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Primitive Ideals and Duflo Theorem

1 · Prerequisites

2 · Summary

This page develops the algebraic prefix of the primitive-ideal theory of the universal enveloping algebra: annihilators of modules, primitive ideals and their primeness, the central character attached to a primitive ideal by Dixmier's lemma, the central reduction at a central character, and the associated variety of a two-sided ideal together with its conicality and coadjoint invariance. The ${s l}_{2}$ central reduction is analysed completely away from the finite-dimensional central characters, giving the smallest instance of a Duflo annihilator.

The page deliberately stops before Duflo surjectivity, the Kazhdan-Lusztig and Joseph fibre theory, and the localisation architecture of the geometric proof; those remain prose-level targets and no item on this page consumes them. The items using the Harish-Chandra parametrisation or the structure of the semisimple centre assume the Axiom of Choice explicitly. The annihilator, primeness, Dixmier and associated-variety arguments are choice-free.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6.1-sol)Open item page →

The annihilator of a module over an enveloping algebra

Definition

Let $g$ be a complex Lie algebra and let $M$ be a nonzero left $U (g)$ -module with action $ρ_{M} ⁣ : U (g) \to {End}_{C} (M)$ , the unital extension of the $g$ -action supplied by Lie algebra actions extend to unital actions of the enveloping algebra. The annihilator of $M$ is

${Ann}_{U (g)} (M) = {u \in U (g) : u m = 0 for every m \in M} = \ker ρ_{M},$

and for $m \in M$ one writes $Ann (m) = {u \in U (g) : u m = 0}$ . Then ${Ann}_{U (g)} (M)$ is a two-sided ideal of $U (g)$ (Left, right and two-sided ideals), equal to the intersection of the left ideals $Ann (m)$ , $m \in M$ ; the action descends to a faithful action of the quotient algebra $U (g) / {Ann}_{U (g)} (M)$ on $M$ .

Remarks

The annihilator is the kernel of the action. $ρ_{M}$ is a $C$ -algebra homomorphism, so its kernel is a two-sided ideal by The kernel of a ring homomorphism is a two-sided ideal; unwinding definitions, this kernel is exactly the set of $u$ annihilating every $m \in M$ .
Pointwise annihilators are left ideals. For fixed $m$ , the map $u \mapsto u m$ is $C$ -linear, so $Ann (m)$ is an additive subgroup; and $u \in Ann (m)$ implies $v u \in Ann (m)$ for every $v \in U (g)$ because $(v u) m = v (u m) = 0$ . Thus each $Ann (m)$ is a left ideal, and ${Ann}_{U (g)} (M) = ⋂_{m \in M} Ann (m)$ because a $u$ annihilating every $m$ is exactly one lying in every pointwise annihilator.
Faithfulness of the quotient action. If $u + Ann M$ acts as zero on $M$ , then $u m = 0$ for all $m \in M$ , so $u \in Ann M$ and the class is zero. The quotient therefore acts faithfully, and $M$ is a left module over it by the same formula (Unital left and right modules over a ring; unqualified module means left module).

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6.1-sol)Open item page →

Primitive ideals of an enveloping algebra

Definition

Let $g$ be a complex Lie algebra. A two-sided ideal $I ⊴ U (g)$ (Left, right and two-sided ideals) is primitive if there exists a simple left $U (g)$ -module $M$ (Simple module: a nonzero module with no proper nonzero submodule) with

$I = {Ann}_{U (g)} (M) = {u \in U (g) : u m = 0 for every m \in M}$

(The annihilator of a module over an enveloping algebra). Equivalently: a two-sided ideal is primitive if it is the kernel of the action of $U (g)$ on some simple module. Every primitive ideal is proper, and $U (g) / I$ admits the faithful simple module $M$ ; no highest-weight hypothesis is imposed on $M$ .

Remarks

Properness. The module $M$ is nonzero, and $1 \in U (g)$ acts as the identity, so $1 \notin {Ann}_{U (g)} (M)$ ; a two-sided ideal containing $1$ is all of $U (g)$ , so a primitive ideal is a proper ideal. This is the only place nonzero-ness of $M$ is used in the definitional consequences.
Faithfulness after quotienting. By The annihilator of a module over an enveloping algebra the action of $U (g) / I$ on $M$ is faithful, so the equivalent kernel formulation and the quotient statement describe the same situation.
No highest-weight hypothesis. The simple module $M$ in the definition is arbitrary; in particular a primitive ideal need not be realised by a highest weight module in the definition itself. For finite-dimensional complex semisimple $g$ , realization by a simple highest-weight module is Duflo's theorem, not part of this definition.

PropositionStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

Annihilators of simple highest-weight modules are primitive

Statement

Let $g$ be a finite-dimensional complex semisimple Lie algebra, let $λ \in h^{*}$ , let $M (λ)$ be the Verma module of highest weight $λ$ , and let $L (λ)$ be its unique simple quotient. Then $I (λ) : = {Ann}_{U (g)} (L (λ))$ is a primitive ideal of $U (g)$ ; moreover ${Ann}_{U (g)} M (λ) \subseteq I (λ)$ , with equality whenever $M (λ)$ is simple (in which case $L (λ) ≅ M (λ)$ ).

Facts & Assumptions

Given: A finite-dimensional complex semisimple Lie algebra $g$ , a weight $λ \in h^{*}$ , the Verma module $M (λ)$ , and its unique simple quotient $L (λ)$ .

[F1]

$M (λ)$ has a unique maximal submodule $J (λ)$ , and $L (λ) = M (λ) / J (λ)$ is simple; it is the unique simple quotient (A Verma module has a unique simple quotient, Verma modules).

[F2]

A two-sided ideal is primitive when it is the annihilator of some simple module (Primitive ideals of an enveloping algebra).

[F3]

The annihilator of a module is a two-sided ideal, annihilators grow when passing to quotients, and the annihilator of a quotient $M / N$ contains the annihilator of $M$ (The annihilator of a module over an enveloping algebra).

Proof

technique · direct

1.1F1F2given

By [F1], $L (λ)$ is a simple $U (g)$ -module, so by [F2] its annihilator $I (λ) = {Ann}_{U (g)} (L (λ))$ is a primitive ideal.

1.2F1F3algebra

The natural surjection $M (λ) ↠ L (λ)$ has kernel $J (λ)$ ; if $u$ annihilates every element of $M (λ)$ , then it annihilates the image of every element in $L (λ)$ , so $u \in {Ann}_{U (g)} L (λ)$ . Hence ${Ann}_{U (g)} M (λ) \subseteq I (λ)$ by [F3].

2.1step 1.2F1algebra∎

If $M (λ)$ is simple, then its unique maximal submodule $J (λ)$ is a proper submodule by [F1] and therefore must be $0$ , since a simple module has no nonzero proper submodule; hence $L (λ) = M (λ) / J (λ) ≅ M (λ)$ . The two modules then have the same annihilator, and by step 1.2 the inclusion is an equality.

PropositionStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

Primitive ideals are prime in the noncommutative sense

Statement

Let $I$ be a primitive ideal of $U (g)$ and let $A, B$ be two-sided ideals of $U (g)$ with $A B \subseteq I$ . Then $A \subseteq I$ or $B \subseteq I$ . The claim holds for every complex Lie algebra $g$ , and equivalently the quotient ring $U (g) / I$ is prime.

Facts & Assumptions

Given: A complex Lie algebra $g$ , a primitive ideal $I$ , a simple left $U (g)$ -module $M$ with $I = {Ann}_{U (g)} (M)$ , and two-sided ideals $A, B ⊴ U (g)$ with $A B \subseteq I$ .

[F1]

$I$ is the annihilator of the simple module $M$ ; the annihilator of a module is a two-sided ideal (Primitive ideals of an enveloping algebra, The annihilator of a module over an enveloping algebra).

[F2]

The product $A B$ consists of finite sums $\sum_{k} a_{k} b_{k}$ and is a two-sided ideal (The sum $I + J$ and product $I J$ of two-sided ideals, The sum and product of two-sided ideals are two-sided ideals). For an ideal $C$ and submodule $N$ , write $C N$ for the finite sums $\sum_{k} c_{k} n_{k}$ ; these form a submodule since $u (c_{k} n_{k}) = (u c_{k}) n_{k}$ and $u c_{k} \in C$ (Left, right and two-sided ideals). Distributing finite sums and using associativity gives $A (B M) = (A B) M$ .

[F3]

$M$ is nonzero and its only submodules are $0$ and $M$ ; in particular a submodule $N \subseteq M$ with $N \neq 0$ equals $M$ (Simple module: a nonzero module with no proper nonzero submodule).

Proof

technique · direct

1.1F1F2F3given

Suppose $B \subseteq̸ I$ , i.e. $B \subseteq̸ {Ann}_{U (g)} (M)$ . Then some $b \in B$ has $b M \neq 0$ , and $B M$ is a nonzero submodule of $M$ : for $u \in U (g)$ , one has $u \sum_{k} b_{k} m_{k} = \sum_{k} (u b_{k}) m_{k} \in B M$ because $u b_{k} \in B$ . By [F3], $B M = M$ .

2.1step 1.1F1F2algebra

Using the associativity of the action and the containment $A B \subseteq I$ : $A M = A (B M) = (A B) M \subseteq I M = 0$ , so every $a \in A$ annihilates $M$ and therefore $A \subseteq {Ann}_{U (g)} (M) = I$ .

3.1step 1.1step 2.1F2algebra∎

The argument shows that $B \subseteq̸ I$ forces $A \subseteq I$ ; contrapositively, if $A \subseteq̸ I$ then $B \subseteq I$ . Hence $A \subseteq I$ or $B \subseteq I$ , and no finite-dimensionality of $g$ was used. Passing to $U (g) / I$ , two-sided ideals of the quotient correspond to two-sided ideals of $U (g)$ containing $I$ , and the product condition becomes $\overset{ˉ}{A} \overset{ˉ}{B} = 0$ ; the displayed alternative is exactly the primeness of $U (g) / I$ .

