Smooth manifolds · The Lumenary

MotivationH

Calculus without global coordinates

The surface of the Earth is a sphere, and no single map of it is faithful everywhere: Mercator's projection sends the poles to infinity, and every flat map tears the sphere somewhere. Yet navigators have always worked happily with local maps that overlap, and so does every GPS receiver today. Wherever two maps overlap, there is a rule converting one set of coordinates into the other.

The same situation appears all over mathematics and physics. The positions of a double pendulum form a torus $S^1 \times S^1$ (two angles); the orientations of a rigid body — a satellite, a drone, a molecule — form the group $SO(3)$ of rotations, which is not a subset of any $\R^3$ in any natural way; the solutions of $x^2 + y^2 - z^2 = 1$ form a hyperboloid; and in general relativity spacetime is a four-dimensional manifold with no preferred coordinates at all. In each case we want to differentiate (velocities, forces, gradients) and integrate (work, flux, volume), but there are no global coordinates in which to do it.

A smooth manifold is the minimal structure that makes this possible: a space covered by overlapping coordinate charts whose changes of coordinates are smooth. Everything in this lesson — tangent vectors, vector fields, differential forms, Stokes's theorem and de Rham cohomology — is calculus made independent of the chart. The payoff is two-fold: calculus on curved and abstract spaces, and a bridge from calculus to topology, since de Rham cohomology uses derivatives to count holes.

This lesson builds on point-set topology, multivariable calculus (the inverse function theorem in particular), dual spaces and multilinear algebra, and it upgrades the calculus of differential forms on open subsets of ℝⁿ to manifolds.

IntuitionH

An atlas of the Earth

An atlas of the Earth is a book of maps. Each page shows a region; neighbouring pages overlap; and in the overlap a point has two pairs of coordinates, related by a conversion formula. Three observations drive the whole theory.

Local questions are chart questions. Whether a curve on the Earth is smooth, how fast a ship is moving, whether a function has a maximum — these can be decided on any single page that contains the point.
The answer must not depend on the page. This is guaranteed if the conversion formulas are smooth: a curve that is smooth on one page is then smooth on the other, by the chain rule. That is the entire content of the definition of a smooth atlas.
Global questions are not chart questions. Whether every closed curve can be shrunk to a point, or whether the wind can blow everywhere at once without a calm spot (the hairy ball theorem), depends on how the pages fit together.

Stereographic projection gives an atlas of the sphere with just two pages: project from the North Pole onto the equatorial plane, and from the South Pole.

Stereographic projection from the north pole $N$: the point $x$ of the sphere is sent to the point $u$ where the line $Nx$ meets the equatorial plane. Every point except $N$ is covered; a second chart, projecting from the south pole, covers everything except $S$.

On the overlap (everything except the poles) the conversion from north-chart coordinates $u$ to south-chart coordinates is $u \mapsto u/|u|^2$ , an inversion in the unit circle. It is smooth away from $u = 0$ , so the two charts form a smooth atlas. We will prove this, and Lean will check it.

DefinitionsH

Charts, atlases and smooth structures

Definition— topological manifold

A topological $n$ -manifold is a topological space $M$ that is Hausdorff, second countable (its topology has a countable base) and locally Euclidean of dimension $n$ : every point has an open neighbourhood homeomorphic to an open subset of $\R^n$ .

Definition— chart, atlas, smooth structure

A chart on $M$ is a pair $(U, \varphi)$ with $U \subseteq M$ open and $\varphi : U \to \varphi(U)$ a homeomorphism onto an open subset of $\R^n$ ; the components $x^i = r^i \circ \varphi$ are local coordinates. Two charts $(U, \varphi)$ and $(V, \psi)$ are smoothly compatible if the transition map

\psi \circ \varphi^{-1} : \varphi(U \cap V) \to \psi(U \cap V)

is $C^\infty$ (its inverse $\varphi \circ \psi^{-1}$ is then $C^\infty$ too, by symmetry, so it is a diffeomorphism). A smooth atlas is a collection of pairwise compatible charts whose domains cover $M$ . Every smooth atlas is contained in a unique maximal one (add every chart compatible with all charts of the atlas). A smooth structure is a maximal smooth atlas, and a smooth manifold is a topological manifold together with a smooth structure.

Definition— smooth map, diffeomorphism

A map $F : M \to N$ between smooth manifolds is smooth if for every $p \in M$ there are charts $(U, \varphi)$ at $p$ and $(V, \psi)$ at $F(p)$ with $F(U) \subseteq V$ such that the local representative $\psi \circ F \circ \varphi^{-1}$ is $C^\infty$ . By the chain rule and the smoothness of transition maps this does not depend on the charts chosen. A diffeomorphism is a smooth bijection with smooth inverse. Write $C^\infty(M)$ for the algebra of smooth functions $M \to \R$ .

