$v$ is locally potential iff $\text{d}\omega = 0$, where $\omega (u) = \left<v,u\right>$ is a $1-$form.

Question

This is a question from Do Carmo's Differential Forms and Applications. A vector $v$ is called locally potential if, locally, there is a function $g$ where $v = \text{grad} \ g$.

This looks simple, but I can't understand one thing: $\omega (u) = \left<v,u\right>$ is supposed to be a $1-$form, but $u$ is a vector on $\Bbb{R}^{n}$. Writing $u=(u_{1},\dots,u_{n})$, the $\text{d}x_{i}$ parts won't appear. Then, I can't calculate de differential of the $1-$form (which is the way I thinks this question is supposed to be answeared).

Answer 1

The equation $v=\text{grad}(g)$ is an equality of vector fields. By using the musical isomorphism given by the metric tensor field, the associated covector field equation is \begin{align} \langle v,\cdot\rangle&= \langle\text{grad}(g),\,\,\cdot\,\, \rangle \end{align} The LHS is what the book calls $\omega$, and the RHS, by definition of the gradient vector field is simply the exterior derivative $dg$.

i.e your question once reformulated from the vector field language (which is unnecessarily complicated) into one about covector fields/1-forms says

Prove that a 1-form $\omega$ is locally exact (i.e every point has an open neighborhood $A$ on which there is a smooth $0$-form i.e a function $g$ such that $\omega=dg$ on $A$) if and only if $d\omega =0$.

Surely you can prove this statement/ refer to a very well-known theorem.