Pulling back forms computation

Question

Let $f:\mathbb{R}^2 \to \mathbb{R}^2$ be a function mapping $$(x, y) \to \left(e^{2x}, xy\right).$$

How do you compute pulling back form of a 2-form $$\alpha(x, y)=xy(dxdy),$$ in other words, $f^*(\alpha)$?

Answer 1

Pullback commutes with the wedge product, and $xy \, dx \, dy$ is the wedge product of $x, y, dx$ and $dy$ (two $0$-forms and two $1$-forms respectively). So it suffices to compute the pullback of each of these.

The pullback of $x$ is the $x$-component of your function, and similarly for $y$. So $f^{\ast}(x) = e^{2x}$ and $f^{\ast}(y) = xy$.

Pullback also commutes with the exterior derivative, so the pullback of $dx$ is the exterior derivative of the pullback of $x$, and similarly for $y$. So

$$f^{\ast}(dx) = d e^{2x} = 2 e^{2x} dx$$ $$f^{\ast}(dy) = d (xy) = x \, dy + y \, dx.$$

So in total we get

$$f^{\ast}(xy \, dx \, dy) = e^{2x} (xy)(2e^{2x} \, dx)(x \, dy + y \, dx) = 2x^2 y e^{4x} \, dx \, dy.$$