I get $d\phi^*\alpha = 0$ for any $1$-form $\alpha$ on $\mathbb{R}^2$. This cannot be correct.

Question

I'm trying to get a hold on differential forms; I'm not sure about the following calculation. Let $\alpha$ be a 1-form on $\mathbb{R}^2$, i.e. \begin{equation} \alpha(x,y) = a\ dx + b\ dy \end{equation} and let $\phi:\mathbb{R^2}\rightarrow\mathbb{R}^2$ be a change of coordinates, with $\phi(u,v)=(x,y)$.

I want to explicitly check that $d\phi^*\alpha = \phi^*d\alpha$. When doing the calculations, I get \begin{align} \phi^*\alpha(u,v) &= a(\phi(u,v))[\partial_ux\ du + \partial_vx\ dv]+b(\phi(u,v))[\partial_uy\ du + \partial_vy\ dv] = \\ &=[(a\circ\phi)\partial_ux+(b\circ\phi)\partial_uy]\ du + [(a\circ\phi)\partial_vx+(b\circ\phi)\partial_vy]\ dv = \\ &= [\bar{a}\partial_ux+\bar{b}\partial_uy]\ du + [\bar{a}\partial_vx+\bar{b}\partial_vy]\ dv \end{align} where $\bar{a}=a\circ\phi$ and $\bar{b}=b\circ\phi$. Then, I have \begin{align} d\phi^*\alpha &= d([\bar{a}\partial_ux+\bar{b}\partial_uy]\ du + [\bar{a}\partial_vx+\bar{b}\partial_vy]\ dv) = \\ &= d([\bar{a}\partial_ux+\bar{b}\partial_uy]\ du) + d([\bar{a}\partial_vx+\bar{b}\partial_vy]\ dv) =\\ &= d\bar{a}[\partial^2_{uu}x\ du+\partial^2_{uv}x\ dv]\ du +d\bar{b}[\partial^2_{uu}y\ du+\partial^2_{uv}y\ dv]\ du\ + \\ &\quad+d\bar{a}[\partial^2_{vu}x\ du+\partial^2_{vv}x\ dv]\ dv +d\bar{b}[\partial^2_{vu}y\ du+\partial^2_{vv}y\ dv]\ dv = \\ &= d\bar{a}\ \partial^2_{uv}x\ dv\ du +d\bar{b}\ \partial^2_{uv}y\ dv\ du\ +d\bar{a}\ \partial^2_{vu}x\ du\ dv +d\bar{b}\ \partial^2_{vu}y\ du\ dv = \\ &= -d\bar{a}\ \partial^2_{uv}x\ du\ dv -d\bar{b}\ \partial^2_{uv}y\ du\ dv\ +d\bar{a}\ \partial^2_{vu}x\ du\ dv +d\bar{b}\ \partial^2_{vu}y\ du\ dv = \\ &=0 \end{align} But I guess it cannot be correct, since a $2$-form on $\mathbb{R}^2$ (a volume form) should not be automatically zero. What am I getting wrong?

Answer 1

\begin{align} d\phi^*\alpha &= d([\bar{a}\partial_ux+\bar{b}\partial_uy]\ du) + d([\bar{a}\partial_vx+\bar{b}\partial_vy]\ dv) \\ &= d\bar{a}[\partial^2_{uu}x\ du+\partial^2_{uv}x\ dv]\ du +d\bar{b}[\partial^2_{uu}y\ du+\partial^2_{uv}y\ dv]\ du\ + \\ &\quad+d\bar{a}[\partial^2_{vu}x\ du+\partial^2_{vv}x\ dv]\ dv +d\bar{b}[\partial^2_{vu}y\ du+\partial^2_{vv}y\ dv]\ dv = \\ &= d\bar{a}\ \partial^2_{uv}x\ dv\ du +d\bar{b}\ \partial^2_{uv}y\ dv\ du\ +d\bar{a}\ \partial^2_{vu}x\ du\ dv +d\bar{b}\ \partial^2_{vu}y\ du\ dv = \\ &= -d\bar{a}\ \partial^2_{uv}x\ du\ dv -d\bar{b}\ \partial^2_{uv}y\ du\ dv\ +d\bar{a}\ \partial^2_{vu}x\ du\ dv +d\bar{b}\ \partial^2_{vu}y\ du\ dv = \\ &=0 \end{align}

I believe you totally blundered taking the derivative of product in second equality. Correct way:

$$ d([\bar{a}\partial_ux+\bar{b}\partial_uy]\ du)= \left[\frac{\partial}{\partial u} ( \tilde{a} \partial_u x + \tilde{b} \partial _u y) du + \frac{\partial}{\partial v} ( \tilde{a} \partial_u x + \tilde{b} \partial _u y)dv \right] \wedge du = \partial_v ( \tilde{a} \partial_u x + \tilde{b} \partial_u y) dv \wedge du$$

You need to keep the wedge inserted because if you switch around the order of the term being wedged you get a negative sign. I.e:

$$ dv \wedge du = - du \wedge dv$$