Part E · curved space · branch

Differential forms and Cartan: Stokes' theorem was always one theorem

One algebra with a sign in it, one derivative, one theorem. Maxwell's four equations become two, the four integral theorems become one, and unit 04's hole acquires a number.

needs 31, 24, 33~3 h slides + 6 h problemsproblems: ./mr problems 34

This unit is a branch, not a spine unit: unit 35 derives the field equations without it. It is here because three separate things this course has already done turn out to be one thing, and because the tool that shows it — the exterior derivative — is the fastest way to compute curvature, which units 36 and 39 will need at length.

The three things. Unit 04 proved the gradient theorem and left open why the vortex, with vanishing curl, had circulation $2\pi$. Unit 05 proved the divergence theorem and unit 23 the curl theorem, each with its own proof and its own orientation convention. Unit 24 wrote four Maxwell equations and noted that two of them are identities once the potentials are introduced. All three are statements about one operator.

How to read this deck

The running example is one field and one hole: Maxwell's equations, rewritten and expanded back into components so that nothing is taken on trust; and unit 04's vortex on the punctured plane, followed until it is a cohomology class with a number attached. Sections 1 and 2 build the algebra and the derivative, section 3 integrates, section 4 asks what closed forms fail to be exact, section 5 brings the metric in, and section 6 recomputes unit 33's curvature in a fraction of the arithmetic and proves the Gauss–Bonnet formula unit 32 left owing.

Proofs are on the slides where they fit and in these notes under Full proof where they do not; every result imported from elsewhere is marked cited and says where its proof is.

What this unit does not assume. Its needs are units 31, 24 and 33, and through them unit 26's Minkowski metric. It does not need unit 28: the field two-form is built here from unit 24's potentials, and the two field invariants arrive without a single boost. If you have taken unit 28, the notes say where the two accounts meet.

Conventions, fixed once: Greek indices $\mu,\nu$ run $0,1,2,3$ with $x^0 = ct$ and $\eta_{\mu\nu} = \operatorname{diag}(-1,1,1,1)$ as in unit 26; Latin $i,j,k$ run over space; Latin $a,b,c$ label an orthonormal frame; Einstein summation throughout; SI units, so $\varepsilon_0$ and $\mu_0$ appear and $\mu_0\varepsilon_0 c^2 = 1$.

34 · overview

One algebra, one derivative, one theorem

1 · the algebra

A k-form is an alternating multilinear map on tangent vectors. The wedge product makes them an algebra; antisymmetry is what makes a determinant.

2 · the derivative

One operator $\mathrm d$, fixed by three axioms, with $\mathrm d^2 = 0$. Grad, curl and divergence are $\mathrm d$ in degrees $0, 1, 2$.

3 · the theorem

$\int_M \mathrm d\omega = \int_{\partial M}\omega$, proved once on a square by Fubini. Unit 04's gradient theorem, Green, unit 23's curl theorem and unit 05's divergence theorem are its cases $n = 1, 2, 2, 3$.

4 · the holes

$\mathrm d\omega = 0$ does not give $\omega = \mathrm d\alpha$. The failure is de Rham cohomology, and it is a number: unit 04's $2\pi$, the circulation of a closed form that no $\mathrm d\alpha$ could have.

5 · the metric

The Hodge star is the only place a metric is needed. With it, $F$ obeys $\mathrm dF = 0$ and $\mathrm d\star F = \mu_0\star J$, with $J$ the one-form built from charge and current density.

6 · the payoff

Cartan's structure equations: unit 33's curvature, from a coframe, in a fraction of the arithmetic.

Dotted underlines like this are glossary terms — click one, or press G for the index and N for the notes, where the long proofs are. A solid underline like the vortex $\theta$ shows a piece of the running example on hover.

The order is forced by what each step needs. The algebra needs only a vector space. The exterior derivative needs a smooth manifold and nothing more — no metric, no connection, which is exactly why $\mathrm dF = 0$ will survive unit 32's move to curved spacetime untouched. Integration needs an orientation. Cohomology needs the previous three and no more. Only in section 5 does a metric arrive, and it arrives for one reason: to define $\star$, which is what turns the second pair of Maxwell equations into a statement about $F$.

Keep score of which structure each statement uses. It is the single most useful habit this unit teaches, and it pays off immediately in unit 32's remark that the comma-goes-to-semicolon rule is unnecessary for $\mathrm dF = 0$: the exterior derivative never needed a connection, so there is nothing to upgrade.

The symbols in boxes 3–5 are defined where they are used: $\partial M$ is the boundary of $M$, $\star$ is the Hodge star of section 5, and $J$ is the current one-form of the slide $\star F$, computed — unit 24's $\rho$ and $\vec J$ packed into one object, as $F$ packs $\vec E$ and $\vec B$. The next slide sketches how each half of the running example is reached.

34 · overview · the running example

One field and one hole

the field · four equations into two

1. Pack unit 24's potentials into one one-form $A$ and set $F = \mathrm dA$: $\vec E$ and $\vec B$ become one two-form.

2. $\mathrm dF = \mathrm d\mathrm dA = 0$ by Clairaut. Expanded, it is $\nabla\cdot\vec B = 0$ and Faraday.

3. The metric enters through $\star$ alone: $\mathrm d\star F = \mu_0\star J$ expands to Gauss and Ampère–Maxwell.

the hole · a number for unit 04's vortex

1. $\theta$ is closed: $\mathrm d\theta = 0$ at every point of the punctured plane $U$.

2. $\oint\theta = 2\pi$ round the hole; if $\theta$ were $\mathrm df$, Stokes would make it $0$.

3. Every closed one-form on $U$ is $c\,\theta + \mathrm df$, with $c$ its circulation over $2\pi$: one number per hole.

The vortex $\theta$ on $U$. No curl anywhere, yet the circulation round the hole is $2\pi$; a loop that misses the hole gets $0$. Arrow length falls like $1/r$, capped near the hole.

Where each step is done. The field: step 1 is Worked example 1 — the field as a two-form, step 2 is Worked example 2 — $\mathrm dF = 0$, expanded, and step 3 is the pair of slides $\star F$, computed and Worked example 4, after section 5 has defined the star. The hole: step 1 is a two-line computation done in example 3, step 2 is its step 1 and needs only Stokes' theorem from section 3, and step 3 is the rest of example 3, which unwraps the plane by polar coordinates.

What to look for in the figure. Away from the origin the arrows turn but do not swirl about any point you could stand on: a small loop anywhere in $U$ that does not enclose the origin collects nothing, which is $\mathrm d\theta = 0$ read through Stokes. Only loops that go round the missing point collect anything, and they all collect the same $2\pi$. The two halves of the running example meet in section 4: a magnetic monopole would make $F$ the same kind of object one dimension up.

section 1

The exterior algebra

Antisymmetry, and why the object that measures oriented volume is the only one worth integrating.

34 · algebra

Why antisymmetric, and not something else

Unit 16 defined a $(0,k)$-tensor as a multilinear map $V^k\to\mathbb R$ and showed that symmetry and antisymmetry are the two pieces that survive a change of basis. Here is the physical reason to keep the antisymmetric piece.

A thing to be integrated over a $k$-dimensional patch must eat $k$ edge vectors and return a number that scales like the oriented volume they span. Two demands follow at once: it is multilinear — doubling an edge doubles the patch — and it is zero when two edges coincide — a degenerate patch has no volume. And "zero on repeats" is antisymmetry: expand $T(u+v,u+v) = 0$ to get $T(u,v) = -T(v,u)$.

One patch, three readings. Listing the edges in the other order reverses the orientation and the sign; repeating an edge flattens the patch to nothing.

the precedent

$\det$ is the unique alternating $n$-linear function of $n$ column vectors with $\det\mathbb 1 = 1$. Unit 05's change-of-variables Jacobian was already this object; so was unit 08's cross product, and unit 16's angular velocity, which is an antisymmetric matrix wearing a vector's clothes.

The expansion, in full. Suppose $T$ is bilinear and $T(w,w) = 0$ for every $w$. Then $0 = T(u+v,u+v) = T(u,u) + T(u,v) + T(v,u) + T(v,v) = T(u,v) + T(v,u)$. Conversely if $T(u,v) = -T(v,u)$ then $T(w,w) = -T(w,w)$, so $2T(w,w) = 0$ and $T(w,w) = 0$ — over $\mathbb R$, where $2 \ne 0$. Over a field of characteristic 2 the two conditions part company and "alternating" means the stronger one; this course never leaves $\mathbb R$.

Reading the figure. In the plane, take $T(u,v) = u^1v^2 - u^2v^1$, the determinant of the two edge vectors' components. It is positive exactly when turning from the first edge to the second is anticlockwise. Its absolute value is the area. The sign is not a nuisance to be removed by an absolute value, as unit 05's Jacobian did: it is the orientation, and section 3 needs it.

Where the symmetric piece went. It is not lost, it is elsewhere. The metric of unit 32 is a symmetric $(0,2)$-tensor, the inertia tensor of unit 16 and the Maxwell stress tensor of unit 24 are symmetric, and none of them is integrated over a surface as such — what gets integrated is a contraction of them with a normal, which is a one-form, which is antisymmetric trivially. The division of labour is clean: symmetric tensors measure states, antisymmetric ones measure fluxes through patches.

History. Grassmann published the Ausdehnungslehre in 1844 with essentially this algebra and almost no readers; the 1862 rewrite fared little better. Poincaré (1899) and then Cartan (1899, 1901) made the exterior derivative the centre of the subject, and de Rham's 1931 thesis proved the theorem connecting it to topology. Physics adopted it late: Misner, Thorne and Wheeler (1973) is where most physicists first met it.

34 · algebra

$\Lambda^k V^*$, defined

definition

Let $V$ be a real vector space of dimension $n$. A k-covector on $V$ is a map $\alpha : V^k \to \mathbb R$ that is linear in each argument and satisfies $\alpha(v_{\sigma(1)},\dots,v_{\sigma(k)}) = \operatorname{sgn}(\sigma)\,\alpha(v_1,\dots,v_k)$ for every permutation $\sigma$. They form a vector space $\Lambda^kV^*$, the k-th exterior power. By convention $\Lambda^0V^* = \mathbb R$ and $\Lambda^1V^* = V^*$.

basis and dimension

If $e^1,\dots,e^n$ is a basis of $V^*$ then $\{\,e^{i_1}\wedge\cdots\wedge e^{i_k} : i_1 < \cdots < i_k\,\}$ is a basis of $\Lambda^kV^*$, so $$\dim\Lambda^kV^* = \binom nk ,$$ zero for $k > n$.

check it

$n = 4$: dimensions $1,4,6,4,1$, total $16 = 2^4$. The six is the number of independent components of an antisymmetric $4\times4$ array — three of $\vec E$ and three of $\vec B$, which example 1 assembles into one object.

Full proof of the basis claim (the wedge is defined on the next slide; take its properties as given for a moment). Spanning. A $k$-covector is determined by its values on basis $k$-tuples $(e_{i_1},\dots,e_{i_k})$, by multilinearity. Antisymmetry makes the value vanish when two indices repeat and fixes it up to a sign under reordering, so the values on increasing tuples determine $\alpha$. If $\alpha_{i_1\cdots i_k} := \alpha(e_{i_1},\dots,e_{i_k})$ then $\alpha = \sum_{i_1<\cdots<i_k}\alpha_{i_1\cdots i_k}\, e^{i_1}\wedge\cdots\wedge e^{i_k}$, because both sides agree on every increasing tuple: $(e^{i_1}\wedge\cdots\wedge e^{i_k})(e_{j_1},\dots,e_{j_k}) = \delta^{i_1\cdots i_k}_{j_1\cdots j_k}$, which is $1$ if the increasing tuples match and $0$ otherwise. Independence. Pair a vanishing combination with the tuple $(e_{j_1},\dots,e_{j_k})$ and read off that the coefficient of $e^{j_1}\wedge\cdots\wedge e^{j_k}$ is zero.

Why $k > n$ gives zero. Any $k$ vectors with $k > n$ are linearly dependent, so one is a combination of the others; expand by multilinearity and every term repeats an argument. This is the statement that an $n$-dimensional space has no oriented $(n+1)$-volume, and on spacetime it is why there are no five-forms and why $\mathrm d$ of a four-form is automatically zero.

Dimension check. $\sum_k\binom nk = 2^n$: the exterior algebra $\Lambda V^* = \bigoplus_k\Lambda^kV^*$ has dimension $2^n$, which for $n = 4$ is 16 — the same as the $4\times4$ array of a general $(0,2)$-tensor, a coincidence of small numbers and nothing more.

34 · algebra

The wedge product

definition · determinant convention

For $\alpha\in\Lambda^kV^*$ and $\beta\in\Lambda^lV^*$, $$(\alpha\wedge\beta)(v_1,\dots,v_{k+l}) = \sum_{\sigma\in\mathrm{Sh}(k,l)}\operatorname{sgn}(\sigma)\, \alpha(v_{\sigma(1)},\dots,v_{\sigma(k)})\,\beta(v_{\sigma(k+1)},\dots,v_{\sigma(k+l)}),$$ the sum running over $(k,l)$-shuffles: permutations increasing on $1..k$ and on $k{+}1..k{+}l$.

bilinear, associative

$(\alpha\wedge\beta)\wedge\gamma = \alpha\wedge(\beta\wedge\gamma)$, so brackets may be dropped.

graded commutative

$\alpha\wedge\beta = (-1)^{kl}\beta\wedge\alpha$. One-forms anticommute; a two-form commutes with everything of even degree.

the normalisation

$(e^1\wedge\cdots\wedge e^n)(v_1,\dots,v_n) = \det[\,v^i_j\,]$, with no factorial. That is what the shuffle sum buys.

Full proof of graded commutativity. It suffices to check it on basis elements, by bilinearity. Moving $e^{j_1}\wedge\cdots\wedge e^{j_l}$ past $e^{i_1}\wedge\cdots\wedge e^{i_k}$ means moving each of the $l$ factors past each of the $k$ factors, one transposition at a time, which is $kl$ transpositions and hence a sign $(-1)^{kl}$. For $k = l = 1$ this is $e^i\wedge e^j = -e^j\wedge e^i$, and setting $i = j$ gives $e^i\wedge e^i = 0$.

An immediate corollary. If $k$ is odd, $\alpha\wedge\alpha = -\alpha\wedge\alpha = 0$. If $k$ is even this is false: on $\mathbb R^4$, $(e^1\wedge e^2 + e^3\wedge e^4)^{\wedge 2} = 2\,e^1\wedge e^2\wedge e^3\wedge e^4 \ne 0$. Section 5's invariant $F\wedge F$ is exactly a case where this matters, and section 6's symplectic form is another: $\omega^{\wedge n}$ being nonzero is the definition of nondegeneracy.

Associativity is the one tedious check. With the shuffle convention, both $(\alpha\wedge\beta)\wedge\gamma$ and $\alpha\wedge(\beta\wedge\gamma)$ equal the sum over $(k,l,m)$-shuffles of $\operatorname{sgn}(\sigma)\alpha(\cdot)\beta(\cdot)\gamma(\cdot)$; the combinatorial identity needed is that every $(k+l,m)$-shuffle composed with a $(k,l)$-shuffle of the first block arises exactly once as a $(k,l,m)$-shuffle, and signs multiply. Cited: Lee, Introduction to Smooth Manifolds, 2nd ed., ch. 14, the proposition listing the properties of the wedge product.

34 · algebra

The wedge product, drawn

On $\mathbb R^3$ take the dual basis $\mathrm dx, \mathrm dy, \mathrm dz$. For $k = l = 1$ the shuffle sum has two terms:

$$(\mathrm dx\wedge\mathrm dy)(u,v) = u^xv^y - u^yv^x ,$$

the signed area of the parallelogram's shadow on the $xy$-plane. The other two, $\mathrm dy\wedge\mathrm dz$ and $\mathrm dz\wedge\mathrm dx$, give the shadows on the other two planes, and the three numbers are the components of unit 08's $u\times v$.

two conventions exist

Some texts define $\alpha\wedge\beta = \frac{(k+l)!}{k!\,l!}\operatorname{Alt}(\alpha\otimes\beta)$ — the same thing — and others drop the factorial, which puts a $1/k!$ into every determinant. This course uses the determinant convention, always. A formula copied from elsewhere may differ by $k!\,l!/(k+l)!$.

A two-form measures a patch by its shadows. Here $u = (1.5, 0.2, 1.0)$ and $v = (0.3, 1.5, 0.6)$, so the shadow's signed area is $1.5\cdot1.5 - 0.2\cdot0.3 = 2.19$.

The normalisation, checked. For $k = l = 1$ the shuffles of $(1,1)$ are the identity and the transposition, so $(e^1\wedge e^2)(v_1,v_2) = v^1_1v^2_2 - v^1_2v^2_1 = \det$ of the $2\times 2$ matrix of components. The general case is the Leibniz expansion of a determinant, term by term.

Why the shadow. $\mathrm dx\wedge\mathrm dy$ only sees the $x$ and $y$ components of its arguments, so it cannot tell $u$ from its vertical projection $(u^x, u^y, 0)$: the patch and its shadow give the same number. On the $xy$-plane itself the number is the signed area, by the previous slide's determinant. This is the picture behind every flux integral: a two-form $B_x\,\mathrm dy\wedge\mathrm dz + B_y\,\mathrm dz\wedge\mathrm dx + B_z\,\mathrm dx\wedge\mathrm dy$ weighs the three shadows of a small patch by the three components of $\vec B$, which is $\vec B\cdot(u\times v)$, the flux through it.

Checking against the cross product. $u\times v = (u^yv^z - u^zv^y,\ u^zv^x - u^xv^z,\ u^xv^y - u^yv^x)$, and the three entries are $(\mathrm dy\wedge\mathrm dz)(u,v)$, $(\mathrm dz\wedge\mathrm dx)(u,v)$, $(\mathrm dx\wedge\mathrm dy)(u,v)$ in that order. For the vectors in the figure, $u\times v = (-1.38, -0.60, 2.19)$.

