15.3 Tangent, cotangent and tensor spaces

中文

Tangent vectors

Let $x^i=x^i(t)$ be a curve in ${\mathbb{R}}^n$ passing through the point ${\mathbf{x}}_0=\mathbf{x}(t_0)$ . In elementary mathe matics it is common to define the ‘tangent’ to the curve, or ‘velocity’, at ${\mathbf{X}}_0$ as the $n$-vector $\mathbf{v}=(\dot{\mathbf{x}})=\left({\dot{x}}^1,{\dot{x}}^2,\dots ,{\dot{x}}^n\right)$ where ${\dot{x}}^i=(\mathrm{d}{x}^i/\mathrm{d}t{)}_{t=t_0}$ . In an $n$-dimensional manifold it is not satisfactory to define the tangent by its components, since general coordinate transforma tions are permitted. For example, by a rotation ofaxes in ${\mathbb{R}}^n$ it is possible to achieve that the tangent vector has components $\mathbf{v}=(v,0,0,\dots ,0)$ ). A coordinate-independent, or invari ant, approach revolves around the concept of the directional derivative of a differentiable function $f:{\mathbb{R}}^n\to \mathbb{R}$ along the curve at ${\mathbf{X}}_0$

$$Xf=\frac{\mathrm{d}f(\mathbf{x}(t))}{\mathrm{d}t}{|}_{t=t_0}=\frac{\mathrm{d}{x}^i(t)}{\mathrm{d}t}{|}_{t=t_0}\frac{\partial f(\mathbf{x})}{\partial x^i}{|}_{\mathbf{x}={\mathbf{x}}_0},$$

where $X$ is the linear differential operator

$$X={\frac{\mathrm{d}{x}^i(t)}{\mathrm{d}t}|}_{t=t_0}{\frac{\partial }{\partial x^i}|}_{\mathbf{x}={\mathbf{x}}_0}.$$

The value of the operator $X$ when applied to a function $f$ only depends on the values taken by the function in a neighbourhood of ${\mathbf{X}}_0$ along the curve in question, and is independen of coordinates chosen for the space ${\mathbb{R}}^n$ . The above expansion demonstrates, however, that the components of the tangent vector in any coordinates on ${\mathbb{R}}^n$ can be extracted from the directional derivative operator from its coefficients of expansion in terms of coordinate partial derivatives.

The directional derivative operator $X$ is a real-valued map on the algebra ofdifferentiable functions at ${\mathbf{X}}_0$ . Two important properties hold for the map $X:{\mathcal{F}}_{{\mathbf{x}}_0}({\mathbb{R}}^n)\mathbb{R}:$

(i) It is linear on the vector space ${\mathcal{F}}_{{\mathbf{x}}_0}({\mathbb{R}}^n)$ ; that is, for any pair of functions $f,g$ and rea numbers $a,b$ we have $X(af+bg)=aXf+bXg$

(ii) The application of $X$ on any product offunctions $fg$ in the algebra ${\mathcal{F}}_{{\mathbf{x}}_0}({\mathbb{R}}^n)$ ) is determined by the Leibnitz rule, $X(fg)=f({\mathbf{x}}_0)Xg+g({\mathbf{x}}_0)Xf$

These two properties completely characterize the class of directional derivative operators (see Theorem 15.1), and will be used to motivate the definition ofa tangent vector at a poin of a general manifold.

A tangent vector $X_p$ at any point $p$ of a differentiable manifold $M$ is a linear map from the algebra of differentiable functions at $p$ to the real numbers, $X_p:{\mathcal{F}}_p(M)\to \mathbb{R}$ , which satisfies the Leibnitz rule for products:

$$X_p(af+bg)=a{X}_{p}f+b{X}_{p}g(\text{ linearity }),\text{\hspace{1em}(15.3)}$$

$$X_p(fg)=f(p)X_{p}g+g(p)X_{p}f(\text{ Leibnitz rule }).\text{\hspace{1em}(15.4)}$$

