# Pullback function on basis 1-forms

For an assignment in a Differential Geometry class I am asked to proved the following...

Let $\phi :\mathbb{R}^2\rightarrow \mathbb{R}^2$ denote rotation counterclockwise about the origin by angle $\theta$. Let $dx$ and $dy$ denote the usual basis 1-forms on $\mathbb{R}^2$. Prove that:

$$\phi^*dx=\cos \theta dx-\sin \theta dy$$ $$\phi^*dy=\sin \theta dx+\cos \theta dy$$

$\phi^*$ is a pullback on one forms such that $\phi^*\mu(\overrightarrow{v})=\mu(\phi_*\overrightarrow{v})$ where $\overrightarrow{v}$ is vector, $\mu$ is a 1-form and $\phi_*\overrightarrow{v} = (\overrightarrow{v}[\phi_1],\overrightarrow{v}[\phi_2])$

I am having a hard time proving this. I am not sure that I defining $\phi :\mathbb{R}^2\rightarrow \mathbb{R}^2$ the right way.. I don't really understand what a rotation like this would look like with two parameters.

Any help would be greatly appreciated!

The map $\phi$ is just the linear map

$$\phi \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \cos \theta - y \sin \theta \\ x \sin \theta + y \cos \theta\end{pmatrix}.$$

Since $\phi$ is linear, we have $d\phi|_{p} = \phi$ for all $p \in \mathbb{R}^2$. Denote by $(e_1,e_2)$ the standard basis of $\mathbb{R}^2$ and compute:

$$\phi^{*}(dx)(e_1) = dx(d\phi(e_1)) = dx(\phi(e_1)) = dx \begin{pmatrix} \cos \theta \\ \sin \theta \end{pmatrix} = \cos \theta, \\ \phi^{*}(dx)(e_2) = dx(d\phi(e_2)) = dx(\phi(e_2)) = dx \begin{pmatrix} -\sin \theta \\ \cos \theta \end{pmatrix} = -\sin \theta.$$

This implies that

$$\phi^{*}(dx) = \cos \theta \, dx - \sin \theta \, dy.$$

• This makes total sense, thank you! I was trying to use theta as the independent variable. – DeathByTensors Apr 15 '17 at 20:59