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

Dixmier's lemma: endomorphisms of a simple module over a countable-dimensional algebra

Statement

Let $A$ be a unital $C$ -algebra which admits a countable $C$ -basis, and let $M$ be a simple left $A$ -module. Then every $A$ -endomorphism of $M$ is multiplication by a scalar: ${End}_{A} (M) = C \cdot {id}_{M}$ . Consequently the center $Z (A)$ acts on $M$ by scalars and there is a unital $C$ -algebra homomorphism $χ ⁣ : Z (A) \to C$ with $z m = χ (z) m$ for all $z \in Z (A)$ , $m \in M$ . In particular this applies to $A = U (g)$ for a finite-dimensional complex Lie algebra $g$ .

Facts & Assumptions

Given: A unital $C$ -algebra $A$ with a countable $C$ -basis, a simple left $A$ -module $M$ , and $D : = {End}_{A} (M)$ .

[F1]

A nonzero homomorphism between simple modules is an isomorphism, and the endomorphism ring of a simple module is a division ring (Schur's lemma for simple modules, Division ring: a ring with $1 \neq 0$ in which every nonzero element is a unit).

[F2]

$D = {End}_{A} (M)$ is a unital ring under pointwise addition and composition, and it is a $C$ -algebra whose scalars $λ {id}_{M}$ are central: $λ {id}_{M}$ is $A$ -linear because the scalar action of $A$ is $C$ -linear (The endomorphism ring ${End}_{R} (M)$ under addition and composition, Division ring: a ring with $1 \neq 0$ in which every nonzero element is a unit).

[F3]

$M$ is simple and nonzero, so for every $0 \neq v \in M$ the submodule $A v$ is nonzero, hence equal to $M$ (Simple module: a nonzero module with no proper nonzero submodule).

[F4]

$C$ is algebraically closed, so every nonconstant polynomial in $C [t]$ is a product of linear factors; in a division ring a product of nonzero elements is nonzero, so a product of nonzero factors is zero only if one factor is zero (The complex numbers are algebraically closed, Division ring: a ring with $1 \neq 0$ in which every nonzero element is a unit).

[F5]

The rational function field $C (t) = Frac (C [t])$ consists of the fractions $f / g$ with $f, g \in C [t]$ , $g \neq 0$ (For a field $F$ , $F (t) = Frac (F [t])$ is its rational function field; in particular $R (t) = Frac (R [t])$ , The field of fractions $Frac (D) = (D ∖ {0})^{- 1} D$ of an integral domain).

[F6]

If a vector space over a field is spanned by $n$ vectors, then every linearly independent subset is finite with at most $n$ elements (If $V$ has a spanning set with $n$ elements, then every linearly independent subset of $V$ is finite with at most $n$ elements; in particular $V$ has no linearly independent subset equinumerous with $N$ ); $N \times N \approx N$ ( $N \times N \approx N$ ); every subset of an at most countable set is at most countable (Every subset of an at most countable set is at most countable); $R$ is uncountable ( $R$ is uncountable (Cantor's nested intervals, 1874)) and embeds in $C$ as the constant classes, whence $C$ is uncountable (The complex numbers as $R [x] / (x^{2} + 1)$ , with the real embedding and imaginary unit $i$ , Finite, countably infinite, countable, uncountable).

[F7]

For a finite-dimensional complex Lie algebra $g$ with ordered basis $x_{1}, \dots, x_{r}$ , the ordered monomials $x_{1}^{a_{1}} \dots x_{r}^{a_{r}}$ form a basis of $U (g)$ (PBW gives an ordered monomial basis for the enveloping algebra, The universal enveloping algebra as a tensor quotient); the set of exponent tuples is a finite product of copies of $N$ , hence at most countable (A product of two at most countable sets is at most countable, Finite, countably infinite, countable, uncountable).

Proof

technique · contradiction

1.1F1F2given

By [F1], $D$ is a division ring. The map $C \to D$ , $λ \mapsto λ {id}_{M}$ , is an injective unital ring homomorphism onto a central copy of $C$ by [F2], so $D$ is a $C$ -division algebra containing $C$ in its center.

1.2F6construct

Let $V$ be a $C$ -vector space spanned by a sequence $b_{1}, b_{2}, \dots$ and let $L \subseteq V$ be linearly independent. Then $L$ is at most countable. Indeed, put $V_{k} = span (b_{1}, \dots, b_{k})$ ; the recursive rule that retains $b_{i}$ exactly when $b_{i} \notin span (retained b_{j}, j < i)$ is a definition by recursion on $i$ and is a canonical ordered basis $(c_{1}, \dots, c_{d_{k}})$ of $V_{k}$ with $d_{k} \leq k$ : retained elements are independent and each $b_{i}$ lies in the span of the retained elements up to $i$ . Hence every $v \in V_{k}$ has a unique coordinate vector in $C^{d_{k}}$ . Order $C$ by real part and then imaginary part, and $C^{d_{k}}$ lexicographically. Since $V = ⋃_{k} V_{k}$ , for $v \in L$ the set of $k$ with $v \in V_{k}$ is nonempty and $n (v) : = \min {k : v \in V_{k}}$ is well defined; let $L_{k} = {v \in L : n (v) = k}$ . Each $L_{k}$ is linearly independent in $V_{k}$ , hence finite of size at most $d_{k} \leq k$ by [F6], so the set ${w \in L_{k} : w < v}$ is finite. Now $f (v) : = (n (v), # {w \in L_{n (v)} : w < v})$ lies in $N \times N$ and is injective: if $n (v) = n (w)$ and the two counts agree, then the coordinate vectors of $v$ and $w$ are not distinct, because if one preceded the other lexicographically the corresponding counts would differ; totality of the lexicographic order gives $v = w$ . Thus $L$ injects into the at most countable set $N \times N$ and is at most countable.

1.3F7

For $g$ finite-dimensional, $U (g)$ has a countable $C$ -basis by [F7].

2.1F3step 1.2given

Since $M$ is a quotient of $A$ as a $C$ -vector space — the map $a \mapsto a v$ is a surjective $C$ -linear map for any $0 \neq v \in M$ by [F3] — it is spanned by the images of a countable basis of $A$ . Applying step 1.2 with this spanning sequence, every linearly independent subset of $M$ is at most countable.

2.2F4step 1.1assume-contragiven

Suppose, for contradiction, that $D \neq C$ and choose $x \in D ∖ C$ ; then $x$ is transcendental over $C$ . For otherwise $p (x) = 0$ for a nonzero $p \in C [t]$ ; by [F4] write $p = c \prod_{i} (t - a_{i})$ with $c \neq 0$ , so $c \prod_{i} (x - a_{i}) = 0$ and one factor $x - a_{i}$ is zero, giving $x = a_{i} \in C$ , a contradiction.

3.1F5step 2.2algebra

Let $x$ be transcendental as in step 2.2. Evaluation $C [t] \to D$ , $p \mapsto p (x)$ , is an injective unital homomorphism whose image is commutative because $x$ commutes with the central copy of $C$ . Every nonzero $p$ has $p (x) \neq 0$ and $D$ is a division ring, so $p (x)$ is a unit; the formula $f / g \mapsto f (x) g (x)^{- 1}$ is therefore well defined — two representations of the same fraction cross-multiply, and multiplying the resulting identity by the inverses gives equality — and defines an injective field homomorphism $C (t) \to D$ by [F5]. In particular, for $a \in C$ the elements $u_{a} : = (x - a)^{- 1}$ exist in $D$ . They are $C$ -linearly independent: if $\sum_{i} λ_{i} u_{a_{i}} = 0$ with distinct $a_{i}$ , multiplying by $\prod_{i} (x - a_{i})$ gives $p (x) = 0$ for $p (t) = \sum_{i} λ_{i} \prod_{j \neq i} (t - a_{j})$ ; injectivity of $p \mapsto p (x)$ forces $p = 0$ , and evaluating at $t = a_{j}$ gives $λ_{j} \prod_{i \neq j} (a_{j} - a_{i}) = 0$ , so $λ_{j} = 0$ because the factors $a_{j} - a_{i}$ are nonzero.

4.1F3F6step 3.1algebra

Fix $0 \neq v \in M$ . The evaluation map ${ev}_{v} ⁣ : D \to M$ , $T \mapsto T v$ , is $C$ -linear and injective: if $T v = 0$ then $T (a v) = a (T v) = 0$ for every $a \in A$ because $T$ is $A$ -linear, and $A v = M$ by [F3], so $T = 0$ . Hence the vectors $({ev}_{v} (u_{a}))_{a \in C} = ((x - a)^{- 1} v)_{a \in C}$ form a $C$ -linearly independent subset $S \subseteq M$ by step 3.1. The assignment $a \mapsto (x - a)^{- 1} v$ is a bijection $C \to S$ — injective because its values are linearly independent — and $C$ is uncountable by [F6], so $S$ is uncountable.

5.1step 1.3step 2.1step 4.1givenalgebradischarge-contradiction∎

Step 2.1 makes every linearly independent subset of $M$ , in particular $S$ , at most countable, while step 4.1 makes $S$ uncountable; this contradiction forces $D = C$ , so every $A$ -endomorphism of $M$ is a scalar. Consequently, for $z \in Z (A)$ the map $μ_{z} ⁣ : M \to M$ , $m \mapsto z m$ , is $A$ -linear because $μ_{z} (a m) = z (a m) = (a z) m = a (z m)$ for $a \in A$ , so $μ_{z} = χ (z) {id}_{M}$ for some $χ (z) \in C$ ; the assignment $χ ⁣ : Z (A) \to C$ is a unital algebra homomorphism since $μ_{z z^{'}} = μ_{z} \circ μ_{z^{'}}$ and $μ_{1} = {id}_{M}$ with $M \neq 0$ . By step 1.3 the hypotheses hold for $A = U (g)$ with $g$ finite-dimensional.

PropositionStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

A primitive ideal determines a central character

Statement

Let $g$ be a finite-dimensional complex Lie algebra and let $I$ be a primitive ideal of $U (g)$ , say $I = {Ann}_{U (g)} (M)$ with $M$ simple. Then $Z (U (g))$ acts on $M$ by a central character $χ_{I} ⁣ : Z (U (g)) \to C$ , and

$I \cap Z (U (g)) = \ker χ_{I} .$

In particular $I \cap Z (U (g))$ is a maximal ideal of the commutative algebra $Z (U (g))$ , and $U (g) \ker χ_{I} \subseteq I$ .

