Part III · General Relativity — Chapter 3.5

Forms, Lie Derivatives, Killing Vectors

A chapter that mostly collects. Three promises made earlier in the book fall due at once, and the tool that makes the next chapter but one solvable is built at the end.

Tools you'll need.  Chapter 0.7 heavily: §5 (the three integral theorems and the table in §5.4), §7.1 and §7.2 (the two second-derivative identities), §7.3 (the vector potential and the quoted Poincaré lemma), and §2.4's closed-but-not-exact counterexample on the punctured plane. Chapter 2.6 §2.3 (the field tensor), §3.2 (the homogeneous Maxwell pair) and §3.4 (why two of Maxwell's equations are bookkeeping). Chapter 0.6 §8 (change of variables and the Jacobian determinant) and §8.3, which already told you the answer to §6 below. Chapter 0.4 §5, the determinant as the unique alternating multilinear function of the columns. Section 1 is that theorem doing physics. Chapter 3.2 §6 (one-forms and the basis $\dd x^{\mu}$) and §8 (the commutator of vector fields). Chapter 3.3 §5 (the covariant derivative), §7 (metric compatibility and the symmetry of the connection) and §8 (geodesics). Chapter 1.4 §2 in full. Section 9 below is that theorem, restated.

1 · What integration wants, and why it wants antisymmetry

Here is the destination, stated before we set out. We want objects that can be integrated over a $p$-dimensional surface sitting inside a manifold, with the answer depending on the surface and not on how the surface was labelled. We will find that this single demand forces the integrand to be totally antisymmetric, and that nothing else is forced.

1.1 · The case that already works: a line integral

Chapter 3.2 §6 built one-forms, so take one: $\omega=\omega_{\mu}\,\dd x^{\mu}$. Take also a curve $C$, given in some chart by $x^{\mu}(\lambda)$ for $\lambda\in[a,b]$. Define

Two things about (3.5.1) deserve to be said out loud, because the whole chapter is built on them.

No metric was used. There is no length, no angle and no volume anywhere in (3.5.1). A one-form eats a vector and returns a number, and the tangent $\dd x^{\mu}/\dd\lambda$ is a vector. The pairing between them is the duality of Chapter 3.2 §6, which predates any metric.

The answer does not depend on $\lambda$. Let us check that. Reparametrise the curve, $\lambda=\lambda(\sigma)$, with $\dd\lambda/\dd\sigma>0$. Two factors in the integrand respond. The chain rule turns the tangent into $\dd x^{\mu}/\dd\sigma = (\dd x^{\mu}/\dd\lambda)(\dd\lambda/\dd\sigma)$. The ordinary change-of-variable rule for a single integral turns the measure into $\dd\lambda=(\dd\lambda/\dd\sigma)\,\dd\sigma$. Put the two together:

Look at what did the work there. The Jacobian of the reparametrisation appeared twice: once from the tangent vector and once, inverted, from the measure. The two cancelled. That cancellation is the entire mechanism, and everything in §1.2 is the same cancellation in more dimensions.

1.2 · Two dimensions, where something has to give

Now try a surface. Parametrise a two-dimensional patch by $(u^{1},u^{2})\mapsto x^{\mu}(u)$, and suppose the integrand is some rank-2 tensor $T_{\mu\nu}$. The only expression available with the right index structure is

Ask what happens under a change of parameters $u\mapsto u'$. Write the Jacobian matrix $J^{a}{}_{b}=\partial u^{a}/\partial u'^{b}$, with $a,b\in\{1,2\}$. Two things change, and we take them one at a time.

Change one: the measure. By Chapter 0.6 §8, $\dd u^{1}\dd u^{2}=\abs{\det J}\,\dd u'^{1}\dd u'^{2}$. For an oriented integral we drop the absolute value and keep $\det J$, agreeing to count a patch negatively if the new parameters reverse its handedness. That is a definition, and it is the same choice made in Chapter 0.7 when Stokes' theorem acquired a right-hand rule.

Change two: the two tangent vectors. By the chain rule, $\partial x^{\mu}/\partial u'^{b}=(\partial x^{\mu}/\partial u^{a})J^{a}{}_{b}$. So the pair of tangents feeding $T$ becomes

Abbreviate $T_{ab}\equiv T_{\mu\nu}(\partial x^{\mu}/\partial u^{a})(\partial x^{\nu}/\partial u^{b})$. That is just $T$ evaluated on the two coordinate tangents, a $2\times2$ array of numbers. Our demand is that (3.5.3) come out unchanged. Put beside change one, that demand says we need

Expand both sides. The left has four terms. The right has two, once we write out $\det J=J^{1}{}_{1}J^{2}{}_{2}-J^{2}{}_{1}J^{1}{}_{2}$:

The four products $J^{1}{}_{1}J^{1}{}_{2}$, $J^{1}{}_{1}J^{2}{}_{2}$, $J^{2}{}_{1}J^{1}{}_{2}$ and $J^{2}{}_{1}J^{2}{}_{2}$ can be varied independently, so their coefficients must match one at a time. Reading them off in order:

Let's read what those four conditions say. The first and last say the diagonal vanishes. The third says the off-diagonal entries are opposite. Both statements are contained in the single condition $T_{ab}=-T_{ba}$, since setting $a=b$ in it gives $T_{aa}=-T_{aa}$ and hence $T_{aa}=0$.

One step remains, from the array $T_{ab}$ back to the tensor. By choosing the patch we can make $\partial x^{\mu}/\partial u^{a}$ point in any two independent directions we like. So antisymmetry of $T_{ab}$ for every patch forces antisymmetry of $T_{\mu\nu}$ itself.

1.3 · Differential forms, and the wedge

Time for the definition. A $p$-form is a totally antisymmetric tensor of type $(0,p)$: an object $\omega_{\mu_{1}\cdots\mu_{p}}$ that changes sign whenever any two of its indices are exchanged. The first few degrees are things we already have names for.

• A $0$-form is a function.
• A one-form is a covector field.
• A $2$-form is what §1.2 just showed a surface integral wants.

Two pieces of notation, both used constantly from here on.

Antisymmetrisation brackets. For any array,

with $\operatorname{sgn}(\pi)=\pm1$ the sign of the permutation (Chapter 0.4 §5). The $1/p!$ makes the operation idempotent: antisymmetrising something already antisymmetric returns it unchanged, since all $p!$ terms are then equal to the original. For two indices, $T_{[\mu\nu]}=\half(T_{\mu\nu}-T_{\nu\mu})$, which is Chapter 2.4 §7.1's antisymmetric part.

The wedge product. Given a $p$-form $\alpha$ and a $q$-form $\beta$, their tensor product $\alpha_{\mu_{1}\cdots\mu_{p}}\beta_{\nu_{1}\cdots\nu_{q}}$ is not antisymmetric across the two groups. Antisymmetrise it and normalise:

The combinatorial factor is chosen so that basis elements multiply without stray numbers left over. The smallest case shows what it does. For two one-forms, $(p,q)=(1,1)$ and the factor is $2!/1!1!=2$, so

The sign rule, derived. Swap the roles of $\alpha$ and $\beta$ in (3.5.9). To turn $\beta_{[\mu_{1}\cdots\mu_{q}}\alpha_{\mu_{q+1}\cdots\mu_{p+q}]}$ back into the original ordering, every one of the $p$ indices of $\alpha$ must be moved past every one of the $q$ indices of $\beta$. That is $pq$ transpositions, each costing a minus sign inside a totally antisymmetric bracket. Hence

Read off the two cases we will use. Two one-forms anticommute, $\alpha\wedge\beta=-\beta\wedge\alpha$, and in particular $\alpha\wedge\alpha=0$ for any one-form. A one-form and a $2$-form commute. Nothing here is a convention we adopted: (3.5.11) was counted.

The basis. Every $p$-form can be written in the basis built by wedging coordinate differentials,

the $1/p!$ compensating for the fact that the sum runs over all orderings of each index set and each ordering contributes the same thing. Equivalently, sum over strictly increasing index sets with no factorial. For example a $2$-form in four dimensions is

six terms in all.

The count. A $p$-form in $n$ dimensions has one independent component for each way of choosing $p$ distinct indices from $n$. They must be distinct, because a repeated index makes the component vanish by antisymmetry. And only the choice matters, not the order, because reordering changes only the sign. So the number of components is $\binom{n}{p}$, which is Chapter 2.4 §7.2's count. In four dimensions:

$p$	$0$	$1$	$2$	$3$	$4$	$5$
components $\binom{4}{p}$	$1$	$4$	$6$	$4$	$1$	$0$

Two entries in that row are worth pausing on. The $6$ is the number of independent components of the electromagnetic field tensor (Chapter 2.6 §2.3). The field tensor is therefore a $2$-form, and not merely an antisymmetric array, which is what §10 collects.

