Pullback of $1$-form in coordinates

Question

Let $$\theta(p) = \sum_{i=1}^n f_i(p) \, dx_i$$ be a $1$-form in local coordinates. then we define $F^*(\omega(p))(X_1,\ldots,X_n) = \omega(F(p))(DF(p)(X_1),\ldots,DF(p)(X_n))$ as the pullback of a general n-form

My question is now, can we also express the pullback of the $1$-form $\theta$ in local coordinates?

I suspect that it would be something like $$F^*(\theta(p)) = \sum_{i=1}^n f_i(F(p)) \, d(F(x_i)),$$

but I am particularly uncertain how the coordinates $dx_i$ transform under pullback.

Answer 1

Write $\theta = \sum f_i \,{\rm d}x_i$, so: $$F^\ast\theta = F^\ast\left(\sum_{i=1}^n f_i\,{\rm d}x_i\right) = \sum_{i=1}^n \left(f_i \circ F\right) F^\ast {\rm d}x_i.$$ Now we apply a point $p$: $$(F^\ast\theta)_p = \sum_{i=1}^n f_i(F(p)) (F^\ast{\rm d}x_i)_p.$$ Now take a tangent vector $X$: $$(F^\ast\theta)_p(X) = \sum_{i=1}^n f_i(F(p)) (F^\ast{\rm d}x_i)_p(X) = \sum_{i=1}^nf_i(F(p))\,({\rm d}x_i)_{F(p)}({\rm d}F_p(X)).$$

In details: $F^\ast$ takes a $1$-form $\theta$ to a $1$-form $F^\ast\theta$. The $1$-form $F^\ast\theta$ takes a point $p$ to a linear functional $(F^\ast\theta)_p$. This linear functional acts on a vector $X$ to produce a number $(F^\ast\theta)_p(X)$. Writing ${\rm d}x_i$ instead of $({\rm d}x_i)_p$, confusing the $1$-form ${\rm d}x_i$ with the linear functional $({\rm d}x_i)_p$ is only ok because the value does not depend on $p$, but it is good to keep this in mind. Do not confuse the map taking points to multilinear mappings with the linear mappings themselves.

Answer 2

To be explicit, suppose $F : U \rightarrow V$ is a smooth map for $U \subset \mathbb{R}^m, V \subset \mathbb{R}^n$ open sets with coordinates $(\tilde{x}_1,\ldots,\tilde{x}_m)$ and $(x_1,\ldots,x_n)$ respectively.

A 1-form $\theta \in \Omega^1(V)$ can be written as $\theta = \sum_if_i(x_1,\ldots,x_m)dx_i$ for functions $f_i : V\rightarrow \mathbb{R}$. The pullback of $\theta$ along $F$ at a point $p \in U$ can then be defined by $$(F^*\theta)_p(X) = \theta_p(dF_p(X)) = \sum_i f_i(F(p))(dx_i)_{F(p)}(dF_p(X))$$ for $X \in T_pU$.

In the special case that the 1-form is the differential of a coordinate function, this says that $(F^*dx_i)_p(X) = (dx_i)_p(dF_p(X))$ which you'll notice is equal to $((dx_i)_p\circ dF_p)(X)$. The key to any calculation here comes from the chain rule which tells us that the differential operator $d$ distributes over composition of maps: $d(x_i\circ F)_p = (dx_i)_{F(p)}\circ dF_p$ (here the coordinate function $x_i$ is a special kind of smooth map whose target space is the field $\mathbb{R}$).

So ultimately, the answer is that $F^*dx_i = d(x_i\circ F) = dF_i$ (where $F = (F_1,\ldots,F_n)$ is the coordinate representation of $F$ with $F_i : U \rightarrow \mathbb{R}$ smooth for each $i$).