This is exercise 8-2.3 from Ted Shifrin's Multivariable Mathematics (a good book).
Suppose $\mathbf{v},\mathbf{w}\in\mathbb{R}^{3}.$ Show that
\begin{align*}
dx\left(\mathbf{v}\times\mathbf{w}\right)= & dy\wedge dz\left(\mathbf{v},\mathbf{w}\right)\\
dy\left(\mathbf{v}\times\mathbf{w}\right)= & dz\wedge dx\left(\mathbf{v},\mathbf{w}\right)\text{ and }\\
dz\left(\mathbf{v}\times\mathbf{w}\right)= & dx\wedge dy\left(\mathbf{v},\mathbf{w}\right).
\end{align*}
In my interpretation of the notation, this isn't correct. (See the
end of this post for a proposed fix.) The cross product is actually
a rank-2 anti-symmetric tensor in this context, so we need three more
components.
I am going to use the notation I typically use so that I don't get
(more?) confused, and so I can copy and paste. Following Wrede (and Schouten), the
symbol $\mathcal{E}_{ijk}$ is the completely anti-symmetric covariant
tensor density ($\approx$Levi-Civita tensor). $E^{ijk}=\mathcal{E}_{ijk}$
is the contravariant counterpart. Indices are raised on basis one-forms
and on contravariant vector components $v^{i}=dx^{i}\left\langle \mathfrak{v}\right\rangle .$
The angle bracket pair is called an "applicator". It specifically
means to apply a vector or form to its argument. It is my own invention (because I don't like hungry operators). Einstein summation is intended.
I am almost certain that the argument list for the two-form above
can be treated as a single tensor product of the two vectors
\begin{align*}
dx\left( \mathbf{v},\mathbf{w}\right) = & dy\wedge dz\left\langle \mathfrak{v}\otimes\mathfrak{w}\right\rangle \\
= & dx^{2}\wedge dx^{3}\left\langle \mathfrak{v}\mathfrak{w}\right\rangle \\
= & dx^{2}\wedge dx^{3}\left\langle \mathfrak{e}_{k}\mathfrak{e}_{l}\right\rangle v^{k}w^{l}\\
= & dx^{2}\wedge dx^{3}\left\langle \mathfrak{e}_{2}\mathfrak{e}_{3}\right\rangle v^{2}w^{3}\\
= & dy\wedge dz\left\langle \mathfrak{e}_{2}\mathfrak{e}_{3}\right\rangle v^{2}w^{3}\\
= & v^{2}w^{3}.
\end{align*}
Notice that when there is no summation, there is nothing to distinguish $dx\wedge dy$ from $dxdy.$
As an axial vector, the cross product is
\begin{align*}
\mathfrak{v}\times\mathfrak{w}= & \mathfrak{e}_{l}\mathcal{E}_{ijk}v^{i}w^{j}\delta^{kl}\\
= & \mathfrak{e}_{i}d\mathit{x}^{i}\langle\mathfrak{v}\times\mathfrak{w}\rangle\\
= & \mathfrak{e}_{m}d\mathit{x}^{m}\left\langle \mathfrak{e}_{l}\right\rangle \mathcal{E}_{ijk}v^{i}w^{j}\delta^{kl}\\
= & \mathfrak{e}_{m}\delta_{l}^{m}\mathcal{E}_{ijk}v^{i}w^{j}\delta^{kl}\\
= & \mathcal{E}_{ijk}v^{i}w^{j}dx^{k}\\
= & \begin{pmatrix}+\left(v^{2}w^{3}-v^{3}w^{2}\right)dx^{1}\\
+\left(v^{3}w^{1}-v^{1}w^{3}\right)dx^{2}\\
+\left(v^{1}w^{2}-v^{2}w^{1}\right)dx^{3}
\end{pmatrix}.
\end{align*}
Note that applying $\mathfrak{e}_{i}d\mathit{x}^{i}$ to a vector
simply returns the vector. This is what Menzel calls the identity
dyadic (tensor). It handles transformations smartly, but can be deceptive. The use of the basis one-forms in the result is algebraically
correct, but should not be taken to mean the cross product is natively
a one-form. I contend it is natively a bivector.
In order to get the full rank-two tensor equivalent to the cross product
we have to toggle the wedge product by running through the indices. So we write:
\begin{align*}
\mathfrak{T}= & \mathfrak{e}_{i}\mathfrak{e}_{j}T^{ij}\\
= & \mathfrak{e}_{i}\mathfrak{e}_{j}dx^{i}\wedge dx^{j}\left\langle \mathfrak{v}\mathfrak{w}\right\rangle \\
= & \mathfrak{e}_{i}\mathfrak{e}_{j}dx^{i}\wedge dx^{j}\left\langle \mathfrak{e}_{k}\mathfrak{e}_{l}\right\rangle v^{k}w^{l}\\
= & \mathfrak{e}_{i}\wedge\mathfrak{e}_{j}v^{i}w^{j}\\
= & \mathfrak{v}\wedge\mathfrak{w}\\
= & \begin{pmatrix}+\mathfrak{e}_{1}\mathfrak{e}_{2}\left(v^{1}w^{2}-v^{2}w^{1}\right)\\
+\mathfrak{e}_{2}\mathfrak{e}_{3}\left(v^{2}w^{3}-v^{3}w^{2}\right)\\
+\mathfrak{e}_{3}\mathfrak{e}_{1}\left(v^{3}w^{1}-v^{1}w^{3}\right)
\end{pmatrix}.
