A manifold question: Why smooth functions and what is a Jacobian?

Question

My question is what does a Jacobian have to do with the change of coordinates (coordinate transformation). Why do we care about this notion to start with? Also, why should it be non-singular?

Answer 1

1. The functions are "smooth" because we want to be able to speak about derivatives on our manifold $M$, and for that, it is convenient to have a smooth structure on $M$ (one could settle for a $C^k$-structure with $k$ as needed, but physicists rarely care for such details). And of course we want to be able to take derivative because we might be interested in the divergence of a vector field, for instance, or the Laplacian of some function, etc... Physics needs the notion of (spatial) change, and that's what the derivative gives. If nothing else is said, all manifolds that appear in physics are silently assumed to be smooth manifolds.

2. That the $x'(x)$, or rather, the $\phi_j\circ\phi_i^{-1}$, are smooth follows from the definition of a smooth atlas. That the Jacobian is non-singular follows from the $\phi_i,\phi_j$ being diffeomorphisms, and so $\phi_j\circ\phi_i^{-1}$ is also a diffeomorphism, and diffeomorphisms have non-singular Jacobians.

Answer 2

1) If the transformation itself is smooth, then the Jacobian will be smooth. This is a desirable property because...

2) The coordinate transformation tells us how the coordinates change, but the Jacobian tells us directly how the coordinate basis vectors change. For example, a transformation from Cartesian to polar coordinates would use the Jacobian to relate the basis vectors $\partial_x, \partial_y$ to the basis vectors $\partial_r, \partial_\theta$.

Ensuring that the Jacobian is invertible (non-singular) as well means that that non-degenerate volumes in one coordinate system will still be non-degenerate (i.e. nonzero, actual volumes instead of planes) in another, and that a nonzero vector field in one coordinate system will not be mapped to a zero vector field in another.