Vector field and 2-form correspondence - is this explanation correct?

Question

I'm trying to understand – at my High School maths level – why a vector field$$\mathbf{w}\left(x,y,z\right)=f_{1}\hat{\mathbf{e}}_{x}+f_{2}\hat{\mathbf{e}}_{y}+f_{3}\hat{\mathbf{e}}_{z}$$can be associated with the 2-form$$\omega_{2}=f_{1}\,dy\wedge dz+f_{2}\,dz\wedge dx+f_{3}\,dx\wedge dy.$$ Is it OK to say this works because the wedge product of the 1-forms $\omega=u^{1}dx+u^{2}dy+u^{3}dz$ and $\nu=v^{1}dx+v^{2}dy+v^{3}dz:$ $$\omega\wedge\nu=\left(u^{2}v^{3}-u^{3}v^{2}\right)dy\wedge dz+\left(u^{3}v^{1}-u^{1}v^{3}\right)dz\wedge dx+\left(u^{1}v^{2}-u^{2}v^{1}\right)dx\wedge dy$$ has the same components as the cross product of the vectors $\mathbf{u}=u^{1}\hat{\mathbf{e}}_{x}+u^{2}\hat{\mathbf{e}}_{y}+u^{3}\hat{\mathbf{e}}_{z}$ and $\mathbf{v}=v^{1}\hat{\mathbf{e}}_{x}+v^{2}\hat{\mathbf{e}}_{y}+v^{3}\hat{\mathbf{e}}_{z}$?

Answer 1

Yes, the wedge product of two vectors in $\Bbb R^3$ corresponds to the cross product. But what's going on here is that to a vector field $\mathbf w=(f_1,f_2,f_3)$ you can associate both a $1$-form $$\omega = f_1\,dx + f_2\,dy + f_3\,dz$$ and a $2$-form $$\eta=\star\omega = f_1\,dy\wedge dz + f_2\,dz\wedge dx + f_3\,dx\wedge dy.$$ If you have a curve $\gamma$, then $\int_\gamma\omega$ gives the work done by $\mathbf w$ along $\gamma$. If you have an oriented surface $\Sigma$, then $\int_\Sigma\eta$ gives the flux of $\mathbf w$ across $\Sigma$.

Without any discussions of Stokes's (or the Divergence) Theorem, the reason for the flux interpretation is as follows. Let $\mathbf u,\mathbf v$ span an oriented parallelogram $\mathscr P$ in $\Bbb R^3$, and let $\mathbf n$ the unit vector normal to the parallelogram. Then \begin{align*} \eta(\mathbf u,\mathbf v) &= \left|\begin{matrix} f_1 & f_2 & f_3 \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3\end{matrix}\right| = \mathbf w\cdot (\mathbf u\times\mathbf v)\\ &= (\mathbf w\cdot\mathbf n)(\mathbf n\cdot(\mathbf u\times\mathbf v)) = (\mathbf w\cdot\mathbf n)\|\mathbf u\times\mathbf v\| \\ &= (\mathbf w\cdot\mathbf n)\text{area}(\mathscr P). \end{align*}

Answer 2

I understand that what you need is some more (rigorous) details. There are a number of identifications implicitly assumed in all of this.

1) Every vector is a (unique) linear combination of $i$,$j$, and $k$. We make a bijection between space of vectors and 1-forms by sending $i$ to $dx$, $j$ to $dy$, and $k$ to $dz$. This is a choice really, but a natural one.

2) We also "borrow from" the vectors a dot product between 1-forms. More precisely, we declare $$ \langle dx, dy \rangle =\langle dx,dz\rangle=\langle dy,dz \rangle =0 $$ and $$ \langle dx,dx \rangle = \langle dy,dy\rangle = \langle dz,dz\rangle =1 $$ And on arbitrary 1-forms by bi-linearity and symmetry.

3) If we have any inner product on space of 1-forms, we can also define a dot product on the space of $n$-forms for any $n$ as follows: If $w_i$'s and $v_i$'s are 1-forms then define $$ \langle w_1 \wedge w_2\wedge \cdots \wedge w_n,v_1 \wedge v_2\wedge \cdots \wedge v_n \rangle := \det (\langle w_i,v_j \rangle) \, , $$ that is, dot product one by one and and arrange them in an $n\times n$ matrix, then take determinant. One then extends this to arbitrary $n$-forms by linearity. Every $n$-form is a linear combination of such terms.

4) Hodge star operator. Let me do it is $\mathbb{R}^3$ for simplicity. It is a linear operator from $i$-forms to $(3-i)$-forms. It is thus determined by what it does to a basis. On 1-forms $$ \star dx = dy\wedge dz, \quad \star dy = -dx \wedge dz, \quad \star dz = dx\wedge dy \, , $$ the signs are determined by the requirement that for any form $w$ of any degree, $(\star w) \wedge w = \langle w,w\rangle \, dx\wedge dy \wedge dz $, where the dot product is explained above -- which for 1-forms agrees with the dot product of their corresponding vectors. Under these definitions it holds that $$ \langle \star w, \star v \rangle = \langle w, v \rangle $$ for any pair of (same dimensional) forms.

As a result of all of this, especially 4, in identifying vectors with either 1-forms or with 2-forms, via Hodge, the inner product of vectors is also transformed the same way. This is the identification lying behind all: we can freely see the same object as a vector, a 1-form, or a 2-form!

Now the identification between the cross products of vectors (by the way this ONLY makes sense in $\mathbb{R}^3$) and the forms is as follows (care needed): Given vectors $v$ and $w$, you turn them into 1-forms, see (1) above. Then you wedge them, the result is of course a 2-form. Then you take $\star$ of it which brings it back to a 1-form. Then you turn this 1-form back to a 1-form, using (1) again. Then you look closely and you see a happy coincidence: The vector you have at hand is what you would have gotten by cross product of $v$ and $w$ as vectors. That is it,the end!