Bourbaki

Proposition 54

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

Lie Groups and Lie Algebras, chapter III, § 3. Passage from a Lie group to its Lie algebra

Let G be a Lie group of finite dimension n, ω a left invariant differential form of degree n on G and φ an endomorphism of G. Then

ϕ∗(ω)=(det⁡L(ϕ))ω. \phi^*(\omega) = (\det L(\phi)) \omega.

We write L(ϕ)=uL(\phi) = u, ωe=f\omega_e = f and ϕ∗(ω)e=g\phi^*(\omega)_e = g. For all x1,…,xnx_1, \ldots, x_n in L(G)L(G),

g(x1,…,xn)=f(ux1,…,uxn)=(det⁡u)f(x1,…,xn) g(x_1, \ldots, x_n) = f(ux_1, \ldots, ux_n) = (\det u)f(x_1, \ldots, x_n)

and hence ϕ∗(ω)e=det⁡L(ϕ)⋅ωe\phi^*(\omega)_e = \det L(\phi) \cdot \omega_e. On the other hand, if g∈Gg \in G,

ϕ∘γ(g)=γ(ϕ(g))∘ϕ \phi \circ \gamma(g) = \gamma(\phi(g)) \circ \phi

and hence γ(g)∗ϕ∗(ω)=ϕ∗(ω)\gamma(g)^*\phi^*(\omega) = \phi^*(\omega). Thus ϕ∗(ω)\phi^*(\omega) is left invariant, whence the proposition.

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

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Transcribed from the printed volume. Not translated.