Proposition 12

Tag 09G3, permanent. Cite it as [Bourbaki, Tag 09G3].

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

— With the foregoing notations,

\int_{X/H} (\beta/\beta')_u d(\mu/\beta)(u) = \mu/\beta'.

Let $f \in \mathscr{K}(X)$ , and let $f^b \in \mathscr{K}(X/H')$ , defined by

f^b(\pi'(x)) = \int_{H'} f(x\xi') d\beta'(\xi').

It suffices (cf. No. 2) to prove that $f^b$ has the same integral with respect to the two members of (22). Now, $\langle \mu/\beta', f^b \rangle = \langle \mu, f \rangle$ . On the other hand,

\left\langle \int_{X/H} (\beta/\beta')_u d(\mu/\beta)(u), f^b \right\rangle = \int_{X/H} \langle (\beta/\beta')_u, f^b \rangle d(\mu/\beta)(u).

Now, let $x \in X$ and $u = \pi(x)$ . We have

\langle (\beta/\beta')_u, f^b \rangle = \langle \omega_x(\beta/\beta'), f^b \rangle = \int_{H/H'} f^b(\omega_x(\dot{\xi})) d(\beta/\beta')(\dot{\xi})

= \int_{H/H'} f^b(\pi'(x\xi)) d(\beta/\beta')(\dot{\xi})

= \int_{H/H'} d(\beta/\beta')(\dot{\xi}) \int_{H'} f(x\xi\xi') d\beta'(\xi')

= \int_H f(x\xi) d\beta(\xi).

Therefore

\left\langle \int_{X/H} (\beta/\beta')_u d(\mu/\beta)(u), f^b \right\rangle = \int_{X/H} d(\mu/\beta)(u) \int_H f(x\xi) d\beta(\xi) = \langle \mu, f \rangle,

which proves the proposition.

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

Cited by

Exercise 7, § 3

Cites

Nothing.