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Enveloping quotient kernels and augmentation intersections

Statement

For countably based complex Lie algebras with supplied compatible bases, a surjection $θ : L \to L / R$ with ideal kernel $R$ induces $\ker U (θ) = R U (L)$ . For any subalgebra $R \subset L$ with such a compatible basis, $R \cap R U_{0} (L) = [R, R]$ . Here $U_{0} (L)$ is the kernel of the augmentation $U (L) \to C$ . These hypotheses hold for the homogeneous subalgebras used in this page by finite-degree elimination.

Facts & Assumptions

Given: The indicated bases; R is an ideal for the first claim and only a subalgebra for the second.

[F1]

PBW gives the compatible ordered monomial bases. (PBW for countably presented Kac Moody Lie algebras).

[F2]

The tensor quotient realizes Lie homomorphisms as associative homomorphisms. (The universal enveloping algebra as a tensor quotient).

Proof

1.1

If $R$ is an ideal, $R U (L)$ is two-sided: $x r = r x + [x, r]$ with $[x, r] \in R$ lets every left generator pass it. It is killed by $U (θ)$ . The class of $x$ in $U (L) / R U (L)$ depends only on $x + R$ , giving a Lie map $L / R \to U (L) / R U (L)$ . F2 extends it to an associative inverse of the map $U (L) / R U (L) \to U (L / R)$ : both composites fix all Lie generators. This proves the kernel equality.

F2given

1.2

For the subalgebra $R$ , order its basis before a complement and let $W$ span the nonempty ordered complement monomials. Multiplication and F1 identify $U (L) = U (R) \oplus (U (R) \otimes W)$ as left $U (R)$ -modules. Augmentation then gives $U_{0} (L) = U_{0} (R) \oplus (U (R) \otimes W)$ . Nonempty words yield $R U (R) = U_{0} (R)$ and $R U_{0} (R) = U_{0} (R)^{2}$ : in a product of two nonempty words the first letter lies in $R$ , and the remaining word is nonempty, and conversely. Left multiplication by $R$ therefore yields $R U_{0} (L) = U_{0} (R)^{2} \oplus (U_{0} (R) \otimes W)$ .

F1F2given

2.1

For any algebra $K$ with these bases, $[K, K] \subset K \cap U_{0} (K)^{2}$ because $[x, y] = x y - y x$ . Mapping to $U (K / [K, K])$ sends $U_{0} (K)^{2}$ into the square of its augmentation ideal. This enveloping algebra is the polynomial algebra on a basis of the abelian quotient by F1: ordered words commute and have independent monomials. Its degree-one subspace has zero intersection with the ideal of polynomials of degree at least two. Thus an element of $K \cap U_{0} (K)^{2}$ maps to zero in $K / [K, K]$ , proving equality.

F1F2step 1.1

3.1

The subspace $R$ lies in the first summand of step 1.2. Intersecting gives $R \cap R U_{0} (L) = R \cap U_{0} (R)^{2} = [R, R]$ by step 2.1. This calculation retains commutators that can have PBW length one; it makes no false assertion that the augmentation square has only ordered monomials of length at least two. In the homogeneous applications, finite-degree echelon bases of F1 supply all compatible bases used above.

F1step 2.1step 1.2∎

Sources

Source comparison: Kleshchev, Lemmas 9.3.1–9.3.3, pp.122–124; corrected left U(R)-module proof for Lemma 9.3.3.

Depends on

Used by

Kac moody relation module embeds in verma modules and obeys the casimir constraint Lemma

Dependency tree · one level

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Sources

Kleshchev, Lectures on Infinite Dimensional Lie Algebras — Lemmas 9.3.1–9.3.3, pp.122–124; corrected left U(R)-module proof for Lemma 9.3.3 (standard reference, not scraped)