Definition— manifold with boundary

Let $\mathbb{H}^n = \{x \in \R^n : x^n \ge 0\}$ be the closed upper half-space. A smooth manifold with boundary is defined in the same way, but with charts onto open subsets of $\mathbb{H}^n$ ; a map on a subset of $\mathbb{H}^n$ is smooth if it extends locally to a smooth map on an open subset of $\R^n$ . The points sent to $\partial \mathbb{H}^n = \{x^n = 0\}$ by some (equivalently, every) chart form the boundary $\partial M$ , a smooth $(n-1)$ -manifold without boundary. The closed disc $\overline{D}{}^n$ is a manifold with boundary $\partial \overline{D}{}^n = S^{n-1}$ .

Why Hausdorff and second countable? Hausdorffness makes limits and integral curves unique; second countability guarantees partitions of unity — smooth functions $\psi_\alpha \ge 0$ with locally finite supports, each inside a chart, summing to $1$ . Partitions of unity are how local constructions (integrals, Riemannian metrics, embeddings) are glued into global ones, and they are why both axioms appear in the definition.

Worked examplesH

Examples, non-examples and boundary cases

Example— Euclidean space, open subsets and graphs

$\R^n$ with the single chart $\mathrm{id}$ is a smooth manifold. Any open subset of a smooth manifold is one, by restricting charts. So $GL(n, \R) = \{A : \det A \ne 0\}$ , an open subset of the space $M_n(\R) \cong \R^{n^2}$ of matrices, is a smooth manifold of dimension $n^2$ . The graph of a smooth function $f : U \to \R^k$ ( $U \subseteq \R^n$ open) is a smooth $n$ -manifold with the single chart $(x, f(x)) \mapsto x$ .

Example— The sphere, with two stereographic charts

On $S^n = \{x \in \R^{n+1} : |x| = 1\}$ write points as $(x, t)$ with $x \in \R^n$ , $t \in \R$ . Stereographic projection from the north pole $N = (0, 1)$ is $\sigma_N(x, t) = x/(1 - t)$ , with inverse

\sigma_N^{-1}(u) = \left( \frac{2u}{|u|^2 + 1}, \; \frac{|u|^2 - 1}{|u|^2 + 1} \right),

and from the south pole it is $\sigma_S(x, t) = x/(1 + t)$ . With $r = |u|^2$ ,

\sigma_S \circ \sigma_N^{-1}(u) = \frac{2u/(r+1)}{1 + (r-1)/(r+1)} = \frac{2u/(r+1)}{2r/(r+1)} = \frac{u}{|u|^2} \qquad (u \ne 0).

The inversion $u \mapsto u/|u|^2$ is smooth on $\R^n \setminus \{0\}$ and is its own inverse, so $\{\sigma_N, \sigma_S\}$ is a smooth atlas of $S^n$ . Its Jacobian determinant is $-|u|^{-2n}$ , which is negative: the two charts induce opposite orientations, a fact we return to below.

Example— Real projective space

$\R P^n$ is the set of lines through $0$ in $\R^{n+1}$ , i.e. $(\R^{n+1} \setminus \{0\})/\R^\times$ , with the quotient topology (it is Hausdorff and second countable). Write $[x^0 : \dots : x^n]$ for the line through $x$ . For each $i$ , the set $U_i = \{x^i \ne 0\}$ is open and

\varphi_i [x^0 : \dots : x^n] = \left( \tfrac{x^0}{x^i}, \dots, \widehat{\tfrac{x^i}{x^i}}, \dots, \tfrac{x^n}{x^i} \right) \in \R^n

is a chart (the hat means "omit"). The transition maps divide by one coordinate and are smooth where that coordinate is non-zero. For $n = 1$ : $\varphi_0[x^0 : x^1] = x^1/x^0 = t$ and $\varphi_1 \circ \varphi_0^{-1}(t) = 1/t$ on $\R \setminus \{0\}$ , and $\R P^1$ is diffeomorphic to a circle.

Example— Products, tori and configuration spaces

If $M$ and $N$ are smooth manifolds, products of charts form an atlas of $M \times N$ , of dimension $\dim M + \dim N$ . The torus $T^n = S^1 \times \dots \times S^1$ is the configuration space of $n$ independent rotors; the double pendulum lives on $T^2$ , and its phase space (positions and momenta) is the $4$ -manifold $T^*T^2 \cong T^2 \times \R^2$ .

Example— Non-examples

A cross. The union of the two coordinate axes in $\R^2$ is not a topological manifold: a small connected neighbourhood of the origin falls into four pieces when the origin is removed, while a connected open subset of $\R^1$ falls into two and one of $\R^n$ ( $n \ge 2$ ) stays connected. The double cone $x^2 + y^2 = z^2$ fails at its apex for the same reason (two pieces).
The line with two origins. Glue two copies of $\R$ along $\R \setminus \{0\}$ . Every point has a neighbourhood homeomorphic to $\R$ , and the space is second countable, but the two origins cannot be separated by open sets: it is not Hausdorff, and the sequence $1/k$ converges to both origins.
The cone $z = \sqrt{x^2 + y^2}$ (one nappe) is a topological manifold, homeomorphic to $\R^2$ by vertical projection, and it can be given a smooth structure by declaring that projection a chart. But it is not a smooth embedded submanifold of $\R^3$ : no plane is tangent at the apex. A manifold is not "a nice subset of $\R^N$ "; smoothness is extra structure.

