# Differential forms and determinants

2-forms are defined as $du^{j} \wedge du^{k}(v,w) = v^{j}w^{k}-v^{k}w^{j} = \begin{vmatrix} du^{j}(v) & du^{j}(w) \\ du^{k}(v) & du^{k}(w) \end{vmatrix}$

But what if I have two concret 1-forms in $R^{3}$? For example $(2dx-3dy+dz)\wedge (dx+2dy-dz)$ which gives

$(2dx-3dy+dz)\wedge (dx+2dy-dz)=-7dy \wedge dx +3dz \wedge dx - dy \wedge dz= 7 dx \wedge dy + 3 dz \wedge dx + dy \wedge dz$ I know this is the same as the vector product between $(2,-3,1)^{T}$ and $(1,2,-1)^{T}$.

What is the relationship with the determinant? Because when I calculate the wedge product between two 1-forms in $R^{2}$ then I get the value of the determinant $(2dx+4dx)\wedge (3dx+9dy) = -18 dx\wedge dy +12 dx \wedge dy = 6 dx \wedge dy$, but for 1-forms in $R^{3}$ I get the vector product. And is the interpretation right that parts of the area spanned by the 2-form above (the vector product) is that 7 get's projected onto the $xy$ plane, 3 onto $zx$ and 1 onto $yz$? Or is it in this case another coordinate system with planes $dxdy$, $dzdx$ and $dydz$? And since all the differential forms are functions of vectors, what happens when they come into the picture? Because doesn't $7 dx \wedge dy + 3 dz \wedge dx + dy \wedge dz$ looks like this?

$7 \begin{vmatrix} dx(v) & dx(w) \\ dy(v) & dy(w) \end{vmatrix} + 3 \begin{vmatrix} dz(v) & dz(w) \\ dx(v) & dx(w) \end{vmatrix} + \begin{vmatrix} dy(v) & dy(w) \\ dz(v) & dz(w) \end{vmatrix}$ For some vectors $v,w \in R^{3}$?

• If you imagine a third vector so that you have a $3\times 3$ matrix, then you have the cofactors associated with the imaginary new vector. So, once you wedge again, you get a determinant. This is definitely related to the permutation definition of determininant, which happens naturally for forms when swapping terms like $dy\wedge dx$ to get $dx\wedge dy$ in order to collect like terms. Any multilinear anti-symmetric scalar form $f(\cdot,\cdot,\cdots,\cdot)$ on $\mathbb{R}^{n}$ is a scalar multiple of the determinant. Jul 30, 2014 at 20:46
• Hi, thx for the comment. Ok but why do I have to choose a third vector? I know that the $e_{i}$ would pick out subdeterminants and that you can only build the determinant if k=n. But isn't it ok to just let it be the cross product if k=2 and n=3 ? Or is this just a textbook example for what you can do with the wedge product? Aug 1, 2014 at 11:52
• Just to get you thinking about cofactor expansions. No real reason. It's the same reason people will put $\hat{x}$, $\hat{y}$, $\hat{z}$ into a determinant when teaching people to do cross product. Aug 1, 2014 at 13:43
• Ok, thank you, I'll think about it. Aug 1, 2014 at 17:30
• You might find some of my YouTube lectures (linked in my profile) helpful. Dec 30, 2020 at 18:37

Let's focus on a particular example: how the 2-form $$dx\wedge dy$$ on $$\mathbb{R}^3$$ is related to determinants. Well, given two three-dimensional vectors $$\mathbf{u}=[u_1, u_2, u_3]^t$$ and $$\mathbf{v}=[v_1, v_2, v_3]^t$$, we have
$$dx\wedge dy(\mathbf{u},\mathbf{v})=dx(\mathbf{u})dy(\mathbf{v})-dy(\mathbf{u})dx(\mathbf{v})=u_1v_2-u_2v_1=\det\begin{pmatrix}u_1 & v_1\\u_2 & v_2\end{pmatrix}$$
where the final matrix is the matrix we get by stacking $$\mathbf{u}$$ and $$\mathbf{v}$$ into a $$3\times 2$$ matrix and then selecting the rows corresponding to the $$x$$ and $$y$$ coordinates. We get a square matrix we can take the determinant of by selecting the appropriate rows.
More generally, consider a $$k$$-form $$d_{i_1}\wedge\cdots\wedge d_{i_k}$$ on $$\mathbb{R}^n$$. The value of this form on a $$k$$-tuple of $$n$$-dimensional vectors can be found first by stacking the $$k$$ vectors into an $$n\times k$$ matrix, then selecting the $$k$$ rows corresponding to the indices $$i_1, \ldots, i_k$$, and finally taking the determinant of the resulting $$k\times k$$ matrix.