The $0$ is not an accident of arithmetic either. A $5$-form in four dimensions must have a repeated index somewhere, by the pigeonhole principle, so every one of its components vanishes. In $n$ dimensions there are no forms of degree above $n$, which is why the tower above stops.

2 · The exterior derivative, which needs no connection

Again the destination first. We want a derivative that turns a $p$-form into a $(p+1)$-form and is a tensor without any connection being chosen. We will find that antisymmetrising the ordinary partial derivative does exactly this, and that the reason it works is the same symmetry of the connection that Chapter 3.3 §7.2 called the torsion-free condition.

2.1 · The definition, and the three familiar cases

Define the exterior derivative of a $p$-form $\omega$ by

That is the component form. We will want the same definition written in the wedge-product basis of (3.5.12) as well, because that is the version §5's grind box computes with, so here it is:

The factor $(p+1)$ in (3.5.14) cancels the $1/(p+1)!$ hidden in the bracket, leaving a plain alternating sum. Take the three smallest cases one at a time.

$p=0$. A $0$-form is a function $f$, the bracket does nothing, and

That is the gradient one-form of Chapter 0.6 §4. It is the object that chapter insisted was a row and not a column, and which needed a metric before it could be turned into the arrow called $\nabla f$. Here it appears with no metric at all. That is the first sign that $\dd$ is cheaper than $\nabla$.

$p=1$. With two indices the bracket is $\half(\;\cdot\;-\;\cdot\;)$ and the prefactor is $2$:

In flat three-dimensional space with Cartesian coordinates, identify a vector field $\vv F$ with the one-form $\omega_{i}=F^{i}$. Then $(\dd\omega)_{12}=\partial_{x}F^{y}-\partial_{y}F^{x}$, which is $(\nabla\times\vv F)_{z}$, and the other two independent components are the other two components of the curl. The exterior derivative of a one-form is the curl.

$p=2$ in three dimensions. Identify a vector field with a $2$-form by $\omega_{jk}=\epsilon^{ijk}F^{i}$. This is the same identification Chapter 0.7 §3.1 made implicitly when it wrote a flux as $\vv F\cdot\dd\vv A$, since a flux integrand eats two edge vectors. Then

where the three-term alternating sum has collapsed to three terms rather than six because the bracket's other three terms are equal to these by antisymmetry of $\omega$. The exterior derivative of a $2$-form is the divergence.

2.2 · Why $\dd\omega$ is a tensor, with no connection anywhere

Chapter 3.3 §4 proved that $\partial_{\mu}V^{\nu}$ is not a tensor, and the whole of that chapter's §5 was the repair. Antisymmetrising escapes the problem entirely, and the reason is one line.

Write the covariant derivative of a one-form, from Chapter 3.3 (3.3.30):

Our move is to antisymmetrise both sides in $\mu\nu$ and see what survives. On the left the result is $\nabla_{[\mu}\omega_{\nu]}$. On the right, the first term gives $\partial_{[\mu}\omega_{\nu]}$ and the second gives $-\Gamma^{\lambda}{}_{[\mu\nu]}\omega_{\lambda}$.

Now look at that second term. $\Gamma^{\lambda}{}_{\mu\nu}$ is symmetric in its two lower indices. That is the torsion-free condition, imposed in Chapter 3.3 §7.2 and derived there to be equivalent to the vanishing of $\Gamma^{\lambda}{}_{[\mu\nu]}$. So the second term is antisymmetrising something symmetric, which gives zero. What is left is

The same argument runs at every $p$. Each of the $p$ correction terms in the covariant derivative of a $p$-form carries a $\Gamma$ with the differentiating index and one form index sitting on its two lower slots, and antisymmetrising over those two kills it.

Let's look at what (3.5.20) is actually saying. It is worth stating twice, once in each direction. Read left to right, it says that the plain partial derivative, antisymmetrised, is secretly covariant and is therefore a tensor. Read right to left, it says that $\dd$ can be computed with partial derivatives and no Christoffel symbols at all, even on a curved manifold, and that the answer does not depend on which connection you would have used.

The exterior derivative exists before the metric does. Chapter 3.6 will lean on this without saying so. Chapter 6.3, where a connection on an internal space is built from scratch, leans on it harder.

3 · $\dd^{2}=0$, and three debts collected at once

This is the section the chapter exists for. Here is the destination and the collection together. We will prove one line, that applying the exterior derivative twice always gives zero. Then we show that this single line is Chapter 0.7's identity that the curl of a gradient vanishes, Chapter 0.7's identity that the divergence of a curl vanishes, and Chapter 2.6's discovery that two of Maxwell's four equations carry no physical content. Those are not three analogous facts. They are one fact seen three times.

3.1 · The proof

Take any $p$-form $\omega$ and apply (3.5.14) twice. The result has $p+2$ indices, and every term in it carries two partial derivatives:

Look only at the two derivative indices. They sit inside a total antisymmetrisation, so the expression changes sign when they are exchanged. But $\partial_{\mu_{0}}\partial_{\mu_{1}}=\partial_{\mu_{1}}\partial_{\mu_{0}}$ by Clairaut's theorem (Chapter 0.6 §6.1), so the expression is unchanged when they are exchanged. A quantity equal to minus itself is zero. Hence

That is the whole proof, and its brevity is the point rather than a shortcut. Nothing was assumed except that $\omega$ is twice continuously differentiable, which is Clairaut's hypothesis and was stated out loud in Chapter 0.6 §6.1.

3.2 · Debt one — the two vector identities of Chapter 0.7

The way to collect a debt is to look at $\dd\dd=0$ at one degree at a time. Start at the bottom. Set $p=0$, so that $\omega=f$ is a function and $\dd f$ is the gradient one-form (3.5.16). Then (3.5.22) reads, using the $p=1$ expansion (3.5.17),

In flat three-dimensional space the three independent components of a $2$-form are the three components of a vector, and §2.1 identified $\dd$ on a one-form with the curl. So (3.5.23) is $\nabla\times\nabla f=\vv 0$, which is (0.7.54). That was proved there in exactly this way, by writing out one component and cancelling two terms.

Now set $p=1$. Then $\dd\omega$ is the curl, $\dd\dd\omega$ is the divergence of it, and (3.5.22) is $\nabla\cdot(\nabla\times\vv A)=0$, which is (0.7.56). Chapter 0.7 proved that one by writing out six terms and cancelling them in pairs. Here the six terms are the six permutations inside the bracket of (3.5.21), and they cancel for the same reason.

So the two identities of Chapter 0.7 §7 are (3.5.22) at $p=0$ and $p=1$. Chapter 0.7 §7.4 asserted this and flagged it as a promise. It is now a theorem.

3.3 · Debt two — half of Maxwell's equations

Now four dimensions. Assemble the electromagnetic potential of Chapter 2.6 §2.2 into a one-form,

Our aim is to recognise something familiar, so the next step is to take the exterior derivative of that one-form and read off what the components are. The $p=1$ case of the definition, (3.5.17), gives them straight away:

That is Chapter 2.6's definition (2.6.9) of the field tensor, character for character. So $F=\dd A$. The electromagnetic field is the exterior derivative of the potential, and its six components are exactly the six independent components a $2$-form in four dimensions is allowed by §1.3's table.

Now apply (3.5.22) to it:

Write out what $\dd F=0$ says in components. $F$ is a $2$-form, so $\dd F$ is a $3$-form with $\binom{4}{3}=4$ independent components, and by the same collapse as in (3.5.18) each is a three-term cyclic sum:

That is (2.6.25) exactly. Chapter 2.6 §3.2 expanded its four components and found $\nabla\cdot\vv B=0$ together with the three components of Faraday's law. Chapter 2.6 §3.4 then substituted $F=\dd A$, watched six terms cancel in pairs by Clairaut, and concluded that given potentials, those two Maxwell equations cannot fail and carry no information about how electromagnetism works.

Here is the same conclusion in one symbol. The six terms Chapter 2.6 cancelled are the six permutations in (3.5.21) at $p=1$, and the cancellation is (3.5.22). Chapter 2.6's warning box promised precisely this and called it a promise rather than a derivation. The promise is now kept.

3.4 · What the fact actually is

Three results have just been shown to be one result. It is worth saying what the one result is, because it is not a fact about derivatives at all.