Facts & Assumptions

Given: A finite-dimensional complex Lie algebra $g$ , a simple left $U (g)$ -module $M$ , and the primitive ideal $I = {Ann}_{U (g)} (M)$ .

[F1]

$U (g)$ is countable-dimensional for finite-dimensional $g$ , so Dixmier's lemma applies: ${End}_{U (g)} (M) = C$ , every central element acts on $M$ by a scalar, and these scalars form a unital $C$ -algebra homomorphism $χ_{I} ⁣ : Z (U (g)) \to C$ with $z m = χ_{I} (z) m$ (Dixmier's lemma: endomorphisms of a simple module over a countable-dimensional algebra, Central character of a Lie algebra module).

[F2]

A unital homomorphism $χ ⁣ : Z (U (g)) \to C$ has image $C$ , so $Z (U (g)) / \ker χ ≅ C$ is a field and $\ker χ$ is a maximal ideal; maximal means maximal among proper ideals ( $R / M$ is a field if and only if $M$ is a maximal ideal, Prime ideals and maximal ideals in a commutative ring).

[F3]

$I$ is a two-sided ideal equal to the kernel of the action; $M \neq 0$ and $1$ acts as the identity, so $I$ is proper (The annihilator of a module over an enveloping algebra, Primitive ideals of an enveloping algebra, Simple module: a nonzero module with no proper nonzero submodule).

Proof

technique · direct

1.1F1F3given

By [F1] there is a central character $χ_{I} ⁣ : Z (U (g)) \to C$ with $z m = χ_{I} (z) m$ for all $z \in Z (U (g))$ , $m \in M$ . If $z \in I \cap Z (U (g))$ , then $z$ acts on $M$ as $0$ and as the scalar $χ_{I} (z)$ ; since $M \neq 0$ this forces $χ_{I} (z) = 0$ , so $I \cap Z (U (g)) \subseteq \ker χ_{I}$ .

1.2F1F3given

Conversely, if $z \in \ker χ_{I}$ , then $z m = χ_{I} (z) m = 0$ for every $m \in M$ , so $z \in {Ann}_{U (g)} (M) = I$ , hence $z \in I \cap Z (U (g))$ . Thus $\ker χ_{I} \subseteq I \cap Z (U (g))$ .

2.1step 1.1step 1.2F2

Steps 1.1 and 1.2 give $I \cap Z (U (g)) = \ker χ_{I}$ , which is a maximal ideal of $Z (U (g))$ by [F2] because $χ_{I}$ is a unital homomorphism onto $C$ .

3.1step 2.1F3algebra∎

Every $z \in \ker χ_{I}$ lies in the two-sided ideal $I$ , so every product $u z$ with $u \in U (g)$ lies in $I$ ; as these products generate the two-sided ideal $U (g) \ker χ_{I}$ , one has $U (g) \ker χ_{I} \subseteq I$ .

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6.1-sol)Open item page →

The central reduction of the enveloping algebra at a central character

Definition

Let $g$ be a complex Lie algebra, let $Z (U (g))$ be the center of its enveloping algebra (The universal enveloping algebra as a tensor quotient), and let $χ ⁣ : Z (U (g)) \to C$ be a unital $C$ -algebra homomorphism, that is, a central character (Central character of a Lie algebra module). Write $\ker χ = {z \in Z (U (g)) : χ (z) = 0}$ and let

$U (g) \ker χ = {\sum_{k = 1}^{m} u_{k} z_{k} : m \geq 0, u_{k} \in U (g), z_{k} \in \ker χ}$

be the two-sided ideal of $U (g)$ generated by $\ker χ$ (The sum $I + J$ and product $I J$ of two-sided ideals, The sum and product of two-sided ideals are two-sided ideals). The central reduction of $U (g)$ at $χ$ is the quotient algebra

$U_{χ} : = U (g) / U (g) \ker χ .$

It is a unital associative $C$ -algebra, the natural map $U (g) \to U_{χ}$ is a surjective algebra homomorphism with kernel $U (g) \ker χ$ , the image of $Z (U (g))$ in $U_{χ}$ is $C \cdot 1$ , and for a $U (g)$ -module $M$ the following are equivalent:

(i) $M$ has central character $χ$ , i.e. every $z \in Z (U (g))$ acts on $M$ by the scalar $χ (z)$ ;
(ii) $\ker χ$ annihilates $M$ ;
(iii) the action of $U (g)$ on $M$ factors uniquely through $U_{χ}$ .

Thus the $U (g)$ -modules with central character $χ$ are exactly the $U_{χ}$ -modules.

Remarks

The generating set is central, so the ideal is two-sided. Every $z \in \ker χ$ commutes with $U (g)$ , hence $u_{k} z_{k} = z_{k} u_{k}$ and the set displayed above is already closed under left and right multiplication; it is an additive subgroup because finite sums of such terms are such terms, and it contains $0$ as the empty sum. This is the ideal generated by $\ker χ$ (Left, right and two-sided ideals).
Every central element is scalar modulo the ideal. For $z \in Z (U (g))$ one has $z - χ (z) \cdot 1 \in \ker χ \subseteq U (g) \ker χ$ , so $z$ and the scalar $χ (z)$ have the same image in $U_{χ}$ . Thus the image of the center consists of scalar classes; this does not imply that every element of $U_{χ}$ is scalar.
The equivalences. Condition (i) says $z$ acts by $χ (z)$ for every central $z$ , which is equivalent to $(z - χ (z)) m = 0$ for all such $z$ and all $m$ , hence to $\ker χ$ annihilating $M$ because $\ker χ$ consists exactly of the elements $z$ with $χ (z) = 0$ . If $ρ ⁣ : U (g) \to {End}_{C} (M)$ denotes the action, then (ii) says $\ker χ \subseteq \ker ρ$ , so $U (g) \ker χ \subseteq \ker ρ$ because $\ker ρ$ is a two-sided ideal; the quotient universal property (A ring homomorphism whose kernel contains a two-sided ideal factors uniquely through the quotient ring) then produces a unique factorization of $ρ$ through $U_{χ}$ , which is (iii). Conversely a factorization through $U_{χ}$ kills $U (g) \ker χ$ and hence $\ker χ$ , which is (ii).

PropositionStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

The Verma annihilator contains the central-character ideal

Statement

Let $λ \in h^{*}$ , let $χ_{λ} ⁣ : Z (U (g)) \to C$ be the central character determined by $λ$ through the Harish-Chandra projection, so that every $z \in Z (U (g))$ acts on the Verma module $M (λ)$ by the scalar $χ_{λ} (z) = pr (z) (λ)$ . Then

$U (g) \ker χ_{λ} \subseteq {Ann}_{U (g)} M (λ) \subseteq {Ann}_{U (g)} L (λ) = I (λ),$

and every element of the two-sided ideal generated by $\ker χ_{λ}$ annihilates every cyclic highest-weight module of highest weight $λ$ .

Facts & Assumptions

Given: A weight $λ \in h^{*}$ , the Verma module $M (λ)$ with its unique simple quotient $L (λ)$ , and the central character $χ_{λ}$ with $χ_{λ} (z) = pr (z) (λ)$ .

[F1]

Every cyclic highest-weight module $M = U (g) v$ of highest weight $λ$ has a central character, and every $z \in Z (U (g))$ acts on $M$ by the scalar $pr (z) (λ)$ (Central elements act by scalars on cyclic highest-weight modules, The Harish-Chandra projection computes the highest-weight scalar, Central character of a Lie algebra module, Highest-weight vectors and cyclic highest-weight modules).

[F2]

The Verma module $M (λ)$ has a unique simple quotient $L (λ) = M (λ) / J (λ)$ , and $L (λ)$ is simple (A Verma module has a unique simple quotient, Verma modules).

[F3]

The annihilator of a module is a two-sided ideal, and annihilators grow when passing to quotients: if $M \to Q$ is a surjection of $U (g)$ -modules and $u M = 0$ , then $u Q = 0$ (The annihilator of a module over an enveloping algebra).

Proof

technique · direct

1.1F1given

By [F1], $z \in Z (U (g))$ acts on $M (λ)$ by the scalar $pr (z) (λ) = χ_{λ} (z)$ . Hence $z \in \ker χ_{λ}$ acts on $M (λ)$ by $0$ , that is, $\ker χ_{λ} \subseteq {Ann}_{U (g)} M (λ)$ .

1.2F1algebra

More generally, let $M = U (g) v$ be any cyclic highest-weight module of highest weight $λ$ and let $u \in U (g)$ , $z \in \ker χ_{λ}$ . For $a \in U (g)$ one has $(u z) (a v) = u (z a v) = u (a z v) = u a χ_{λ} (z) v = 0$ by [F1] and centrality of $z$ ; since every element of $M$ has the form $a v$ , the element $u z$ annihilates $M$ . As products $u z$ span $U (g) \ker χ_{λ}$ , every element of that ideal annihilates every cyclic highest-weight module of highest weight $λ$ .

2.1step 1.1F3algebra

Since ${Ann}_{U (g)} M (λ)$ is a two-sided ideal by [F3], it contains all products $u z$ with $u \in U (g)$ and $z \in \ker χ_{λ}$ , hence contains the two-sided ideal $U (g) \ker χ_{λ}$ generated by $\ker χ_{λ}$ . Thus $U (g) \ker χ_{λ} \subseteq {Ann}_{U (g)} M (λ)$ .