The set of tangent vectors at $p$ form a vector space $T_p(M)$ , since any linear combination $a{X}_p+b{Y}_p$ of tangent vectors at $p,$ , defined by

$$(a{X}_p+b{Y}_p)f=a{X}_{p}f+b{Y}_{p}f,$$

is a tangent vector at $p$ since it satisfies Eq. (15.3) and Eq. (15.4). It is called the tangent space at $p.$ . If $(U,\varphi )$ is any chart at $p$ with coordinate functions $x^i$ , define the operators

$${\left({\partial }_{x^i}\right)}_p\equiv \frac{\partial }{\partial x^i}{|}_p:{\mathcal{F}}_p(M)\to \mathbb{R}$$

by

$${\left({\partial }_{x^i}\right)}_{p}f\equiv \frac{\partial }{\partial x^i}{|}_{p}f=\frac{\partial \hat{f}(x^1,\dots ,x^n)}{\partial x^i}{|}_{\mathbf{x}=\varphi (p)},\text{\hspace{1em}(15.5)}$$

where $\hat{f}=f\circ {\varphi }^{-1}:{\mathbb{R}}^n\to \mathbb{R}$ . These operators are clearly tangent vectors since they satisfy Eq. (15.3) and Eq. (15.4). Thus any linear combination

$$X_p=X^i\frac{\partial }{\partial x^i}{|}_p\equiv \sum _{i=1}^n{X}^i\frac{\partial }{\partial x^i}{|}_p\text{ where }{X}^i\in \mathbb{R}$$

is a tangent vector. The coefficients $X^j$ can be computed from the action of $X$ on the coordinate functions $x^j$ themselves:

$$X_p{x}^j=X^i\frac{\partial x^j}{\partial x^i}{|}_{\mathbf{x}=\varphi (p)}=X^i{\delta }_i^j=X^j.$$

Theorem 15.1 · A coordinate basis of the tangent space

$H(U,\varphi ;x^i)$ is a chart at $p\in M,$ , then the operators ${\left({\partial }_{x^i}\right)}_p$ defined by Eq. (15.5)form a basis ofthe tangent space $T_p(M)$ , and its dimension is $n=\dim \mathcal{M}$

Proof

Let $X_p$ be a tangent vector at the given fixed point $p.$ Firstly, it follows by the Leibnitz rule Eq. (15.4) that $X_p$ applied to a unit constant function $f=1$ always results in zero, $X_{p}1=0$ , for

$$X_{p}1=X_p(1.1)=1.X_{p}1+1.X_{p}1=2X_{p}1.$$

By linearity, $X_p$ applied to any constant function $f=c$ results in zero, $Xc=X(c.1)=cX1=0$

Set the coordinates of $p$ to be $\varphi (p)=\mathbf{a}=(a^1,a^2,...,a^n)$ , and let $\mathbf{y}=\varphi (q)$ be any point in a neighbourhood ball $B_r(\mathbf{a})\subseteq \varphi (U)$ . The function $\hat{f}=f\circ {\varphi }^{-1}$ can be written as

$$\begin{array}{l}\hat{f}(y^1,y^2,\dots ,y^n)=\hat{f}(y^1,y^2,\dots ,y^n)-\hat{f}(y^1,\dots ,y^{n-1},a^n)\\ +\hat{f}(y^1,\dots ,y^{n-1},a^n)-\hat{f}(y^1,\dots ,y^{n-2},a^{n-1},a^n)+\dots \\ +\hat{f}(y^1,a^2,\dots ,a^n)-\hat{f}(a^1,a^2,\dots ,a^n)+\hat{f}(a^1,a^2,\dots ,a^n)\\ =\hat{f}(a^1,a^2,\dots ,a^n)\\ +{\sum }_{i=1}^n{\int }_0^1\frac{\partial \hat{f}(y^1,\dots ,y^{i-1},a^i+t(y^i-a^i),a^{i+1},\dots ,a^n)}{\partial t}dt\\ =\hat{f}(\mathbf{a})+{\sum }_{i=1}^n{\int }_0^1\frac{\partial \hat{f}}{\partial x^i}(y^1,\dots ,y^{i-1},a^i+t(y^i-a^i),a^{i+1},\dots ,a^n)\\ \times dt(y^i-a^i).\end{array}$$

