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Let $M$ be a smooth manifold with volume measure $\operatorname{vol}$. And let $T:M\to M$ be a (global) diffeomorphism.

I know by change of variables formula that $$\int_M f(T(x))\operatorname{vol}(dx)=\int_M f(x)\mathcal{J}_{T^{-1}}(x)\operatorname{vol}(dx);$$ with $$\mathcal{J}_{T^{-1}}(x)=\lim_{r\to 0^+}\frac{\operatorname{vol}(T^{-1}(B_r(x)))}{\operatorname{vol}(B_r(x))}.$$

But since $T$ is a diffeomorphism, the mean of the Jacobian $$\int_M \mathcal{J}_{T^{-1}}(x)\operatorname{vol}(dx)=\int_M\operatorname{vol}(dx),$$ since $T_*\mu(M)=\mu(T^{-1}(M))=\mu(M)$.

Intuitively, I think that if $T$ streches some regions of $M$ it must compress other regions to compensate the volumes.

With that said, my geometric intuition leads me to think that for every integrable function $f: M\to \mathbb{R}$ do we have $\int_M f(T(x))\operatorname{vol}(dx)=\int_M f(x)\operatorname{vol}(dx)$.

On the other hand, thinking analytically, a sequence of measures in the weak* topology converges to a given measure if the integrals of bounded continuous functions converges. So it seems to me that as the topological space of Borel measures with the weak* toplogy is Hausdorff then this equality is wrong.

Why is my geometrical intuition wrong?

Gomes93
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    btw the determinant you speak of doesn’t really make sense, because the tangent map $Tf_p:T_pM\to T_{f(p)}N$ is a linear map between different vector spaces (even if you set $M=N$ and assume $f$ is a diffeomorphism). The correct Jacobian term you’re after is a Radon-Nikodym derivative of two measures. See here for a precise statement of the change-of-variables theorem on semi-Riemannian manifolds. You could of course also assume $M$ is orientable and oriented and formulate everything using differential forms. – peek-a-boo Aug 21 '24 at 19:32
  • @peek-a-boo consider then $M=\mathbb{R}^n$ to make sense of this Jacobian. Then the Radon-Nikodym derivative is the Jacobian of the inverse when $\mu$ is the Lebesgue measure. – Gomes93 Aug 21 '24 at 19:43
  • Moerover, one could change it by the Jacobian defined as the the limit of the ratio $\mu(T(B_r(x))}/\mu(B_r(x))$ as $r\to 0$. I'll edit the question to make it right. – Gomes93 Aug 21 '24 at 19:49
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    yes of course in $\Bbb{R}^n$ this reduces to the usual Jacobian determinant. My first comment was merely to point out that writing $\det$ abstractly like this doesn’t make sense (unless we redefine what we mean by $\det$ in some fashion so as to to make things work out). Also, I think the place you’re going wrong is that $(T_*\mu)(M)=\mu(M)$ is indeed true (and in many practical applications both sides are $\infty$), but in order for $T$ to preserve volumes, this equation has to hold for all (say Borel) sets $A\subset M$, not just for $M$. – peek-a-boo Aug 21 '24 at 19:55

1 Answers1

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My intuition was wrong because the mean of the Jacobian compensates on the whole $M$. But the volume of the support of $f$ may be changed by $T$.

For example, if $M= \mathbb{R}$ and $\operatorname{vol}$ is simply the Lebesgue measure, $T(x)=x/2$ and $f(x)=1_{[0,1]}$ is the characteristic function of the set $[0,1]$, then $f\circ T= 1_{[0,2]}$. Thus, $\int_\mathbb{R} f(T(x))dx=2\int_\mathbb{R}f(x)dx$.

Gomes93
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    This still isn’t the whole issue. The support of $f$ could be left the same (it could even be all of $\Bbb{R}$), but the integral might still be different. The issue really is that $T$ needs to preserve volumes (of ‘every’ set). – peek-a-boo Aug 21 '24 at 20:30