Part III · General Relativity — Chapter 3.2

Manifolds

The chapter that takes the arrow away, and puts something better in its place. It is the hardest reframe in the book, so we go slowly.

Tools you'll need.  Chapter 0.4 §1: the axiomatic definition of a vector space, in which no arrow appears anywhere. That section was written for this chapter, and 2.4 §3 for covectors. Also 0.4 §4 (change of basis, components transforming with the inverse of the basis), and 2.4 §3 for covectors and the dual space. Chapter 0.6 §2 (the derivative as a linear map), §5 (the multivariable chain rule, which does most of the work below), §6.1 (Clairaut's theorem on mixed partials) and §8 (the Jacobian). Chapter 0.2 §2, the fundamental theorem of calculus, used once in a grind box. Chapter 0.7 §1.2, which warned that a vector field is not a vector and named the cheque this chapter has to cash. Chapter 2.4 in its entirety. Its §2, §3, §5 and §5.1 are re-used here with one alteration, and identifying that alteration is the point of §7.

1 · Why the old arena will not do

Two separate things go wrong with $\R^{n}$, and it is worth keeping them apart because they have different remedies.

1.1 · The global objection

Physics has no licence to assume that spacetime, taken as a whole, is $\R^{4}$. Nothing we can measure is a global statement. Observations are made in a finite region over a finite time, and Chapter 3.1 §5 quantified just how finite. Whether the universe closes up on itself, or has more than one connected piece, or has a shape that no single set of four coordinates can cover, is a question to be answered rather than assumed. Chapter 3.9 will ask it seriously.

So the arena must be built to allow shapes other than $\R^{4}$ while still supporting calculus. The recipe is the one you would guess: look like $\R^{n}$ in patches, and be honest about the seams.

1.2 · The local objection, which is worse

Even in a single patch there is trouble, and it is not about shape at all. Consider the sphere, and a vector tangent to it at a point $p$. The picture in your head is an arrow lying in the plane tangent to the sphere at $p$. That arrow is drawn in the surrounding three-dimensional space.

That picture uses $\R^{3}$. And $\R^{3}$ is not part of the sphere. It is extra scaffolding, and three separate objections attach to leaning on it.

(i) The scaffolding is not unique. ⚑ Whitney's embedding theorem states that any smooth $n$-dimensional manifold can be placed inside $\R^{2n}$. We quote it without proof and will not use it. So an ambient space always exists. The trouble is that many of them exist, and a given surface can sit inside one in genuinely different ways. Any object defined by reference to the embedding would depend on a choice nobody made on physical grounds. Chapter 2.4 spent a whole chapter teaching us what to do with quantities that depend on an arbitrary choice, and the answer was never "keep them".

(ii) Physically there is no outside. Spacetime is not observed sitting inside anything, and no experiment reaches out of it. A definition requiring an exterior is a definition that cannot be checked from where we are standing. It also imports structure that no measurement constrains: the extra dimensions, their geometry, and the way the surface sits in them.

(iii) The arrow was never a subtraction of points anyway. In $\R^{n}$ the arrow from $p$ to $q$ is the difference $q-p$, and that difference is meaningful because $\R^{n}$ is a vector space as well as a place: points can be added and subtracted. A sphere has no addition. There is no point $p+q$ on a sphere, so there is no arrow from $p$ to $q$ in the sense that made the picture work.

2 · Charts, atlases, and why one chart is not enough

Start with the shape problem, which is the easy one. The fix is to stop demanding one description of the whole space and settle for a set of local ones that agree where they overlap.

Notice what is and is not in that definition. There are coordinates, but no preferred ones. There is no distance, no angle, no straight line, no notion of one point being "between" two others. A manifold is the least structure on which calculus can be done. That austerity is deliberate. We will add the metric in Chapter 3.3 as a separate object, because in general relativity the metric is dynamical and must be able to vary. Structure baked into the arena cannot vary.

2.1 · One chart cannot cover a sphere

It is worth proving rather than asserting that the patchwork is necessary, because the proof is short and it explains exactly what goes wrong.

Suppose there were: a map $\varphi:S^{2}\to\R^{2}$, one-to-one, continuous, with continuous inverse, whose image $V=\varphi(S^{2})$ is an open subset of $\R^{2}$. We derive a contradiction in four steps.

