Proposition 15

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

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

— Assume K to be commutative. Let u be an automorphism of the vector space E = K^n. Then

\mathrm{mod}_E u = \mathrm{mod}_K(\det u).

It suffices to verify the formula when $u$ runs over a system of generators of $\mathbf{GL}(E)$ . Now, $\mathbf{GL}(E)$ is generated by the following elements (A, II, §10, No. 13, Cor. 2 of Prop. 14): (a) The elements $u_1$ of the form

(x_1, \ldots, x_n) \mapsto (x_{\sigma(1)}, \ldots, x_{\sigma(n)}),

where $\sigma \in \mathfrak{S}_n$ ; (b) the elements $u_2$ of the form

(x_1, \ldots, x_n) \mapsto (ax_1, x_2, \ldots, x_n)

with $a \in K^*$ ; (c) the elements $u_3$ of the form

(x_1, \ldots, x_n) \mapsto (x_1 + \sum_{i=2}^n c_i x_i, x_2, \ldots, x_n).

If $f \in \mathscr{H}(E)$ then, denoting by $\alpha$ a Haar measure on $K^+$ ,

\int \cdots \int_{K^n} f(x_1 + \sum_{i=2}^n c_i x_i, x_2, \ldots, x_n) d\alpha(x_1) d\alpha(x_2) \ldots d\alpha(x_n)

= \int \cdots \int_{K^{n-1}} d\alpha(x_2) \ldots d\alpha(x_n) \int_K f(x_1 + \sum_{i=2}^n c_i x_i, x_2, \ldots, x_n) d\alpha(x_1)

= \int \cdots \int_{K^{n-1}} d\alpha(x_2) \ldots d\alpha(x_n) \int_K f(x_1, x_2, \ldots, x_n) d\alpha(x_1)

= \int \cdots \int_{K^n} f(x_1, \ldots, x_n) d\alpha(x_1) \ldots d\alpha(x_n),

and on the other hand $\mathrm{mod}_K(\det u_3) = \mathrm{mod}_K(1) = 1$ , whence the result for $u_3$ . It is established in an analogous manner for $u_1$ and $u_2$ .

Let K be a commutative locally compact field, E a vector space of finite dimension n over K. If $\varphi$ is an isomorphism of the vector space $K^n$ onto the vector space E, $\varphi$ transforms the topology of $K^n$ into a topology on E that makes E a locally compact vector space. This topology (called canonical) is independent of $\varphi$ since every automorphism of the vector space $K^n$ is bicontinuous. Absent express mention to the contrary, when we speak of E as a topological vector space, the topology will always be understood to be the one just defined. Every automorphism u of the vector space E is bicontinuous, therefore $\mathrm{mod}_E u$ is defined. If, on the other hand, u is a noninvertible endomorphism of E, one sets $\mathrm{mod}_E u = 0$ . Then:

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

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Cites

Prop. 14