Why is it true that $ \frac{\partial}{\partial x_i}(q)=dx_q(e_i)? $

Question

The following is the definition of Riemannian metrics in Riemannian Geometry by do Carmo:

I don't quite understand the underlined sentence.

In the book, the following is the definition of "differentials":

Here is my question:

How does one get $$ \frac{\partial}{\partial x_i}(q)=dx_q(0,...,1,...,0)? $$

Answer 1

The map $x:U\subseteq \mathbb R^n\longrightarrow \phi(U)$ is a diffeomorphism with respect to the smooth structure $U$ inherits from $\mathbb R^n$ and the smooth structure $\phi(U)$ inherits from $M$. Hence: $$dx_p:T_pU=\mathbb R^n\longrightarrow T_{x(p)}\phi(U)=T_{x(p)}M,$$ is an isomorphism for every $p\in U$. Therefore you can transport the cannonical basis of $\mathbb R^n$ to a basis of $T_{x(p)}M$ setting $$\frac{\partial}{\partial x_j}(p):=dx_p(e_j),$$ where $e_j$ is the $j$-th cannonical vector in $\mathbb R^n$.

Answer 2

It is only a notation, you are defining the vector $\frac{\partial}{\partial x_i}(q)$, not saying that it is equal to something. If you prefer you can simply say that the map $q\mapsto\langle dx_qv,dx_qw\rangle$ is differentiable for any $v$ and $w$.

Answer 3

I take it that you understand what the derivative matrix of a multivariable, multidimensional map between Euclidian spaces looks like: http://mathinsight.org/derivative_matrix

$\frac{\partial}{\partial x_i}(q)=dx_q(0,...,1,...,0)$ is basically saying take the $i$-th "column" of the "derivative matrix" $dx_{q}$ (which would represent a basis vector in this isomorphism), except we are not necessarily embedding $M$ in a suitably large Euclidean space with the map $x$, so one shouldn't think of it as the derivative matrix, but instead as a mapping on the tangent space (that's why I put quotes on "derivative matrix").