Step 1. Write $\varphi=(\varphi^{1},\varphi^{2})$ and look at the first component alone. It is a continuous real-valued function on $S^{2}$.

Step 2. The sphere, sitting in $\R^{3}$ as the set of points with $\abs{\vv x}=1$, is closed and bounded. ⚑ The extreme value theorem therefore guarantees that $\varphi^{1}$ attains a largest value, at some point $p\in S^{2}$. That theorem is the multivariable form of a familiar one-dimensional statement, namely that a continuous function on a closed interval attains its maximum.

Step 3. Now look at where $p$ goes. Its image $\varphi(p)=(a,b)$ lies in $V$, and $V$ is open, so some small disc of radius $\varepsilon$ around $(a,b)$ lies entirely inside $V$. In particular the point $(a+\varepsilon/2,\;b)$ is in $V$.

Step 4. Since $V$ is the image of $\varphi$, there is a point $q\in S^{2}$ with $\varphi(q)=(a+\varepsilon/2,\;b)$. But then $\varphi^{1}(q)=a+\varepsilon/2 \gt a=\varphi^{1}(p)$, contradicting Step 2's statement that $a$ was the largest value. $\blacksquare$

Let's look at what did the work. The sphere is compact, meaning closed and bounded, so continuous functions on it attain their extremes. An open subset of $\R^{2}$ has no extremes, because around every one of its points there is room to go further in every direction. Compactness and openness are incompatible under a continuous bijection, and the sphere is the smallest interesting example. The identical argument, run with $S^{1}$ and $\R^{1}$, shows that a circle needs at least two charts. The same argument again shows that no compact manifold is covered by one chart.

2.2 · An explicit two-chart atlas

Two charts do suffice for the sphere, and the standard pair is worth computing in full because its transition map is used repeatedly below. Realise $S^{2}$ as $\{(x,y,z):x^{2}+y^{2}+z^{2}=1\}$ and project stereographically.

The north chart. Delete the north pole $N=(0,0,1)$. For any other point $P$, draw the straight line from $N$ through $P$ and record where it meets the equatorial plane $z=0$. Parametrise that line as $N+t\,(P-N)$, whose third component is $1+t(z-1)$. Setting that component to zero gives $t=1/(1-z)$, which is finite exactly because $z\neq1$. What we want is where the line lands, so substitute that $t$ into the first two components:

The south chart. Now run the same construction from the other pole. Deleting $S=(0,0,-1)$ and projecting from there gives $t=1/(1+z)$ instead, and the second chart is:

Their union is all of $S^{2}$, since no point is both poles, so this is an atlas. Each chart maps onto the whole of $\R^{2}$, which is open. The overlap $U_{N}\cap U_{S}$ is the sphere minus both poles, and its image under $\varphi_{N}$ is $\R^{2}$ minus the origin.

The transition map. Take $(u,v)$ in the overlap and compute $\varphi_{S}$ of the same point. One preliminary identity does the whole job. From (3.2.1),

where the second equality used $x^{2}+y^{2}+z^{2}=1$ and the third factorised the difference of squares. Now write $\tilde u$ in terms of $u$ by inserting a factor of one:

where the last step used (3.2.3). The same computation gives $\tilde v$. Putting the two components together, here is the dictionary between the charts:

This is inversion in the unit circle: it keeps the direction from the origin and replaces the distance $\varrho$ by $1/\varrho$. It is defined and infinitely differentiable everywhere except at the origin. And the origin is exactly the point that is not in the overlap, since it is the image of the south pole. The seam is clean.

3 · Smoothness, and the only place it can live

Calculus needs derivatives, and derivatives need smoothness. Where does smoothness live on a manifold?

Not on $M$ itself. There is nothing there to differentiate, since $M$ is just a set of points with charts attached. Only one kind of object in the construction is a map between pieces of $\R^{n}$, which means only one kind of object is something Chapter 0.6's machinery can be applied to. Those are the transition maps, and so that is where the requirement goes.

The definition of a smooth function comes with an obligation, and discharging it is the point of requiring smooth transition maps. Suppose $f$ looks smooth in chart $\alpha$. Does it look smooth in chart $\beta$? Write the chart-$\beta$ representative in terms of the chart-$\alpha$ one by inserting the identity $\varphi_{\alpha}^{-1}\circ\varphi_{\alpha}$:

A composition of smooth maps is smooth, which is the chain rule of Chapter 0.6 §5 applied repeatedly. So the answer is yes. Smoothness of a function on $M$ is a chart-independent property precisely because the transition maps are smooth. Let's see what that requirement is buying. Had we allowed a transition map with a kink, a perfectly innocent function could be smooth in one chart and not in another, and the word would mean nothing.

