Vrftjkry

How do I convert a 2-form $x dy wedge dz + y^2 dx wedge dz$ on $mathbb{R^{3}}$ to a vector field on $mathbb{R^{3}}$?

Attempt:

Suppose we have two vectors fields a and b in $mathbb{R^{3}}$ such that $a=(a_{1},a_{2},a_{3})$ and $b=(b_{1},b_{2},b_{3})$ .

Now,

$(dy wedge dz)(a,b) = det begin{pmatrix} a_{2} & a_{3} \ b_{2} & b_{3} end{pmatrix} = a_{2}b_{3}-a_{3}b_{2}$

$(dx wedge dz)(a,b) = det begin{pmatrix} a_{1} & a_{3} \ b_{1} & b_{3} end{pmatrix} =a_{1}b_{3}-a_{3}b_{1}$

I ended up getting

$x (a_{2}b_{3}-a_{3}b_{2}) + y^2 (a_{1}b_{3}-a{3}b_{1})$

How does the $x$ and the $y^{2}$ convert?

Edit:
The user Travis just pointed to an more general way in the comments.

With the volume from
$$Omega = mathbf e_1 wedge mathbf e_2 wedge mathbf e_3$$
we get an isomorphism via
$$
T mathbb R^3 to Lambda^2 mathbb R^3 : mathbf v mapsto Omega( mathbf v, cdot, cdot).
$$

It's inverse yields exactly the same map as described below, i.e.
$$mathrm{d}y wedge mathrm{d}z mapsto mathbf e_1,
quad
mathrm{d}z wedge mathrm{d} x mapsto mathbf e_2 quad text{and} quad
mathrm{d}x wedge mathrm{d} y mapsto mathbf e_3.$$

Old, less general reply:

There is the Hodge star operator, which assigns
$$star(mathrm{d}y wedge mathrm{d}z) mapsto mathrm{d} x,
quad
star(mathrm{d}z wedge mathrm{d} x) mapsto mathrm dy quad text{and} quad
star(mathrm{d}x wedge mathrm{d} y) mapsto mathrm d z.$$

You can look up the exact definition of the Hodge star operator in the link above or in books. It is a map $Lambda^k mathcal M to Lambda^{m-k} mathcal M$. In our case it allows us to convert a 2-from into a 1-from.

Then we are left with the task to convert a 1-from (or a co-tangential vector) into a tangential vector. Such a map is for example provided, if there is a scalar product on the tangential spaces, i.e. a Riemannian metric $langle cdot , cdot rangle$.

At a point $p in mathcal M$, the map is given by
$$T_p mathcal M to T^*_p mathcal M: mathbf v mapsto langle mathbf v, cdot rangle_p.$$

In the case of $mathbb R^3$ the inverse of this map is given by
$$mathrm dx mapsto mathbf e_1 = begin{pmatrix} 1 \ 0 \ 0 end{pmatrix},
quad
mathrm dy mapsto mathbf e_2 = begin{pmatrix} 0 \ 1 \ 0 end{pmatrix} quad text{and} quad
mathrm dz mapsto mathbf e_3 = begin{pmatrix} 0 \ 0 \ 1 end{pmatrix}.$$

Putting those two isomorphism together yields a map $Lambda^2_p mathbb R^3 to T_p mathbb R^3$.

Why?
For an intuition, I always think of $mathrm{d} x wedge mathrm{d} y$ as something which measures two dimensional volumes in $mathbb R^3$ and the vector $mathbf e_1$ is somehow a vector which is needed to span up the rest or $mathbb R^3$, something like a normal vector to the surface.

Example

In your case, we get the 1-from
$$star(x mathrm dy wedge mathrm dz + y^2 mathrm dx wedge mathrm dz) = x mathrm d x - y^2 mathrm d y =: omega.$$

Following the steps from above, the vector field is then defined as
$$
mathbf v = omega(mathbf e_1) mathbf e_1 + omega(mathbf e_2) mathbf e_2 + omega(mathbf e_3) mathbf e_3.
$$

Hence, the corresponding vector field is
$$
begin{pmatrix} x \ -y^2 \ 0 end{pmatrix}.
$$

While the question is about vector fields, the map $phi : Gamma(bigwedge^2 T^* Bbb R^3) stackrel{cong}{to} Gamma(bigwedge^1 T Bbb R^3) = Gamma(T Bbb R^3)$ is tensorial, so it suffices to describe it at a single point, say, $phi_0 : Lambda^2 T^*_0 Bbb R^3 to T_0 Bbb R^3$. Using the canonical identification $T_a Bbb R^3 cong Bbb R^3$ and suppressing the subscript, we'll write $$phi : textstyle{bigwedge^2 (Bbb R^3)^*} stackrel{cong}{to} textstyle{bigwedge^1 Bbb R^3} = Bbb R^3 .$$

First, since this map is linear, so as with any linear map the coefficients come along for the ride:
$$phi(x ,dy wedge dz + y^2 dx wedge dz) = x phi(dy wedge dz) + y^2 phi(dx wedge dz) .$$

We get such a map from any volume form, that is, any nonvanishing top form $omega in bigwedge^3 (Bbb R^3)^*$. With this in hand, it's easier to describe the inverse, which is just interior multiplication,

$$phi^{-1} : Bbb R^3 to textstyle{bigwedge^2 (Bbb R^3)^*}, qquad X mapsto iota_X omega = omega(X,,cdot,,,cdot,) .$$

Since it's not stated otherwise, we'll take the standard volume form, $dx wedge dy wedge dz$. So, for example, $$phi^{-1}(partial_x) = iota_{partial_x} (dx wedge dy wedge dz) = dy wedge dz .$$ Thus,
$$boxed{phi(dy wedge dz) = partial_x} ,$$
and by symmetry cyclic permutation gives that image under $phi$ of the other two basis elements.

We can also write down the inverse in an invariant (i.e., coordinate-independent) way: Since the space $bigwedge^3 (Bbb R^3)^*$ is $1$-dimensional, there is a unique element $Omega$ of its dual, $bigwedge^3 Bbb R^3$, that satisfies $omega(Omega) = 3! = 6$. Here, the notation on the l.h.s. just means the full contraction of $omega$ and $Omega$.) Then, checking on a basis (again, by symmetry, it's enough to check a single basis element), we find that
$$phi(mu) = frac{1}{2} Omega(mu,,cdot,),$$
where the notation on the r.h.s. just means that we contract $mu$ with the first two slots of $Omega$. Computing gives in our case that $$Omega = partial_x wedge partial_y wedge partial_z .$$