34 · algebra

Forms on a manifold

definition

A differential k-form on a smooth manifold $M$ is a smooth section of $\Lambda^kT^*M$: an assignment $p\mapsto\omega_p\in\Lambda^kT_p^*M$ with smooth components in every chart. Write $\Omega^k(M)$ for the space of them. In a chart, $$\omega = \tfrac1{k!}\,\omega_{i_1\cdots i_k}\,\mathrm dx^{i_1}\wedge\cdots\wedge\mathrm dx^{i_k} = \!\!\sum_{i_1<\cdots<i_k}\!\!\omega_{i_1\cdots i_k}\,\mathrm dx^{i_1}\wedge\cdots\wedge\mathrm dx^{i_k}.$$

Unit 31 built $T^*_pM$, the dual basis $\mathrm dx^i$, and what "smooth section" means. A $k$-form is a $(0,k)$-tensor field that happens to be antisymmetric — nothing new is being assumed, only a subspace being named.

A one-form drawn as a stack of sheets: $\mathrm df$ for $f = 2x + y^2/4$, sheets at $f = 1, 2, \dots$. Its value on a vector is how many sheets the vector pierces.

degrees you have already met

$\Omega^0$: functions. $\Omega^1$: $\mathrm df$ from unit 31, and $\sum p_i\,\mathrm dq_i$ from unit 14. $\Omega^2$: unit 14's $\omega = \sum\mathrm dp_i\wedge\mathrm dq_i$ — written with a wedge, thirty units before the wedge was defined. $\Omega^n$: the thing you integrate.

The two displayed sums are the same form. In the first, the indices run independently and the $1/k!$ compensates for counting each increasing tuple $k!$ times; in the second they increase and there is no factor. Physics papers use the first, mathematics books the second. Mixing them silently is the commonest source of stray factorials in this subject.

What is not new. Unit 31 defined tensor fields as sections of bundles of multilinear maps and proved that "transforms like a tensor" is a theorem about change of basis. Everything here inherits that. In particular a $k$-form's components transform with $k$ copies of the inverse Jacobian, and the antisymmetry is basis-independent because permuting arguments commutes with changing basis.

Reading the figure. $\mathrm df(v) = v^x\partial_xf + v^y\partial_yf$ is the change of $f$ along $v$ to first order, and the sheets are one unit of $f$ apart, so the value counts sheets crossed. At $p = (1.0, 0.6)$ with $v = (1.2, 0.5)$: $\mathrm df(v) = 2(1.2) + \tfrac12(0.6)(0.5) = 2.55$. A vector lying along a sheet, like $w$, crosses none: $\mathrm df(w) = 0$. Where the sheets crowd together the form is large. This is Misner, Thorne and Wheeler's picture of a one-form, and it needs no metric: there is no arrow "perpendicular to the sheets" anywhere in it — that arrow, the gradient, is $\mathrm df$ raised by a metric.

Unit 14, in hindsight. That unit wrote the symplectic form as $\omega = \sum_i\mathrm dp_i\wedge\mathrm dq_i$ and defined it by the bilinear formula $\omega(u,v) = \sum_i(u_{p_i}v_{q_i} - u_{q_i}v_{p_i})$, which is exactly the shuffle definition for $k = l = 1$ summed over $i$. Nothing said there needs revising; section 6 only replaces a two-page computation by one line.

34 · algebra

Worked example 1 — the field as a two-form

example 1 · building $F$

Unit 24's potentials are $\varphi$ and $\vec A$, with $\vec E = -\nabla\varphi - \partial_t\vec A$ and $\vec B = \nabla\times\vec A$. Assemble the potential one-form $A = A_i\,\mathrm dx^i - \varphi\,\mathrm dt$ and set $F = \mathrm dA$. Then

$$\mathrm dA = \underbrace{\partial_jA_i\,\mathrm dx^j\wedge\mathrm dx^i}_{\text{space--space}} \;+\;\underbrace{\partial_tA_i\,\mathrm dt\wedge\mathrm dx^i - \partial_i\varphi\,\mathrm dx^i\wedge\mathrm dt}_{\text{space--time}} .$$

Space–space. $\partial_jA_i\,\mathrm dx^j\wedge\mathrm dx^i = \tfrac12(\partial_jA_i-\partial_iA_j)\mathrm dx^j\wedge\mathrm dx^i$, whose $\mathrm dx\wedge\mathrm dy$ coefficient is $\partial_xA_y-\partial_yA_x = B_z$; likewise for the other two.

Space–time. $\partial_tA_i\,\mathrm dt\wedge\mathrm dx^i - \partial_i\varphi\,\mathrm dx^i\wedge\mathrm dt = -(\partial_tA_i+\partial_i\varphi)\,\mathrm dx^i\wedge\mathrm dt = E_i\,\mathrm dx^i\wedge\mathrm dt .$

the field two-form $$F = B_x\,\mathrm dy\wedge\mathrm dz + B_y\,\mathrm dz\wedge\mathrm dx + B_z\,\mathrm dx\wedge\mathrm dy + (E_x\,\mathrm dx+E_y\,\mathrm dy+E_z\,\mathrm dz)\wedge\mathrm dt$$

Dimensions. $[B\,\mathrm dy\wedge\mathrm dz] = \mathrm{T\,m^2}$ and $[E\,\mathrm dx\wedge\mathrm dt] = (\mathrm{V/m})\,\mathrm{m\,s} = \mathrm{V\,s} = \mathrm{Wb} = \mathrm{T\,m^2}$: the two halves agree, as they must to be added. Special case. Electrostatics, $\vec B = 0$ and $\vec E$ static: $F = E_i\,\mathrm dx^i\wedge\mathrm dt$, six components down to three.

The step that looks like a trick. $\partial_jA_i\,\mathrm dx^j\wedge\mathrm dx^i$ — the indices are summed independently, and $\mathrm dx^j\wedge\mathrm dx^i$ is antisymmetric in $(j,i)$, so only the antisymmetric part of $\partial_jA_i$ survives. Replacing a coefficient by its antisymmetric part when it is contracted with something antisymmetric is the single manipulation this whole unit runs on; it is unit 16's symmetric/antisymmetric decomposition, used as a computational rule.

Checking $B_z$. The $\mathrm dx\wedge\mathrm dy$ terms of $\partial_jA_i\,\mathrm dx^j\wedge\mathrm dx^i$ are $\partial_xA_y\,\mathrm dx\wedge\mathrm dy$ and $\partial_yA_x\,\mathrm dy\wedge\mathrm dx = -\partial_yA_x\,\mathrm dx\wedge\mathrm dy$, so the coefficient is $\partial_xA_y - \partial_yA_x = (\nabla\times\vec A)_z = B_z$. The cyclic pattern $\mathrm dy\wedge\mathrm dz,\ \mathrm dz\wedge\mathrm dx,\ \mathrm dx\wedge\mathrm dy$ carries $B_x, B_y, B_z$ — the same cyclic pattern as the flux integrals of unit 05.

In index form, for readers who will meet it elsewhere: $F = \tfrac12F_{\mu\nu}\mathrm dx^\mu\wedge\mathrm dx^\nu$ with $F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$, $A_0 = -\varphi/c$, $A_i$ the vector potential, and $x^0 = ct$. This gives $F_{0i} = -E_i/c$ and $F_{ij} = \epsilon_{ijk}B_k$. Everything on this slide can be read off that, and the slide's version is what a computation actually uses.

A remark worth keeping. $F$ is defined here without any metric, without any orientation, and without any choice of frame. Nothing so far distinguishes the electric from the magnetic part — that split is a choice of time coordinate, and section 5 will show it is exactly the choice that $\star$ depends on. Unit 28, which this unit does not need, reaches the same object as the tensor $F_{\mu\nu}$ from the Lorentz transformation of fields; here it is assembled from potentials and a wedge, and its six components are the ones listed above.

section 2

One derivative

Three axioms, and grad, curl and divergence stop being three things.

34 · the derivative

The exterior derivative, axiomatically

theorem · existence and uniqueness

There is exactly one family of $\mathbb R$-linear maps $\mathrm d:\Omega^k(M)\to\Omega^{k+1}(M)$, the exterior derivative, with

on functions, $\mathrm df$ is the differential of unit 31;
$\mathrm d(\alpha\wedge\beta) = \mathrm d\alpha\wedge\beta + (-1)^k\alpha\wedge\mathrm d\beta$ for $\alpha\in\Omega^k$;
$\mathrm d\circ\mathrm d = 0$.

In any chart it is $\;\mathrm d\big(f\,\mathrm dx^{i_1}\wedge\cdots\wedge\mathrm dx^{i_k}\big) = \mathrm df\wedge\mathrm dx^{i_1}\wedge\cdots\wedge\mathrm dx^{i_k}$.

uniqueness, on the slide

$\mathrm d$ is local (notes). In a chart write $\omega = \sum f_I\,\mathrm dx^I$. Axiom 2 gives $\mathrm d(f_I\,\mathrm dx^I) = \mathrm df_I\wedge\mathrm dx^I + f_I\,\mathrm d(\mathrm dx^I)$, and $\mathrm d(\mathrm dx^I) = 0$ by axioms 2 and 3 applied factor by factor. So the chart formula is forced.

$\mathrm d$ is dimensionless

Count each $\mathrm dx$ as a length: $\mathrm d = \partial_\mu\,\mathrm dx^\mu$ divides by a length and multiplies by one. So $[\mathrm d\omega] = [\omega]$ — and $[F] = [\mathrm dA] = [A] = \mathrm{Wb}$, which is the first check on every equation in this unit.

Full proof of existence. Define $\mathrm d$ in a chart by the displayed formula. It is $\mathbb R$-linear. Axiom 1 holds by inspection ($k = 0$). Axiom 3: for $\omega = f\,\mathrm dx^I$, $\mathrm d\mathrm d\omega = \mathrm d(\partial_jf\,\mathrm dx^j\wedge\mathrm dx^I) = \partial_i\partial_jf\,\mathrm dx^i\wedge\mathrm dx^j\wedge\mathrm dx^I$, and $\partial_i\partial_jf$ is symmetric in $(i,j)$ by Clairaut (unit 03) while $\mathrm dx^i\wedge\mathrm dx^j$ is antisymmetric, so the contraction vanishes term by term. Axiom 2: for $\alpha = f\,\mathrm dx^I$, $\beta = g\,\mathrm dx^J$, $\mathrm d(\alpha\wedge\beta) = \mathrm d(fg)\wedge\mathrm dx^I\wedge\mathrm dx^J = (g\,\mathrm df + f\,\mathrm dg)\wedge\mathrm dx^I\wedge\mathrm dx^J$, and moving $\mathrm dg$ past $\mathrm dx^I$ costs $(-1)^k$. Finally, two charts give the same answer because both satisfy the axioms on the overlap and uniqueness applies there.

Why $\mathrm d$ is local. If $\omega$ vanishes on an open $W$ and $p\in W$, pick a bump function $\chi$ supported in $W$ with $\chi(p) = 1$ (unit 31, partitions of unity). Then $\chi\omega = 0$, so $0 = \mathrm d(\chi\omega) = \mathrm d\chi\wedge\omega + \chi\,\mathrm d\omega$; evaluate at $p$, where $\omega_p = 0$ and $\chi(p) = 1$, to get $(\mathrm d\omega)_p = 0$. Hence $\mathrm d\omega$ on $W$ depends only on $\omega$ on $W$, which is what "compute it in a chart" requires.

Axiom 3 is not an extra assumption in disguise. It is equivalent, given the others, to $\mathrm d\mathrm df = 0$ for functions, which is Clairaut. That is the whole content: the single most consequential identity in this unit is the symmetry of second partial derivatives. Magnetic flux conservation, charge conservation and the existence of gauge freedom all come from it, as the rest of this deck shows.

34 · the derivative

Grad, curl and divergence are one operator

On $\mathbb R^3$ with the Euclidean metric and the standard orientation, write $\vec v^\flat = v_i\,\mathrm dx^i$ for the one-form of a vector field (unit 32's lowering) and $\star$ for the linear map that swaps each basis form with its complement: $\mathrm dx\leftrightarrow\mathrm dy\wedge\mathrm dz$, $\mathrm dy\leftrightarrow\mathrm dz\wedge\mathrm dx$, $\mathrm dz\leftrightarrow\mathrm dx\wedge\mathrm dy$, $1\leftrightarrow\mathrm dx\wedge\mathrm dy\wedge\mathrm dz$ (section 5 builds it from the metric). Then the whole of unit 04's and unit 05's differential calculus is:

degree	$\mathrm d$ does	classical name
$\Omega^0\to\Omega^1$	$f\mapsto\mathrm df$	gradient, $(\mathrm df)^\sharp = \nabla f$
$\Omega^1\to\Omega^2$	$\vec v^\flat\mapsto\mathrm d\vec v^\flat$	curl, $\ \nabla\times\vec v = (\star\,\mathrm d\,\vec v^\flat)^\sharp$
$\Omega^2\to\Omega^3$	$\star\vec v^\flat\mapsto\mathrm d{\star}\vec v^\flat$	divergence, $\ (\nabla\cdot\vec v)\,\mathrm{vol} = \mathrm d{\star}\vec v^\flat$

two identities, one line

$\nabla\times\nabla f = 0$ is $\mathrm d\mathrm df = 0$.
$\nabla\cdot(\nabla\times\vec v) = 0$ is $\mathrm d(\star\star\mathrm d\vec v^\flat) = \mathrm d\mathrm d\vec v^\flat = 0$, using $\star\star = 1$ in Euclidean $\mathbb R^3$.

the accident of three dimensions

Curl exists because $\binom32 = 3$: a two-form on $\mathbb R^3$ has as many components as a vector. In four dimensions $\binom42 = 6$ and there is no curl of a vector field — only $\mathrm d$, which is why $\vec E$ and $\vec B$ had to merge.

Verifying the curl entry. $\vec v^\flat = v_1\mathrm dx + v_2\mathrm dy + v_3\mathrm dz$, so $\mathrm d\vec v^\flat = (\partial_1v_2-\partial_2v_1)\mathrm dx\wedge\mathrm dy + (\partial_2v_3-\partial_3v_2)\mathrm dy\wedge\mathrm dz + (\partial_3v_1-\partial_1v_3)\mathrm dz\wedge\mathrm dx$. The Euclidean star sends $\mathrm dy\wedge\mathrm dz\mapsto\mathrm dx$ and cyclically, so $\star\mathrm d\vec v^\flat = (\nabla\times\vec v)_i\,\mathrm dx^i$, and raising gives the vector.

Verifying the divergence entry. $\star\vec v^\flat = v_1\,\mathrm dy\wedge\mathrm dz + v_2\,\mathrm dz\wedge\mathrm dx + v_3\,\mathrm dx\wedge\mathrm dy$ — the flux two-form of unit 05. Its exterior derivative is $(\partial_1v_1+\partial_2v_2+\partial_3v_3)\,\mathrm dx\wedge\mathrm dy\wedge\mathrm dz$. Unit 05 wrote the same object as $\vec v\cdot\mathrm d\vec S$ and called the integral a flux; the two-form is what $\mathrm d\vec S$ always meant.

Why the identities were ever separate. Unit 04 proved $\nabla\times\nabla f = 0$ by Clairaut and unit 23 proved $\nabla\cdot(\nabla\times\vec v) = 0$ by Clairaut, twice, with different index bookkeeping. They are the same theorem in degrees 0 and 1. A useful rule of thumb: any vector identity whose proof is "the mixed partials cancel" is $\mathrm d^2 = 0$ in disguise, and the problem sheet's d squared in disguise makes that precise for Maxwell.

The curl is a two-form, and that settles unit 04's anomaly. Unit 04 noticed that the curl does not transform like a vector under a reflection. Here is why: the object $\mathrm d\vec v^\flat$ is a two-form and transforms as one under every smooth map, reflections included. Turning it into a vector takes $\star$, and $\star$ depends on the orientation (section 5): reverse the orientation and $\star$ changes sign. So $\nabla\times\vec v$ picks up an extra minus under a reflection, which is all "pseudovector" ever meant. The cross product and angular momentum are the same story.

Laplacian. On functions, $\star\,\mathrm d\star\mathrm df = \nabla^2f$, unit 05's operator. The generalisation to forms is the Hodge Laplacian $\Delta = \mathrm d\delta + \delta\mathrm d$ with $\delta = \pm\star\mathrm d\star$; it is not needed here and is named only so that the notation is not a surprise later.

34 · the derivative

Pullback, and why forms are worth the trouble

definition

For a smooth $\phi:N\to M$ and $\omega\in\Omega^k(M)$, the pullback $\phi^*\omega\in\Omega^k(N)$ is $$(\phi^*\omega)_p(v_1,\dots,v_k) = \omega_{\phi(p)}\big(\mathrm d\phi_p v_1,\dots,\mathrm d\phi_p v_k\big).$$ Unit 31 defined it for tensors; nothing is new except the name for the antisymmetric case.

naturality

$\phi^*(\alpha\wedge\beta) = \phi^*\alpha\wedge\phi^*\beta$ and $$\boxed{\ \phi^*(\mathrm d\omega) = \mathrm d(\phi^*\omega)\ }$$ for every smooth $\phi$ — no assumption that $\phi$ is a diffeomorphism, and no metric anywhere.

proof

On functions, $\phi^*(\mathrm df) = \mathrm d(f\circ\phi)$ is the chain rule. Both $\omega\mapsto\phi^*\mathrm d\omega$ and $\omega\mapsto\mathrm d\phi^*\omega$ are linear and satisfy the same Leibniz rule, and every form is locally a sum of $f\,\mathrm dg^1\wedge\cdots\wedge\mathrm dg^k$; on such a form both sides give $\mathrm d(f\circ\phi)\wedge\mathrm d(g^1\circ\phi)\wedge\cdots$. $\square$

what it buys

Change of variables costs nothing: pull back and differentiate in either order. Compare unit 05, where every change of coordinates produced a Jacobian to be tracked by hand, and unit 31, where the same content needed the chain rule written out per index.

