[Math] Partial derivative of a function on manifold

differential-geometrymanifoldssmooth-manifolds

Bishop and Goldberg define ("Tensor analysis on manifolds") the partial derivative of a smooth function on a manifold $M$ in the following way:
$\partial_i f= \frac{\partial f}{\partial{x^i}}=\frac{\partial g}{\partial{u^i}}\circ\mu$, where $\mu= (x^1,…,x^d)$ is the coordinate system at a point $m\in M$, $g=f\circ\mu^{-1}$ is the coordinate expression for $f$, and $u^i$ are the cartesian coordinates on $R^d$. Notice the difference: $f$ is differentiated with respect to $x^i$ and $g$ with respect to $u^i$. By definition $x^i=u^i\circ\mu$.

In many cases $x^i=u^i$ and I don't have any problems. But what if we consider polar coordinates ($x^1=r, x^2=\phi$) instead of standard ($x,y$)? I can't make sense of the formula defining the partial derivative.

Best Answer

I personally find your book a bit confusing. You shouldn't try to think about two sets of coordinates, one for $f$ and one for $g$. This does not make much sense. $f$ is defined on the manifold $M$ so it has a priori no relation with any $d$-tuple $(x^1, \cdots, x^d)$. Expressing $f$ in a given chart/coordinate system $\mu : M' \subset M \rightarrow U \subset \mathbb{R}^d$ is defined as considering the function $g = f \circ \mu^{-1} : U \subset \mathbb{R}^d \rightarrow \mathbb{R}$ instead of $f$. In a way notation $\frac{\partial f}{\partial x^i}$ is purely symbolic. It is by definition equal to $\frac{\partial (f \circ \mu^{-1})}{\partial x^i} \circ \mu$. So first I don't see so much value in using different names $u^i$ and $x^i$ and second equality $x^i=u^i\circ\mu$ seems meaningless to me.

Now the fact you are considering "polar coordinates" is certainly due to the phrase "cartesian coordinates" in the book. This is again terribly confusing. Expressions like "polar coordinates" and "cartesian coordinates" belong to the domain of finite dimensional real vector spaces (or more precisely finite dimensional real affine spaces). It is perfectly true that $\mathbb{R}^d$ can be given a natural structure of $d$-dimensional real affine space. It is also true that one of the key ideas of differential geometry is that one can always identify a piece of a manifold $M$ with a piece of $\mathbb{R}^d$. But this identification has nothing to do with this affine structure. A chart is a diffeomorphism. That is to say we want to identify the local "smoothness structures" of $M$ and $\mathbb{R}^d$ because functions on $\mathbb{R}^d$ has a natural concept of partial derivative with respect to each one of its $d$ parameters. Any additional structure on $\mathbb{R}^d$ is of little importance.

That said, you can indeed interprete "polar coordinates" and "cartesian coordinates" using differential geometry but in this context you have to consider a 2-dimensional affine space $A$ as the manifold itself, not as the space of coordinates. Suppose we have an affine frame $(O, e_x, e_y)$, we can define

  • a chart $\mu_\textrm{cart} : P \in A \mapsto (x,y) \in \mathbb{R}^2$ such that $P = O + x \cdot e_x + y \cdot e_y$

  • a chart $\mu_\textrm{pol} : P \in A_\textrm{pol} \subset A \mapsto (r,\theta) \in \mathbb{R}^2$ such that $P = O + r \cdot \cos(\theta) \cdot e_x + r \cdot \sin(\theta) \cdot e_y$

It is the forms of $\mu_\textrm{cart}$ and of $\mu_\textrm{pol}$ that allow us to speak of "cartesian coordinates" and "polar coordinates", but the fact that $\mathbb{R}^2$ could be seen itself as an affine space is not very important.