Bourbaki

Proposition 14

Tag 09G8, permanent. Cite it as [Bourbaki, Tag 09G8].

Integration, chapter VII, § 2. Quotient of a space by a group; homogeneous spaces

— Suppose that the conditions of Prop. 13 are fulfilled and that, in addition, Y is normal.

a) The restriction of μG\mu_G to Ω\Omega is, up to a constant factor, the image of μX⊗μY\mu_X \otimes \mu_Y under the mapping (x,y)↦xy(x, y) \mapsto xy of X×YX \times Y onto Ω\Omega.

b) For x∈Xx \in X and y∈Yy \in Y,

ΔG(xy)=ΔX(x)ΔY(y)mod⁡(ix), \Delta_G(xy) = \Delta_X(x)\Delta_Y(y) \operatorname{mod}(i_x),

where ixi_x denotes the automorphism v↦x−1vxv \mapsto x^{-1}vx of Y.

We have ΔG=ΔY\Delta_G = \Delta_Y on Y (Prop. 10 b)), therefore a) follows from (23). Let x0∈X, y0∈Yx_0 \in X,\ y_0 \in Y. Denote by pp the mapping (x,y)↦xy(x, y) \mapsto xy of X×YX \times Y onto Ω\Omega. Since

xy(x0y0)−1=xx0−1(x0yy0−1x0−1)=xx0−1ix0−1(yy0−1), xy(x_0y_0)^{-1} = xx_0^{-1}(x_0yy_0^{-1}x_0^{-1}) = xx_0^{-1}i_{x_0^{-1}}(yy_0^{-1}),

we have

ΔG(x0y0)p(μX⊗μY)=δ(x0y0)p(μX⊗μY)=p(δ(x0)μX⊗ix0−1δ(y0)μY)=p(ΔX(x0)μX⊗ΔY(y0)(mod⁡ ix0)μY)=ΔX(x0)ΔY(y0)(mod⁡ ix0)p(μX⊗μY), \begin{align*} \Delta_G(x_0y_0)p(\mu_X \otimes \mu_Y) &= \delta(x_0y_0)p(\mu_X \otimes \mu_Y) \\ &= p(\delta(x_0)\mu_X \otimes i_{x_0^{-1}}\delta(y_0)\mu_Y) \\ &= p(\Delta_X(x_0)\mu_X \otimes \Delta_Y(y_0)(\operatorname{mod}\ i_{x_0})\mu_Y) \\ &= \Delta_X(x_0)\Delta_Y(y_0)(\operatorname{mod}\ i_{x_0})p(\mu_X \otimes \mu_Y), \end{align*}

whence b).

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

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Cites

Transcribed from the printed volume. Not translated.