Invariance of measure

Question

I was trying to speak about multidimensional integration with a friend with a no advanced knowledge about differential geometry, when I sumbled to a basic problem. How to convince someone that the determinant of the Jacobian is the right factor to add in a change of variable in dimension $n>1$. Suppose you have $$\int_A f(x,y)dxdy$$ where $f(x,y)$ is a scalar function and $A\in\mathbb{R}^2$. This quantity must be (?) independent from the choice of the coordinate system for fixed $A$, and so $$\int_{A'}f(x',y')dx'dy'$$ must be a way to define the same thing valid as the other. To demonstrate this I'll try to calculate $dxdx$ in term of $dx'dy'$ because one can imagine to have a map $(x',y')\mapsto(x(x',y'),y(x',y'))$. $A'$ is the redefinition of $A$ in the new coordinate system. \begin{array}{lcl} dx'dy' & \mapsto & dxdy=\frac{\partial x}{\partial x'}\frac{\partial y}{\partial x'}(dx')^2+\frac{\partial x}{\partial x'}\frac{\partial y}{\partial y'}dx'dy'+ \\ & &+\frac{\partial x}{\partial y'}\frac{\partial y}{\partial x'}dy'dx'+\frac{\partial x}{\partial y'}\frac{\partial y}{\partial y'}(dy')^2. \quad(\star)\\ \end{array} It is clear to me that the first and the last are different object from my purpose, but why on earth one would set $$dxdy=-dydx$$ on first place in order to obtain the invariance of (or the right correction to) the integral. I am aware of the differential geometry of manifolds ((co)tangent spaces, differential forms, Cartan exterior algebra, Hodge and Poincaré duality, (co)comology, etc..). In the language of differential forms it is the wedge product to give rise to $dx^\mu dx^\mu=0$ and $dx^\mu dx^\nu=-dx^\nu dx^\mu$. It is easy to see that the minus sign can be related to the determinant of the Jacobian $$\det\Big( \begin{array}{cc} \frac{\partial x}{\partial x'}&\frac{\partial x}{\partial y'}\\ \frac{\partial y}{\partial x'}&\frac{\partial y}{\partial y'} \end{array} \Big)$$ but why plus is wrong while minus is right? I am searching for answers or references.

Answer 1

You could start by pointing out to your friend that the area of the parallelogram spanned by vectors $v,w$ in the plane is (up to sign) the determinant of the matrix $[v\;w]$.

Then you could point out that under a linear map, the said area transforms by the determinant of the linear map.

Once he absorbed that information, you could point out that the integral is the infinite sum of areas of infinitesimal parallelograms into which one can partition a domain, and therefore it transforms by the Jacobian which is the infinitesimal version of the function at each infinitesimal parallelogram.