Chapter 0.7 §5.4 made the observation, filed it, and moved on: a boundary has no boundary of its own. The boundary of a disc is a circle, and a circle has no endpoints. The boundary of a ball is a sphere, and a sphere has no edge. In symbols, $\partial(\partial M)=\emptyset$.

Section 5 below proves the general Stokes theorem, $\int_{M}\dd\omega=\oint_{\partial M}\omega$, which pairs the operator $\dd$ with the operation of taking a boundary. Grant it for a moment and run the pairing twice. For any $(p-2)$-form $\omega$ and any $p$-dimensional region $M$,

The last step holds because there is nothing left to integrate over. And since the chain holds for every region $M$, however small, the integrand itself must vanish: $\dd\dd\omega=0$.

Read that chain in reverse and it says the same thing. The statement that applying the derivative twice gives nothing is the algebraic shadow of the statement that a boundary has no boundary. A fact about regions is showing up as a fact about derivatives.

That is why the two vanishings of Chapter 0.7 felt like a coincidence, and why it felt like a second coincidence that two of Maxwell's equations turned out to be empty. The real situation is this. A single geometric triviality about shapes has three separate algebraic disguises, and in each case we met the disguise before we met the shape.

4 · Closed, exact, and the shape of the region

Section 3 answered one direction of a question. Here is the other, and it is the direction where the topology of the region stops being a technicality.

Two definitions. A form $\omega$ is closed if $\dd\omega=0$. It is exact if $\omega=\dd\alpha$ for some form $\alpha$ of one degree lower. Equation (3.5.22) says exact implies closed, always, with no hypotheses. The question is the converse.

You have met this question three times already, without its name attached.

• Chapter 0.7 §2.3 asked when a curl-free field is a gradient. In the new language, that is asking when a closed one-form is exact.
• Chapter 0.7 §7.3 asked when a divergence-free field is a curl, which is a closed $2$-form being exact. It quoted the answer as the Poincaré lemma and said this chapter would prove it.
• Chapter 3.4 §8 met the same distinction in geometric dress, as the gap between having vanishing curvature and being a piece of flat space.

All three are the same question, and in each of them the obstruction is the shape of the region.

4.1 · The counterexample, so that the theorem has something to exclude

Chapter 0.7 §2.4 built the counterexample in full and it is worth having in the new notation. On the plane with the origin removed, define

It is closed. Compute $(\dd\omega)_{xy}=\partial_{x}\omega_{y}-\partial_{y}\omega_{x}$ with $\omega_{x}=-y/(x^{2}+y^{2})$ and $\omega_{y}=x/(x^{2}+y^{2})$. Differentiating the first by the quotient rule,

and the identical computation on $\omega_{x}$ gives the same thing, so the difference is zero everywhere the form is defined.

It is not exact. If $\omega$ were $\dd f$ then $\oint\omega$ around any closed loop would be $f(\text{end})-f(\text{start})=0$. But on the unit circle, parametrised by $(x,y)=(\cos\theta,\sin\theta)$, the definition (3.5.1) gives $\omega_{\mu}\dd x^{\mu}/\dd\theta = (\cos\theta)(\cos\theta)-(\sin\theta)(-\sin\theta)=1$, so $\oint\omega=2\pi\neq0$.

Nothing is wrong with $\omega$. Something is wrong with the region. The punctured plane has a loop that cannot be shrunk to a point without crossing the missing origin, and that loop is exactly what $\oint\omega=2\pi$ detects. Closed but not exact is a measurement of the hole.

4.2 · The Poincaré lemma, for one-forms, proved

Call a region star-shaped about a point if the straight segment from that point to any point of the region stays inside. A ball is star-shaped. The punctured plane is not.

Claim. On a star-shaped region about the origin, every closed one-form is exact.

The construction. Given a closed $\omega$, define a function by integrating $\omega$ out along the straight ray from the origin to $x$:

This is legitimate precisely because the region is star-shaped: the ray $tx$ for $t\in[0,1]$ stays inside, so $\omega$ is defined all along it.

Line 1: differentiate under the integral sign. Chapter 0.2 §4.4 licensed this move for continuous integrands. We want $\partial_{\nu}f$, and there are two places for the derivative to land. It hits the argument $tx$ of $\omega$ by the chain rule, and it hits the explicit factor $x^{\mu}$ by the product rule:

Two small things to account for there. The factor $t$ in the first term is the derivative of the argument $tx^{\nu}$ with respect to $x^{\nu}$. The second term came from $\partial_{\nu}x^{\mu}=\delta^{\mu}{}_{\nu}$.

Line 2: use closedness, and only here. Closed means $\partial_{\nu}\omega_{\mu}=\partial_{\mu}\omega_{\nu}$. That lets us swap the two labels inside the first term, which is the one move the hypothesis buys us:

Line 3: recognise a product rule in $t$. The bracket in (3.5.33) looks like the output of a derivative, so let's go looking for the thing it is the derivative of. Consider the function $t\mapsto t\,\omega_{\nu}(tx)$ and differentiate it with respect to $t$. The product rule gives one term from the explicit $t$ and one from the argument, and the chain rule supplies $x^{\mu}$ in the second:

The right-hand side of (3.5.34) is exactly the bracket in (3.5.33).

Line 4: integrate. The integrand is now a total derivative in $t$, so the Fundamental Theorem of Calculus (Chapter 0.2 §2) evaluates it at the two endpoints and we are done:

So $\omega=\dd f$. $\blacksquare$

Read line by line, the proof is short enough to hold in the head. Integrate the form radially, differentiate, use closedness once to swap two derivative labels, notice that what remains is a total derivative in the radial parameter, and apply the Fundamental Theorem. The star-shaped hypothesis is spent in exactly one place, and that place is writing down (3.5.31) at all.

Grind box — the same proof for a $p$-form, and hence the vector potential

The construction generalises with one extra factor of $t$ per index. For a $p$-form $\omega$ on a region star-shaped about the origin, define the $(p-1)$-form

$$\big(K\omega\big)_{\mu_{2}\cdots\mu_{p}}(x) \;\equiv\; \int_{0}^{1}t^{\,p-1}\;x^{\mu_{1}}\,\omega_{\mu_{1}\mu_{2}\cdots\mu_{p}}(tx)\;\dd t.$$

The claim is the homotopy formula

$$\dd\big(K\omega\big) \;+\; K\big(\dd\omega\big) \;=\; \omega,$$

from which the lemma is immediate: if $\dd\omega=0$ the second term dies and $\omega=\dd(K\omega)$ is exact.

The verification is the four lines of the main text with more indices. Rather than sketch that, we do one case completely and explicitly, namely the case the promise was about: $p=2$ in three dimensions, where the claim is the existence of the vector potential.

Setting up. Under §2.1's identification a $2$-form corresponds to a vector field $\vv B$, and $K$ applied to it reads, in components,

$$A_{k}(\vv x) \;=\; \int_{0}^{1} t\;\epsilon_{kij}\,B_{i}(t\vv x)\,x_{j}\;\dd t, \qquad\text{that is}\qquad \vv A(\vv x) \;=\; \int_{0}^{1} t\;\vv B(t\vv x)\times\vv x\;\dd t.$$

Take the curl, differentiating under the integral sign. The derivative hits the argument $t\vv x$ by the chain rule, producing a factor $t$, and separately hits the explicit $x_{j}$:

$$\big(\nabla\times\vv A\big)_{m} \;=\; \int_{0}^{1}\epsilon_{mnk}\epsilon_{kij}\Big[\,t^{2}\,\big(\partial_{n}B_{i}\big)(t\vv x)\,x_{j} \;+\; t\,B_{i}(t\vv x)\,\delta_{jn}\Big]\dd t.$$

Use the epsilon identity $\epsilon_{mnk}\epsilon_{kij}=\delta_{mi}\delta_{nj}-\delta_{mj}\delta_{ni}$, which Chapter 0.7 §4.4's grind box established. The first bracket becomes $t^{2}\big[(\partial_{j}B_{m})x_{j}-(\partial_{i}B_{i})x_{m}\big]$, and here is the one place the hypothesis is spent: $\partial_{i}B_{i}=\nabla\cdot\vv B=0$, so the second piece dies and the first is $t^{2}(\vv x\cdot\nabla)B_{m}$. The second bracket becomes $t\big[3B_{m}-B_{m}\big]=2tB_{m}$, the $3$ coming from $\delta_{nn}=3$.

