Differential topology · The Lumenary

MotivationH

Why must the wind stop somewhere?

At every moment, somewhere on the Earth the horizontal wind is calm. Stir a cup of coffee and let it settle: in an idealised model, some point of the surface ends up exactly where it started. Crumple a map of Nairobi and drop it on the floor of a room in Nairobi: some point of the map lies exactly over the place it represents. These are three faces of two theorems — the hairy ball theorem and the Brouwer fixed-point theorem — and neither can be proved by calculus alone, because neither is about a formula. They are about the shape of the sphere and the disc.

Differential topology studies smooth manifolds up to diffeomorphism, using smooth maps as probes. Its central trick is to count: given a smooth map $f : M \to N$ between manifolds of the same dimension, count the solutions of $f(x) = y$ , with a sign telling whether $f$ preserves or reverses orientation near $x$ . For a generic $y$ the count is finite, and — this is the miracle — it does not depend on $y$ , nor on deforming $f$ . That number, the degree, is a topological invariant built from derivatives, and with it the theorems above take a page each.

This lesson builds on smooth manifolds (regular values, orientation, vector fields), and it meets the fundamental group (the degree of a map of circles is a winding number) and the Euler characteristic, which reappears as the sum of the indices of a vector field.

IntuitionH

Counting preimages, with signs

Wrap a rubber band round a pencil three times and look at it as a map from the circle to the circle. A typical point of the target is covered three times. Now push a fold into the band: near the fold, some points are covered five times — the band goes forward, back, forward. The unsigned count jumped by $2$ . But the extra two passes go in opposite directions, so if we count forward passes as $+1$ and backward ones as $-1$ , the signed count is still $3$ .

Part of the graph of a map that folds back. The level $y$ is met three times; the middle crossing (the differently coloured dot) runs backwards, so the signed count is $1 - 1 + 1 = 1$ — the same as for a level above the peak or below the valley, which is met only once. The unsigned count jumps only at the critical values, the heights of the peak and the valley.

Three ideas make this precise. Regular values are the levels that avoid the tips of folds; Sard's theorem says almost every level is regular. The degree is the signed count at a regular value; it does not depend on the level because, as the level moves, preimages are born and die in pairs of opposite sign (exactly at the critical values). And it does not change under deformation, because a deformation sweeps out a surface whose preimage of a level is a union of arcs and circles — and arcs have two ends.

DefinitionsH

Regular values, degree, index and transversality

Throughout, manifolds are smooth, Hausdorff and second countable; a closed manifold is compact without boundary.

Definition— critical point, regular value, measure zero

Let $f : M \to N$ be smooth. A point $x \in M$ is critical if $df_x : T_xM \to T_{f(x)}N$ is not surjective; its image $f(x)$ is a critical value. Every other point of $N$ — including every point not in the image — is a regular value. A subset $A \subseteq N$ has measure zero if $\varphi(A \cap U)$ has Lebesgue measure zero for every chart $(U, \varphi)$ ; this is well defined because diffeomorphisms of open subsets of $\R^n$ preserve null sets.

Definition— degree

Let $f : M \to N$ be smooth, with $M$ closed, $N$ connected and $\dim M = \dim N = n$ . For a regular value $y$ , the preimage $f^{-1}(y)$ is finite (it is compact, and discrete by the inverse function theorem). The mod 2 degree is $\deg_2 f = \# f^{-1}(y) \bmod 2$ . If $M$ and $N$ are oriented, each $x \in f^{-1}(y)$ has a sign $\operatorname{sign} df_x = \pm 1$ according as the isomorphism $df_x$ preserves or reverses orientation, and the degree is

\deg f = \sum_{x \in f^{-1}(y)} \operatorname{sign} df_x \in \Z .

That these numbers do not depend on $y$ is Theorem 3 below.

Definition— index of a zero of a vector field

Let $v$ be a vector field with an isolated zero at $p$ . In a chart centred at $p$ , $v$ becomes a map $\R^n \supseteq U \to \R^n$ with $v(0) = 0$ and no other zeros in a small ball $B_\varepsilon$ . The index $\operatorname{ind}_p(v)$ is the degree of $u \mapsto v(u)/|v(u)|$ from the sphere $\partial B_\varepsilon$ to $S^{n-1}$ . It does not depend on $\varepsilon$ or on the chart. In the plane it is the number of times $v$ turns round as one walks once anticlockwise round $p$ .