Why this is the load-bearing property. A vector field cannot be pulled back — you can push it forward only along a diffeomorphism — and a $(1,1)$-tensor cannot be pulled back at all. Forms can, always, along any smooth map: restriction to a submanifold, projection, a flow, a coordinate change, a degenerate map that collapses dimensions. That, plus naturality of $\mathrm d$, is the technical reason Stokes' theorem can be proved once on a cube and exported everywhere: the cube maps into the manifold, and both $\mathrm d$ and $\int$ commute with the map.

Full proof that pullback respects the wedge. Both sides are evaluated on $k+l$ vectors; the right side's shuffle sum has $\alpha(\mathrm d\phi\,v_{\sigma(1)},\dots)$ where the left has $\alpha$ applied to the same pushed-forward vectors. The two expressions are term-by-term identical because $\mathrm d\phi_p$ is a single linear map applied to each argument. No smoothness beyond $C^1$ is used pointwise.

A caution about "locally a sum of $f\,\mathrm dg^1\wedge\cdots$". In a chart take $g^i = x^i$; that is the whole content. The proof above is a genuine proof, not a sketch, provided one also knows $\mathrm d$ is local, which the previous slide's notes established.

34 · the derivative

$\mathrm d$ without coordinates, and where structure equations come from

theorem · the invariant formula

For $\alpha\in\Omega^1(M)$ and vector fields $X, Y$, $$\mathrm d\alpha(X,Y) = X\big(\alpha(Y)\big) - Y\big(\alpha(X)\big) - \alpha\big([X,Y]\big),$$ with $[X,Y]$ unit 31's Lie bracket.

proof

Both sides are $\mathbb R$-bilinear and antisymmetric, and both are local, so check on $\alpha = f\,\mathrm dg$. Left: $(\mathrm df\wedge\mathrm dg)(X,Y) = (Xf)(Yg) - (Yf)(Xg)$. Right: $X(f\,Yg) - Y(f\,Xg) - f[X,Y]g = (Xf)(Yg) + fXYg - (Yf)(Xg) - fYXg - f(XY-YX)g$, and the four $f$-terms cancel. $\square$

read it twice

Left to right: $\mathrm d$ needs no coordinates.
Right to left: the bracket term is the price of using a non-coordinate frame — and it is precisely the term that becomes Cartan's connection one-form in section 6.

The general formula, for a $k$-form, is $\mathrm d\omega(X_0,\dots,X_k) = \sum_i(-1)^iX_i\big(\omega(\dots\widehat{X_i}\dots)\big) + \sum_{i<j}(-1)^{i+j}\omega([X_i,X_j],\dots\widehat{X_i}\dots\widehat{X_j}\dots)$. Cited: Lee, Introduction to Smooth Manifolds, 2nd ed., ch. 14, the invariant formula for the exterior derivative; the proof is the same check on $f\,\mathrm dg^1\wedge\cdots$, longer. Only the $k = 1$ case is used in this course.

The test that makes it believable. Take a coordinate frame, $X = \partial_i$, $Y = \partial_j$. Then $[X,Y] = 0$ by Clairaut (unit 31 proved coordinate fields commute) and the formula reduces to $\mathrm d\alpha(\partial_i,\partial_j) = \partial_i\alpha_j - \partial_j\alpha_i$ — the chart formula. In an orthonormal frame, which is what section 6 uses, the brackets are not zero, and the extra term is exactly the structure functions $c^a{}_{bc}$ that appear when $\mathrm de^a$ is computed.

Why this matters for the physics. The formula says $\mathrm d\alpha$ measures the failure of a circulation to be recoverable from directional derivatives alone — the leftover is the bracket, the failure of the flows to commute. Unit 31 interpreted $[X,Y]$ as the gap in a commutator square; this says the curl is that gap, paired with $\alpha$.

34 · the derivative

Worked example 2 — $\mathrm dF = 0$, expanded

example 2 · the homogeneous pair

Take $F$ from example 1 and differentiate, in two pieces.

The magnetic piece. $\mathrm d\big(B_x\,\mathrm dy\wedge\mathrm dz + B_y\,\mathrm dz\wedge\mathrm dx + B_z\,\mathrm dx\wedge\mathrm dy\big) = (\nabla\cdot\vec B)\,\mathrm dx\wedge\mathrm dy\wedge\mathrm dz + \partial_tB_i\,\mathrm dt\wedge\varepsilon^i$, writing $\varepsilon^1 = \mathrm dy\wedge\mathrm dz$, $\varepsilon^2 = \mathrm dz\wedge\mathrm dx$, $\varepsilon^3 = \mathrm dx\wedge\mathrm dy$.

The electric piece. $\mathrm d\big(E_i\,\mathrm dx^i\wedge\mathrm dt\big) = \big(\mathrm d(E_i\,\mathrm dx^i)\big)\wedge\mathrm dt = (\nabla\times\vec E)_i\,\varepsilon^i\wedge\mathrm dt = (\nabla\times\vec E)_i\,\mathrm dt\wedge\varepsilon^i$, since $\varepsilon^i\wedge\mathrm dt = \mathrm dt\wedge\varepsilon^i$ (two transpositions).

$$\mathrm dF = (\nabla\cdot\vec B)\,\mathrm dx\wedge\mathrm dy\wedge\mathrm dz + \big[\partial_t\vec B + \nabla\times\vec E\big]_i\,\mathrm dt\wedge\varepsilon^i$$

the four basis three-forms are independent

$\mathrm dF = 0\iff\nabla\cdot\vec B = 0\ $ and $\ \nabla\times\vec E + \partial_t\vec B = 0$. Unit 24's homogeneous pair, both signs, nothing left over.

And since $F = \mathrm dA$, this is $\mathrm d\mathrm dA = 0$: two of Maxwell's equations are Clairaut's theorem.

The magnetic piece, in full. $\mathrm d(B_x\,\mathrm dy\wedge\mathrm dz) = (\partial_xB_x\,\mathrm dx + \partial_yB_x\,\mathrm dy + \partial_zB_x\,\mathrm dz + \partial_tB_x\,\mathrm dt)\wedge\mathrm dy\wedge\mathrm dz$. The $\mathrm dy$ and $\mathrm dz$ terms die on repeats, leaving $\partial_xB_x\,\mathrm dx\wedge\mathrm dy\wedge\mathrm dz + \partial_tB_x\,\mathrm dt\wedge\mathrm dy\wedge\mathrm dz$. The other two terms give $\partial_yB_y$ and $\partial_zB_z$ with the same orientation — check: $\mathrm dy\wedge\mathrm dz\wedge\mathrm dx = \mathrm dx\wedge\mathrm dy\wedge\mathrm dz$, two transpositions — so the first sum is the divergence.

The sign that people get wrong. $\varepsilon^i\wedge\mathrm dt$ versus $\mathrm dt\wedge\varepsilon^i$: moving a one-form past a two-form costs $(-1)^{1\cdot2} = +1$. Moving it past a one-form would cost a minus. Writing all four three-forms in the same order before comparing coefficients is not pedantry; it is the only way the signs come out.

What has actually been shown. Not that Maxwell's homogeneous pair is true — unit 24 established that from experiment and from the potentials — but that it is a single coordinate-free statement, and that given $F = \mathrm dA$ it is an identity with no physical content at all. The physics is in the existence of $A$, which section 4 shows is not automatic when space has a hole in it.

Dimensions. $[\mathrm dF] = [F] = \mathrm{Wb}$, and $[(\nabla\cdot\vec B)\mathrm dx\wedge\mathrm dy\wedge\mathrm dz] = (\mathrm{T/m})\,\mathrm m^3 = \mathrm{T\,m^2} = \mathrm{Wb}$; $[(\nabla\times\vec E)\,\mathrm dt\wedge\varepsilon] = (\mathrm{V/m^2})\,\mathrm{s\,m^2} = \mathrm{V\,s} = \mathrm{Wb}$. Both terms match.

34 · the derivative

Gauge freedom is the freedom to add an exact form

Unit 24's gauge transformation is $\vec A\mapsto\vec A + \nabla\lambda$, $\varphi\mapsto\varphi - \partial_t\lambda$. In one line:

$$A \longmapsto A + \mathrm d\lambda,\qquad F = \mathrm dA \longmapsto \mathrm dA + \mathrm d\mathrm d\lambda = F .$$

Check the components: $\mathrm d\lambda = \partial_i\lambda\,\mathrm dx^i + \partial_t\lambda\,\mathrm dt$, so $A_i\mapsto A_i + \partial_i\lambda$ and $-\varphi\mapsto-\varphi+\partial_t\lambda$ — unit 24's pair, including the sign nobody remembers.

the converse

If $\mathrm dA' = \mathrm dA$ then $A'-A$ is closed. On a contractible region section 4's Poincaré lemma makes it $\mathrm d\lambda$, so the gauge orbit is exactly the set of potentials. Not on the punctured plane, and that is the entire content of unit 23's solenoid.

Lorenz gauge is not a form statement

$\partial_\mu A^\mu = 0$ raises an index, so it needs a metric. Gauge freedom is metric-free; gauge fixing is not. In the language of section 5, the Lorenz condition is $\mathrm d\star A = 0$.

Why the sign works out. With $A = A_i\,\mathrm dx^i - \varphi\,\mathrm dt$, adding $\mathrm d\lambda$ adds $\partial_t\lambda\,\mathrm dt$ to the last slot, so $-\varphi\mapsto-\varphi+\partial_t\lambda$, i.e. $\varphi\mapsto\varphi-\partial_t\lambda$. Unit 24 derived that sign by demanding $\vec E$ be unchanged; here it is forced by the single requirement that the change be $\mathrm d$ of something.

Lorenz gauge in forms. $\star A$ is a three-form; $\mathrm d\star A$ is a four-form, a single function times the volume form, and computing it in Minkowski coordinates with $x^0 = ct$ gives exactly $(\partial_\mu A^\mu)\,\mathrm{vol}$: from the frame formulas of the Hodge-star slide, $\mathrm d\star(A_0\,\mathrm dx^0) = -\partial_0A_0\,\mathrm{vol} = \partial_0A^0\,\mathrm{vol}$ and $\mathrm d\star(A_1\,\mathrm dx^1) = \partial_1A_1\,\mathrm{vol}$, and likewise for $x^2, x^3$. Stating it as $\mathrm d\star A = 0$ makes obvious what unit 24 had to argue: the condition is Lorentz invariant, because $\star$ and $\mathrm d$ are built from the metric and the orientation, both of which a Lorentz transformation preserves.

The claim that this is the content of the solenoid. Outside an infinite solenoid, $\vec B = 0$, so the spatial one-form $\vec A^\flat$ is closed there. It is not exact: $\oint\vec A\cdot\mathrm d\vec\ell = \Phi_B \ne 0$. So there is no gauge function $\lambda$ with $\vec A = \nabla\lambda$ outside, even though $\vec B$ vanishes there. Section 4 returns to this with a number.

section 3

One theorem

Orientation, integration, and the four classical theorems collapsing into a single line.

34 · integration

Orientation and the volume form

definition

An orientation of an $n$-manifold is a choice of nowhere-vanishing $\mu\in\Omega^n(M)$, two such being equivalent if $\mu' = f\mu$ with $f > 0$. A manifold admitting one is orientable; $M$ connected and orientable has exactly two orientations.

the metric volume form

Given a metric $g$ and an orientation, there is a unique $n$-form $\mathrm{vol}_g$, the volume form, equal to $+1$ on every positively oriented orthonormal frame, and in a chart $$\mathrm{vol}_g = \sqrt{|\det g_{ij}|}\;\mathrm dx^1\wedge\cdots\wedge\mathrm dx^n .$$

dimension check

On $\mathbb R^3$ in spherical coordinates $g = \operatorname{diag}(1, r^2, r^2\sin^2\theta)$, so $\sqrt{\det g} = r^2\sin\theta$ and $\mathrm{vol} = r^2\sin\theta\,\mathrm dr\wedge\mathrm d\theta\wedge\mathrm d\phi$ — unit 05's Jacobian, derived rather than remembered.

Full proof of the chart formula. Let $(e_a)$ be a positively oriented orthonormal frame and $\partial_i = \Lambda^a{}_i e_a$. Then $g_{ij} = \Lambda^a{}_i\Lambda^b{}_j\,\eta_{ab}$, so $\det g = \det(\Lambda)^2\det\eta$ and $|\det\Lambda| = \sqrt{|\det g|}$. Evaluating the top form on $(\partial_1,\dots,\partial_n)$ gives $\det\Lambda$, positive for a positively oriented chart. Hence $\mathrm{vol}_g = \det\Lambda\;\mathrm dx^1\wedge\cdots\wedge\mathrm dx^n$, which is the displayed formula.

Why $|\det g|$ and not $\det g$. In Lorentzian signature $\det g < 0$: for $\eta = \operatorname{diag}(-1,1,1,1)$, $\det\eta = -1$ and $\sqrt{|\det\eta|} = 1$, so in inertial coordinates $\mathrm{vol} = \mathrm dx^0\wedge\mathrm dx^1\wedge\mathrm dx^2\wedge\mathrm dx^3$, as one would want. In Schwarzschild coordinates (unit 36) the same formula gives $r^2\sin\theta\,\mathrm dt\wedge\mathrm dr\wedge\mathrm d\theta\wedge\mathrm d\phi$, the metric functions cancelling — a fact worth remembering before that unit.

What "exactly two orientations" means. If $\mu$ and $\mu'$ are both nowhere zero, then $\mu' = f\mu$ for a nowhere-zero function $f$, which on a connected manifold has one sign throughout (intermediate value theorem along a path). So every orientation is either $[\mu]$ or $[-\mu]$. The next slide is the manifold where no $\mu$ exists at all.

34 · integration

The Möbius band has no orientation

the Möbius band

Not orientable: carry a frame once around and it comes back reversed. There is no nowhere-zero top form, so no consistent notion of "flux through it" — which is why unit 05 quietly required every surface to have a chosen normal.

On a surface in $\mathbb R^3$ an orientation is the same thing as a continuous choice of unit normal $n$: the two-form $\mu(u,v) = n\cdot(u\times v)$ is then nowhere zero, and conversely. The figure tries to make that choice and fails.

The normal carried once round the centre line (dashed) turns over as it goes and returns to its starting point pointing the other way. The two shades show which face of the strip is towards you: they swap where the band twists.

The computation behind the figure. Parametrise the band by $P(t,s) = \big((1 + s\cos\tfrac t2)\cos t,\ (1 + s\cos\tfrac t2)\sin t,\ s\sin\tfrac t2\big)$ with $t\in[0,2\pi]$ and $|s| \le 0.44$. On the centre line $s = 0$ the normal $\partial_tP\times\partial_sP$ is $(0,0,-1)$ at $t = 0$ and $(0,0,+1)$ at $t = 2\pi$ — the same point of the band, since $P(2\pi, s) = P(0, -s)$. A continuous unit normal along the closed centre line would have to agree with itself there, and no choice does better than this one: each point has only two unit normals, so any continuous choice is $\pm$ this one throughout, and the sign cannot jump.

Why a normal is an orientation. Given a unit normal field $n$, the two-form $\mu(u,v) = n\cdot(u\times v)$ is nonzero on any basis of the tangent plane. Conversely, given a nowhere-zero $\mu$, there is exactly one unit normal $n$ with $n\cdot(u\times v)$ of the same sign as $\mu(u,v)$, and it varies continuously. So "no continuous normal" is "no orientation".

Orientability is a real restriction, but not in relativity. A spacetime is normally required to be orientable and time-orientable, because without the first there is no consistent volume element to integrate an action over and without the second no consistent distinction between future and past. Both are assumptions, stated as such; nothing in the field equations forces them.

34 · integration

Integrating a form, and where the Jacobian went

definition

For $\omega = f\,\mathrm dx^1\wedge\cdots\wedge\mathrm dx^n$ compactly supported in a positively oriented chart, $\displaystyle\int_M\omega := \int_{\mathbf x(U)}f\circ\mathbf x^{-1}\,\mathrm d^n x$, an ordinary multiple integral (unit 05). For general $\omega$, chop it up with a partition of unity and add. Cited: independence of the partition, Lee, ch. 16.

theorem · no Jacobian appears

If $\phi: U\to V$ is an orientation-preserving diffeomorphism, then $\int_V\omega = \int_U\phi^*\omega$.

proof

$\phi^*(\mathrm dy^1\wedge\cdots\wedge\mathrm dy^n) = \phi^*\mathrm dy^1\wedge\cdots\wedge\phi^*\mathrm dy^n = \frac{\partial\phi^1}{\partial x^{j_1}}\cdots\frac{\partial\phi^n}{\partial x^{j_n}}\, \mathrm dx^{j_1}\wedge\cdots\wedge\mathrm dx^{j_n} = \det(D\phi)\,\mathrm dx^1\wedge\cdots\wedge\mathrm dx^n$, the last step being the Leibniz formula for a determinant. So $\int_U\phi^*\omega = \int(f\circ\phi)\det(D\phi)\,\mathrm d^nx$, and unit 05's change-of-variables theorem says that is $\int f\,\mathrm d^ny$ provided $\det D\phi > 0$, which orientation-preserving means. $\square$

The Jacobian did not vanish: it is inside $\phi^*$, put there by antisymmetry. That is the whole reason $n$-forms, and not functions, are the things one integrates.

Read the proof backwards. Unit 05 had to carry $|\det D\phi|$ explicitly and to argue about the absolute value. Here the sign is carried by the orientation and the magnitude by the form, and the two bookkeeping problems become one. If $\phi$ reverses orientation both sides change sign, and the identity still holds with a minus — which is the correct statement of what unit 05 handled by insisting on $|\cdot|$.

The determinant step in detail. In the wedge of $n$ one-forms $\sum_j(\partial\phi^a/\partial x^j)\mathrm dx^j$, the only surviving terms have $j_1,\dots,j_n$ a permutation of $1,\dots,n$, and reordering to increasing order contributes $\operatorname{sgn}$ of that permutation. The sum $\sum_\sigma\operatorname{sgn}(\sigma)\prod_a \partial\phi^a/\partial x^{\sigma(a)}$ is $\det D\phi$ by definition.