Recognise a product rule in $t$. Differentiating $t^{2}B_{m}(t\vv x)$ with respect to $t$ gives $2tB_{m}(t\vv x)+t^{2}x_{j}(\partial_{j}B_{m})(t\vv x)$, which is exactly the integrand. So by the Fundamental Theorem of Calculus

$$\big(\nabla\times\vv A\big)_{m} \;=\; \Big[t^{2}B_{m}(t\vv x)\Big]_{t=0}^{t=1} \;=\; B_{m}(\vv x). \qquad\blacksquare$$

What this settles. A divergence-free field on a star-shaped region is the curl of something, and the something is written down above. That is the existence of the vector potential, quoted at Chapter 0.7 §7.3 with the note that Chapter 3.5 would prove it. ⚑ The general $p$ runs the same way. The epsilon identity is replaced by the expansion of $\dd\omega$, the hypothesis $\dd\omega=0$ is spent at the same single step, and what survives is again a total derivative in $t$ whose endpoints give $\omega$. But that is a sketch of a proof and not a proof. What is derived here is the case $p=2$ in three dimensions. The general statement is quoted, and this is the one place in this chapter where that happens.

4.3 · What is left over is topology

Put the two halves together. Exact always implies closed, by (3.5.22). Closed implies exact on a star-shaped region, by §4.2. On a region that is not star-shaped the implication can fail, by §4.1. So the collection of closed forms that are not exact, counted degree by degree, measures how far the region is from being simple. It is a measurement made with calculus that returns an answer about shape.

That measurement has a name, de Rham cohomology, promised by that name at Chapter 0.7 §7.3. We will not develop it. What matters here is that the same distinction has now appeared four times in this book, and in every case the local information is impeccable while the global conclusion fails.

• The curl-free field with a non-zero circulation.
• The divergence-free field with no potential.
• Chapter 3.4 §8's cone, flat everywhere and yet not a plane.
• The angle form (3.5.29) above.

Chapter 3.9 meets the same distinction once more, on a universe that can be flat everywhere and still closed. Part VI meets it as the reason certain field configurations cannot be smoothly undone.

5 · One theorem instead of four

Chapter 0.7 §5.4 lined up the Fundamental Theorem of Calculus, the gradient theorem, Stokes' theorem and the divergence theorem in a table, observed that every row says the same thing, wrote down $\int_{M}\dd\omega=\oint_{\partial M}\omega$ as (0.7.46), and said plainly: we are quoting the general form, not deriving it. The machinery is Chapter 3.5's job. The machinery now exists. This section pays.

Here is the destination. We define what it means to integrate a $p$-form over a $p$-dimensional oriented region, check that the definition does not depend on the labelling, prove the theorem on a cube by the Fundamental Theorem of Calculus applied one face-pair at a time, and extend it to general regions by the chopping-and-cancelling argument of Chapter 0.7. This time that argument cancels exactly rather than approximately, because the antisymmetry of §1 does the work that hand-waving did before.

5.1 · The integral of a $p$-form

Let $M$ be a $p$-dimensional region inside the manifold, parametrised by $(u^{1},\ldots,u^{p})\mapsto x^{\mu}(u)$ over some region $U$ of parameter space. Define

We still owe the check that this is well defined. Section 1.2 did the two-dimensional case, and the general case is the same three sentences. Changing parameters multiplies the product of tangent vectors by $p$ copies of the Jacobian matrix, contracted into a totally antisymmetric object. By Chapter 0.4 §5 that is precisely $\det J$ times the original. The measure supplies $\det J$ the other way. They cancel. The integral belongs to the region.

Two remarks that will be needed later.

First, no metric appears in (3.5.36), so this notion of integration exists on a bare manifold. Section 6 is where a metric is finally wanted, and it is wanted there for a different job.

Second, reversing the orientation of the region reverses the sign of the integral, since it swaps two of the $u$'s and the integrand is antisymmetric. That second remark is what makes the whole of §5.3 work.

5.2 · The theorem on a square, in full

Take $M$ to be the unit square $0\le u^{1},u^{2}\le1$ and $\omega$ a one-form on it, written $\omega=P\,\dd u^{1}+Q\,\dd u^{2}$. Then by (3.5.17) the only independent component of $\dd\omega$ is $\partial_{1}Q-\partial_{2}P$, so

The $Q$ term. Do the $u^{1}$ integral first and apply the Fundamental Theorem of Calculus in that variable alone, holding $u^{2}$ fixed:

The $P$ term. Do the $u^{2}$ integral first, same move, and keep the minus sign:

The boundary. The square's boundary is four edges. Traverse them counterclockwise, which is the orientation induced by the square's own, and evaluate (3.5.1) on each. On the bottom edge $u^{2}=0$ with $u^{1}$ running from $0$ to $1$, the tangent has only a $u^{1}$ component, so only $P$ contributes and the edge gives $+\int_{0}^{1}P(u^{1},0)\,\dd u^{1}$. On the right edge only $Q$ contributes, upward, giving $+\int_{0}^{1}Q(1,u^{2})\,\dd u^{2}$. The top edge runs backwards and gives $-\int_{0}^{1}P(u^{1},1)\,\dd u^{1}$. The left edge runs downward and gives $-\int_{0}^{1}Q(0,u^{2})\,\dd u^{2}$. Adding the four:

which is (3.5.38) plus (3.5.39) term for term. Hence $\int_{M}\dd\omega=\oint_{\partial M}\omega$ on the square. $\blacksquare$

Notice what proved it: the Fundamental Theorem of Calculus, used twice, in one variable at a time. Everything else was bookkeeping about which edge runs which way, and that bookkeeping is what the antisymmetry of §1 automates.

Grind box — the same argument on a $p$-dimensional cube, with the signs counted

Let $M$ be the unit cube $0\le u^{a}\le1$ in $p$ dimensions and let $\omega$ be a $(p-1)$-form. By linearity it is enough to treat one basis term at a time, so take

$$\omega \;=\; f\;\dd u^{1}\wedge\cdots\wedge\widehat{\dd u^{k}}\wedge\cdots\wedge\dd u^{p},$$

the hat meaning that the $k$-th differential is omitted.

The left side. Applying (3.5.15), every term of $\dd\omega$ except the one carrying $\partial_{k}f$ contains a repeated differential and vanishes. Moving $\dd u^{k}$ from the front into its home slot costs $k-1$ transpositions, so

$$\dd\omega \;=\; \pdv{f}{u^{k}}\;\dd u^{k}\wedge\dd u^{1}\wedge\cdots\widehat{\dd u^{k}}\cdots \;=\; (-1)^{k-1}\,\pdv{f}{u^{k}}\;\dd u^{1}\wedge\cdots\wedge\dd u^{p}.$$

Integrating over the cube and doing the $u^{k}$ integral by the Fundamental Theorem,

$$\int_{M}\dd\omega \;=\; (-1)^{k-1}\int\Big[f\big|_{u^{k}=1} - f\big|_{u^{k}=0}\Big]\;\dd u^{1}\cdots\widehat{\dd u^{k}}\cdots\dd u^{p}.$$

The right side. The cube has $2p$ faces. On a face $u^{j}=\text{const}$ with $j\neq k$, the parametrisation of the face has $\dd u^{j}=0$, and $\omega$ contains $\dd u^{j}$, so $\omega$ restricted to that face is zero and it contributes nothing. Only the two faces $u^{k}=0$ and $u^{k}=1$ survive, and on them $\omega$ restricts to $f$ times the volume element of the face. The orientation induced on the face $u^{k}=1$ by the cube is the one for which the outward direction comes first, which is $(-1)^{k-1}$ times the natural ordering $u^{1},\ldots,\widehat{u^{k}},\ldots,u^{p}$. The face $u^{k}=0$ has outward direction reversed and so carries the opposite sign. Adding the two gives exactly the previous display. $\blacksquare$

At $p=2$, $k=1$ gives the $Q$ term of the main text and $k=2$ gives the $P$ term, minus sign and all.

5.3 · General regions, and the cancellation made exact

Assume $M$ admits such a decomposition. By that we mean it can be chopped into finitely many cells, each smoothly mapped to a cube, with piecewise-smooth boundary. That is a real hypothesis, and the callout below says what it costs.

Then here is the argument. Chop $M$ into small cells, each of which can be mapped to a cube. Write the theorem for each cell and add. On the right, every interior face is shared by exactly two neighbouring cells. The orientation each cell induces on that shared face is the outward one for that cell, so the two neighbours induce opposite orientations on it. By the second remark of §5.1 the two contributions are exact negatives, and they cancel. What survives is the sum over faces lying on the outer boundary. Hence

5.4 · The four rows, recovered

Read (3.5.41) at each degree and each dimension.

