# Computing (and evaluating) the pullback of a differential form explicitly

It seems that this is a very simple exercise, but now I'm confused because I've rarely seen detailed examples of this particular kind.

Let $\omega$ be the $2$-form in $\mathbb{R}^2$ given by $\omega=\omega(x, y)=dx\wedge dy$, and consider $\Phi=(\Phi_1, \Phi_2) \colon (0, \infty) \times (0, 2\pi)\to \mathbb R^2,\qquad (r, \theta)\mapsto (r\cos \theta, r\sin \theta)$ . I'm asked to find $\Phi^*\omega$. Moreover, if $p=\left(1, \frac\pi2\right), v=(1,0)$ and $w=(3,1)$, I must compute $(\Phi^*\omega )_p(v, w)$.

At first, it is straightforward to find out that $\Phi^*\omega =r\,dr\wedge d\theta$, I don't have any problem at this point. (This is what everyone learns to do effortlesstly)

On the other side, one learns that $(\Phi^*\omega )_p(v, w):=\omega_{\Phi(p)}\big(d\Phi_p(v), d\Phi_p(w)\big)$. Also, $$d\Phi_p(v)=\begin{bmatrix} \frac{\partial \Phi_1}{\partial r}(p) & \frac{\partial \Phi_1}{\partial \theta}(p) \\ \frac{\partial \Phi_2}{\partial r}(p) & \frac{\partial \Phi_2}{\partial \theta}(p)\end{bmatrix} \begin{bmatrix}1 \\ 0 \end{bmatrix} =\cdots = \begin{bmatrix}0 & -1 \\ 1& 0 \end{bmatrix} \begin{bmatrix}1 \\ 0 \end{bmatrix}= \begin{bmatrix}0 \\ 1 \end{bmatrix}.$$ Similarly, $d\Phi_p(w)= \begin{bmatrix}-1 \\3\end{bmatrix}$.

So now, $(\Phi^*\omega )_p(v, w)= \omega_{(0, 1)}\left( \begin{bmatrix}0 \\ 1 \end{bmatrix}, \begin{bmatrix}-1 \\3\end{bmatrix} \right)$.

What's next, who is $\omega_{(0, 1)}$? It should be $dx_{(0, 1)}\wedge dy_{(0, 1)}$, but again, that doesn't tells much to me (and how to obtain $\omega_{(0, 1)}\left( \begin{bmatrix}0 \\ 1 \end{bmatrix}, \begin{bmatrix}-1 \\3\end{bmatrix} \right)$ afterwards?) [there must appear a determinant thingy, but I'm not sure if mine is right]

I'm a bit rust with this things so I started (re-)doing some exercises, but I bet almost no text ever does such an explicit calculation, so this one was a bit confusing for me.

The wedge product of two $1$-forms $\alpha$, $\beta$ is given by $$\alpha\!\wedge\!\beta(v,w) = \alpha(v)\beta(w) - \alpha(w)\beta(v).$$ Thus \begin{align} \omega_{q}\left( \begin{bmatrix}a \\ b \end{bmatrix}, \begin{bmatrix}c \\d\end{bmatrix} \right) &= dx_q\left(\begin{bmatrix} a \\ b \end{bmatrix}\right) dy_q\left(\begin{bmatrix} c \\ d \end{bmatrix}\right) - dx_q\left(\begin{bmatrix} c \\ d \end{bmatrix}\right) dy_q\left(\begin{bmatrix} a \\ b \end{bmatrix}\right) \\ &= ad - cb. \end{align} In particular, we have $$\omega_{(0,1)}\left( \begin{bmatrix}0 \\ 1 \end{bmatrix}, \begin{bmatrix}-1 \\3\end{bmatrix} \right) = 1.$$