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?

Thanks in advance,



1 Answer 1


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 :)

  • $\begingroup$ Thanks for your reply! That would definitely do it, but I fear I must be lacking in background in some area, because that equality doesn't seem 'obvious' to me. Is there a rigorous proof of it? It certainly reminds me of some kind structure preserving map like a homomorphism for rings/groups, but my topological background is limited. $\endgroup$
    – Jerome
    Commented Jun 8, 2020 at 0:51
  • $\begingroup$ @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. $\endgroup$
    – peek-a-boo
    Commented Jun 8, 2020 at 0:56

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