$p=1$, $\omega$ a function. The region is a curve, its boundary is two points, and the right-hand side is the difference of $f$ between them. That is the gradient theorem of Chapter 0.7 §2.2. On a straight interval in one dimension it is the Fundamental Theorem of Calculus itself. So this one degree covers rows one and two of Chapter 0.7's table.

$p=2$, $\omega$ a one-form, in three dimensions. By §2.1, $\dd\omega$ is the curl. The region is a surface, its boundary a closed curve, and (3.5.41) reads $\int_{S}(\nabla\times\vv F)\cdot\dd\vv A=\oint_{\partial S}\vv F\cdot\dd\vv r$. That is (0.7.41), Stokes' theorem, which is row three. Restricting to a flat region in the plane gives Green's theorem (0.7.42), which Chapter 0.7 had already shown was not independent.

$p=3$, $\omega$ a $2$-form, in three dimensions. By (3.5.18), $\dd\omega$ is the divergence, and (3.5.36) with the identification $\omega_{jk}=\epsilon^{ijk}F^{i}$ turns $\oint_{\partial V}\omega$ into the flux $\oint\vv F\cdot\dd\vv A$. So (3.5.41) is $\int_{V}(\nabla\cdot\vv F)\dd V=\oint_{\partial V}\vv F\cdot\dd\vv A$, the divergence theorem (0.7.44), which is row four.

Count what has just happened. Four theorems, four names, four sets of hypotheses and four right-hand rules have become one line, proved once, in any number of dimensions, on any manifold, with no metric anywhere. That was Chapter 0.7's promise, and this is it.

6 · The volume element, and a divergence theorem that works on a manifold

Here is the destination. Chapter 3.6 has to integrate a scalar over all of spacetime and then discard a total derivative. Neither operation is available yet: "all of spacetime" needs a chart-independent notion of volume, and "discard a total derivative" needs a divergence theorem. This section builds both, and the tool it uses is Chapter 0.6 §8's Jacobian determinant, which will then be used a second time in Chapter 3.6 for a different purpose.

6.1 · There is essentially one top-degree form

In $n$ dimensions an $n$-form has $\binom{n}{n}=1$ independent component, by §1.3's count. So any two top-degree forms are proportional, and the whole question is what the single component does under a change of chart.

Take $\Omega=\Omega_{01\cdots(n-1)}\,\dd x^{0}\wedge\cdots\wedge\dd x^{n-1}$. Under a chart change $x\to x'$, the components of a $(0,n)$ tensor pick up $n$ Jacobian factors:

Take the one independent component, $\mu_{1}\cdots\mu_{n}=01\cdots(n-1)$. The right-hand side is then a totally antisymmetric contraction of $n$ copies of the Jacobian matrix, which by Chapter 0.4 §5's characterisation of the determinant is $\det(\partial x/\partial x')$ times $\Omega_{01\cdots(n-1)}$. So the single component of a top-degree form transforms by one factor of the Jacobian determinant, and by nothing else.

6.2 · The metric determinant supplies exactly that factor

So what we need is a quantity that carries exactly one factor of the Jacobian determinant, to be cancelled against the one the coordinate volume carries. The metric supplies it, and the rest of this section is the check.

Write $g\equiv\det g_{\mu\nu}$, a number at each point, negative in Lorentzian signature because the metric has one positive and three negative eigenvalues. Chapter 3.3 §1 derived the transformation law of the metric components, (3.3.3):

Read (3.5.43) as a matrix equation: $g'=J^{\!\top}gJ$ with $J^{\alpha}{}_{\mu}=\partial x^{\alpha}/\partial x'^{\mu}$. Take determinants, using $\det(AB)=\det A\det B$ and $\det J^{\!\top}=\det J$ from Chapter 0.4 §5:

Two details there are worth naming. The square is why the sign of $\det J$ drops out, and the modulus is why taking the square root is harmless.

Now we have half of what we want. The other half is the measure. Chapter 0.6 §8 established $\dd^{n}x=\abs{\det J}^{-1}\dd^{n}x'$, so the coordinate volume shrinks by exactly the factor the metric determinant grows by. That is the pairing we were hoping for, so multiply the two together:

This is the invariant volume element. For any scalar $f$, the quantity $\int f\sqrt{-g}\,\dd^{n}x$ is a number belonging to the region, not to the chart. Chapter 0.6 §8.3 told you this was coming and even checked the answer in two flat cases: plane polar coordinates give $\sqrt{\det g_{ij}}=\sqrt{1\cdot r^{2}}=r$, which is (0.6.52), and spherical coordinates give $r^{2}\sin\theta$, which is (0.6.53). Both were obtained there by cutting up regions by hand. (3.5.45) is the reason they came out that way. The sphere of radius $a$ gives $\sqrt{a^{2}\cdot a^{2}\sin^{2}\theta}=a^{2}\sin\theta$, which is the area element Chapter 3.4 §1 used to compute the region enclosed by its loop.

It is convenient to have the corresponding tensor. Define

Here $[\mu\nu\rho\sigma]$ is the permutation symbol, $+1$ for an even permutation of $0123$, $-1$ for an odd one, $0$ if any index repeats. The symbol on its own is not a tensor. It has the same entries in every chart, which no tensor with four indices does. But multiplying by $\sqrt{-g}$ supplies exactly the one Jacobian factor §6.1 showed a top-degree form needs, so $\varepsilon_{\mu\nu\rho\sigma}$ is a tensor. The volume element is then the $4$-form $\varepsilon=\sqrt{-g}\;\dd x^{0}\wedge\dd x^{1}\wedge\dd x^{2}\wedge\dd x^{3}$.

6.3 · A formula for the derivative of a determinant

One more tool is needed, here and again in Chapter 3.6 §4. Here is what we are after: we want the change in $\det M$ produced by a small change in the entries of $M$.

Line 1: the cofactor expansion. Expanding a determinant along its $a$-th row gives $\det M=\sum_{b}M_{ab}C_{ab}$, where the cofactor $C_{ab}$ contains no entry from row $a$. Because the cofactor does not contain $M_{ab}$, differentiating with respect to a single entry is immediate: $\partial(\det M)/\partial M_{ab}=C_{ab}$.

Line 2: the cofactors are the inverse. We would rather not carry cofactors around, and we do not have to. The adjugate formula of Chapter 0.4 says $(M^{-1})_{ba}=C_{ab}/\det M$, so $C_{ab}=\det M\,(M^{-1})_{ba}$.

Line 3: chain rule. Now put the two lines together for a general variation. For any $\delta M_{ab}$ of the entries,

This is Jacobi's formula. The case we will use is the metric, so apply it with $M=g_{\mu\nu}$ and with $\delta$ standing for any derivation whatever. That gives the version §6.4 reaches for:

6.4 · The divergence, and the theorem Chapter 3.6 needs

The contracted connection. Take the Christoffel formula (3.3.50) and contract the upper index with the first lower one:

The first and third terms cancel, and the cancellation is worth seeing rather than taking on trust. Relabel the two summed indices in the third term, swapping the names $\mu$ and $\sigma$, which is legitimate because both are dummies. It becomes $-g^{\sigma\mu}\partial_{\mu}g_{\sigma\lambda}$, and $g^{\sigma\mu}=g^{\mu\sigma}$ by symmetry of the metric, so it is minus the first term exactly. What is left is

the middle step being (3.5.48) with $\delta=\partial_{\lambda}$, and the last being $\partial_{\lambda}\ln\sqrt{-g}=\half\partial_{\lambda}\ln(-g)=\partial_{\lambda}g/(2g)$, where the minus signs inside the logarithm cancel between numerator and denominator.

The divergence. Now write out $\nabla_{\mu}V^{\mu}$ from Chapter 3.3 §5 and substitute:

the last step being the product rule read backwards. The covariant divergence of a vector field is an ordinary divergence in disguise, with the volume factor tucked inside. No Christoffel symbol survives, which is why this identity is worth having.

The theorem. Multiply (3.5.51) by $\sqrt{-g}$ and integrate. The result is an ordinary partial derivative under an ordinary integral sign, and §5 knows how to convert one of those into a boundary term. To do that we need something for §5 to act on, so build the $3$-form $\Theta_{\nu\rho\sigma}\equiv\varepsilon_{\mu\nu\rho\sigma}V^{\mu}$ and take its exterior derivative. Using (3.5.15) and the permutation-symbol identity $\tfrac{1}{3!}[\mu\nu\rho\sigma][\lambda\nu\rho\sigma]=\delta^{\mu}{}_{\lambda}$, every index sum collapses and

Feeding that into (3.5.41) gives the divergence theorem for a curved manifold:

The only use Chapter 3.6 makes of (3.5.53) is this: if a variation of the metric is required to vanish near the boundary, a term of the form $\nabla_{\mu}V^{\mu}$ in the integrand contributes nothing at all. That single sentence is what turns a four-line variation into a field equation, and §4 of the next chapter also asks what it costs when the variation does not vanish there.