Unit 32's volume element, as promised. For a function $f$ on an oriented Riemannian or Lorentzian manifold, $\int_Mf\,\mathrm{vol}_g = \int f\sqrt{|g|}\,\mathrm d^nx$ in any positively oriented chart — the definition above applied to the previous slide's chart formula. That is unit 32's integral of a scalar against its volume element. The two agree on an oriented manifold; the form carries a sign that the element's absolute value threw away, and the orientation is what fixes it.

Compactly supported is not a technicality to skip. Without it the integral need not converge, and Stokes' theorem fails: $\omega = \mathrm dx$ on $\mathbb R$ has $\int_{\mathbb R}\mathrm d(x) $ divergent, and $\partial\mathbb R = \emptyset$. Every statement of Stokes in this unit carries the hypothesis.

34 · integration

The generalised Stokes theorem

theorem · Stokes, in Cartan's form

Let $M$ be an oriented smooth $n$-manifold with boundary, $\partial M$ carrying the induced orientation, and let $\omega\in\Omega^{n-1}(M)$ be compactly supported. Then $$\int_M\mathrm d\omega \;=\; \int_{\partial M}\omega .$$

degree bookkeeping

$\mathrm d$ raises degree by one; $\partial$ lowers dimension by one. The theorem says these two are adjoint — and $\partial\partial = \emptyset$ mirrors $\mathrm d\mathrm d = 0$, which is not a coincidence but de Rham's theorem in embryo.

The proof of Stokes' theorem has two halves: the cube, which is a computation, and the passage to a general manifold, which is a partition of unity and the naturality of $\mathrm d$. The proof, on the square does the first.

The one convention in the statement is how $\partial M$ is oriented. The next slide draws it.

The second half, in outline. Cover $M$ by charts, each mapping to either an open subset of $\mathbb R^n$ or a half-space; take a partition of unity $\{\chi_a\}$ subordinate to the cover, with finitely many terms on the compact support of $\omega$. Then $\omega = \sum_a\chi_a\omega$ and both sides are linear, so it suffices to treat each $\chi_a\omega$, which lives in one chart. Pull it back to the model cube by the chart map; naturality of $\mathrm d$ and the previous slide's change-of-variables theorem say both integrals are unchanged. What is left is the cube computation. A loose end: $\sum_a\mathrm d(\chi_a\omega) = \mathrm d\omega$ because $\sum_a\mathrm d\chi_a = \mathrm d(1) = 0$. Cited for the partition of unity itself: unit 31, which cited it in turn to Lee, ch. 2.

Who proved what. "Stokes' theorem" is a historical accident: the curl theorem appeared as a problem set by Stokes in the 1854 Smith's Prize examination, having been told to him by Kelvin in an 1850 letter. The general statement is Poincaré's and Cartan's, and its modern form and name settled after Cartan's 1945 Les systèmes différentiels extérieurs. Élie Cartan is also the Cartan of section 6; his son Henri is the Cartan of several other theorems.

Manifolds with boundary were not defined in unit 31. The definition needed here is the obvious one: charts map to open subsets of the half-space $\{x^n\ge0\}$, and $\partial M$ is the set of points mapping to $x^n = 0$, which is an $(n-1)$-manifold. Smoothness of transition maps is required up to the boundary. Nothing else in this deck depends on the details.

34 · integration

Outward normal first, drawn

the induced orientation

"Outward normal first": a frame $(v_1,\dots,v_{n-1})$ of $T_p\partial M$ is positive when $(\nu, v_1,\dots,v_{n-1})$ is positive in $T_pM$, $\nu$ outward. Every sign in unit 05's and unit 23's theorems is this convention and nothing else.

In the plane ($n = 2$) with $\mathrm dx\wedge\mathrm dy$ positive, "$(\nu, v)$ positive" means that turning from $\nu$ to $v$ is a quarter turn anticlockwise — so $M$ is on your left as you walk along $\partial M$ in the direction $v$.

Anticlockwise round the outside, clockwise round a hole: the same rule, "$M$ on the left", at every boundary point.

Reading the figure. At each boundary point the outward normal $\nu$ and the tangent $v$ make a pair with $(\mathrm dx\wedge\mathrm dy)(\nu, v) > 0$: turning from $\nu$ to $v$ is a quarter turn anticlockwise. On the outer curve that sends $v$ anticlockwise round $M$; on the inner curve $\nu$ points into the hole, and the same rule sends $v$ clockwise. So for a region with a hole, Stokes' boundary integral is the outer circulation minus the inner one, both measured anticlockwise — exactly the combination worked example 3 needs in section 4.

In three dimensions the same rule says: for a solid region, a pair $(v_1, v_2)$ tangent to the boundary surface is positive when $(\nu, v_1, v_2)$ is a right-handed frame, with $\nu$ pointing out. That is unit 05's outward normal. For a surface $S$ with boundary curve, applied one dimension down, it gives unit 23's right-hand rule between the normal of $S$ and the direction round $\partial S$.

34 · integration

The proof, on the square

Let $M = [0,1]^2$ with $\mathrm dx\wedge\mathrm dy$ positive, and $\omega = P\,\mathrm dx + Q\,\mathrm dy$. Then $\mathrm d\omega = (\partial_xQ - \partial_yP)\,\mathrm dx\wedge\mathrm dy$ and, by Fubini (unit 05),

$$\begin{aligned}\int_M\!\partial_xQ &= \int_0^1\!\!\big[Q(1,y)-Q(0,y)\big]\mathrm dy,\\ \int_M\!\partial_yP &= \int_0^1\!\!\big[P(x,1)-P(x,0)\big]\mathrm dx .\end{aligned}$$

Run the boundary anticlockwise, as "outward normal first" requires. On the horizontal edges $\mathrm dy$ pulls back to zero, on the vertical ones $\mathrm dx$ does, so

$$\begin{aligned}\oint_{\partial M}\!\!\omega = {}&\int_0^1\!\!P(x,0)\,\mathrm dx + \int_0^1\!\!Q(1,y)\,\mathrm dy\\ &- \int_0^1\!\!P(x,1)\,\mathrm dx - \int_0^1\!\!Q(0,y)\,\mathrm dy ,\end{aligned}$$

which is the previous line. $\square$ This is Green's theorem, and it is also the whole of Stokes: the general case is the same computation in $n$ slots.

Each edge's contribution. The top and left edges run against $x$ and $y$, hence their minus signs.

the sign, checked on the vortex

$\oint_{S^1}\theta$ anticlockwise is $2\pi$; clockwise, $-2\pi$. "Outward normal first" picks one.

Full proof on the $n$-cube. Write $\omega = \sum_i f_i\,\mathrm dx^1\wedge\cdots\widehat{\mathrm dx^i}\cdots\wedge\mathrm dx^n$ (hat = omitted). Then $\mathrm d\omega = \sum_i\partial_if_i\,\mathrm dx^i\wedge\mathrm dx^1\wedge \cdots\widehat{\mathrm dx^i}\cdots = \sum_i(-1)^{i-1}\partial_if_i\,\mathrm dx^1\wedge\cdots\wedge\mathrm dx^n$, the sign coming from moving $\mathrm dx^i$ past $i-1$ factors. Integrating over $[0,1]^n$ and doing the $x^i$ integral first, $\int_{[0,1]^n}\mathrm d\omega = \sum_i(-1)^{i-1}\int\big[f_i|_{x^i=1} - f_i|_{x^i=0}\big]\,\mathrm d^{n-1}x$. On the face $x^i = c$ the pullback of $\omega$ keeps only the $i$-th term, since every other term contains $\mathrm dx^i$, which pulls back to zero. The outward normal on $x^i=1$ is $+\partial_i$ and on $x^i=0$ is $-\partial_i$, and moving $\partial_i$ to the front of $(\partial_1,\dots,\widehat{\partial_i},\dots,\partial_n)$ costs $(-1)^{i-1}$; so the induced orientation on the two faces contributes exactly $(-1)^{i-1}$ and $-(-1)^{i-1}$. Summing gives $\int_{\partial[0,1]^n}\omega$. The corners and edges are a measure-zero set and contribute nothing. $\square$

Where compact support is used. On the cube it is not: the cube is compact. It is used when the chart is a half-space, where the missing faces are at infinity and must carry no integral.

The vortex check, in full. $\oint_{S^1}\theta$ round the unit circle anticlockwise is $2\pi$; clockwise it is $-2\pi$. The orientation convention is the difference between the two, and it is fixed once by "outward normal first".

The four edges. Anticlockwise means bottom ($y=0$, $x$ increasing), right ($x=1$, $y$ increasing), top ($y=1$, $x$ decreasing), left ($x=0$, $y$ decreasing). The top and left edges run backwards, which is where the two minus signs in the boundary sum come from.

The anticlockwise claim. On the bottom edge the outward normal is $-\partial_y$, so a positive frame $(v)$ of the edge satisfies $(-\partial_y, v)$ positive, i.e. $-\mathrm dx\wedge\mathrm dy(\partial_y, v) > 0$, giving $v = +\partial_x$: $x$ increasing. Run the same argument on the other three edges.

34 · integration

Four theorems, one theorem

$n$	$\omega$	$\int_M\mathrm d\omega = \int_{\partial M}\omega$ reads	met in
1	$f\in\Omega^0$	$\int_a^b\mathrm df = f(b)-f(a)$ — the gradient theorem, and the fundamental theorem of calculus	unit 04
2, plane	$P\,\mathrm dx + Q\,\mathrm dy$	$\iint(\partial_xQ-\partial_yP) = \oint(P\,\mathrm dx + Q\,\mathrm dy)$ — Green	unit 04
2, in $\mathbb R^3$	$\vec v^\flat$	$\iint_S(\nabla\times\vec v)\cdot\mathrm d\vec S = \oint_{\partial S}\vec v\cdot\mathrm d\vec\ell$ — the curl theorem	units 04, 23
3	$\star\vec v^\flat$	$\iiint_V(\nabla\cdot\vec v)\,\mathrm dV = \oiint_{\partial V}\vec v\cdot\mathrm d\vec S$ — the divergence theorem	unit 05

what was actually different

Nothing but the degree of $\omega$ and the dimension of $M$. The four proofs in units 04, 05 and 23 were four instances of the cube computation, each obscured by a different notation for the same boundary orientation.

Row three deserves a word. Here $M = S$ is a surface with boundary sitting in $\mathbb R^3$, and the integrand is the pullback of $\mathrm d\vec v^\flat$ to $S$ — which is exactly $(\nabla\times\vec v)\cdot\mathrm d\vec S$, because the pullback of a two-form to a surface is its contraction with the surface's oriented area element. This is the one row where naturality of $\mathrm d$ is doing visible work: $\mathrm d$ computed upstairs in $\mathbb R^3$ and then restricted equals $\mathrm d$ of the restriction.

Ampère's law, one line. Unit 23 obtained $\oint\vec B\cdot\mathrm d\vec\ell = \mu_0I_{\text{enc}}$ by applying the curl theorem to $\nabla\times\vec B = \mu_0\vec J$. In this language that is $\oint_{\partial S}B^\flat = \int_S\mathrm dB^\flat = \mu_0\int_S\star J^\flat$, with the Euclidean star of $\mathbb R^3$, since $\mathrm dB^\flat = \star(\nabla\times\vec B)^\flat$. The left side depends only on the loop, so the right side must not depend on which surface $S$ spans it. Two spanning surfaces together bound a region $V$, and Stokes on $V$ says their integrals differ by $\mu_0\int_V\mathrm d\star J^\flat = \mu_0\int_V(\nabla\cdot\vec J)\,\mathrm{vol}$. So the two-form $\star J^\flat$ must be closed. For a steady current $\nabla\cdot\vec J = 0$ and it is; for a charging capacitor it is not, and unit 24's displacement current is exactly the term that restores closedness: $\nabla\cdot(\vec J + \varepsilon_0\partial_t\vec E) = 0$. The requirement itself is $\mathrm d\mathrm d = 0$ once more: $\mathrm dB^\flat$ is exact, so whatever it is set equal to had better be closed.

The cases not in the table. $n = 4$ gives the statement that the integral of $\mathrm d\star F$ over a spacetime region equals the flux of $\star F$ through its boundary — Gauss' law in four dimensions, which is where the conserved charge of section 5 comes from. On a $2n$-dimensional phase space, Stokes applied to unit 14's symplectic form gives Poincaré's integral invariants, which section 6 reaches by another road.

section 4

Closed, exact, and the difference

Unit 04 left a question open on purpose. Thirty units later, it has an answer with a number in it.

34 · cohomology

Closed does not mean exact

definition

$\omega\in\Omega^k$ is closed if $\mathrm d\omega = 0$ and exact if $\omega = \mathrm d\alpha$ for some $\alpha\in\Omega^{k-1}$. Exact $\Rightarrow$ closed, because $\mathrm d^2 = 0$. The converse is the question.

Poincaré lemma

If $U\subseteq\mathbb R^n$ is star-shaped about $0$, every closed $k$-form on $U$ with $k\ge1$ is exact.

proof for $k = 1$, in full

Let $\alpha = a_i\,\mathrm dx^i$ be closed, so $\partial_ja_i = \partial_ia_j$. Put $f(x) = \int_0^1 a_i(tx)\,x^i\,\mathrm dt$. Then $$\begin{aligned}\partial_jf &= \int_0^1\!\big[t\,x^i\partial_ja_i(tx) + a_j(tx)\big]\mathrm dt\\ &= \int_0^1\!\frac{\mathrm d}{\mathrm dt}\big[t\,a_j(tx)\big]\mathrm dt = a_j(x),\end{aligned}$$ using closedness in the middle step. So $\alpha = \mathrm df$. $\square$

The proof integrates along rays from the origin — which is exactly the "build the potential by integrating along a path" of unit 04, with star-shapedness as the hypothesis that the path exists and is canonical.

The middle step, spelled out. $\frac{\mathrm d}{\mathrm dt}[t\,a_j(tx)] = a_j(tx) + t\,x^i\,\partial_ia_j(tx)$. Closedness replaces $\partial_ia_j$ by $\partial_ja_i$, which is the integrand's first term. Evaluating $[t\,a_j(tx)]_0^1 = a_j(x) - 0$ finishes it. Differentiating under the integral sign is legitimate: the integrand is smooth in $(t,x)$ and $[0,1]$ is compact.

General $k$. Define $(K\omega)(x)(v_1,\dots,v_{k-1}) = \int_0^1t^{k-1}\, \omega(tx)(x, v_1,\dots,v_{k-1})\,\mathrm dt$. Then $\mathrm dK + K\mathrm d = \mathrm{id}$ on $\Omega^k(U)$ for $k\ge1$, so $\mathrm d\omega = 0$ gives $\omega = \mathrm d(K\omega)$. The verification is the same computation with more indices. Cited: Lee, Introduction to Smooth Manifolds, 2nd ed., ch. 17, the Poincaré lemma for star-shaped sets; Bott and Tu, §4. The statement generalises: on any contractible manifold — star-shaped is a special case — closed implies exact, because homotopic maps induce the same map on cohomology (cited, Lee ch. 17; the degree-one case is proved on the homotopy-invariance slide, which comes next).

The hypothesis is not decorative. On $U = \mathbb R^2\setminus\{0\}$, which is not star-shaped about any point, $\theta$ is closed and not exact. That single example is the reason the lemma is stated with a hypothesis rather than as "closed equals exact", and it is what unit 04 ran into.

34 · cohomology

Homotopy invariance: unit 04's promise kept

theorem

Let $\alpha$ be a closed one-form on $U$, and $\gamma_0,\gamma_1:[0,1]\to U$ loops joined by a smooth homotopy of loops $H:[0,1]^2\to U$, $H(\cdot,0) = \gamma_0$, $H(\cdot,1) = \gamma_1$, $H(0,t) = H(1,t)$. Then $\oint_{\gamma_0}\alpha = \oint_{\gamma_1}\alpha$.

proof · pull back, then Stokes on the square

$H^*\alpha$ is a one-form on the square, and it is closed: $\mathrm d(H^*\alpha) = H^*(\mathrm d\alpha) = 0$ by naturality. Stokes on $[0,1]^2$ gives $0 = \oint_{\partial[0,1]^2}H^*\alpha$. Round the boundary anticlockwise: the bottom edge is $\oint_{\gamma_0}\alpha$, the top edge run backwards is $-\oint_{\gamma_1}\alpha$, and the two side edges are the same path $t\mapsto H(0,t) = H(1,t)$ run in opposite directions, so they cancel. $\square$

$H$ carries the square onto the band between the loops. Its bottom edge goes to $\gamma_0$, its top edge to $\gamma_1$, and both dashed sides to the same path, once each way.

Why this needed forms. Unit 04 proved the rectangle case and sketched a grid argument for a general homotopy, noting that making "the image of each small square lies in a ball" precise for a merely continuous $H$ takes care. Here all the care goes into one operation, pullback, which works along any smooth map: $H$ need not be injective, and its image can fold over itself or collapse onto a point. The square does not care what $H$ does to it. This is the concrete meaning of "forms can always be pulled back", and it is worth comparing with how awkward the same argument is for a vector field, which cannot be pulled back along a map that is not a diffeomorphism.

The side edges, carefully. The anticlockwise boundary of $[0,1]^2$ runs up the right side ($s = 1$, $t$ increasing) and down the left side ($s = 0$, $t$ decreasing). Both sides are mapped by $H$ to the same curve $t\mapsto H(0,t) = H(1,t)$, so their pulled-back integrals are $+\int$ and $-\int$ of the same thing. For loops with a fixed base point the sides are constant and each contributes zero; the theorem as stated covers the free case too.

What the figure does not need. It is drawn with the band embedded and the two loops nested round the hole, because that is the case unit 04 cared about. The proof never looks at the band: if $H$ folds the square over itself, or squashes it onto a curve, $H^*\alpha$ is still a closed one-form on the square and Stokes on the square still applies. The hole is never inside the band, because $H$ maps into $U$; that is the whole of the hypothesis.

