This is exercise V 5.8 of Edwards's Advanced Calculus of Several Variables. I thought I had cracked the code, but my result seems a bit strange.
Let $\varphi:\mathbb{R}^{m}\to\mathbb{R}^{n}$ be a $\mathscr{C}^{1}$ mapping. If $\alpha$ is a $k$--form on $\mathbb{R}^{n},$ prove that $$ \left(\varphi^{*}\alpha\right)_{\mathbf{u}}\left(\mathbf{v}_{1},\dots,\mathbf{v}_{k}\right)=\alpha_{\varphi\left(\mathbf{u}\right)}\left(d\varphi_{\mathbf{u}}\left(\mathbf{v}_{1}\right),\dots,d\varphi_{\mathbf{u}}\left(\mathbf{v}_{k}\right)\right). $$ This fact, that the value of $\varphi^{*}\alpha$ on the vectors $\mathbf{v}_{1},\dots,\mathbf{v}_{k}$ is equal to the value of $\alpha$ on their images under the induced linear mapping $d\varphi,$ is often taken as the definition of the pull-back $\varphi^{*}\alpha.$
I use the following notation:
A segment of natural numbers$\mathbb{S}_{n}=\left\{ 1,\dots,n\right\}.$
The $k^{th}$ Cartesian product of a segment is$\mathbb{S}_{n}^{k}.$ Elements $\mathfrak{n},\mathfrak{i}\in\mathbb{S}_{n}^{k}$ are $k$-tuples of natural numbers up to $n$, which are used as multi-indices.
$\left\lfloor \mathfrak{n}\right\rfloor $ restricts the $k$-tuples to increasing sets. Similarly for $\mathfrak{m}\in\mathbb{S}_{m}^{k}.$
Raised indices indicate contravariant components and objects. Lowered indices indicate covariant components and objects.
I have chosen to rename, $\varphi$ to $\vec{x}:\mathbb{R}^{m}\to\mathbb{R}^{n}.$
The derivative matrix of $\vec{x}$ is written $\frac{d\vec{x}}{d\mathbf{u}},$ and submatrices are selected using multi-indexing $$\frac{d\vec{x}^{\mathfrak{n}}}{d\mathbf{u}^{\mathfrak{m}}}.$$
Einstein summation is essentially the same for multi-indexing as for single indexing.
Using an over-tilde indicating alternating, a multi-wedge $k$-form basis element is written $$d\tilde{x}^{\mathfrak{i}}=dx^{i_{1}}\wedge\dots\wedge dx^{i_{k}}.$$
So, in my notation the target equation of the problem is
$$ \left(\vec{x}^{*}\tilde{\alpha}\right)_{\mathbf{u}}\left(\mathbf{v}_{1},\dots,\mathbf{v}_{k}\right)=\tilde{\alpha}_{\vec{x}\left(\mathbf{u}\right)}\left(d\vec{x}_{\mathbf{u}}\left(\mathbf{v}_{1}\right),\dots,d\vec{x}_{\mathbf{u}}\left(\mathbf{v}_{k}\right)\right). $$
Without a pullback our $k$-form is written
$$ \tilde{\alpha}=\mathit{a_{\left\lfloor \mathfrak{n}\right\rfloor }}d\tilde{x}^{\mathfrak{n}}. $$
My first question is how many components should be ascribed to the $\mathbf{v}_{i}?$ I infer the number should be $m$ since they appear in the pull-back, and the differential of the transformation is $d\vec{x}_{\mathbf{u}}:\mathbb{R}^{m}\to\mathbb{R}^{n}.$ To put those vectors in the original $k$-form apparently requires the transformed images to be $n$-component vectors.