Example— Different smooth structures, same manifold

On $\R$ , the single chart $\psi(x) = x^3$ is a smooth atlas. It is not compatible with the identity chart, since the transition map $\mathrm{id} \circ \psi^{-1}(y) = y^{1/3}$ is not smooth at $0$ ; so it defines a different maximal atlas. Nevertheless $F(x) = x^3$ is a diffeomorphism from $(\R, \psi)$ to $(\R, \mathrm{id})$ : its local representative $\mathrm{id} \circ F \circ \psi^{-1}(y) = y$ is smooth with smooth inverse. "Different smooth structures" and "non-diffeomorphic smooth manifolds" are different statements. Genuinely exotic structures exist — Milnor's exotic $7$ -spheres (1956), and uncountably many smooth structures on $\R^4$ — but every topological manifold of dimension at most $3$ has a smooth structure, unique up to diffeomorphism (Radó, Moise). See differential topology.

DefinitionsH

Tangent vectors, differentials and vector fields

There is no ambient space in which to draw arrows, so we define a tangent vector by what it does: it differentiates functions.

Definition— tangent space

A derivation at $p \in M$ is a linear map $v : C^\infty(M) \to \R$ satisfying the Leibniz rule $v(fg) = f(p)\, v(g) + g(p)\, v(f)$ . The derivations at $p$ form a vector space, the tangent space $T_pM$ .

Two consequences of the Leibniz rule: $v(1) = v(1 \cdot 1) = 2v(1)$ , so $v$ kills constants; and $v$ is local — if $f = g$ near $p$ then $v(f) = v(g)$ (take a bump function $\chi$ with $\chi(p) = 1$ supported where $f = g$ ; then $\chi (f - g) = 0$ , so $0 = v(\chi(f-g)) = \chi(p) v(f - g) + (f-g)(p) v(\chi) = v(f - g)$ ). By locality, $T_pM$ depends only on a neighbourhood of $p$ , and a chart $(U, \varphi)$ with coordinates $x^i$ gives derivations

\left.\frac{\partial}{\partial x^i}\right|_p f = \frac{\partial (f \circ \varphi^{-1})}{\partial r^i}\big(\varphi(p)\big).

Theorem— Theorem 1 (dimension of the tangent space)

If $\dim M = n$ , the derivations $\partial/\partial x^1|_p, \dots, \partial/\partial x^n|_p$ form a basis of $T_pM$ . Every $v \in T_pM$ equals $\sum_i v(x^i)\, \partial/\partial x^i|_p$ .

Proof

By locality we may work in the chart and assume $p = a \in \R^n$ , with $f$ smooth on a ball $B$ around $a$ . Hadamard's lemma: for $x \in B$ ,

f(x) - f(a) = \int_0^1 \frac{d}{dt} f\big(a + t(x - a)\big)\, dt = \sum_{i=1}^n (x^i - a^i)\, g_i(x), \qquad g_i(x) = \int_0^1 \partial_i f\big(a + t(x-a)\big)\, dt,

and the $g_i$ are smooth with $g_i(a) = \partial_i f(a)$ . Apply $v$ , using that it kills constants and the Leibniz rule:

v(f) = \sum_i \Big( (a^i - a^i)\, v(g_i) + g_i(a)\, v(x^i - a^i) \Big) = \sum_i \partial_i f(a)\, v(x^i).

So $v = \sum_i v(x^i)\, \partial_i|_a$ : the $\partial_i|_a$ span. They are independent because $\partial_i|_a (x^j) = \delta_i^j$ . ∎

∎

Definition— differential

The differential of a smooth map $F : M \to N$ at $p$ is the linear map $dF_p : T_pM \to T_{F(p)}N$ , $\; dF_p(v)(f) = v(f \circ F)$ .

The chain rule is now a one-line computation: $d(G \circ F)_p(v)(f) = v(f \circ G \circ F) = dF_p(v)(f \circ G) = dG_{F(p)}\big(dF_p(v)\big)(f)$ , so $d(G \circ F)_p = dG_{F(p)} \circ dF_p$ , and $d(\mathrm{id})_p = \mathrm{id}$ . In coordinates $x$ on $M$ and $y$ on $N$ , $dF_p(\partial/\partial x^i) = \sum_j (\partial \hat F^j / \partial x^i)\, \partial/\partial y^j$ : the matrix of $dF_p$ is the Jacobian of the local representative $\hat F$ .

Vectors under a change of chart. If $x$ and $\tilde x$ are two coordinate systems at $p$ , the chain rule gives $\partial/\partial x^i = \sum_j (\partial \tilde x^j/\partial x^i)\, \partial/\partial \tilde x^j$ . So if $v = \sum_i v^i \partial/\partial x^i = \sum_j \tilde v^j \partial/\partial \tilde x^j$ , then

\tilde v^j = \sum_i \frac{\partial \tilde x^j}{\partial x^i} v^i, \qquad \text{i.e.} \quad \tilde v = J v, \quad J = \frac{\partial \tilde x}{\partial x}.

