# Derivation of dual vector component transformation. Wald's book

I'm working through the second chapter of General Relativity by Robert Wald, and I'm getting stuck on something that should be simple, so I'm probably missing something obvious.

Wald writes that the transformation law for the components of a dual vector is

$$\omega'_{\mu'} = \omega_\mu \frac{\partial x^\mu}{\partial x'^{\mu'}},\tag{2.3.7}$$ with summation over repeated indices. No surprises there, but he claims that this follows from only the corresponding transformation law for vectors,

$$v'^{\mu'} = v^\mu \frac{\partial x'^{\mu'}}{\partial x^\mu},\tag{2.3.6}$$

and the definition of the dual basis,

$$e^{\mu}(e_\nu) = \delta_\nu^{\mu}.\tag{2.3.1}$$

It is when I try to verify this that I run into trouble. Let us use coordinate bases with the conventional notations $$\partial_\mu$$ and $$dx^\mu$$. My attempt starts by using (2.3.1) to write

$$\omega'_{\mu'} = \omega'_{\nu'} \delta_{\mu'}^{\nu'} = \omega'_{\nu'} dx'^{\nu'}(\partial'_{\mu'}) = \omega'_{\nu'} dx'^{\nu'}(\delta_{\mu'}^{\rho'} \partial'_{\rho'}).\tag{2}$$

Then, for fixed $$\mu'$$ thinking of the $$\delta_{\mu'}^{\rho'}$$ as the components of the vector $$\partial'_{\mu'}$$, I use (2.3.6) to write

$$\delta_{\mu'}^{\rho'} \partial'_{\rho'} = \delta_{\mu'}^{\rho'} \frac{\partial x^\rho}{\partial x'^{\rho'}} \partial_{\rho} = \frac{\partial x^\rho}{\partial x'^{\mu'}} \partial_{\rho},\tag{3}$$

and hence, using the linearity of the action of the dual vector, I get

$$\omega'_{\mu'} = \omega'_{\nu'} \frac{\partial x^\rho}{\partial x'^{\mu'}} dx'^{\nu'}(\partial_{\rho}).\tag{4}$$

But I still cannot resolve $$dx'^{\nu'}(\partial_{\rho})$$ into anything useful, so this is where I am stuck.

It feels like I am close. Any tips?

Wait, I think I've got it! I can simply write $$\omega'_{\nu'} dx'^{\nu'}$$ in the unprimed basis of $$V^*$$,
$$\omega'_{\nu'} dx'^{\nu'} = \omega_\mu dx^\mu.$$
$$\omega'_{\mu'} = \omega_{\mu} \frac{\partial x^\rho}{\partial x'^{\mu'}} dx^{\mu}(\partial_\rho) = \omega_{\mu} \frac{\partial x^\rho}{\partial x'^{\mu'}} \delta_\rho^\mu = \omega_{\mu} \frac{\partial x^\mu}{\partial x'^{\mu'}},$$