Definition— transversality

A smooth map $f : X \to Y$ is transverse to a submanifold $Z \subseteq Y$ if $\operatorname{im} df_x + T_{f(x)}Z = T_{f(x)}Y$ whenever $f(x) \in Z$ . Two submanifolds are transverse if the inclusion of one is transverse to the other. A regular value is the special case $Z = \{y\}$ .

Definition— Morse function

A critical point $p$ of $h : M \to \R$ is non-degenerate if the Hessian of $h$ at $p$ (in any chart) is invertible; its Morse index $\lambda$ is the number of negative eigenvalues. $h$ is a Morse function if all its critical points are non-degenerate.

Worked examplesH

Examples

Example— Degrees of maps of circles and spheres

$z \mapsto z^k$ on $S^1 \subset \C$ has degree $k$ : for $k > 0$ each $w$ has $k$ preimages, all orientation-preserving; for $k < 0$ all $|k|$ preimages reverse orientation. The degree of a map $S^1 \to S^1$ is its winding number.
Complex conjugation $z \mapsto \bar z$ has degree $-1$ .
The antipodal map $a(x) = -x$ of $S^n$ is the restriction of $-I$ on $\R^{n+1}$ ; it sends the outward normal at $x$ to the outward normal at $-x$ , so it preserves the orientation of $S^n$ exactly when $\det(-I) = (-1)^{n+1} > 0$ . Hence $\deg a = (-1)^{n+1}$ .
A constant map, or any map that is not onto, has degree $0$ (a point outside the image is a regular value with no preimages). So a map of non-zero degree is surjective.
A non-constant polynomial $p$ of degree $d$ extends to the Riemann sphere $\C \cup \{\infty\}$ with $p(\infty) = \infty$ . Being holomorphic, it preserves orientation at every regular point (its Jacobian determinant is $|p'(z)|^2 > 0$ ), so its degree is $d$ , the number of solutions of $p(z) = w$ for a regular value $w$ . In particular every $w$ has a preimage: this is a proof of the fundamental theorem of algebra.

Example— Indices of zeros in the plane

Think of a planar field as a complex function $v(z)$ .

field	picture	index
$v = z$	source	$+1$
$v = -z$	sink	$+1$
$v = iz$	centre (rotation)	$+1$
$v = \bar z$	saddle	$-1$
$v = z^k$	$k-1$ "petals"	$k$
$v = \bar z^k$		$-k$

A source and a sink both have index $+1$ : the index sees how the arrow turns, not whether the flow goes in or out. For a linear field $v(x) = Ax$ with $\det A \ne 0$ , the index is $\operatorname{sign}\det A$ . In particular the gradient of $h = -x_1^2 - \dots - x_\lambda^2 + x_{\lambda+1}^2 + \dots + x_n^2$ has index $(-1)^\lambda$ at $0$ .

Example— Vector fields on the sphere and the torus

On $S^2$ , the rotation field $(-y, x, 0)$ vanishes at the two poles, each of index $+1$ ; the gradient of the height function $z$ (projected onto the sphere) has a source at the south pole and a sink at the north pole, again $+1 + 1 = 2$ . On the torus, the field $\partial_\theta$ along the parallels vanishes nowhere: total index $0$ . The gradient of the height of an upright torus has a maximum, a minimum and two saddles: $+1 - 1 - 1 + 1 = 0$ . In every case the total is the Euler characteristic.

Example— Critical values that cannot be avoided

$f(x) = x^2$ on $\R$ has the single critical point $0$ and the critical value $0$ . The height function on the upright torus has exactly four critical values. Sard's theorem does not say there are no critical values; it says they form a set of measure zero. (A cautionary example: a smooth function $\R \to \R$ can have uncountably many critical values — integrate a smooth function that vanishes exactly on a Cantor set of positive measure — yet they still have measure zero.)

ProofH

Sard's theorem

Theorem— Theorem 1 (Sard, 1942)

For every smooth map $f : M \to N$ , the set of critical values has measure zero in $N$ . Consequently the regular values are dense.

The general proof is an induction on dimension that stratifies the critical set by the order to which derivatives vanish (Milnor, Topology from the Differentiable Viewpoint, §3); it needs $f$ of class $C^k$ with $k \ge \max(1, \dim M - \dim N + 1)$ . Degree theory only needs the case of equal dimensions, which has a short complete proof.

Theorem— Theorem 2 (Sard in equal dimensions)

Let $U \subseteq \R^n$ be open, $f : U \to \R^n$ of class $C^1$ , and $C = \{x \in U : \det Df(x) = 0\}$ . Then $f(C)$ has Lebesgue measure zero.