The tangent bundle $TM = \bigsqcup_p T_pM$ is a smooth $2n$ -manifold, with charts $(p, v) \mapsto (x^1(p), \dots, x^n(p), v^1, \dots, v^n)$ ; the transition maps are $(x, v) \mapsto (\tilde x(x), J(x) v)$ , which are smooth. A vector field is a smooth section $X : M \to TM$ . Equivalently, it is a derivation of the algebra $C^\infty(M)$ , acting by $(Xf)(p) = X_p f$ . The Lie bracket $[X, Y] f = X(Yf) - Y(Xf)$ is again a vector field (the second derivatives cancel), with $[X, Y] = \sum_j \big(X(Y^j) - Y(X^j)\big) \partial_j$ in coordinates; it makes the vector fields a Lie algebra.

Theorem— Theorem 2 (fundamental theorem on flows; stated)

Every smooth vector field $X$ on $M$ has a unique maximal flow: a smooth map $\theta$ , defined on an open subset of $\R \times M$ containing $\{0\} \times M$ , such that $t \mapsto \theta_t(p)$ is the maximal integral curve of $X$ starting at $p$ . If $M$ is compact (or $X$ has compact support), the flow is defined on all of $\R \times M$ , and $t \mapsto \theta_t$ is a one-parameter group of diffeomorphisms: $\theta_{s+t} = \theta_s \circ \theta_t$ , $\theta_0 = \mathrm{id}$ .

This is the existence and uniqueness theorem for ODEs, applied in charts and glued using uniqueness (and Hausdorffness); see Lee, Introduction to Smooth Manifolds, Chapter 9. Example: on $S^2 \subset \R^3$ the field $X(x, y, z) = (-y, x, 0)$ is tangent to the sphere, its flow rotates the sphere about the vertical axis, and it vanishes at the two poles. On $S^3 \subset \C^2$ the field $X(z) = iz$ vanishes nowhere; its flow $z \mapsto e^{it}z$ is the Hopf flow.

DefinitionsH

Covectors and differential forms

Definition— cotangent space, differential of a function

The cotangent space is the dual $T_p^*M = (T_pM)^*$ . A smooth function $f$ has a differential $df_p \in T_p^*M$ , $df_p(v) = v(f)$ . In a chart, the $dx^i|_p$ form the basis dual to the $\partial/\partial x^i|_p$ , and $df = \sum_i (\partial f/\partial x^i)\, dx^i$ .

Covectors under a change of chart. From $d\tilde x^j = \sum_i (\partial \tilde x^j / \partial x^i)\, dx^i$ , a covector $\omega = \sum_i \omega_i\, dx^i = \sum_j \tilde \omega_j\, d\tilde x^j$ has $\omega_i = \sum_j \tilde\omega_j\, \partial \tilde x^j/\partial x^i$ , that is

\omega = J^{\mathsf T} \tilde\omega, \qquad \tilde \omega = (J^{\mathsf T})^{-1} \omega .

Vectors transform by $J$ , covectors by the inverse transpose, and the pairing $\omega(v) = \sum_i \omega_i v^i = \tilde\omega^{\mathsf T} J v$ is the same in both charts. This is why the classical literature calls vectors "contravariant" and covectors "covariant", and it is why the gradient of a function is not a vector until a metric is chosen: $df$ is a covector, and turning it into a vector needs an inner product on each tangent space (see Riemannian geometry).

Definition— differential forms

A differential $k$ -form is a smooth section of the bundle $\Lambda^k T^*M$ of alternating $k$ -linear forms on tangent spaces. In a chart, $\omega = \sum_{i_1 < \dots < i_k} \omega_{i_1 \dots i_k}\, dx^{i_1} \wedge \dots \wedge dx^{i_k}$ with smooth coefficients. Write $\Omega^k(M)$ for the space of $k$ -forms. The wedge product is graded commutative: $\alpha \wedge \beta = (-1)^{kl}\beta \wedge \alpha$ for $\alpha \in \Omega^k$ , $\beta \in \Omega^l$ .

Definition— exterior derivative and pullback

The exterior derivative is the unique family of linear maps $d : \Omega^k(M) \to \Omega^{k+1}(M)$ such that $df$ is the differential of $f$ on functions, $d(\alpha \wedge \beta) = d\alpha \wedge \beta + (-1)^k \alpha \wedge d\beta$ , and $d \circ d = 0$ . In a chart, $d\big(\sum_I \omega_I\, dx^I\big) = \sum_I d\omega_I \wedge dx^I$ . For smooth $F : M \to N$ the pullback $F^* : \Omega^k(N) \to \Omega^k(M)$ is $(F^*\omega)_p(v_1, \dots, v_k) = \omega_{F(p)}(dF_p v_1, \dots, dF_p v_k)$ .

