Contravariant form on a triangular module

Let $𝔤 = 𝔫 - \oplus 𝔥 \oplus 𝑛 +$ be a triangular Lie algebra over $𝕂$ with an involutive antiautomorphism $𝜔 : 𝔤 𝐨 𝐩 \to 𝔤$ so that

𝜔 ([𝑥, 𝑦]) = [𝜔 (𝑦), 𝜔 (𝑥)] 𝜔 𝔥 = 𝔥 𝜔 𝔫 \pm = 𝔫 \pm 𝜔 2 = 1

and let $𝜆 : 𝔥 \to 𝕂$ be an $𝜔$ -invariant¹ linear form. Then the corresponding Triangular module $𝑀 (𝜆)$ has a ^symmetric unique contravariant form, a bilinear form $𝑏 : 𝑀 (𝜆) \times 𝑀 (𝜆) \to 𝕂$ satisfying lie

$𝑏 (𝑥 ⊙ 𝑣, 𝑤) = 𝑏 (𝑣, 𝜔 (𝑥) ⊙ 𝑤)$ for all $𝑥 \in 𝔤$ and $𝑣, 𝑤 \in 𝑀 (𝜆)$
$𝑏 (𝑣 𝜆, 𝑣 𝜆) = 1$

Proof

Note that $𝜔$ extends to an involutive antiautomorphism of the universal enveloping algebra $𝜔 : 𝑈 (𝔤) 𝐨 𝐩 \to 𝑈 (𝔤)$ so that
$𝜔 (𝑥 𝑦) = 𝜔 (𝑦) 𝜔 (𝑥)$
for $𝑥, 𝑦 \in 𝑈 (𝔤)$ .

First we prove uniqueness of $𝑏$ . Clearly ^F1 extends to
$𝑏 (𝑥 ⊙ 𝑣, 𝑤) = 𝑏 (𝑣, 𝜔 (𝑥) 𝑤)$
for $𝑥 \in 𝑈 (𝔤)$ and $𝑣, 𝑤 \in 𝑀 (𝜆)$ .

Note by the Poincaré-Birkhoff-Witt theorem
$𝑈 (𝔤) = 𝖵 𝖾 𝖼 𝗍 𝕂 𝑈 (𝔫 -) \otimes 𝑈 (𝔥) \otimes 𝑈 (𝔫 +) = 𝖵 𝖾 𝖼 𝗍 𝕂 (𝕂 \oplus 𝔫 - 𝑈 (𝔫 -)) \otimes 𝑈 (𝔥) \otimes (𝕂 \oplus 𝑈 (𝔫 +) 𝔫 +) = 𝖵 𝖾 𝖼 𝗍 𝕂 𝔫 - 𝑈 (𝔫 -) \otimes 𝑈 (𝔥) \otimes 𝑈 (𝔫 +) 𝔫 + \oplus 𝔫 - 𝑈 (𝔫 -) \otimes 𝑈 (𝔥) \oplus 𝑈 (𝔥) \otimes 𝑈 (𝔫 +) 𝔫 + \oplus 𝑈 (𝔥) = 𝖵 𝖾 𝖼 𝗍 𝕂 (𝔫 - 𝑈 (𝔫 -) 𝑈 (𝔥) 𝑈 (𝔫 +) + 𝔫 - 𝑈 (𝔫 -) 𝑈 (𝔥) + 𝑈 (𝔥) 𝑈 (𝔫 +) 𝔫 +) \oplus 𝑈 (𝔥) = 𝖵 𝖾 𝖼 𝗍 𝕂 (𝔫 - 𝑈 (𝔤) + 𝑈 (𝔤) 𝔫 +) \oplus 𝑈 (𝔥)$
so we may define the projection operator
$𝑃 : 𝑈 (𝔤) \to 𝑈 (𝔥) = 𝖴 𝖠 𝗌 𝖠 𝗅 𝗀 𝕂 𝑆 ∙ (𝔥)$
Given any $𝑣, 𝑤 \in 𝑀 (𝜆)$ by irreducibility
$𝑣 = 𝑥 ⊙ 𝑣 𝜆 𝑤 = 𝑦 ⊙ 𝑣 𝜆$
for some $𝑥, 𝑦 \in 𝑈 (𝔤)$ so
$𝑏 (𝑣, 𝑤) = 𝑏 (𝑥 ⊙ 𝑣 𝜆, 𝑦 ⊙ 𝑣 𝜆) = 𝑏 (𝑣 𝜆, 𝜔 (𝑥) 𝑦 ⊙ 𝑣 𝜆)$
Since $𝑣 𝜆$ is a vacuum vector and
$𝑏 (𝑣 𝜆, 𝔫 - 𝑈 (𝔤) ⊙ 𝑣 𝜆) = 𝑏 (𝔫 + ⊙ 𝑣 𝜆, 𝑈 (𝔤) ⊙ 𝑣 𝜆) = 0$
it follows
$𝑏 (𝑣, 𝑤) = 𝑏 (𝑣 𝜆, 𝜔 (𝑥) 𝑦 ⊙ 𝑣 𝜆) = 𝑏 (𝑣 𝜆, 𝑃 (𝜔 (𝑥) 𝑦) ⊙ 𝑣 𝜆) = 𝑏 (𝑣 𝜆, 𝜆 (𝑃 (𝜔 (𝑥) 𝑦)) 𝑣 𝜆) = 𝜆 (𝑃 (𝜔 (𝑥) 𝑦))$
Therefore the behaviour of $𝑏$ is completely determined by properties ^F1 and ^F2, so if $𝑏$ exists it is unique.