Proof

$U$ is a countable union of closed cubes, and a countable union of null sets is null, so it suffices to show that $f(C \cap Q)$ is null for a closed cube $Q \subseteq U$ of side $\ell$ . Let $M = \max_Q \|Df\|$ . Since $Df$ is uniformly continuous on $Q$ , there is a function $\varepsilon(\eta) \to 0$ as $\eta \to 0$ with $\|Df(y) - Df(x)\| \le \varepsilon(\eta)$ whenever $x, y \in Q$ , $|x - y| \le \eta$ . By the mean value inequality applied to $y \mapsto f(y) - Df(x)(y - x)$ on the segment $[x, y] \subseteq Q$ ,

\big| f(y) - f(x) - Df(x)(y - x) \big| \le \varepsilon(\eta)\, |y - x| \qquad (|y - x| \le \eta).

Divide $Q$ into $N^n$ subcubes of side $\ell/N$ and diameter $\eta = \sqrt n\, \ell / N$ . Suppose a subcube $Q'$ contains a critical point $x$ . For $y \in Q'$ , the point $f(x) + Df(x)(y - x)$ lies in the affine subspace $f(x) + \operatorname{im} Df(x)$ , of dimension at most $n - 1$ , within distance $M\eta$ of $f(x)$ ; and $f(y)$ lies within $\varepsilon(\eta)\eta$ of it. Choosing orthonormal coordinates in which a hyperplane containing $\operatorname{im}Df(x)$ is $\{y_n = 0\}$ , we see that $f(Q')$ lies in a box with $n - 1$ sides of length $2(M + \varepsilon(\eta))\eta$ and one side of length $2\varepsilon(\eta)\eta$ . So

\operatorname{vol} f(Q') \le 2^n \big(M + \varepsilon(\eta)\big)^{n-1} \varepsilon(\eta)\, \eta^n .

There are at most $N^n$ such subcubes and $N^n \eta^n = (\sqrt n\, \ell)^n$ , so the outer measure of $f(C \cap Q)$ is at most $2^n (M + \varepsilon(\eta))^{n-1} \varepsilon(\eta) (\sqrt n\,\ell)^n$ , which tends to $0$ as $N \to \infty$ . ∎

∎

For manifolds of the same dimension, apply Theorem 2 in countably many charts. Mathlib proves an even stronger version — differentiability suffices, no continuity of $Df$ is needed — and we cite it in the formal section.

ProofH

The degree is well defined

We use one classification theorem without proof (see Milnor, Appendix, or Guillemin–Pollack, Appendix 2).

Theorem— Fact (compact 1-manifolds)

Every compact connected $1$ -manifold with boundary is diffeomorphic to a circle or to a closed interval. Hence a compact $1$ -manifold with boundary has an even number of boundary points; if it is oriented, its boundary points (with their Stokes signs) sum to $0$ .

We also use the regular value theorem for manifolds with boundary: if $y$ is a regular value of $F : W \to N$ and of $F|_{\partial W}$ , then $F^{-1}(y)$ is a manifold with boundary $F^{-1}(y) \cap \partial W$ , of dimension $\dim W - \dim N$ .

Theorem— Theorem 3 (invariance of the degree)

Let $M$ be closed, $N$ connected, $\dim M = \dim N$ . Then (a) $\#f^{-1}(y) \bmod 2$ is the same for all regular values $y$ of $f$ ; (b) homotopic maps have the same mod 2 degree. If $M$ and $N$ are oriented, the same holds for the signed count $\deg f$ .

Proof

Local constancy. If $y$ is regular with $f^{-1}(y) = \{x_1, \dots, x_k\}$ , the inverse function theorem gives disjoint neighbourhoods $U_i \ni x_i$ mapped diffeomorphically onto a common neighbourhood $V$ of $y$ ; shrinking $V$ to avoid the compact set $f(M \setminus \bigcup U_i)$ , every $y' \in V$ is regular with exactly $k$ preimages, of the same signs.

Homotopy invariance (b). Let $F : M \times [0, 1] \to N$ be a smooth homotopy from $f$ to $g$ , and let $y$ be a regular value of $f$ and $g$ . By local constancy and Sard's theorem we may replace $y$ by a nearby value $z$ that is regular for $F$ too, without changing the counts for $f$ and $g$ . Then $F^{-1}(z)$ is a compact $1$ -manifold with boundary

F^{-1}(z) \cap \big(M \times \{0, 1\}\big) = f^{-1}(z) \times \{0\} \;\sqcup\; g^{-1}(z) \times \{1\}.