Three facts do most of the work: $F^*(\alpha \wedge \beta) = F^*\alpha \wedge F^*\beta$ ; $F^*(df) = d(f \circ F)$ ; and $F^* \circ d = d \circ F^*$ . The identity $d(df) = \sum_{i, j} \partial_j \partial_i f\, dx^j \wedge dx^i = 0$ is the symmetry of second derivatives meeting the antisymmetry of $\wedge$ .

Example— Polar coordinates

Let $F(r, \theta) = (r\cos\theta, r\sin\theta)$ . Then $F^*dx = \cos\theta\, dr - r\sin\theta\, d\theta$ and $F^*dy = \sin\theta\, dr + r\cos\theta\, d\theta$ , so $F^*(dx \wedge dy) = r\, dr \wedge d\theta$ (the Jacobian appears automatically) and

F^*\left( \frac{x\, dy - y\, dx}{x^2 + y^2} \right) = \frac{r^2\, d\theta}{r^2} = d\theta .

The form $\omega = (x\,dy - y\,dx)/(x^2+y^2)$ on $\R^2 \setminus \{0\}$ is closed ( $d\omega = 0$ ) and is locally $d$ of an angle function — but there is no global angle function, and we will see that $\omega$ is not exact.

Definition— orientation and integration

An atlas is oriented if all its transition maps have positive Jacobian determinant; $M$ is orientable if it has one (equivalently, a nowhere-vanishing $n$ -form). For an $n$ -form $\omega$ with compact support in an oriented chart $(U, \varphi)$ , write $(\varphi^{-1})^*\omega = f\, dx^1 \wedge \dots \wedge dx^n$ and set $\int_M \omega = \int_{\R^n} f\, dx^1 \cdots dx^n$ . In general, $\int_M \omega = \sum_\alpha \int_M \psi_\alpha \omega$ for a partition of unity $\{\psi_\alpha\}$ subordinate to an oriented atlas.

This is well defined because, under a change of chart, the form's coefficient changes by $\det J$ while the change-of-variables formula for integrals uses $|\det J|$ ; the two agree exactly when $\det J > 0$ . That is why we need an oriented atlas to integrate forms. The sphere is orientable: the stereographic charts have $\det J < 0$ , so compose one of them with a reflection. $\R P^2$ is not orientable, since it contains a Möbius band; in general $\R P^n$ is orientable exactly when $n$ is odd.

ProofH

The regular value theorem and embeddings

Definition— regular value

Let $F : M \to N$ be smooth. A point $p \in M$ is a regular point if $dF_p$ is surjective; $c \in N$ is a regular value if every point of $F^{-1}(c)$ is regular (in particular if $F^{-1}(c) = \varnothing$ ). A subset $S \subseteq M$ is an embedded $k$ -dimensional submanifold if each $p \in S$ has a chart $(U, \varphi)$ of $M$ with $\varphi(U \cap S) = \varphi(U) \cap (\R^k \times \{0\})$ (a slice chart).

Theorem— Theorem 3 (regular value theorem)

Let $F : M \to N$ be smooth, $\dim M = m$ , $\dim N = n$ , and let $c$ be a regular value. Then $F^{-1}(c)$ is an embedded submanifold of $M$ of dimension $m - n$ (or empty), and $T_p F^{-1}(c) = \ker dF_p$ .

Proof

The statement is local. Fix $p \in F^{-1}(c)$ and charts centred at $p$ and at $c$ ; in them $F$ becomes a smooth map $\hat F : W \to \R^n$ on an open $W \ni 0$ of $\R^m$ , with $\hat F(0) = 0$ and $D\hat F(0)$ of rank $n$ (so $m \ge n$ ). After reordering coordinates, write $x = (x', x'') \in \R^{m-n} \times \R^n$ with $\partial \hat F/\partial x''(0)$ invertible. Define $G(x', x'') = \big(x', \hat F(x', x'')\big)$ . Its Jacobian at $0$ is block lower-triangular,

DG(0) = \begin{pmatrix} I & 0 \\ \partial \hat F/\partial x' & \partial \hat F/\partial x'' \end{pmatrix},

hence invertible, so by the inverse function theorem $G$ is a diffeomorphism from a neighbourhood $W_0$ of $0$ onto a neighbourhood of $0$ . In the new chart $G$ (composed with the old one) the map $F$ reads $\hat F \circ G^{-1}(y', y'') = y''$ , a linear projection, and $F^{-1}(c) \cap W_0$ corresponds to the slice $\{y'' = 0\}$ . So $F^{-1}(c)$ has a slice chart of dimension $m - n$ at every point. Finally $F$ is constant on $S = F^{-1}(c)$ , so $dF_p$ vanishes on $T_pS$ ; thus $T_pS \subseteq \ker dF_p$ , and both have dimension $m - n$ (rank–nullity), so they are equal. ∎

∎

Theorem— Corollary 4 (spheres and matrix groups are manifolds)

$S^n$ is an embedded $n$ -dimensional submanifold of $\R^{n+1}$ , and $T_xS^n = x^\perp$ .
$O(n) = \{A : A^{\mathsf T} A = I\}$ is an embedded submanifold of $M_n(\R)$ of dimension $n(n-1)/2$ , and $T_I O(n)$ is the space of skew-symmetric matrices.
$SL(n, \R) = \{A : \det A = 1\}$ is an embedded submanifold of dimension $n^2 - 1$ , and $T_I SL(n, \R)$ is the space of trace-zero matrices.

