Chapter 5: Cotangent Fields

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Chapter 5: Cotangent Fields

Contents
Contents
 1.  Covectors
 2.  The differential of a function
 3.  Pullbacks of covector fields

In this chapter a relatively new concept is introduced, named tangent covectors or cotangent vectors. In next chapter on differential forms, we will see differential 1-forms are covector fields, which consists of tangent covectors.

If you have learned some theory of linear functional analysis, then some concepts, such as linear functionals and dual space, may be familiar to you. Here we consider only vector spaces with finite dimension, whereas in linear functional analysis we are primarily interested in normed vector space which can be infinite dimensional.
1. Covectors Let V be a finite-dimensional vector space.

Since V has finite dimension, the dual space has a simple structure. Given a basis (E _ i) for V, any vector in V can be written as v=v^iE _ i. We take n covectors in V^\ast which are \varepsilon^1,\dots,\varepsilon^n\in V^\ast defined by \varepsilon^j(v)=v^j. We can show these covectors form a basis for V^\ast. Let \omega be any covectors on V. Then \omega(v)=v^i\omega(E _ i)=\omega(E _ i)\varepsilon^i(v), indicating \omega=\omega(E _ i)\varepsilon^i, so \omega can be represented by these covectors. Moreover they are linearly independent. If k _ i\varepsilon^i=0 then k _ iv^i=0 for all v\in V, so all k _ i=0. Thus (\varepsilon^i) form a basis for V^\ast, called dual basis to (E _ i). We also have \dim V^\ast=\dim V.

With dual basis, we can write
\[
\omega=\omega _ i \varepsilon^i,\quad \omega(v)=\omega _ iv^i.
\]
Suppose A:V\to W is a linear map. We define a linear map A^\ast:W^\ast\to V^\ast, called the dual map or transpose of A, by
\[
(A^\ast \omega)(v)=\omega(Av),\quad \omega\in W^\ast,\, v\in V.
\]It is easy to see A^\ast\omega\in V^\ast and then A^\ast is a linear map on W^\ast.

Consider the dual of a dual space, V^{\ast\ast}. For each v\in V, we can define a linear map in V^{\ast\ast}:
\begin{align*}
\xi(v):V^\ast & \to\mathbb R \\
\omega & \mapsto \xi(v)(\omega)=\omega(v).
\end{align*}It can be easily checked \xi\in V^{\ast\ast}.

Such a \xi has an interesting property. The map \xi:V\to V^{\ast\ast} is actually an isomorphism that is canonically defined, so we can identify V^{\ast\ast} with V itself. To see why \xi is an isomorphism, it suffices to show it is injective since \dim V^{\ast\ast}=\dim V, by the knowledge of linear algebra. If v\neq0 we just need to show \xi(v)\neq0. Extend v to a basis (E _ i) for V and let (\varepsilon^i) be the corresponding dual basis. Then \xi(v)(\varepsilon^1)=\varepsilon^1(E _ 1)=1, which shows \xi(v) is not zero.

Tangent covectors

The tangent space about a point of a manifold is a linear space, so it makes sense to consider its dual space. The dual space is called the cotangent space.

Let (U,(x^i)) be a chart of M. For each p\in U the coordinate basis (\frac{\partial}{\partial x^i}| _ p) can give us a dual basis for T _ p^\ast M, which we denote it by (\mathrm dx^i)| _ p for the moment. This is only a notation here but we will assign a meaning to it later so that this makes good sense.

Similar to Chapter 2, the cotangent bundle of M is the disjoint union
\[
T^\ast M=\coprod _ {p\in M}T _ p^\ast M.
\]In a chart, for each p the dual basis for T^\ast _ pM dual to (\frac{\partial}{\partial x^i}|p) is (\mathrm dx^i| _ p). Then we obtain n maps from U to T^\ast M, called coordinate covector field.

A covector field on M is a map from M to TM that assigns to each point p a tangent covector at p. We will see in next chapter a covector field can also be called a differential 1-form.

Change of coordinates

In Chapter 2 we have derived the transformation rule for coordinate vector fields under the change of coordinates:
\[
\frac{\partial}{\partial x^i}\Big| _ p=\frac{\partial\tilde x^j}{\partial x^i}(p)\frac{\partial}{\partial\tilde x^j}\Big| _ p.
\]Here we are sloppy so p is to denote either a point in M or its coordinate representation as appropriate. In the case of covector field, we write \omega in coordinate in both charts: \omega=\omega _ i\mathrm dx^i| _ p=\tilde \omega _ j\mathrm d\tilde x^j| _ p. Then
\[
\omega _ i=\omega\Big(\frac{\partial}{\partial x^i}\Big| _ p\Big)=\omega\Big(\frac{\partial\tilde x^j}{\partial x^i}(p)\frac{\partial}{\partial\tilde x^j}\Big| _ p\Big)=\frac{\partial\tilde x^j}{\partial x^i}(p)\tilde\omega _ j.
\]
Recall that for a tangent vector v, the transformation is
\[
\tilde v^j=\frac{\partial\tilde x^j}{\partial x^i}(p)v^i.
\]Compare these two transformations. We can find they transform in the opposite direction, and it is customary to call tangent vectors contravariant vectors and cotagent vectors covariant vectors. Of course these terminologies do not make a lot of sense. They are provided for informational purposes only.
2. The differential of a function In Chapter 2 we defined the differential of a smooth map F:M\to N to be a linear map between the tangent spaces. Here we introduce a concept in terms of tangent covectors which is also called differential. We will soon discuss the relationship between these two concepts.

