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Let $F$ be a local field, then the additive Haar measure is easy to relate to the multiplicative Haar measure:

$$ \int_F f(x) d^+ x = \int_{F^\times} f(x) |x| d^+ x.$$

I know that the ideles have zero measure in the adeles, so there is no way in comparing the additive Haar measure with the multiplicative measure of the ring of adeles.

Can we relate the measure on the compact quotients $F^\times \backslash \mathbb{A}^1$ and $F \backslash \mathbb{A}$?

Strong approximation comes to my mind, but I am not sure how to apply it in a reasonable manner. Any suggestions?

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I don't know why you need this, but you could try thinking of the ideles as a hyperbola in the square of the adeles. In a local sense it is not very hard to see the relationshop of the measures. –  Charles Matthews Mar 10 '12 at 12:21
    
I do not get, what you mean by hyperbolas in the squares? Could you please elaborate or indicate, where this idea originates from? $x \mapsto x^2$ is also not well behaved in resiude characteristic $2$. Of course, if I work in characteristic zero, then this is no problem, but for positive characteristic.... –  plusepsilon.de Mar 10 '12 at 17:04
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pm, Charles might mean that the ideles are the solution to the "hyperbola" $xy=1$ in $\mathbb A\times\mathbb A$. This is where you get the topology of the ideles from; I don't know if it helps in relating the measures on the quotients (because I haven't thought about it). –  B R Mar 10 '12 at 18:10
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Weil's book "Basic Number Theory" is a good reference for such questions. –  GH from MO Mar 13 '12 at 22:00
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1 Answer

up vote 4 down vote accepted

I think it is slightly misleading to compare the additive and multiplicative Haar measures on local fields, although it is possible. Less tangibly, but more indicative of the nature of the situation, is the fact that the multiplicative group of a non-archimedean local field has a unique maximal compact (and open) subgroup, the local units, which (correctly) suggests the normalization of the Haar measure, to give the local units measure 1. In contrast, the additive group of a non-archimedean local field is an ascending union of compact open subgroups.

The story of the measure on the compact A/k is perhaps the story that the absolutely ramified non-archimedean places have their additive measure normalized in a certain way depending on the local "different", and because the global different is the product of the local, and the discriminant-squared is the ideal-norm of the different, amazingly the inherited measure on the quotient A/k is 1.

Rather than choose fundamental domains, the measure on a quotient $G/H$ of abelian topological groups is completely determined by the measure on $G$ and that on $H$, by $\int_G f = \int_{G/H} \int_H f(gh)\,dh\;dg$ for compactly supported continuous $f$.

The measure on the ideles is determined by the local measures everywhere, which at finite places give the local units measure 1. Then $J/k^\times$ has a uniquely-determined measure determined by counting measure on $k^\times$ and the measure on $J$. Then $(J/k^\times)/(J^1/k^\times)\approx (0,+\infty)$ is given the usual measure $dx/x$ via that natural isomorphism. By the relation of measures, as above, this uniquely specifies the measure on $J^1/k^\times$.

It is a further exercise to show that with this canonically defined measure the total measure of $J^1/k^\times$ is the usual $2^{r_1}(2\pi)^{r_2}hR/D^{1/2}w$.

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Okay, if there was such a formula, the class number formula obviously would need to play the role of a scaling factor. Actually, a possible integral formula should have simplified a computation involving intertwiner of Eisensteinseries. Your (topological) arguments bet against such a formula. Thanks. –  plusepsilon.de Mar 13 '12 at 8:14
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