The Unapologetic Mathematician

Mathematics for the interested outsider

Identifying Vector Fields

We know what vector fields are on a region U\subseteq M, but to identify them in the wild we need to verify that a given function sending each p\in U to a vector in \mathcal{T}_pM is smooth. This might not always be so easy to check directly, so we need some equivalent conditions. First we need to define how vector fields act on functions.

If X\in\mathfrak{X}U is a vector field and f\in\mathcal{O}U is a smooth function then we get another function Xf by defining Xf(p)=\left[X(p)\right](f). Indeed, X(p)\in\mathcal{T}_pM, so it can take (the germ of) a smooth function at p and give us a number. Essentially, at each point the vector field defines a displacement, and we ask how the function f changes along this displacement. This action is key to our conditions, and to how we will actually use vector fields.

Firstly, if X is a vector field — a differentiable function — and if (V,x) is a chart with V\subseteq U, then Xx^i is always smooth. Indeed, remember that (V,x) gives us a coordinate patch (\pi^{-1}(V),\bar{x}) on the tangent bundle. Since \bar{x} is smooth and X is smooth, the composition

\displaystyle\bar{x}\circ X\vert_V=(x\circ I\vert_V;X\vert_V(x^1),\dots,x\vert_V(x^n))

is also smooth. And thus each component Xx^i is smooth on V.

Next, we do not assume that X 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 (V,x) with V\subseteq U each Xx^i is smooth. Now we will show that Xf is smooth for every smooth f\in\mathcal{O}V, not just those that arise as coordinate functions. To see this, we use the decomposition of X into coordinate vector fields:

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

which didn’t assume that X was smooth, except to show that the coefficient functions were smooth. We can now calculate that X^i=Xx^i, since

\displaystyle Xx^i=\sum\limits_{j=1}^nX^j\frac{\partial x^i}{\partial x^j}=\sum\limits_{j=1}^nX^j\delta^i_j=X^i

But this means we can write

\displaystyle Xf=\sum\limits_{i=1}^nXx^i\frac{\partial f}{\partial x^i}

which makes Xf a linear combination of the smooth (by assumption) functions Xx^i with the coefficients \frac{\partial f}{\partial x^i}, proving that it is itself smooth.

Okay, now I say that if Xf is smooth for every smooth function f\in\mathcal{O}V on some region V\subseteq U, then X is smooth as a function, and thus is a vector field. In this case around any p\in U we can find some coordinate patch (V,x). Now we go back up to the composition above:

\displaystyle\bar{x}\circ X\vert_V=(x\circ I\vert_V;X\vert_V(x^1),\dots,x\vert_V(x^n))

Everything in sight on the right is smooth, and so the left is also smooth. But this is exactly what we need to check when we’re using the local coordinates (V,x) and (\pi^{-1}(V),\bar{x}) to verify the smoothness of X at p.

The upshot is that when we want to verify that a function X really is a smooth vector field, we take an arbitrary smooth “test function” and feed it into X. If the result is always smooth, then X is smooth. In fact, some authors take this as the definition, regarding the action of X on functions as fundamental, and only later talking in terms of its “value at a point”.


May 25, 2011 - Posted by | Differential Topology, Topology


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