Hence, in a neighbourhood of $\mathbf{a}=\varphi (vp)$ , any function $\hat{f}$ can be written in the form

$$\hat{f}(\mathbf{y})=\hat{f}(\mathbf{a})+{\hat{f}}_i(\mathbf{y})(y^i-a^i)\text{\hspace{1em}(15.6)}$$

where the functions ${\hat{f}}_i(y^1,y^2,\dots ,y^n)$ are differentiable at $a$. Thus, in a neighbourhood of $p_{:}$ ,

$$f(q)=\hat{f}\circ \varphi =f(p)+f_i(q)(x^i(q)-a^i)$$

where $f_i={\hat{f}}_i\circ \varphi \in {\mathcal{F}}_p(M)$ . Using the linear and Leibnitz properties of $X_p$

$$X_{p}f=X_{p}f(p)+X_p{f}_i\left(x^i(p)-a^i\right)+f_i(p)\left(X_p{x}^i-X_p{a}^i\right)=f_i(p)X_p{x}^i$$

since $X_{p}c=0$ for any constant $c,$ , and $x^i(p)=a^i$ . Furthermore,

$$f_i(p)=\frac{\partial \hat{f}}{\partial x^i}\left(a^1,\dots ,a^n\right){\int }_0^{1}dt={\frac{\partial }{\partial x^i}|}_{p}f,$$

and the tangent vectors ${\left({\partial }_{x^i}\right)}_p$ span the tangent space $T_p(M)$ ,

$$X_p=X^i\frac{\partial }{\partial x^i}{|}_p=X^i{\left({\partial }_{x^i}\right)}_p\text{ where }{X}^i=X_p{x}^i.\text{\hspace{1em}(15.7)}$$

To show that they form a basis, we need linear independence. Suppose

$$A^i\frac{\partial }{\partial x^i}{|}_p=0,$$

then the action on the coordinate functions $f=x^j$ gives

$$0=A^i\frac{\partial }{\partial x^i}{|}_p{x}^j=A^i\frac{\partial x^j}{\partial x^i}{|}_{\mathbf{a}}=A^i{\delta }_i^j=A^j$$

as required.

This proof shows that, for every tangent vector $X_p,$ , the decomposition given by Eq. (15.7) is unique. The coefficients $X^i=X_p{x}^i$ are said to be the components of the tangent vector $X_p$ in the chart $(U;x^i)$ .

How does this definition of tangent vector relate to that given earlier for a curve in ${\mathbb{R}}^{n_{\cdot }}$ Let $\gamma :(a,b)M$ be a smooth parametrized curve passing through the point $p\in M$ at $t=t_0$ . Define the tangent vector to the curve at $p$ to be the operator ${\dot{\gamma }}_p$ defined by the action on an arbitrary differentiable function $f$ at $p,$

$${\dot{\gamma }}_{p}f=\frac{\mathrm{d}f\circ \gamma (t)}{\mathrm{d}t}{|}_{t=t_0}.$$

It is straightforward to verify that the ${\dot{\gamma }}_p$ is a tangent vector at $p_{:}$ , as it satisfies Eqs. Eq. (15.3) and Eq. (15.4). In a chart with coordinate functions $x^i$ at $p_{:}$ , let the coordinate representation of the curve be $\hat{\gamma }=\varphi \circ \gamma =\left({\gamma }^1(t),\dots ,{\gamma }^n(t)\right)$ . Then

$${\dot{\gamma }}_{p}f=\frac{\mathrm{d}\hat{f}\circ \hat{\gamma }(t)}{\mathrm{d}t}{|}_{t=t_0}=\frac{\partial \hat{f}}{\partial x^i}{|}_{\varphi (p)}\frac{\mathrm{d}{\gamma }^i(t)}{\mathrm{d}t}{|}_{t=t_0}$$

and

$${\dot{\gamma }}_p={\dot{\gamma }}^i(t_0)\frac{\partial }{\partial x^i}{|}_p\text{ where }{\dot{\gamma }}^i(t)=\frac{\mathrm{d}{\gamma }^i(t)}{\mathrm{d}t}.$$

In the case $M={\mathbb{R}}^n$ the operator ${\dot{\gamma }}_p$ is precisely the directional derivative of the curve.

