# vector field

## Vector fields in $\mathbb{R}^{n}$

###### Definition.

A vector field $v$ on some open set $U\subset\mathbb{R}^{n}$ is a function which associates a vector $v(x)$ to each point $x$ of $U$. That is, $v$ is a function from $\mathbb{R}^{n}$ to $\mathbb{R}^{n}$. However, we give it a rather special interpretation   . This distinction will become clear when we generalize. If $v$ is differentiable   , then we say the vector field is differentiable.

Here and throughout this entry we use “differentiable” to mean “infinitely differentiable”. Other smoothness criteria (continuity, $C^{r}$, real analytic) can of course be substituted in the obvious way.

###### Example.

Let $v(x,y)=(-y,x)$. At the point $(x,y)$, we view this as an arrow pointing counterclockwise around the origin, with a length proportional to (in fact equal to) the distance from the origin. So this might be the field of velocity vectors of a fluid rotating en masse.

## Vector fields on manifolds

Suppose first that $M$ is a manifold  of dimension  $n$ embedded in $\mathbb{R}^{m}$. Then we would like to define a vector field on $M$ to be a function as before. But $M$ has a natural notion of tangent space  at each point, so now we would like all the vectors to be tangent   to $M$. If we are to define a vector field as before, as simply a function from some open set $U$ of $M$ to $\mathbb{R}^{n}$, we must pick a basis for the tangent space at each point; the basis elements must, however, be differentiable. It is not obvious that we can always pick such a differentiable basis (think of the tangent spaces to the Möbius strip). The problem is that the tangent spaces form a fiber bundle  , and this may not be trivial. We could get around this by shrinking $U$ until we could always construct such a basis, but then we would have to describe how to convert bases on the overlap. This can be done, and it is one way to approach the theory of differential manifolds; at this point one might as well do away with the ambient space $\mathbb{R}^{n}$.

Instead of doing this, we will take a coordinate-free approach. The first thing to notice is that given a tangent vector to a manifold, it makes sense to take a differentiable function on the manifold and ask what the directional derivative   along the vector is. If we have two different tangent vectors, then we can find a function whose directional derivative along each vector is different. So we could identify tangent vectors with directional derivative operators. This is how we will define them in a general setting.

###### Definition.

Let $M$ be a differential manifold of dimension $n$, and let $x$ be a point on $M$. Let $V$ be the space of differentiable functions defined in some neighborhood of $x$. Then a tangent vector $X$ at $x$ is a linear operator on $V$ such that

1. 1.

$X(fg)=fX(g)+gX(f)$, and

2. 2.

if $f$ and $g$ agree on some neighborhood of $x$, then $X(f)=X(g)$.

We write the set of tangent vectors at $x$ as $T_{x}M$; it is a vector space  of dimension $n$.

A vector field on an open set $U\subset M$ is a family of tangent vectors $X_{x}$ at each $x$ such that for every differentiable function $f$ on an open set $V\subset U$, the function $x\mapsto X_{x}(f)$ is differentiable.

What does this definition actually mean? Suppose we have a coordinate chart $\phi\colon V\to\mathbb{R}^{n}$ on some open set $V\subset M$. Then we can define a differentiable function $x_{i}$ on $V$ by applying $\phi$ followed by extracting the $i$th coordinate   . So the function $x_{i}$ really extracts the $x_{i}$ coordinate in this coordinate system  . Now, let $X$ be a vector field on $V$. With some thought, we see that

 $X(f)=\sum_{i=1}^{n}X(x_{i})\frac{\partial}{\partial x_{i}}f,$

or, in some sense

 $X=\sum_{i=1}^{n}X(x_{i})\frac{\partial}{\partial x_{i}}.$

Each $X(x_{i})$ is by definition just a differentiable function on $V$.

If we had chosen a different coordinate chart $\psi$ on an open set $W\subset M$, we would obtain coordinate functions $y_{i}$ by analogy  with the $x_{i}$. Then in this coordinate system we would have

 $X=\sum_{i=1}^{n}X(y_{i})\frac{\partial}{\partial y_{i}}.$

If $V$ and $W$ overlap, then on their overlap we can compare the components of $X$ in these two coordinate systems. A calculation will reveal

 $X=\sum_{i=1}^{n}X(y_{i})\frac{\partial}{\partial y_{i}}=\sum_{i=1}^{n}\sum_{j=% 1}^{n}X(x_{j})\frac{\partial y_{i}}{\partial x_{j}}\frac{\partial}{\partial y_% {i}}.$

Observe that this transformation law means that we can’t compare vectors at different points in a coordinate-independent way, or at least, it will require some significant cleverness to transport a vector from one point to another. This is in fact possible with some extra information in the form of a connection   on $M$, allowing parallel transport of vectors along curves. The result will continue to depend on the curve, as you can imagine if you imagine trying to parallel-transport a vector from the North Pole  to Baghdad to Mexico City and then back to the North Pole: it will have rotated. This is in fact a direct result of the curvature  of the Earth.

###### Example.

Let $M$ be $\mathbb{R}^{2}$, and let

 $\nu_{t}(x,y):=\begin{pmatrix}\cos t&-\sin t\\ \sin t&\cos t\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix}.$

Then define a vector field $X$ by

 $f\mapsto\frac{\partial}{\partial t}(f\circ\nu_{t}).$

We have a natural coordinate patch defined by the identity function; in this coordinate system, we can easily calculate that

 $X=-y\frac{\partial}{\partial x}+x\frac{\partial}{\partial y};$

this $X$ is precisely the vector field we had before, viewed in a new and more confusing light. Now, with a little imagination, we can see that even for fixed $t$, the function $\nu_{t}$ is a different kind of object than $X$; while $\nu_{t}$ represents a rotation  , a smooth map from $M$ to itself, while $X$ is a piece of information attached in an essential way to each point, perhaps representing a velocity vector field.

## The Tangent Bundle

###### Definition.

Let $TM$ denote the vector bundle  with base space $M$ having fiber over $x$ given by $T_{x}M$, and having local trivializations over coordinate charts with change of coordinates given by the formulas   in the previous section       . Then the bunlde $TM$ is called the tangent bundle  of $M$.

## References

 Title vector field Canonical name VectorField Date of creation 2013-03-22 11:59:27 Last modified on 2013-03-22 11:59:27 Owner mathcam (2727) Last modified by mathcam (2727) Numerical id 16 Author mathcam (2727) Entry type Definition Classification msc 46E40 Synonym vector-valued function Related topic Gradient  Related topic LeibnizNotationForVectorFields Related topic TangentBundle Related topic TangentSpace Related topic PrimitiveRecursiveVectorValuedFunction