We can transform them as one rank-$k$ contravariant tensor
$$ \mathfrak{V}=\mathbf{e}_{\mathfrak{n}}v^{\mathfrak{n}}=\mathbf{e}_{\mathfrak{n}}\frac{d\vec{x}^{\mathfrak{n}}}{d\mathbf{u}^{\mathfrak{m}}}v^{\mathfrak{m}}, $$
where $\mathbf{e}_{\mathfrak{n}}$ is the tensor product of $k$ standard basis vectors. So contracting this tensor with $\tilde{\alpha}$ gives
$$ \tilde{\alpha}\left(\mathfrak{V}\right)=\mathit{a_{\left\lfloor \mathfrak{n}\right\rfloor }}d\tilde{x}^{\mathfrak{n}}\left(\mathbf{e}_{\mathfrak{i}}\right)v^{\mathfrak{i}}=\mathit{a_{\left\lfloor \mathfrak{n}\right\rfloor }}\delta_{\mathfrak{i}}^{\mathfrak{n}}v^{\mathfrak{i}}=\mathit{a_{\left\lfloor \mathfrak{n}\right\rfloor }}v^{\mathfrak{n}}=\mathit{a_{\left\lfloor \mathfrak{n}\right\rfloor }}\frac{d\vec{x}^{\mathfrak{n}}}{d\mathbf{u}^{\mathfrak{m}}}v^{\mathfrak{m}} $$
$$ =\tilde{\alpha}_{\vec{x}\left(\mathbf{u}\right)}\left(d\vec{x}_{\mathbf{u}}\left(\mathbf{v}_{1}\right),\dots,d\vec{x}_{\mathbf{u}}\left(\mathbf{v}_{k}\right)\right). $$
I believe this is correct so far. MTW do things a bit differently, by using alternating $p$-vectors; and quite honestly the method used by Edwards, et al., seems far superior.
Now for the fun part. The pullback
$$ \vec{x}^{*}\tilde{\alpha}=\left(\mathit{a_{\left\lfloor \mathfrak{n}\right\rfloor }}\circ\vec{x}\right)\vec{x}^{*}d\tilde{x}^{\mathfrak{n}}. $$
To indicate we want a wedge product in the multi-differential transformation, we replace the over-arrow with a tilde
$$ \vec{x}^{*}d\tilde{x}^{\mathfrak{n}}=d\tilde{x}^{\mathfrak{n}}=\frac{d\tilde{x}^{\mathfrak{n}}}{d\mathbf{u}^{\mathfrak{m}}}d\vec{u}^{\mathfrak{m}}=\left|\frac{d\vec{x}^{\mathfrak{n}}}{d\mathbf{u}^{\mathfrak{m}}}\right|d\tilde{u}^{\left\lfloor \mathfrak{m}\right\rfloor }. $$
$$ \vec{x}^{*}\tilde{\alpha}=\left(\mathit{a_{\left\lfloor \mathfrak{n}\right\rfloor }}\circ\vec{x}\right)\left|\frac{d\vec{x}^{\mathfrak{n}}}{d\mathbf{u}^{\mathfrak{m}}}\right|d\tilde{u}^{\left\lfloor \mathfrak{m}\right\rfloor }. $$
I'm leaving the point of evaluation implied because it doesn't seem to impact my question, which is:
Is this correct so far?
$$ \vec{x}^{*}\tilde{\alpha}\left(v^{\mathfrak{m}}\right)=\mathit{a_{\left\lfloor \mathfrak{n}\right\rfloor }}\left|\frac{d\vec{x}^{\mathfrak{n}}}{d\mathbf{u}^{\mathfrak{m}}}\right|v^{\left\lfloor \mathfrak{m}\right\rfloor } $$
It would mean $$\frac{d\vec{x}^{\mathfrak{n}}}{d\mathbf{u}^{\mathfrak{m}}}v^{\mathfrak{m}}=\left|\frac{d\vec{x}^{\mathfrak{n}}}{d\mathbf{u}^{\mathfrak{m}}}\right|v^{\left\lfloor \mathfrak{m}\right\rfloor }.$$
Which is a bit hard to understand.
