On an $n$-dimensional real manifold $M$, we cannot integrate functions in a chart-independent way. Rather, suppose we have a chart $(\varphi, U)$ and an $n$-form $\omega=f\;dx^1\wedge\ldots\wedge dx^n$ with compact support in $U$. Then we define its integral as

$$ \int_U \omega = \int_{\varphi(U)} (\varphi^{-1})^{\ast} \omega = \int_{\varphi(U)} f(x) \;dx^1\ldots dx^n$$

which ends up being coordinate independent. Explicitly, this happens because when changing variables, we get a jacobian factor which compensates the change in differential area:

$$ dx^\mu = \frac{\partial x^\mu}{\partial \tilde{x}^\nu}d\tilde x^\nu \implies dx^1\wedge\ldots\wedge dx^n = \det\left(\frac{\partial x}{\partial \tilde x}\right) d\tilde x^1\wedge\ldots\wedge d\tilde x^n$$

On a Riemannian manifold we can integrate functions. Since we have a canonical volume form $v=\sqrt{g}\;dx^1\wedge\ldots\wedge dx^n$, we can define $\int fv$. However, here the argument is different; the metric tensor components change as

$$ g_{\mu\nu} = \frac{\partial \tilde x^\rho}{\partial x^\mu} \frac{\partial \tilde x^\sigma}{\partial x^\nu}\;\tilde{g}_{\rho\sigma}\implies \sqrt{g} = \det\left(\frac{\partial \tilde x}{\partial x}\right)\sqrt{\tilde g}$$

So the determinants from changing charts cancel each other, and $v$ is a well-defined (not chart-dependent) object, since we have

$$ \sqrt{g}\;dx^1\wedge\ldots\wedge dx^n = \sqrt{\tilde g}\;d\tilde x^1\wedge\ldots\wedge d\tilde x^n$$

However, now our integration loses its jacobian factor, which is what we wanted in the first place. Why is the integral still chart independent?


I think I figured it out. When changing variables while integrating, the jacobian doesn't disappear; it is just a happy coincidence that when changing variables, the new functional form of the integral (almost) coincides with the old one

$$ \int f(x)\sqrt{g}\;dx^1\ldots dx^n = \int f(x(\tilde x))\sqrt{g}\;\det\left(\frac{dx}{d\tilde{x}}\right)d\tilde x^1 \ldots d\tilde x^n = \int\tilde{f}(\tilde{x})\sqrt{\tilde{g}}\;d\tilde x^1\ldots d\tilde x^n$$

where $\tilde f(\tilde x) = f(x(\tilde x))$ (the reason I wrote "almost"). So the jacobian factor is not lost; rather, it is absorbed by notation.


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