34 · cohomology

Simply connected: closed means exact

Left: a loop in a region with no holes shrinks through loops to a point, so by the previous slide its circulation equals that of a constant loop, $0$. Right: round a hole the shrinking stops, and so does the argument.

corollary · unit 04's cited theorem

On a simply connected open $U$, every closed one-form is exact.

proof

Every loop is homotopic to a constant loop, whose integral is $0$. So every circulation vanishes, integrals are path-independent, and unit 04's $f(x) = \int_{x_0}^x\alpha$ has $\mathrm df = \alpha$. $\square$

only one thing cited

A continuous homotopy can be replaced by a smooth one with the same ends.

What is cited. The smoothing step: if two smooth loops are homotopic through continuous loops, they are homotopic through smooth ones. Cited: Lee, ch. 6, the Whitney approximation theorem for homotopies. Everything else on the slide is proved.

The corollary's last step is unit 04's own construction: fix $x_0$, define $f(x)$ as the integral of $\alpha$ along any path from $x_0$ to $x$ (well defined because two paths differ by a loop, and loops give zero), and differentiate along a short straight segment at $x$ to get $\partial_if = \alpha_i$.

Simply connected, said precisely. $U$ is connected and every loop in it is homotopic, through loops, to a constant loop. A disc, a star-shaped set and all of $\mathbb R^n$ are simply connected; the punctured plane is not, and the right-hand picture is why — though proving that no homotopy at all removes the loop needs an invariant, and $\oint\theta = 2\pi$ is the one this section supplies. The punctured space $\mathbb R^3\setminus\{0\}$ is simply connected: a loop can slip round the missing point. Its hole shows up one degree higher, on the monopole slides.

34 · cohomology

de Rham cohomology

definition

$$H^k_{\mathrm{dR}}(M) \;=\; \frac{\ker\big(\mathrm d:\Omega^k\to\Omega^{k+1}\big)}{\operatorname{im}\big(\mathrm d:\Omega^{k-1}\to\Omega^k\big)} \;=\; \frac{\text{closed }k\text{-forms}}{\text{exact }k\text{-forms}} .$$ The quotient makes sense precisely because $\mathrm d^2 = 0$: the image sits inside the kernel. $\dim H^k$ is the $k$-th Betti number $b_k$.

$H^0$ counts pieces

A closed $0$-form is a locally constant function, and there are no exact ones. So $b_0$ is the number of connected components.

Poincaré restated

$M$ contractible $\Rightarrow$ $H^k(M) = 0$ for $k\ge1$. Cohomology vanishes where there is nothing to detect.

de Rham's theorem · cited

$H^k_{\mathrm{dR}}(M)\cong H^k_{\text{sing}}(M;\mathbb R)$: a purely analytic object computes a purely topological one. de Rham 1931.

an invariant of the space, not of the field

$H^k$ depends on $M$ alone. Changing $\vec E$ and $\vec B$ moves $F$ around inside a class; changing the region the fields live on can change which classes exist at all. Unit 04 hit this exactly: the same formula, two different domains, two different verdicts.

Why the quotient is the right object. Two closed forms differing by an exact form have the same integral over every closed submanifold, by Stokes: $\int_Z\mathrm d\alpha = \int_{\partial Z}\alpha = 0$ when $\partial Z = \emptyset$. So the class, not the form, is what an integral can see. Conversely, de Rham's theorem says the integrals over cycles see the class completely. The next slide is that statement with numbers.

What "cited" covers here. de Rham's theorem needs singular homology, which this course does not build. Everything else in this section — the computation of $H^1$ of the punctured plane, of the torus, and the monopole obstruction — is proved from the Poincaré lemma and Stokes, with no algebraic topology at all.

Finite dimensionality is a theorem for compact manifolds (cited: Bott and Tu, §5, by the Mayer–Vietoris argument on a finite good cover) and false in general for the naive reason: $H^0$ of an infinite discrete set is infinite dimensional. For the spaces in this unit — punctured plane, punctured space, torus — the Betti numbers are small and computed explicitly.

34 · cohomology

Worked example 3 — $H^1$ of the punctured plane

example 3 · unit 04's question, closed

On $U = \mathbb R^2\setminus\{0\}$, the vortex $\theta = (x\,\mathrm dy - y\,\mathrm dx)/(x^2+y^2)$ has $\mathrm d\theta = 0$ (compute: both terms give $\pm(y^2-x^2)/r^4\,\mathrm dx\wedge\mathrm dy$ and cancel) and $\oint_{S^1}\theta = 2\pi$.

Claim. $H^1(U) = \mathbb R$, generated by $[\theta]$.

Step 1 — $[\theta]\ne0$. If $\theta = \mathrm df$ then $\oint_{S^1}\theta = 0$ by Stokes, since $S^1$ has no boundary. But it is $2\pi$.

Step 2 — the circle integral does not depend on the circle. For radii $r_1<r_2$ the annulus $\mathcal A$ between them lies in $U$, and $\int_{\mathcal A}\mathrm d\alpha = 0$ for closed $\alpha$. With the induced orientation, $\partial\mathcal A$ is the outer circle anticlockwise and the inner one clockwise, so Stokes reads $\oint_{r_2}\alpha - \oint_{r_1}\alpha = 0$.

Step 3, on the next slide: every closed $\alpha$ is a multiple of $\theta$ plus an exact form.

The annulus never meets the hole, so Stokes applies to it. Its boundary is two circles run in opposite senses.

$\mathrm d\theta = 0$, in full. $\mathrm d\!\left(\frac{x}{r^2}\mathrm dy\right) = \partial_x\!\left(\frac{x}{r^2}\right)\mathrm dx\wedge\mathrm dy = \frac{r^2-2x^2}{r^4}\mathrm dx\wedge\mathrm dy = \frac{y^2-x^2}{r^4}\mathrm dx\wedge\mathrm dy$, and $\mathrm d\!\left(-\frac{y}{r^2}\mathrm dx\right) = -\partial_y\!\left(\frac{y}{r^2}\right)\mathrm dy\wedge\mathrm dx = \frac{r^2-2y^2}{r^4}\mathrm dx\wedge\mathrm dy = \frac{x^2-y^2}{r^4}\mathrm dx\wedge\mathrm dy$. They cancel. Unit 04 computed the same thing as "the curl of the vortex vanishes".

Step 2 is the generalised Stokes theorem's region with a hole. The slide The generalised Stokes theorem drew exactly this situation: outward normal first makes the inner boundary run clockwise. Measuring both circle integrals anticlockwise, as one naturally does, turns that into the difference on the slide. Nothing about circles was used: any two loops that together bound a region inside $U$ have the same integral of a closed form.

34 · cohomology

Worked example 3, continued — unwrapping the plane

example 3 · step 3, every class is a multiple

Let $\alpha$ be closed on $U$, and put $c = \frac1{2\pi}\oint_{S^1}\alpha$. Pull back by the polar map $\Phi(r,\phi) = (r\cos\phi, r\sin\phi)$ to the convex half-plane $\Pi = (0,\infty)\times\mathbb R$. There $\Phi^*\alpha = \mathrm dg$ by the Poincaré lemma, and $\Phi^*\theta = \mathrm d\phi$. By step 2, $g(r,\phi+2\pi) - g(r,\phi) = \oint_{r}\alpha = 2\pi c$ for every $r$. So $g - c\phi$ is $2\pi$-periodic in $\phi$, hence equals $f\circ\Phi$ for a smooth $f$ on $U$, and $\alpha - c\,\theta = \mathrm df$. $\square$

On $\Pi$ the angle is an honest function, so closed forms have potentials there. A segment of height $2\pi$ maps once round a circle, and the potential gains that circle's integral, $2\pi c$.

Why $\Phi^*\theta = \mathrm d\phi$. $\Phi^*\mathrm dx = \cos\phi\,\mathrm dr - r\sin\phi\,\mathrm d\phi$ and $\Phi^*\mathrm dy = \sin\phi\,\mathrm dr + r\cos\phi\,\mathrm d\phi$, so $\Phi^*(x\,\mathrm dy - y\,\mathrm dx) = r\cos\phi(\sin\phi\,\mathrm dr + r\cos\phi\,\mathrm d\phi) - r\sin\phi(\cos\phi\,\mathrm dr - r\sin\phi\,\mathrm d\phi) = r^2\,\mathrm d\phi$, and dividing by $\Phi^*r^2$ leaves $\mathrm d\phi$. The point of the whole argument is that $\phi$ exists upstairs and not downstairs.

The descent step. $g - c\phi$ being $2\pi$-periodic means it is constant on the fibres of $\Phi$, so it defines a function $f$ on $U$; smoothness is inherited because $\Phi$ is a local diffeomorphism. Then $\mathrm d(f\circ\Phi) = \Phi^*\alpha - c\,\mathrm d\phi = \Phi^*(\alpha - c\theta)$, and since $\Phi^*$ is injective on forms (it is a submersion onto $U$), $\alpha - c\theta = \mathrm df$.

What unit 04 could and could not do. It proved the vortex was not conservative by exhibiting a loop, and it proved that on a simply connected domain vanishing curl implies a potential. What it could not say was how much obstruction there is. The answer is: one real number per hole, and the number attached to a given field is its circulation.

The name $\Pi$. The half-plane is called $\Pi$ here, not $H$, because $H$ already names the homotopy of the previous slides and $H^1$ the cohomology being computed.

34 · cohomology

The pairing: a class against a cycle

periods

If $Z\subseteq M$ is a compact oriented $k$-manifold without boundary, then $$\big([\omega], Z\big)\longmapsto\int_Z\omega$$ is well defined on classes: changing $\omega$ by $\mathrm d\alpha$ changes nothing, by Stokes. The numbers so obtained are the periods of the class.

On the punctured plane there is one independent cycle, the circle, and the period of the generator is

$$\oint_{S^1}\theta = \int_0^{2\pi}\mathrm d\phi = 2\pi = 6.283185307\ldots$$

Every closed one-form on $U$ has period $2\pi c$ for exactly one $c\in\mathbb R$, and $c$ is its class. Unit 04's $2\pi$ was a measurement of the hole all along.

$\oint\theta$ on three loops: the shape is irrelevant, only how many times the loop goes round the hole counts — $2\pi$ per turn.

the dictionary

$b_1 = 1$: one hole · generator: $[\theta]$ · period of generator: $2\pi$ · physical content: an irrotational field can have circulation, and no local measurement can predict it.

Why $2\pi$ and not $1$. Nothing forces the normalisation. $[\theta/2\pi]$ generates just as well and has period 1; that is the convention under which the monopole charge later comes out an integer. The number $2\pi$ is an artefact of writing $\theta$ in the form that makes $\mathrm d\theta = 0$ obvious. What is not an artefact is that the set of periods of all closed one-forms is $\mathbb R$, one dimension, not two and not zero.

Independence of the cycle's shape. Any loop in $U$ winding once about the origin gives $2\pi$ — deform it to a circle through loops in $U$ and apply the homotopy-invariance slide. Unit 04 found $2\pi$ on circles of every radius and called $\frac1{2\pi}$ of the circulation the winding number, unchanged by continuous deformation; the reason it is unchanged is here.

Reading the figure. The wobbly loop and a small circle inside it bound a ring-shaped region in $U$, so they have the same integral, $2\pi$, as in step 2 of example 3. The loop that goes twice round can be cut at its crossing point into two loops, each once round, so it gets $4\pi$. The loop that misses the hole bounds a region inside $U$, and Stokes there gives $\int\mathrm d\theta = 0$. The integer multiplying $2\pi$ is unit 04's winding number.

Looking ahead. de Rham's theorem says the period map is a perfect pairing between $H^k_{\mathrm{dR}}$ and $H_k$ with real coefficients. That is why a statement about integrals of fields — a physicist's statement — and a statement about holes — a topologist's — are the same statement. The monopole slide is the case $k = 2$.

34 · cohomology

The solenoid, as a class

Unit 23's infinite solenoid: inside, $\vec B = \mu_0nI\hat z$; outside, $\vec B = 0$. On the exterior $W$ — an open set that deformation-retracts to a circle — the one-form $A^\flat = \vec A\cdot\mathrm d\vec r$ is closed, since $\mathrm dA^\flat$ is the magnetic two-form and $\vec B = 0$ there. Its period is the enclosed flux:

$$\oint_{\partial S}A^\flat = \int_S\mathrm dA^\flat = \Phi_B \quad\Longrightarrow\quad [A^\flat] = \frac{\Phi_B}{2\pi}\,[\theta] .$$

Outside, $\vec B = 0$ but $\vec A$ circulates as $\Phi_B/2\pi r$: every loop round the solenoid collects $\Phi_B$.

a number

$n = 1000$ turns m$^{-1}$, $I = 2.00$ A, radius $a = 2.00$ cm: $B = \mu_0nI = 2.5133\times10^{-3}$ T, $\Phi_B = B\pi a^2 = 3.1583\times10^{-6}$ Wb. So $[A^\flat] = 5.0265\times10^{-7}\,[\theta]$ Wb.

what is and is not gauge

$A^\flat\mapsto A^\flat + \mathrm d\lambda$ moves the form and fixes the class. So the class is gauge invariant and equals the flux — a quantity in a region where the field is zero. Classically this is bookkeeping; quantum mechanically it is measurable, which is the Aharonov–Bohm effect, outside this course.

The numbers, computed. $\mu_0 = 4\pi\times10^{-7}\ \mathrm{T\,m\,A^{-1}}$ exactly in the convention unit 23 used, so $B = 4\pi\times10^{-7}\times1000\times2.00 = 2.5133\times10^{-3}$ T. Then $\Phi_B = B\,\pi a^2 = 2.5133\times10^{-3}\times\pi\times(0.0200)^2 = 3.1583\times10^{-6}$ Wb, and $\Phi_B/2\pi = 5.0265\times10^{-7}$ Wb. Dimensions: $[\theta]$ is dimensionless — it is $\mathrm d(\text{angle})$ — so the coefficient carries the weber, as it must for $A^\flat$, whose dimension is $\mathrm{T\,m^2} = \mathrm{Wb}$.

Why the exterior retracts to a circle. The exterior of an infinite cylinder is $\{(x,y,z) : x^2+y^2 > a^2\}$, which is $(\mathbb R^2\setminus \bar D_a)\times\mathbb R$; the disc and the $z$-line both contract, leaving a circle. So its $H^1$ is $\mathbb R$, computed by exactly the argument of example 3 with the puncture fattened into a disc — the proof never used that the removed set was a point, only that the complement is covered by the polar map.

The potential in the figure. One gauge choice outside is $A^\flat = \frac{\Phi_B}{2\pi}\theta$ itself, with $\theta$ read as a form on space that ignores $z$: in components $\vec A = \frac{\Phi_B}{2\pi r}\hat\phi$, circling the axis and falling off like $1/r$, which is what the arrows show. Its circulation round a circle of radius $r$ is $\frac{\Phi_B}{2\pi r}\cdot2\pi r = \Phi_B$ for every $r > a$, and its curl is zero there — unit 04's vortex, scaled by the flux. Any other gauge adds $\mathrm d\lambda$ and changes the arrows, not the circulation.

The honest statement of the classical content. Nothing measurable in classical electrodynamics depends on $[A^\flat]$: the force on a charge depends on $\vec E$ and $\vec B$, which vanish outside. What the class does is record that no global gauge function exists there — a statement about the description, not about a force. Unit 23 noted the strangeness and left it; this is the precise form of it.

34 · cohomology

Monopoles: the vortex, one dimension up

Suppose a magnetic charge $q_m$ at the origin, so that on $\mathbb R^3\setminus\{0\}$ $$\begin{aligned}\vec B &= \frac{\mu_0q_m}{4\pi}\frac{\hat r}{r^2},\\ F &= \frac{\mu_0q_m}{4\pi}\,\frac{x\,\mathrm dy\wedge\mathrm dz + y\,\mathrm dz\wedge\mathrm dx + z\,\mathrm dx\wedge\mathrm dy}{r^3}.\end{aligned}$$

$\mathrm dF = 0$ away from the origin — the same computation as $\nabla\cdot\vec B = 0$ for the inverse-square field in unit 05. But

$$\oint_{S^2}F = \mu_0q_m \neq 0,$$

so $F$ is not exact: if $F = \mathrm dA$ then Stokes on the closed surface $S^2$ gives zero. There is no global potential, and $[F]\ne0$ in $H^2(\mathbb R^3\setminus\{0\})$, which is $\mathbb R$ (cited, notes).

The same story, one degree higher. Both: a puncture, a sphere around it (a circle is the one-dimensional sphere), and a period that no local field measurement gives.

$\oint_{S^2}F$, computed. Parametrise the sphere of radius $R$; the two-form restricted to it is $\frac{\mu_0q_m}{4\pi}\frac{1}{R^2}\times(\text{area form})$, because $x\,\mathrm dy\wedge\mathrm dz + y\,\mathrm dz\wedge\mathrm dx + z\,\mathrm dx\wedge\mathrm dy$ pulls back to $R\cdot R^2\sin\theta\,\mathrm d\theta\wedge\mathrm d\phi$ on $r = R$. Hence $\oint_{S^2}F = \frac{\mu_0q_m}{4\pi R^3}\cdot R^3\cdot4\pi = \mu_0q_m$, independent of $R$ — the same independence as the annulus argument, one dimension up, and the same computation as unit 05's Gauss law for a point source.

What is cited. That $H^2(\mathbb R^3\setminus\{0\})$ is exactly one-dimensional, so that $[F]$ spans it: cited, Lee, ch. 17, the de Rham cohomology of punctured $\mathbb R^n$, computed there by Mayer–Vietoris. What is proved on the slide, and all this section uses, is that $[F]\ne0$.

Reading the figure. On the left the arrows run along the circle, because a one-form is integrated along a curve; on the right they cross the sphere, because a two-form is integrated as a flux through a surface. In both cases the quantity is closed away from the puncture — no curl, no divergence — so any loop or sphere round the puncture gives the same number, by Stokes on the region between two of them.