Check the sphere. Its transition map (3.2.5) is built from polynomials divided by $u^{2}+v^{2}$, which is nonzero throughout the overlap, so it is infinitely differentiable there. The two-chart atlas is smooth, and $S^{2}$ is a smooth two-dimensional manifold.

4 · The tangent vector, rebuilt from the inside

This is the hard step, and everything after it is comparatively easy. Read §1.2 again if the motivation has faded: the arrow has to go, because it was drawn in a space that is not there. Here is where we are heading.

4.1 · What a direction is for

Before building anything, ask what the arrow was ever used for. Not for decoration. In every physical application, a direction at a point is fed into something that measures a rate of change: the rate at which temperature rises as you walk that way, the rate at which a potential falls, the rate at which any measurable quantity varies. A measurable quantity spread over spacetime is a function on $M$. So:

a direction at $p$ is a device that assigns, to each smooth function on $M$, the rate at which it changes at $p$ in that direction.

That sentence mentions only $M$, functions on $M$, and real numbers. Nothing outside. If we can make it a definition, the ambient space is not needed, and the pieces we would have thrown away are exactly the pieces that were never measurable in the first place.

4.2 · What is available from inside

Exactly two kinds of object are available on a bare smooth manifold, and both are defined without reference to anything external.

Curves. A smooth map $\gamma$ from an interval of $\R$ into $M$, with $\gamma(0)=p$. Concretely, a worldline. In a chart it has coordinate representatives $x^{\mu}(\lambda)\equiv\varphi^{\mu}\big(\gamma(\lambda)\big)$, which are ordinary functions of one real variable.

Functions. A smooth map $f:M\to\R$, in the sense of §3. Concretely, any scalar measurement made at each event.

That is the whole toolbox. Everything below is built from those two.

4.3 · The construction

Take a curve $\gamma$ through $p$ and a function $f$ on $M$. Compose them: $f\circ\gamma$ takes a real number $\lambda$ and returns a real number $f(\gamma(\lambda))$. That is a function of one real variable, and Chapter 0.1 tells us exactly what to do with one. Differentiate it at $\lambda=0$.

Now compute it in a chart, to see what it is made of. Write $\hat f\equiv f\circ\varphi^{-1}$ for the function expressed in coordinates, so that $f\big(\gamma(\lambda)\big)=\hat f\big(x^{1}(\lambda),\ldots,x^{n}(\lambda)\big)$. Differentiating a function of several variables each of which depends on one parameter is precisely the multivariable chain rule of Chapter 0.6 §5:

summed over $\mu=1,\ldots,n$. Look at how that separates. The first factor knows only about the curve. The second knows only about the function. Since $f$ was arbitrary, we may strip it off and read (3.2.7) as an identity between operators:

Let's stop and take stock, because the whole reframe is visible in that line. The numbers $V^{\mu}$ are the coordinate velocities of the curve. They are exactly the components you would have written down for the old arrow. Nothing quantitative has been lost. What has changed is what those numbers are components of: no longer of an arrow in a surrounding space, but of an operator, expanded in the $n$ operators $\partial_{\mu}|_{p}$.

Two properties of $X_{\gamma}$ are worth isolating, because they are about to become the definition.

Linearity. $X_{\gamma}(af+bg)=aX_{\gamma}(f)+bX_{\gamma}(g)$ for constants $a,b$, because differentiation is linear (Chapter 0.1 §4).

The Leibniz rule at $p$. Apply the product rule of Chapter 0.1 §4 to $(fg)\circ\gamma=(f\circ\gamma)\,(g\circ\gamma)$ and evaluate at $\lambda=0$, where $\gamma(0)=p$:

Now turn these into the definition. This is the same manoeuvre Chapter 2.4 §2 made when it defined a vector by the transformation law it obeys rather than by a picture: point at the properties, and admit everything that has them.