It is also true that every tangent vector is tangent to some curve. For example, the basis vectors ${\left({\partial }_{x^i}\right)}_p$ are tangent to the ‘coordinate lines’ at $p={\varphi }^{-1}(\mathbf{a})$ ,

$${\gamma }^i:t\mapsto {\varphi }^{-1}((a^1,a^2,\dots ,x^i=a^i+t-t_0,\dots ,a^n)).$$

An arbitrary tangent vector $X_p=X^i{\left({\partial }_{x^i}\right)}_p$ at $p$ is tangent to the curve

$$\gamma :t\mapsto {\varphi }^{-1}((a^1+X^1(t-t_0),\dots ,x^i=a^i+X^i(t-t_0),\dots ,a^n+X^n(t-t_0))).$$

Example 15.7

The curves $\alpha ,\beta$ and $\gamma$ on ${\mathbb{R}}^2$ given respectively by

$$\begin{array}{ll}{\alpha }^1(t)=1+\sin t\cos t& {\alpha }^2(t)=1+3t\cos 2t,\\ {\beta }^1(t)=1+t& {\beta }^2(t)=1+3t{\mathrm{e}}^{3t},\\ {\gamma }^1(t)={\mathrm{e}}^t& {\gamma }^2(t)={\mathrm{e}}^{3t},\end{array}$$

all pass through the point $p=(1,1)$ at $t=0$ and are tangent to each other there,

$${\dot{\alpha }}_p={\dot{\beta }}_p={\dot{\gamma }}_p={\left({\partial }_{x^1}\right)}_p+3{\left({\partial }_{x^2}\right)}_p.$$

Cotangent and tensor spaces

The dual space $T_p^{\ast }(M)$ associated with the tangent space at $p\in M$ is called the cotangent space at $p.$ . It consists of all linear functionals on $T_p(M)$ , also called covectors or 1-forms at $p.$ . The action of a covector ${\omega }_p$ at $p$ on a tangent vector $X_p$ will be denoted by ${\omega }_p(X_p)⟨{\omega }_p,X_p⟩$ or $⟨X_p,{\omega }_p⟩$ . From Section 3.7 we have that $\dim$ $T_p^{\ast }(M)=n=\dim T_p(M)=$ $\dim$ $M$.

If $f$ is any function that is differentiable at $p_{:}$ , we define its differential at $p$ to be the covector $(\mathrm{d}f{)}_p$ whose action on any tangent vector $X_p$ at $p$ is given by

$$⟨(\mathrm{d}f{)}_p,X_p⟩=X_{p}f.\text{\hspace{1em}(15.8)}$$

This is a linear functional since, for any tangent vectors $X_p,Y_p$ and scalars $a,b\in \mathbb{R}$

$$⟨(\mathrm{d}f{)}_p,a{X}_p+b{Y}_p⟩=(a{X}_p+b{Y}_p)f=a{X}_{p}f+b{Y}_{p}f=a⟨(\mathrm{d}f{)}_p,X_p⟩+b⟨(\mathrm{d}f{)}_p,Y_p⟩.$$

Given a chart $(U,\varphi ;x^i)$ at $p,$ the differentials of the coordinate functions have the property

$$⟨(\mathrm{d}{x}^i{)}_p,X_p⟩=X_p{x}^i=X^i$$

where $X^i$ are the components of the tangent vector, $X_p=X^i{\left({\partial }_{x^i}\right)}_p$ . Applying $(\mathrm{d}{x}^i{)}_p$ to the basis tangent vectors, we have

$$⟨(\mathrm{d}{x}^i{)}_p,{\left({\partial }_{x^j}\right)}_p⟩=\frac{\partial }{\partial x^j}{|}_p{x}^i=\frac{\partial x^i}{\partial x^j}{|}_{\varphi (p)}={\delta }_j^i.$$

Hence the linear functionals $(\mathrm{d}{x}^1{)}_p,(\mathrm{d}{x}^2{)}_p,\dots ,(\mathrm{d}{x}^n{)}_p$ are the dual basis, spanning the cotangent space, and every covector at $p$ has a unique expansion

$${\omega }_p=w_i(\mathrm{d}{x}^i{)}_p\text{ where }{w}_i=⟨{\omega }_p,{\left({\partial }_{x^i}\right)}_p⟩.$$

The $w_i$ are called the components of the linear functional ${\omega }_p$ in the chart $(U;x^i)$ .