I just stumbled upon what appears to be essentially the same type of sum of determinants on a $k$-form basis, (but not using "grunge" notation)
https://youtu.be/lpicAI53e04?t=2160
A two-by-two example of how the determinant came about is
$$ d\tilde{x}=dx\wedge dy=\frac{d\tilde{x}}{d\mathbf{u}}d\vec{u}=\begin{pmatrix}\begin{bmatrix}\frac{\partial x}{\partial\mathrm{u}} & \frac{\partial x}{\partial\mathrm{v}}\end{bmatrix}\\ \wedge\\ \begin{bmatrix}\frac{\partial y}{\partial\mathrm{u}} & \frac{\partial y}{\partial\mathrm{v}}\end{bmatrix} \end{pmatrix}\begin{bmatrix}du\\ dv \end{bmatrix} $$
$$ =\left(\frac{\partial x}{\partial\mathrm{u}}du+\frac{\partial x}{\partial\mathrm{v}}dv\right)\wedge\left(\frac{\partial y}{\partial\mathrm{u}}du+\frac{\partial y}{\partial\mathrm{v}}dv\right) $$
$$ =\left(\frac{\partial x}{\partial\mathrm{u}}\frac{\partial y}{\partial\mathrm{v}}-\frac{\partial x}{\partial\mathrm{v}}\frac{\partial y}{\partial\mathrm{u}}\right)du\wedge dv $$
$$ =\left|\frac{d\vec{x}}{d\mathbf{u}}\right|du\wedge dv $$
where we are wedging the rows of the derivative matrix.
About my notation
I've been playing with mixing Einstein summation and multi-indexing for a while. I am not 100% confident in how I use it. When something like this actually works, it has the power to isolate parts of the conceptual structure being communicated, so that distinct ideas appear as distinct components of an expression.
Using raised and lowered indices comes from literature on the theory of relativity, but can typically be a very powerful means of keeping equations balanced, and seeing what tensors (vectors, k-forms, etc) should or shouldn't be contracted with others. Even in rectangular Cartesian coordinates with the Pythagorean metric where toggling indices has no effect on the value of the indicated component, the technique is extremely useful (to me).
Using the tilde to indicate that something is alternating (as in wedge product) came to me while posting this question, but it sure feels right.
Here's why I use "full" derivatives in
$$\vec{x}^{\prime}=\frac{d\vec{x}}{d\mathbf{u}},$$
rather than partials, which might seem more appropriate, since it is a matrix of partials. From https://www.scribd.com/read/282634061/Advanced-Calculus-of-Several-Variables?mode=standard#
My mistake started with the correct interpretation of the left-hand side of the following equation as the right-hand side:
$$\mathit{a_{\left\lfloor \mathfrak{n}\right\rfloor }}d\tilde{x}^{\mathfrak{n}}= \mathit{a_{\mathfrak{n}}d\tilde{x}^{\left\lfloor \mathfrak{n}\right\rfloor }}.$$
I mistakenly applied to increasing $k$-tuple restriction to the multi-index on $\mathbf{e}_{\mathfrak{i}}.$ So this is incorrect:
$$d\tilde{x}^{\left\lfloor \mathfrak{n}\right\rfloor }\left(\mathbf{e}_{\mathfrak{i}}\right)=\delta_{\mathfrak{i}}^{\mathfrak{n}}$$
This time I think I really have it
Applying $\tilde\alpha$ to the transformed vectors
The the $k$ vectors are given implicitly in $m$-space, which I will indicate using an over-bar on objects and indices. The multi-index $\mathfrak{n}\in\mathbb{S}_{n}^{k}$ has $k$-tuples ranging to $n$. Similarly for $\bar{\mathfrak{m}}\in\mathbb{S}_{m}^{k}.$ Writing the vectors as one tensor product gives
$$ \mathfrak{V}=\left\{ V^{\mathfrak{n}}\right\} =\frac{d\vec{x}^{\mathfrak{n}}}{d\mathbf{u}^{\bar{\mathfrak{m}}}}V^{\bar{\mathfrak{m}}}=\frac{d\vec{x}}{d\mathbf{u}}\bar{\mathfrak{V}.} $$
Feeding that to our $k$-form on $\mathbb{R}^{n}$ means taking sub-matrix determinants of the matrix product of the derivative $\vec{x}^{\prime}$ and vector matrix. Since $d\tilde{x}^{\mathfrak{n}}$ selects rows, we apply the selection to the derivative alone $$ \tilde{\alpha}\left(\mathfrak{V}\right)=\mathit{a_{\left\lfloor \mathfrak{n}\right\rfloor }}d\tilde{x}^{\mathfrak{n}}\left(\frac{d\vec{x}}{d\mathbf{u}}\bar{\mathfrak{V}}\right)=\mathit{a_{\left\lfloor \mathfrak{n}\right\rfloor }}\left|\frac{d\vec{x}^{\mathfrak{n}}}{d\mathbf{u}}\bar{\mathfrak{V}}\right|. $$
So the next task is to transform the $k\times k$ determinants $\left|\frac{d\vec{x}^{\mathfrak{n}}}{d\mathbf{u}}\bar{\mathfrak{V}}\right|$ arising from products of $k\times m$ and $m\times k$ matrices.
