What does it mean to "identify" a function $f$ with a $0$-form $\omega_0$ and a $3$-form $\omega_3$?

Question

In class we wrote down the following:

We identify a function $f$ in $U \subset \mathbb{R}^3$ with a $0$-form $\omega_0$ and a $3$-form $\omega_3$ and a vector field $u$ with a $1$-form $\omega_1$ and a $2$-form $\omega_2$.

I do not understand what the identify party means. (I know the definition of $k$-forms). Could you please explain this to me?

Answer 1

Let $(x,y,z)$ be the usual Cartesian coordinate system on $\Bbb{R}^3$ (and since $U$ is open, we can restrict these to get a coordinate system on $U$ as well), and $\Omega^k(U)$ the space of smooth $k$-forms on $U$.

• The mapping $f\mapsto f\,dx\wedge dy\wedge dz$ is an isomorphism of $C^{\infty}(U)=\Omega^0(U)$ onto $\Omega^3(U)$. In more abstract notation, this is the effect of the Hodge star acting on 0-forms: $\star f=f\,dx\wedge dy\wedge dz$, because $dx\wedge dy\wedge dz$ is the volume form of $\Bbb{R}^3$ (corresponding to the Riemannian metric).
• The mapping $u=(u_1,u_2,u_3)\mapsto u_1\,dx+u_2\,dy+u_3\,dz$ is an isomorphism of the space, $\mathfrak{X}(U)$, of smooth vector fields on $U$, onto $\Omega^1(U)$. In fancier terms, this is the result of the musical isomorphism $g^{\flat}$ corresponding to the standard Riemannian metric, $g=dx\otimes dx+dy\otimes dy+dz\otimes dz$, on $\Bbb{R}^3$
• The mapping $F\,dx+G\,dy+H\,dz\mapsto F\,dy\wedge dz+G\,dz\wedge dx+H\,dx\wedge dy$ is an isomorphism of $\Omega^1(U)$ onto $\Omega^2(U)$. This is the effect of the Hodge star acting on 1-forms.

This is the explicit description of how we “identify” (in $\Bbb{R}^3$ only!) smooth functions with 3-forms, and vector fields, 1-forms, 2-forms. More generally, we have an isomorphism $\Omega^k(M)\cong \Omega^{n-k}(M)$ given by the Hodge-star, and $\mathfrak{X}(M)\cong \Omega^1(M)$ given by the musical isomorphism.