The differential of any function at $p$ has a coordinate expansion

$$(\mathrm{d}f{)}_p=f_i(\mathrm{d}{x}^i{)}_p$$

where

$$f_i=⟨(\mathrm{d}f{)}_p,{\left({\partial }_{x^i}\right)}_p⟩=\frac{\partial }{\partial x^i}{|}_{p}f=\frac{\partial \hat{f}}{\partial x^i}{|}_{\varphi (p)}.$$

A common way of writing this is the ‘chain rule

$$(\mathrm{d}f{)}_p=f_{,i}(p)(\mathrm{d}{x}^i{)}_p\text{\hspace{1em}(15.9)}$$

where

$$f_{,i}=\frac{\partial \hat{f}}{\partial x^i}\circ \varphi .$$

These components are often referred to as the gradient ofthe function at $p.$ . Differentials have never found a comfortable place in calculus as non-vanishing quantities that are ‘arbitrarily small’. The concept of differentials as linear functionals avoids these problems, yet has al the desired properties such as the chain rule of multivariable calculus.

As in Chapter 7, a tensor of type $(r,s)$ at $p$ is a multilinear functional

$$A_p:\underset{r}{\underbrace{T_p^{\ast }(M)\times T_p^{\ast }(M)\times \cdots \times T_p^{\ast }(M)}}\underset{s}{\underbrace{\times T_p(M)\times \cdots \times T_p(M)}}\to \mathbb{R},$$

We denote the tensor space of type $(r,s)$ at $p$ by $T_p^{(r,s)}(M)$ ). It is a vector space of dimension $n^{r+s}$

Vector and tensor fields

A vector field $X$ is an assignment of a tangent vector $X_p$ at each point $p\in M$ . In other words, $X$ is a map from $M$ to the set ${\bigcup }_{p\in M}{T}_p(M)$ with the property that the image of every point, $X(p)$ , belongs to the tangent space $T_p(M)$ at $p$ . We may thus write $X_p$ in place of $X(p)$ . The vector field is said to be differentiable or smooth if for every differentiable function $f\in \mathcal{F}(M)$ the function $Xf$ defined by

$$(Xf)(p)=X_{p}f$$

is differentiable, $Xf\in \mathcal{F}(M)$ . The set of all differentiable vector fields on $M$ is denoted $\tau (M)$

Exercise

Show that $\mathcal{T}(M)$ forms a module over the ring of functions $\mathcal{F}(M)!$ : if $X$ and $Y$ are vecto fields, and $f\in \mathcal{F}(M)$ then $X+fY$ is a vector field.

Every smooth vector field defines a map $X:\mathcal{F}(M)\mathcal{F}(M)$ , which is linea

$$X(af+bg)=aXf+bXg\text{ for all }f,g\in \mathcal{F}(M)\text{ and all }a,b\in \mathbb{R},$$

and satisfies the Leibnitz rule for products

$$F(fg)=fXg+gXf.$$

Conversely, any map $X$ with these properties defines a smooth vector field, since for each point $p$ the map $X_p:{\mathcal{F}}_p(M)\to {\mathcal{F}}_p(M)$ defined by $X_{p}f=(Xf)(p)$ satisfies Eqs. Eq. (15.3) and Eq. (15.4) and is therefore a tangent vector at $p$ .

We may also define vector fields on any open set $U$ in a similar way as an assignment of a tangent vector at every point of $U$ such that $Xf\in \mathcal{F}(U)$ for all $f\in \mathcal{F}(U)$ . By the term local basis of vector fields at $p$ we will mean an open neighbourhood $U$ of $p$ and a set of vector fields $\{e_1,e_2,\dots ,e_n\}$ on $U$ such that the tangent vectors $(e_i{)}_q$ span the tangen space $T_q(M)$ at each point $q\in U$ . For any chart $(U,\varphi ;x^i)$ , define the vector fields on the domain $U$

$${\partial }_{x^i}\equiv \frac{\partial }{\partial x^i}:\mathcal{F}(U)\to \mathcal{F}(U)$$

by

$${\partial }_{x^i}f=\frac{\partial }{\partial x^i}f=\frac{\partial f\circ {\varphi }^{-1}}{\partial x^i}.$$