4.4 · The coordinate operators are a basis — proved, not assumed

The definition would be worthless if $T_{p}M$ turned out to be the wrong size. We now show it has dimension exactly $n$, by exhibiting a basis. Here are the two halves of the job. The $n$ operators $\partial_{\mu}|_{p}$ span $T_{p}M$, and they are linearly independent.

Independence, which is three lines. Suppose some combination vanishes as an operator: $a^{\mu}\partial_{\mu}|_{p}=0$, meaning it gives zero on every function. Feed it the coordinate function $x^{\nu}$, which is a perfectly good smooth function, namely "the $\nu$-th coordinate of the point". The partial derivative of one coordinate with respect to another is $\partial x^{\nu}/\partial x^{\mu}=\delta^{\nu}{}_{\mu}$, because the coordinates are independent of each other: differentiating $x^{2}$ with respect to $x^{2}$ gives one, and with respect to $x^{1}$ gives zero. So

where the last step is the Kronecker delta doing its only job, which is to rename an index and collapse a sum. Every coefficient vanishes, so the operators are independent.

Spanning, which needs a genuine theorem. The claim is that every derivation at $p$ is of the form $a^{\mu}\partial_{\mu}|_{p}$, and that has to include the derivations which arise from no curve at all. The coefficients are given by what the derivation does to the coordinate functions:

The argument has two ingredients, and both are short. One is that a derivation annihilates constants. The other is a factorisation of smooth functions due to Hadamard. The logic runs like this: subtract the constant $f(p)$, factor what is left as coordinates times smooth functions, apply the Leibniz rule, and watch all but one term die because the coordinates vanish at $p$. The algebra is in the grind box, and the result is (3.2.11).

Grind box — proof that every derivation is $X(x^{\mu})\,\partial_{\mu}$

Step 1. A derivation kills constants. Let $\mathbf 1$ denote the function that is equal to $1$ everywhere. Since $\mathbf 1\cdot\mathbf 1=\mathbf 1$, the Leibniz rule gives

$$X(\mathbf 1) \;=\; X(\mathbf 1\cdot\mathbf 1) \;=\; \mathbf 1(p)\,X(\mathbf 1) + \mathbf 1(p)\,X(\mathbf 1) \;=\; 2X(\mathbf 1),$$

so $X(\mathbf 1)=0$. By linearity $X(c\,\mathbf 1)=cX(\mathbf 1)=0$ for any constant $c$. Derivations do not see constants, exactly as ordinary derivatives do not.

Step 2. Hadamard's lemma. Choose a chart around $p$ and translate it so that $p$ sits at the origin, $\varphi(p)=0$. That is legitimate, because composing a chart with a translation of $\R^{n}$ gives another chart. Let $\hat f$ be the coordinate representative of $f$ on a ball around the origin. Then there exist smooth functions $h_{\mu}$ on that ball with

$$\hat f(x) \;=\; \hat f(0) \;+\; x^{\mu}\,h_{\mu}(x), \qquad h_{\mu}(0) \;=\; \partial_{\mu}\hat f(0).$$

Proof. Consider $t\mapsto \hat f(tx)$ for $t\in[0,1]$, which walks along the straight segment from the origin to $x$. By the fundamental theorem of calculus (Chapter 0.2 §2),

$$\hat f(x)-\hat f(0) \;=\; \int_{0}^{1}\dv{}{t}\,\hat f(tx)\;\dd t.$$

Evaluate the integrand with the chain rule (Chapter 0.6 §5): the $\mu$-th coordinate of the point $tx$ is $tx^{\mu}$, whose $t$-derivative is $x^{\mu}$, so

$$\dv{}{t}\hat f(tx) \;=\; x^{\mu}\,\big(\partial_{\mu}\hat f\big)(tx).$$

The factor $x^{\mu}$ does not depend on $t$ and comes out of the integral:

$$\hat f(x)-\hat f(0) \;=\; x^{\mu}\underbrace{\int_{0}^{1}\big(\partial_{\mu}\hat f\big)(tx)\;\dd t}_{\displaystyle \equiv\;h_{\mu}(x)}.$$

Each $h_{\mu}$ is smooth, because differentiating under the integral sign (the move Chapter 1.2 §3.1 used) turns derivatives of $h_{\mu}$ into integrals of derivatives of $\hat f$, all of which exist. And setting $x=0$ gives $h_{\mu}(0)=\int_{0}^{1}\partial_{\mu}\hat f(0)\,\dd t=\partial_{\mu}\hat f(0)$. $\square$

