I am currently reading a course on microlocal analysis and I do not understand how they find some relations involving vector fields, after a change of coordinates.
Let $ (\xi_1, \ldots, \xi_n) $ be the standard coordinates in $\mathbb{R}^n$, and we put : $$ \rho = \frac{1}{\xi_1}, \, \, \, \nu_j = \frac{\xi_j}{\xi_1}$$ for $j = 2,\ldots, n$. Then, the following holds :
$$ \xi_1 \partial_{\xi_1} = - \rho \partial_{\rho} - \sum_{k = 2}^n \nu_k \partial_{\nu_k},$$
and there are three more but I do not see why that first relation is correct. Because it seems to me that $d\rho = d \left( \frac{1}{\xi_1}\right) = - \frac{1}{\xi^2_1} d \xi_1$, thus $\partial_{\rho} = - \xi^2_1 \partial_{\xi_1} $, and then $\rho \partial_{\rho} = -\xi_1 \partial_{\xi_1} $. Any help would be nice, thanks !
If you have coordinates $(x^1, \dots, x^n)$ and $(y^1, \dots, y^n)$, then \begin{align*} \frac{\partial}{\partial x^i} = \frac{\partial y^j}{\partial x^i}\frac{\partial}{\partial y^j}. \end{align*}
Therefore, if in your example, we set $x = \xi$ and $y = (\rho, \nu^2, \dots, \nu^n)$, then \begin{align*} \frac{\partial}{\partial\xi^i} &= \frac{\partial\rho}{\partial\xi^i}\frac{\partial}{\partial\rho} + \frac{\partial\nu^j}{\partial\xi^i}\frac{\partial}{\partial\nu^j}. \end{align*} Therefore, \begin{align*} \frac{\partial}{\partial\xi^1} &= -\frac{1}{(\xi^1)^2}\left(\frac{\partial}{\partial\rho} + \xi^j\frac{\partial}{\partial\nu^j}\right)\\ &= -\frac{1}{\xi^1}\left(\rho\frac{\partial}{\partial\rho} + \nu^j\frac{\partial}{\partial\nu^j}\right). \end{align*} It follows that $$ \xi^1\frac{\partial}{\partial\xi^1} = -\left(\rho\frac{\partial}{\partial\rho} + \nu^j\frac{\partial}{\partial\nu^j}\right). $$ I'll let you do the rest of the computation..
As for doing it via the dual basis, it is important to note that the formula for a given dual basis element depends on the entire basis. In particular, you cannot infer the formula for $\partial_\rho$ using only the formula for $d\rho$. So you have to compute the differential of all of the coordinates before doing anything else: \begin{align*} d\rho &= -\frac{d\xi^1}{(\xi^1)^2}\\ d\nu^j &= \frac{d\xi^j}{\xi^1} - \frac{\xi^j\,d\xi^1}{(\xi^1)^2}. \end{align*} If $$ \partial_\rho = a^i\partial_{\xi^i}, $$ then \begin{align*} 1&= \langle d\rho,\partial_\rho\rangle\\ &= \left\langle -\frac{d\xi^1}{(\xi^1)^2},a^i\partial_{\xi^i}\right\rangle\\ &= -\frac{a^1}{(\xi^1)^2}\\ 0 &= \langle d\eta^j,\partial_\rho\rangle\\ &= \left\langle \frac{d\xi^j}{\xi^1} - \frac{\xi^j\,d\xi^1}{(\xi^1)^2}, a^i\partial_{\xi^i}\right\rangle\\ &= \frac{a^j}{\xi^1} - \frac{\xi^ja^1}{(\xi^1)^2}. \end{align*} This leads to the same answer as above.