The obstruction is cohomological, and that is the point of the section. "Can a closed form be written as $\mathrm d$ of something" is not a question about the form; it is a question about the space, answered by one number per dimension. Unit 04's field, unit 23's solenoid and the hypothetical monopole are three instances of one statement.

34 · cohomology

The monopole needs two potentials

Remove the lower half of the $z$-axis: what is left is star-shaped about $(0,0,1)$, so the Poincaré lemma gives a potential $A^+$ there. Removing the upper half instead gives $A^-$.

Where both exist — space minus the whole axis — $A^+ - A^-$ is closed, and its period round the axis is the flux through the sphere: $$\oint(A^+ - A^-) = \int_{\text{upper}}F + \int_{\text{lower}}F = \mu_0q_m .$$ So $A^+ - A^- = \frac{\mu_0q_m}{2\pi}\,\theta + \mathrm d\lambda$: not exact, and no gauge change makes the two agree.

what this does and does not say

It says: if monopoles exist, $F = \mathrm dA$ fails globally, and the whole potential formalism must be patched together chart by chart. It does not say monopoles exist. None has been observed.

Each potential fails only on a half-axis (dashed). The upper hemisphere uses $A^+$, the lower $A^-$, and on the equator, where both work, their difference carries the whole flux.

The patching that is actually used. Write $A^{+}$ on the complement of the closed half-axis $\{x = y = 0,\ z\le0\}$ and $A^{-}$ on the complement of $\{x = y = 0,\ z\ge0\}$; each is star-shaped (about $(0,0,1)$ and $(0,0,-1)$), so the Poincaré lemma applies. On the overlap, space minus the $z$-axis, $A^+ - A^-$ is closed, and its period round the axis is the whole flux $\mu_0q_m$ — so it is $\frac{\mu_0q_m}{2\pi}\theta$ plus an exact form, not exact. That is $H^2$ of the sphere read off from $H^1$ of the overlap of two hemispheres, the statement unit 23 promised; problem P9 carries it out with explicit potentials. Requiring a quantum phase built from the would-be gauge function $\frac{\mu_0q_m}{2\pi}\phi$ to be single-valued is Dirac's argument and yields quantised $q_m$ (cited: P. A. M. Dirac, "Quantised singularities in the electromagnetic field", Proc. R. Soc. A 133, 60, 1931). That argument needs quantum mechanics and is outside this course — it is mentioned so that the phrase "Dirac quantisation" is not a surprise, not used.

The display, step by step. Orient the equator anticlockwise seen from above. It bounds the upper hemisphere $S_+$, with the outward normal of the sphere, and — with the opposite orientation — the lower hemisphere $S_-$. Stokes on each: $\oint A^+ = \int_{S_+}\mathrm dA^+ = \int_{S_+}F$ and $\oint A^- = -\int_{S_-}F$. Subtracting, $\oint(A^+ - A^-) = \int_{S_+}F + \int_{S_-}F = \oint_{S^2}F = \mu_0q_m$. Here $\theta$ is the vortex of the $xy$-plane, read as a form on space that ignores $z$; on the region where both potentials live it is closed and has period $2\pi$ round the axis, so the class of $A^+ - A^-$ is $\frac{\mu_0q_m}{2\pi}[\theta]$ by the argument of example 3.

section 5

The metric enters, once

Everything so far used no metric. The second pair of Maxwell equations needs one, and exactly one operator carries it.

34 · the star

The Hodge star

definition

A metric $g$ induces an inner product on $\Lambda^kT_p^*M$ by $\langle\alpha^1\wedge\cdots\wedge\alpha^k,\ \beta^1\wedge\cdots\wedge\beta^k\rangle = \det\big[\langle\alpha^i,\beta^j\rangle\big]$. With an orientation, the Hodge star $\star:\Lambda^k\to\Lambda^{n-k}$ is the unique linear map with $$\alpha\wedge\star\beta = \langle\alpha,\beta\rangle\,\mathrm{vol}_g\qquad\text{for all }\alpha,\beta\in\Lambda^k .$$

the double star

$\star\star = (-1)^{k(n-k)}s$ on $\Lambda^k$, where $s = \operatorname{sign}\det g$. Euclidean $\mathbb R^3$: $s=1$ and $\star\star = 1$ in every degree. Minkowski, $n=4$, $k=2$: $\star\star = (-1)^{4}\cdot(-1) = -1$.

worked in an orthonormal coframe

$\star(e^0\wedge e^i) = -\tfrac12\epsilon_{ijk}\,e^j\wedge e^k$, $\star(e^i\wedge e^j) = \epsilon_{ijk}\,e^0\wedge e^k$, $\star e^0 = -e^1\wedge e^2\wedge e^3$, $\star e^i = -e^0\wedge\star_3 e^i$, with $\mathrm{vol} = e^0\wedge e^1\wedge e^2\wedge e^3$ and $\star_3$ the star of $\mathbb R^3$ from the grad–curl–divergence slide ($\star_3e^1 = e^2\wedge e^3$, cyclically).

this is where the metric lives

$\wedge$, $\mathrm d$, pullback, closed, exact, cohomology, Stokes: no metric. $\star$: metric. So every metric-dependence of Maxwell's equations sits in one symbol — and in unit 32's curved spacetime only that symbol changes.

Deriving one of the frame formulas. Take $\beta = e^0\wedge e^1$ in $\eta = \operatorname{diag}(-1,1,1,1)$. Then $\langle\beta,\beta\rangle = \eta^{00}\eta^{11} = -1$. We need $\beta\wedge\star\beta = -\mathrm{vol}$. Writing $\star\beta = c\,e^2\wedge e^3$ gives $\beta\wedge\star\beta = c\,e^0\wedge e^1\wedge e^2\wedge e^3 = c\,\mathrm{vol}$, so $c = -1$ and $\star(e^0\wedge e^1) = -e^2\wedge e^3$, which is the stated formula at $i = 1$. The others go the same way; the only care needed is reordering the wedge to $e^0e^1e^2e^3$ and counting transpositions.

Full proof of the double-star formula. By linearity check on a basis element $e^I$ with $|I| = k$. Then $\star e^I = \pm e^{I^c}$ with the sign determined by $e^I\wedge e^{I^c} = \varepsilon(I)\,\mathrm{vol}$ and by the product of the $\eta^{ii}$ for $i\in I$. Applying $\star$ again reverses the roles of $I$ and $I^c$, and $e^{I^c}\wedge e^{I} = (-1)^{k(n-k)}e^I\wedge e^{I^c}$; the metric factors from $I$ and $I^c$ multiply to $\det\eta^{-1}$, whose sign is $s$. Collecting, $\star\star e^I = (-1)^{k(n-k)}s\,e^I$.

Why it matters that $\star\star = -1$ here. In Euclidean signature, $\star$ on two-forms in four dimensions squares to $+1$ and splits $\Lambda^2$ into self-dual and anti-self-dual pieces. In Lorentzian signature it squares to $-1$: no real eigenvectors, no such split, and every attempt to decompose the electromagnetic field into two independent halves over the reals fails. Complexifying is the standard way round it: $F\mp\mathrm i\star F$ are eigenforms of $\star$ with eigenvalues $\pm\mathrm i$ (apply $\star$ and use $\star\star = -1$), and each is determined by the complex vector $\vec E + \mathrm ic\vec B$ or its conjugate.

Full proof that $\star$ exists and is unique. The inner product on $\Lambda^k$ is well defined because the determinant is multilinear and alternating in the $\alpha^i$ and in the $\beta^j$, so it extends bilinearly from decomposable forms; in an orthonormal coframe the $e^I$ are orthogonal with $\langle e^I,e^I\rangle = \pm1$. The pairing $\Lambda^k\times\Lambda^{n-k}\to\Lambda^n$, $(\alpha,\gamma)\mapsto\alpha\wedge\gamma$, is nondegenerate, since $e^I\wedge e^J$ is nonzero only for $J = I^c$. So for each $\beta$ the linear functional $\alpha\mapsto\langle\alpha,\beta\rangle$ is $\alpha\mapsto\alpha\wedge\gamma/\mathrm{vol}$ for exactly one $\gamma$, and $\star\beta := \gamma$; linearity in $\beta$ follows from the uniqueness.

34 · the star

$\star F$, computed

Apply the frame formulas to $F$, with $e^0 = c\,\mathrm dt$ and $e^i = \mathrm dx^i$:

$$\begin{aligned}\star F \;=\; {}&\frac1c\big(E_x\,\mathrm dy\wedge\mathrm dz + E_y\,\mathrm dz\wedge\mathrm dx + E_z\,\mathrm dx\wedge\mathrm dy\big)\\ &-\; c\,\big(B_x\,\mathrm dx + B_y\,\mathrm dy + B_z\,\mathrm dz\big)\wedge\mathrm dt\end{aligned}$$

So $\star$ acts on the pair as $(\vec B,\ \vec E)\mapsto(\vec E/c,\ -c\vec B)$ — the duality rotation. Applying it twice gives $(-\vec B, -\vec E)$, confirming $\star\star F = -F$.

dimensions

$[\frac1cE\,\mathrm dy\wedge\mathrm dz] = \frac{\mathrm s}{\mathrm m}\frac{\mathrm V}{\mathrm m}\mathrm m^2 = \mathrm{V\,s} = \mathrm{Wb}$ and $[cB\,\mathrm dx\wedge\mathrm dt] = \frac{\mathrm m}{\mathrm s}\,\mathrm T\,\mathrm m\,\mathrm s = \mathrm{T\,m^2} = \mathrm{Wb}$. Same as $F$, as $\star$ preserves.

the current one-form

$J = -c^2\rho\,\mathrm dt + J_i\,\mathrm dx^i$, so that $$\star J = c\rho\,\mathrm dx\wedge\mathrm dy\wedge\mathrm dz - c\,J_i\,\mathrm dt\wedge\varepsilon^i .$$ This is unit 24's $\rho$ and $\vec J$ assembled with the index lowered by $\eta$.

Doing the computation. Write $F = \tfrac12F_{\mu\nu}e^\mu\wedge e^\nu$ with $F_{0i} = -E_i/c$ and $F_{ij} = \epsilon_{ijk}B_k$, apply the two frame formulas, and use $\epsilon_{ijk}\epsilon_{ijl} = 2\delta_{kl}$. The $F_{0i}$ term produces $\frac1{2c}\epsilon_{ijk}E_i\,e^j\wedge e^k$, which is the $\vec E$ flux two-form over $c$; the $F_{ij}$ term produces $B_l\,e^0\wedge e^l = c\,B_l\,\mathrm dt\wedge\mathrm dx^l = -c\,B_l\,\mathrm dx^l\wedge\mathrm dt$. That is the displayed answer.

The current one-form's odd-looking $c^2$. $J$ is defined on the slide. Raising its index with $\eta$ and $x^0 = ct$ gives the vector $J^\mu = (c\rho, \vec J)$ — the object unit 28 calls the four-current — so $J_0 = -c\rho$ and $J_0\,\mathrm dx^0 = -c\rho\cdot c\,\mathrm dt$. The factor is an artefact of writing the answer in $\mathrm dt$ rather than $\mathrm dx^0$; in $x^0$ coordinates $J = -c\rho\,\mathrm dx^0 + J_i\,\mathrm dx^i$, which is tidier and less useful for reading off components.

A check worth doing by hand. A plane wave has $\vec E\perp\vec B$ and $|\vec E| = c|\vec B|$. Then $\star F$ has electric part $\vec E' = -c\vec B$ and magnetic part $\vec B' = \vec E/c$, which again satisfies $|\vec E'| = c|\vec B'|$ and $\vec E'\perp\vec B'$: the duality maps plane waves to plane waves, rotating the polarisation by a quarter turn. Such a field has both invariants zero — the last slide of this section computes them.

34 · the star

Worked example 4 — $\mathrm d\star F = \mu_0\star J$, expanded

example 4 · the inhomogeneous pair

Differentiate the two pieces of $\star F$, exactly as in example 2:

$$\mathrm d\star F = \frac{\nabla\cdot\vec E}{c}\,\mathrm dx\wedge\mathrm dy\wedge\mathrm dz + \Big[\frac1c\partial_t\vec E - c\,\nabla\times\vec B\Big]_i\,\mathrm dt\wedge\varepsilon^i$$

and compare with $\mu_0\star J = \mu_0c\rho\,\mathrm dx\wedge\mathrm dy\wedge\mathrm dz - \mu_0c\,J_i\,\mathrm dt\wedge\varepsilon^i$.

Volume part. $\dfrac{\nabla\cdot\vec E}{c} = \mu_0c\rho \iff \nabla\cdot\vec E = \mu_0c^2\rho = \dfrac{\rho}{\varepsilon_0}$, using $\mu_0\varepsilon_0c^2 = 1$.

Mixed part. $\dfrac1c\partial_tE_i - c(\nabla\times\vec B)_i = -\mu_0cJ_i$; divide by $-c$: $\ (\nabla\times\vec B)_i - \dfrac1{c^2}\partial_tE_i = \mu_0J_i$.

Gauss and Ampère–Maxwell, both signs, nothing left over

Including unit 24's displacement current, which here is not an extra term but the $\mathrm dt$ component of a single four-dimensional equation.

The two derivatives. $\mathrm d$ of the flux two-form $\frac1c E_i\varepsilon^i$ gives $\frac1c(\nabla\cdot\vec E)\mathrm dx\wedge\mathrm dy\wedge\mathrm dz + \frac1c\partial_tE_i\,\mathrm dt\wedge\varepsilon^i$, by the computation in example 2's notes with $\vec B$ replaced by $\vec E/c$. And $\mathrm d$ of $-cB_i\mathrm dx^i\wedge\mathrm dt = -c\,(\nabla\times\vec B)_i\,\varepsilon^i\wedge\mathrm dt = -c(\nabla\times\vec B)_i\,\mathrm dt\wedge\varepsilon^i$.

Dimensions. $[\mathrm d\star F] = [\star F] = \mathrm{Wb} = \mathrm{T\,m^2}$; $[\star J] = (\mathrm{m/s})(\mathrm{C/m^3})\mathrm m^3 = \mathrm{A\,m}$, and $\mu_0[\star J] = (\mathrm{T\,m\,A^{-1}})(\mathrm{A\,m}) = \mathrm{T\,m^2}$. They agree.

Special case. Statics with no current: $\mathrm d\star F = 0$ with $\star F = \frac1c\,\vec E$-flux, giving $\nabla\cdot\vec E = 0$ — unit 23's Laplace equation for the potential, since $F = \mathrm dA$ makes $\vec E = -\nabla\varphi$. And with no charge anywhere, $F$ closed and co-closed is the four-dimensional harmonic condition whose solutions are unit 24's plane waves.

Why "four into two" is more than counting. The first pair, $\mathrm dF = 0$, is a statement about a two-form on a bare manifold. The second, $\mathrm d\star F = \mu_0\star J$, is the same statement about the dual — and the dual is the only place the metric, and hence the causal structure, appears. That is the cleanest available answer to the question of which part of Maxwell's equations knows about the geometry of spacetime — the geometry unit 26 found hiding in the wave operator. Only the star does.

34 · the star

Charge conservation is $\mathrm d^2 = 0$, again

theorem

$\mathrm d\star F = \mu_0\star J$ forces $\mathrm d\star J = 0$, which in components is $\partial_t\rho + \nabla\cdot\vec J = 0$.

proof

$\mu_0\,\mathrm d\star J = \mathrm d\mathrm d\star F = 0$. $\square$

In components: $\mathrm d\star J = c\big(\partial_t\rho + \nabla\cdot\vec J\big)\, \mathrm dt\wedge\mathrm dx\wedge\mathrm dy\wedge\mathrm dz$, and the coefficient must vanish.

the symmetry of the argument

$\mathrm dF = 0$ is $\mathrm d\mathrm dA = 0$: no magnetic charge, by construction.
$\mathrm d\star J = 0$ is $\mathrm d\mathrm d\star F = 0$: electric charge conserved, by construction. Two identities, one theorem of Clairaut's, applied to the two halves.

unit 24's argument, inverted

There, charge conservation was the input and the displacement current the output. Here the displacement current is inside $\star F$ from the start and conservation is the output. Same content, opposite direction — and here it is forced, not chosen.

Stokes on the slab: $\mathrm d\star J = 0$ inside, so equal charge crosses $\Sigma_1$ and $\Sigma_2$.

The component computation. $\mathrm d(c\rho\,\mathrm dx\wedge\mathrm dy\wedge\mathrm dz) = c\,\partial_t\rho\;\mathrm dt\wedge\mathrm dx\wedge\mathrm dy\wedge\mathrm dz$. And $\mathrm d(-cJ_i\,\mathrm dt\wedge\varepsilon^i) = -c\,\partial_jJ_i\,\mathrm dx^j\wedge\mathrm dt\wedge\varepsilon^i$; only $j = i$ survives, $\mathrm dx^i\wedge\mathrm dt\wedge\varepsilon^i = -\mathrm dt\wedge\mathrm dx^i\wedge\varepsilon^i = -\mathrm dt\wedge\mathrm dx\wedge\mathrm dy\wedge\mathrm dz$, so the term is $+c(\nabla\cdot\vec J)\,\mathrm dt\wedge\mathrm dx\wedge\mathrm dy\wedge\mathrm dz$. Adding gives the stated result.

The conserved charge, properly. Integrate $\mathrm d\star J = 0$ over the region between two spacelike slices and apply Stokes: the flux of $\star J$ through the two slices agrees, and $\int_{\Sigma}\star J = c\int_\Sigma\rho\,\mathrm dV = cQ$. So $Q$ is the period of a closed three-form — a conserved charge is a cohomological statement, which is the form-language version of unit 05's local-to-global conversion by the divergence theorem.

