# Partial Derivatives on Manifolds as Derivations

first time poster, finally decided to take the plunge and not just lurk anonymously. I'm just an experimental physicist with a desire to know some math beyond my few semesters of abstract algebra.

Just a quick question regarding the set of point derivations $$D: C_p^\infty (M) \rightarrow \mathbb{R}$$ on a smooth manifold $$M$$. I'm currently working through tangent bundles, and while I understand $$T_p\mathbb{R}^n$$ as the point derivations on $$C_p^\infty(\mathbb{R}^n)$$, I'm realizing I'm missing some of the subtleties of $$C_p^\infty (M)$$, and therefore $$T_pM$$. For some $$f \in C_p^\infty (M)$$, a chart $$(U,\phi) = (U,x^1,...,x^n)$$, and $$x^i = r^i \circ \phi$$, by definition

$$\frac{\partial f}{\partial x^i}(p) := \frac{\partial (f \circ \phi^{-1})}{\partial r^i} \big(\phi(p)\big)$$

from Tu's Intro to Manifolds on page 87. However, he claims that it's "easy to check" that this is indeed a derivation, but I'm having difficulty getting the nitty gritty to work. Say $$f,g \in C_p^\infty(M)$$. My first thought was to introduce a second chart $$(V,\psi) = (V,y^1,...y^n)$$ on $$p \in U \cap V$$,

$$\frac{\partial(fg)}{\partial x_i} (p) = \frac{\partial(fg \circ \phi^{-1})}{\partial r_i} \big(\phi(p)\big) = \frac{\partial\big(( fg \circ \psi^{-1} ) \circ ( \psi \circ \phi^{-1}) \big)}{\partial r_i} \big(\phi(p)\big) = \sum_j^n \frac{\partial ( fg \circ \psi^{-1} )}{\partial r^j} \big(\psi(\phi^{-1}(\phi(p)))\big) \frac{\partial(\psi \circ \phi^{-1})^j}{\partial r^i} \big( \phi(p)\big) = \sum_j^n \frac{\partial(fg)}{\partial y^j}(p) \cdot \frac{\partial y^j}{\partial x^i}(p)$$

then setting the chart $$(V,i_{\mathbb{R}^n}) = (V,r^1,..,r^n)$$ so that the partials $$\partial(fg) / \partial r^j$$ can be manipulated by standard calculus product rule, then work it backwards with two terms up to the definitions of partial derivatives,

$$\frac{\partial f}{\partial x^i}(p) \cdot g(p) + f(p) \frac{\partial g}{\partial x^i}(p)$$

and I felt good about this, but realized that it only works on $$M = \mathbb{R}^n$$, else $$fg:M \rightarrow \mathbb{R}$$ differentiated by $$r^i$$ is nonsensical. Is there some easy way of proving this Liebniz rule that's flown way over my head? Maybe something elegant, like the chain rule with the differential map?

There's no need to introduce a second chart. Just note that \begin{align} (f \cdot g) \circ \phi^{-1} = (f \circ \phi^{-1}) \cdot (g \circ \phi^{-1}) \end{align} So, now you can use the standard product rule which you know is true for maps from $$\Bbb{R}^n \to \Bbb{R}$$ (or between open subsets), and of course, unwind the meaning of all the symbols present :)
• @AtomJZ sure, there is a rigorous proof of it, and it's a very simple consequence of the definitions. Just evaluate both sides on an arbitary element $a$ in the domain of $\phi^{-1}$. $((f\cdot g) \circ \phi^{-1})(a) := (f \cdot g)(\phi^{-1}(a)) := f(\phi^{-1}(a)) \cdot g(\phi^{-1}(a)) = (f \circ \phi^{-1})(a) \cdot (g \circ \phi^{-1})(a) = [(f \circ \phi^{-1}) \cdot (g \circ \phi^{-1})](a)$. Most of these equal signs are simply true by definition of how composition,$\circ$ and multiplication $\cdot$, of functions is defined. Commented Jun 8, 2020 at 0:56