Proof

(1) $F(x) = |x|^2$ has $dF_x(h) = 2\langle x, h\rangle$ , which is onto $\R$ whenever $x \ne 0$ ; so $1$ is a regular value, and $T_x S^n = \ker dF_x = x^\perp$ .

(2) Let $\mathrm{Sym}_n$ be the space of symmetric matrices, of dimension $n(n+1)/2$ , and $F : M_n(\R) \to \mathrm{Sym}_n$ , $F(A) = A^{\mathsf T}A$ . Expanding,

(A + tH)^{\mathsf T}(A + tH) = A^{\mathsf T}A + t\big(A^{\mathsf T}H + H^{\mathsf T}A\big) + t^2 H^{\mathsf T}H,

so $dF_A(H) = A^{\mathsf T}H + H^{\mathsf T}A$ . If $A \in O(n)$ and $S \in \mathrm{Sym}_n$ , then $H = \tfrac12 AS$ satisfies $A^{\mathsf T}H + H^{\mathsf T}A = \tfrac12 S + \tfrac12 S^{\mathsf T}A^{\mathsf T}A = S$ . So $I$ is a regular value, $\dim O(n) = n^2 - n(n+1)/2 = n(n-1)/2$ , and $T_I O(n) = \ker dF_I = \{H : H + H^{\mathsf T} = 0\}$ .

(3) For $A \in SL(n, \R)$ , $\det\big((1+t)A\big) = (1 + t)^n$ , so $d(\det)_A(A) = n \ne 0$ and $d(\det)_A$ is onto $\R$ . Hence $1$ is a regular value and $\dim SL(n, \R) = n^2 - 1$ . Expanding $\det(I + tH) = 1 + t \operatorname{tr} H + O(t^2)$ shows $d(\det)_I(H) = \operatorname{tr} H$ , so $T_I SL(n, \R)$ consists of the trace-zero matrices. ∎

∎

Note the codomain in (2): $A^{\mathsf T}A$ is always symmetric, so as a map into all of $M_n(\R)$ the differential is never surjective, and the theorem would not apply. These tangent spaces at the identity are the Lie algebras $\mathfrak{so}(n)$ and $\mathfrak{sl}(n, \R)$ of Lie groups.

Theorem— Theorem 5 (Whitney embedding, compact case)

Every compact smooth manifold $M$ admits a smooth embedding into some $\R^N$ .

Proof

For each $p$ choose a chart $(U_p, \varphi_p)$ and a smooth bump function $\rho_p : M \to [0, 1]$ supported in $U_p$ and equal to $1$ on an open $V_p \ni p$ . By compactness finitely many $V_1, \dots, V_k$ cover $M$ . Let $F = (\rho_1\varphi_1, \dots, \rho_k\varphi_k, \rho_1, \dots, \rho_k) : M \to \R^{k(n+1)}$ , where $\rho_i\varphi_i$ is extended by $0$ outside $U_i$ (it is smooth). Injective: if $F(p) = F(q)$ and $p \in V_i$ , then $\rho_i(q) = \rho_i(p) = 1$ , so $q \in U_i$ , and $\varphi_i(p) = \varphi_i(q)$ , so $p = q$ . Immersion: near $p \in V_i$ the component $\rho_i\varphi_i$ equals the chart $\varphi_i$ , whose differential is injective, so $dF_p$ is injective. An injective continuous map from a compact space to a Hausdorff space is a homeomorphism onto its image, so $F$ is an embedding. ∎

∎

Whitney (1936, 1944) proved much more: every smooth $n$ -manifold (second countable and Hausdorff, compact or not) embeds in $\R^{2n}$ , and $2n$ cannot be lowered in general ( $\R P^2$ does not embed in $\R^3$ ). But no embedding is canonical, and nothing intrinsic depends on one.

ProofH

Stokes's theorem

Let $M$ be an oriented $n$ -manifold with boundary. Give $\partial M$ the Stokes orientation: a basis $(v_1, \dots, v_{n-1})$ of $T_p\partial M$ is positive when $(\nu, v_1, \dots, v_{n-1})$ is positive in $T_pM$ for an outward-pointing $\nu$ . For $\mathbb{H}^n$ the outward vector is $-\partial_n$ , and since $(-\partial_n, \partial_1, \dots, \partial_{n-1})$ has orientation sign $(-1)\cdot(-1)^{n-1} = (-1)^n$ , the Stokes orientation of $\partial\mathbb{H}^n = \R^{n-1}$ is $(-1)^n$ times the standard one.

Theorem— Theorem 6 (Stokes's theorem)

Let $M$ be an oriented smooth $n$ -manifold with boundary, with $\partial M$ carrying the Stokes orientation, and let $\omega$ be a compactly supported smooth $(n-1)$ -form on $M$ . Then

\int_M d\omega = \int_{\partial M} \omega ,

where the right-hand side is $0$ if $\partial M = \varnothing$ .

Proof

Step 1: the half-space. Let $M = \mathbb{H}^n$ and $\omega = \sum_{i=1}^n \omega_i\, dx^1 \wedge \dots \wedge \widehat{dx^i} \wedge \dots \wedge dx^n$ , supported in $[-R, R]^{n-1} \times [0, R)$ . Then $d\omega = \sum_i (-1)^{i-1} \partial_i\omega_i\, dx^1 \wedge \dots \wedge dx^n$ , so

\int_{\mathbb{H}^n} d\omega = \sum_{i=1}^n (-1)^{i-1} \int_{\mathbb{H}^n} \partial_i \omega_i \, dx .

For $i < n$ , integrate first in $x^i$ over $[-R, R]$ (Fubini): by the fundamental theorem of calculus the result is $\omega_i(\dots, R, \dots) - \omega_i(\dots, -R, \dots) = 0$ . For $i = n$ , $\int_0^R \partial_n\omega_n\, dx^n = -\omega_n(x', 0)$ since $\omega_n$ vanishes at $x^n = R$ . Hence

\int_{\mathbb{H}^n} d\omega = (-1)^{n-1}\cdot(-1) \int_{\R^{n-1}} \omega_n(x', 0)\, dx' = (-1)^n \int_{\R^{n-1}} \omega_n(x', 0)\, dx' .

On $\partial \mathbb{H}^n$ the form $dx^n$ restricts to $0$ , so only the $i = n$ term survives: $\omega|_{\partial\mathbb{H}^n} = \omega_n(x', 0)\, dx^1 \wedge \dots \wedge dx^{n-1}$ . Integrating with the Stokes orientation, which is $(-1)^n$ times the standard one, gives exactly the same number. The same computation with $\R^n$ in place of $\mathbb{H}^n$ gives $\int_{\R^n} d\omega = 0$ , because now every term vanishes.

Step 2: gluing. Cover the compact set $\operatorname{supp}\omega$ by finitely many oriented charts, each onto an open subset of $\R^n$ or of $\mathbb{H}^n$ , and let $\{\psi_\alpha\}$ be a subordinate partition of unity. Since $\sum_\alpha \psi_\alpha = 1$ near $\operatorname{supp}\omega$ , we have $\sum_\alpha d\psi_\alpha = 0$ there, so

\sum_\alpha d(\psi_\alpha\omega) = \sum_\alpha d\psi_\alpha \wedge \omega + \sum_\alpha \psi_\alpha\, d\omega = d\omega .

Each $\psi_\alpha\omega$ is supported in a single chart. Pulling back by that chart preserves integrals (the chart is orientation-preserving), commutes with $d$ , and carries $\partial M$ to $\partial\mathbb{H}^n$ with matching Stokes orientations, so Step 1 gives $\int_M d(\psi_\alpha \omega) = \int_{\partial M} \psi_\alpha\omega$ . Summing over $\alpha$ proves the theorem. ∎

∎

For $n = 1$ and $M = [a, b]$ this is the fundamental theorem of calculus (the boundary is $\{b\}$ with sign $+$ and $\{a\}$ with sign $-$ ). For domains in $\R^2$ and $\R^3$ it contains Green's theorem, the classical Stokes theorem and the divergence theorem. On a closed manifold ( $\partial M = \varnothing$ , $M$ compact), every exact $n$ -form integrates to $0$ ; so a volume form on a closed oriented manifold is never exact.

ProofH

de Rham cohomology

Since $d \circ d = 0$ , every exact form ( $\omega = d\eta$ ) is closed ( $d\omega = 0$ ). The failure of the converse measures topology.

Definition— de Rham cohomology

$H^k_{\mathrm{dR}}(M) = \dfrac{\ker\big(d : \Omega^k(M) \to \Omega^{k+1}(M)\big)}{\operatorname{im}\big(d : \Omega^{k-1}(M) \to \Omega^k(M)\big)}$ . A smooth $F : M \to N$ induces $F^* : H^k_{\mathrm{dR}}(N) \to H^k_{\mathrm{dR}}(M)$ , since $F^*$ commutes with $d$ ; and the wedge product makes $H^\bullet_{\mathrm{dR}}(M)$ a graded-commutative algebra.

$H^0_{\mathrm{dR}}(M)$ is the space of locally constant functions ( $df = 0$ means $f$ is constant on each component), so $\dim H^0_{\mathrm{dR}}(M)$ is the number of connected components.

Theorem— Theorem 7 (the circle)

Integration over the circle induces an isomorphism $H^1_{\mathrm{dR}}(S^1) \cong \R$ , $[\omega] \mapsto \int_{S^1}\omega$ . In particular $d\theta$ (the form $x\,dy - y\,dx$ restricted to $S^1$ ) is closed but not exact.

Proof

Let $\pi : \R \to S^1$ , $\pi(\theta) = (\cos\theta, \sin\theta)$ ; it is a local diffeomorphism. Every $1$ -form on the $1$ -manifold $S^1$ is closed, and $\pi^*\omega = f(\theta)\, d\theta$ with $f$ smooth and $2\pi$ -periodic; $\omega = dg$ for some $g \in C^\infty(S^1)$ exactly when $f = G'$ for the $2\pi$ -periodic function $G = g \circ \pi$ (a smooth periodic $G$ descends to a smooth $g$ on $S^1$ , because $\pi$ is a local diffeomorphism). Also $\int_{S^1}\omega = \int_0^{2\pi} f(\theta)\, d\theta$ .

Exact forms integrate to zero: if $f = G'$ with $G$ periodic, then $\int_0^{2\pi} f = G(2\pi) - G(0) = 0$ .

Forms with zero integral are exact: if $\int_0^{2\pi} f = 0$ , let $G(\theta) = \int_0^\theta f$ . Then $G' = f$ , and $G(\theta + 2\pi) - G(\theta) = \int_\theta^{\theta + 2\pi} f = \int_0^{2\pi} f = 0$ , because the integral of a periodic function over a full period does not depend on where the period starts. So $G$ is periodic and $\omega = dg$ .

Thus the linear map $\omega \mapsto \int_{S^1}\omega$ has kernel exactly the exact forms. It is onto, since $\int_{S^1} d\theta = 2\pi \ne 0$ , so it induces an isomorphism $H^1_{\mathrm{dR}}(S^1) \to \R$ . ∎

∎

The same argument applied to $\omega = (x\,dy - y\,dx)/(x^2+y^2)$ on $\R^2 \setminus \{0\}$ shows it is not exact: its integral round the unit circle is $2\pi$ , whereas an exact form integrates to $0$ round every closed curve. More generally (proofs in Bott–Tu or Lee, Chapters 17–18):

Poincaré lemma and homotopy invariance: homotopic maps induce the same map on $H^\bullet_{\mathrm{dR}}$ ; hence $H^k_{\mathrm{dR}}(\R^n) = 0$ for $k \ge 1$ , and every closed form is locally exact.
Computations (by Mayer–Vietoris): $H^k_{\mathrm{dR}}(S^n) \cong \R$ for $k = 0, n$ and $0$ otherwise; $H^k_{\mathrm{dR}}(T^n) \cong \R^{\binom nk}$ ; $H^1_{\mathrm{dR}}(\R^2 \setminus \{p_1, \dots, p_m\}) \cong \R^m$ , detected by the periods around the $m$ punctures.
Top degree: for a compact connected oriented $n$ -manifold without boundary, $\int_M : H^n_{\mathrm{dR}}(M) \to \R$ is an isomorphism (Stokes gives well-definedness; surjectivity is clear; injectivity is the real content).

Theorem— Theorem 8 (de Rham's theorem; stated)

For every smooth manifold $M$ , integration of forms over smooth singular chains induces an isomorphism $H^k_{\mathrm{dR}}(M) \cong H^k(M; \R)$ with singular cohomology with real coefficients, compatible with products.

Stokes's theorem, $\int_c d\omega = \int_{\partial c}\omega$ , is precisely the statement that integration is a chain map from the de Rham complex to the singular cochain complex; the theorem (de Rham, 1931) says it is a quasi-isomorphism. So de Rham cohomology is a topological invariant, even though it is defined with derivatives. See algebraic topology.

Formal mathematicsF

Smooth manifolds in Lean and Mathlib

Mathlib's manifolds follow the definitions above, with one generalisation: a manifold is modelled on a model with corners $I : H \to E$ , which lets the same definitions cover manifolds without boundary (𝓡 n, modelled on $\R^n$ ), with boundary (𝓡∂ n, on a half-space) and with corners. ChartedSpace H M is an atlas, IsManifold I n M says the transition maps are $C^n$ (with ω meaning analytic), ContMDiff is smoothness of maps, and mfderiv is the differential.

First, the stereographic charts. We define the inverse projection from the north pole and the projection from the south pole on $E \times \R$ for any real inner product space $E$ , check that the first lands on the sphere, and prove that the transition map is the inversion, which is smooth on $E \setminus \{0\}$ and its own inverse. These are the computations of the sphere example, line by line:

stereoInvNorthLean 4Formally verified

/-- **Inverse stereographic projection from the north pole.** The point `u` of the equatorial
space `E` is sent to the point `(x, t) ∈ E × ℝ` where the line from the north pole `(0, 1)`
through `(u, 0)` meets the unit sphere again:
`x = 2u / (‖u‖² + 1)`, `t = (‖u‖² - 1) / (‖u‖² + 1)`. -/
noncomputable def stereoInvNorth (u : E) : E × ℝ :=
  ((2 / (‖u‖ ^ 2 + 1)) • u, (‖u‖ ^ 2 - 1) / (‖u‖ ^ 2 + 1))