The Unapologetic Mathematician

Mathematics for the interested outsider

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 ith 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 ith 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 - Posted by | Differential Topology, Topology

4 Comments »

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