# The Unapologetic Mathematician

## Coordinate Vector Fields

If we consider an open subset $U\subseteq M$ along with a suitable map $x:U\to\mathbb{R}^n$ such that $(U,x)$ is a coordinate patch, it turns out that we can actually give an explicit basis of the module $\mathfrak{X}U$ of vector fields over the ring $\mathcal{O}U$.

Indeed, at each point $p\in U$ we can define the $n$ coordinate vectors:

$\displaystyle\frac{\partial}{\partial x^i}(p)\in\mathcal{T}_pM$

Thus each $\frac{\partial}{\partial x^i}$ itself qualifies as a vector field in $U$ as long as the map $p\mapsto\frac{\partial}{\partial x^i}(p)$ is smooth. But we can check this using the coordinates $(U,x)$ on $M$ and the coordinate patch induced by $(U,x)$ on the tangent bundle. With this choice of source and target coordinates the map is just the inclusion of $U$ into the subspace
$\displaystyle U\times\left\{(0,\dots,0,1,0,\dots,0)\right\}\subseteq U\times\mathbb{R}^n$

where the $1$ occurs in the $i$th place. This is clearly smooth.

Now we know at each point that the coordinate vectors span the tangent space. So let’s take a vector field $X\in\mathfrak{X}U$ and break up the vector $X(p)$. We can write

$\displaystyle X(p)=\sum\limits_{i=1}^nX^i(p)\frac{\partial}{\partial x^i}(p)$

which defines the $X^i(p)$ as real-valued functions on $U$. It’s also smooth; we know that $X:U\to U\times\mathbb{R}^n$ is smooth by the definition of a vector field and the same choice of local coordinates as above, and passing from $X(p)$ to $X^i(p)$ is really just the projection onto the $i$th component of $\mathbb{R}^n$ in these local coordinates.

Since this now doesn’t really depend on $p$ we can write

$\displaystyle X=\sum\limits_{i=1}^nX^i\frac{\partial}{\partial x^i}$

which describes an arbitrary vector field $X$ as a linear combination of the coordinate vector fields times “scalar coefficient” functions $X^i\in\mathcal{O}U$, showing that these coordinate vector fields span the whole module $\mathfrak{X}U$. It should be clear that they’re independent, because if we had a nontrivial linear combination between them we’d have one between the coordinate vectors at at least one point, which we know doesn’t exist.

We should note here that just because $\mathfrak{X}U$ is a free module — not a vector space since $\mathcal{O}U$ might have a weird structure — in the case where $(U,x)$ is a coordinate patch does not mean that all the $\mathfrak{X}U$ are free modules over their respective rings of smooth functions. But in a sense every “sufficiently small” open region $U$ can be contained in some coordinate patch, and thus $\mathfrak{X}U$ will always be a free module in this case.

May 24, 2011