Partial Derivatives on Manifolds - Is this conclusion right?

Question

I'm self-studying Differential Geometry and I've asked here about how to describe functions on a manifold, and now that I'm pretty sure that my conclusions about that are correct I've started to think on how do we compute partial derivatives. Well, Spivak defines on his book that if $(x,U)$ is a chart on a smooth manifold $M$ and if $f : U \to \Bbb R$ is differentiable, then we can define the $i$-th partial with respect to this chart as:

$$\frac{\partial f}{\partial x^i}(p)=D_i(f\circ x^{-1})(x(p))$$

This is very natural and very good, but I started to think on ways to compute with this. Indeed, I'm studying Spivak's Differential Geometry books, but in addition to the theory, I'm trying to get the way to compute things. So my thought was: following the question I've refered, if I can express any function $f : U \to \Bbb R$ as combination of the coordinate functions with usual functions defined on the real line, then I can take any partials if I know the partials of the coordinate functions. Indeed, I computed as follows:

$$\frac{\partial x^i}{\partial x^j}(p)=D_j(x^i \circ x^{-1})(x(p))$$

But I've defined $x^i=I^i \circ x$ where $I:\Bbb R^n \to \Bbb R$ is the identity. So we have that:

$$x^i\circ x^{-1}=(I^i\circ x)\circ x^{-1}$$

But composition is associative, and since $x : U\to \Bbb R^n$ and $x^{-1} : x(U)\subset \Bbb R^n \to U $ we have that $x \circ x^{-1} : x(U)\subset \Bbb R^n \to \Bbb R^n$ and this is just the identity $I$, so that $(I^i \circ x)\circ x^{-1} = I^i$ and so

$$\frac{\partial x^i}{\partial x^j}(p)=D_jI^i(x(p))$$

But we know that $D_jI^i(q) = \delta_j^i$ independent of the point, where $\delta^i_j$ is the Kronecker Delta. So we have that:

$$\frac{\partial x^i}{\partial x^j}(p)=\delta_j^i$$

I've shown similarly that the partial derivative on a manifold is linear, obeys the product rule and that it obeys the chain rule. So suppose now $M$ is a manifold of dimension $2$, $U\subset M$ and that $(x,U)$ is a chart. Then we have coordinate functions $x^1$ and $x^2$ and we can express for instance the following function $f : U \to \Bbb R$ as:

$$f = \sin \circ (x^1 x^2)$$

Then we have that the partial with respect to $x^1$ for instance is:

$$\frac{\partial f}{\partial x^1}(p)=\cos(x^1(p)x^2(p))\frac{\partial (x^1 x^2)}{\partial x^1}(p)=\cos(x^1(p)x^2(p))\left(\frac{\partial x^1}{\partial x^1}(p)x^2(p)+x^1(p)\frac{\partial x^2}{\partial x^1}(p)\right)$$

And using the result I've shown above I would get:

$$\frac{\partial f}{\partial x^1}(p)=x^2(p)\cos(x^1(p)x^2(p))$$

So is really like this that we compute partial derivatives in practice on Manifolds? Are all my conclusions correct?

Thanks very much!

Answer 1

Your calculations appear on the mark. The idea of differentiating with respect to manifold coordinates is necessarily abstract. However, at the end of the discussion the point to remember is simply this:

a manifold is a curved space which allows a local calculus

The interesting thing about the derivatives you're working through is that we may take them on a sphere, cylinder, projective space, three-dimensional Thurston model geometry, whatever. The method to calculate is the same. The partial derivative of $f$ with respect to $x^j$ measures the change in $f$ along the $x^j$-th coordinate direction on the manifold at the point in question. Notice this is the same as it was in calculus, just now the space is (possibly) curved.

Furthermore, yes, once you have manifold coordinates and functions expressed in manifold coordinates it is just like what we already know from multivariate calculus. It has to be. Remember $\mathbb{R}^n$ is the essential and trivial example of an $n$-dimensional manifold. Therefore, you should expect to recover all the usual theorems of calculus for manifolds; linearity, the Leibniz rule, the chain rule, implicit and inverse function theorems and so forth. Even integration, once given a suitable way of patching things together can be discussed. Of course, sometimes it's easy to get lost in all the notation, but it seems you're finding your way just fine.