Step 3. Apply $X$. Transporting Hadamard's identity back to $M$, we have $f = f(p)\,\mathbf 1 + x^{\mu}h_{\mu}$ as functions near $p$, where now $x^{\mu}$ means the coordinate function and $h_{\mu}$ the corresponding function on $M$. Apply $X$, using linearity on the sum and the Leibniz rule on each product:

$$X(f) \;=\; \underbrace{f(p)\,X(\mathbf 1)}_{=\;0\ \text{by Step 1}} \;+\; \underbrace{x^{\mu}(p)}_{=\;0\ \text{since } \varphi(p)=0}\,X(h_{\mu}) \;+\; h_{\mu}(p)\,X(x^{\mu}).$$

Two of the three terms vanish outright. The first goes because derivations kill constants, and the second because we placed $p$ at the coordinate origin. What is left, using $h_{\mu}(p)=\partial_{\mu}\hat f(0)$ from Step 2, is

$$X(f) \;=\; X(x^{\mu})\;\partial_{\mu}\hat f\big|_{p},$$

which is (3.2.11), since $f$ was arbitrary. $\blacksquare$

One honest caveat. Hadamard's factorisation holds on a ball around $p$, not on all of $M$, so strictly the argument shows that a derivation is determined by the behaviour of $f$ near $p$. That is a feature rather than a gap: it says a tangent vector is a local object, which is exactly what one wants of a direction. Making it airtight requires a bump function that is $1$ near $p$ and $0$ far away, whose existence is a standard construction we shall not need again.

So $T_{p}M$ has dimension $n$, and $\{\partial_{\mu}|_{p}\}$ is a basis for it, called the coordinate basis of that chart. Every tangent vector is written

The components are recovered by feeding the vector the coordinate functions. That is a pleasant inversion of the usual order: instead of the coordinates telling you the vector's components, the vector tells you its components by acting on the coordinates.

4.5 · Changing chart, and Chapter 2.4 collected

Two charts, coordinates $x^{\mu}$ and $x'^{\mu}$, overlapping near $p$. How do the two bases relate? Apply the chain rule (0.6 §5) to any function, differentiating with respect to a primed coordinate by going through the unprimed ones:

with $\Lambda^{\mu}{}_{\nu}\equiv\partial x'^{\mu}/\partial x^{\nu}$ exactly as in Chapter 2.4 §2. The basis vectors transform with the inverse Jacobian.

Now impose the one thing we insist on: the vector itself is a single operator and does not care which chart you expand it in. So $V'^{\mu}\partial'_{\mu}=V^{\nu}\partial_{\nu}$. Substitute (3.2.13) into the left side:

The $\partial_{\nu}$ are linearly independent by §4.4, so their coefficients must agree index by index:

What we want is $V'$ alone on the left, so multiply both sides by $\Lambda^{\rho}{}_{\nu}$ and sum over $\nu$. The identity $\Lambda^{\rho}{}_{\nu}(\Lambda^{-1})^{\nu}{}_{\mu}=\delta^{\rho}{}_{\mu}$ does the work, and it is the chain-rule identity of Chapter 2.4 §3 rather than a new assumption. The delta then renames $\mu$ into $\rho$ on the left:

5 · A different vector space at every point

First, $T_{p}M$ really is a vector space in the sense of Chapter 0.4 §1, the axiomatic sense, with no arrow in it. Given derivations $X,Y$ at $p$ and reals $a,b$, define $(aX+bY)(f)\equiv aX(f)+bY(f)$. Linearity is immediate. What has to be checked is that the Leibniz rule survives the combination, and it does:

where the second line is nothing but a regrouping of the four terms by whether they carry $f(p)$ or $g(p)$. So sums and multiples of derivations are derivations. The remaining axioms are inherited from the reals, and $T_{p}M$ is an $n$-dimensional real vector space. Chapter 0.4's decision to define a vector space by axioms rather than by arrows is what makes this paragraph possible at all.

5.1 · The sentence

Why it is true is immediate from the definition, and this is the payoff for having built tangent vectors the way we did. A derivation at $p$ is an operator that reports rates of change at $p$. A derivation at $q$ reports rates of change at $q$. They act on the same functions but return different kinds of information, and they satisfy different Leibniz rules. Look at (3.2.9) again: it evaluates $f$ at $p$ in one case and at $q$ in the other. There is no operation that turns one into the other, because there is no operation in the construction that relates the two points at all.

