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


  1. […] Next, we do not assume that is a vector field — it is a function but not necessarily a differentiable one — but we assume that it satisfies the conclusion of the preceding paragraph. That is, for every chart with each is smooth. Now we will show that is smooth for every smooth , not just those that arise as coordinate functions. To see this, we use the decomposition of into coordinate vector fields: […]

    Pingback by Identifying Vector Fields « The Unapologetic Mathematician | May 25, 2011 | Reply

  2. […] a special case, if is a coordinate patch then we have the coordinate vector fields . The fact that partial derivatives commute means that the brackets […]

    Pingback by Brackets and Flows « The Unapologetic Mathematician | June 18, 2011 | Reply

  3. […] I left off last time by pointing out that coordinate vector fields […]

    Pingback by Building Charts from Vector Fields « The Unapologetic Mathematician | June 22, 2011 | Reply

  4. […] to return to differential geometry, let’s say we have a coordinate patch . We get a basis of coordinate vector fields, which let us define the matrix-valued […]

    Pingback by Inner Products on 1-Forms « The Unapologetic Mathematician | October 1, 2011 | Reply

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s

%d bloggers like this: