February 23, 2020

# Vector, Covector and Tensor Fields

Last time I said that our motivation for defining the tangent bundle was so that we'd be able to define smooth vector fields on a manifold, but I didn't quite get there! Let's do that now, since we are already equipped to do so!

Definition. Let $M$ be a smooth manifold and $TM$ its tangent bundle. A smooth vector field on $M$ is a smooth section of the bundle projection map $\pi:TM\to M$.

We write $\Gamma(TM)$ to denote the set of all vector fields on $M$.

This may seem like gibberish at first, so let's parse the definition a bit. If $X$ is a vector field on $M$, then as a section of $\pi$ it must be a map $X:M\to TM$ for which $\pi\circ X=1_M$. So it has the correct domain and codomain, since we want a vector field to be an assignment of a tangent vector to each point of our manifold. Furthermore, it's a smooth map, which makes sense because $M$ and $TM$ are both manifolds. Lastly, it assigns to each point $p\in M$ a tangent vector in $T_p M$ based at $p$. That's because, as we saw last time, sections provide representatives for the level sets of a function. More formally, since $\pi(X(p))=p$, it must be that $X(p)$ is of the form $(Y, p)\in TM$ for some vector $Y\in T_pM$, so that $\pi\big((Y, p)\big)=p$.

This is only part of the picture, though, because we want to be able to work with vector fields algebraically. One might think to turn $\Gamma(TM)$ into a vector field over $\R$. This would allow us to scale vector fields and add them together in the natural ways. However, scaling an entire vector field by a fixed amount is pretty boring. We can do much better!

Notice that the set of smooth functions $C^\infty(M)$ is a commutative ring with unity under the usual addition and multiplication of functions. It fails to be a field, though, because it lacks inverses for certain nonzero elements (a nonzero function which vanishes anywhere has no multiplicative inverse). If we use smooth functions as our scalars by turning $\Gamma(TM)$ into a module over $C^\infty(M)$, we gain the power to scale vector fields by smooth functions. That is, the amount of scaling will depend on where in the manifold we look.

We think of tangent vectors as maps that act on smooth functions and spit out real numbers. Unfortunately, vector fields as we defined them do not act on smooth functions, they act on points in $M$ and produce tangent vectors. The following trick allows us to think of them instead as objects which act on smooth functions again.

Definition. Given a smooth manifold $M$ and a vector field $X\in\Gamma(TM)$, the associated derivation of $X$ is the map $\hat{X}:C^\infty(M)\to C^\infty(M)$ defined by

$$(\hat{X}f)(p)=X(p)f$$

for any point $p\in M$.

The fact that the assoiated derivation of a vector field is actually a derivation follows from the fact that tangent vectors are derivations.

Theorem. The associated derivation of a vector field is a derivation on $C^\infty(M)$.

Proof. Let $M$ be a smooth manifold and choose a vector field $X\in\Gamma(TM)$. If $f$ and $g$ are smooth functions in $C^\infty(M)$ then for any point $p\in M$,

\begin{align} (\hat{X}(fg))(p) &= X(p)(fg) \\ &= fX(p)g + gX(p)f \\ &= f(\hat{X}(g))(p) + g(\hat{X}(f))(p) \\ &= (f\hat{X}g + g\hat{X}f)(p) \end{align}

because $X(p)$ is a tangent vector in $T_pM$ which we know to be a derivation. It follows that $\hat{X}(fg) = f\hat{X}g + g\hat{X}f$ and thus $\hat{X}$ is a derivation of smooth functions.

Because we like to think of vector fields as objects which act on smooth functions, we will henceforth treat a vector field and its associated derivation as the same object and never speak of this again. Just remember what's really going on behind the scenes, so that the definitions all line up properly.

It is convenient now to make the following definition, which extends our previous definition of partial derivative operators with respect to a chart component at a point:

Definition. Let $M$ be a smooth manifold and let $(U, x)$ be a chart on $M$. The partial derivative vector field with respect to the $i$th component of the chart map $x$ is the vector field

$$\frac{\partial}{\partial x^i}:C^\infty(M)\to C^\infty(M)$$

defined by

$$\frac{\partial}{\partial x^i}f = \partial_i(f\circ x^{-1})$$

for any smooth function $f\in C^\infty(M)$, where $\partial_i$ represents the normal partial derivative operator with respect to the $i$th component in the sense of multivariable calculus.

