Bourbaki

Proposition 2

Tag 0A1T, permanent. Cite it as [Bourbaki, Tag 0A1T].

Lie Groups and Lie Algebras, chapter II, § 1. Enveloping bigebra of a Lie algebra

We have

(4)

(πu⊗πu)∘c=cu+∘πu. (\pi_u \otimes \pi_u) \circ c = c_u^+ \circ \pi_u.

Let xx be in EE; then

(πu∘πu)(c(x))=((1−ηu)⊗(1−ηu))(c(x))=c(x)−(1⊗ηu)(c(x))−(ηu⊗1)(c(x))+(ηu⊗ηu)(c(x)). \begin{align*} (\pi_u \circ \pi_u)(c(x)) &= ((1 - \eta_u) \otimes (1 - \eta_u))(c(x)) \\ &= c(x) - (1 \otimes \eta_u)(c(x)) - (\eta_u \otimes 1)(c(x)) + (\eta_u \otimes \eta_u)(c(x)). \end{align*}
As ε\varepsilon is counit of EE,
(1⊗ηu)(c(x))=x⊗u,(ηu⊗1)(c(x))=u⊗x (1 \otimes \eta_u)(c(x)) = x \otimes u, \qquad (\eta_u \otimes 1)(c(x)) = u \otimes x
whence
(ηu⊗ηu)(c(x))=(ηu⊗1)((1⊗ηu)(c(x)))=ε(x).u⊗u; (\eta_u \otimes \eta_u)(c(x)) = (\eta_u \otimes 1)((1 \otimes \eta_u)(c(x))) = \varepsilon(x).u \otimes u;
from this we conclude
(πu⊗πu)(c(x))=c(x)−x⊗u−u⊗x+ε(x).u⊗u. (\pi_u \otimes \pi_u)(c(x)) = c(x) - x \otimes u - u \otimes x + \varepsilon(x).u \otimes u.
On the other hand,
cu+(πu(x))=c(x)−x⊗u−u⊗x+ε(x).u⊗u, c_u^+(\pi_u(x)) = c(x) - x \otimes u - u \otimes x + \varepsilon(x).u \otimes u,
whence formula (4).

As E+E^+ is a direct factor submodule of EE, E+⊗E+E^+ \otimes E^+ can be identified with a direct factor submodule of E⊗EE \otimes E. With this identification, πu⊗πu\pi_u \otimes \pi_u is a projector of E⊗EE \otimes E onto E+⊗E+E^+ \otimes E^+. By formula (4), cu+c_u^+ maps E+E^+ into E+⊗E+E^+ \otimes E^+ and πu\pi_u is a morphism of the cogebras (E,c)(E, c) into the cogebras (E+,cu+)(E^+, c_u^+).

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

Cites

Nothing.

Transcribed from the printed volume. Not translated.