7
$\begingroup$

I have a doubt about the fact that a derivative of $f:M\to \mathbb R$ of a $\mathcal C^1$ manifold is well defined... Indeed, let $a\in M$ and $(U,\varphi )$ a chart from a $\mathbb C^1$ atlas s.t. $a\in A$. Then, $$f'(a)=\left.\frac{d}{dx}\right|_{x=\varphi (a)}f(\varphi ^{-1}(x)).$$

But if $(\psi,V)$ is an other chart s.t. $a\in V$, then, $$f'(a)=\left.\frac{d}{dx}\right|_{x=\psi(a)}f(\psi^{-1}(x)).$$

The thing is that I don't see any reason to have $$\left.\frac{d}{dx}\right|_{x=\varphi (a)}f(\varphi ^{-1}(x))=\left.\frac{d}{dx}\right|_{x=\psi(a)}f(\psi^{-1}(x)).$$

$\endgroup$
  • 1
    $\begingroup$ Once you take a differential topology class, you learn that the "right notion" of a derivative is a differential 1-form. $\endgroup$ – Moishe Kohan May 11 at 17:55
5
$\begingroup$

While the derivative of $f$ in a coordinate system is map-dependent, there's a way to define it without using any map or coordinates.

In such a definition, the derivative of a function $f: M\rightarrow \mathbb R$ at point $p\in M$ is a covector $(df)_p \in {T_p}^*M$ such that for any curve $\gamma : \mathbb R \rightarrow M$, such that $\gamma(0) = p$ the following holds: $$ \left.\frac{d}{dt}\right|_{t=0}f(\gamma(t)) = \langle (df)_p, \gamma'(0) \rangle $$ where $\gamma'(0) \in T_pM$ is the vector tangent to curve $\gamma$ at point $p$.

You can check that this defines $(df)_p$ in a unique way, and if you choose any coordinate system, then the coefficients of $(df)_p$ in the induced basis in ${T_p}^*M$ are equal to the partial derivatives of $f$ in respect to chosen coordinates.

$\endgroup$
1
$\begingroup$

Using the notation $f'(a)$ here is not a good idea, since the partial derivatives do depend on the chosen coordinate system.

Take for example $\mathbb{R}^2\setminus(-\infty,0]$ as your manifold and consider the function $f$ that maps a point to its squared distance to the origin. In Cartesian coordinates, this function is given by $$ f(x,y) = x^2+y^2, $$ while in polar coordinates, $$ f(r,\theta) = r^2. $$ Then differentiating with respect to the first coordinate gives $$ \frac{\partial f}{\partial x}(x,y) = 2x, \quad\text{while}\quad \frac{\partial f}{\partial r}(r,\theta) = 2r, $$ but these two are not equal e.g. at the point $(0,1)$. Note that this does not mean that the partial derivatives on a manifold are not well-defined, but only that they depend on the chosen frame of reference (i.e. the chosen coordinate system, or local chart if you wish).

To see how they are related, note that by the chain rule, \begin{align} \frac{\partial}{\partial x_j}\bigg|_{x = \phi(a)}(f\circ \phi^{-1})(x) &= \frac{\partial }{\partial x_j}\bigg|_{x = \phi(a)}(f \circ \psi^{-1}\circ \psi\circ \phi^{-1})(x)\\ &= \sum_{k=1}^n\frac{\partial}{\partial y_k}\bigg|_{y = \psi(a)}(f \circ \psi^{-1})(y) \cdot \frac{\partial}{\partial x_j}\bigg|_{x = \phi(a)}(\psi \circ \phi^{-1})_k(x), \end{align} so the partial derivatives are related by the "usual" Jacobian change of variables matrix, i.e. $$ \left.\begin{pmatrix} \partial_1 (\psi\circ \phi^{-1})_1 & \ldots & \partial_n (\psi\circ \phi^{-1})_1\\ \vdots & & \vdots \\ \partial_1 (\psi\circ \phi^{-1})_n & \ldots & \partial_n (\psi\circ \phi^{-1})_n \end{pmatrix}\right|_{\phi(a)}. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.