7 · Dragging a tensor along a flow

The rest of the chapter changes subject, so here is the new destination. Chapter 3.3 §4 showed that the ordinary derivative of a tensor is not a tensor and needed a whole chapter of repair. Section 2 found one escape from that, available only to antisymmetric objects. Here is a second escape, available to everything, at the price of choosing a vector field first. It costs no connection, and in §8 it becomes the definition of a symmetry.

7.1 · The flow of a vector field

A vector field $X$ assigns a direction to every point, so it defines a system of ordinary differential equations,

The solutions of that system are the integral curves of $X$, the paths that follow the arrows. Chapter 0.8 §1 guarantees a unique solution through each point for smooth $X$. Following every point along its own curve for a parameter distance $\varepsilon$ defines a map of the manifold to itself, $\phi_{\varepsilon}$, called the flow. Solving (3.5.54) to first order by Taylor's theorem (Chapter 0.3 §1),

The idea. A tensor at the point $\phi_{\varepsilon}(x)$ still cannot be compared with a tensor at $x$. That was Chapter 3.2 §5's central prohibition and nothing has repealed it. But the flow supplies a map between the two points, and any smooth map between points carries tensors along with it, by contracting each index with the map's Jacobian. So here is the recipe: carry the tensor at $\phi_{\varepsilon}(x)$ back to $x$ along the flow, subtract the tensor already at $x$, divide by $\varepsilon$, and let $\varepsilon\to0$. That is the Lie derivative $\mathcal{L}_{X}$.

The price is visible in the recipe. The answer depends on the whole field $X$ near $x$, not just on $X$ at $x$, because a nearby direction is needed to define the map. So $\mathcal{L}_{X}T$ is not a derivative "in the direction $X$" in the way $\nabla_{X}T$ is. It is a comparison of $T$ with the version of itself that the flow of $X$ would produce.

7.2 · Scalars, then vectors, with every index step shown

A scalar. Carrying a number requires no Jacobian, so the recipe reads

which is the directional derivative, and also the way Chapter 3.2 §4 defined a tangent vector in the first place.

A vector. Now the Jacobian is needed. From (3.5.55), writing $x'=\phi_{\varepsilon}(x)$,

Carrying a vector forward along the flow multiplies its components by (3.5.57). We want to carry backward, so we need the inverse matrix, which to first order is obtained by flipping the sign of the small term:

as one checks by multiplying the two and seeing the $\varepsilon$ terms cancel.

Now assemble. The vector to be carried back is $Y$ evaluated at the shifted point, and Taylor's theorem gives $Y^{\nu}(x+\varepsilon X)=Y^{\nu}(x)+\varepsilon X^{\lambda}\partial_{\lambda}Y^{\nu}+O(\varepsilon^{2})$. Contract with (3.5.58), keeping terms to first order only:

Two things happened in the second line and both are worth naming. The $\delta^{\mu}{}_{\nu}$ hit both factors, producing the first and second terms. The $-\varepsilon\partial_{\nu}X^{\mu}$ hit only $Y^{\nu}$, because its other product is second order.

That is the dragged-back vector. The recipe now says to finish the job: subtract $Y^{\mu}$, divide by $\varepsilon$ and take the limit.

Let's look at that last equality, because it is not something we arranged. It is (3.2.32), the components of the commutator of two vector fields, which Chapter 3.2 §8 built by an entirely different route: it demanded that the composition of two derivations be a derivation. Two independent constructions have produced the same object, and the bracket has now acquired a second meaning worth carrying. $[X,Y]$ measures the failure of $Y$ to be carried into itself by the flow of $X$.

7.3 · Covectors, and then everything

A covector is carried the other way, so its components are contracted with (3.5.57) rather than its inverse. Repeating the computation with the sign of the Jacobian term flipped and the index in the other position,

The pattern is now visible and it is the same one Chapter 3.3 §5.4 found for the covariant derivative: one transport term, plus one correction per index, with the sign set by whether the index is up or down. For the case this chapter needs, a $(0,2)$ tensor,

7.4 · The partial derivatives may be replaced by covariant ones

Something in the last three equations should be bothering you. Equations (3.5.60) to (3.5.62) contain plain partial derivatives, and by Chapter 3.3 §4 a plain partial derivative of a tensor is not a tensor. The combinations above are tensors anyway, because the construction that produced them was manifestly tensorial. But there is a cleaner way to see it, and §8 needs the result.

Claim. In (3.5.60) and (3.5.62), every $\partial$ may be replaced by $\nabla$ without changing the answer, for any torsion-free connection.

Proof for the vector case. Write the two covariant derivatives out using Chapter 3.3 §5:

Look only at the two $\Gamma$ terms. In the second of them, relabel the two dummy indices $\lambda$ and $\nu$, swapping their names. That is legitimate, since both are summed. It becomes $-X^{\lambda}\Gamma^{\mu}{}_{\nu\lambda}Y^{\nu}$. Now use $\Gamma^{\mu}{}_{\nu\lambda}=\Gamma^{\mu}{}_{\lambda\nu}$, the torsion-free symmetry of Chapter 3.3 §7.2, and it is exactly minus the first. They cancel, and what remains is (3.5.60). $\blacksquare$

The $(0,2)$ case is the identical two moves applied to three $\Gamma$ terms, and it is done in the next section where it is used.

8 · Killing vectors: the directions in which nothing changes

Here is the destination. We ask which vector fields have flows that leave the metric exactly as they found it, translate that demand into an equation on the field, and check it on two spaces. The equation is the geometric form of the word "symmetry", and §9 turns each solution of it into a conserved quantity.

8.1 · The definition, and Killing's equation

A vector field $\xi$ is a Killing vector field if dragging the metric along its flow changes nothing:

That is compact, and compact is no use for computing with. Our next goal is a condition on the components of $\xi$, so write the Lie derivative out with the $(0,2)$ formula (3.5.62), taking $T$ to be the metric itself:

Now simplify, one move at a time.

Move 1: replace the partial derivatives by covariant ones. Section 7.4 licensed this. To see it explicitly here, expand the three covariant derivatives: $\xi^{\lambda}\nabla_{\lambda}g_{\mu\nu}$ contributes $-\xi^{\lambda}\Gamma^{\sigma}{}_{\lambda\mu}g_{\sigma\nu}-\xi^{\lambda}\Gamma^{\sigma}{}_{\lambda\nu}g_{\mu\sigma}$ beyond the partial derivative, while $g_{\lambda\nu}\nabla_{\mu}\xi^{\lambda}$ contributes $+g_{\lambda\nu}\Gamma^{\lambda}{}_{\mu\sigma}\xi^{\sigma}$ and $g_{\mu\lambda}\nabla_{\nu}\xi^{\lambda}$ contributes $+g_{\mu\lambda}\Gamma^{\lambda}{}_{\nu\sigma}\xi^{\sigma}$. Relabelling the dummies $\lambda\leftrightarrow\sigma$ in the last two and using the symmetry of $\Gamma$ in its lower indices makes them cancel the first two exactly.

Move 2: kill the first term. Metric compatibility says $\nabla_{\lambda}g_{\mu\nu}=0$. That was imposed in Chapter 3.3 §7.1 as Demand 1, and written out in components there as (3.3.41). So the first term is gone altogether.

Move 3: pull the metric inside the derivative. The same fact, $\nabla g=0$, lets the metric pass through a covariant derivative, so $g_{\lambda\nu}\nabla_{\mu}\xi^{\lambda}=\nabla_{\mu}\big(g_{\lambda\nu}\xi^{\lambda}\big)=\nabla_{\mu}\xi_{\nu}$, which is the index-lowering of Chapter 3.3 §1.1 performed under the derivative. The third term becomes $\nabla_{\nu}\xi_{\mu}$ by the same move. What is left is

This is Killing's equation. It says the covariant derivative of $\xi$, with both indices down, is antisymmetric.

Count the equations against the unknowns, because the count is the interesting part. In four dimensions Killing's equation is ten equations, one for each symmetric index pair, imposed on four unknown functions. That is heavily overdetermined. It is why most metrics have no Killing vectors at all, and why the ones that do are special.