To prove existence, consider the annihilator of $𝑣 𝜆$
$𝔍 = 𝑈 (𝔤) (𝔫 + + \sum ℎ \in 𝔥 𝕂 (ℎ - 𝜆 (ℎ) 1))$
which is a left-ideal $𝑈 (𝔤)$ . Thus
$𝜑 : 𝑈 (𝔤) / 𝔍 \to 𝑀 (𝜆) 𝑥 + ℑ \mapsto 𝑥 ⊙ 𝑣 𝜆$
is a $𝔤$ -module isomorphism. Now
$𝑃 (𝑥 𝑦) = 𝑃 (𝑥) 𝑦 𝑃 (𝑦 𝑥) = 𝑦 𝑃 (𝑥)$
for $𝑥 \in 𝑈 (𝔤)$ and $𝑦 \in 𝑈 (𝔥)$ , whence
$𝜆 (𝑃 (𝔍)) = 𝜆 (𝑃 (𝜔 (𝔍))) = 0$
Thus the above formula
$𝑏 (𝑥 ⊙ 𝑣 𝜆, 𝑦 ⊙ 𝑣 𝜆) = 𝜆 (𝑃 (𝜔 (𝑥) 𝑦))$
for $𝑥, 𝑦 \in 𝑈 (𝔤)$ is well-defined, since
$𝜔 (𝑥 𝑦) 𝑧 = 𝜔 (𝑦) 𝜔 (𝑥) 𝑧 𝜆 (𝑃 (𝜔 (1) 1)) = 1$
for $𝑥, 𝑦, 𝑧 \in 𝑈 (𝔤)$ . Therewithal since
$𝑃 \circ 𝜔 = 𝜔 \circ 𝑃$
it follows
$𝜆 (𝑃 (𝜔 (𝑦) 𝑥)) = 𝜆 (𝑃 (𝜔 (𝜔 (𝑥) 𝑦))) = 𝜆 (𝜔 (𝑃 (𝜔 (𝑥) 𝑦))) = 𝜆 (𝑃 (𝜔 (𝑥) 𝑦))$
for $𝑥, 𝑦 \in 𝑈 (𝔤)$ , so $𝑏$ is ^symmetric.

See also the special case of a Hermitian contravariant form on a complex triangular module.

tidy | en | SemBr

i.e. $𝜆 𝜔 ℎ = 𝜆 ℎ$ for any $ℎ \in 𝔥$ . ↩

Contravariant form on a triangular module

Graph View

Backlinks

Contravariant form on a triangular module

Footnotes

Graph View

Backlinks