- #1
etotheipi
I was reading these notes and then on page 23 I saw something a bit weird. Back in this thread I learned that ##\{ \partial_i \}## form a basis of ##T_p M##, and that a tangent vector can be written ##X = X^i \partial_i##, and it's not too difficult to show that components transform like ##\bar{X}^j = \frac{\partial \bar{x}^j}{\partial x^i} X^i##. But this guy says that
The most alarming part of that is that he seems to suggest that ##\frac{\partial}{\partial x^i}## is just shorthand/notation for ##\frac{\partial \vec{x}}{\partial x^i}##. That seems quite wrong, since I thought the operators ##\frac{\partial}{\partial x^i}## are basis vectors already, in their own right. And ##\partial_i## and ##\partial_i \vec{x}## are completely different objects. Plus, the base space is not even necessarily a vector space, in which case I thought the position vector ##\vec{x}## is not even defined!
So I wondered if this section is wrong, or if I am missing something
. Thanks!
The most alarming part of that is that he seems to suggest that ##\frac{\partial}{\partial x^i}## is just shorthand/notation for ##\frac{\partial \vec{x}}{\partial x^i}##. That seems quite wrong, since I thought the operators ##\frac{\partial}{\partial x^i}## are basis vectors already, in their own right. And ##\partial_i## and ##\partial_i \vec{x}## are completely different objects. Plus, the base space is not even necessarily a vector space, in which case I thought the position vector ##\vec{x}## is not even defined!
So I wondered if this section is wrong, or if I am missing something
![Beaming face with smiling eyes :grin: 😁](https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f601.png)
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