3.1step 2.1step 1.2F2F3∎

The Verma module $M (λ)$ has its unique simple quotient $L (λ)$ by [F2], and an operator annihilating $M (λ)$ annihilates each quotient by [F3], so ${Ann}_{U (g)} M (λ) \subseteq {Ann}_{U (g)} L (λ) = I (λ)$ . Together with step 2.1 this gives the displayed chain, and step 1.2 gives the final assertion.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

The adjoint action preserves the associated graded of a two-sided ideal

Statement

Let $g$ be a finite-dimensional complex Lie algebra with the PBW filtration $F^{n} U (g)$ , $n \geq 0$ , and the identification $gr U (g) = S (g)$ . For every two-sided ideal $I \subseteq U (g)$ the associated graded subspace

$gr I = ⨁_{n \geq 0} (I \cap F^{n} U (g)) / (I \cap F^{n - 1} U (g)) \subseteq S (g)$

is a graded ideal of $S (g)$ . Moreover, writing $D_{x}$ for the derivation of $S (g)$ that extends the linear map ${ad}_{x} ⁣ : g \to g$ , $y \mapsto [x, y]$ (Derivations of Lie algebras), one has $D_{x} (gr I) \subseteq gr I$ for every $x \in g$ , and, with $σ_{n} ⁣ : F^{n} U (g) \to F^{n} U (g) / F^{n - 1} U (g)$ the degree- $n$ quotient map, the induced derivation satisfies:

$σ_{n} ({ad}_{x} u) = D_{x} (σ_{n} (u)), {ad}_{x} u = x u - u x,$

for every $n \geq 0$ and $u \in F^{n} U (g)$ . If the commutator has degree less than $n$ , its degree- $n$ symbol is zero.

Facts & Assumptions

Given: A finite-dimensional complex Lie algebra $g$ , a two-sided ideal $I ⊴ U (g)$ , and an element $x \in g$ .

[F1]

$F^{∙} U (g)$ is the PBW filtration by tensor degree with $F^{- 1} = 0$ ; multiplication in $U (g)$ induces a product on $gr U (g)$ for which the symbol of a product of elements of $F^{m}$ and $F^{n}$ is the product of their symbols (The PBW filtration by tensor degree on the enveloping algebra).

[F2]

An ordered basis of $g$ has its ordered monomials as a basis of $U (g)$ , and multiplication identifies $gr U (g)$ with the symmetric algebra $S (g)$ ; in particular $F^{1} U (g) = C \oplus g$ and the symbol of $y \in g$ is $y$ (PBW gives an ordered monomial basis for the enveloping algebra).

[F3]

$gr U (g)$ is commutative: $[F^{m} U (g), F^{n} U (g)] \subseteq F^{m + n - 1} U (g)$ (The associated graded algebra of the PBW filtration is commutative).

[F4]

$I$ is an additive subgroup closed under left and right multiplication by $U (g)$ (Left, right and two-sided ideals); ${ad}_{x}$ denotes the linear map $y \mapsto [x, y]$ of $g$ (Derivations of Lie algebras). Any $C$ -linear map $g \to g$ extends uniquely to a derivation of the symmetric algebra $S (g)$ , by declaring the Leibniz rule on monomials in a basis; this extension is $D_{x}$ .

Proof

technique · direct

1.1F1F3F4given

$I \cap F^{n} U (g)$ defines an increasing filtration of $I$ with $I \cap F^{- 1} U (g) = 0$ , so $gr I$ is a graded subspace of $gr U (g)$ by construction. It is an ideal: if $a \in I \cap F^{m} U (g)$ and $b \in F^{n} U (g)$ , then $a b \in I \cap F^{m + n} U (g)$ by [F4], and $b a \in I \cap F^{m + n} U (g)$ likewise; passing to symbols with [F1] exhibits every product of a symbol of $I$ with a symbol of $U (g)$ as a symbol of an element of $I$ . Since $gr U (g) = S (g)$ is commutative by [F3], one-sided closure suffices and $gr I$ is a graded ideal.

1.2F1F4givenalgebra

The commutator map ${ad}_{x} ⁣ : U (g) \to U (g)$ , $u \mapsto x u - u x$ , is a derivation of $U (g)$ : ${ad}_{x} (u v) = (x u - u x) v + u (x v - v x) = {ad}_{x} (u) v + u {ad}_{x} (v)$ . For a word $a_{1} \dots a_{m}$ with $a_{i} \in g$ , the derivation rule gives $[x, a_{1} \dots a_{m}] = \sum_{i = 1}^{m} a_{1} \dots a_{i - 1} [x, a_{i}] a_{i + 1} \dots a_{m}$ , and each $[x, a_{i}]$ lies in $g$ ; hence every term still has PBW degree at most $m$ , so ${ad}_{x} (F^{n} U (g)) \subseteq F^{n} U (g)$ . It maps $I$ into itself because $I$ is two-sided.

2.1F1step 1.1step 1.2algebra

By step 1.2, ${ad}_{x}$ preserves each $I \cap F^{n} U (g)$ , so it induces a graded linear map $gr ({ad}_{x})$ of $gr U (g)$ that sends $gr I$ into itself: the induced map is $σ_{n} (u) \mapsto σ_{n} ({ad}_{x} u)$ , well defined because ${ad}_{x} (F^{n - 1}) \subseteq F^{n - 1}$ . The derivation identity of step 1.2 passes to symbols via [F1], so $gr ({ad}_{x})$ is a derivation of $gr U (g) = S (g)$ .

3.1F2F4step 2.1

On degree one, $gr ({ad}_{x}) (y) = {ad}_{x} (y) = [x, y]$ for $y \in g$ , by [F2] and [F4]; that is, the induced derivation restricts on $g$ to the given linear map ${ad}_{x}$ .

4.1F2F4step 2.1step 3.1algebra∎

A derivation of $S (g)$ is determined by its values on $g$ : on a monomial $y_{1} \dots y_{n}$ the Leibniz rule forces $\sum_{i} y_{1} \dots {ad}_{x} (y_{i}) \dots y_{n}$ , and a monomial basis of $S (g)$ extends these values linearly. Hence the derivation $gr ({ad}_{x})$ of step 2.1, whose degree-one restriction is the given map ${ad}_{x}$ by step 3.1, equals $D_{x}$ . Therefore $D_{x} (gr I) = gr ({ad}_{x}) (gr I) \subseteq gr I$ and $σ_{n} ({ad}_{x} u) = D_{x} (σ_{n} (u))$ for every $n \geq 0$ and $u \in F^{n} U (g)$ , which is the assertion.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6.1-sol)Open item page →

The associated graded variety of a two-sided ideal

Definition

Let $g$ be a finite-dimensional complex Lie algebra, with PBW filtration $F_{n} U (g)$ and associated graded algebra $gr U (g)$ (The PBW filtration by tensor degree on the enveloping algebra), and let $I ⊴ U (g)$ be a two-sided ideal (Left, right and two-sided ideals). For each $n \geq 0$ put $F_{n} I : = I \cap F_{n} U (g)$ with $F_{- 1} I = 0$ , and define the associated graded ideal

$gr I : = ⨁_{n \geq 0} F_{n} I / F_{n - 1} I \subseteq gr U (g) .$

Fix an ordered basis $x_{1}, \dots, x_{n}$ of $g$ . By PBW gives an ordered monomial basis for the enveloping algebra, its ordered monomials form a basis of $U (g)$ and multiplication identifies $gr U (g)$ with the symmetric algebra $S (g)$ , which under this basis is the polynomial algebra $C [x_{1}, \dots, x_{n}]$ on the symbols; by The associated graded algebra of the PBW filtration is commutative this algebra is commutative, so $gr I$ is an ideal of it. The dual basis of $x_{1}, \dots, x_{n}$ identifies $g^{*}$ with $C^{n}$ , so elements of $S (g)$ are polynomial functions on $g^{*}$ . The associated variety of $I$ is the classical affine algebraic set (Classical affine algebraic sets, including the empty boundaries)

$V (I) : = {f \in g^{*} : p (f) = 0 for every p \in gr I} .$

It is the zero locus of the family $gr I$ in the polynomial ring on $g^{*}$ ; equivalently $V (I) = V (gr I)$ in the notation of the classical zero loci, a Zariski closed subset of $g^{*}$ (Classical affine zero loci form the Zariski closed sets).

Remarks

$gr I$ is a graded ideal, not merely a graded subspace. If $u \in F_{m} U (g)$ and $v \in F_{n} I$ , then $u v \in I \cap F_{m + n} U (g)$ , and the symbol of $u v$ is the product of the symbols of $u$ and $v$ ; hence $gr I$ is closed under multiplication by the whole of $gr U (g)$ .
The definition does not depend on the ordered basis. The subspaces $F_{n} U (g)$ are defined by tensor degree with no reference to a basis, so $gr I$ is intrinsic; changing the ordered basis changes the identification of $S (g)$ with a polynomial ring by an invertible linear change of variables, whose induced map on $C^{n}$ is a linear isomorphism carrying one zero locus onto the other. The associated variety $V (I)$ as a subset of $g^{*}$ is therefore well defined (The classical vanishing ideal, The coordinate ring of a classical affine algebraic set).
Proper ideals and the empty case. If $I = U (g)$ then $F_{n} I = F_{n} U (g)$ for all $n$ , so $gr I = gr U (g)$ and $V (I) = \emptyset$ ; if $I = 0$ then $gr I = 0$ and $V (I) = g^{*}$ . Both boundary cases are allowed by the definition. Primitive ideals are proper, but may be zero; for example, when $g = 0$ , the simple $U (0) = C$ -module $C$ has annihilator $I = 0$ .

PropositionStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

The associated variety is a closed conical coadjoint-invariant cone

Statement

Let $g$ be a finite-dimensional complex Lie algebra and let $I \subseteq U (g)$ be a two-sided ideal. Then $V (I) \subseteq g^{*}$ is a closed cone: for $ξ \in V (I)$ and $t \in C$ one has $t ξ \in V (I)$ . Moreover $V (I)$ is invariant under the coadjoint action: for every $x \in g$ and $ξ \in V (I)$ the curve $t \mapsto ξ \circ e^{- t {ad}_{x}}$ (using the matrix exponential on $g$ , whose dual maps form a one-parameter group of linear automorphisms of $g^{*}$ ) stays inside $V (I)$ . In particular the same conclusions hold for every primitive ideal.

Facts & Assumptions

Given: A finite-dimensional complex Lie algebra $g$ , a two-sided ideal $I ⊴ U (g)$ , the associated graded ideal $gr I \subseteq S (g) = C [g^{*}]$ , and the associated variety $V (I) = {ξ \in g^{*} : f (ξ) = 0 for all f \in gr I}$ .