We now compute the coordinate representation of \mathrm df. In a chart suppose \mathrm df _ p in coordinates is A _ i(p)\mathrm dx^i| _ p for some functions A _ i:U\to\mathbb R. Then these coefficient functions can be computed:
\[
A _ i(p)=\mathrm df _ p\Big(\frac{\partial}{\partial x^i}\Big| _ p\Big)=\frac{\partial}{\partial x^i}\Big| _ pf=\frac{\partial\hat f}{\partial x^i}(\hat p).
\]Thus,
\[
\mathrm df _ p=\frac{\partial f}{\partial x^i}(p)\mathrm dx^i\big| _ p.
\]In terms of covector fields,
\[
\mathrm df=\frac{\partial f}{\partial x^i}\mathrm dx^i.
\]
Suppose the coordinate map is \varphi=(r^1,\dots,r^n). We consider f=r^i:U\to\mathbb R, that is, f is one of the coordinate functions. Then
\[
\mathrm dr^j\big| _ p=\frac{\partial r^j}{\partial x^i}(p)\mathrm d x^i\big| _ p=\mathrm d x^j\big| _ p.
\]This tells us that the coordinate covector field \mathrm dx^j is none other than the differential \mathrm d r^j! This is the reason why we adopt the notation \mathrm dx^j| _ p,

Equivalence of definitions

In Chapter 2 we defined \mathrm df _ p as a linear map from T _ p(M) to T _ {f(p)}\mathbb R. In this section we defined \mathrm df _ p as a covector in T^\ast(M), that is, a linear map from T _ p(M) to \mathbb R. These two are actually the same thing.

Recall that T _ {f(p)}\mathbb R\cong\mathbb R. If v\in T _ {f(p)}\mathbb R is written in terms of coordinate basis: v=\tilde v\frac{\mathrm d}{\mathrm dt} where \tilde v\in\mathbb R, then an isomorphism is given by v\mapsto \tilde v.

If \mathrm df _ p is treated as the differential of a linear map from T _ p(M) to T _ {f(p)}\mathbb R, then for any v=v^i\frac{\partial}{\partial x^i}| _ {p}\in T _ pM, by the differentials in coordinates, we have
\[
\mathrm df _ p(v)=\Big(v^i\frac{\partial\hat f}{\partial x^i}(\hat p)\Big)\frac{\mathrm d}{\mathrm dt}\Big| _ {\hat p}.
\]As usual \hat p is f(p). Now if \mathrm df _ p is a covector in T^\ast(M), then
\[
\mathrm df _ p(v)=\frac{\partial f}{\partial x^i}(p)\mathrm dx^i\big| _ p(v)=v^i\frac{\partial\hat f}{\partial x^i}(\hat p).
\]From the canonical isomorphism T _ {f(p)}\mathbb R\cong\mathbb R, the equivalence of the definition is clear.
3. Pullbacks of covector fields A smooth map yields a linear map on tangent vectors called the differential. Dualizing this leads to a linear map on covectors going in the opposite direction.

The pullback \mathrm dF _ p^\ast is characterized by
\[
\mathrm dF _ p^\ast(\omega)(v)=\omega(\mathrm d F _ p(v)),\quad \omega\in T^\ast _ {F(p)}N,\, v\in T _ pM.
\]
In the case of vector fields, we only define the pushforward under smooth maps when the maps are diffeomorphisms. A surprising fact of covector fields is that covector fields always pull back to covector fields.

In contrast to the vector field case, there is no ambiguity here about what point to pull back from: the value of F^\ast \omega at p is the pullback of \omega at F(p).

Proof. By definition and the linearity of \mathrm dF _ p^\ast,
\begin{align*}
&(F^{\ast}(u \omega)) _ {p} =\mathrm d F _ {p}^{\ast}((u \omega) _ {F(p)})=\mathrm d F _ {p}^{\ast}(u(F(p)) \omega _ {F(p)}) \\
={}&u(F(p))\,\mathrm d F _ {p}^{\ast}(\omega _ {F(p)})=u(F(p))(F^{\ast} \omega) _ {p}=((u \circ F) F^{\ast} \omega) _ {p}.
\end{align*}Suppose u is smooth. By definition,
\begin{align*}
(F^{\ast}\mathrm d u) _ {p}(v) & =(\mathrm d F _ {p}^{\ast}(\mathrm d u _ {F(p)}))(v)
=\mathrm d u _ {F(p)}(\mathrm d F _ {p}(v)) \\
& =\mathrm d F _ {p}(v) u =v(u \circ F) =\mathrm d(u \circ F) _ {p}(v).
\end{align*}The proof is done.

.

With this proposition, the computation of pullbacks in coordinates is exceedingly simple. Let p\in M and (V,(y^i)) be a chart containing F(p) for N. Suppose \omega is a covector field on N and write it in coordinates: \omega=\omega _ i\, \mathrm dy^i. Then,
\[
F^\ast\omega=F^\ast(\omega _ i\, \mathrm dy^i)=(\omega _ i\circ F)F^\ast\, \mathrm dy^i=(\omega _ i\circ F)\, \mathrm d(y^i\circ F).
\]In other words, to compute F^\ast\omega, all we need to do is substitute the component functions of F for the coordinate functions of N everywhere they appear in \omega! For example, let F:\mathbb R^3\to\mathbb R^2 be given by (u,v)=F(x,y,z)=(x^2y,y\sin z) and let \omega=u\, \mathrm dv+v\mathrm du. Then the pullback F^\ast\omega is (u\circ F)\, \mathrm d(v\circ F)+(v\circ F)\, \mathrm d(u\circ F)=(x^2y)\, \mathrm d(y\sin z)+(y\sin z)\, \mathrm d(x^2y)=\cdots