as their annihilator. Their unique simple quotients are the Verma modules themselves, so $I (1 / 2) = I (- 5 / 2)$ although $1 / 2 \neq - 5 / 2$ .

1.3F1F4assume-contra

Suppose $I (n) = J (n) = : K$ . By [F4] the algebra $A : = U (g) / K$ embeds into ${End}_{C} L (n)$ , because $K$ is the kernel of the action on $L (n)$ ; hence $A$ is finite-dimensional.

2.1step 1.3F2F4algebra

Also by [F4] $A$ acts faithfully on $M (- n - 2)$ . Fix $0 \neq m \in M (- n - 2)$ and consider the map $A \to M (- n - 2)$ , $a \mapsto a m$ : it is $A$ -linear because $b (a m) = (b a) m$ , and its image is a nonzero $A$ -submodule of $M (- n - 2)$ , hence equals $M (- n - 2)$ because $M (- n - 2)$ is simple by [F2]. The images of a finite basis of $A$ thus span $M (- n - 2)$ , which is finite-dimensional.

3.1step 2.1F2F3discharge-contradiction∎

But $M (- n - 2)$ is infinite-dimensional by [F2], since its weights $(- n - 2) - 2 k$ , $k \in Z_{\geq 0}$ , are infinitely many distinct weights with finite-dimensional weight spaces. This contradiction shows $I (n) \neq J (n)$ ; the two distinct primitive ideals share the central character $χ_{n}$ , so a central character does not determine the primitive ideal. Taking $n = 0$ exhibits two points in the fibre over the regular integral central character $χ_{0}$ .

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Sources

P. Etingof, Representations of Lie Groups (18.757, MIT OCW 2023 full notes) (standard reference, not scraped)
R. E. Block, The irreducible representations of the Lie algebra sl(2) and of the Weyl algebra, Adv. Math. 39 (1981) 69-110 (standard reference, not scraped)