There's just one minor issue, though. Only modules over division rings are guaranteed to have a basis. Since $C^\infty(M)$ is not a division ring, it is possible that $\Gamma(TM)$ might not have a basis for some manifolds $M$.

Example. Let $M=\R$. Vector fields in $\R$ are pretty boring. They're all of the form

$$X=f\frac{\partial}{\partial x}$$

for some smooth function $f$, where $(\R, x)$ is the global chart on $\R$. That's because at each point, a tangent vector can only point in one of two directions: negative or positive. Thus, $\Gamma(T\R)$ in this case has $\Big(\dfrac{\partial}{\partial x}\Big)$ as a basis, since every vector field is a $C^\infty(M)$-multiple of $\dfrac{\partial}{\partial x}$.

Example. Let $M=S^2$, the two-dimensional unit sphere. It is a theorem of algebraic topology (the hairy ball theorem) that there is no nonvanishing continuous tangent vector field on spheres of even dimension. Thus, there is certainly no nonvanishing smooth tangent vector field on $S^2$. Basis vector fields can't vanish, and so it follows that there is no basis for $\Gamma(TS^2)$.

This doesn't really turn out to be a problem though, because it turns out that $\Gamma(TM)$ always has a local basis on any chart. That is, if we choose a chart $(U, x)$ on $M$ then $\Big(\dfrac{\partial}{\partial x^i}\Big)_{i=1}^n$ is a basis for $\Gamma(TU)$. Since we usually work in terms of charts anyway, this is good enough for our purposes.

Next we turn to the cotangent bundle, so we can develop covector fields.

Definition. Given a smooth manifold $M$, its cotangent bundle is the set

$$T^*M = \bigsqcup_{p\in M}T_p^* M.$$

Just as we did for the tangent bundle, we can turn the cotangent bundle into a topological space by defining a bundle projection map $\pi^*:T^*M\to M$ and using the induced topology. We can then give $T^*M$ a smooth manifold structure exactly as we did for $TM$, and of course it will also have twice the dimension of the original manifold. I will omit these details, though, so that this post is not an almost exact duplicate of the previous post.

This allows us to define smooth covector fields in the obvious way:

Definition. Let $M$ be a smooth manifold and $T^*M$ its cotangent bundle. A smooth covector field on $M$ is a smooth section of the bundle projection map $\pi^*:T^*M\to M$.

We write $\Gamma(T^*M)$ to denote the set of all covector fields on $M$.

While vector fields act on functions, covector fields act on vector fields. However, we give them the same $C^\infty(M)$-module structure.

Definition. Let $M$ be a smooth manifold and let $f\in C^\infty(M)$ be a smooth function. The differential of $f$ is the covector field

$$\d f:\Gamma(TM)\to C^\infty(M)$$

defined by

$$(\d f)X = Xf$$

for any smooth vector field $X\in\Gamma(TM)$.

Just like with vector fields, given a chart $(U, x)$, the covector fields $(\d x^i)_{i=1}^n$ form a local basis for $\Gamma(T^*U)$.

Now that we have a grip of smooth vector and covector fields on a smooth manifold, along with their $C^\infty(M)$-module structures, we can define smooth tensor fields — one of the most important objects in differential geometry.

Definition. Let $M$ be a smooth manifold. A smooth tensor field on $M$ is a $C^\infty(M)$-multilinear map

$$T:\prod_{i=1}^j \Gamma(T^*M)\times\prod_{i=1}^k \Gamma(TM)\to\R.$$

We say that the rank of this tensor field is $(j,k)$.

In any chart $(U, x)$, a smooth tensor field is locally a tensor product of the form

$$T = \sum_{\substack{i\in I\\ j\in J}}\bigotimes_{i\in I}f^i\frac{\partial}{\partial x^i}\otimes\bigotimes_{j\in J}g_j\d x^j.$$

There's a lot more to be said about tensor fields, but I'll leave it at that for now.

One last thing I'll mention, but won't go into too much detail on, is that the pushforward and pullback can be extended to vector and covector fields, respectively. Hence, the pushforward acts as a covariant functor from the category of smooth manifolds with smooth maps into the category of vector bundles. Likewise, the pullback acts as a contravariant functor between the same categories.