Define pull-back on a manifold by pull-back on the linear space.

Question

It appears to me that pull-back on a manifold

If $f: X \to Y$ is a smooth map and $\omega$ is a $p$-form on $Y$, define a $p$-form $f^*\omega$ on $X$ as follows: $$f^*\omega(x) = (df_x)^*\omega[f(x)].$$

is defined by pull-back on the linear space:

Suppose $A: V \to W$ is a linear map. Then the transpose map $A^*: W^* \to V^*$ extends to the exterior algebras, $A^*: \Lambda^p(W^*) \to \Lambda^p(V^*)$ for all $p>0$. If $T \in \Lambda^p(W^*)$, just define $A^* T \in \Lambda^p(V^*)$ by $$A^*T(v_1, \dots, v_p) = T(Av_1, \dots, Av_p)$$ for all vectors $v_1, \dots, v_p \in V$.

So my observation is to put in $df^*$ as $A^*$, $\omega$ as $T$, and $f(x) = (f^1(x), \dots, f^p(x))$ as $v_1, \dots, v_p$: $$f^*\omega(x) = (df_x)^*\omega[f(x)] = \omega(df_x f^1(x), \dots, df_x f^p(x)) = \omega \circ df_x(f^1(x), \dots, f^p(x)).$$

I know it should result in $\omega \circ df_x(x_1, \dots, x_q)$, where $p$ is the dimension of $Y$, and $q$ is the dimension of $X$. But where I messed it up?

Why $\omega[f(x)]$ is $\omega \circ df_x(x_1, \dots, x_q)$, instead of $\omega \circ df_x(f^1(x), \dots, f^p(x))$?

---

Here's an example showing my confusion, from James S. Cook's answer to the question Intuition about pullbacks in differential geometry

So in his answer, he states:

$$ \Psi^*(\omega)(\partial_1,\partial_2,\partial_3,\partial_4)=\omega (\Psi_*(\partial_1), \Psi_*(\partial_2),\Psi_*(\partial_2),\Psi_*(\partial_4)).$$

I see it reasonable. However, in my textbook, pull-back is defined to be $$f^*\omega(x) = (df_x)^*\omega[f(x)].$$

Therefore, I understand it as: $$ \Psi^*(\omega)(\partial_1,\partial_2,\partial_3,\partial_4)=\omega (\Psi_*(f^1(\partial_1,\partial_2,\partial_3,\partial_4), \cdots, \Psi_*(f^4(\partial_1,\partial_2,\partial_3,\partial_4)).$$

Answer 1

My answer there simply omits the point-dependence of the form. Let me put in the hideous point-dependence. Let us suppose $\Psi: M \rightarrow N$ and $\omega$ is a $4$-form at $\Psi(x) \in N$. The pull-back of $\omega$ will be a $4$-form at $x \in M$. I wrote: $$ \Psi^*(\omega)(\partial_1,\partial_2,\partial_3,\partial_4)=\omega (\Psi_*(\partial_1), \Psi_*(\partial_2),\Psi_*(\partial_2),\Psi_*(\partial_4)).$$ However, explicitly: $$ (\Psi^*(\omega))_x(\partial_1|x,\partial_2|x,\partial_3|x,\partial_4|x)=\omega[f(x)] (d\Psi_x(\partial_1|x), d\Psi_x(\partial_2|x),d\Psi_x(\partial_3|x),d\Psi_x(\partial_4|x)).$$ Again, the notation $\omega[f(x)]$ indicated we a selecting the $4$-form $\omega[f(x)]$ at the point $f(x)$ of the 4-form-field $\omega$. Unfortunately, we use the term differential form for both the field and the object fixed at a point. This may be part of the confusion. Hope this helps.

---

The secondary question of what $\partial_j$ denotes requires some discussion. I'll define it by a formula which borrows the differentiation process from the parameter space, however there are several other ways. To be clear, we consider a smooth manifold $M$ with coordinate chart $(x,U)$ and we'll focus on a point $p$. This means $x: U \rightarrow V \subseteq \mathbb{R}^m$ and $x^{-1}: V \rightarrow U$. Let $f \in C^{\infty} (p)$ which means that $f \circ \gamma$ is a smooth curve in $\mathbb{R}^m$ for all curves $\gamma$ through $p \in M$ (for our purposes here this means the derivative I'm about to write exists). We define $$ \frac{\partial}{\partial x^j}{\bigg|}_p (f) = \biggl[ \frac{\partial}{\partial u^j}(f \circ x^{-1}) (u^1, \dots u^m) \biggr]\bigg|_{u=x(p)} $$ where $u^1,\dots,u^m$ are cartesian coordinates of $V$. In words, you take the function $f$ near $p$ and pull it down to a function $f \circ x^{-1}$ on $\mathbb{R}^m$ near $x(p) \in \mathbb{R}^m$. Then do plain-old partial differentiation with respect to $u^j$ and once that is done, plug $u = x(p)$. Of course, this can be phrased in terms of directional derivatives as is done: Partial Derivatives on Manifolds - Is this conclusion right? .

The neat thing about the manifold partial is that the notation hides this subtlety, but is honest for the uniformed. For example, $\frac{\partial x^j}{\partial x^i} = \delta_{ij}$ and the chain-rule (supposing $y$ is another coordinate system at the point considered) is $\frac{\partial f}{\partial x^i} = \sum_{j=1}^m \frac{\partial y^j}{\partial x^i}\frac{\partial f}{\partial y^j}$. These manifold partial derivatives are more commonly called the coordinate derivations because they satisfy the Leibniz rule: $$ \partial_j|p(fg) = \partial_j|p(f)g(p)+f(p)\partial_j|p(g) $$ It is also important to know these form a natural basis for $T_pM$. There is more to say, but I think this suffices for our purposes here.