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Let $M$ be a (real) manifold and let $(U , \varphi)$ and $(U , \psi)$ be two charts on $M$ with $\varphi = (x_1 , \ldots , x_n)$ and $\psi = (y_1 , \ldots , y_n)$. If $j \in \{1 , \ldots , n\}$, $p \in U$ and $x_j = y_j$, is ${\left(\frac{\partial}{\partial x_j}\right)}_p = {\left(\frac{\partial}{\partial y_j}\right)}_p$ true? I think it should not be necessarily true, but I do not know how to show it.

joseabp91
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1 Answers1

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No. Partial derivative $\frac{\partial f}{\partial x_j}$ depends not only on $x_j$ but the other variables as well.

Example.
Let's do change of variables from $(x_1,x_2)$ to $(y_1,y_2)$ like this: $$ x_1 = y_1+y_2, \\ x_2 = y_2, $$ or in reverse $$ y_1 = x_1-x_2, \\ y_2 = x_2. $$ So if your function is $f(x_1,x_2) = x_1$, which is the same function as $f(y_1,y_2) = y_1+y_2$, then we get $$ \frac{\partial f}{\partial x_2} = 0, \\ \frac{\partial f}{\partial y_2} = 1. $$ despite the fact that $x_2 = y_2$.

GEdgar
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