5.2 · Why this feels wrong, and what the flat case was hiding

The objection writes itself: in $\R^{n}$ we slide arrows around all the time and nobody complains. Let's diagnose that carefully, because it identifies precisely what a general manifold is missing.

$\R^{n}$ is two things at once. It is a manifold, which is to say a set with a chart. It is also a vector space, in which points can be added and subtracted. That second structure is extra, and it is what supplies the identification. Given a tangent vector at $p$ with components $V^{\mu}$, one declares it "the same as" the tangent vector at $q$ with the same components. The declaration is legitimate there because $\R^{n}$'s vector-space structure singles out one way of doing it, and Cartesian coordinates make it invisible.

A sphere has no addition of points, so no such declaration is available. And the failure is not a formality. It has a concrete measurable signature, and the following figure exhibits it.

5.3 · "Are these two vectors parallel?" has no answer

Take two points $p$ and $q$ on a sphere, with a tangent vector at each. Ask whether they point the same way.

To answer, you must bring one of them to the other, and the only honest way to do that on a surface is to carry it along a path, keeping it as unchanged as the surface allows at every step. Along a great circle there is a rule for doing exactly that: keep the vector at a constant angle to the direction you are walking. That rule is entirely intrinsic, since it refers only to angles measured within the surface and never to any surrounding space. Build any path out of great-circle arcs and the rule extends to it.

That rule is well defined, natural, and the answer it gives depends on which path you took.

separation

100°

detour

60°

start bearing

0°

make the two paths coincide

route A (equator): arrives on bearing 0.00°

route B (detour): arrives on bearing 60.27°

the two answers differ by 60.27° [tangency check at q: 0.0e+0]

5.4 · The consequence, which is why Chapter 3.3 exists

Everything so far has been about comparing two vectors. Now notice that differentiating a vector field is exactly that comparison, and therefore inherits the whole problem.

To differentiate a field $V$ along a curve you would write, following Chapter 0.1 §2 to the letter,

Look at what the numerator is asking for. It subtracts an element of $T_{\gamma(\varepsilon)}M$ from an element of $T_{\gamma(0)}M$, and by §5.1 that subtraction does not exist. Equation (3.2.18) is not a difficult limit. It is not a limit at all, because its numerator is not defined for any $\varepsilon\neq0$.

One escape suggests itself, and it is worth seeing why it fails. Work in a chart, where the field is $n$ ordinary functions $V^{\mu}(x)$, and differentiate those. The array $\partial_{\nu}V^{\mu}$ is a perfectly good collection of $n^{2}$ numbers. The trouble is that those numbers describe the chart as much as they describe the field. Differentiating the transformation law (3.2.16) requires the product rule, so the derivative lands on the Jacobian as well as on the components. Unlike in Chapter 2.4, the Jacobian is no longer constant, so that extra term does not vanish. The result is that $\partial_{\nu}V^{\mu}$ fails the tensor transformation law. Chapter 3.3 §4 does that computation in full and displays the offending term. Chapter 3.3 §5 then defines a correction whose only job is to cancel it, and the correction turns out to be the gravitational field.

5.5 · The tangent bundle

One piece of bookkeeping before we leave this section. All those separate tangent spaces are worth collecting into a single object, so that a vector field becomes a map into one place rather than a family of maps into many:

whose points are pairs: a place, and a vector there. It has $2n$ coordinates, $n$ of them to say where and $n$ to say which vector, and it is itself a smooth manifold. A vector field is a rule assigning to each $p$ an element of $T_{p}M$, smoothly.

Two debts are settled by that paragraph. Chapter 0.7 §1.2 named $TM$ as the object it was deferring. And Chapter 1.2 §4 observed that a Lagrangian $L(x,\dot x)$ is properly a function on the tangent bundle of the configuration space, with position and velocity as independent coordinates. There the constraint "$\dot x$ is the derivative of $x$" was imposed only when a path was substituted. That is exactly (3.2.19): the pair (place, vector there), with both free.

