Converting a 2-form on $\mathbb{R}^{3}$ to a vector field on $\mathbb{R}^{3}$

Question

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?

Answer 1

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 $$ \begin{pmatrix} x \\ -y^2 \\ 0 \end{pmatrix}. $$

Answer 2

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 $$\Omega = \partial_x \wedge \partial_y \wedge \partial_z .$$