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Let $F$ be a local field of characteristic 0, and $f:F\rightarrow \mathbb{C}$ be an integrable function. Is the following formulation valid?

$ \int_{F^\times}f(x^2) d^\times x=\int_{F^{\times 2}}f(x) d^\times x $

where $d^\times x$ is a chosen multiplicative measure on $F^\times$.

It seems if $F=\mathbb{R}$, this is true. But is it true for any local field?

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It seems I didn't state the question properly. First of all, I had to make sure what the measure on $F^{\times 2}$ is. If we simply choose the measure to be the restriction of the measure on $F^\times$. This doesn't seem to be true. For example if $F=\mathbb{C}$, then $F^{\times 2}=F^\times$ and so it shouldn't be true. –  Windi Jun 1 '13 at 20:57

3 Answers 3

up vote 2 down vote accepted

Since the subgroup of squares is open, the restriction of Haar measure from the full multiplicative group to the squares is a Haar measure on the subgroup of squares, and Haar measure is unique up to scalars.

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This is false for non-archimedean local fields. On the other hand, as Paul Garrett explained, the two sides are equal up to an absolute constant.

Let $F=\mathbb{Q}_2$, and let $f$ be the characteristic function of $1+8\mathbb{Z}_2$. Then the left hand side is the measure of $1+2\mathbb{Z}_2$, while the right hand side is the measure of $1+8\mathbb{Z}_2$, hence they differ by a factor of $4$.

Let $F=\mathbb{Q}_p$ for $p>2$, and let $f$ be the characteristic function of $1+p\mathbb{Z}_p$. Then the left hand side is the measure of $\cup(\pm 1+p\mathbb{Z}_p)$, while the right hand side is the measure of $1+p\mathbb{Z}_p$, hence they differ by a factor of $2$.

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Isn't the only discrepancy a uniform constant? –  paul garrett Jun 1 '13 at 19:43
    
@paul: Yes, the discrepancy is a uniform constant. And my response was also slightly in error, so let me update it. –  GH from MO Jun 1 '13 at 20:55

I think for an appropriate choice of the measure on $F^{\times 2}$, I still believe this is true, which is essentially what Paul Garret mentioned. Indeed $F^{\times 2}\cong F^{\times}/ \{\pm1\}$, and the function $x↦f(x^2)$ can be viewed as a function on this quotient. Then \begin{equation} \int_{F^\times}f(x^2)d^\times x=\int_{F^{\times}/{\pm 1}}\sum_{\{\pm 1\}}f(y^2)d^\times y \end{equation}

where $d^\times y$ is the quotient measure. Then by identifying $F^{\times}/{±1}$ with $F^{\times 2}$, one can see that the left hand side is 2 times $\int_{F^{\times 2}}f(y)d^\times y$.

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Yes, but this is not what you originally asked. Your question involved $F^{\times 2}$ as a subset of $F^\times$ (not just as an abstract group), hence the correction factor you need (in your original equation) will also depend on the index $(F^\times:F^{\times 2})$. This, on the other hand, depends on whether $2$ is a unit in the maximal compact subring of $F$ or not. Read my response. –  GH from MO Jun 1 '13 at 21:45

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