6 · Covectors, and what a differential has been all along

Chapter 0.6 §4 named, and Chapter 2.4 §3 built, the dual $V^{*}$ of a vector space $V$: the space of linear maps from $V$ to $\R$. Neither chapter said how big that space is, and we will need to know, so this section settles it a few paragraphs below. Now apply that construction at each point of the manifold.

There is one natural way to manufacture covectors, and it needs nothing but a function.

Now feed the machine the coordinate functions themselves and see what falls out. Apply $\dd x^{\mu}$ to the basis vector $\partial_{\nu}$:

the same three-step computation used in §4.4's independence proof, run in the other direction. Equation (3.2.20) is exactly the condition that defines a dual basis: each member of the one set reads off one component of the other, and ignores the rest. So the $n$ covectors $\{\dd x^{\mu}\}$ form the basis of $T^{*}_{p}M$ dual to $\{\partial_{\mu}\}$, and every covector expands as $\omega=\omega_{\mu}\,\dd x^{\mu}$ with $\omega_{\mu}=\langle\omega,\partial_{\mu}\rangle$.

The covector we most want in that basis is $\dd f$ itself, so expand it. Its components are what it returns on the basis vectors, $\langle \dd f,\partial_{\mu}\rangle=\partial_{\mu}f$, which gives

6.1 · How covectors transform

The basis covectors follow from the chain rule, differentiating each new coordinate through the old ones:

Compare with (3.2.13): the basis vectors carried $\Lambda^{-1}$ and the basis covectors carry $\Lambda$. The two bases transform oppositely, which is what "dual" means. Demanding that the covector $\omega=\omega_{\mu}\dd x^{\mu}$ be one object independent of chart then forces its components the other way, by the same three-line argument as §4.5:

which is Chapter 2.4 §3's definition of a covariant vector, unaltered. As a check, the pairing is invariant:

where the two Jacobians met and collapsed to a Kronecker delta, which then renamed an index. That is Chapter 2.4 §3.1's calculation, character for character, and it still works because the collapse $\Lambda\Lambda^{-1}=\delta$ holds at each point separately and we never moved between points.

7 · Tensors, unchanged

With vectors and covectors in place at each point, Chapter 2.4 §5's definition can be transferred without modification.

Everything Chapter 2.4 proved about tensors used only two facts: that the law is linear and homogeneous in the components, and that $\Lambda\Lambda^{-1}=\delta$. Both hold here at each point. So the three operations of Chapter 2.4 §5.1 survive with their proofs intact. Those were linear combination of tensors of the same type, outer product, and contraction of an upper index against a lower one. So does the theorem of Chapter 2.4 §6, that an equation between tensors of the same type, true in one chart, is true in every chart. None of it needs redoing.

Three things are new, and they are the whole difference between Part II and Part III.

(a) $\Lambda$ depends on position. Stated already in §4.5, and it is the only alteration.

(b) Adding tensors at different points is forbidden. The definition is pointwise. Two tensors of the same type at the same point may be added. Two at different points may not, for exactly the reason of §5.1. In Chapter 2.4 this restriction was invisible, because the constant $\Lambda$ made components at different events directly comparable.

(c) Differentiation leaves the class. In Chapter 2.4's table, $\partial_{\nu}V^{\mu}$ was listed as a type $(1,1)$ tensor. On a manifold it is not, by §5.4. That single entry going wrong is what Chapter 3.3 is for, and the repaired derivative is the connection.

8 · Vector fields, and the one new operation

A vector field $X$ assigns a tangent vector to each point, so in a chart it is $X=X^{\mu}(x)\,\partial_{\mu}$ with the components now functions rather than numbers. A vector field therefore acts on a function to give another function: $(Xf)(p)\equiv X_{p}(f)$. This is new. A single tangent vector produced a number, but a field produces a function, and functions can be differentiated again.

Which raises the question: if $X$ and $Y$ are vector fields, is the composite $XY$, meaning "apply $Y$, then apply $X$", a vector field?

8.1 · It is not, and the failure is instructive

Test the Leibniz rule. Apply $XY$ to a product, expanding one operator at a time and using Leibniz for $Y$ first, then for $X$:

The second and fourth terms are what the Leibniz rule requires. The first and third are surplus:

So $XY$ is not a derivation, and correspondingly not a vector field. In components the reason is visible: expanding $XY(f)=X^{\mu}\partial_{\mu}\big(Y^{\nu}\partial_{\nu}f\big)$ with the product rule gives

and the second term contains second derivatives of $f$. A tangent vector is a first-order object, and second derivatives have no business in it.

