Bourbaki

Proposition 16

Tag 09E1, permanent. Cite it as [Bourbaki, Tag 09E1].

Integration, chapter VII, § 1. Construction of a Haar measure

— Let α\alpha be a Haar measure of the additive group of AA. The measures

(modKNA/K(a))−1dα(a),(modKNA0/K(a))−1dα(a) (\mathrm{mod}_K N_{A/K}(a))^{-1} d\alpha(a), \quad (\mathrm{mod}_K N_{A^0/K}(a))^{-1} d\alpha(a)

on A∗A^* are, respectively, left and right Haar measures of the multiplicative group A∗A^*.

Let α′\alpha' be the restriction of α\alpha to the open set A∗A^*. For a∈A∗a \in A^*, one has La(α′)=(modKNA/K(a))−1α′L_a(\alpha') = (\mathrm{mod}_K N_{A/K}(a))^{-1} \alpha', therefore (modKNA/K(a))−1dα′(a)(\mathrm{mod}_K N_{A/K}(a))^{-1} d\alpha'(a) is a left Haar measure on A∗A^* (No. 8, Cor. 1 of Prop. 10). Passing to the opposite algebra, one sees that (modKNA0/K(a))−1dα′(a)(\mathrm{mod}_K N_{A^0/K}(a))^{-1} d\alpha'(a) is a right Haar measure on A∗A^*.

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.