I want to show that:

Given $f,g \in C^\infty(M)$ defined in a differential manifold of dimension $n$ and $a \in M$, we have $$(dfg)_a=f(a)(dg)_a+g(a)(df)_a,$$

using the following proposition:

Let $M$ be a $n$-dimensional manifold.

Let $(U,\varphi)$ be a coordinate chart around $a$ with coordinates $x_1,\ldots,x_n$.

Then, if $f \in C^\infty(M)$ and in the coordinate chart $f \varphi^{-1} = \phi(x_1,\ldots,x_n)$, we have $$(df)_a=\sum_{i=1}^n \frac{\partial \phi}{\partial x_i}(\varphi(a))(dx_i)_a$$

My atempt: $$f(a)(dg)_a + g(a)(df)_a = f(a)\sum_{i=1}^n \frac{\partial \phi_2}{\partial x_i}(\varphi(a))(dx_i)_a + g(a) \sum_{i=1}^n \frac{\partial \phi_1}{\partial x_i}(\varphi(a))(dx_i)_a$$

So, $$f(a)(dg)_a + g(a)(df)_a = \sum_{i=1}^n [f(a)\frac{\partial \phi_2}{\partial x_i}(\varphi(a)) + \frac{\partial \phi_1}{\partial x_i}(\varphi(a))](dx_i)_a$$

Now confusion begins, I can't proceed but I know that:

  1. $f\varphi^{-1}=\phi_1(x_1,\ldots,x_n)$

  2. $g\varphi^{-1}=\phi_2(x_1,\ldots,x_n)$

I would greatly appreciate it if you give me some hint or solution.

  • $\begingroup$ Oh, I'm seeing that $fg=(\phi_1 \circ \varphi)(\phi_2 \circ \varphi) = \phi_1 \phi_2 \circ \varphi$ $\endgroup$ – Leafar Mar 7 '15 at 19:52

We have $$(dfg)_a = \sum_{i=1}^n \frac{\partial(\phi_1 \phi_2)}{\partial x_i}(\varphi(a))(dx_i)_a,$$

so $$(dfg)_a = \sum_{i=1}^n [\frac{\partial\phi_1}{\partial x_i}(\varphi(a))\phi_2(\varphi(a)) + \phi_1(\varphi(a))\frac{\partial \phi_2}{\partial x_i}(\varphi(a))](dx_i)_a,$$

then $$(dfg)_a = \sum_{i=1}^n [\frac{\partial\phi_1}{\partial x_i}(\varphi(a))g(a) + f(a)\frac{\partial \phi_2}{\partial x_i}(\varphi(a))](dx_i)_a,$$

which matches the attempt.


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