8.2 · Flat spacetime has ten

Take $g_{\mu\nu}=\eta_{\mu\nu}$ in Cartesian coordinates, where every component is constant and every $\Gamma$ vanishes, so $\nabla=\partial$ and (3.5.66) reads $\partial_{\mu}\xi_{\nu}+\partial_{\nu}\xi_{\mu}=0$.

The four translations. $\xi=\partial_{a}$ for each $a=0,1,2,3$ has constant components, so both terms vanish. Four solutions.

The six rotations and boosts. Try $\xi_{\mu}=\omega_{\mu\nu}x^{\nu}$ with $\omega$ a constant array. Then $\partial_{\mu}\xi_{\nu}=\omega_{\nu\mu}$, and (3.5.66) becomes $\omega_{\nu\mu}+\omega_{\mu\nu}=0$: the array must be antisymmetric. An antisymmetric $4\times4$ array has $\binom{4}{2}=6$ independent entries, by §1.3's count. Six more solutions.

Ten in all, and they are exactly the ten generators of the Poincaré group: four translations, three rotations and three boosts. Six of those ten are the Lorentz group, which Chapter 2.3 obtained by demanding that the interval be preserved and counted by hand in its Problem 4. The four translations are new here, and they arrive on the same footing as the rest.

Take a moment over what that means. Killing vectors are the infinitesimal version of "transformations that leave the geometry alone". In flat spacetime that turns out to be the whole of special relativity's symmetry group, recovered from a differential equation.

8.3 · The sphere has three, and one obvious candidate is not among them

Take the sphere of radius $a$, with $g_{\theta\theta}=a^{2}$ and $g_{\phi\phi}=a^{2}\sin^{2}\theta$.

The easy one. No metric component mentions $\phi$, so by the practical test $\xi=\partial_{\phi}$ is Killing. It is the rotation about the polar axis.

The two hidden ones. The sphere is no more special about its poles than about anywhere else, so rotations about the other two axes must also be symmetries. In this chart they are not coordinate directions, so the practical test cannot find them and we have to write them down ourselves. They are

and substituting either into (3.5.65) gives zero in all three components. (Checked symbolically: for both fields all three components of $\mathcal{L}_{\xi}g$ vanish identically, and the covariant form (3.5.66) gives the same array as the partial-derivative form (3.5.65), which is Move 1 verified rather than trusted.)

The candidate that fails. Try $\xi=\partial_{\theta}$, "push everything south". Its components are constant, so (3.5.65) reduces to its first term, $\partial_{\theta}g_{\mu\nu}$, and only one metric component depends on $\theta$:

So $\partial_{\theta}$ is not a Killing vector. But (3.5.68) does more than say so. It says precisely how the field fails. The flow leaves north–south distances untouched and stretches east–west distances, at a rate that vanishes at the equator, where $\sin2\theta=0$, and is largest at colatitude $45^{\circ}$. That is a sharper statement than "the sphere is not symmetric under pushing south", and the figure measures it.

flow parameter s

0.0°

start colatitude θ₀

35°

run the flow

reset

left ξ=∂/∂φ : N–S edge 0.383972435 → 0.383972435 | E–W edge 0.280302322 → 0.280302322

right ξ=∂/∂θ : N–S edge 0.383972435 → 0.383972435 | E–W edge 0.280302322 → 0.280302322

at colatitude θ₀+s = 35.0° : measured d(ℓ²_EW)/ds ÷ (Δφ)² = 0.939691395 predicted a² sin 2(θ₀+s) = 0.939692621

9 · A conserved quantity along every geodesic

This is the section Chapter 3.7 spends. The destination has to be stated twice, because the result is obtained twice. First: for any Killing vector $\xi$ and any geodesic with tangent $u$, the single number $\xi_{\mu}u^{\mu}$ is the same at every point of the geodesic. Four lines. Second: that number is the Noether charge of Chapter 1.4 for the symmetry generated by $\xi$, so the theorem is not an analogue of Noether's but an instance of it.

9.1 · The four-line derivation

Let $x^{\mu}(\tau)$ be a geodesic with tangent $u^{\mu}=\dd x^{\mu}/\dd\tau$. The geodesic equation of Chapter 3.3 §8 then reads $u^{\nu}\nabla_{\nu}u^{\mu}=0$, which says the curve parallel-transports its own tangent. Let $\xi$ satisfy Killing's equation (3.5.66). We want to show that a particular number does not change along the curve, so define $Q\equiv\xi_{\mu}u^{\mu}$ and differentiate it.

Line 1. $Q$ is a scalar, so its rate of change along the curve is the directional derivative of it, and for a scalar the covariant and ordinary derivatives agree:

Line 2: Leibniz. There are two factors inside that derivative, so split it. The covariant derivative obeys the product rule (Chapter 3.3 §5.4), which gives

Line 3: the second term is the geodesic equation. Look at it: it is exactly $u^{\nu}\nabla_{\nu}u^{\mu}$, which is zero. So it vanishes and only the first survives:

Line 4: only the symmetric part survives, and it is zero. This is the index step to spell out. The factor $u^{\mu}u^{\nu}$ is symmetric under exchanging $\mu$ and $\nu$, because it is a product of two copies of the same object. Contracting a symmetric array with any array picks out only the symmetric part of the second, since the antisymmetric part contributes equal and opposite amounts under the exchange. Concretely, relabel the dummies $\mu\leftrightarrow\nu$ in (3.5.71), add the result to the original and divide by two:

And the bracket in that last line vanishes, because it is the left-hand side of Killing's equation (3.5.66). So the rate of change of $Q$ is zero at every point of the curve, which is what we set out to prove:

$\blacksquare$ Four lines, and the only inputs were the geodesic equation and Killing's equation. Note where each was spent: the geodesic equation killed the second term of (3.5.70), and Killing's equation killed the first. Neither could have done the other's job.

9.2 · The same result as Noether's theorem

Now the identification the plan of this book has been pointing at. Chapter 1.4 proved that every continuous symmetry of an action yields a conserved quantity, and gave it as (1.4.13): $Q=\sum_{i}p_{i}K_{i}-F$, with $K_{i}$ the infinitesimal change in the coordinates and $p_{i}$ the canonical momentum. We now run that theorem on a particle in a curved spacetime and show that it returns (3.5.73).

The action. Chapter 3.3 §3 gave $S=-mc^{2}\int\dd\tau$ as (3.3.15), and Chapter 3.3 §8.2 showed that extremising it produces the geodesic equation, working with the integrand $F=\sqrt{g_{\alpha\beta}\dot x^{\alpha}\dot x^{\beta}}$. For the present purpose the square root is a nuisance, and it can be removed. Compare the Euler–Lagrange equations of $F$ with those of $\half F^{2}$: by the product and chain rules,

Along a timelike geodesic parametrised by proper time, $F^{2}=g_{\mu\nu}u^{\mu}u^{\nu}=c^{2}$ is constant by (3.3.58), so $\dd F/\dd\tau=0$ and the last term drops. The two brackets then vanish together. So on proper-time-parametrised curves the two Lagrangians have the same extremals, and we may use

The canonical momentum. Differentiate (3.5.75) with respect to $\dot x^{\lambda}$. The velocity appears twice, so the product rule gives two terms, and the symmetry of $g$ makes them equal:

So the momentum conjugate to a coordinate is the four-velocity with its index lowered, times the mass. That is Chapter 2.5's four-momentum, now with a position-dependent metric doing the lowering.

The symmetry. Take the transformation $x^{\mu}\to x^{\mu}+\epsilon\,\xi^{\mu}(x)$, so that Chapter 1.4's $K$ is $\xi^{\mu}$. Compute the change in $L$. Two things vary: the metric, because it is evaluated at a shifted point, and the velocities, because the shift depends on position:

Handle the second term. By the chain rule $\dd\xi^{\nu}/\dd\tau=\dot x^{\lambda}\partial_{\lambda}\xi^{\nu}$, so that term is $2g_{\mu\nu}\dot x^{\mu}\dot x^{\lambda}\partial_{\lambda}\xi^{\nu}$. Relabel its dummy indices so that the two velocities carry the labels $\mu$ and $\nu$: swapping the names $\nu$ and $\lambda$ turns it into $2g_{\mu\lambda}\dot x^{\mu}\dot x^{\nu}\partial_{\nu}\xi^{\lambda}$. Using the symmetry of $g\dot x\dot x$ in $\mu\nu$, that equals $\big(g_{\lambda\nu}\partial_{\mu}\xi^{\lambda}+g_{\mu\lambda}\partial_{\nu}\xi^{\lambda}\big)\dot x^{\mu}\dot x^{\nu}$. Substituting back,

Look at the underbraced bracket. It is (3.5.65) exactly. The change in the Lagrangian under a coordinate shift is the Lie derivative of the metric, contracted with two velocities, which is the cleanest possible statement of what a Killing vector is for. If $\xi$ is Killing then $\delta L=0$, so the symmetry hypothesis of Chapter 1.4 §1.2 is satisfied with the allowed total derivative $F$ equal to zero.