These vector fields assign the basis tangent vectors ${\left({\partial }_{x^i}\right)}_p$ at each point $p\in U$ , and form a local basis of vector fields at any point of $U$. When it is restricted to the coordinate domain $U_{:}$ , every differentiable vector field $X$ on $M$ has a unique expansion in terms of these vector fields

$$X{|}_U=X^i\frac{\partial }{\partial x^i}=X^i{\partial }_{x^i}$$

where the components $X^i:U\in \mathbb{R}$ are differentiable functions on $U.$ . The local vector fields ${\partial }_{x^i}$ form a module basis on $U$, but they are not a vector space basis since as a vector space $\mathcal{T}(U)$ is the direct sum of tangent spaces at all points $p\in U$ , and is infinite dimensional.

In a similar way we define a covector field or differentiable 1-form $\omega$ as an assignment ofa covector ${\omega }_p$ at each point $p\in M$ , such that the function $⟨\omega ,X⟩$ defined by $⟨\omega ,X⟩(\mathbf{p})=⟨{\omega }_p,X_p⟩$ is differentiable for every smooth vector field $X$. The space of differentiable 1- forms will be denoted ${\mathcal{T}}^{\ast }(M)$ . Given any smooth function $f,$ , let $\mathrm{d}f$ be the differentiable 1-form defined by assigning the differential $d$ $f_p$ at each point $p,$ , so that

$$⟨\mathrm{d}f,X⟩=Xf\text{ for all }X\in \mathcal{T}(M).$$

We refer to this covector field simply as the differential of $f$. A local module basis on any chart $(U,\varphi ;x^i)$ consists of the 1-forms $\mathrm{d}{x}^i$ , which have the property

$$⟨\mathrm{d}{x}^i,{\partial }_{x^j}⟩=\frac{\partial x^i}{\partial x^j}={\delta }_j^i.$$

Every differential can be expanded locally by the chain rule,

$$\mathrm{d}f=f_{,i}\mathrm{d}{x}^i\text{ where }{f}_{,i}=\frac{\partial }{\partial x^i}f.\text{\hspace{1em}(15.10)}$$

Tensor fields are defined in a similar way, where the differentiable tensor field $A$ of type $(r,s)$ has a local expansion in any coordinate chart

$$A=A_{j_1\dots j_s}^{i_1{i}_2\dots i_r}\frac{\partial }{\partial x^{i_1}}\otimes \frac{\partial }{\partial x^{i_2}}\otimes \cdots \otimes \frac{\partial }{\partial x^{i_r}}\otimes d{x}^{j_1}\otimes \cdots \otimes d{x}^{j_s}.\text{\hspace{1em}(15.11)}$$

The components are differentiable functions over the coordinate domain $U$ given by

$$A_{j_1\dots j_s}^{i_1{i}_2\dots i_r}=A\left(\mathrm{d}{x}^{i_1},\mathrm{d}{x}^{i_2},\dots ,\mathrm{d}{x}^{i_r},\frac{\partial }{\partial x^{j_1}},\dots ,\frac{\partial }{\partial x^{j_s}}\right).$$

Coordinate transformations

Let $(U,\varphi ;x^i)$ and $(U^{\prime },{\varphi }^{\prime };x^{\prime j})$ be any two coordinate charts. From the chain rule of partia differentiation

$$\frac{\partial }{\partial x^{\prime j}}=\frac{\partial x^i}{\partial x^{\prime j}}\frac{\partial }{\partial x^i},\frac{\partial }{\partial x^i}=\frac{\partial x^{\prime j}}{\partial x^i}\frac{\partial }{\partial x^{\prime j}}.\text{\hspace{1em}(15.12)}$$

Exercise

Show these equations by applying both sides to an arbitrary differentiable function $f$ on $M$.

