Consider the differential form $f dx$ in $\Omega^1 (\mathbb{R}^2)$, where $f \in C^{\infty }(M)$. $\Omega^1 (\mathbb{R}^2)$ has basis $\{ dx, dy \}$.

I am looking for a rigorous calculation of the pullback of $dx$ to $\mathbb{R}$ by some smooth map $\phi : \mathbb{R} \rightarrow \mathbb{R}^2$. Its basic but I want to do it rigorously.

I think $f dx$ should pull back to $p \mapsto f \circ \phi (p) \frac{\partial \phi_1 } {\partial t}(p) dt$ in $\Omega^1 (\mathbb{R})$, but I'm not sure.

If someone gives me a brief outline of how to do this calculation, I can fill in the details for myself.


Your guess is correct. If you just want a hint, try to use the following:

$1.$ Think about how the pullback interacts with an expression like $fdx$.

$2.$ Think about how the pullback acts on smooth functions.

$3.$ Think about how the pullback interacts with the exterior derivative $d$.

$4.$ Think about what the exterior derivative of a function locally looks like.

Spoiler below:

If we call $\omega=f dx,$ then \begin{align*}\phi^*\omega&=\phi^*(f dx)=(\phi^* f)(\phi^* dx)\\ &=(f\circ \varphi)d( \varphi^* x)=(f\circ \varphi)d (x\circ\varphi)\\ &=(f\circ \varphi)\frac{\partial \varphi^1}{\partial t}dt, \end{align*} where $\varphi^1=x\circ\varphi.$ Here, we used that $(1)$ $\varphi^*(g\omega)=(\varphi^*g)(\varphi^* \omega)$ for functions $g$ and forms $\omega$, $(2)$ $\varphi^*f=f\circ\varphi$ for functions $f$, $(3)$ $\varphi^*$ commutes with $d$, and $(4)$ a local expression for $d$: $dg=\sum_i\frac{\partial g}{\partial x^i}dx^i$. It's especially easy to verify in this case. $$ $$ In general, you can show that, if $\omega=\sum\limits_{j} a_jdx^j$ in a chart, then $$F^*\omega=\sum\limits_{i,j} (a_i\circ F)\frac{\partial F^i}{\partial x^j}dx^j.$$


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