[F1]

$V (I)$ is the zero locus of the family $gr I$ in the polynomial algebra on $g^{*}$ , hence a Zariski closed classical affine algebraic set; $gr I$ is a graded ideal of $S (g)$ (The associated graded variety of a two-sided ideal, Classical affine zero loci form the Zariski closed sets, Classical affine algebraic sets, including the empty boundaries).

[F2]

For every $x \in g$ , the derivation $D_{x}$ of $S (g)$ extending $y \mapsto [x, y]$ satisfies $D_{x} (gr I) \subseteq gr I$ ; hence $D_{x}^{k} (gr I) \subseteq gr I$ for all $k \geq 0$ (The adjoint action preserves the associated graded of a two-sided ideal).

[F3]

Fix a basis of $g$ and use its maximum-coordinate norm and the induced submultiplicative operator norm on $End (g)$ . The exponential series for $- t {ad}_{x}$ is dominated on every compact $t$ -disc by the scalar series $\sum_{k \geq 0} (∣ t ∣ ∥ {ad}_{x} ∥)^{k} / k!$ , which converges for every $t$ (The exponential series converges absolutely for every real argument); it therefore converges locally uniformly, may be differentiated termwise, and has entire matrix entries (A complex power-series sum has complex derivatives of every order, obtained by repeated termwise differentiation, The sum of a complex power series is analytic throughout its open disc of convergence, Every complex analytic function is holomorphic). Composing those entries with the polynomial $f$ and evaluating at $ξ$ shows $φ (t) = f (ξ \circ e^{- t {ad}_{x}})$ is entire (Complex analytic functions are closed under finite linear combinations, products, quotients with nonzero denominator, and composition). Put $E (t) = e^{- t {ad}_{x}}$ and $ξ_{t} = ξ \circ E (t)$ . Termwise differentiation gives $E^{'} (t) = - E (t) {ad}_{x}$ . On a linear polynomial $y \in g$ , this yields $\frac{d}{d t} y (ξ_{t}) = - (D_{x} y) (ξ_{t})$ ; the product and sum rules extend the identity to every polynomial. Iteration gives $φ^{(k)} (0) = (- 1)^{k} (D_{x}^{k} f) (ξ)$ (Linearity, product, reciprocal, and quotient rules for complex derivatives). Absolute convergence permits multiplying the exponential series entrywise; collecting total degree and using the binomial formula gives $E (s) E (t) = E (s + t)$ , so $E (0) = 1$ and $E (- t) = E (t)^{- 1}$ . Thus the dual maps $ξ \mapsto ξ_{t}$ form the asserted one-parameter group. Finally, an entire function is equal on $C$ to its Taylor series at $0$ , so if all these derivatives vanish then $φ \equiv 0$ (A holomorphic function equals its Taylor series throughout the largest centred disc in its domain).

Proof

technique · direct

1.1F1givenalgebra

That $V (I)$ is closed is [F1]. For the cone property, write an arbitrary $f \in gr I$ as a finite sum $f = \sum_{d} f_{d}$ of its homogeneous components $f_{d} \in gr I$ of degree $d$ , which lie in $gr I$ because it is graded. For $t \in C$ and $ξ \in V (I)$ one has $f_{d} (t ξ) = t^{d} f_{d} (ξ) = 0$ : by homogeneity for $d > 0$ , while for $d = 0$ the degree-zero part of $gr I$ is $0$ whenever $I$ is proper, since $(I \cap F^{0} U (g)) / 0 = I \cap C \cdot 1 = 0$ ; if $I = U (g)$ then $gr I = gr U (g)$ contains $1$ and $V (I) = \emptyset$ , so the assertion is vacuous. Hence $f (t ξ) = 0$ for every $f$ , that is $t ξ \in V (I)$ .

1.2F2F3algebra

Let $x \in g$ , $ξ \in V (I)$ and $f \in gr I$ homogeneous, and put $φ (t) = f (ξ \circ e^{- t {ad}_{x}})$ . By [F2] each $D_{x}^{k} f$ lies in $gr I$ , so [F3] gives $φ^{(k)} (0) = (- 1)^{k} (D_{x}^{k} f) (ξ) = 0$ for every $k \geq 0$ . By [F3] again $φ$ is entire, so $φ \equiv 0$ ; as $f$ was an arbitrary homogeneous element of $gr I$ , the whole curve $t \mapsto ξ \circ e^{- t {ad}_{x}}$ lies in $V (I)$ .

2.1givenstep 1.1step 1.2∎

Every primitive ideal of $U (g)$ is a two-sided ideal, so steps 1.1 and 1.2 apply to it; no primitivity-specific hypothesis is used in the argument, and the same conclusions therefore hold for every primitive ideal.

Remarks

Only the ideal property and the derivation invariance enter. Closedness comes from the definition of the associated variety as a zero locus, conicality from gradedness of $gr I$ , and coadjoint invariance from the derivation invariance $D_{x} (gr I) \subseteq gr I$ . Nothing about primitivity, semisimplicity, or nilpotent orbits is used; the deep Borho-Brylinski/Joseph statement that this cone is a single nilpotent orbit closure is not asserted here.
The exponential curve. The statement is about the orbit of $ξ$ under the one-parameter group $t \mapsto e^{- t {ad}_{x}}$ in the linear automorphism group of $g$ , transported to $g^{*}$ by duality; the analytic input is the convergence of the finite-dimensional exponential series, recorded in [F3].

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

Every central character of a semisimple enveloping algebra arises from a weight

Statement

Assume the Axiom of Choice. Let $g$ be a finite-dimensional complex semisimple Lie algebra with Cartan subalgebra $h$ and a fixed positive system. Then every unital $C$ -algebra homomorphism $χ ⁣ : Z (U (g)) \to C$ equals $χ_{λ}$ for some $λ \in h^{*}$ , where $χ_{λ} (z) = pr (z) (λ)$ ; equivalently, the maximal ideals of $Z (U (g))$ are exactly the kernels $\ker χ_{λ}$ , and $λ \mapsto \ker χ_{λ}$ induces a bijection $h^{*} / W \to Max Z (U (g))$ , so that $χ_{λ} = χ_{μ}$ if and only if $μ \in W \cdot λ$ . Here $w \cdot λ = w (λ + ρ) - ρ$ , and $h^{*} / W$ denotes the quotient for this dot action.

Facts & Assumptions

Given: The Axiom of Choice, a finite-dimensional complex semisimple Lie algebra $g$ with Cartan $h$ and positive system, and a unital $C$ -algebra homomorphism $χ ⁣ : Z (U (g)) \to C$ .

[F1]

The shifted Harish-Chandra map ${HC}_{ρ} ⁣ : Z (U (g)) \to S (h)^{W}$ , ${HC}_{ρ} (z) (λ) = pr (z) (λ - ρ)$ , is an algebra isomorphism under AC (Harish-Chandra isomorphism for the center); the projection computes the highest-weight scalars, so $χ_{λ} (z) = pr (z) (λ)$ is the scalar by which $z$ acts on $M (λ)$ (The Harish-Chandra projection computes the highest-weight scalar, The Harish-Chandra projection, Central character of a Lie algebra module).

[F2]

Under AC, $S (h)$ is a free $S (h)^{W}$ -module of rank $∣ W ∣$ , and $S (h)^{W}$ is a polynomial algebra on $rank g$ homogeneous generators; hence $S (h)$ is a finite $S (h)^{W}$ -module and $Z (U (g)) ≅ S (h)^{W}$ is a finitely generated $C$ -algebra (Chevalley shephard todd for finite weyl groups, The center of the enveloping algebra is polynomial on rank-many generators).

[F3]

Determinant trick (Nakayama): if $R$ is commutative, $I ⊴ R$ and $M$ a finitely generated $R$ -module with $I M = M$ , then $(1 - a) M = 0$ for some $a \in I$ (Determinant trick for Nakayama).

[F4]

Under AC every proper ideal of a nonzero commutative ring is contained in a maximal ideal; maximal means maximal among proper ideals (In a nonzero commutative ring, every proper ideal is contained in a maximal ideal, Prime ideals and maximal ideals in a commutative ring, The Axiom of Choice).

[F5]

Over the algebraically closed field $C$ , every maximal ideal of a finitely generated commutative $C$ -algebra is the kernel of a unital $C$ -algebra homomorphism to $C$ (Over an algebraically closed field, maximal ideals of an affine algebra are kernels of points); an ideal is maximal exactly when the quotient is a field ( $R / M$ is a field if and only if $M$ is a maximal ideal).

[F6]

$C$ is algebraically closed, so it has no nontrivial finite-dimensional field extension (The complex numbers are algebraically closed).

[F7]

Under AC, $χ_{λ} = χ_{μ}$ if and only if $μ \in W \cdot λ$ (Central characters are dot-Weyl orbits).

Proof

technique · direct

1.1F1F5given

Define $ψ : = χ \circ {HC}_{ρ}^{- 1} ⁣ : S (h)^{W} \to C$ . It is a unital $C$ -algebra homomorphism by [F1], and $ψ ({HC}_{ρ} (z)) = χ (z)$ for every $z$ . Its image contains $1$ and is closed under the $C$ -scalars, so the image is all of $C$ ; hence $S (h)^{W} / \ker ψ ≅ C$ is a field and $\ker ψ$ is a maximal, in particular proper, ideal of $S (h)^{W}$ .

2.1F2F3step 1.1algebra

By [F2], $S (h)$ is a finitely generated $S (h)^{W}$ -module, so [F3] applies with $R = S (h)^{W}$ , $I = \ker ψ$ and $M = S (h)$ : if $(\ker ψ) S (h) = S (h)$ , then $(1 - a) S (h) = 0$ for some $a \in \ker ψ$ , and evaluating at $1 \in S (h)$ gives $1 = a$ , contradicting $ψ (1) = 1$ . Therefore $(\ker ψ) S (h) \neq S (h)$ , and this ideal is proper.

