# Exterior derivative well defined on vector fields

Let $M$ be a manifold and $d$ the exterior derivative on forms. We extend $d$ to vector fields as follows: if $X = \sum_i X_i \partial_i$ in local coordinates, then

$$dX = \sum_i (dX_i) \partial_i$$

i.e. if $Y$ is another vector field then $dX(Y)$ is the vector field $\sum_i (dX_i)(Y) \partial_i$. I am trying to show that this is well defined, independent of coordinate changes. When I calculate it out explicitly I get a rather nasty expression which does not seem to simplify. A coordinate-free definition of $d$ would be ideal, but I would also be satisfied with a coordinate based calculation.

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You are trying to define an object $dX$ that, given a vector field $Y$, can eat $1$-forms $\omega$. This is the same as feeding both $X$ and $Y$ into the $2$-form $d\omega$. – user53153 Dec 26 '12 at 5:08
Am I missing something in the definition then? It does not appear to be antisymmetric in $X$ and $Y$ for fixed $\omega$: namely, we only have derivatives on $X$, not $Y$. – user15464 Dec 26 '12 at 13:52
I was trying to suggest a coordinate-free definition of $dX$, rather than checking yours. I should have written "This should be the same as ..." – user53153 Dec 28 '12 at 5:20

This operation is not well defined. One way to see this is to note that, according to your definition, a vector field $X$ satisfies $dX=0$ if and only if it is a linear combination of the coordinate vector fields $\partial_i$ with constant coefficients (at least when the coordinate domain is connected). When you change to a different coordinate system, the same vector fields will typically not have constant coefficients.
The map $X\mapsto dX$ you've defined is an example of a connection. Given any smooth local frame for $TM$ on some subset $U\subset M$ (for example, a coordinate frame), there is a unique connection defined on $U$ for which the basis vector fields are parallel; but the connection depends heavily on the choice of frame.