8.2 · The repair: antisymmetrise

Look at (3.2.26) again. Its right-hand side is $(Xf)(Yg)+(Xg)(Yf)$, which is symmetric under exchanging $X$ and $Y$: swapping the two names turns the first term into the second and the second into the first. So computing $YX(fg)-f\,YX(g)-g\,YX(f)$ gives the identical expression, and subtracting the two statements makes the surplus cancel exactly. Define

Then $[X,Y](fg)-f[X,Y](g)-g[X,Y](f)=0$, so $[X,Y]$ obeys the Leibniz rule. It is linear because $X$ and $Y$ are. So it is a derivation at each point. The commutator of two vector fields is a vector field. This object is called the Lie bracket.

8.3 · Its components, with every index step named

What we want are the components $[X,Y]^{\nu}$, and (3.2.27) already supplies half of them. Write that half out again as the first term of the commutator:

and that is the first of the two terms we need. The second is the same line with the letters $X$ and $Y$ exchanged throughout, since that exchange is exactly what $YX$ asks for:

Now deal with the two second-derivative terms, one manipulation at a time.

Move 1. Relabel the dummy indices in (3.2.30). In the second term of (3.2.30) both $\mu$ and $\nu$ are summed over, so the letters are private and may be exchanged with each other. Swapping $\mu\leftrightarrow\nu$ throughout that one term turns $Y^{\mu}X^{\nu}\partial_{\mu}\partial_{\nu}f$ into $Y^{\nu}X^{\mu}\partial_{\nu}\partial_{\mu}f$. Nothing has happened except a change of alphabet.

Move 2. Use the symmetry of mixed partials. Clairaut's theorem (Chapter 0.6 §6.1) says $\partial_{\nu}\partial_{\mu}f=\partial_{\mu}\partial_{\nu}f$ for smooth $f$. The two numerical factors $X^{\mu}$ and $Y^{\nu}$ are ordinary functions and multiply in either order, so they may be swapped as well. The term therefore becomes $X^{\mu}Y^{\nu}\partial_{\mu}\partial_{\nu}f$.

Move 3. Subtract. The two second-derivative terms are now literally identical and cancel in (3.2.29) minus (3.2.30), leaving only the first-derivative terms:

Since $f$ was arbitrary, strip it off and read the components straight from the coefficient of $\partial_{\nu}$:

8.4 · What the bracket measures

Ask first what the bracket does on the basis itself. Set $X=\partial_{\mu}$ and $Y=\partial_{\nu}$ in (3.2.32). Their components are constants, namely $X^{\rho}=\delta^{\rho}{}_{\mu}$ and $Y^{\rho}=\delta^{\rho}{}_{\nu}$, so both derivative terms vanish and

A coordinate basis commutes. Geometrically, that says stepping $\varepsilon$ along the $x^{1}$ direction and then $\varepsilon$ along $x^{2}$ lands you at exactly the same point as doing it in the other order. That is precisely what it means for the two to be coordinates of one grid. So the bracket measures the failure of two flows to commute, and it vanishes for coordinate directions by construction.

Non-coordinate frames are a different story. Take plane polar coordinates and compare the coordinate basis $\{\partial_{r},\partial_{\theta}\}$ with the normalised directions $\hat e_{r}=\partial_{r}$ and $\hat e_{\theta}=\tfrac1r\,\partial_{\theta}$, which is the basis of unit vectors used in every mechanics course. Working out $[\hat e_{r},\hat e_{\theta}]$ requires the product rule once, since $\tfrac1r$ is a function:

What we want on the right is a single vector field, so apply Clairaut once more. The last two terms cancel against each other, leaving

Non-zero, so the unit vectors of elementary mechanics are not a coordinate basis: there are no coordinates whose partial derivatives they are. That single fact is what Chapter 1.1 §4 was struggling with when it had to differentiate rotating unit vectors by hand. That is what made the elementary calculation awkward. It is not yet where the centrifugal and Coriolis terms come from, and Worked example 2 below separates the two. ⚑ The general statement is Frobenius's theorem, which we quote and will not use: $n$ independent vector fields form a coordinate basis if and only if all their brackets vanish.

9 · Worked examples

10 · Your turn