3.1F2F4F5step 1.1step 2.1

By [F4] the proper ideal $(\ker ψ) S (h)$ is contained in a maximal ideal $n$ of $S (h)$ ; in particular $\ker ψ \subseteq n$ . The contraction $n \cap S (h)^{W}$ is proper because $1 \notin n$ , and contains the maximal ideal $\ker ψ$ ; hence $n \cap S (h)^{W} = \ker ψ$ . The quotient $S (h) / n$ is a field by [F5] containing the field $S (h)^{W} / \ker ψ ≅ C$ of step 1.1, and it is finite-dimensional over it because $S (h)$ is a finite $S (h)^{W}$ -module.

4.1F6step 3.1algebra

A finite-dimensional field extension of the algebraically closed field $C$ is $C$ itself by [F6], so the quotient map of step 3.1 is a unital $C$ -algebra homomorphism $φ ⁣ : S (h) \to C$ extending $ψ$ : for $a \in S (h)^{W}$ the class $a + n$ depends only on $a + \ker ψ$ and equals $ψ (a)$ .

5.1step 4.1construct

The homomorphism $φ$ is evaluation at a weight: fix a $C$ -basis $h_{1}, \dots, h_{r}$ of $h$ ; since $S (h) = C [h_{1}, \dots, h_{r}]$ with the $h_{i}$ as coordinate functions, $φ$ is determined by the scalars $φ (h_{i})$ , which define a unique $λ \in h^{*}$ by $λ (h_{i}) : = φ (h_{i})$ , and $φ (p) = p (λ)$ for every polynomial $p$ .

6.1F1step 1.1step 4.1step 5.1algebra

Combining the identities, for every $z \in Z (U (g))$ : $χ (z) = ψ ({HC}_{ρ} (z)) = φ ({HC}_{ρ} (z)) = {HC}_{ρ} (z) (λ) = pr (z) (λ - ρ)$ by [F1]; writing $μ : = λ - ρ$ gives $χ = χ_{μ}$ . Thus every unital homomorphism $Z (U (g)) \to C$ is a $χ_{μ}$ .

7.1F1F5F7step 6.1∎

By [F2], $Z (U (g)) ≅ S (h)^{W}$ is a finitely generated commutative $C$ -algebra, so by [F5] every maximal ideal of $Z (U (g))$ is the kernel of a unital homomorphism to $C$ , hence by step 6.1 of the form $\ker χ_{λ}$ . Conversely each $χ_{λ}$ is a surjective unital homomorphism onto the field $C$ , so $\ker χ_{λ}$ is maximal by [F5], and $\ker χ_{λ} = \ker χ_{μ}$ implies $χ_{λ} = χ_{μ}$ because both factor through the common quotient, which is $C$ via the unital structure map. By [F7], $χ_{λ} = χ_{μ}$ exactly when $μ \in W \cdot λ$ ; hence $λ \mapsto \ker χ_{λ}$ induces a bijection $h^{*} / W \to Max Z (U (g))$ , and the stated equivalences follow.

CorollaryStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

Primitive ideals are partitioned by their dot-orbit central character

Statement

Assume the Axiom of Choice. Let $g$ be a finite-dimensional complex semisimple Lie algebra with Cartan subalgebra $h$ and a fixed positive system. If $I$ is a primitive ideal of $U (g)$ , then $I \cap Z (U (g)) = \ker χ_{I}$ for a unique central character $χ_{I}$ , and there is a unique dot-Weyl orbit $W \cdot λ \subseteq h^{*}$ with $χ_{I} = χ_{λ}$ ; this orbit is determined by $I$ alone. Consequently, for primitive ideals $I, J$ : $I \cap Z = J \cap Z$ if and only if $χ_{I} = χ_{J}$ , if and only if $W \cdot λ = W \cdot μ$ for any $λ, μ$ with $χ_{λ} = χ_{I}$ and $χ_{μ} = χ_{J}$ . Thus the primitive ideals are partitioned by their central characters, and the central character of a primitive ideal is exactly a dot-Weyl orbit under the Harish-Chandra isomorphism.

Facts & Assumptions

Given: The Axiom of Choice, a finite-dimensional complex semisimple Lie algebra $g$ with Cartan $h$ and positive system, and primitive ideals $I, J$ of $U (g)$ .

[F1]

For a primitive ideal $I$ , the center acts on a simple module with annihilator $I$ through a central character $χ_{I}$ , and $I \cap Z (U (g)) = \ker χ_{I}$ ; a unital homomorphism $Z (U (g)) \to C$ is determined by its kernel, because the quotient by the kernel is $C$ via the unital structure map (A primitive ideal determines a central character, Primitive ideals of an enveloping algebra).

[F2]

Under AC, every central character $χ ⁣ : Z (U (g)) \to C$ equals $χ_{λ}$ for some $λ \in h^{*}$ (Every central character of a semisimple enveloping algebra arises from a weight, Central character of a Lie algebra module, The Axiom of Choice).

[F3]

Under AC, $χ_{λ} = χ_{μ}$ if and only if $μ \in W \cdot λ$ (Central characters are dot-Weyl orbits).

Proof

technique · direct

1.1F1given

By [F1], $I \cap Z (U (g)) = \ker χ_{I}$ for a central character $χ_{I}$ ; any central character with the same kernel equals $χ_{I}$ , because the kernel determines the unital homomorphism through the quotient $Z (U (g)) / \ker χ_{I} ≅ C$ . Hence $χ_{I}$ is unique, and it is determined by $I$ .

2.1F2F3step 1.1

By [F2] there is $λ \in h^{*}$ with $χ_{I} = χ_{λ}$ . If also $χ_{I} = χ_{μ}$ , then $χ_{λ} = χ_{μ}$ , so $μ \in W \cdot λ$ by [F3]; hence the dot orbit $W \cdot λ$ is independent of the choice of $λ$ and determined by $I$ .

3.1F1F3step 1.1step 2.1∎

For primitive ideals $I, J$ , the identity $I \cap Z = J \cap Z$ is equivalent to $\ker χ_{I} = \ker χ_{J}$ by [F1], which is equivalent to $χ_{I} = χ_{J}$ because a central character is determined by its kernel; and by step 2.1 with [F3] this is equivalent to $W \cdot λ = W \cdot μ$ for any $λ, μ$ realizing $χ_{I}, χ_{J}$ . Thus the fibres of $I \mapsto χ_{I}$ on primitive ideals are exactly the sets of primitive ideals with a common dot-Weyl orbit, i.e. the primitive ideals are partitioned by their central characters.

LemmaStatement: Literature-sourcedProof: AI-adaptedOpen item page →

The central reduction of U(sl2) is simple away from the finite-dimensional central characters

Statement

Assume the Axiom of Choice. Let $g = {s l}_{2} (C)$ with standard basis $e, f, h$ , Casimir element $Ω$ , and center $Z (U (g)) = C [Ω]$ . Let $χ$ be a central character, determined by the scalar $χ (Ω)$ . Then the central reduction $U_{χ}$ is a simple ring if and only if $χ$ is not the central character of any nonzero finite-dimensional $U (g)$ -module, equivalently if and only if $χ (Ω)$ is not the eigenvalue of $Ω$ on any finite-dimensional simple module $L (n)$ , $n \geq 0$ (these eigenvalues are pairwise distinct). When $U_{χ}$ is simple, every nonzero $U (g)$ -module with central character $χ$ is faithful over $U_{χ}$ , every Verma module $M (λ)$ with $χ_{λ} = χ$ is simple, and ${Ann}_{U (g)} M (λ) = U (g) \ker χ$ .

Facts & Assumptions

Given: The Axiom of Choice, $g = {s l}_{2} (C)$ with basis $e, f, h$ and Casimir $Ω = e f + f e + \frac{1}{2} h^{2}$ (a nonzero scalar multiple of the library Casimir), a central character $χ$ , the scalar $c : = χ (Ω) \in C$ , and $U_{χ} = U (g) / U (g) \ker χ$ .

[F1]

The Axiom of Choice is assumed (The Axiom of Choice). The PBW symbol of a central element is invariant under the adjoint action: for each $x \in g$ , The adjoint action preserves the associated graded of a two-sided ideal gives $σ_{n} ([x, z]) = D_{x} σ_{n} (z)$ for $z \in F^{n} U (g)$ , so $D_{x} σ (z) = 0$ . Chevalley restriction identifies $S (g)^{g}$ with $S (h)^{W}$ (Chevalley restriction for symmetric invariants); for ${s l}_{2}$ , $h = C h$ is the Cartan subalgebra, the roots are $\pm α$ with $α (h) = 2$ , and the reflection sends the coordinate $h$ to $- h$ , so $S (h)^{W} = C [h^{2}]$ (The special linear Lie algebra sl_2, Cartan subalgebra, Normalizer of a Lie subalgebra, Root and root space, Root reflections and the Weyl group action). The library Casimir $C$ of The quadratic Casimir element has PBW symbol restricting to $h^{2} / 8$ on $h$ , by the Killing-form calculation in step 1.1 (The Killing form of a semisimple Lie algebra); hence $σ (C)$ generates $S (g)^{g}$ . Subtracting the matching scalar multiple of $C^{k}$ from any central element of degree $2 k$ lowers its PBW degree, so induction and centrality of $C$ give $Z (U (g)) = C [C]$ . Step 1.1 gives $Ω = 4 C$ , hence $Z (U (g)) = C [Ω]$ and $\ker χ = (Ω - c)$ as an ideal of this polynomial algebra.

[F2]

$U_{χ}$ is the quotient of $U (g)$ by the two-sided ideal $U (g) \ker χ = (Ω - c) U (g)$ ; a $U (g)$ -module has central character $χ$ exactly when it is a $U_{χ}$ -module, and $U_{χ}$ is a unital associative $C$ -algebra (The central reduction of the enveloping algebra at a central character, Central character of a Lie algebra module).

[F3]

Ordered monomials in a basis of $g$ form a basis of $U (g)$ ; the PBW filtration by tensor degree has $gr U (g) = C [e, f, h]$ , a polynomial algebra; symbols multiply and the symbol of a product of nonzero symbols is nonzero (PBW gives an ordered monomial basis for the enveloping algebra, The PBW filtration by tensor degree on the enveloping algebra, The associated graded algebra of the PBW filtration is commutative).