Transforming the $\left|\frac{d\vec{x}^{\mathfrak{n}}}{d\mathbf{u}}\bar{\mathfrak{V}}\right|$
For every alternating $k$-linear function $\mathcal{M}$ on $\mathbb{R}^{n}$ its components for permutations $\mathfrak{p}$ of fixed a set of basis vectors whose increasing order is $\left\lfloor \mathfrak{n}\right\rfloor $ are $M_{\mathfrak{p}}=\pm\mathcal{M}\left(\hat{\mathfrak{e}}_{\left\lfloor \mathfrak{n}\right\rfloor }\right)$ according to the parity of $\mathfrak{p}.$ This means it is sufficient to write $\mathcal{M}=M_{\left\lfloor \mathfrak{n}\right\rfloor }d\tilde{x}^{\mathfrak{n}}.$ Because the $d\tilde{x}^{\mathfrak{n}}$ produce the requisite sign-toggling as all sets of $k$ basis vectors and permutations thereof are visited during evaluation.
To obtain the transformation of our determinants, we will need the following fact. Multiplying a $k\times m$ matrix $\mathfrak{A}$ by the $m\times k$ matrix $\hat{\mathfrak{e}}_{\bar{\mathfrak{m}}}=\mathcal{I}_{\bar{\mathfrak{m}}}$ produces the $k\times k$ matrix$\mathfrak{A}_{\bar{\mathfrak{m}}}.$
Given an additional $m\times k$ matrix $\mathfrak{B},$ the determinant of the matrix product $\mathfrak{A}\mathfrak{B}$ may be written
$$ \left|\mathfrak{A}\mathfrak{B}\right|=\left|\mathfrak{A}_{\bar{\mathfrak{m}}}\right|\left|\mathfrak{B}^{\left\lfloor \bar{\mathfrak{m}}\right\rfloor }\right|. $$
This is show by considering the consequence of holding $\mathfrak{A}$ fixed while permuting the $k$ columns of $\mathfrak{B},$ which toggles the sign of the determinant. Thus we have the alternating $k$-linear function on $\mathbb{R}^{m}$ $$ \mathcal{A}\left(\mathfrak{B}\right)=\left|\mathfrak{A}\mathfrak{B}\right|=A_{\bar{\mathfrak{m}}}d\tilde{u}^{\bar{\mathfrak{m}}}\left(\mathfrak{B}\right) $$ Evaluating this on the standard basis vectors gives its components, so
$$ \mathcal{A}\left(\hat{\mathfrak{e}}_{\left\lfloor \bar{\mathfrak{m}}\right\rfloor }\right)=\left|\mathfrak{A}\hat{\mathfrak{e}}_{\left\lfloor \bar{\mathfrak{m}}\right\rfloor }\right|=\left|\mathfrak{A}{}_{\left\lfloor \bar{\mathfrak{m}}\right\rfloor }\right|=A_{\left\lfloor \bar{\mathfrak{m}}\right\rfloor }. $$
Thus we have the advertised result $$ \mathcal{A}\left(\mathfrak{B}\right)=\left|\mathfrak{A}\mathfrak{B}\right|=A_{\left\lfloor \bar{\mathfrak{m}}\right\rfloor }d\tilde{u}^{\bar{\mathfrak{m}}}\left(\mathfrak{B}\right)=\left|\mathfrak{A}_{\bar{\mathfrak{m}}}\right|\left|\mathfrak{B}^{\left\lfloor \bar{\mathfrak{m}}\right\rfloor }\right|. $$
