Bourbaki

Proposition 23

Tag 0AA5, permanent. Cite it as [Bourbaki, Tag 0AA5].

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

(i) The mapping t↦Ltt \mapsto L_t (resp. t↦Rtt \mapsto R_t) is an isomorphism of the vector space U(G)U(G) onto the vector field of left (resp. right) invariant distributions on GG. (ii) For t,t′t, t' in U(G)U(G), Lt∗t′=Lt∘Lt′L_{t*t'} = L_t \circ L_{t'}, Rt∗t′=Rt′∘RtR_{t*t'} = R_{t'} \circ R_t, Lt∘Rt′=Rt′∘LtL_t \circ R_{t'} = R_{t'} \circ L_t (with the abuse of notation of no. 5). (iii) If θ\theta is the mapping g↦g−1g \mapsto g^{-1} of GG onto GG, then θ(Lt)=Rt∨\theta(L_t) = R_{t^\vee}. (iv) If t∈U(G)t \in U(G) and g∈Gg \in G, then (Lt)g=(Rεg∗t∗εg−1)g(L_t)_g = (R_{\varepsilon_g * t * \varepsilon_g^{-1}})_g.

In GG every right translation commutes with every left translation. By Proposition 21 of no. 5, LtL_t is therefore left invariant. As (Lt)e=t(L_t)_e = t, the mapping t↦Ltt \mapsto L_t is injective. Let Δ\Delta be a field of left invariant distributions on GG; let t=Δet = \Delta_e; then Δ\Delta and LtL_t have the same value at ee and are left invariant and hence Δ=Lt\Delta = L_t. This proves (i) for LtL_t and the argument is similar for RtR_t. The formulae Lt∗t′=Lt∘Lt′L_{t*t'} = L_t \circ L_{t'}, Rt∗t′=Rt′∘RtR_{t*t'} = R_{t'} \circ R_t follow from (21) and (18). Let t∈Us(G)t \in U_s(G), t′∈Us′(G)t' \in U_{s'}(G), f∈Cr(Ω,F)f \in C^r(\Omega, F), where Ω\Omega is open in GG and s+s′≤rs + s' \leq r; then

LtRt′f=Lt(t′∨∗f)=(t′∨∗f)∗t=t′∨∗(f∗t∨)=Rt′Ltf \begin{align*} L_t R_{t'} f &= L_t ({t'}^\vee * f) = ({t'}^\vee * f) * t \\ &= {t'}^\vee * (f * t^\vee) \tag{Proposition 20} \\ &= R_{t'} L_t f \end{align*}
and hence Lt∘Rt′=Rt′∘LtL_t \circ R_{t'} = R_{t'} \circ L_t. As θ\theta is an isomorphism of GG onto G∨G^\vee, θ(Lt)\theta(L_t) is a field of right invariant distributions on GG; its value at ee is θ(t)=t∨\theta(t) = t^\vee; hence θ(Lt)=Rt∨\theta(L_t) = R_{t^\vee}. Finally,
(Lt)g=εg∗t=(εg∗t∗εg−1)∗εg=(Rεg∗t∗εg−1)g. (L_t)_g = \varepsilon_g * t = (\varepsilon_g * t * \varepsilon_g^{-1}) * \varepsilon_g = (R_{\varepsilon_g * t * \varepsilon_g^{-1}})_g.

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.