2
$\begingroup$

The following is a description of the partial derivative on smooth manifolds from John Lee's Introduction to Smooth Manifolds.

Suppose $M$ is a smooth manifold and let $(U,\phi)$ be a smooth coordinate chart on $M$. Then $\phi$ is a diffeomorphism from $U$ to an open subset $\hat{U}\subset \mathbb{R}^n$. So $d\phi_p : T_p M \to T_{\phi(p)} \mathbb{R}^n$ is an isomorphism.

Let the standard coordinate frame be $\partial/\partial x^1 |_{\phi(p)},\dots \partial / \partial x^n|_{\phi(p)}$ for $T_{\phi(p)} \mathbb{R}^n$. Then we use the notation $\partial/ \partial x^i|_p$ as the preimages of these vectors under the isomorphism $d\phi_p$ that form a basis for $T_p M$ characterized by either of the following expressions: $$\frac{\partial}{\partial x^i}|_p = (d\phi_p)^{-1}(\frac{\partial}{\partial x^i}|_{\phi(p)}) = d(\phi^{-1})_{\phi(p)}(\frac{\partial}{\partial x^i}|_{\phi(p)}).$$

Unwinding the definitions, we see that $\partial/ \partial x^i|_p$ acts on a function $f\in C^\infty(U)$ by $$\frac{\partial}{\partial x^i}|_p f = \frac{\partial}{\partial x^i}|_{\phi(p)} (f\circ \phi^{-1}) = \frac{\partial \hat{f}}{\partial x^i}(\hat{p}),$$ where $\hat{f}=f\circ \phi^{-1}$ is the coordinate representation of $f$, and $\hat{p}=(p^1, \dots, p^n)= \phi(p)$ is the coordinate representation of $p$. In other words, $\partial / \partial x^i|_p$ is just the derivation that takes the $i$th partial derivative of the coordinate representation of $f$ at the coordinate representation of $p$. In the special case of standard coordinates on $\mathbb{R}^n$, the vectors $\partial/ \partial x^i|_p$ are literally the partial derivative operators.

Now, for the general smooth manifold $M$ how do we guarantee that that mixed partial derivatives of a smooth function can be taken in any order, i.e. $\frac{\partial^2 f}{\partial x^i \partial x^j}=\frac{\partial^2 f}{\partial x^j \partial x^i}$? I think this follows from the fact that this is true for smooth functions on the Euclidean space, but how do we compute the second order derivative for smooth functions on manifolds as in the single derivative case above?

I cannot figure out how to define $\frac{\partial}{\partial x^i}|_p \frac{\partial f}{\partial x^i}= \frac{\partial}{\partial x^j}|_p \frac{\partial \hat{f}}{\partial x^j}(\hat{p})$ using the definition above. Can we put the second order derivative of $f\in C^\infty(M)$ in terms of second order derivative of the coordinate representation of $f$ in $C^\infty(\mathbb{R}^n)$ so we can use the result in $\mathbb{R}^n$ to conclude the result? Or do we rely on a different result to conclude that mixed partial derivatives can take any order? I would greatly appreciate some help as I have been puzzled about the precise form of mixture of partial derivatives for smooth manifolds and why the order does not matter as for smooth functions on $\mathbb{R}^n$.

$\endgroup$
1
  • 2
    $\begingroup$ Partial derivative only makes sense in the context of a smooth chart. It has no intrinsic meaning on a manifold. $\endgroup$ Commented Dec 17, 2023 at 3:25

1 Answer 1

4
$\begingroup$
  • $\frac{\partial f}{\partial x^i}$ by definition means $[D_i(f\circ \phi^{-1})]\circ \phi$.
  • So, $\frac{\partial}{\partial x^j}\frac{\partial f}{\partial x^i}$ means $\left[D_j\left(\frac{\partial f}{\partial x^i}\circ\phi^{-1}\right)\right]\circ\phi=\left[D_j(D_i(f\circ\phi^{-1}))\right]\circ\phi=[D_jD_i(f\circ\phi^{-1})]\circ\phi$. The expression on the right is symmetric under exchange of $i,j$ by basic multivariable calculus.

Continue inductively to see that for all multindices, $\frac{\partial^{|I|}f}{\partial x^I}=[D_I(f\circ \phi^{-1})]\circ\phi$; the expression on the right is symmetric under all permutations of $I$, hence so is the left.


Warning: if you mix coordinate systems, the symmetry no longer holds

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .