Part E · curved space · spine

Manifolds: doing all of it without coordinates

Coordinates stop being data. What survives is a space, the vectors tangent to it, the tensor fields on it, and one derivative — the Lie derivative — that needs nothing but a flow.

needs 16, 30~3 h slides + 6 h problemsproblems: ./mr problems 31

Unit 30 ended with a proof and a demand. The proof: a freely falling frame kills the first derivatives of the metric at a point, but cannot kill the second, and what survives is unit 05's tidal tensor. The demand: build the mathematics that measures exactly that surviving object. This unit is the first of the three that do it, and it is the one that removes coordinates from the list of things a physical theory may depend on.

Three consecutive units of mathematics is unusual for this course, and it is worth saying why it is not a lapse. Every earlier unit could pull in its mathematics on demand because the physical question named the tool: work wanted line integrals, oscillators wanted linear ODE theory. Here the physical question — what is left of gravity when no coordinate change removes it — has no meaning at all until there is an object that coordinate changes act on. The object has to come first.

What is genuinely new, and what is not

Unit 16 already defined a tensor properly: a multilinear map on a vector space and its dual, with the transformation law as a theorem about change of basis. Nothing about that changes here. What changes is that the vector space is no longer $\mathbb R^3$ sitting in the room with you: it is built at each point of a space that has no room to sit in. Once that is done, unit 16's algebra runs unaltered, one point at a time.

How to read this deck

One example runs through every section: the two-sphere with two stereographic charts. Section 1 builds it, section 2 puts a vector on it, section 3 a metric, section 4 a Lie algebra of flows, section 5 the statement that the rotation field moves the metric nowhere. Every component computation is checked at one point $P = (2/3, 1/3, 2/3)$, so there is always a number to compare against.

Proofs are on the slides where they are short and in these notes under Full proof where they are not; nine results are imported and marked cited. Conventions: indices $i, j, k$ for a chart on an $n$-manifold, Einstein summation as in unit 16, $\partial_i$ for $\partial/\partial x^i$, and $\mathbf x = (x^1,\dots,x^n)$ for a chart map.

31 · overview

One sphere, six sections

the space

$S^2$ with two charts. Charts are labels, not data: the transition between them is a diffeomorphism and nothing physical can see it.

the vectors

A tangent vector, defined three ways — curve, derivation, component law — and the three shown to be one space.

the fields

Tensor fields, with unit 16's transformation law recovered as nothing but a change of basis, chart by chart.

the flows

Vector fields are flows; the Lie bracket measures their failure to commute, and the three rotations of the sphere close into $\mathfrak{so}(3)$.

the derivative

The Lie derivative: a derivative of tensors needing no extra structure. Fields whose flow leaves the metric $g$ — the rule that measures lengths — unchanged, $\mathcal L_Xg = 0$, are Killing vector fields.

the caution

Submanifolds, Frobenius — and an honest account of how little general covariance says on its own.

Words with a dotted underline are glossary terms: click for a definition, G for the index, N for the notes, where the long proofs are. A solid underline like the test point $P$ shows a piece of the running example on hover.

The order is forced. You cannot define a tangent vector before you have a smooth structure, a tensor field before tangent vectors, a Lie bracket before vector fields, or a Killing field before a Lie derivative and a metric. The only free choice was section 6, which could have gone anywhere; it is last because its warning lands hardest after the machinery is built.

Unit 32 adds the connection, unit 33 the curvature. This unit deliberately stops short of both: everything here is available on a bare smooth manifold, with a metric brought in only where a metric is genuinely needed. Knowing which statements need which structure is worth more later than it looks now — the Lie derivative needs no connection, and that is why it, not the covariant derivative, is what defines a symmetry.

section 1

Coordinates stop being data

A space, an atlas, and the rule that nothing may depend on which page you opened.

31 · manifolds

Three warnings this course has already given

unit 05

Spherical coordinates go singular on the polar axis. Nothing is wrong with the sphere there.

unit 16

Euler angles lock at $\theta = 0,\pi$. Not carelessness: $SO(3)$ is compact, so no three coordinates can cover it.

unit 30

Rindler coordinates have a non-constant metric and a horizon — on flat spacetime. Coordinate effects can look exactly like physics.

the common cause

All three are the same statement: a space need not be covered by one coordinate system, and the failures of a coordinate system are facts about the system, not about the space. The fix is to allow several coordinate systems at once and to demand that everything said be independent of which one is used.

The historical order was the reverse of the logical one. Gauss (1827) studied surfaces in $\mathbb R^3$ and discovered with the theorema egregium that curvature is intrinsic — computable from the metric alone, with no reference to the ambient space. Riemann's Habilitationsvortrag (1854) then proposed $n$-dimensional manifolds with no ambient space at all, in prose, with almost no formulas. Weyl (1913) gave the first definition in today's shape, with charts and transition maps; Whitney (1936) tidied up the point-set hypotheses. Einstein in 1915 had Riemann's calculus but not Weyl's definition, which is part of why the hole argument cost him two years — the slide General covariance says less than it sounds like returns to it.

A remark about the $SO(3)$ claim, which unit 16 stated and this unit can now prove. If a single chart covered $SO(3)$ it would be a homeomorphism onto an open subset of $\mathbb R^3$. But $SO(3)$ is compact — it is a closed bounded subset of the $3\times3$ matrices, by $R^{\mathsf T}R = \mathbb 1$ — and a non-empty open subset of $\mathbb R^3$ is never compact. The argument is the same one the slide No one chart covers the sphere makes for $S^2$.

31 · manifolds

What a manifold is

definition

A topological manifold of dimension $n$ is a topological space $M$ that is Hausdorff, second countable, and locally Euclidean: every $p\in M$ lies in an open $U$ with a homeomorphism $\mathbf x: U\to \mathbf x(U)\subseteq\mathbb R^n$ onto an open set. The pair $(U,\mathbf x)$ is a chart, and $\mathbf x = (x^1,\dots,x^n)$ its coordinates.

why Hausdorff is not free

The line with two origins — two copies of $\mathbb R$ glued along $\mathbb R\setminus\{0\}$ — is locally Euclidean and not Hausdorff. An ODE through the junction has two solutions, so unit 01's uniqueness theorem fails on it.

why second countable is not free

With Hausdorff it buys paracompactness, hence partitions of unity — glued together out of bump functions — hence the existence of the metrics and connections of units 32–35. Without it, an uncountable discrete union of lines is a one-manifold.

The line with two origins, drawn split open at $0$ (schematic). Every neighbourhood of $0$ and every neighbourhood of $0'$ share a punctured interval on each side, so the two points cannot be separated — and a solution of $\dot x = 1$ arriving from $x=-1$ may pass through either.

Reading the figure. Away from $0$ there is one line; only the origin is doubled, and the picture pulls the two copies apart so you can see both. The shaded bands are a neighbourhood of each origin: an interval around it on its own strand, which necessarily includes points of the common line on both sides. Those common points are where the two neighbourhoods meet, however small they are made.

Second countable means the topology has a countable base. Both conditions are inherited by subspaces of $\mathbb R^N$, so every surface, every sphere and every matrix group you will meet satisfies them without checking. They are in the definition to exclude pathologies, not because they are ever in doubt in physics.

The two-origins line, in detail. Take $\mathbb R\times\{0\}$ and $\mathbb R\times\{1\}$ and identify $(t,0)\sim(t,1)$ for every $t\ne0$. The two images of $0$ have no disjoint neighbourhoods: any neighbourhood of either contains a punctured interval around $0$, and those always intersect. Now run $\dot x = 1$ from $x = -1$. At $t = 1$ the solution reaches the junction and may continue into either origin, so there are two solutions with the same initial data. Picard–Lindelöf is a local statement in a chart and survives; what fails is the global uniqueness that unit 01 relied on. A spacetime is required to be Hausdorff for exactly this reason: determinism.

Dimension is well defined. If $p$ has a chart to $\mathbb R^n$ and another to $\mathbb R^m$ then the transition is a homeomorphism between open subsets of $\mathbb R^n$ and $\mathbb R^m$, and Brouwer's invariance of domain forces $n = m$. Cited: Brouwer 1912; it needs algebraic topology this course does not build. For smooth manifolds the transition is a diffeomorphism, its differential is a linear isomorphism, and $n = m$ follows from linear algebra alone — which is the version we actually use.

31 · manifolds

Atlases, and what smooth can mean with no ambient space

definition

Charts $(U,\mathbf x)$ and $(V,\tilde{\mathbf x})$ are smoothly compatible if the transition map $$\tilde{\mathbf x}\circ\mathbf x^{-1}:\ \mathbf x(U\cap V)\to\tilde{\mathbf x}(U\cap V)$$ and its inverse are $C^\infty$ — a statement about a map between open subsets of $\mathbb R^n$, which unit 03 already knows how to read.

An atlas is a family of pairwise compatible charts covering $M$; a smooth structure is a maximal one.

Two charts on $M$ (schematic). The overlap $U\cap V$ has two pictures, one in each copy of $\mathbb R^n$, and the transition map carries one to the other. It is the only map in sight whose smoothness already means something.

Reading the figure. Neither $U$ nor $V$ is a subset of anything Euclidean: $M$ is drawn as a blob only because paper needs a shape. The two lower boxes are genuine copies of $\mathbb R^n$, and everything calculus can say is said there. Follow a point of the overlap down the left arrow, then back up and down the right arrow: that round trip is $\tilde{\mathbf x}\circ\mathbf x^{-1}$, and smooth compatibility asks it — and its inverse — to be $C^\infty$ on the shaded set.

Why $C^\infty$ and not $C^k$. Convenience, chiefly: $C^\infty$ is closed under everything we do, and the derivation definition of a tangent vector gives the wrong answer for finite $k$ (an infinite-dimensional space of derivations). Physics has no reason to prefer it; it simply costs nothing.

31 · manifolds

One atlas, one smooth structure

the whole trick

There is no notion of a smooth map $M\to\mathbb R$ to begin with. So smoothness is defined in charts, and compatibility is exactly the condition making the definition chart-independent.

maximal, not minimal

Every atlas sits in exactly one maximal atlas, so you specify two charts and get the structure. Two atlases give the same structure iff their union is an atlas.

A trap. On $\mathbb R$, the chart $\mathbf y(t) = t^3$ is a homeomorphism, but it is not compatible with the identity chart: one transition is $s\mapsto s^3,$ the other $s\mapsto\sqrt[3]{s}$, and the second is not differentiable at $0$.

The two transition maps between the identity chart and $\mathbf y(t) = t^3$ on $\mathbb R$. One is smooth; its inverse has a vertical tangent at $0$. So $\{\mathrm{id}\}$ and $\{\mathbf y\}$ are two different smooth structures on one line.

Why maximal. Given an atlas $\mathcal A$, let $\bar{\mathcal A}$ be every chart compatible with every chart of $\mathcal A$. The only thing to check is that two such charts are compatible with each other, and that is local: near a point of the overlap, pass through a chart of $\mathcal A$ containing it and compose two smooth maps. So $\bar{\mathcal A}$ is an atlas, is maximal, and is the unique maximal atlas containing $\mathcal A$. This is why nobody ever writes a maximal atlas down: you hand over two charts and the rest is forced.

A trap worth meeting once. On $M = \mathbb R$ the chart $\mathbf y(t) = t^3$ is not compatible with the identity chart, because $\mathbf y\circ\mathrm{id}^{-1}$ has inverse $s\mapsto s^{1/3}$, which is not differentiable at $0$. So these are two different smooth structures on the same topological space. They are nevertheless diffeomorphic — the map $t\mapsto t^{1/3}$ is a diffeomorphism from one to the other — so nothing is lost. Distinguishing structures up to diffeomorphism is a deep matter: $\mathbb R^n$ has exactly one smooth structure up to diffeomorphism for every $n$ except $n = 4$, where there are uncountably many. Cited: Freedman and Donaldson, 1982–83, for the first exotic $\mathbb R^4$; Taubes 1987 for the uncountable family. The figure on the slide is this pair of maps, drawn. Problem P4 works the $t^3$ example through.

31 · manifolds · example

The sphere, twice

example 1 · the running example is built

$S^2 = \{(x,y,z): x^2+y^2+z^2 = 1\}$, with the subspace topology. Project from the north pole $N = (0,0,1)$ onto the plane $z = 0$, and from $S = (0,0,-1)$: $$\mathbf x(x,y,z) = \Big(\tfrac{x}{1-z},\ \tfrac{y}{1-z}\Big),\qquad \tilde{\mathbf x}(x,y,z) = \Big(\tfrac{x}{1+z},\ \tfrac{y}{1+z}\Big),$$ on $U_N = S^2\setminus\{N\}$ and $U_S = S^2\setminus\{S\}$. Each is a bijection onto $\mathbb R^2$, with inverse $$\mathbf x^{-1}(u,v) = \frac{(2u,\ 2v,\ u^2+v^2-1)}{u^2+v^2+1}.$$

The plane through the $z$-axis and the test point $P$, to scale. The line from $N$ through $P$ lands at radius $\sqrt5$; the line from $S$ through $P$ at radius $1/\sqrt5$. One point, two labels, and the radii multiply to $1$.

Reading the figure. Both charts land in the same plane $z = 0$, which is why one drawing can show both. $P = (\tfrac23,\tfrac13,\tfrac23)$ is at height $\tfrac23$ and at distance $\sqrt{x^2+y^2} = \sqrt5/3$ from the axis. Similar triangles from $N$: the drop from $N$ to $P$ is $1 - \tfrac23 = \tfrac13$ over a run of $\sqrt5/3$, so the drop of $1$ to the plane takes a run of $\sqrt5$ — that is $|\mathbf x(P)| = |(2,1)| = \sqrt5$. From $S$ the rise to $P$ is $\tfrac53$, so the rise of $1$ to the plane takes $\sqrt5/5$. The equator, where the sphere meets the plane, is fixed by both projections.

Deriving $\mathbf x^{-1}$. Given $(u,v)$ we know $r^2 = (1+z)/(1-z)$, which inverts to $z = (r^2-1)/(r^2+1)$, and then $1-z = 2/(1+r^2)$, so $x = u(1-z) = 2u/(1+r^2)$ and similarly for $y$. That the result lies on the sphere is one line: $4u^2 + 4v^2 + (r^2-1)^2 = 4r^2 + r^4 - 2r^2 + 1 = (r^2+1)^2$.

Why stereographic and not latitude–longitude. Spherical coordinates give one chart with two bad points and a seam; stereographic gives two clean charts with a smooth overlap. The first is what a physicist reaches for, the second is what a manifold actually is.

31 · manifolds · example

The transition is inversion in the unit circle

example 1, continued · the transition map

The transition. On the sphere $x^2+y^2 = 1-z^2$, so with $r^2 = u^2+v^2$, $$r^2 = \frac{1-z^2}{(1-z)^2} = \frac{1+z}{1-z}.$$ Hence $\dfrac{x}{1+z} = \dfrac{x}{1-z}\cdot\dfrac{1-z}{1+z} = \dfrac{u}{r^2}$, and likewise for $v$: $$\boxed{\ (\tilde u,\tilde v) = \frac{(u,v)}{u^2+v^2}\ }$$ — inversion in the unit circle, defined and smooth on $\mathbb R^2\setminus\{0\} = \mathbf x(U_N\cap U_S)$, and its own inverse.

Both charts drawn in one plane. $\mathbf x(P)$ and $\tilde{\mathbf x}(P)$ lie on one ray from the origin at reciprocal distances; the unit circle (the equator) is fixed, and inside and outside trade places. The origin — a pole — is the one point with no partner.

Check. $U_N\cup U_S = S^2$; the transition is smooth both ways; so $\{(U_N,\mathbf x),(U_S,\tilde{\mathbf x})\}$ is a smooth atlas and $S^2$ is a smooth 2-manifold.

Geometry of the inversion. A point of the southern hemisphere lands inside the unit circle under $\mathbf x$ and outside it under $\tilde{\mathbf x}$, with $|u|\,|\tilde u| = 1$. The equator is fixed by both and by the transition. The hover figure is drawn to scale: the point $Q$ maps to radius $\tfrac12$ from one pole and radius $2$ from the other.

Orientation, and a sign some books insert. Writing $w = u + \mathrm iv$, the transition is $w\mapsto 1/\bar w$, which is anti-holomorphic; its Jacobian determinant is $-1/r^4 \lt 0$, so this atlas is not oriented. Replacing $\tilde{\mathbf x}$ by $(x/(1+z),\,-y/(1+z))$ makes the transition $w\mapsto 1/w$, holomorphic, with positive Jacobian determinant $+1/r^4$ — the Riemann sphere of unit 20, and an oriented atlas. Orientation plays no part in this unit, so we keep the unreflected charts, whose rotation fields come out in the same form in both charts. Unit 34 will need the oriented version for integration.

31 · manifolds

No one chart covers the sphere

theorem

A compact $n$-manifold $M$ with $n\ge1$ admits no chart whose domain is all of $M$. In particular two charts are the minimum for $S^2$.

proof

Suppose $\mathbf x: M\to W\subseteq\mathbb R^n$ is a homeomorphism onto an open $W$. Then $W = \mathbf x(M)$ is compact, being a continuous image of a compact space, hence closed in $\mathbb R^n$. So $W$ is open, closed and non-empty in the connected space $\mathbb R^n$, forcing $W = \mathbb R^n$ — which is not compact. Contradiction. $\square$

what this settles

The atlas of the running example is optimal. And every statement about the sphere must therefore be made in a way that survives the change of chart — there is no privileged page.

The proof uses only metric-space topology: compactness is preserved by continuous maps, compact subsets of $\mathbb R^n$ are closed and bounded, and $\mathbb R^n$ is connected. No algebraic topology, and none of the hypotheses is doing secret work — $n\ge1$ is needed only because a zero-manifold is a point, which is compact and covered by a chart to $\mathbb R^0$.

The same argument disposes of $SO(3)$, of the torus $T^2$, and of any closed universe: no global coordinate system exists, as a matter of topology, before any physics is done. Note the two things it does not say. It gives a lower bound of two and never an upper bound — some compact manifolds need many more. And it says nothing at all about non-compact manifolds: the punctured plane is non-compact, has a hole, and is nevertheless covered by a single chart, which problem P3 asks you to exhibit. Compactness is the operative property, not connectivity.

The practical form of the warning in relativity is the other way round: a coordinate system that looks global may be covering only part of the manifold. Unit 36's Schwarzschild coordinates stop at $r = 2GM/c^2$ and the Kruskal chart shows the spacetime continues past it — a chart running out, mistaken for the world running out.

31 · manifolds

Smooth maps, and the one theorem that makes them well defined

definition

$F: M\to N$ is smooth at $p$ if for some charts $(U,\mathbf x)$ around $p$ and $(V,\mathbf y)$ around $F(p)$ with $F(U)\subseteq V$, the coordinate representation $\hat F = \mathbf y\circ F\circ\mathbf x^{-1}$ is $C^\infty$ near $\mathbf x(p)$. A bijective $F$ with $F$ and $F^{-1}$ smooth is a diffeomorphism.

theorem · and the proof is the point

If $\hat F$ is $C^\infty$ for one such pair of charts, it is for every such pair.

proof

For other charts $(U',\mathbf x{}'),(V',\mathbf y{}')$, $\ \mathbf y{}'\circ F\circ\mathbf x{}'^{-1} = (\mathbf y{}'\circ\mathbf y^{-1})\circ\hat F\circ(\mathbf x\circ\mathbf x{}'^{-1})$ on the overlap: a composite of three $C^\infty$ maps between open subsets of Euclidean spaces, the outer two being transition maps. $\square$

That three-line proof is the reason the definition of a smooth structure demands compatible charts and nothing else. Compatibility is exactly what is needed, and nothing more is needed: every later definition made in a chart — tangent vectors, tensor fields, the Lie derivative — is legitimised by a proof with this shape. The pattern is: define in a chart, then conjugate by transition maps and watch the conjugation be harmless.

Smooth functions on $M$ are the case $N = \mathbb R$ with the identity chart, giving $C^\infty(M)$, a commutative algebra under pointwise operations. Section 2 will recover the tangent space entirely from this algebra, which is the first hint of a much larger idea: the algebra of functions remembers the space. Milnor's exercise, and Gelfand's theorem in the continuous category, make that precise; we use only the smooth, finite-dimensional corner.

Diffeomorphic, not equal. Two manifolds can be the same set with the same topology and different smooth structures, and still be diffeomorphic (the $t^3$ example). "The same manifold" always means "diffeomorphic", and that is the equivalence general relativity treats as physical identity — which is the content of the hole argument at the end of this deck.

31 · manifolds · physics

Every configuration space you have used was one

Unit 12 called $Q$ a "configuration space" and changed coordinates on it freely. It was a manifold, and the standard examples are not $\mathbb R^n$:

system	$Q$	$\dim$
pendulum	$S^1$	1
double pendulum	$T^2 = S^1\times S^1$	2
spherical pendulum	$S^2$	2
rigid body	$SO(3)\times\mathbb R^3$	6
$N$ points in space	$\mathbb R^{3N}$	$3N$

what this explains

Unit 12's theorem — the Euler–Lagrange equations are form-invariant under any invertible change of $q$ — was the statement that the action is a function on a manifold and $q$ was only ever a chart.

and what it promises

Unit 14's phase space is the cotangent bundle $T^*Q$: momenta are covectors, which is why $p_i$ carries a lower index and transforms oppositely to $\dot q^i$. Unit 34 makes $\mathrm{d}p_i\wedge\mathrm{d}q^i$ a form on it.

$T^2$ is a manifold because a product of manifolds is: charts multiply, transition maps multiply, smoothness is componentwise. Problem P2 builds the atlas explicitly for the double pendulum and writes its kinetic energy as a metric in each chart — four charts in the product atlas, since $S^1$ needs two.

$SO(3)$ is a manifold by the regular value theorem later in this deck, applied to $F(A) = A^{\mathsf T}A$ as a map into symmetric matrices: $\dim = 9 - 6 = 3$. It is also a group whose operations are smooth — a Lie group — which is why unit 16's exponential map behaved as it did.

Why the cotangent bundle and not the tangent bundle. $p_i = \partial L/\partial\dot q^i$, and under $q\to Q$ the velocities transform with $\partial Q/\partial q$ while the momenta transform with its inverse transpose — the covector law of unit 16. So the Legendre transform of unit 14 was a map $TQ\to T^*Q$ all along, and Hamilton's equations live on the cotangent bundle, which carries a canonical symplectic form that $TQ$ does not. Liouville's theorem is then the statement that the flow preserves that form's top power. None of this is needed here; it is recorded so that the earlier units read correctly in hindsight.

section 2

What is a vector, with nothing to point into?

Three definitions — a velocity, a derivation, a list of numbers with a law — and one theorem saying they are the same.

31 · tangent vectors

The problem, stated precisely

In $\mathbb R^3$ a vector at $p$ is an arrow from $p$, and "arrow from $p$" means a point of $\mathbb R^3$ with a new name. On $S^2$ there is no room for the arrow: it would leave the sphere.

the temptation, and why it fails

Embed $S^2$ in $\mathbb R^3$ and use the tangent plane. That works, and it is useless: spacetime is not given inside anything, and Whitney's theorem shows an embedding exists but never canonically. A definition depending on a choice nobody made is not a definition.

what is available instead

curves through $p$, and their coordinate velocities;
the algebra $C^\infty(M)$, and directional differentiation of it;
charts, and the Jacobians of their transitions.

Each gives a definition. They are the same, and physics uses all three without saying so.

Reading the literature is the practical reason for doing all three. A relativist writes $u^\mu$ and means definition C; a differential geometer writes $X = X^i\partial_i$ and means definition B; a mechanician writes $\dot\gamma(0)$ and means definition A. Papers switch between them silently, mid-paragraph, and the switch is invisible only if you have seen the isomorphisms once.

There is a fourth definition, useful in algebraic geometry and worth knowing exists: a tangent vector at $p$ is an element of $(\mathfrak m_p/\mathfrak m_p^2)^*$, where $\mathfrak m_p$ is the ideal of germs vanishing at $p$. Hadamard's lemma, which appears in the notes of The three definitions describe one space, is exactly the statement that this agrees with definition B in the smooth category.

31 · tangent vectors

Definitions A and C: a velocity, and a list with a law

A · kinematic

On smooth curves $\gamma:(-\varepsilon,\varepsilon)\to M$ with $\gamma(0)=p$, declare $$\gamma\sim\sigma \iff (\mathbf x\circ\gamma)'(0) = (\mathbf x\circ\sigma)'(0)$$ for some chart at $p$. A tangent vector is an equivalence class $[\gamma]$.

Chart-independent? Yes. If $\tilde{\mathbf x} = \tau\circ\mathbf x$ with $\tau$ the transition, the chain rule of unit 03 gives $(\tilde{\mathbf x}\circ\gamma)'(0) = D\tau\,(\mathbf x\circ\gamma)'(0)$, and $D\tau$ is invertible — so equality in one chart is equality in every chart.

C · component

A tangent vector assigns to each chart at $p$ an $n$-tuple $\xi^i$, so that any two assignments are related by $$\tilde\xi^{\,a} = \frac{\partial\tilde x^a}{\partial x^i}\Big|_{\mathbf x(p)}\,\xi^i .$$ Addition and scalars act chartwise.

what C is not

It is not "a thing that transforms like a vector" as a slogan. It is a function on the set of charts, constrained by a cocycle condition — and the condition holds because Jacobians of composed transitions multiply.

The cocycle condition. For three charts, $\tau_{31} = \tau_{32}\circ\tau_{21}$, so by the chain rule $D\tau_{31} = D\tau_{32}\,D\tau_{21}$ and $D\tau_{11} = \mathbb 1$. That is what makes definition C consistent: you may hop from chart 1 to chart 3 directly or via chart 2 and get the same tuple. Without it the definition would be empty, and checking it is the only content in C.

Why this is the physicist's definition. It is the one that generalises without thought to tensors of any type, and it is what "$u^\mu$ transforms as a contravariant vector" means. Its weakness is that it makes the transformation law look like the definition of the object, which unit 16 spent a section arguing against, and which this deck's section 3 settles for good.

Smoothness of $\gamma$ means smooth as a map of manifolds, an interval being a manifold with the identity chart; concretely, $\mathbf x\circ\gamma$ is $C^\infty$ near $0$. Note that $\gamma$ need not be injective, and the zero vector is the class of the constant curve.

31 · tangent vectors

Definition B: a derivation of germs

B · algebraic

A germ at $p$ is an equivalence class of smooth functions defined near $p$, two being identified if they agree on some neighbourhood; they form an algebra $C^\infty_p$. A derivation at $p$ is a linear $D: C^\infty_p\to\mathbb R$ with $$D(fg) = f(p)\,Dg + g(p)\,Df .$$ $T_pM$ is the vector space of derivations at $p$.

Immediate consequence. $D(1) = D(1\cdot1) = 1\cdot D1 + 1\cdot D1 = 2D1$, so $D1 = 0$ and $D$ kills constants: a derivation sees only how a function changes.

the appeal

No charts, no curves, no choices. A vector is the operation "differentiate in my direction", and the Leibniz rule is the whole of what a direction is.

Why germs and not $C^\infty(M)$. Either works on a smooth manifold, because a bump function lets you modify a global function away from $p$ without changing $Df$ — but that argument has to build the bump function first, and then argue that the modification is invisible. Germs make the locality manifest and need neither step. The price is one more equivalence relation.

Locality, proved. If $f$ vanishes on a neighbourhood $W$ of $p$, then $Df = 0$: pick a smooth $\chi$ with $\chi(p) = 1$ and $\chi = 0$ outside $W$; then $\chi f \equiv 0$, so $0 = D(\chi f) = \chi(p)Df + f(p)D\chi = Df$. This is the statement that a derivation cannot see distant behaviour, and it is why the germ and global definitions agree.

The finite-smoothness trap. On $C^k$ germs with $k\lt\infty$ the space of derivations is infinite-dimensional — the proof on the next slide breaks because Hadamard's remainders are only $C^{k-1}$. This is the concrete reason the definition of a smooth manifold says $C^\infty$.

31 · tangent vectors

The three definitions describe one space

theorem

Let $A$, $B$, $C$ be the sets defined by A, B and C. The maps $$\Phi: A\to B,\quad \Phi[\gamma](f) = (f\circ\gamma)'(0);\qquad \Psi: B\to C,\quad \Psi(D) = \big(D(x^1),\dots,D(x^n)\big)$$ are bijections, $\Psi$ is a linear isomorphism, and $\dim T_pM = n$. ($B$ and $C$ are vector spaces by construction; $A$ receives its structure through $\Phi$.)

proof · the two easy halves

$\Phi[\gamma]$ is linear and obeys Leibniz by the product rule; it depends only on $[\gamma]$ because $(f\circ\gamma)'(0) = \partial_i\hat f\cdot(\mathbf x\circ\gamma)'(0)$ by the chain rule, and $\Phi$ is injective for the same reason. $\Psi$ lands in $C$ because $\tilde x^a$ is a germ and $D(\tilde x^a) = \frac{\partial\tilde x^a}{\partial x^i}D(x^i)$ — the Hadamard argument in the notes.

proof · surjectivity of $\Phi$

Given $D$, put $\xi^i = D(x^i)$ and take $\gamma(t) = \mathbf x^{-1}(\mathbf x(p) + t\xi)$. Hadamard's lemma gives $D = \xi^i\partial_i|_p$, and $\Phi[\gamma] = \xi^i\partial_i|_p$ too. $\square$

Full proof

Hadamard's lemma. Let $\hat f$ be smooth on a convex neighbourhood of $a = \mathbf x(p)$ in $\mathbb R^n$. Then $$\hat f(s) = \hat f(a) + (s^i - a^i)\,h_i(s),\qquad h_i(s) = \int_0^1 \partial_i\hat f\big(a + t(s-a)\big)\,\mathrm{d}t,$$ with each $h_i$ smooth and $h_i(a) = \partial_i\hat f(a)$. Proof: the fundamental theorem of calculus applied to $t\mapsto\hat f(a+t(s-a))$ on $[0,1]$, whose derivative is $(s^i-a^i)\partial_i\hat f(a+t(s-a))$ by unit 03's chain rule. Smoothness of $h_i$ is differentiation under the integral sign, legitimate because the integrand is smooth in $(t,s)$ on a compact $t$-interval.

Consequence. For any derivation $D$, write $f = f(p) + (x^i - a^i)h_i$ as germs. $D$ kills the constant $f(p)$; by Leibniz, $D\big((x^i-a^i)h_i\big) = (a^i - a^i)Dh_i + h_i(p)\,D(x^i) = \partial_i\hat f(a)\,D(x^i)$. So $Df = \xi^i\,\partial_i\hat f(a)$ with $\xi^i = D(x^i)$: every derivation is a directional derivative, and $\{\partial_i|_p\}$ spans. They are independent because $\partial_i|_p(x^j) = \delta_i^j$. Hence $\dim T_pM = n$.

Injectivity of $\Psi$ is that same formula: the numbers $D(x^i)$ determine $D$. Surjectivity of $\Psi$: given a compatible family $\xi$, the derivation $\xi^i\partial_i|_p$ maps to it, and the transformation law in C is exactly the chain-rule identity $\tilde\xi^a = \partial\tilde x^a/\partial x^i\,\xi^i$, so the family is determined by one chart's tuple. Linearity of all the maps is immediate from the formulas.

Where the smoothness is spent. Only in Hadamard's lemma. That single integral is what makes the derivation definition finite-dimensional, and it is why the theorem is false for $C^k$ germs.

31 · tangent vectors

The coordinate basis, and the count that follows

definition

In a chart $(U,\mathbf x)$ the coordinate basis of $T_pM$ is $$\partial_i\big|_p : f\ \longmapsto\ \frac{\partial (f\circ\mathbf x^{-1})}{\partial s^i}\Big|_{\mathbf x(p)} .$$ Every $X\in T_pM$ is $X = X^i\,\partial_i|_p$ with $X^i = X(x^i)$.

The tangent bundle $TM = \bigsqcup_p T_pM$ is itself a smooth $2n$-manifold, with charts $(x^i, X^i)$.

the notation trap

$\partial_i|_p$ depends on all $n$ coordinates, not just $x^i$. Changing $x^2$ alone changes $\partial_1$, because $\partial_1$ means "differentiate along $x^1$ holding the others fixed". Unit 16 made the same point for a dual basis.

the change of basis

$\displaystyle\partial_i = \frac{\partial\tilde x^a}{\partial x^i}\,\tilde\partial_a$, so components go the other way: $\tilde X^a = \frac{\partial\tilde x^a}{\partial x^i}X^i$. That is the whole content of definition C.

Proof of the change-of-basis formula. Apply $\partial_i|_p$ to the germ of a function $f$ and use the chain rule on $f\circ\mathbf x^{-1} = (f\circ\tilde{\mathbf x}^{-1})\circ\tau$ with $\tau = \tilde{\mathbf x}\circ\mathbf x^{-1}$: the derivative of the outer map times $D\tau$, whose entries are $\partial\tilde x^a/\partial x^i$. The basis vectors carry the Jacobian, so the components carry its inverse — the pattern unit 16 called "upper indices by $S^{-1}$, lower by $S$", now with $S$ varying from point to point.

Why $TM$ is a manifold. Given a chart $(U,\mathbf x)$, map $(p, X)\mapsto(\mathbf x(p), X^1,\dots,X^n)$ into $\mathbb R^{2n}$. The transition between two such is $(s, \xi)\mapsto(\tau(s), D\tau(s)\,\xi)$, smooth because $\tau$ is smooth and $D\tau$ depends smoothly on $s$; it is linear in $\xi$, which is the extra structure making $TM$ a vector bundle rather than merely a manifold. A vector field is then a smooth map $M\to TM$ that is the identity after projection — a section.

The $\partial$ notation. Writing a tangent vector as $\partial/\partial x^i$ is Cartan's, and it is genuinely an identification of the vector with an operator, not an abuse. Physicists who write $\hat e_i$ and geometers who write $\partial_i$ mean the same element of the same vector space.

31 · tangent vectors · example

One vector, two charts, three descriptions

example 2 · at the test point $P$

Take $P = (\tfrac23,\tfrac13,\tfrac23)$, so $\mathbf x(P) = (2,1)$ — check: $u = x/(1-z) = (2/3)/(1/3) = 2$ — and $\tilde{\mathbf x}(P) = (2,1)/5 = (\tfrac25,\tfrac15)$.

A · a curve. Rotate about the $z$-axis: $\gamma(t) = R_z(t)P$. Then $\dot\gamma(0) = (-y,x,0)|_P = (-\tfrac13,\tfrac23,0)$ in $\mathbb R^3$.

C · components. In the first chart $u = x/(1-z),\ v = y/(1-z)$ with $z$ fixed along the rotation, so $\dot u = -v$, $\dot v = u$: at $P$, $\ \xi = (-1, 2)$.

The Jacobian check. From $(\tilde u,\tilde v) = (u,v)/r^2$, $$\frac{\partial\tilde x^a}{\partial x^i} = \frac{1}{r^4}\begin{pmatrix} v^2-u^2 & -2uv\\ -2uv & u^2-v^2\end{pmatrix} \ \overset{P}{=}\ \frac{1}{25}\begin{pmatrix}-3 & -4\\ -4 & 3\end{pmatrix}.$$ Then $\ \tilde\xi = \tfrac1{25}\begin{pmatrix}-3&-4\\-4&3\end{pmatrix}\begin{pmatrix}-1\\2\end{pmatrix} = \begin{pmatrix}-1/5\\ 2/5\end{pmatrix}$, which is exactly $(-\tilde v,\tilde u)$ at $\tilde{\mathbf x}(P)$.

B · a derivation. The same object is $-v\,\partial_u + u\,\partial_v$, and applying it to the germ of $z = \frac{r^2-1}{r^2+1}$ gives $0$: rotation about $z$ does not change $z$.

Deriving $\dot u = -v,\ \dot v = u$ honestly. Under $R_z(t)$ the sphere point has $\dot x = -y$, $\dot y = x$, $\dot z = 0$. Then $\dot u = \frac{\dot x(1-z) + x\dot z}{(1-z)^2} = \frac{-y}{1-z} = -v$, and similarly $\dot v = u$. The $\dot z$ term is absent only because this is the $z$-rotation; for the other two generators it survives and produces the quadratic fields of example 4.

The Jacobian, computed. $\partial_u(u/r^2) = 1/r^2 - 2u^2/r^4 = (v^2-u^2)/r^4$, and $\partial_v(u/r^2) = -2uv/r^4$; the second row is the same with $u\leftrightarrow v$. At $r^2 = 5$: $r^4 = 25$, $v^2-u^2 = -3$, $-2uv = -4$. Its determinant is $-1/r^4 = -1/25$, never zero on the overlap, as a transition map's Jacobian must be.

The check that matters. Both chart descriptions had to be computed independently and then compared: $(-1,2)$ pushed through the Jacobian gives $(-1/5,2/5)$, and $(-\tilde v, \tilde u)$ at $(\tfrac25,\tfrac15)$ gives $(-1/5, 2/5)$. Equal. That the field has the same form in both charts is a happy accident of the unreflected stereographic pair — inversion commutes with rotation about the axis through the two poles.

31 · tangent vectors

The differential of a smooth map

definition

For smooth $F: M\to N$ and $p\in M$, the differential (or pushforward) $\mathrm{d}F_p: T_pM\to T_{F(p)}N$ is $$\big(\mathrm{d}F_p(X)\big)(f) = X(f\circ F),\qquad f\in C^\infty_{F(p)} .$$ It is linear, and in charts its matrix is the Jacobian $\partial\hat F^a/\partial x^i$.

theorem · the chain rule, finally in its right form

$\mathrm{d}(G\circ F)_p = \mathrm{d}G_{F(p)}\circ\mathrm{d}F_p$, and $\mathrm{d}(\mathrm{id})_p = \mathrm{id}$.

proof

$\mathrm{d}(G\circ F)_p(X)(f) = X(f\circ G\circ F) = \mathrm{d}F_p(X)(f\circ G) = \mathrm{d}G_{F(p)}\big(\mathrm{d}F_p(X)\big)(f)$. $\square$

In definition A this is $[\gamma]\mapsto[F\circ\gamma]$; in definition C it is multiplication by the Jacobian. The transition-map formula of the last slide is the special case $F = \mathrm{id}$, read in two charts.

Unit 03 defined the total derivative of $\mathbb R^n\to\mathbb R^m$ as the linear map best approximating it, and said the chain rule is composition of linear maps rather than a page of subscripts. This slide is that sentence, now true on manifolds, where there is no other option: "best linear approximation" has no meaning without a vector space, but "acts on derivations by precomposition" always does.

Corollary, used constantly. If $F$ is a diffeomorphism then $\mathrm{d}F_p$ is a linear isomorphism, since $\mathrm{d}(F^{-1})\circ\mathrm{d}F = \mathrm{id}$. The converse is local only: the inverse function theorem of unit 04 says $\mathrm{d}F_p$ invertible implies $F$ is a diffeomorphism of some neighbourhood of $p$, not of $M$. The map $\theta\mapsto(\cos\theta,\sin\theta)$ from $\mathbb R$ to $S^1$ has invertible differential everywhere and is not injective.

Rank. $\operatorname{rank}\mathrm{d}F_p$ is chart-independent, because changing charts multiplies the Jacobian by invertible matrices on both sides. Section 6's immersions, submersions and regular values are all rank conditions, and this is why they are legitimate statements about $F$ rather than about coordinates.

section 3

Covectors, and tensor fields

Unit 16's algebra, run once at every point — and the transformation law demoted to a theorem for the second time.

31 · tensor fields

The cotangent space, and $\mathrm{d}f$

definition

$T_p^*M = (T_pM)^*$, the cotangent space: linear maps $T_pM\to\mathbb R$. For $f\in C^\infty_p$ the differential of a function is $$\mathrm{d}f_p(X) = X(f),$$ manifestly linear in $X$, and defined with no metric and no choice.

Applying this to the coordinate functions gives $\mathrm{d}x^i(\partial_j) = \partial_j(x^i) = \delta^i_j$: the $\mathrm{d}x^i$ are the dual basis of unit 16, and $\mathrm{d}f = \partial_i\hat f\,\mathrm{d}x^i$.

Level curves of the height $f = z$ in the first chart, and two vectors at $P$ (drawn at $0.4\times$). $\mathrm{d}f$ eats a vector and returns how fast it climbs through the levels: zero for $R_z = (-1,2)$, which runs along its level curve; $\tfrac59$ for the radial $u\,\partial_u + v\,\partial_v = (2,1)$.

Reading the figure. A covector at a point is pictured best as its level lines, not as an arrow: $\mathrm{d}f_P(X)$ is the rate at which $X$ crosses the levels of $f$. The circles are $z = \text{const}$, which in this chart is $r^2 = (1+z)/(1-z)$; equal steps in $z$ give circles that crowd outwards, because the far north of the sphere is squeezed into the far reaches of the plane. With $\mathrm{d}f_P = \tfrac29\mathrm{d}u + \tfrac19\mathrm{d}v$ (the notes of the next slide derive it), the rotation vector gives $\tfrac29(-1) + \tfrac19(2) = 0$ and the radial one $\tfrac29(2) + \tfrac19(1) = \tfrac59$. No metric was used: "perpendicular to the level curve" never entered, only "along it" or "across it".

$\mathrm{d}f = 0$ where? Only at $u = v = 0$ in the first chart — the south pole — and at $\tilde u = \tilde v = 0$ in the second — the north pole. Two critical points on a compact surface, which is the minimum, and their count is not an accident: it is the Morse-theoretic shadow of $\chi(S^2) = 2$ — the Euler characteristic, vertices minus edges plus faces of any triangulation, $4 - 6 + 4 = 2$ for a tetrahedron drawn on the sphere — the same number the hairy ball theorem will charge us later in this deck.

31 · tensor fields

Covector components, and the gradient that is not there

gradient is not $\mathrm{d}f$

Unit 04 planted this and unit 16 made it structural. $\mathrm{d}f$ exists on any manifold; $\nabla f$ needs a metric to raise the index, $(\nabla f)^i = g^{ij}\partial_jf$. On a bare manifold there is no gradient at all.

the law, by the same chain rule

$\mathrm{d}\tilde x^a = \frac{\partial\tilde x^a}{\partial x^i}\mathrm{d}x^i$, so components of a covector go with $\partial x^i/\partial\tilde x^a$ — opposite to a vector's, as they must be for $\omega(X)$ to be a number.

checked at $P$

$\mathrm{d}z = \tfrac29\,\mathrm{d}u + \tfrac19\,\mathrm{d}v$ in the first chart, $-\tfrac{10}9\,\mathrm{d}\tilde u - \tfrac59\,\mathrm{d}\tilde v$ in the second. The law with $\partial x^i/\partial\tilde x^a = \begin{pmatrix}-3&-4\\-4&3\end{pmatrix}$ at $P$: $-3\cdot\tfrac29 - 4\cdot\tfrac19 = -\tfrac{10}9$ ✓, $-4\cdot\tfrac29 + 3\cdot\tfrac19 = -\tfrac59$ ✓.

Example, computed, for the running example. Let $f$ be the height $z$, restricted to the sphere. In the first chart $f = (r^2-1)/(r^2+1)$, so $$\partial_u f = \frac{2u(r^2+1) - (r^2-1)2u}{(r^2+1)^2} = \frac{4u}{(1+r^2)^2},$$ and likewise for $v$: $\ \mathrm{d}f = \frac{4(u\,\mathrm{d}u + v\,\mathrm{d}v)}{(1+u^2+v^2)^2}$. At $P$, where $r^2 = 5$: $\mathrm{d}f = \tfrac29\mathrm{d}u + \tfrac19\mathrm{d}v$. In the second chart $f = (1-\tilde r^2)/(1+\tilde r^2)$ — note the sign, since that chart projects from the other pole — giving $\mathrm{d}f = -\tfrac{10}{9}\mathrm{d}\tilde u - \tfrac59\mathrm{d}\tilde v$ at $P$. The covector law with the inverse Jacobian $\begin{pmatrix}-3&-4\\-4&3\end{pmatrix}$ reproduces it: $-3\cdot\tfrac29 - 4\cdot\tfrac19 = -\tfrac{10}9$ and $-4\cdot\tfrac29 + 3\cdot\tfrac19 = -\tfrac59$. Problem P6 asks for this from scratch.

31 · tensor fields

Tensor fields: unit 16, one point at a time

definition

A tensor field of type $(p,q)$ on $M$ assigns to each point $m$ a tensor $T_m\in T^p_q(T_mM)$ — a multilinear map $(T_m^*M)^p\times(T_mM)^q\to\mathbb R$, exactly unit 16's definition — smoothly, meaning its components $$T^{i_1\dots i_p}{}_{j_1\dots j_q} = T\big(\mathrm{d}x^{i_1},\dots,\partial_{j_1},\dots\big)$$ are smooth functions in every chart. Equivalently: a smooth section of the bundle $T^p_qM$.

nothing is new

Tensor products, contraction, symmetric and antisymmetric parts: all of unit 16, applied in the vector space $T_mM$, for each $m$ separately. Smoothness is the only new word.

one thing is new

The basis $\{\partial_i\}$ now varies with the point, so the change of basis is a function of position. Everything that will go wrong in unit 32 goes wrong here.

Examples already in hand: a Riemannian or Lorentzian metric is $(0,2)$, symmetric and non-degenerate; the electromagnetic field tensor $F_{\mu\nu}$ of unit 28 is $(0,2)$ and antisymmetric; unit 22's stress tensor is $(0,2)$ symmetric; a vector field is $(1,0)$; unit 16's inertia tensor is $(0,2)$ on a single tangent space, not a field. Unit 33's Riemann tensor will be $(1,3)$.

Why "smooth in every chart" is not a burden. Smoothness in one chart implies it in every overlapping chart, because the components are related by products of $\partial\tilde x/\partial x$ entries, which are smooth. This is the fourth instance in this deck of the same argument; after this one we stop remarking on it.

Tensor fields versus pointwise assignments. A subtlety unit 32 will need: a $C^\infty(M)$-multilinear map on vector fields is the same thing as a tensor field. That is the tensor characterisation lemma, and it is what one checks to prove that some expression built from derivatives is nevertheless tensorial — which is exactly how the Riemann tensor is certified. The Lie derivative slide two sections on is the first place the question bites.

31 · tensor fields

"Transforms like a tensor" is a theorem, again

theorem

Let $T$ be a $(p,q)$-tensor field and let two charts overlap. Write $A^a{}_i = \partial\tilde x^a/\partial x^i$ and $S^i{}_a = \partial x^i/\partial\tilde x^a$, so $A^a{}_iS^i{}_b = \delta^a_b$. Then on the overlap $$\tilde T^{a\dots}{}_{\dots b} = A^a{}_i\cdots S^j{}_b\cdots\,T^{i\dots}{}_{\dots j}.$$

proof

Evaluate $T$ on the tilded basis and expand it in the untilded one: $\mathrm{d}\tilde x^a = A^a{}_i\,\mathrm{d}x^i$ and $\tilde\partial_b = S^j{}_b\,\partial_j$. Multilinearity pulls every coefficient out of the slots. $\square$

what has changed since unit 16

There, $A$ and $S$ were constant matrices from a change of basis of one vector space. Here they are functions of position, being Jacobians of a transition map. The proof is character-for-character the same; the consequences in unit 32 are not.

Unit 16's version was $T'^{a\dots}{}_{\dots b} = A_{ai}\cdots S_{jb}\cdots T^{i\dots}{}_{\dots j}$ with $A = S^{-1}$ for a change of basis $\vec f_a = S_{ia}\vec e_i$. Identical. The point of repeating it is that the reader should feel no new content — because there is none. What makes the manifold case worth stating separately is only that $A$ and $S$ vary, so that $\partial_k A^a{}_i \ne 0$, and every derivative of a tensor field will inherit those extra terms. That single observation is the seed of the connection in unit 32, of the Lie derivative two sections on, and of the exterior derivative in unit 34.

Second derivatives of the transition. Writing $\tau = \tilde{\mathbf x}\circ\mathbf x^{-1}$, the terms that spoil naive differentiation are $\partial^2\tilde x^a/\partial x^i\partial x^k$. They are symmetric in $i$ and $k$ by Clairaut's theorem (unit 03), and that symmetry is precisely why two different repairs work: antisymmetrising kills them, which gives the exterior derivative and the Lie bracket, and subtracting a compensating field kills them, which gives the connection. Both routes appear in this course, and both exist because of Clairaut.

31 · tensor fields

Components are not the tensor

the counterfeit

Define a rule: "in every chart, the array $T_{ij} = \delta_{ij}$". That is $n^2$ perfectly good smooth functions per chart. It is not a tensor field, because applying the theorem to the first chart predicts $\tilde T_{ab} = S^i{}_aS^j{}_b\,\delta_{ij}$, which is $\delta_{ab}$ only if $S$ is orthogonal — and transition Jacobians are not.

convicted on the sphere

At $P$, $S = \begin{pmatrix}-3&-4\\-4&3\end{pmatrix}$, so $S^{\mathsf T}\delta S = \begin{pmatrix}25&0\\0&25\end{pmatrix}$. The rule says $\tilde T_{11} = 1$; the law says $25$. One of them is wrong, and it is the rule.

the honest test

An assignment of arrays to charts is a tensor field iff it is equivariant — unit 16's word: it must commute with every change of chart. One chart's array then determines all the others, and there is nothing left to choose.

The same trap in more dangerous dress: "$T_{ij} = \partial_i\partial_j f$ for a fixed smooth $f$". Those are smooth functions in every chart, and they are not the components of a tensor field: under a change of chart, $\tilde\partial_a\tilde\partial_b f = S^i{}_aS^j{}_b\,\partial_i\partial_j f + \frac{\partial^2x^i}{\partial\tilde x^a\partial\tilde x^b}\partial_i f$, and the second term is absent from the tensor law. It vanishes exactly at a critical point of $f$, which is why the Hessian at a critical point is well defined — the fact Morse theory runs on, and the fact that lets unit 32 say a freely falling frame kills first derivatives but not second. Problem P5 works this through on the sphere.

Where the $25$ came from. $S^{\mathsf T}S$ with $S = \begin{pmatrix}-3&-4\\-4&3\end{pmatrix}$ is $\begin{pmatrix}9+16 & 12-12\\ 12-12 & 16+9\end{pmatrix} = 25\,\mathbb 1$: this particular $S$ is $5$ times an orthogonal matrix, which is what makes the arithmetic clean and the verdict unambiguous. It is not orthogonal, so the rule is convicted.

Unit 16 convicted the rule "return $\mathrm{diag}(I_{11},I_{22},I_{33})$" with the book: $0.00288$ against $0.00864$. This is the same conviction, on a manifold, with the extra feature that $S$ now varies from point to point, so a rule can even be equivariant at one point and not at another.

31 · tensor fields

A metric is a bilinear form on each tangent space

definition

A Riemannian metric on $M$ is a $(0,2)$-tensor field $g$ that at every point is symmetric and positive definite. A Lorentzian metric is the same with signature $(-,+,+,+)$ at every point. Components $g_{ij} = g(\partial_i,\partial_j)$; length $\lVert X\rVert^2 = g_{ij}X^iX^j$.

The physics is unit 26's and unit 30's: $g$ is what a clock and a ruler measure. The mathematics is unit 16's: a symmetric bilinear form, nothing more, now varying smoothly from point to point.

unit 30's inoculation, restated

Non-constant $g_{ij}$ is not gravity. Polar coordinates on the flat plane give $g = \mathrm{d}r^2 + r^2\mathrm{d}\theta^2$; Rindler coordinates give a position-dependent lapse on flat spacetime. What cannot be removed is unit 33's business.

Existence. Every second-countable smooth manifold carries a Riemannian metric: take a metric in each chart, and glue with a partition of unity — positive definiteness survives a convex combination. Cited: partitions of unity, next-but-one section. Lorentzian metrics are different: $S^2$ admits none, because a Lorentzian metric determines a line field, which on a compact manifold forces $\chi = 0$, and $\chi(S^2) = 2$. So topology can forbid a spacetime structure outright. That is the first place in this course where the shape of a manifold constrains the physics that can live on it.

Signature is well defined pointwise by Sylvester's law of inertia, and locally constant by continuity of eigenvalues; on a connected manifold it is therefore constant. Unit 16's spectral theorem is what makes "positive definite" checkable, and unit 26's $\eta$ is the constant-coefficient case.

Raising and lowering is the isomorphism $T_mM\to T^*_mM$, $X\mapsto g(X,\cdot)$, which is where a bare manifold acquires a gradient. Unit 16 called this the choice physics makes; it is a choice of metric, and on a spacetime it is the dynamical field itself.

31 · tensor fields · example

The round metric, in both charts at once

example 3 · one tensor, two arrays

Pull the Euclidean $\mathrm{d}x^2+\mathrm{d}y^2+\mathrm{d}z^2$ back through $\mathbf x^{-1}$. With $\mathbf x^{-1} = (2u, 2v, r^2-1)/(1+r^2)$ a short computation gives the round metric $$g = \frac{4\,(\mathrm{d}u^2+\mathrm{d}v^2)}{(1+u^2+v^2)^2},\qquad \tilde g = \frac{4\,(\mathrm{d}\tilde u^2+\mathrm{d}\tilde v^2)}{(1+\tilde u^2+\tilde v^2)^2}.$$

Do they agree? Under $\tilde{\mathbf x} = \mathbf x/r^2$, inversion is conformal with $\mathrm{d}\tilde u^2+\mathrm{d}\tilde v^2 = (\mathrm{d}u^2+\mathrm{d}v^2)/r^4$, and $1+\tilde r^2 = (r^2+1)/r^2$. So $$\tilde g = \frac{4}{r^4}\cdot\frac{r^4}{(1+r^2)^2}(\mathrm{d}u^2+\mathrm{d}v^2) = g .\ \checkmark$$

At the test point. $r^2 = 5$: $\ g_{11} = g_{22} = 4/36 = \tfrac19$. And $\tilde r^2 = 1/5$: $\ \tilde g_{11} = \tilde g_{22} = 4/(6/5)^2 = \tfrac{25}9$. Different arrays.

One number, three routes. For $R_z$ at $P$: $$\lVert R_z\rVert^2 = \tfrac19\big(1+4\big) = \tfrac59,$$ $$= \tfrac{25}9\big(\tfrac1{25}+\tfrac4{25}\big) = \tfrac59,$$ $$= \big\lVert(-\tfrac13,\tfrac23,0)\big\rVert^2_{\mathbb R^3} = \tfrac59 .$$

Full derivation of the round metric

With $\rho = 1+r^2$, $\ x = 2u/\rho$ gives $\mathrm{d}x = \frac{2\,\mathrm{d}u}{\rho} - \frac{2u(2u\,\mathrm{d}u+2v\,\mathrm{d}v)}{\rho^2} = \frac{2}{\rho^2}\big[(\rho - 2u^2)\mathrm{d}u - 2uv\,\mathrm{d}v\big]$, and similarly $\mathrm{d}y = \frac{2}{\rho^2}\big[-2uv\,\mathrm{d}u + (\rho-2v^2)\mathrm{d}v\big]$, $\mathrm{d}z = \frac{4}{\rho^2}(u\,\mathrm{d}u+v\,\mathrm{d}v)$ using $z = (r^2-1)/(r^2+1)$ and the derivative computed on the $\mathrm{d}f$ slide.

The $\mathrm{d}u^2$ coefficient. Squaring and adding gives $\frac{4}{\rho^4}\big[(\rho-2u^2)^2 + 4u^2v^2 + 4u^2\big]$, and the bracket expands to $\rho^2 - 4u^2\rho + 4u^4 + 4u^2v^2 + 4u^2 = \rho^2 - 4u^2\big(\rho - r^2 - 1\big) = \rho^2$, because $\rho = 1 + r^2$. So the coefficient is $4/\rho^2$, and the $\mathrm{d}v^2$ coefficient is the same by the symmetry $u\leftrightarrow v$.

The cross term vanishes. Collecting the coefficient of $\mathrm{d}u\,\mathrm{d}v$, each square contributes twice its product of coefficients: $$\frac{8}{\rho^4}\Big[(\rho-2u^2)(-2uv) + (-2uv)(\rho-2v^2) + 4uv\Big] = \frac{8}{\rho^4}\Big[-4uv\big(\rho - r^2\big) + 4uv\Big] = \frac{8}{\rho^4}\cdot 0,$$ since $\rho - r^2 = 1$. Hence $g = 4(\mathrm{d}u^2+\mathrm{d}v^2)/\rho^2$: the metric is conformally flat, as every surface metric is locally.

Why the agreement check is not circular. The two formulas were written down independently — each is the pullback of the same Euclidean metric through a different parametrisation — and the transition map was derived earlier from the geometry. That they agree on the overlap is the content of "$g$ is one tensor field", and it was verified symbolically as well as at $P$.

The three routes to $5/9$. The third uses the embedding, so it is available only because this sphere happens to sit in $\mathbb R^3$; the first two do not, and they are the ones that generalise. Agreement is the point: the number belongs to the vector and the metric, not to a chart.

section 4

Vector fields, flows, and the bracket

A vector field is a flow. Two flows rarely commute, and the bracket says by how much.

31 · flows

A vector field is a flow

definition

A vector field $X$ is a smooth section of $TM$; equivalently a derivation of $C^\infty(M)$, $\ f\mapsto Xf = X^i\partial_if$. An integral curve through $m$ solves $$\dot\gamma(t) = X_{\gamma(t)},\qquad \gamma(0) = m,$$ which in a chart is $\dot s^i = X^i(s)$ — an autonomous system exactly as in unit 02.

Unit 01's Picard–Lindelöf gives existence and uniqueness, in a chart; uniqueness patches the chart-local solutions into one maximal curve.

the flow

$\varphi^X_t(m) = \gamma_m(t)$. Where defined, $\varphi^X_{t+s} = \varphi^X_t\circ\varphi^X_s$ and $\varphi^X_0 = \mathrm{id}$: a local one-parameter group of diffeomorphisms, unit 02's group property on a manifold.

complete, and when

$X$ is complete if every integral curve runs for all $t$. Every vector field on a compact manifold is — so all three rotation fields on $S^2$ are, and their flows are genuine rotations.

Why compactness gives completeness. Around each point there is a chart and an $\varepsilon\gt0$ such that every integral curve starting in a smaller neighbourhood exists for $|t|\lt\varepsilon$ — Picard–Lindelöf with a uniform Lipschitz constant on a compact closure. Cover $M$ by finitely many such neighbourhoods and take the least $\varepsilon$; then every curve extends by $\varepsilon$ from wherever it is, indefinitely. The same argument fails on $\mathbb R$ with $X = x^2\partial_x$, whose flow $x_0/(1-tx_0)$ escapes in finite time — unit 01's maximal interval of existence, now a statement about incompleteness.

Incompleteness is how singularities are defined. A spacetime is called singular when a timelike geodesic is incomplete — it ends after finite proper time with nowhere to continue. Unit 36's Schwarzschild interior and unit 40's Penrose theorem are both statements of this shape, and they are statements about vector fields, not about curvature blowing up.

Patching. Two chart-local solutions with the same initial condition agree on the overlap of their domains, because the set where they agree is non-empty, closed by continuity and open by local uniqueness, and the domain interval is connected. So the maximal integral curve is well defined — this is exactly unit 01's continuation argument, with "interval" and "connected" doing the work.

31 · flows · example

Two rotation fields, drawn in the first chart

$R_z = -v\,\partial_u + u\,\partial_v$ generates rotation about the $z$-axis (example 2). Rotation about the $x$-axis — $\dot x = 0,\ \dot y = -z,\ \dot z = y$ in $\mathbb R^3$ — gives the field $$R_x = uv\,\partial_u + \tfrac12\big(1-u^2+v^2\big)\,\partial_v,$$ derived on the $\mathfrak{so}(3)$ slide. At $P$: $R_z = (-1,2)$, $R_x = (2,-1)$.

The orbits. $R_z$ keeps $z$ fixed, so its integral curves are the latitudes, which the chart shows as circles $r = \text{const}$. $R_x$ keeps $x = 2u/(1+u^2+v^2)$ fixed; $x = c$ is $$\big(u - \tfrac1c\big)^2 + v^2 = \tfrac1{c^2} - 1,$$ a circle about $(1/c,0)$. They shrink onto $u = \pm1$, the images of $(\pm1,0,0)$, where $R_x$ vanishes.

Integral curves of $R_z$ (teal) and $R_x$ (orange) in the first chart, computed from the formulas on the left; the thick ones pass through $P$, with both field vectors drawn there at $\tfrac13$ scale. $R_z$ vanishes at the origin, $R_x$ at the two orange dots.

Reading the figure. Every orbit of either field is a circle on the sphere — a cut by a plane perpendicular to the rotation axis — and stereographic projection sends these circles to circles in the chart, as the one-line computation on the slide shows for $R_x$. One member of the $R_x$ family passes through the point at infinity: $c = 0$ is the great circle $x = 0$, which passes through the north pole and so appears as the straight line $u = 0$. Through every point other than the zeros passes one circle of each family, and The bracket measures the failure of flows to commute asks whether following one and then the other is the same as the reverse order.

Completeness, seen. Each orbit is a closed circle traversed in time $2\pi$, so both flows run for all $t$, as compactness promised on the previous slide. In the chart the one $R_x$ orbit $u = 0$ leaves every bounded region — it passes through the north pole, which this chart does not contain — and the second chart is where that orbit stays finite. Incompleteness in one chart is not incompleteness of the field.

Checking $R_x$ at $P$. $uv = 2\cdot1 = 2$ and $\tfrac12(1 - 4 + 1) = -1$. In $\mathbb R^3$ the velocity is $(0,-z,y) = (0,-\tfrac23,\tfrac13)$ at $P$, and its squared length, $\tfrac59$, matches $g_{ij}R_x^iR_x^j = \tfrac19(4+1)$ in the round metric of section 3.

31 · flows

The Lie bracket

definition

For vector fields $X, Y$, the Lie bracket is the commutator of derivations $$[X,Y]f = X(Yf) - Y(Xf).$$

theorem

$[X,Y]$ is a derivation, hence a vector field, with components $$[X,Y]^i = X^j\partial_jY^i - Y^j\partial_jX^i .$$

proof

In a chart, $X(Yf) = X^j\partial_j(Y^i\partial_if) = X^jY^i\partial_j\partial_if + X^j(\partial_jY^i)\partial_if$. Subtract the same with $X\leftrightarrow Y$: the second-derivative terms are equal by Clairaut (unit 03) and cancel, leaving a first-order operator with the stated components. $\square$

properties

Bilinear, antisymmetric, and the Jacobi identity $$[X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]]=0$$ holds because commutators of associative operators always satisfy it. So the vector fields on $M$ form a Lie algebra, infinite-dimensional.

not $C^\infty$-bilinear

$[X, fY] = f[X,Y] + (Xf)\,Y$. So the bracket is not a tensor in $Y$ — it genuinely differentiates.

Cancellation of the second-order terms, in full. $X^jY^i\partial_j\partial_if$ minus $Y^jX^i\partial_j\partial_if$: relabel $i\leftrightarrow j$ in the second term to get $Y^iX^j\partial_i\partial_jf$, which equals the first by Clairaut. So the difference is zero without any assumption on $X$ or $Y$. Clairaut's theorem needs $f$ to be $C^2$ with continuous second derivatives, which is free here.

Proof of the Jacobi identity. Write $L_X$ for the operator $f\mapsto Xf$. Then $[X,Y] \leftrightarrow L_XL_Y - L_YL_X$, and for any associative algebra the double commutator identity $[a,[b,c]]+[b,[c,a]]+[c,[a,b]] = 0$ holds by expanding all twelve products, which cancel in pairs. Nothing about vector fields is used.

Proof of the Leibniz property. $[X,fY]g = X(fYg) - fY(Xg) = (Xf)(Yg) + fX(Yg) - fY(Xg) = (Xf)Yg + f[X,Y]g$. This is the formal statement that the bracket is a derivative, not a pointwise operation: $[X,Y]_m$ depends on $Y$ near $m$, not only at $m$. Contrast $g(X,Y)_m$, which depends only on the values at $m$ — that is what being a tensor means.

Coordinate fields commute. $[\partial_i,\partial_j] = 0$ by Clairaut. The converse is true and useful: $n$ pointwise-independent commuting fields are the coordinate fields of some chart. That is the canonical form theorem, and it is the $n$-dimensional case of Frobenius in section 6.

31 · flows

The bracket measures the failure of flows to commute

theorem

In any chart, for small $t, s$, $$\begin{aligned}&\varphi^Y_{-s}\circ\varphi^X_{-t}\circ\varphi^Y_{s}\circ\varphi^X_{t}(m)\\ &\qquad= m + ts\,[X,Y]_m + O\big((|t|+|s|)^3\big),\end{aligned}$$ and $\ \varphi^X_t\circ\varphi^Y_s = \varphi^Y_s\circ\varphi^X_t$ near $m$ for all small $t,s$ if and only if $[X,Y] = 0$ there.

checked numerically

On the sphere at $P$ with $X = $ $R_z$, $Y = R_x$ (both drawn in Two rotation fields) and $t=s=0.02$: the square closes short by $(7.8804,\ 8.1238)\times10^{-4}$ in the first chart, against the predicted $ts[X,Y]_P = (8,8)\times10^{-4}$ — ratios $0.985$ and $1.015$, and both $\to1$ as $t\to0$.

The square, computed in the first chart from $P$ with $t = s = 0.2$ — large, so the gap shows. Solid legs go forward, dashed legs come back; the red arrow is where the path ends, $(0.067,\,0.096)$ from $P$. The ring is the prediction $P + ts\,[X,Y]_P = P + (0.08,\,0.08)$; at this size the $O(t^3)$ term still shows.

Reading the figure. The four arcs are pieces of the circles drawn in Two rotation fields, drawn in the first chart: an $R_z$ circle about the origin, then an $R_x$ circle, then the same two kinds backwards. Each arc is traced for the same parameter time $0.2$, but the field's speed differs from place to place, so the return arcs are not the outgoing arcs reversed. The mismatch is the red arrow. Its leading part is $ts[X,Y]_P$ with $[R_z,R_x] = -R_y$, and $-R_y = \big(\tfrac12(1+u^2-v^2),\,uv\big) = (2,2)$ at $P$ — the formula for $R_y$ is on the next slide.

Full proof of the leading term

Work in a chart and expand each flow to second order: $\varphi^X_t(s) = s + tX(s) + \tfrac{t^2}{2}(\partial X)X(s) + O(t^3)$, where $(\partial X)X$ has components $X^j\partial_jX^i$ — differentiate $\dot\gamma^i = X^i(\gamma)$ once more and use the chain rule. Compose the four maps, keeping every term of total degree $\le2$. The pure-$t$ and pure-$s$ terms cancel exactly, since the first and third factors are inverse flows of the same field and likewise the second and fourth. What survives at degree two is the single cross term $ts\big(X^j\partial_jY^i - Y^j\partial_jX^i\big) = ts\,[X,Y]^i$. The bookkeeping is four compositions of quadratics; it is mechanical, and problem P12 asks for it only numerically.

The "if and only if". If the flows commute, the left side is $m$ for all small $t,s$, so the $ts$ coefficient vanishes and $[X,Y]_m = 0$. Conversely, $[X,Y] = 0$ implies $Y$ is invariant under the flow of $X$ — by the Lie derivative theorem of the next section, $\mathcal L_XY = [X,Y] = 0$ — so $\varphi^X_t$ carries integral curves of $Y$ to integral curves of $Y$, and the two flows commute. Cited in that direction: the uniqueness of integral curves does the work, and the argument is in Lee, Introduction to Smooth Manifolds, 2nd ed., chapter 9.

The numbers. Computed with exact rotation matrices in $\mathbb R^3$ and the endpoint mapped into the first chart. At $t = s = 0.2$ the ratios are $0.839$ and $1.202$; at $0.1$, $0.925$ and $1.087$; at $0.05$, $0.963$ and $1.041$; at $0.02$, $0.985$ and $1.015$. The error halves as $t$ halves, which is the $O(t^3)$ against $O(t^2)$ signature.

Physical reading. "Move east, then north, then west, then south, and you are not where you started" is the bracket of two vector fields on the sphere. It is not curvature — that is unit 33, and it is the failure of a different square to close, one built from parallel transport rather than from flows. Keeping the two apart is worth the effort now.

31 · flows · example

$\mathfrak{so}(3)$, realised by flows

example 4 · the three rotations of the sphere

Differentiate $u = x/(1-z)$, $v = y/(1-z)$ along each rotation of $\mathbb R^3$, using $z/(1-z) = (r^2-1)/2$:

$$\begin{aligned} R_z &= -v\,\partial_u + u\,\partial_v,\\ R_x &= uv\,\partial_u + \tfrac12(1-u^2+v^2)\,\partial_v,\\ R_y &= -\tfrac12(1+u^2-v^2)\,\partial_u - uv\,\partial_v . \end{aligned}$$

Apply $[X,Y]^i = X^j\partial_jY^i - Y^j\partial_jX^i$:

$$[R_x,R_y] = -R_z,\ \ [R_y,R_z] = -R_x,$$ $$[R_z,R_x] = -R_y .$$

A three-dimensional Lie subalgebra, and $J_k\mapsto -R_k$ is an isomorphism onto unit 16's $\mathfrak{so}(3)$: $[-R_x,-R_y] = [R_x,R_y] = -R_z$.

the minus sign is not a slip

Unit 16's matrices obey $[J_x,J_y] = +J_z$. Fields generated by a left action obey the opposite sign: $A\mapsto R_A$ is an anti-homomorphism. Stated once, here.

Deriving $R_x$. Rotation about the $x$-axis has $\dot x = 0$, $\dot y = -z$, $\dot z = y$ at $t=0$. Then $\dot u = \frac{\dot x(1-z)+x\dot z}{(1-z)^2} = \frac{xy}{(1-z)^2} = uv$, and $\dot v = \frac{\dot y(1-z)+y\dot z}{(1-z)^2} = \frac{-z(1-z)+y^2}{(1-z)^2} = -\frac{z}{1-z} + v^2 = \frac{1-r^2}{2} + v^2 = \tfrac12(1-u^2+v^2)$, using $z/(1-z) = (r^2-1)/2$, which follows from $1-z = 2/(1+r^2)$. $R_y$ is the same computation with $\dot x = z$, $\dot z = -x$.

A bracket, in full. $[R_z,R_x]^1 = R_z^j\partial_jR_x^1 - R_x^j\partial_jR_z^1$ $= (-v)(v) + (u)(u) - \big[uv\cdot 0 + \tfrac12(1-u^2+v^2)(-1)\big] = u^2 - v^2 + \tfrac12(1-u^2+v^2) = \tfrac12(1+u^2-v^2) = -R_y^1$. The second component is $[R_z,R_x]^2 = (-v)\cdot(-u) + u\cdot v - \big[uv\cdot 1 + \tfrac12(1-u^2+v^2)\cdot0\big] = uv = -R_y^2$. All six brackets were verified symbolically.

Why an anti-homomorphism. The flow of $R_A$ is $m\mapsto \exp(tA)\,m$, left multiplication. Composing left multiplications reverses order relative to composing the corresponding operators on functions, and the sign is the residue of that reversal. Using the right action instead, or defining $R_A$ with a minus, restores $+$. Nothing physical depends on it; a great deal of confusion does.

The field in the other chart. Because inversion commutes with rotation about the polar axis, $\tilde R_z = -\tilde v\,\partial_{\tilde u} + \tilde u\,\partial_{\tilde v}$ — the same form. $R_x$ and $R_y$ do not have that luxury. $R_z$ vanishes at the origin of the first chart (the south pole) and at the origin of the second (the north pole): two zeros, which the hairy ball theorem later in this deck says cannot be avoided.

section 5

The Lie derivative, and the fields that move a metric nowhere

A derivative of tensors that needs no connection — only a flow to drag them along.

31 · Lie derivative

Differentiating a tensor field along a flow

definition · the pullback

For a diffeomorphism $F$ and a $(0,q)$-tensor $T$, the pullback is $$(F^*T)_m(X_1,\dots,X_q) = T_{F(m)}\big(\mathrm{d}F_mX_1,\dots\big).$$ For $(p,0)$ use $(F^{-1})^*$ on the covector slots; mixed types slot by slot.

definition · the derivative

$$\mathcal L_XT\ =\ \frac{\mathrm{d}}{\mathrm{d}t}\Big|_{t=0}\big(\varphi^X_t\big)^*T .$$

A symmetric $(0,2)$-tensor drawn as the ellipse $\{X : T(X,X) = 1\}$ in each tangent space (schematic). $T$ at $\varphi_t(m)$ cannot be compared with $T$ at $m$; its pullback (dashed) can. The difference, divided by $t$, tends to $\mathcal L_XT$ at $m$.

Reading the figure. The ellipse is the set of vectors of unit "$T$-length", which for a metric is the unit circle of that tangent space; it is the most faithful picture of a symmetric positive $(0,2)$-tensor. The flow carries $m$ to $\varphi_t(m)$, and its differential carries vectors at $m$ to vectors there; the pullback asks which vectors at $m$ land on the far ellipse, and that set is the dashed ellipse. If the flow is an isometry the dashed and solid ellipses coincide for every $t$, and $\mathcal L_XT = 0$: that is the Killing condition at the end of this section, seen in advance.

On functions ($q = 0$): $(\varphi_t^*f)(m) = f(\varphi_t(m))$, so $\mathcal L_Xf = \frac{\mathrm{d}}{\mathrm{d}t}\big|_0 f(\varphi_t(m)) = X_mf$ — the Lie derivative of a function is just $Xf$, and the definition is consistent with everything so far.

Well-definedness. The derivative exists because $t\mapsto(\varphi_t^*T)_m$ is a smooth curve in the fixed finite-dimensional vector space $T^p_q(T_mM)$ — smooth because the flow depends smoothly on $t$ and on the initial point, which is the smooth-dependence half of unit 01's theorem. That is where the differentiability comes from; it is not assumed.

31 · Lie derivative

Why the flow, and what it costs

why this is the right idea

$T_{\varphi_t(m)}$ and $T_m$ live in different vector spaces and cannot be subtracted. The flow supplies the only canonical map between them — so the difference quotient becomes legal, and no extra structure is invented.

the price

$\mathcal L_XT$ at $m$ depends on $X$ near $m$, not only on $X_m$. It is therefore not a directional derivative: there is no "$\mathcal L_vT$" for a single vector $v$. That is exactly the gap unit 32's connection fills.

Two fields equal at $m$, different nearby: translation, and rotation about $(0,1)$. Their Lie derivatives of the covector $\mathrm du$ at $m$ are $\mathrm d(Xu) = \mathrm d(1) = 0$ and $\mathrm d(X'u) = \mathrm d(1-v) = -\mathrm dv$. Same vector at $m$, different answer.

Reading the figure. For the differential of a function, $\mathcal L_X\,\mathrm df = \mathrm d(Xf)$ — the Lie derivative commutes with $\mathrm d$, which follows from the covector formula on the next slide: $(\mathcal L_X\mathrm df)_i = X^k\partial_k\partial_if + \partial_kf\,\partial_iX^k = \partial_i(X^k\partial_kf)$. With $f = u$ that gives $\mathrm d(X^1)$: $0$ for the translation, whose first component is the constant $1$, and $-\mathrm dv$ for the rotation, whose first component is $1-v$. Both fields are $\partial_u$ at $m$. The rotation's flow turns the level lines of $u$ as it carries them, the translation's does not, and no information at the single point $m$ can tell the two apart.

Properties, all immediate from the definition. $\mathcal L_X$ is $\mathbb R$-linear, obeys Leibniz over tensor products, $\mathcal L_X(fT) = (Xf)T + f\mathcal L_XT$, and commutes with contraction — because the pullback does all four, and the derivative at $t = 0$ inherits them. These four properties characterise it uniquely given its action on functions and vector fields, which is the cleanest way to remember the coordinate formulas rather than memorising them.

History. Named for Sophus Lie, who studied continuous transformation groups in the 1870s and 80s, but written in this form by Ślebodziński in 1931; van Dantzig named it in 1932 and Schouten made it standard. Lie himself had the bracket; the derivative of a general tensor along a flow is later, and is the object that lets "symmetry" be defined without any metric or connection.

31 · Lie derivative

The formulas, and the theorem behind the first one

theorem

$\mathcal L_XY = [X,Y]$ for vector fields.

in components

$$\mathcal L_Xf = X^k\partial_kf$$ $$(\mathcal L_XY)^i = X^k\partial_kY^i - Y^k\partial_kX^i$$ $$(\mathcal L_X\omega)_i = X^k\partial_k\omega_i + \omega_k\,\partial_iX^k$$ $$(\mathcal L_XT)_{ij} = X^k\partial_kT_{ij} + T_{kj}\partial_iX^k + T_{ik}\partial_jX^k$$

the pattern

Transport term $X^k\partial_k$, then one correction per slot: $+$ and $\partial_{\text{slot}}X^k$ for each lower index, $-$ and $\partial_kX^{\text{slot}}$ for each upper one. Count the slots and you have the formula for any type.

no metric anywhere

Every formula above uses only $\partial$ and the chart. That is the whole appeal: a symmetry can be stated before any geometry is chosen.

Full proof that $\mathcal L_XY = [X,Y]$

Work in a chart with $\varphi_t$ the flow of $X$. For a vector field the pullback is $(\varphi_t^*Y)_m = \mathrm{d}(\varphi_{-t})_{\varphi_t(m)}\,Y_{\varphi_t(m)}$. Expand each factor to first order in $t$. Write $s = \mathbf x(m)$. First, $\varphi_t(s) = s + tX(s) + O(t^2)$, so $Y(\varphi_t(s)) = Y(s) + t\,(\partial_kY)(s)X^k(s) + O(t^2)$. Second, the Jacobian of $\varphi_{-t}$ at $\varphi_t(s)$ is $\mathbb 1 - t\,\partial X(s) + O(t^2)$, since $\partial\varphi_{-t}^i/\partial s^j = \delta^i_j - t\,\partial_jX^i + O(t^2)$. Multiplying, $$(\varphi_t^*Y)^i(s) = Y^i + t\big(X^k\partial_kY^i - Y^k\partial_kX^i\big) + O(t^2),$$ and differentiating at $t=0$ gives $[X,Y]^i$. $\square$

The covector formula, derived the same way. $(\varphi_t^*\omega)_m(v) = \omega_{\varphi_t(m)}(\mathrm{d}\varphi_t v)$; now the Jacobian is $\mathbb 1 + t\,\partial X$ and it sits on the argument, so the correction changes sign relative to the vector case: $X^k\partial_k\omega_i + \omega_k\partial_iX^k$. The $(0,2)$ formula is the same computation with two slots, and the general formula follows by the Leibniz rule over tensor products.

A sanity check worth doing once. Contract $\omega_iY^i$, a function. The Lie derivative of the contraction must be $X^k\partial_k(\omega_iY^i)$. Adding the covector and vector formulas: $Y^i(X^k\partial_k\omega_i + \omega_k\partial_iX^k) + \omega_i(X^k\partial_kY^i - Y^k\partial_kX^i)$. The second and fourth terms cancel after relabelling $i\leftrightarrow k$, leaving $X^k\partial_k(\omega_iY^i)$. ✓ The signs are forced by this consistency, which is a better way to remember them than the mnemonic.

31 · Lie derivative

Why $\mathcal L_XT$ is a tensor and $\partial_kT_{ij}$ is not

Differentiate $\tilde T_{ab} = S^i{}_aS^j{}_b\,T_{ij}$ in the tilded chart:

$$\begin{aligned}\tilde\partial_c\tilde T_{ab} = \ &S^k{}_cS^i{}_aS^j{}_b\,\partial_kT_{ij}\\ +\ &\big(\tilde\partial_cS^i{}_a\big)S^j{}_bT_{ij} + S^i{}_a\big(\tilde\partial_cS^j{}_b\big)T_{ij}.\end{aligned}$$

The last two terms appear in no tensor law: they carry $\partial^2x/\partial\tilde x\,\partial\tilde x$, second derivatives of the transition map.

what the extra terms in $\mathcal L_X$ do

In $X^k\partial_kT_{ij} + T_{kj}\partial_iX^k + T_{ik}\partial_jX^k$, the corrections carry the same second derivatives with the opposite sign, because $\partial_iX^k$ is non-tensorial in the mirrored way. They cancel.

but only together

No term alone is a tensor. Splitting the formula across charts is the commonest way to get a wrong answer here.

Full proof, for the $(0,1)$ case, which carries the whole idea

Let $\omega$ be a covector field, $X$ a vector field, $A^a{}_i = \partial\tilde x^a/\partial x^i$, $S^i{}_a$ its inverse. Then $\tilde\omega_a = S^i{}_a\omega_i$ and $\tilde X^a = A^a{}_iX^i$. Compute the two pieces in the tilded chart:

$\tilde X^c\tilde\partial_c\tilde\omega_a = A^c{}_kX^k\,S^l{}_c\partial_l\big(S^i{}_a\omega_i\big) = X^l\partial_l\big(S^i{}_a\omega_i\big) = S^i{}_aX^l\partial_l\omega_i + \big(X^l\partial_lS^i{}_a\big)\omega_i,$ using $A^c{}_kS^l{}_c = \delta^l_k$.

$\tilde\omega_c\,\tilde\partial_a\tilde X^c = S^i{}_c\omega_i\,S^l{}_a\partial_l\big(A^c{}_kX^k\big) = S^i{}_cA^c{}_kS^l{}_a\omega_i\partial_lX^k + S^i{}_c\big(S^l{}_a\partial_lA^c{}_k\big)X^k\omega_i = S^l{}_a\omega_k\partial_lX^k + S^i{}_cS^l{}_a\big(\partial_lA^c{}_k\big)X^k\omega_i .$

The two unwanted terms are $\big(X^l\partial_lS^i{}_a\big)\omega_i$ and $S^i{}_cS^l{}_a\big(\partial_lA^c{}_k\big)X^k\omega_i$. Differentiate $S^i{}_cA^c{}_k = \delta^i_k$ along $\partial_l$: $\ (\partial_lS^i{}_c)A^c{}_k + S^i{}_c\partial_lA^c{}_k = 0$, so $S^i{}_c\partial_lA^c{}_k = -(\partial_lS^i{}_c)A^c{}_k$. Substituting turns the second unwanted term into $-S^l{}_a(\partial_lS^i{}_c)A^c{}_kX^k\omega_i = -X^l\partial_lS^i{}_a\,\omega_i$ after using $A^c{}_kX^k = \tilde X^c$ and $S^l{}_a\partial_l = \tilde\partial_a$. It is the negative of the first. They cancel, so $\tilde X^c\tilde\partial_c\tilde\omega_a + \tilde\omega_c\tilde\partial_a\tilde X^c = S^i{}_a\big(X^l\partial_l\omega_i + \omega_k\partial_iX^k\big)$, the covector law. $\square$

The $(0,2)$ and general cases are the same manipulation once per slot, and the cancellation is slot by slot. The identity that does all the work — differentiate $SA = \mathbb 1$ — is worth remembering: it reappears in unit 32 when the Christoffel symbols' inhomogeneous term is computed.

31 · Lie derivative

Killing vector fields

definition

$X$ is a Killing vector field for a metric $g$ if $\mathcal L_Xg = 0$, that is $$X^k\partial_kg_{ij} + g_{kj}\,\partial_iX^k + g_{ik}\,\partial_jX^k = 0 .$$

theorem · what it means

$\mathcal L_Xg = 0$ iff every $\varphi^X_t$ is an isometry: $(\varphi^X_t)^*g = g$ for all $t$ in the domain.

proof

$\frac{\mathrm d}{\mathrm dt}(\varphi_t^*g) = \varphi_t^*(\mathcal L_Xg)$ by the group law $\varphi_{t+h} = \varphi_h\circ\varphi_t$. So $\varphi_t^*g$ is constant in $t$ iff $\mathcal L_Xg$ vanishes; at $t=0$ it is $g$. $\square$

why this is planted here

A Killing field is a continuous symmetry of a geometry, stated with no coordinates and no connection. Unit 36 will read Schwarzschild's conserved energy and angular momentum straight off two Killing fields — the same bookkeeping as unit 13's Noether theorem, with the metric as the Lagrangian.

a Killing field is rare

The equation is $n(n+1)/2$ PDEs for $n$ functions: overdetermined. A generic metric has none. The maximum is $n(n+1)/2$, attained only by maximally symmetric spaces — $10$ in four dimensions, as Minkowski has, and $6$ for the maximally symmetric spatial slices unit 39 needs.

Why the equation is overdetermined, precisely. $\mathcal L_Xg$ is symmetric in $ij$, so it has $n(n+1)/2$ independent components, and the unknown is the $n$-tuple $X^i$. For $n = 4$ that is ten equations for four functions. The solution space is nevertheless a finite-dimensional Lie algebra, of dimension at most $n(n+1)/2$: a Killing field is determined by $X_m$ and the antisymmetric part of $\partial X$ at any one point, which is $n + n(n-1)/2$ numbers. Cited: the proof needs unit 32's covariant derivative, where the Killing equation becomes $\nabla_iX_j + \nabla_jX_i = 0$ and its second derivative is determined by curvature.

The bracket of two Killing fields is Killing, since $\mathcal L_{[X,Y]} = \mathcal L_X\mathcal L_Y - \mathcal L_Y\mathcal L_X$ on tensors — the Lie derivative is a representation of the Lie algebra of vector fields. So the Killing fields form a Lie algebra: exactly what example 4 exhibited for the sphere, whose three rotations are all Killing and close into $\mathfrak{so}(3)$, dimension $3 = 2\cdot3/2$. The sphere is maximally symmetric, and unit 39 will use that fact.

The Noether connection, in advance. Once unit 32 supplies geodesics, $g(X,\dot\gamma)$ is constant along every geodesic when $X$ is Killing — one line from the geodesic equation. For Schwarzschild, $\partial_t$ gives energy and $\partial_\phi$ gives angular momentum, and unit 37's Mercury calculation is impossible without them. That is why this object, which belongs to pure differential geometry, is planted in a physics course five units early.

31 · Lie derivative · example

The rotation field is Killing; the dilation is not

example 5 · the running example, closed

Take $g_{ij} = \Omega^2\delta_{ij}$ with $\Omega^2 = 4/(1+r^2)^2$, and $R_z = (-v,\,u)$. Since $X^k\partial_k r^2 = -2uv + 2uv = 0$, the transport term dies: $X^k\partial_kg_{ij} = \frac{\mathrm{d}\Omega^2}{\mathrm{d}(r^2)}\big(X^k\partial_kr^2\big)\delta_{ij} = 0$. With $\partial_1X^1 = \partial_2X^2 = 0$, $\partial_2X^1 = -1$, $\partial_1X^2 = +1$:

$$(\mathcal L_Xg)_{11} = 2\Omega^2\partial_1X^1 = 0,$$ $$(\mathcal L_Xg)_{12} = \Omega^2(\partial_1X^2+\partial_2X^1) = 0,$$ $$(\mathcal L_Xg)_{22} = 2\Omega^2\partial_2X^2 = 0 .$$

So $\mathcal L_{R_z}g = 0$: $R_z$ is Killing, and its flow — rotation by $t$ — is an isometry of the round sphere, as geometry insists.

Latitudes are the orbits of $R_z$. Its flow carries the short arc at $P$ round its latitude without changing its length; $R_z$ vanishes at both poles.

Reading the figure. The flow of $R_z$ is rotation of the sphere about its axis, so every latitude is carried onto itself and every arc onto an arc of the same length — the geometric content of $\mathcal L_{R_z}g = 0$. In the chart, the computation on the slide saw this as two facts: the transport term dies because $R_z$ runs along the circles $r = \text{const}$, on which $\Omega^2$ is constant; and the two correction terms cancel because $\partial_iX^j$ is antisymmetric, an infinitesimal rotation of the chart plane.

All three rotations are Killing — verified symbolically for $R_x$ and $R_y$ too, where the transport term does not vanish and the cancellation against the correction terms is genuine. Problem P10 asks for $R_x$ by hand, which is the version worth doing.

Closing the running example. The sphere has been built as a manifold, given a vector, given a metric, given a Lie algebra of flows, and shown to have a symmetry — all without ever preferring one of the two charts, and with every component statement checked in both. That is the whole method of the unit, and of the two units after it.

31 · Lie derivative · example

The dilation stretches the south and shrinks the north

contrast: the dilation

$D = u\,\partial_u + v\,\partial_v$ has $\partial_iD^j = \delta_i^j$ and $D^k\partial_kr^2 = 2r^2$, giving $$\mathcal L_Dg = \frac{2(1-r^2)}{1+r^2}\,g\ \ne 0 .$$ In terms of the height, $r^2 = \frac{1+z}{1-z}$ makes this $\boxed{\mathcal L_Dg = -2z\,g}$: zero only on the equator. $D$ is conformal Killing: it preserves angles, not lengths. At $P$, where $r^2 = 5$, the factor is $-\tfrac43$.

Left: the meridians, along which the flow of $D$ runs from $S$ to $N$. Right: the rate $\lambda$ in $\mathcal L_Dg = \lambda g$ against height, the straight line $\lambda = -2z$ — lengths grow below the equator, shrink above it, and at $P$ the rate is $-\tfrac43$.

The dilation computation. $(\mathcal L_Dg)_{ij} = D^k\partial_k(\Omega^2)\delta_{ij} + 2\Omega^2\delta_{ij}$, using $\partial_iD^k = \delta_i^k$ for the two correction terms. Now $\partial_{(r^2)}\Omega^2 = -8/(1+r^2)^3$, so $D^k\partial_k\Omega^2 = 2r^2\cdot(-8)/(1+r^2)^3 = -16r^2/(1+r^2)^3$, and $$(\mathcal L_Dg)_{ij} = \Big[\frac{-16r^2}{(1+r^2)^3} + \frac{8}{(1+r^2)^2}\Big]\delta_{ij} = \frac{8(1-r^2)}{(1+r^2)^3}\delta_{ij} = \frac{2(1-r^2)}{1+r^2}\,g_{ij}. \checkmark$$ At $r^2 = 5$ the factor is $2(-4)/6 = -4/3$. Verified symbolically as well.

From $r$ to $z$. With $r^2 = (1+z)/(1-z)$, $1 - r^2 = -2z/(1-z)$ and $1 + r^2 = 2/(1-z)$, so $\frac{2(1-r^2)}{1+r^2} = -2z$. A formula computed in one chart has become a function on the sphere — the height — which the second chart must reproduce, and it does: in the second chart $D = -(\tilde u\,\partial_{\tilde u} + \tilde v\,\partial_{\tilde v})$, since inversion turns $u\mapsto\mathrm e^tu$ into $\tilde u\mapsto\mathrm e^{-t}\tilde u$, and the same computation with the sign reversed gives $-\frac{2(1-\tilde r^2)}{1+\tilde r^2}$. In the second chart $z = (1-\tilde r^2)/(1+\tilde r^2)$ — the sign flipped by projecting from the other pole — so this is $-2z$ again.

What $D$ does geometrically. Its flow is $u\mapsto\mathrm e^tu$, which on the sphere slides every point along a meridian from the south pole towards the north — a "boost" in the Riemann-sphere picture of unit 20. It preserves angles because stereographic projection is conformal, and stretches lengths except on the equator, where the two effects balance. The round sphere has a six-dimensional conformal algebra and only a three-dimensional isometry algebra; $D$ is in the difference.

section 6

Submanifolds, and what covariance does not say

Three tools taken on trust, one theorem about when a bracket integrates, and one warning about a slogan.

31 · submanifolds

Submanifolds, immersions, embeddings

definition

$F: M\to N$ is an immersion if $\mathrm{d}F_m$ is injective everywhere, a submersion if surjective everywhere, an embedding if it is an immersion and a homeomorphism onto its image. An embedded submanifold is the image of an embedding.

regular value theorem · cited

If $F: N\to\mathbb R^k$ is smooth and $c$ is a regular value — $\mathrm{d}F$ surjective at every point of $F^{-1}(c)$ — then $F^{-1}(c)$ is an embedded submanifold of dimension $\dim N - k$. Proved from unit 04's implicit function theorem.

two instances

$S^2 = F^{-1}(1)$ for $F = x^2+y^2+z^2$, with $\mathrm{d}F = 2(x,y,z)\ne0$ there: dimension $3-1 = 2$. ✓

$O(3) = F^{-1}(\mathbb 1)$ for $F(A) = A^{\mathsf T}A$ into the $6$-dimensional symmetric matrices: $9-6 = 3$, and $SO(3)$ is its identity component. ✓

Why $\mathrm{d}F$ is surjective for $SO(3)$. At $A\in SO(3)$, differentiate $F(A) = A^{\mathsf T}A$ in the direction $H$: $\mathrm{d}F_A(H) = H^{\mathsf T}A + A^{\mathsf T}H$. Given any symmetric $\Sigma$, take $H = \tfrac12A\Sigma$; then $\mathrm{d}F_A(H) = \tfrac12(\Sigma A^{\mathsf T}A + A^{\mathsf T}A\Sigma) = \Sigma$. So $\mathrm{d}F_A$ is onto and $\mathbb 1$ is a regular value. Its kernel is $\{H : A^{\mathsf T}H \text{ antisymmetric}\}$, which is $\mathfrak{so}(3)$ translated — unit 16's $\Omega = R^{\mathsf T}\dot R$, now identified as the tangent space to a submanifold.

Whitney's theorem, cited: every smooth $n$-manifold embeds in $\mathbb R^{2n}$. So nothing is lost by refusing to use an ambient space, and nothing is gained by using one, since the embedding is never canonical. Nash's theorem strengthens this to isometric embeddings for Riemannian metrics, at the cost of more dimensions.

31 · submanifolds

Injective and immersed, yet not embedded

the figure eight

$F(t) = (\sin2t,\ \sin t)$ on the open interval $(-\pi,\pi)$. Its velocity $(2\cos2t,\ \cos t)$ never vanishes — where $\cos t = 0$, $\cos 2t = -1$ — so $F$ is an immersion; and $F$ is injective, since $\sin t$ fixes $t$ up to $t\mapsto\pm\pi - t$ and $\sin2t$ then separates the two (notes).

But the image is not a copy of the interval. As $t\to\pm\pi$, $F(t)\to(0,0) = F(0)$: points far from $0$ in the interval come arbitrarily close to $F(0)$ in the plane, so $F^{-1}$ is not continuous on the image.

The image of $(-\pi,\pi)$, computed. It passes through the origin once, at $t = 0$, and both open ends run back towards that point without reaching it. No neighbourhood of the origin in the plane meets the curve in one interval around $t = 0$.

Reading the figure. The strand through the origin is the image of a small interval around $t = 0$, crossing with direction $(2,1)$. The two open ends, approaching along direction $(\mp2,\pm1)$, belong to $t$ near $\pm\pi$. Any disc about the origin contains pieces of all three, so the subspace topology on the image puts $t = 0$ next to $t = \pm\pi$ — a topology the interval does not have. $F$ is a continuous bijection onto its image with a discontinuous inverse, which is exactly what "not an embedding" means.

Checking injectivity. If $\sin t = \sin t'$ with $t\ne t'$ in $(-\pi,\pi)$, then $t' = \pi - t$ (or $-\pi - t$), and then $\sin 2t' = -\sin 2t$; equality of the first components forces $\sin2t = 0$, so $t\in\{0,\pm\tfrac\pi2\}$, and each of those has $t' \notin(-\pi,\pi)$ or $t' = t$. The ends $t = \pm\pi$, which would collide with $t = 0$, are excluded by taking the interval open.

The distinction matters physically. An immersed submanifold can come arbitrarily close to itself; the irrational line on the torus, $t\mapsto(t,\alpha t)$ with $\alpha$ irrational, is injectively immersed and dense. That is exactly unit 17's quasiperiodic orbit on an invariant torus, and the reason a Poincaré section of it fills a curve rather than a finite set of points. Embedded submanifolds are the well-behaved ones, and every level set of a regular value is embedded.

31 · submanifolds

Frobenius: when a bracket closes, a foliation exists

definition

A rank-$k$ distribution $\mathcal D$ assigns to each $m$ a $k$-dimensional subspace $\mathcal D_m\subseteq T_mM$, smoothly. It is involutive if $X, Y$ tangent to $\mathcal D$ implies $[X,Y]$ is too; integrable if every point lies on a $k$-dimensional submanifold whose tangent spaces are the $\mathcal D$'s.

Frobenius' theorem · cited

Involutive $\iff$ integrable. Moreover there are charts in which the leaves are the slices $x^{k+1},\dots,x^n = \text{const}$.

the direction that is easy

Integrable $\Rightarrow$ involutive: on a leaf, $X$ and $Y$ are tangent, so their flows stay on it, so the commutator square stays on it, so $[X,Y]$ is tangent. The converse is the theorem.

31 · submanifolds · example

The parallel-parking distribution

the standard failure

On $\mathbb R^3$: $X = \partial_x$, $Y = \partial_y + x\,\partial_z$. Then $$[X,Y] = \partial_z \notin \mathrm{span}\{X,Y\},$$ so no surface has this plane field as its tangent planes anywhere. Problem P13.

Run the square from the origin. Along $X$ by $t$: $(t,0,0).$ Along $Y$ by $s$, which lifts at rate $x = t$: $(t,s,ts).$ Back along $X$: $(0,s,ts).$ Back along $Y$, now at $x = 0$ where $Y$ does not lift: $(0,0,ts).$ Every leg was tangent to the planes, and the path ends straight above its start — out of the plane there.

The planes $\mathrm{span}\{\partial_x,\ \partial_y + x\,\partial_z\}$ at five points of the $x$-axis (oblique view, schematic). Each contains the $x$-direction; the slope in $y$ is $x$, so they twist as $x$ grows. The dashed path is the square run on the left with $t = s = 0.9$; it ends straight above the origin, along the bracket, out of the plane there.

Reading the figure. Walk along the $x$-axis and watch the planes: each contains the axis itself, and the tilt turns steadily — flat at $x = 0$, rising one way for $x \gt 0$ and the other way for $x \lt 0$. That steady turning is $[X,Y] = \partial_z$ made visible.

Why no surface fits, exactly. Suppose a surface $\Sigma$ through the origin had these planes as its tangent planes. The fields $X$ and $Y$ are tangent to $\Sigma$, so their integral curves starting on $\Sigma$ stay on it, and the square on the slide never leaves $\Sigma$. Its endpoints $(0,0,ts)$ for all small $t,s$ then fill a segment of the $z$-axis inside $\Sigma$, so $\partial_z$ is tangent to $\Sigma$ at the origin — but the plane there is the $(x,y)$-plane. Contradiction. The commutator-square theorem of section 4 predicts the gap $ts[X,Y]$ to leading order; for these polynomial fields it is exact.

Where it was used earlier, unnamed. A constrained mechanical system whose constraints are non-integrable — a rolling ball, a skate — is non-holonomic, and non-holonomic means exactly that the constraint distribution is not involutive. That is why a car can reach any position and orientation despite having only two controls: the bracket of "drive" and "steer" generates the sideways direction the car cannot drive in. The example on this slide is the parallel-parking distribution.

31 · submanifolds

Three results used, not proved

partitions of unity

Every open cover of a smooth manifold admits a subordinate smooth partition of unity. Cited: needs second countability and bump functions.

Buys: existence of Riemannian metrics and of connections; the ability to glue local constructions.

Whitney embedding

Every smooth $n$-manifold embeds in $\mathbb R^{2n}$. Cited: Whitney 1936, 1944.

Buys: the reassurance that the abstract definition adds no exotica — and the point that the embedding is never canonical, so it must not be used.

hairy ball

Every smooth vector field on $S^{2k}$, $k\ge1$, vanishes somewhere. Cited: Poincaré–Hopf — the indices of the zeros sum to the Euler characteristic $\chi(M)$, which is vertices $-$ edges $+$ faces of any triangulation: $\chi(S^2) = 4-6+4 = 2$.

Buys: $R_z$'s two zeros are compulsory; and $S^2$ admits no Lorentzian metric.

the rule this course keeps

Each is labelled where it is used, with what it buys. None of the three is needed to follow a computation in units 32 to 40 — they guarantee that objects exist, and the computations then proceed with the objects in hand.

Partitions of unity, what the statement is. Given an open cover $\{U_\alpha\}$ there are smooth $\psi_\alpha: M\to[0,1]$ with $\operatorname{supp}\psi_\alpha\subseteq U_\alpha$, the supports locally finite, and $\sum_\alpha\psi_\alpha = 1$. The construction starts from the bump function $h(t) = \mathrm e^{-1/t}$ for $t\gt0$ and $0$ otherwise, which is smooth and not analytic — the same function that makes unit 11's fundamental lemma of the calculus of variations work. This is the single place where smooth differs sharply from analytic: a non-zero analytic function cannot vanish on an open set, so there are no analytic partitions of unity, and none of this machinery works in the analytic category.

Poincaré–Hopf, and the $2$. The index of an isolated zero is the degree of $X/\lVert X\rVert$ on a small sphere around it. For $R_z$ near either pole, the field in the local chart is $(-v, u)$, a rotation, whose index is $+1$; two poles give $1 + 1 = 2 = \chi(S^2)$. So the theorem is not merely an obstruction but an exact count, and the field we have been carrying since section 2 realises it with the minimum possible number of zeros.

No Lorentzian metric on $S^2$. A Lorentzian metric gives at each point a timelike direction up to sign — a line field. On a compact surface a line field forces $\chi = 0$ (pass to the orientation double cover if needed to get a vector field). $\chi(S^2) = 2\ne0$, so none exists. A compact spatial universe is fine; a compact spacetime is topologically restricted, which is one reason closed timelike curves are hard to avoid in such models.

31 · covariance · physics

General covariance says less than it sounds like

Kretschmann's objection, 1917

"The laws take the same form in all coordinate systems" is not a restriction on a theory: any theory can be rewritten that way. Newtonian gravity in generally covariant form is Newton–Cartan theory, 1923. So general covariance alone has no physical content.

Einstein conceded the point and reformulated: what matters is which fields appear, not what the equations look like.

what does carry content

No absolute objects. Every field in the equations is dynamical — there is no fixed background metric handed in from outside. Newton–Cartan keeps one; general relativity does not.
Diffeomorphism invariance. If $(M,g,\Phi)$ solves the equations then so does $(M,F^*g,F^*\Phi)$ for any diffeomorphism $F$, and the two are the same physical situation.

The hole argument, which cost Einstein two years. Take a solution and a diffeomorphism $F$ equal to the identity outside a compact "hole" and non-trivial inside. Then $F^*g$ is also a solution, agreeing with $g$ everywhere outside the hole and differing inside. If points of $M$ had identities of their own, the field equations would fail to determine the field at a point inside the hole from data outside — determinism would fail. Einstein took this in 1913 as proof that generally covariant equations were impossible, abandoned the programme, and returned to it only in late 1915.

The resolution. Points of $M$ have no physical identity independent of the fields. "The point where the two worldlines cross" is meaningful; "the point $(t,x,y,z)$" is not. $(M,g)$ and $(M,F^*g)$ are the same spacetime described twice, so nothing is undetermined. This is the sharpest statement of what this unit is for: coordinates are not data, and neither, it turns out, are the points themselves — only the fields and their relations.

Anderson's absolute objects (1967) is the attempt to make "no fixed background" precise: an absolute object is one that is the same, up to diffeomorphism, in every solution. The Minkowski metric of special relativity is absolute; the metric of general relativity is not. The definition has known difficulties and no universally accepted repair, which is worth knowing before treating "background independence" as a settled criterion. It is nevertheless the right instinct, and it is what distinguishes unit 35's field equations from every theory before them.

31 · checkpoint

Checkpoint

you can now do all four

Construct an atlas for a given manifold and verify the transition maps smooth.
Define a tensor field in one chart, transform it to another, and verify the two agree on the overlap.
Compute Lie brackets and Lie derivatives in coordinates, for tensors of any type.
Decide whether a given vector field is Killing for a given metric.

the unit in one line

Coordinates are labels; the transformation law is a theorem; the only derivative available for free is the Lie derivative, and it defines what a symmetry is.

Unit 32 adds what is not free: a connection, so that $\nabla_iV^j$ is a tensor and a straight line has a meaning. Then unit 33 finds the part of it that no chart can remove.

Thirteen problems, about six hours: ./mr problems 31. When the four items above are true: ./mr done 31.

The falsifiable form: take the torus $T^2$ with the double pendulum's kinetic-energy metric (problem P2), a $(0,2)$-tensor field written in one of its four charts, and the field $\partial_{\theta_1}+\partial_{\theta_2}$. If you cannot produce the atlas, transform the tensor onto an overlap and get agreement, compute $\mathcal L_Xg$ in coordinates, and say whether it vanishes — then the checkpoint is false and unit 32 will not be readable.

Two habits to carry forward. First: whenever a formula is written in components, ask which chart, and whether the statement survives a change of it. Second: whenever a derivative of a tensor appears, ask what makes it tensorial — in this unit the answer was always "antisymmetrisation or a flow", and in unit 32 it will always be "a connection".