The charge. Feed $K^{\mu}=\xi^{\mu}$, $p_{\mu}=mu_{\mu}$ and $F=0$ into (1.4.13):

which is $m$ times (3.5.73). The two derivations agree, and they agree for a reason rather than by luck: §9.1's Line 4 discarded the antisymmetric part of $\nabla\xi$, and §9.2's (3.5.78) discarded it too, both times because the velocity appears twice.

10 · Electromagnetism in three symbols

A language is worth having if it says something short that was long. Here is the test, and here is the destination. All of classical electromagnetism, including charge conservation and gauge freedom and the version valid in curved spacetime, written as $F=\dd A$, $\dd F=0$, $\dd\!\star\!F=\mu_{0}\star\!J$. And, more useful than the compression itself, a clean statement of which half of the subject needs a metric and which half does not.

10.1 · The one new operation

Section 1.3's table is symmetric: in four dimensions the counts run $1,4,6,4,1$, so $p$-forms and $(4-p)$-forms have the same number of components. An operation exchanging them therefore has a chance of existing, and the volume tensor (3.5.46) supplies it. Define the Hodge star of a $p$-form by contracting it into the last $p$ slots of $\varepsilon$:

Note what just happened. The indices of $\omega$ were raised to be contracted, and raising needs the inverse metric. So $\star$ is the first operation in this chapter that requires a metric. Everything before it needed none: forms, the wedge, $\dd$, $\dd^{2}=0$, Stokes. Keep that distinction in view for §10.3, where it separates the two halves of Maxwell's equations.

The two cases needed are $p=2$ and $p=1$:

10.2 · The three statements

First. Section 3.3 already established $F=\dd A$, which reproduced Chapter 2.6's definition (2.6.9) character for character. Six components from four.

Second. $\dd F=\dd\dd A=0$ by (3.5.22), and expanding it gave (2.6.25), which is $\nabla\cdot\vv B=0$ and Faraday's law. Four components.

Third. The inhomogeneous half. Chapter 2.6's (2.6.19) is $\partial_{\mu}F^{\mu\nu}=\mu_{0}j^{\nu}$. The claim is that this is

Start with the shape of the claim. Both sides are $3$-forms, with $\binom{4}{3}=4$ components, which matches the four equations we are trying to reproduce. Verifying it is then a matter of expanding both sides in flat coordinates and comparing. That computation is short but sign-infested, so it sits in the grind box. The result is that (3.5.82) holds exactly with the conventions fixed in (3.5.46) and (3.5.80), namely contracted indices last on $\varepsilon$, and $\varepsilon_{0123}=+\sqrt{-g}$.

Grind box — checking $\dd\!\star\!F=\mu_{0}\star\!J$, and why the sign is a convention worth fixing loudly

Work in flat coordinates where $\sqrt{-g}=1$ and $\varepsilon_{\mu\nu\rho\sigma}$ is the plain permutation symbol with $\varepsilon_{0123}=+1$. Take one component of each side, say $(\lambda\rho\sigma)=(0,1,2)$.

Left side. By (3.5.14) at $p=2$, the component is the three-term cyclic sum $\partial_{0}(\star F)_{12}+\partial_{1}(\star F)_{20}+\partial_{2}(\star F)_{01}$. Each entry of $\star F$ is, by (3.5.81), one term of $F$ with both indices up and the complementary index pair: $(\star F)_{12}=\half\varepsilon_{12\mu\nu}F^{\mu\nu}=\varepsilon_{1203}F^{03}=F^{03}$, using $\varepsilon_{1203}=+1$ and the fact that the two surviving orderings $(0,3)$ and $(3,0)$ contribute equally and cancel the $\half$. Likewise $(\star F)_{20}=\varepsilon_{2013}F^{13}=F^{13}$ and $(\star F)_{01}=\varepsilon_{0123}F^{23}=F^{23}$. So the left side is $\partial_{0}F^{03}+\partial_{1}F^{13}+\partial_{2}F^{23}$, which is $\partial_{\mu}F^{\mu3}$ with the $\mu=3$ term absent. That missing term is $\partial_{3}F^{33}=0$ by antisymmetry, so the left side is exactly $\partial_{\mu}F^{\mu3}$.

Right side. $\mu_{0}(\star J)_{012}=\mu_{0}\varepsilon_{012\mu}j^{\mu}=\mu_{0}\varepsilon_{0123}j^{3}=\mu_{0}j^{3}$.

Equating gives $\partial_{\mu}F^{\mu3}=\mu_{0}j^{3}$, which is the $\nu=3$ component of (2.6.19). The other three components come out the same way. The whole system was also verified symbolically, for a general $A_{\mu}(x)$, with all four triples checked and all four agreeing in sign.

Why the loud convention. Had we instead defined $\star$ on a one-form by putting the contracted index first, writing $(\star J)_{\lambda\rho\sigma}=\varepsilon_{\mu\lambda\rho\sigma}j^{\mu}$, every component would flip sign, since moving an index past three others costs $(-1)^{3}$, and (3.5.82) would carry a minus. Books differ here exactly as they differ over the sign of the Riemann tensor. The rule adopted above puts contracted indices last, at every degree. That rule is self-consistent and is what makes the plus sign correct. This is worth knowing when you reach for a formula from another book: check its convention first.

10.3 · What the compression reveals

(i) Charge conservation is free. Apply $\dd$ to (3.5.82). The left side is $\dd\dd(\star F)=0$ by (3.5.22). Hence

That $3$-form has a single independent component, and expanding it the same way the grind box did gives $\partial_{\mu}j^{\mu}=0$. That is the continuity equation of Chapter 0.7 §6 and Chapter 2.6 §1. Charge conservation is not an extra postulate. It is $\dd^{2}=0$ applied to the field equation. Chapter 0.7's Problem 4 found that Maxwell's displacement current was forced by charge conservation. This is that argument run backwards and compressed into one symbol.

(ii) Gauge freedom is free too. Replacing $A\to A+\dd\chi$ for any function $\chi$ leaves $F=\dd A$ unchanged, again by (3.5.22). That is Chapter 2.6 §4.1's gauge transformation, and the reason it exists is the same one line.

Section 4 has something to add here. The potentials giving a particular $F$ differ by a closed one-form, and on a region without holes a closed one-form is exact and hence of the form $\dd\chi$. So §4's topology is the fine print on the phrase "the potential is determined up to a gauge transformation".

(iii) The asymmetry between the two halves, named. This is the observation the whole section was for. Look at what each statement needs:

Statement	Needs a metric?	Status
$F=\dd A$	no	definition
$\dd F=0$	no	identity — cannot fail
$\dd\!\star\!F=\mu_{0}\star\!J$	yes, inside $\star$	field equation — physics

Chapter 2.6 §3.4 noticed that two of Maxwell's equations were bookkeeping and two were physics, and Chapter 2.6's forward-looking warning box predicted that in this language "the visible asymmetry between $\dd F$ and $\dd\!\star\!F$" would be the reason. It is: the star is where the geometry lives, and only the sourced half of Maxwell touches it.

(iv) Curved spacetime costs nothing. Because $\dd$ needs no connection (§2.2) and $\varepsilon$ already carries $\sqrt{-g}$ (§6.2), the three statements above are the equations of electromagnetism on any curved manifold, unchanged. In components the sourced half becomes, using §6.4's trick,

The middle equality is worth its two lines. Expanding $\nabla_{\mu}F^{\mu\nu}$ gives $\partial_{\mu}F^{\mu\nu}+\Gamma^{\mu}{}_{\mu\lambda}F^{\lambda\nu}+\Gamma^{\nu}{}_{\mu\lambda}F^{\mu\lambda}$. The last term vanishes: $\Gamma^{\nu}{}_{\mu\lambda}$ is symmetric in $\mu\lambda$ and $F^{\mu\lambda}$ is antisymmetric, so the contraction is zero. The first two are exactly (3.5.51) applied to each fixed value of $\nu$. So on a curved manifold Maxwell's equations are Maxwell's equations with $\partial\to\nabla$, and the only visible change is a factor of $\sqrt{-g}$. That is the whole of "electromagnetism in a gravitational field", in one line, and we have it before the gravitational field equations have even been written down.

11 · Worked examples

12 · Your turn