\end{align*}
The wedge product of two contravariant vectors is called a bivector.
Clearly both objects have the same information, and since they are
in closed form (all the indices are contracted on a basis) we can
set them equal. But writing the component equations still requires
some work. To convert from the tensor form to the vector form we write
\begin{align*}
T_{k}= & \frac{1}{2}\mathcal{E}_{ijk}T^{ij}\\
= & \mathcal{E}_{ijk}T^{\left\lfloor ij\right\rfloor }.
\end{align*}
The floor brackets on the indices is my way of indicating they are
restricted to increasing order (or any unique selections). The reason
we can now restrict our summation to increasing order is that we have
already done the wedge toggling. We can use the metric $\delta^{ij}$
to raise the index to put it in to contravariant form, if desired.
In general coordinates things are more complicated; so caution is advised.
To get the tensor components from the vector form we simply write
\begin{align*}
T^{ij}= & T_{k}E^{ijk}.
\end{align*}
All that said; I believe this fixes the exercise:
\begin{align*}
dx\left(\mathbf{v}\times\mathbf{w}\right)= & dy\wedge dz\left(\mathbf{v}\mathbf{w}-\mathbf{w}\mathbf{v}\right),\\
dy\left(\mathbf{v}\times\mathbf{w}\right)= & dz\wedge dx\left(\mathbf{v}\mathbf{w}-\mathbf{w}\mathbf{v}\right)\text{and}\\
dz\left(\mathbf{v}\times\mathbf{w}\right)= & dx\wedge dy\left(\mathbf{v}\mathbf{w}-\mathbf{w}\mathbf{v}\right).
\end{align*}
So, am I correct that, as stated, the proposed result of the exercise
is incorrect? Does my suggestion remedy the problem?
Note: since we aren't doing any summation using the wedge product,
we don't really need it here. So maybe
\begin{align*}
dx\left(\mathbf{v}\times\mathbf{w}\right)= & \left(dy\wedge dz+dz\wedge dy\right)\left(\mathbf{v}\mathbf{w}\right),\\
dy\left(\mathbf{v}\times\mathbf{w}\right)= & \left(dz\wedge dx+dx\wedge dz\right)\left(\mathbf{v}\mathbf{w}\right)\text{and}\\
dz\left(\mathbf{v}\times\mathbf{w}\right)= & \left(dx\wedge dy+dy\wedge dx\right)\left(\mathbf{v}\mathbf{w}\right)?
\end{align*}
Best Answer
The exercise is right as it is. Also, the cross product is by definition a (bilinear anti-symmetric) mapping $\Bbb{R}^3\times\Bbb{R}^3\to\Bbb{R}^3$. So, the cross product of two vectors is again a vector in the same space. The cross-product of two vectors is not an antisymmetric rank 2 tensor. What is a rank 2 tensor is $v\wedge w\in \bigwedge^2(\Bbb{R}^3)$. Now, one can certainly argue the merits of whether the cross-product should be taught and whether or not it is simpler to just introduce wedge products (at the expense of leaving $\Bbb{R}^3$). My personal view: cross products are unnatural (requiring an inner product and an orientation to define it abstractly) and that in the long run it is more natural to just introduce the exterior algebra so we can more naturally introduce things (especially things like angular velocity, angular momentum, moment of inertia etc). But regardless of our views, the definition still stands: given $v,w\in\Bbb{R}^3$, the object $v\times w$ is defined to be an element of $\Bbb{R}^3$ (by the usual ad-hoc prescription).
So, you trying to write $dx(\mathbf{v},\mathbf{w})$ is just wrong from the beginning (also, why bother with tensor densities? they serve no purpose here, and also the Levi-Civita tensor is simply another name for the volume form, which in the case of $\Bbb{R}^3$ is just $\det$ thought of as an alternating $3$-multilinear map on $\Bbb{R}^3$). Anyway, here’s how we can calculate things: \begin{align} dx(\mathbf{v}\times \mathbf{w})&=dx\bigg((v^2w^3-v^3w^2)\mathbf{e}_1+ (\cdots)\mathbf{e}_2 + (\cdots)\mathbf{e}_3\bigg)\\ &=(v^2w^3-v^3w^2)\cdot 1+(\cdots)\cdot 0+(\cdots)\cdot 0\\ &=v^2w^3-v^3w^2\\ &=dy(\mathbf{v})\cdot dz(\mathbf{w})-dz(\mathbf{v})\cdot dy(\mathbf{w})\\ &=(dy\otimes dz)(\mathbf{v},\mathbf{w})-(dz\otimes dy)(\mathbf{v},\mathbf{w})\\ &=(dy\otimes dz-dz\otimes dy)(\mathbf{v},\mathbf{w})\\ &=:(dy\wedge dz)(\mathbf{v},\mathbf{w}). \end{align} The last equality is simply a definition of the wedge product of two 1-forms. You seem to think $dy\wedge dz=dy dz=dy\otimes dz$, which couldn’t be further from the truth. So, you really just have to unwind the definitions here. It really is a trivial exercise, and the other two identities follow just as easily.