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”.

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

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