The first half of the problem: $\tilde{\alpha}\left(\mathfrak{V}\right)$
$$ \tilde{\alpha}\left(\mathfrak{V}\right)=\mathit{a_{\left\lfloor \mathfrak{n}\right\rfloor }}d\tilde{x}^{\mathfrak{n}}\left(\frac{d\vec{x}}{d\mathbf{u}}\bar{\mathfrak{V}}\right)=\mathit{a_{\left\lfloor \mathfrak{n}\right\rfloor }}\left|\frac{d\vec{x}^{\mathfrak{n}}}{d\mathbf{u}^{\bar{\mathfrak{m}}}}\right|\left|\bar{\mathfrak{V}}^{\left\lfloor \bar{\mathfrak{m}}\right\rfloor }\right|. $$
The pullback is easier
The pullback is easier because we don't need the complicated determinant manipulation.
The $dx^{i}$ are the inner products of the $du^{\bar{\mu}}$ with the rows of the derivative matrix $\vec{x}^{\prime}.$ Since we are transforming the $k$-form basis elements $d\tilde{x}^{\mathfrak{n}}$, we need wedge products of the images of $du^{\bar{\mathfrak{m}}}$ so we replace the over-arrow with a tilde (``borrowed from'' $d\tilde{x}^{\mathfrak{n}}$). When the $dx^{i}=\partial_{\bar{\mu}}x^{i}du^{\bar{\mu}}$ are wedged together, the right-hand sides result in determinants of the participating sub-matrices when the $d\tilde{u}^{\left\lfloor \mathfrak{m}\right\rfloor }$ are factored out.
$$ \vec{x}^{*}d\tilde{x}^{\mathfrak{n}}=\frac{d\tilde{x}^{\mathfrak{n}}}{d\mathbf{u}^{\bar{\mathfrak{m}}}}du^{\bar{\mathfrak{m}}}=\left|\frac{d\vec{x}^{\mathfrak{n}}}{d\mathbf{u}^{\bar{\mathfrak{m}}}}\right|d\tilde{u}^{\left\lfloor \bar{\mathfrak{m}}\right\rfloor }. $$
$$ \vec{x}^{*}\tilde{\alpha}=\left(\mathit{a_{\left\lfloor \mathfrak{n}\right\rfloor }}\circ\vec{x}\right)\left|\frac{d\vec{x}^{\mathfrak{n}}}{d\mathbf{u}^{\bar{\mathfrak{m}}}}\right|d\tilde{u}^{\left\lfloor \bar{\mathfrak{m}}\right\rfloor }. $$
The second half of the problem: $\vec{x}^{*}\tilde{\alpha}\left(\bar{\mathfrak{V}}\right)$
$$ \vec{x}^{*}\tilde{\alpha}\left(\bar{\mathfrak{V}}\right)=\left|\frac{d\vec{x}^{\mathfrak{n}}}{d\mathbf{u}^{\bar{\mathfrak{m}}}}\right|d\tilde{u}^{\left\lfloor \bar{\mathfrak{m}}\right\rfloor }\left(\bar{\mathfrak{V}}\right)=\mathit{a_{\left\lfloor \mathfrak{n}\right\rfloor }}\left|\frac{d\vec{x}^{\mathfrak{n}}}{d\mathbf{u}^{\bar{\mathfrak{m}}}}\right|\left|\bar{\mathfrak{V}}^{\left\lfloor \bar{\mathfrak{m}}\right\rfloor }\right|. $$
An example of how I really figured out the determinants in the pullback