Substituting the transformations Eq. (15.12) into the expression of a tangent vector with respect to either of these bases

$$X=X^i\frac{\partial }{\partial x^i}=X^{\prime j}\frac{\partial }{\partial x^{\prime j}}$$

gives the contravariant law oftransformation of components

$$X^{\prime j}=X^i\frac{\partial x^{\prime j}}{\partial x^i}.\text{\hspace{1em}(15.13)}$$

The chain rule Eq. (15.10), written in coordinates $x^{\prime j}$ and setting $f=x^i$ , gives

$$\mathrm{d}{x}^i=\frac{\partial x^i}{\partial x^{\prime j}}\mathrm{d}{x}^{\prime j}.$$

Expressing a differentiable 1-form $\omega$ in both coordinate bases,

$$\omega =w_i\mathrm{d}{x}^i=w_j^{\prime }\mathrm{d}{x}^{\prime j},$$

we obtain the covariant transformation law of components

$$w_j^{\prime }=\frac{\partial x^i}{\partial x^{\prime j}}{w}_i.\text{\hspace{1em}(15.14)}$$

The component transformation laws Eq. (15.13) and Eq. (15.14) can be identified with similar formulae in Chapter 3 on setting

$$A_i^j=\frac{\partial x^{\prime j}}{\partial x^i},A_k^{\prime i}=\frac{\partial x^i}{\partial x^{\prime k}}.$$

The transformation law of a general tensor of type ($r$, $s$) follows from Eq. (7.30):

$$T_{j_1^{\prime }\dots j_s^{\prime }}^{\prime i_1^{\prime }\dots i_r^{\prime }}=T_{j_1\dots j_s}^{i_1\dots i_r}\frac{\partial x^{\prime i_1^{\prime }}}{\partial x^{i_1}}\dots \frac{\partial x^{\prime i_r^{\prime }}}{\partial x^{i_r}}\frac{\partial x^{j_1}}{\partial x^{\prime j_1^{\prime }}}\dots \frac{\partial x^{j_s}}{\partial x^{\prime j_s^{\prime }}}.\text{\hspace{1em}(15.15)}$$

Tensor bundles

The tangent bundle $TM$ on a manifold $M$ consists ofthe set-theoretical union ofall tangen spaces at all points

$$TM=\bigcup _{p\in M}{T}_p(M).$$

There is a natural projection map $\pi :TMM$ defined by $\pi (u)=p\mathrm{if}u\in T_p(M)$ , and for each chart $(U,\varphi ;x^i)$ on $M$ we can define a chart $({\pi }^{-1}(U),\tilde{\varphi })$ on $TM$ where the coordinate map $\tilde{\varphi }:{\pi }^{-1}(U)\to {\mathbb{R}}^{2n}$ is defined by

$$\tilde{\varphi }(v)=\left(x^1(p),\dots ,x^n(p),v^1,\dots ,v^n\right)$$ $$\text{ where }p=\pi (v)\text{ and }v=\sum _{i=1}^n{v}^i\frac{\partial }{\partial x^i}{|}_p.$$

The topology on $TM$ is taken to be the coarsest topology such that all sets ${\tilde{\varphi }}^{-1}(A)$ are open whenever $A$ is an open subset of ${\mathbb{R}}^{2n}$ . With this topology these charts generate a maximal atlas on the tangent bundle $TM$, making it into a differentiable manifold of dimension $2n$

Given an open subset $U\subseteq M,$ , a smooth map $X:U\to TM$ is said to be a smooth vector field on $U$ if $\pi \circ X=\mathrm{id}{|}_U$ . This agrees with our earlier notion, since it assigns exactly one tangent vector from the tangent space $T_p(M)$ to the point $p\in U.\mathrm{A}$ similar idea may be used for a smooth vector field along a parametrized curve $\gamma :(a,b)M$ defined to be a smooth curve $X:(a,b)TM$ that lifts $\gamma$ to the tangent bundle in the sense that $\pi \circ X=\gamma$ . Essentially, this defines a tangent vector at each point of the curve, not necessarily tangent to the curve, in a differentiable manner.

The cotangent bundle $T^{\ast }M$ is defined in an analogous way, as the union ofall cotangen spaces $T_p^{\ast }(M)$ at all points $p\in M$ . The generating charts have the form $({\pi }^{-1}(U),\tilde{\varphi })$ on

$T^{\ast }M$ where the coordinate map $\tilde{\varphi }:{\pi }^{-1}(U)\to {\mathbb{R}}^{2n}$ is defined by

$$\tilde{\varphi }({\omega }_p)=\left(x^1(p),\dots ,x^n(p),w_1,\dots ,w_n\right)$$ $$\text{ where }p=\pi (\omega )\text{ and }\omega =\sum _{i=1}^n{w}_i(\mathrm{d}{x}^i{)}_p,$$

making $T^{\ast }M$ into a differentiable manifold ofdimension $2n$ . This process may be extended to produce the tensor bundle of type $T^{(r,s)}{M}_{;}$ , a differentiable manifold of dimension ${}^{n+}{n}^{r+s}$