Reading the figure. The slab $V$ between the slices $t = t_1$ and $t = t_2$ has boundary $\Sigma_2$ minus $\Sigma_1$, plus the far-away sides, where the current is assumed to vanish. On each slice $\mathrm dt$ pulls back to zero, so only the $c\rho\,\mathrm dx\wedge\mathrm dy\wedge\mathrm dz$ part of $\star J$ survives, and its integral is $cQ$. Stokes, $0 = \int_V\mathrm d\star J = \int_{\Sigma_2}\star J - \int_{\Sigma_1}\star J$, says the charge at $t_2$ equals the charge at $t_1$, however the charge has moved in between.

What would break. If one wrote $\mathrm d\star F = \mu_0\star J$ with a $\star J$ that was not closed, the system would be inconsistent — no solution for any $F$. This is the sharpest version of unit 24's discovery: Ampère's law as Ampère left it corresponds to a $\star J$ that is not closed for time-dependent sources, so no $F$ solves it, and the repair is unique.

34 · the star

The two invariants, and a number

$\wedge$ and $\star$ use only the metric and the orientation, both preserved by unit 26's proper Lorentz transformations. So any scalar built from them is frame-independent, and there are exactly two of degree two in $F$:

$$\begin{aligned}F\wedge F &= -\frac2c\,(\vec E\cdot\vec B)\,\mathrm{vol},\\ F\wedge\star F &= \Big(|\vec B|^2 - \frac{|\vec E|^2}{c^2}\Big)\mathrm{vol} .\end{aligned}$$

Obtained without a single boost: the invariance is a property of $\wedge$ and $\star$, not a computation.

a number

$|\vec E| = 2.50\times10^5\ \mathrm{V\,m^{-1}}$ perpendicular to $|\vec B| = 1.00\times10^{-3}$ T. Then $\vec E\cdot\vec B = 0$, and $$\begin{aligned}|\vec B|^2 - |\vec E|^2/c^2 &= 1.000\times10^{-6} - 6.954\times10^{-7}\\ &= 3.046\times10^{-7}\ \mathrm{T^2},\end{aligned}$$ positive: magnetically dominated, in every frame.

limit check

A plane wave has $|\vec E| = c|\vec B|$ and $\vec E\perp\vec B$, so both invariants vanish — a null field, and no frame makes it purely electric or purely magnetic.

Computing $F\wedge F$. With $F = \beta + \epsilon\wedge\mathrm dt$, where $\beta$ is the $\vec B$ flux two-form and $\epsilon = E_i\mathrm dx^i$: $\beta\wedge\beta = 0$ (every term repeats a factor) and $(\epsilon\wedge\mathrm dt)\wedge(\epsilon\wedge\mathrm dt) = 0$ (two $\mathrm dt$s), so $F\wedge F = 2\beta\wedge\epsilon\wedge\mathrm dt = 2(\vec E\cdot\vec B)\, \mathrm dx\wedge\mathrm dy\wedge\mathrm dz\wedge\mathrm dt$. Since $\mathrm{vol} = e^0\wedge e^1\wedge e^2\wedge e^3 = c\,\mathrm dt\wedge\mathrm dx\wedge\mathrm dy\wedge\mathrm dz$ and moving $\mathrm dt$ to the back costs three transpositions, $\mathrm dx\wedge\mathrm dy\wedge\mathrm dz\wedge\mathrm dt = -\mathrm{vol}/c$, giving the stated $-\frac2c(\vec E\cdot\vec B)\mathrm{vol}$.

Computing $F\wedge\star F$. Same decomposition: the flux-times-flux terms vanish, leaving $-c\,\beta\wedge b\wedge\mathrm dt + \frac1c(\epsilon\wedge\mathrm dt)\wedge\eta$ with $b = B_i\mathrm dx^i$ and $\eta$ the $\vec E$ flux two-form. The first is $-c|\vec B|^2\mathrm dx\wedge\mathrm dy\wedge\mathrm dz\wedge\mathrm dt = |\vec B|^2\mathrm{vol}$ and the second is $\frac1c|\vec E|^2\mathrm dx\wedge\mathrm dy\wedge\mathrm dz\wedge\mathrm dt = -\frac{|\vec E|^2}{c^2}\mathrm{vol}$.

The numbers. $c = 2.99792458\times10^8\ \mathrm{m\,s^{-1}}$ exactly, so $|\vec E|/c = 2.50\times10^5/2.99792458\times10^8 = 8.3391\times10^{-4}$ T and $(|\vec E|/c)^2 = 6.9541\times10^{-7}\ \mathrm{T^2}$. Subtracting from $1.00\times10^{-6}$ gives $3.046\times10^{-7}\ \mathrm{T^2}$.

Full proof of the invariance. A proper Lorentz transformation $L$ is a linear map with $L^*\eta = \eta$ and $\det L = +1$, so $L^*\mathrm{vol} = \mathrm{vol}$ (the pullback of a top form multiplies it by $\det L$, as the integration slide showed). The inner product on two-forms is built from $\eta$ alone, so $\langle L^*\alpha, L^*\beta\rangle = \langle\alpha,\beta\rangle\circ L$, and then the defining identity $\alpha\wedge\star\beta = \langle\alpha,\beta\rangle\mathrm{vol}$ forces $L^*(\star\beta) = \star(L^*\beta)$. Hence $L^*(F\wedge\star F) = L^*F\wedge\star L^*F$: the new observer's field gives the same coefficient of $\mathrm{vol}$ at the same event. The same holds for $F\wedge F$ without using $\star$ at all.

Parity. A reflection has $\det L = -1$ and reverses the orientation. $F\wedge\star F$ survives (both $\star$ and $\mathrm{vol}$ flip, and the signs cancel) but the coefficient of $F\wedge F$ changes sign: $\vec E\cdot\vec B$ is a pseudoscalar, for the reason the notes on grad, curl and divergence gave for the curl.

Why only two. $F\wedge F$ and $F\wedge\star F$ are the only scalars quadratic in $F$ that can be built from $\wedge$, $\star$ and nothing else: the other candidates reduce to them, $\star F\wedge F = F\wedge\star F$ (two-forms commute) and $\star F\wedge\star F = -F\wedge F$ (put $\alpha = \star F$, $\beta = F$ in the defining identity, then use $\star\star = -1$). Higher wedge powers vanish in four dimensions because $F\wedge F$ is already a top form. If you have taken unit 28, these are its $F_{\mu\nu}F^{\mu\nu}$ and $F_{\mu\nu}\widetilde F^{\mu\nu}$ up to constant factors — the same objects in index notation.

section 6

Cartan's moving frames

Unit 33 computed curvature from Christoffel symbols — forty of them in four dimensions. Here it takes a coframe and two equations.

34 · Cartan

Coframes and the first structure equation

Take an orthonormal coframe $\{e^a\}$: one-forms with $g = \eta_{ab}\,e^a\otimes e^b$, $\eta$ the flat metric of the right signature. Unit 32's connection appears as one-forms $\omega^a{}_b = \Gamma^a{}_{cb}\,e^c$.

Cartan's first structure equation

A torsion-free metric connection has connection one-forms $\omega^a{}_b$ determined uniquely by $$\mathrm de^a = -\,\omega^a{}_b\wedge e^b,\qquad \omega_{ab} = -\,\omega_{ba},$$ where $\omega_{ab} = \eta_{ac}\omega^c{}_b$.

uniqueness · proof

If $\omega$ and $\omega'$ both work, $\lambda^a{}_b = \omega^a{}_b-\omega'^a{}_b$ has $\lambda^a{}_b\wedge e^b = 0$ and $\lambda_{ab} = -\lambda_{ba}$. Writing $\lambda_{ab} = \lambda_{abc}e^c$, the first gives $\lambda_{abc} = \lambda_{acb}$: symmetric in the last two, antisymmetric in the first two. Then $\lambda_{abc} = \lambda_{acb} = -\lambda_{cab} = -\lambda_{cba} = \lambda_{bca} = \lambda_{bac} = -\lambda_{abc}$, so $\lambda = 0$. $\square$

the arithmetic

In four dimensions: $\Gamma^\lambda{}_{\mu\nu}$ has $4\times10 = 40$ independent components. $\omega_{ab}$, antisymmetric, has $\binom42 = 6$ one-forms. And they are usually read off from $\mathrm de^a$ by inspection, not solved for.

Where the two conditions come from. Antisymmetry of $\omega_{ab}$ is metric compatibility: $\nabla g = 0$ with $g = \eta_{ab}e^a\otimes e^b$ and $\eta$ constant forces $\eta_{cb}\omega^c{}_a + \eta_{ac}\omega^c{}_b = 0$. The first structure equation is vanishing torsion: unit 32 defined $T(X,Y) = \nabla_XY - \nabla_YX - [X,Y]$, and expanding it in the frame with the invariant formula for $\mathrm d$ from section 2 gives $T^a = \mathrm de^a + \omega^a{}_b\wedge e^b$. Setting $T^a = 0$ is the equation on the slide.

The uniqueness chain, read slowly. Each step uses one of the two symmetries alternately: symmetric in $(bc)$, antisymmetric in $(ab)$, symmetric, antisymmetric, symmetric, antisymmetric. Six steps return to $\lambda_{abc}$ with a minus sign. It is the same argument that gives uniqueness of the Christoffel symbols in unit 32 — that proof cycled $\Gamma_{\lambda\mu\nu}$ through its index symmetries in exactly this way.

Existence follows by solving: the same index cycling, run forwards rather than towards a contradiction, gives $\omega_{abc} = \frac12(c_{cab} - c_{abc} - c_{bca})$, where $\mathrm de^a = -\frac12c^a{}_{bc}\,e^b\wedge e^c$, $c_{abc} = \eta_{ad}c^d{}_{bc}$ and $\omega_{ab} = \omega_{abc}e^c$. Check it on example 5 below: there $c^2{}_{12} = -\cot\vartheta/a$, and the formula gives $\omega_{212} = \cot\vartheta/a$, i.e. $\omega^2{}_1 = \cot\vartheta\,e^2/a = \cos\vartheta\,\mathrm d\phi$, which is what inspection finds. In practice nobody uses the formula: one computes $\mathrm de^a$, guesses $\omega$, and checks. The guess is forced because uniqueness has been proved.

34 · Cartan

The second structure equation, and Bianchi for free

Cartan's second structure equation

The curvature two-forms $\Omega^a{}_b := \mathrm d\omega^a{}_b + \omega^a{}_c\wedge\omega^c{}_b$ carry unit 33's Riemann tensor in the frame: $$\Omega^a{}_b = \tfrac12R^a{}_{bcd}\,e^c\wedge e^d .$$

both Bianchi identities, on the slide

Take $\mathrm d$ of the first structure equation: $0 = -\mathrm d\omega^a{}_b\wedge e^b + \omega^a{}_b\wedge\mathrm de^b$, and substituting both structure equations leaves $\ \Omega^a{}_b\wedge e^b = 0$ — the first Bianchi identity.
Take $\mathrm d$ of the second: the cubic terms cancel in pairs and $\ \mathrm d\Omega^a{}_b = \Omega^a{}_c\wedge\omega^c{}_b - \omega^a{}_c\wedge\Omega^c{}_b$ — the second, whose contraction is what makes unit 33's Einstein tensor divergence-free.

the counting

Coordinates: $R^\rho{}_{\sigma\mu\nu}$ from $\partial\Gamma-\partial\Gamma+\Gamma\Gamma-\Gamma\Gamma$, with 40 Christoffels to produce first.
Cartan: $\binom n2$ curvature forms, each one $\mathrm d$ and one wedge. In four dimensions, 6 against 40.

Full proof of the second structure equation. The curvature of unit 33 is $R(X,Y)Z = \nabla_X\nabla_YZ - \nabla_Y\nabla_XZ - \nabla_{[X,Y]}Z$. Write $\nabla e_b = \omega^a{}_b\otimes e_a$ for the frame fields. Then for the frame vectors, $R(X,Y)e_b = \big(\mathrm d\omega^a{}_b(X,Y) + (\omega^a{}_c\wedge\omega^c{}_b)(X,Y)\big)e_a$ — the $\mathrm d\omega$ term coming from the invariant formula of section 2 (this is exactly where the $-\alpha([X,Y])$ term earns its keep) and the wedge term from the two nested connections. Matching against $R(e_c,e_d)e_b = R^a{}_{bcd}e_a$ and the definition $\Omega^a{}_b(e_c,e_d) = R^a{}_{bcd}$ gives the equation, with the $\tfrac12$ from the independent-index convention.

The cubic cancellation, in full. $\mathrm d\Omega^a{}_b = \mathrm d\omega^a{}_c\wedge\omega^c{}_b - \omega^a{}_c\wedge\mathrm d\omega^c{}_b$ (the sign from moving $\mathrm d$ past a one-form). Substitute $\mathrm d\omega = \Omega - \omega\wedge\omega$ in both places: $(\Omega^a{}_c - \omega^a{}_e\wedge\omega^e{}_c)\wedge\omega^c{}_b - \omega^a{}_c\wedge(\Omega^c{}_b - \omega^c{}_e\wedge\omega^e{}_b)$. The two triple products are $-\omega^a{}_e\wedge\omega^e{}_c\wedge\omega^c{}_b$ and $+\omega^a{}_c\wedge\omega^c{}_e\wedge\omega^e{}_b$, identical after relabelling $e\leftrightarrow c$, and they cancel.

The first Bianchi identity in this language is $R^a{}_{[bcd]} = 0$, unit 33's cyclic identity; here it is one line and holds only because the connection is torsion-free, which is visible from the derivation. That dependence is easy to lose in the coordinate proof.

34 · Cartan

Worked example 5 — the two-sphere, twice

example 5 · Cartan against unit 33's Christoffels

Cartan. $\mathrm ds^2 = a^2(\mathrm d\vartheta^2 + \sin^2\!\vartheta\,\mathrm d\phi^2)$, coframe $e^1 = a\,\mathrm d\vartheta$, $e^2 = a\sin\vartheta\,\mathrm d\phi$.

$\mathrm de^1 = 0$; $\mathrm de^2 = a\cos\vartheta\,\mathrm d\vartheta\wedge\mathrm d\phi$.

Need $\mathrm de^2 = -\omega^2{}_1\wedge e^1$. Try $\omega^2{}_1 = \cos\vartheta\,\mathrm d\phi$: $-\cos\vartheta\,\mathrm d\phi\wedge a\,\mathrm d\vartheta = a\cos\vartheta\,\mathrm d\vartheta\wedge\mathrm d\phi$ ✓. And $\mathrm de^1 = -\omega^1{}_2\wedge e^2 = \cos\vartheta\,\mathrm d\phi\wedge a\sin\vartheta\,\mathrm d\phi = 0$ ✓.

In two dimensions the $\omega\wedge\omega$ term vanishes, so $\Omega^2{}_1 = \mathrm d\omega^2{}_1 = -\sin\vartheta\,\mathrm d\vartheta\wedge\mathrm d\phi = -\frac1{a^2}\,e^1\wedge e^2$, giving $R^2{}_{112} = -1/a^2$ and $K = R_{1212} = 1/a^2$.

Unit 33's route. Invert $g$; compute six $\Gamma^a{}_{bc}$, of which $\Gamma^\vartheta{}_{\phi\phi} = -\sin\vartheta\cos\vartheta$ and $\Gamma^\phi{}_{\vartheta\phi} = \cot\vartheta$ survive; then $R^\vartheta{}_{\phi\vartheta\phi} = \partial_\vartheta\Gamma^\vartheta{}_{\phi\phi} - \Gamma^\vartheta{}_{\phi\phi}\Gamma^\phi{}_{\vartheta\phi} = -\cos2\vartheta + \cos^2\!\vartheta = \sin^2\!\vartheta$; then lower and divide: $K = R_{\vartheta\phi\vartheta\phi}/(g_{\vartheta\vartheta}g_{\phi\phi}) = 1/a^2$.

the saving, counted

Cartan: 2 exterior derivatives, 1 inspection, 1 more exterior derivative — 4 steps.
Coordinates: 2 inverse-metric entries, 18 derivative terms across 6 Christoffels, 6 terms in the Riemann formula, 2 index moves — 28.

Dimensions. $[K] = \mathrm m^{-2}$: for the Earth, $a = 6.371\times10^6$ m gives $K = 2.4637\times10^{-14}\ \mathrm m^{-2}$. Limit. $a\to\infty$ gives $K\to0$: a plane.

Reading $K$ off $\Omega$. $\Omega^2{}_1 = \tfrac12R^2{}_{1cd}e^c\wedge e^d = R^2{}_{112}\,e^1\wedge e^2$, so $R^2{}_{112} = -1/a^2$. Lowering with $\delta$: $R_{2112} = -1/a^2$, and antisymmetry in the first pair gives $R_{1212} = +1/a^2$. In an orthonormal frame in two dimensions the Gaussian curvature is exactly $R_{1212}$, which unit 33 proved via the theorema egregium. Unit 33 also verified this against the holonomy of a spherical triangle: the rotation angle equals the enclosed area times $1/a^2$.

Checking the coordinate route's middle step. $\partial_\vartheta(-\sin\vartheta\cos\vartheta) = -(\cos^2\vartheta - \sin^2\vartheta) = -\cos2\vartheta$, and $-\Gamma^\vartheta{}_{\phi\phi}\Gamma^\phi{}_{\vartheta\phi} = \sin\vartheta\cos\vartheta\cdot\cot\vartheta = \cos^2\vartheta$. Sum: $-\cos2\vartheta + \cos^2\vartheta = -\cos^2\vartheta + \sin^2\vartheta + \cos^2\vartheta = \sin^2\vartheta$. Then $R_{\vartheta\phi\vartheta\phi} = g_{\vartheta\vartheta}R^\vartheta{}_{\phi\vartheta\phi} = a^2\sin^2\vartheta$ and dividing by $g_{\vartheta\vartheta}g_{\phi\phi} = a^4\sin^2\vartheta$ leaves $1/a^2$.

