# Is the dual covector field of a nonzero vector field closed?

Given a nowhere vanishing vector field, say $E_1$, on a manifold $M^n$, it should be possible to extend this locally to a basis of vector fields $E_1,\ldots , E_n$ so that $E_i = \frac{\partial}{\partial x_i}$ for some coordinates $x_i$. (For example, locally we can choose a chart where $E_1$ looks like $\frac{\partial}{\partial x_1}$ and then just define $E_i = \frac{\partial}{\partial x_i}$).

For such a frame there we may define a co-frame on $T^*M$, $\varphi^i$ so that $\varphi^i(E_j) = \delta_{ij}$. By the intrinsic definition of the exterior derivative, $$d\varphi^1(E_i,E_j) = E_i(\varphi^1(E_i)) - E_j(\varphi^1(E_j) - \varphi^1([E_i,E_j]) .$$ The first two terms will be zero as they are the derivative of either the constant function 0 or 1. Similarly, $[E_i,E_j] = [\frac{\partial}{\partial x_i},\frac{\partial}{\partial x_j}] \equiv 0$.

Finally, since $d\varphi^1$ is a two form, we see by linearity that $d\varphi^1 \equiv 0$ .

In fact, any nonvanishing 1-form $\omega$ can (locally) be expressed as the dual of a nonvanishing vector in some coordinate framing. This would seem to imply that any nowhere zero 1-form has zero exterior derivative.

Since I do not believe this is true, there must be some flaw to my reasoning, but I cannot see it.

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You should try to find a nowhere zero $1$-form with non-zero exterior derivative! –  Mariano Suárez-Alvarez Feb 6 '11 at 1:58
On the torus, if $\omega = (2+\sin(y))dx$ then $d\omega = -\cos(y)dx\wedge dy$. $\omega$ is nowhere zero and $d\omega$ is not zero. –  Ben McMillan Feb 6 '11 at 2:21
Your assumption that you can get your vector fields as coordinate directions assumes your vector fields commute, that their Lie brackets are zero. –  Ryan Budney Feb 6 '11 at 2:53
@Ryan: given a non-zero vector field $X$, one can always find (locally!) a coordinate patch which has $X$ as $\frac{\partial}{\partial x_1}$, and then it will commute with the other $\frac{\partial}{\partial x_i}$s, as well as they among themselves. –  Mariano Suárez-Alvarez Feb 6 '11 at 3:11

In fact, any nonvanishing 1-form ω can (locally) be expressed as the dual of a nonvanishing vector in some coordinate framing.

This is the part I question. How do you show this? If you run your argument in reverse, you need that any nonvanishing form $\omega$ is $dx^1$ for some coordinate $x^1$, which now implies that $\omega$ is closed. So perhaps while every non-zero vector field can be taken to be a coordinate field, the corresponding statement for one-forms is not true.

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You are right, that part was something of an after thought, and may not be true (I am thinking about that now). –  Ben McMillan Feb 6 '11 at 10:13
What bothered me originally was the question in the title of the post. However, I think the answer to my question is this: When I say the dual of a covector field, I implicitly assume some framing on the tangent bundle (not necessarily one from a coordinate frame.) However, in the proof I choose a different (coordinate) framing, whose dual coframe will be different from the one I am trying to prove is zero. The conclusion only holds for the new framing, and I have proved nothing about the one I started with. –  Ben McMillan Feb 6 '11 at 10:26

The statement of yours is not wrong, indeed, it is right. Differenial geometry bases on these statements. However, the crucial point is that you will have to choose local coordinates (i.e., you obtain a local statement) and then that the statement follows for a linear combinations of the coordinate vector fields. There's some analogy to complex Kähler geometry where you use the (local) $i\partial\partial^{\dagger}$, where the last derivative means to be the complex conjuagte operator to the first one. There does exist also a global version which imposes however much more assumptions on the underlying manifold. This question is a very excellent one, I really like this kind of thoughts. But please be aware that in order to efficiently calculate something interesting, the global aspects will become more important.