Problems

Problem 15.7

Let $\gamma :\mathbb{R}\to {\mathbb{R}}^2$ be the curve $x=2t+1,y=t^2-3t$ . Show that at an arbitrary parameter value $t$ the tangent vector to the curve is $X_{\gamma (t)}=\dot{\gamma }=2{\partial }_x+(2t-3){\partial }_y=2{\partial }_x+(x-4){\partial }_y$ $I$ $f:{\mathbb{R}}^2\to \mathbb{R}$ is the function $f=x^2-y^2$ , write $f$ as a function of $t$ along the curve and verify the identities

$$X_{\gamma (t)}f=\frac{\mathrm{d}f(t)}{\mathrm{d}t}=⟨(\mathrm{d}f{)}_{\gamma (t)},X_{\gamma (t)}⟩=C_1^1(\mathrm{d}f{)}_{\gamma (t)}\otimes X_{\gamma (t)}.$$

Problem 15.8

Let $x^1=x,x^2=y,x^3=z$ be ordinary rectangular cartesian coordinates in ${\mathbb{R}}^3$ , and let $x^{\prime 1}=r,x^{\prime 2}=\theta ,x^{\prime 3}=\varphi$ be the usual transformation to polar coordinates.

(a) Calculate the Jacobian matrices $[\partial x^i/\partial x^{\prime }{}^j]$ and $[\partial x^{\prime i}/\partial x^j]$

(b) In polar coordinates, work out the components of the covariant vector fields having component in rectangular coordinates (i) (0, 0, 1), (ii) (1, 0, 0), (iii) ($x$ , $y$, $z$).

(c) In polar coordinates, what are the components ofthe contravariant vector fields whose component in rectangular coordinates are $(\mathrm{i})\left(x,y,z\right)$ , (ii) (0, 0, 1), (iii) $(-y,x,0)$ .

(d) $\mathrm{If}{g}_{ij}$ is the covariant tensor field whose components in rectangular coordinates are ${\delta }_{ij}$ , what are its components $g_{ij}^{\prime }$ in polar coordinates?

Problem 15.9

Show that the curve

$$2x^2+2y^2+2xy=1$$

can be converted by a rotation of axes to the standard form for an ellipse

$$x^{\prime 2}+3y^{\prime 2}=1.$$

$\begin{array}{r}\mathrm{If}{x}^{\prime }=\cos \psi ,y^{\prime }=\frac{1}{\sqrt{3}}\end{array}$ $\sin$ $\psi$ is used as a parametrization of this curve, show tha

$$x=\frac{1}{\sqrt{2}}\left(\cos \psi +\frac{1}{\sqrt{3}}\sin \psi \right),y=\frac{1}{\sqrt{2}}\left(-\cos \psi +\frac{1}{\sqrt{3}}\sin \psi \right).$$

Compute the components of the tangent vector

$$X=\frac{\mathrm{d}x}{\mathrm{d}\psi }{\partial }_x+\frac{\mathrm{d}y}{\mathrm{d}\psi }{\partial }_y.$$

Show that $X(f)=(2/\sqrt{3})(x^2-y^2)$

Problem 15.10

Show that the tangent space $T_{(p,q)}(M\times N)$ at any poin $(p,q)$ ofa product manifold $M\times N$ is naturally isomorphic to the direct sum of tangent spaces $T_p(M)\oplus T_q(N)$ .

Problem 15.11

On the unit 2-sphere express the vector fields ${\partial }_x$ and ${\partial }_y$ in terms of the pola coordinate basis ${\partial }_{\theta }$ and ${\partial }_{\varphi }$ . Again in polar coordinates, what are the dual forms to these vector fields?

Problem 15.12

Express the vector field ${\partial }_{\varphi }$ in polar coordinates $(\theta ,\varphi )$ on the unit 2-sphere in terms of stereographic coordinates $X$ and $Y$.