How the step counts were made, so you can disagree with them. A "step" is one differentiation or one product that has to be written down. Cartan: $\mathrm de^1$, $\mathrm de^2$, the inspection that fixes $\omega^2{}_1$, $\mathrm d\omega^2{}_1$. Coordinates: $g^{\vartheta\vartheta}$ and $g^{\phi\phi}$; six symbols $\Gamma^a{}_{bc}$ each needing the three-term formula $\frac12g^{ad}(\partial_bg_{cd}+\partial_cg_{bd}-\partial_dg_{bc})$, eighteen derivative terms in all; the Riemann formula, which with $\lambda$ summed over two values has six terms (two derivatives and four products); then two index manipulations. The ratio matters more than the exact count, and it grows with dimension: the problem sheet's Cartan versus Christoffel (P10) repeats the count on the hyperbolic plane, and P11 on the three-sphere, where the gap is wider.

The number. With the Earth's mean radius $a = 6.371\times10^6$ m, $K = 1/a^2 = 2.4637\times10^{-14}\ \mathrm m^{-2}$ — which is also, by Gauss, the quantity a surveyor measures as the excess of a triangle's angle sum over $\pi$ divided by its area.

34 · Cartan

Gauss–Bonnet, from Stokes and one connection form

theorem · local Gauss–Bonnet

Let $D$ be a closed disc in one chart of an oriented surface, bounded by a smooth simple loop run anticlockwise at unit speed, with geodesic curvature $\kappa_g$. Then $$\int_DK\,\mathrm dA + \oint_{\partial D}\kappa_g\,\mathrm ds = 2\pi .$$

proof

Gram–Schmidt on $\partial_1,\partial_2$ gives an orthonormal frame $e_1, e_2$ on $D$. In two dimensions $\omega\wedge\omega = 0$, so $\mathrm d\omega^1{}_2 = \Omega^1{}_2 = K\,\mathrm dA$. Write the unit tangent as $T = \cos\psi\,e_1 + \sin\psi\,e_2$, so $\psi$ is its angle from $e_1$; then $\kappa_g = \psi' - \omega^1{}_2(T)$ (notes). Integrate and use Stokes: $\oint\kappa_g\,\mathrm ds = \Delta\psi - \int_DK\,\mathrm dA$, and $\Delta\psi = 2\pi$. $\square$

The frame of the proof, turning slowly across $D$, and at one boundary point the tangent $T$, its angle $\psi$ from $e_1$, and the inward normal $N$. Once round, $\psi$ gains $2\pi$.

The formula for $\kappa_g$. The geodesic curvature of a unit-speed curve is $\kappa_g = \langle\nabla_TT, N\rangle$ with $N = -\sin\psi\,e_1 + \cos\psi\,e_2$ the inward normal for an anticlockwise loop. Using $\nabla e_b = \omega^a{}_b\otimes e_a$, $\nabla_TT = \psi'N + \cos\psi\,\omega^2{}_1(T)\,e_2 + \sin\psi\,\omega^1{}_2(T)\,e_1 = \big(\psi' + \omega^2{}_1(T)\big)N$, since $\omega^1{}_2 = -\omega^2{}_1$. So $\kappa_g = \psi' - \omega^1{}_2(T)$. A geodesic has $\kappa_g = 0$ — unit 32's auto-parallel condition — and then $\psi' = \omega^1{}_2(T)$ says the tangent turns relative to the frame exactly as fast as the frame turns relative to parallel transport.

What is cited. $\Delta\psi = 2\pi$: the tangent of a simple closed loop turns once, measured against a frame defined on the whole disc. This is Hopf's theorem of turning tangents (cited: do Carmo, Differential Geometry of Curves and Surfaces, §5-7); in the flat plane with the Cartesian frame it is the familiar fact that walking round a simple loop turns you through one full turn. Everything else is Stokes plus the second structure equation.

Reading the figure. $\psi'$ is how fast the tangent turns relative to the frame, and $\omega^1{}_2(T)$ is how fast the frame itself turns relative to parallel transport along the curve; $\kappa_g$, the turning relative to parallel transport, is their difference. Round the whole loop the first adds up to $2\pi$, the second to $\int_DK\,\mathrm dA$ by Stokes, and that is the theorem.

This is the theorem unit 32 cited (do Carmo, §4-5), now proved in its local form. Unit 32 found that a vector transported round a circle of latitude comes back turned by $2\pi(1-\cos\vartheta_0)$ modulo a full turn, and noted that Gauss–Bonnet, with the geodesic curvature of the boundary counted, is what resolves the full-turn ambiguity. Here it is: the rotation relative to the tangent is $\oint\kappa_g\,\mathrm ds$, the tangent itself turns through $2\pi$, and the difference is $\int K\,\mathrm dA$.

34 · Cartan

Gauss–Bonnet on a polar cap

check · a polar cap

Cap $\vartheta\le\vartheta_0$ of a sphere: $\int K\,\mathrm dA = 2\pi(1-\cos\vartheta_0)$; rim $\kappa_g = \cot\vartheta_0/a$ over length $2\pi a\sin\vartheta_0$ gives $2\pi\cos\vartheta_0$. Sum $2\pi$. Limit $\vartheta_0\to\pi/2$: the equator is a geodesic. Number: the Arctic Circle, $\vartheta_0 = 23.44^\circ$, has $\int K\,\mathrm dA = 0.5185$ rad $= 29.71^\circ$ — the turn of a vector parallel-transported once round it.

the vortex again

The flat plane's polar frame has $\omega^2{}_1 = \mathrm d\phi = \theta$. Closed because $K = 0$; period $2\pi$ because the frame turns once round the origin.

The Arctic Circle's cap, to scale. Curvature inside contributes $0.5185$, the rim's geodesic curvature $5.7647$; they sum to $2\pi = 6.2832$.

Checking the cap's rim. On the sphere, example 5's frame gives $\omega^1{}_2 = -\cos\vartheta\,\mathrm d\phi$. That frame is singular at the pole, so it is not the frame of the proof, but $\kappa_g$ does not depend on the frame and can be computed with it. The rim $\vartheta = \vartheta_0$ run with $\phi$ increasing has $T = e_2$, so $\psi = \pi/2$ is constant and $\kappa_g = -\omega^1{}_2(e_2) = \cos\vartheta_0/(a\sin\vartheta_0)$. The cap area is $2\pi a^2(1-\cos\vartheta_0)$ and $K = 1/a^2$.

The number. The Arctic Circle sits at latitude $66.56^\circ$ N, so $\vartheta_0 = 23.44^\circ$ and $2\pi(1-\cos23.44^\circ) = 0.5185$ rad $= 29.71^\circ$. The rim's part is $2\pi\cos23.44^\circ = 5.7647$ rad, and the two sum to $6.2832 = 2\pi$. Dimensions: $K$ in $\mathrm m^{-2}$ times area in $\mathrm m^2$, and $\kappa_g$ in $\mathrm m^{-1}$ times length in $\mathrm m$ — both angles, as $2\pi$ is.

The vortex, as a connection form. On the punctured plane take $e^1 = \mathrm dr$, $e^2 = r\,\mathrm d\phi$. Then $\mathrm de^1 = 0$ and $\mathrm de^2 = \mathrm dr\wedge\mathrm d\phi$, and $\omega^2{}_1 = \mathrm d\phi$ satisfies $\mathrm de^2 = -\omega^2{}_1\wedge e^1$. That $\mathrm d\phi$ is the vortex $\theta$ of example 3. Its closedness is the flatness of the plane, and its period $2\pi$ is the polar frame turning once as you go round the origin — the reason this frame, unlike the Cartesian one, cannot be extended across the puncture.

34 · Cartan

Liouville's theorem, in one line

two operations

The interior product $i_X:\Omega^k\to\Omega^{k-1}$ is $(i_X\omega)(v_1,\dots,v_{k-1}) = \omega(X, v_1,\dots,v_{k-1})$. Cartan's magic formula relates it to unit 31's Lie derivative: $$\mathcal L_X = \mathrm d\circ i_X + i_X\circ\mathrm d .$$

Liouville

Unit 14's $\omega = \sum_i\mathrm dp_i\wedge\mathrm dq_i = \mathrm d\big(\sum_ip_i\,\mathrm dq_i\big)$ is closed, and the Hamiltonian field satisfies $i_{X_H}\omega = -\mathrm dH$. So $$\mathcal L_{X_H}\omega = \mathrm d(i_{X_H}\omega) + i_{X_H}(\mathrm d\omega) = -\mathrm d\mathrm dH + 0 = 0 ,$$ and hence $\mathcal L_{X_H}\,\omega^{\wedge n} = 0$. Since $\omega^{\wedge n}/n!$ is the phase-space volume form, the flow preserves volume. $\square$

against unit 14

There the proof was: compute $\nabla\cdot(\dot q,\dot p) = \partial_q\partial_pH - \partial_p\partial_qH = 0$. Same Clairaut, half a page, and no route to the stronger statement that every $\omega^{\wedge k}$ is preserved — Poincaré's integral invariants, which the divergence argument cannot see.

Full proof of the magic formula. Both sides are $\mathbb R$-linear, degree-preserving, and derivations of the wedge algebra: $\mathcal L_X$ because the flow pullback is an algebra map, and $\mathrm di_X + i_X\mathrm d$ because $i_X$ is an antiderivation of degree $-1$ and $\mathrm d$ one of degree $+1$, so their anticommutator is a derivation of degree 0. Both commute with $\mathrm d$: $\mathcal L_X\mathrm d = \mathrm d\mathcal L_X$ since pullback does, and $(\mathrm di_X + i_X\mathrm d)\mathrm d = \mathrm di_X\mathrm d = \mathrm d(\mathrm di_X + i_X\mathrm d)$. They agree on functions: $\mathcal L_Xf = Xf = i_X\mathrm df$, and $\mathrm di_Xf = 0$ as $i_X$ kills $0$-forms. Two derivations that commute with $\mathrm d$ and agree on functions agree on every $f\,\mathrm dg^1\wedge\cdots\wedge\mathrm dg^k$, hence everywhere, since forms are locally sums of these. $\square$

The sign in $i_{X_H}\omega = -\mathrm dH$. With unit 14's convention $\omega = \sum\mathrm dp_i\wedge\mathrm dq_i$ and Hamilton's equations $\dot q = \partial H/\partial p$, $\dot p = -\partial H/\partial q$, one computes $\omega(X_H, v) = (X_H)_pv_q - (X_H)_qv_p = -H_qv_q - H_pv_p = -\mathrm dH(v)$. Texts that write $\omega = \sum\mathrm dq_i\wedge\mathrm dp_i$ get $+\mathrm dH$. Either way $\mathcal L_{X_H}\omega = 0$, because $\mathrm d$ of a sign is still zero.

Poincaré's integral invariants. $\mathcal L_{X_H}\omega^{\wedge k} = 0$ for every $k$, so the integral of $\omega^{\wedge k}$ over any $2k$-dimensional surface carried along by the flow is constant. $k = n$ is Liouville; $k = 1$ says the sum of the projected areas $\oint\sum p_i\,\mathrm dq_i$ over a loop is conserved (by Stokes, the loop integral is the integral of $\omega$ over a surface it bounds). The area-preserving return maps of unit 17, for those who take that branch, rest on exactly this invariant.

Unit 14's generating functions are $\mathrm d^2 = 0$. Unit 14 built a canonical transformation $(q,p)\mapsto(Q,P)$ from a function $F_1(q,Q)$ by $p_i = \partial F_1/\partial q_i$, $P_i = -\partial F_1/\partial Q_i$. In one line that is $\sum p_i\,\mathrm dq_i - \sum P_i\,\mathrm dQ_i = \mathrm dF_1$. Apply $\mathrm d$: $\sum\mathrm dp_i\wedge\mathrm dq_i - \sum\mathrm dP_i\wedge\mathrm dQ_i = \mathrm d\mathrm dF_1 = 0$, so the map preserves $\omega$ — it is symplectic. Conversely, if a map preserves $\omega$ then $\sum p\,\mathrm dq - \sum P\,\mathrm dQ$ is closed, and on a star-shaped region the Poincaré lemma makes it $\mathrm d$ of something: a generating function exists locally, of type $F_1(q,Q)$ wherever $(q,Q)$ can serve as coordinates (for the identity map they cannot, and one of unit 14's other types is used). Both directions of unit 14's theorem are $\mathrm d^2 = 0$ and its converse.

34 · Cartan

What used a metric, and what did not

statement	needs	survives a change of metric?
$\mathrm dF = 0$, $F = \mathrm dA$, gauge freedom, $H^k$	a smooth manifold	yes — including the move to curved spacetime in unit 32
Stokes, periods	+ an orientation	yes
$\mathcal L_X$, $i_X$, Liouville	a manifold and a symplectic form	yes; no metric ever entered mechanics
$\star$, $\mathrm d\star F = \mu_0\star J$, the invariants	+ a metric	no — this is the equation that feels gravity
$\omega^a{}_b$, $\Omega^a{}_b$, curvature	+ a connection	no

the running example, closed

Maxwell's four equations are $\mathrm dF = 0$ and $\mathrm d\star F = \mu_0\star J$, expanded back to components with every sign intact. The vortex $\theta$ is a generator of $H^1(\mathbb R^2\setminus\{0\}) = \mathbb R$ with period $2\pi$ — unit 04's number, now a measurement of a hole — and the monopole is the same statement in degree two, obstructing $F = \mathrm dA$.

The first row is unit 32's remark, explained. That $\mathrm dF = 0$ needs no metric is why minimal coupling has nothing to do for the homogeneous pair: there is no comma to turn into a semicolon, because there was never a partial derivative in the statement. The inhomogeneous pair does change, and it changes only through $\star$ and $\mathrm{vol}_g$ — a factor of $\sqrt{-g}$, which in components is the curved-space Maxwell equation $\partial_\nu(\sqrt{-g}F^{\mu\nu}) = \mu_0\sqrt{-g}J^\mu$ — with this unit's $F_{0i} = -E_i/c$ the derivative goes on the second index, as the flat case $\partial_iF^{0i} = \nabla\cdot\vec E/c = \mu_0c\rho$ confirms (cited: MTW ch. 22, where it is derived by minimal coupling; the form version makes it one symbol).

The third row is the reason Hamiltonian mechanics never needed a metric. Unit 14's phase space has no natural inner product — $q$ and $p$ have different dimensions, so there is nothing to compare them with — and everything it proved used only $\omega$. Its canonical transformations and Liouville's theorem are corollaries of that one fact.

Where this unit stops. Not covered: the global Gauss–Bonnet theorem with its Euler characteristic, the Hodge decomposition theorem, the Hodge Laplacian's kernel computing $H^k$ on a compact Riemannian manifold, characteristic classes, and the Chern–Weil theory that reads curvature invariants as cohomology classes. Nothing later in this course depends on them.

34 · checkpoint

Done when

checkpoint

You express Maxwell's four equations as $\mathrm dF = 0$ and $\mathrm d\star F = \mu_0\star J$, expand them back into components recovering the originals exactly, compute $H^1(\mathbb R^2\setminus\{0\})$ — closing unit 04's open question — and reproduce one of unit 33's curvature computations by Cartan's structure equations.

Reading

Misner, Thorne and Wheeler, ch. 4 — the classic physicist's introduction, and the source of most of this deck's notation.
Flanders, Differential Forms with Applications to the Physical Sciences, ch. 1–5.
Lee, Introduction to Smooth Manifolds, 2nd ed., ch. 14–17, for the proofs cited here.
Bott and Tu, §1–4, if the cohomology is what appeals; the Mayer–Vietoris sequence of §2 is the tool this deck did without.

35 the field equations, twice — the spine continues, and does not need this unit.
36 Schwarzschild, computed there with Christoffel symbols; redo it with a coframe and compare.
39 cosmology, whose spatial slices include problem P11's three-sphere.

Problems: ./mr problems 34. Then ./mr done 34.

The checkpoint has four clauses and each is testable on paper without notes. Clauses one and two are examples 2 and 4, redone from a blank page, with problems P2 and P3 as the test; clause three is example 3 and problem P7; clause four is problem P10, with P11 for a harder metric. If clause four is slow the first time, that is expected — the inspection step that fixes $\omega^a{}_b$ takes a few metrics before it becomes automatic, and the deck and the sheet between them give four.

A caution on what this unit does not license. Writing an equation in form language is not evidence that it is right; it is evidence that it is coordinate-free. Every compression here was checked by expanding back into components, and that check is the habit worth keeping. When a form identity and a component computation disagree, the component computation is almost always the one that is right, because it has fewer conventions in it.

Differential forms and Cartan: Stokes' theorem was always one theorem

One algebra, one derivative, one theorem

One field and one hole

The exterior algebra

Why antisymmetric, and not something else

$\Lambda^k V^*$, defined

The wedge product

The wedge product, drawn

Forms on a manifold

Worked example 1 — the field as a two-form

One derivative

The exterior derivative, axiomatically

Grad, curl and divergence are one operator

Pullback, and why forms are worth the trouble

$\mathrm d$ without coordinates, and where structure equations come from

Worked example 2 — $\mathrm dF = 0$, expanded

Gauge freedom is the freedom to add an exact form

One theorem

Orientation and the volume form

The Möbius band has no orientation

Integrating a form, and where the Jacobian went

The generalised Stokes theorem

Outward normal first, drawn

The proof, on the square

Four theorems, one theorem

Closed, exact, and the difference

Closed does not mean exact

Homotopy invariance: unit 04's promise kept

Simply connected: closed means exact

de Rham cohomology

Worked example 3 — $H^1$ of the punctured plane

Worked example 3, continued — unwrapping the plane

The pairing: a class against a cycle

The solenoid, as a class

Monopoles: the vortex, one dimension up

The monopole needs two potentials

The metric enters, once

The Hodge star

$\star F$, computed

Worked example 4 — $\mathrm d\star F = \mu_0\star J$, expanded

Charge conservation is $\mathrm d^2 = 0$, again

The two invariants, and a number

Cartan's moving frames

Coframes and the first structure equation

The second structure equation, and Bianchi for free

Worked example 5 — the two-sphere, twice

Gauss–Bonnet, from Stokes and one connection form

Gauss–Bonnet on a polar cap

Liouville's theorem, in one line

What used a metric, and what did not

Done when

Reading

Next