[F4]

$[h, e] = 2 e$ , $[h, f] = - 2 f$ , $[e, f] = h$ , so $h e = e (h + 2)$ , $h f = f (h - 2)$ (The special linear Lie algebra sl_2). The finite-dimensional simple modules $L (n)$ , $n \geq 0$ , have $Ω$ -eigenvalue $n (n + 2) / 2$ , these values are pairwise distinct, and every finite-dimensional simple module is some $L (n)$ (The quadratic Casimir eigenvalue on a highest-weight module is $(λ, λ + 2 ρ)$ , Finite-dimensional simple modules are classified by dominant highest weights).

[F5]

The Verma module $M (λ)$ is simple if and only if $⟨ λ + ρ, α^{\lor} ⟩ \notin Z_{> 0}$ for all positive roots; for ${s l}_{2}$ there is one positive root and $⟨ λ + ρ, α^{\lor} ⟩ = λ + 1$ in the standard identification $λ = λ (h)$ , so $M (λ)$ is simple exactly when $λ \notin Z_{\geq 0}$ , and for $λ \in Z_{\geq 0}$ its central character is that of the finite-dimensional module $L (λ)$ (The Verma irreducibility criterion from Shapovalov determinants, Verma modules, The quadratic Casimir eigenvalue on a highest-weight module is $(λ, λ + 2 ρ)$ ).

[F6]

$C [h]$ has polynomial division with remainder (For every field $F$ , $F [x]$ is a Euclidean domain with degree as Euclidean function). In a nonzero ideal, a nonzero polynomial $P$ of least degree generates the ideal: dividing any member by $P$ leaves a remainder in the ideal of smaller degree, which must be zero.

[F7]

The field $C$ is algebraically closed, so every nonconstant polynomial in $C [h]$ has a root (The complex numbers are algebraically closed).

Proof

technique · direct

1.1F1F2F4algebra

The element $Ω = e f + f e + \frac{1}{2} h^{2}$ is central. Indeed, using $[h, e] = 2 e$ , $[h, f] = - 2 f$ , $[e, f] = h$ : $[h, Ω] = 2 e f - 2 e f - 2 f e + 2 f e = 0$ ; $[e, e f] = [e, e] f + e [e, f] = e h$ , $[e, f e] = [e, f] e + f [e, e] = h e$ , and $[e, h^{2}] = [e, h] h + h [e, h] = - 2 e h - 2 h e$ , so $[e, Ω] = e h + h e - e h - h e = 0$ ; and $[f, e f] = [f, e] f + e [f, f] = - h f$ , $[f, f e] = [f, f] e + f [f, e] = - f h$ , $[f, h^{2}] = [f, h] h + h [f, h] = 2 f h + 2 h f$ , so $[f, Ω] = - h f - f h + f h + h f = 0$ . The Killing form in the basis $e, f, h$ has $B (h, h) = 8$ and $B (e, f) = B (f, e) = 4$ with all other pairings $0$ (the adjoint matrices have columns $a d_{h} (e) = 2 e$ , $a d_{h} (f) = - 2 f$ , $a d_{h} (h) = 0$ and $a d_{e} (f) = h$ , $a d_{e} (h) = - 2 e$ , $a d_{f} (e) = - h$ , $a d_{f} (h) = 2 f$ , giving traces $8$ and $4$ ), so dual bases are $h / 8$ , $f / 4$ , $e / 4$ and the library Casimir is $C = \frac{1}{8} h^{2} + \frac{1}{4} (e f + f e) = \frac{1}{4} Ω$ . To prove that this Casimir generates the center, first $C h$ is a Cartan subalgebra: it is abelian, hence nilpotent, and the displayed brackets show $N_{g} (C h) = C h$ by Cartan subalgebra and Normalizer of a Lie subalgebra. The root spaces are $C e$ , $C f$ with roots $α, - α$ , where $α (h) = 2$ ; therefore the unique root reflection acts by sign on $h^{*}$ , and $S (h)^{W} = C [h^{2}]$ (Root and root space, Root reflections and the Weyl group action). By Chevalley restriction, $S (g)^{g} ≅ S (h)^{W}$ , and the PBW symbol $σ_{2} (C) = \frac{1}{8} h^{2} + \frac{1}{2} e f$ restricts to $h^{2} / 8$ , so it generates $S (g)^{g}$ (Chevalley restriction for symmetric invariants). If $z \in Z (U (g))$ has PBW degree $d > 0$ , the adjoint-symbol identity in The adjoint action preserves the associated graded of a two-sided ideal shows $σ (z)$ is invariant. It is homogeneous; since the invariant ring is the polynomial ring generated by the degree-two element $σ (C)$ , one has $d = 2 k$ and $σ (z) = a σ (C)^{k}$ for some $a \in C^{\times}$ . Thus $z - a C^{k}$ is central of strictly smaller PBW degree. Induction on degree, with degree-zero elements being scalars and $C$ central by the direct commutator calculation above, yields $Z (U (g)) = C [C] = C [Ω]$ . Consequently $\ker χ = (Ω - c)$ . Since $e f = f e + h$ , one has $Ω = 2 f e + h + \frac{1}{2} h^{2}$ in $U (g)$ , hence $f e = \frac{1}{2} (Ω - h - \frac{1}{2} h^{2})$ exactly; in $U_{χ}$ , where $Ω$ acts by $c$ , this reads $f e = p (h) : = \frac{1}{2} (c - h - \frac{1}{2} h^{2})$ and $e f = p (h) + h = : q (h)$ . Equivalently $p (h) = - \frac{1}{4} (h^{2} + 2 h - 2 c)$ and $q (h) = p (h - 2) = - \frac{1}{4} (h^{2} - 2 h - 2 c)$ ; also $h e = e (h + 2)$ and $h f = f (h - 2)$ .

2.1F3step 1.1algebra

The associated graded of $U_{χ}$ for the induced PBW filtration is $C [e, f, h] / (σ)$ , where $σ$ is the symbol of $Ω - c$ , namely $2 e f + \frac{1}{2} h^{2}$ up to a nonzero scalar: the kernel of $U (g) ↠ U_{χ}$ is generated by $Ω - c$ , whose symbol is a nonzerodivisor of the polynomial algebra $C [e, f, h]$ , so an element of that ideal has top-degree part in the ideal $(σ)$ and conversely every element of $(σ)$ occurs as the top-degree part of a product $u (Ω - c) v$ . Multiplying $σ$ by $2$ identifies the quotient with $C [e, f, h] / (h^{2} + 4 e f)$ ; multiplication by $h^{2} + 4 e f$ from degree $d - 2$ to degree $d$ is injective, so the degree- $d$ part has dimension $(\binom{d + 2}{2}) - (\binom{d}{2}) = 2 d + 1$ .

3.1F3step 1.1step 2.1algebra

Let $N$ be the $C$ -span of the normal forms $f^{i} h^{a}$ $(i, a \geq 0)$ and $e^{j} h^{a}$ $(j \geq 1, a \geq 0)$ , filtered by degree $i + a$ , respectively $j + a$ . Left multiplication by the generators preserves $N$ and raises degree by at most one: using $h f = f (h - 2)$ and $h e = e (h + 2)$ from step 1.1 gives $h \cdot f^{i} h^{a} = f^{i} h^{a + 1} - 2 i f^{i} h^{a}$ and $h \cdot e^{j} h^{a} = e^{j} h^{a + 1} + 2 j e^{j} h^{a}$ ; $e \cdot f^{i} h^{a} = f^{i - 1} q (h - 2 i + 2) h^{a}$ for $i \geq 1$ by step 1.1 and $e \cdot h^{a} = e h^{a}$ ; $e \cdot e^{j} h^{a} = e^{j + 1} h^{a}$ ; $f \cdot f^{i} h^{a} = f^{i + 1} h^{a}$ and $f \cdot h^{a} = f h^{a}$ ; $f \cdot e^{j} h^{a} = e^{j - 1} p (h + 2 j - 2) h^{a}$ for $j \geq 1$ . Since $F^{n} U_{χ}$ is spanned by words of length at most $n$ in $e, f, h$ , induction on $n$ gives $F^{n} U_{χ} \subseteq N \cap F^{n} U_{χ}$ , so the normal forms span $U_{χ}$ . Exactly $(n + 1)^{2}$ of them have degree at most $n$ , and their symbols span $(gr U_{χ})_{\leq n}$ , whose dimension is $\sum_{d \leq n} (2 d + 1) = (n + 1)^{2}$ by step 2.1; hence those symbols are a basis of $(gr U_{χ})_{\leq n}$ and, comparing top degrees, the normal forms are linearly independent. They therefore form a $C$ -basis of $U_{χ}$ , so $U_{χ}$ is infinite-dimensional, and the basis exhibits the ${ad}_{h}$ -weight decomposition $W_{2 k} = e^{k} C [h]$ for $k \geq 0$ , $W_{- 2 k} = f^{k} C [h]$ for $k \geq 1$ .

4.1step 3.1algebra

Let $J \neq 0$ be a two-sided ideal of $U_{χ}$ . Since $[h, u] \in J$ for $u \in J$ , and the weight components of $u$ are polynomial combinations of the elements $({ad}_{h})^{i} u$ , the ideal $J$ contains a nonzero weight vector. If it contains $u = e^{k} Q (h)$ with $k \geq 1$ and $Q \neq 0$ , then $u f^{k} = Q (h - 2 k) \prod_{i = 0}^{k - 1} q (h - 2 i) \in J \cap C [h]$ is nonzero, because $e^{k} f^{k} = \prod_{i = 0}^{k - 1} q (h - 2 i)$ and $C [h]$ has no zero divisors and $Q (h - 2 k) \neq 0$ ; if it contains $u = f^{k} Q (h)$ with $k \geq 1$ , then $e^{k} u = Q (h) \prod_{i = 0}^{k - 1} q (h - 2 i)$ is such a nonzero polynomial; and if it contains a nonzero element of weight $0$ , that element already lies in $C [h]$ . Hence $J \cap C [h] \neq 0$ .

