Why aren't integrals invariant over coordinate transformations?

Question

I am reading chapter 16 of Introduction to smooth manifolds by Lee and at the beginning of the chapter it is stated that an integral is clearly not invariant under coordinate transformations.

Why is that? It looks to me that it is the opposite I mean for instance if I change from cartesian to polar coordinates the integral should be the same, no matter what coordinates I use

Answer 1

The point here is that if you have an arbitrary manifold $M$ with no additional structure, and a function $f$ supported in a chart $\varphi \colon U \to \mathbb{R}^n$, then you can attempt to define the integral of $f$ as $$ \mathcal{I}(f; \varphi) = \int_{\varphi(U)} f \circ \varphi^{-1}. $$

However, this is not invariant under coordinate transforms, meaning that this "integral" depends on the chosen chart $\varphi$. For example, if $\psi \colon U \to \mathbb{R}^n$ is a compatible chart and $\tau = \psi \circ \varphi^{-1} \colon \varphi(U) \to \psi(U)$ is the transition map, then

$$\mathcal{I}(f; \psi) = \int_{\psi(U)} f \circ \psi^{-1} = \int_{\varphi(U)} f \cdot |\det J_{\tau}|, $$ where $J_{\tau}$ is the Jacobian of $\tau$. Thus, we expect generally that $\mathcal{I}(f; \varphi) \neq \mathcal{I}(f; \psi)$.

Answer 2

The integral introduced in (multivariable) calculus for real-valued functions $f : T \to \mathbb R$ defined on suitable subsets $T \subset \mathbb R^n$ heavily depends on the standard coordinate system of $\mathbb R^n$. See here. Actually the integral $\int_T f dV$ with "volume element $dV$" is approximated by Riemannian sums of the form $$\sum f(\xi_k)V(C_k)$$ where

1. the $C_k = \prod_{i=1}^n [a_i,b_i]$ are cubes covering $T$ and having disjoint interiors,
2. $\xi_k \in C_k$,
3. $V(C_k) = \prod_{i=1}^n(b_i - a_i)$ is the volume of $C_k$.

The crucial point is that the volume of $C_k$ is defined by 3. This appears to be the "obvious" definition of a volume function in this situation.

Let us now change perspective. Consider an $n$-dimensional vector space $W$. How can we define an expression $$\int_T f dV$$ for real-valued functions defined on $T \subset W$? Here we do not have a canonical coordinate system, no canonical cubes, let alone a volume function. We can of course introduce coordinates by choosing a linear isomorphism $\phi : W \to \mathbb R^n$ and then define $$\int_T fdV_\phi = \int_{\phi(T)}(f \circ \phi^{-1})dV .$$ This integral depends on the choice of $\phi$. If we use for example the isomorphism $\psi(x) = 2\phi(x)$, then it is easy to see that $$\int_T fdV_\phi = 2^n\int_T fdV_\psi .$$ This is what Lee means when he says that the integral is "clearly not invariant under coordinate transformations, even if we just restrict attention to linear ones."

The situation for a general manifold $M$ is much more complicated than in the case $M = W$. We only have local homeomorphisms between open subsets of $M$ to open subsets of $\mathbb R^n$ and hence many local coordinate systems which produce many distinct "local volume elements".

This shows that integration on manifolds needs a more sophisticated approach.