5.1F6step 4.1

By [F6] the nonzero ideal $J \cap C [h]$ is generated by a polynomial $P$ , chosen of least degree; if $J \neq U_{χ}$ then $P$ is nonconstant, since a nonzero constant in $J \cap C [h]$ would put $1$ in $J$ and force $J = U_{χ}$ .

6.1step 1.1step 5.1algebra

For $P \in J \cap C [h]$ one has $[e, P] = e P - P e = (P (h - 2) - P (h)) e \in J$ and $[f, P] = (P (h + 2) - P (h)) f \in J$ , because $e P (h) = P (h - 2) e$ and $f P (h) = P (h + 2) f$ . Multiplying by $f$ respectively $e$ and using $f e = p$ , $e f = q = p + h$ gives the divisibilities $P ∣ p (P (h + 2) - P (h))$ and $P ∣ q (P (h - 2) - P (h))$ in $C [h]$ .

7.1step 6.1algebra

Let $r$ be a root of $P$ . If $r + 2$ is not a root of $P$ , then $(P (h + 2) - P (h)) (r) = P (r + 2) - P (r) = P (r + 2) \neq 0$ , so the first divisibility of step 6.1 forces $p (r) = 0$ . If $r - 2$ is not a root of $P$ , then $(P (h - 2) - P (h)) (r) = P (r - 2) \neq 0$ , so the second divisibility forces $q (r) = p (r) + r = 0$ .

8.1step 7.1algebra

Since $P$ has finitely many roots, for every root $r$ the upward chain $r, r + 2, r + 4, \dots$ has a last member $s = r + 2 k$ with $s$ a root of $P$ and $s + 2$ not a root, and the downward chain has a last member $s^{'} = r - 2 l$ with $s^{'} - 2$ not a root. Step 7.1 then gives $p (s) = 0$ and $q (s^{'}) = 0$ , so $s \in {r_{+}, r_{-}}$ and $s^{'} \in {q_{+}, q_{-}} = {r_{+} + 2, r_{-} + 2}$ , where by step 1.1 $r_{\pm} = - 1 \pm t$ with $t^{2} = 1 + 2 c$ are the roots of $p$ , and $r = - 2 k + s = s^{'} + 2 l$ for some $k, l \geq 0$ .

9.1F7step 8.1algebra

If $t \notin Z$ then the two roots $r_{\pm}$ are distinct and not congruent modulo $2$ (their difference is $2 t$ ). Writing any root $r$ as in step 8.1, the identity $s - s^{'} = 2 (k + l) \geq 0$ forces $s - s^{'} \in {- 2, 2 t - 2, - 2 t - 2}$ to be nonnegative and even; the first value gives $k + l = - 1$ ; the second gives $t = 1 + k + l \in Z$ ; the third gives $t = - (1 + k + l) \in Z$ . All are impossible, so $P$ has no roots and [F7] makes it constant, hence $J = U_{χ}$ by step 5.1. Thus $U_{χ}$ is simple whenever $t \notin Z$ , in particular whenever $1 + 2 c$ is not the square of an integer.

9.2F7step 8.1algebra

If $t = 0$ then $p (h) = - \frac{1}{4} (h + 1)^{2}$ has the single root $r_{+} = - 1$ , while $q (h) = p (h - 2) = - \frac{1}{4} (h - 1)^{2}$ has the single root $q_{+} = 1$ ; step 8.1 would give $- 1 - 2 k = r = 1 + 2 l$ , i.e. $2 (k + l) = - 2$ , which is impossible. So $P$ has no roots and [F7] makes it constant, whence $J = U_{χ}$ ; $U_{χ}$ is simple for $c = - \frac{1}{2}$ as well.

10.1step 3.1step 9.1step 9.2F4algebra

Conversely let $t \in Z ∖ {0}$ , and put $n : = ∣ t ∣ - 1 \geq 0$ ; then $c = (t^{2} - 1) / 2 = n (n + 2) / 2$ is the $Ω$ -eigenvalue of the nonzero finite-dimensional simple module $L (n)$ by [F4], so $L (n)$ is a nonzero $U_{χ}$ -module. The action map $φ ⁣ : U_{χ} \to {End}_{C} L (n)$ is a nonzero algebra homomorphism (it sends $1$ to the identity), and its image is finite-dimensional while $U_{χ}$ is infinite-dimensional by step 3.1, so $\ker φ \neq 0$ and $\ker φ \neq U_{χ}$ : $U_{χ}$ has a nonzero proper two-sided ideal and is not simple. Conversely, if a nonzero finite-dimensional $U (g)$ -module has central character $χ$ , choose a nonzero submodule of least positive dimension; it is simple and $Ω$ acts there by $c$ , so [F4] makes it some $L (m)$ and $c = m (m + 2) / 2$ . Thus $χ$ is a finite-dimensional central character exactly when $1 + 2 χ (Ω)$ is a positive square, equivalently when $χ (Ω) = n (n + 2) / 2$ for some $n \geq 0$ (the integer parameter is $n = ∣ t ∣ - 1$ , since $t^{2} - 1 = (∣ t ∣ - 1) (∣ t ∣ + 1)$ ). Together with steps 9.1 and 9.2 this is the asserted criterion, and the eigenvalues $n (n + 2) / 2$ are pairwise distinct by [F4].

11.1step 10.1F2F5∎

Finally suppose $U_{χ}$ is simple, let $N \neq 0$ be a $U_{χ}$ -module, and let $ψ ⁣ : U_{χ} \to {End}_{C} (N)$ be the action. Its kernel is a two-sided ideal and is proper because $N \neq 0$ makes $ψ (1) = {id}_{N} \neq 0$ ; simplicity forces $\ker ψ = 0$ , so $N$ is faithful over $U_{χ}$ . If $χ_{λ} = χ$ for a weight $λ$ then $λ \notin Z_{\geq 0}$ , because otherwise $χ_{λ}$ would be the central character of the finite-dimensional module $L (λ)$ by [F5], contrary to the established criterion; hence $M (λ)$ is simple by [F5]. The annihilator of $M (λ)$ in $U (g)$ is the preimage of $\ker ψ$ for $N = M (λ)$ , which is $\ker (U (g) \to U_{χ}) = U (g) \ker χ$ by the definition of the annihilator as a kernel; therefore ${Ann}_{U (g)} M (λ) = U (g) \ker χ$ , as claimed.

RemarkRemark: Literature-sourcedProof: Not applicableOpen item page →

Highest weights can have the same primitive ideal

Remark

Assume the Axiom of Choice (The Axiom of Choice) and let $g$ be finite-dimensional complex semisimple. The assignment $λ \mapsto I (λ) = {Ann}_{U (g)} L (λ)$ from weights to primitive ideals (Annihilators of simple highest-weight modules are primitive, Primitive ideals of an enveloping algebra, The annihilator of a module over an enveloping algebra) is not injective in general, and a fixed central character can carry more than one primitive ideal; describing the fibres of this map is the content of Joseph's theory of Goldie rank polynomials and of Kazhdan-Lusztig cell theory, which is not part of the algebraic prefix on this page. For noninjectivity, already in ${s l}_{2}$ the distinct weights $1 / 2$ and $- 5 / 2$ have normalized Casimir value $5 / 8$ , which is not $n (n + 2) / 2$ for any integer $n \geq 0$ : those values are $0$ for $n = 0$ and at least $3 / 2$ for $n \geq 1$ . By The central reduction of U(sl2) is simple away from the finite-dimensional central characters, both Verma modules are simple and have the same annihilator, namely the central ideal. In contrast, the trivial central character carries the distinct primitive ideals $Ann L (0)$ and $Ann L (- 2)$ — the annihilators of the trivial module and of the simple Verma module $M (- 2) = L (- 2)$ — so Duflo's surjectivity is not a bijection between weights and primitive ideals.

Remarks

What is recorded here and what is not. This item records a boundary: the map from weights to primitive ideals has fibres of size greater than one, and their description requires the character-polynomial machinery of Joseph, Barbasch and Vogan. No fibrewise classification is asserted, and the item is not used as a supplier by any proof on this page.
The ${s l}_{2}$ witness. The two annihilators named above are distinct: the trivial module is finite-dimensional with $h$ acting by $0$ , while $h$ acts with nonzero eigenvalue $- 2$ on the highest vector of $M (- 2)$ , so the annihilators differ. Simplicity of $M (- 2)$ follows from the ${s l}_{2}$ irreducibility criterion recorded in The central reduction of U(sl2) is simple away from the finite-dimensional central characters: $- 2 \notin Z_{\geq 0}$ . This is the same witness that the companion page records in full; the comparison is by central character, since $χ_{0} = χ_{- 2}$ on ${s l}_{2}$ .

5 · Examples, counterexamples and false statements

None yet.

Primitive Ideals and Duflo Theorem

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

The annihilator of a module over an enveloping algebra

Definition

Remarks

Primitive ideals of an enveloping algebra

Definition

Remarks

Annihilators of simple highest-weight modules are primitive

Statement

Facts & Assumptions

Proof

Primitive ideals are prime in the noncommutative sense

Statement

Facts & Assumptions

Proof

Dixmier's lemma: endomorphisms of a simple module over a countable-dimensional algebra

Statement

Facts & Assumptions

Proof

A primitive ideal determines a central character

Statement

Facts & Assumptions

Proof

The central reduction of the enveloping algebra at a central character

Definition

Remarks

The Verma annihilator contains the central-character ideal

Statement

Facts & Assumptions

Proof

The adjoint action preserves the associated graded of a two-sided ideal

Statement

Facts & Assumptions

Proof

The associated graded variety of a two-sided ideal

Definition

Remarks

The associated variety is a closed conical coadjoint-invariant cone

Statement

Facts & Assumptions

Proof

Remarks

Every central character of a semisimple enveloping algebra arises from a weight

Statement

Facts & Assumptions

Proof

Primitive ideals are partitioned by their dot-orbit central character

Statement

Facts & Assumptions

Proof

The central reduction of U(sl2) is simple away from the finite-dimensional central characters

Statement

Facts & Assumptions

Proof

Highest weights can have the same primitive ideal

Remark

Remarks

5 · Examples, counterexamples and false statements

Sources