By the Fact the number of boundary points is even, so $\#f^{-1}(z) \equiv \#g^{-1}(z) \pmod 2$ . With orientations, an arc of $F^{-1}(z)$ joining two points of $f^{-1}(z)$ joins points of opposite sign, an arc joining $f^{-1}(z)$ to $g^{-1}(z)$ joins points of the same sign, and so the signed counts agree (Milnor, §5, checks the sign conventions).

Independence of $y$ (a). For regular values $y, z$ of $f$ there is a diffeomorphism $h : N \to N$ , smoothly isotopic to the identity, with $h(y) = z$ (the homogeneity lemma: move points along flows of compactly supported vector fields, and use connectedness of $N$ ). Then $h \circ f \simeq f$ , and $(h \circ f)^{-1}(z) = f^{-1}(y)$ , with the same signs since $h$ preserves orientation. By (b), $\#f^{-1}(y) = \#(h\circ f)^{-1}(z) \equiv \#f^{-1}(z)$ , and likewise for signed counts. ∎

∎

Theorem— Corollary 4 (properties of the degree)

$\deg(g \circ f) = \deg g \cdot \deg f$ ; a diffeomorphism has degree $\pm 1$ ; a map of non-zero degree is surjective; and if $M = \partial W$ for a compact oriented $W$ and $f$ extends to a smooth $F : W \to N$ , then $\deg f = 0$ .

The last statement is proved like (b): for $y$ regular for $F$ and $f$ , the preimage $F^{-1}(y)$ is a compact oriented $1$ -manifold with boundary $f^{-1}(y)$ , so the signs sum to $0$ .

ProofH

Brouwer's fixed-point theorem and the hairy ball theorem

Theorem— Theorem 5 (no retraction)

There is no smooth map $r : \overline{D}{}^n \to S^{n-1}$ with $r(x) = x$ for all $x \in S^{n-1}$ .

Proof

(M. Hirsch.) Suppose $r$ exists. By Sard's theorem it has a regular value $y \in S^{n-1}$ ; $y$ is also regular for $r|_{S^{n-1}} = \mathrm{id}$ . So $r^{-1}(y)$ is a compact $1$ -manifold with boundary $r^{-1}(y) \cap S^{n-1} = \{y\}$ — a single point. This contradicts the Fact. ∎

∎

Theorem— Theorem 6 (Brouwer's fixed-point theorem)

Every continuous map $f : \overline{D}{}^n \to \overline{D}{}^n$ has a fixed point.

Proof

Smooth case. Suppose $f$ is smooth with $f(x) \ne x$ for all $x$ . Let $u(x) = x - f(x) \ne 0$ and send $x$ along the ray from $f(x)$ through $x$ until it meets the sphere: $r(x) = x + t(x)\,u(x)$ , where $t(x) \ge 0$ solves $|x + t u|^2 = 1$ , that is

t(x) = \frac{-\langle x, u\rangle + \sqrt{\langle x, u\rangle^2 + |u|^2 (1 - |x|^2)}}{|u|^2}.

The discriminant is positive on the disc: for $|x| < 1$ because $|u|^2(1 - |x|^2) > 0$ , and for $|x| = 1$ because $\langle x, u\rangle = 1 - \langle x, f(x)\rangle > 0$ , by Cauchy–Schwarz and $f(x) \ne x$ . So $t$ is smooth, and for $|x| = 1$ it gives $t = 0$ , $r(x) = x$ . Thus $r$ is a smooth retraction, contradicting Theorem 5.

Continuous case. If a continuous $f$ has no fixed point, then $|f(x) - x| \ge \mu > 0$ on the compact disc. By the Weierstrass approximation theorem there is a polynomial map $P$ with $|P - f| < \mu/2$ on the disc. Then $P_1 = P/(1 + \mu/2)$ maps the disc into itself, $|P_1 - f| < \mu$ , and so $P_1$ is a smooth map with no fixed point — impossible by the smooth case. ∎

∎

Brouwer's theorem holds for any space homeomorphic to a disc (any compact convex set with non-empty interior), but not for an annulus (rotate it) or for $\R^n$ (translate it). It says nothing about how many fixed points there are, or how to find one.

Theorem— Theorem 7 (hairy ball theorem)

$S^n$ has a continuous tangent vector field without zeros if and only if $n$ is odd.

Proof

If $n = 2m - 1$ is odd, regard $S^n \subset \C^m$ and let $v(z) = iz$ , i.e. $(x_1, y_1, \dots, x_m, y_m) \mapsto (-y_1, x_1, \dots, -y_m, x_m)$ . Then $\langle z, v(z)\rangle = 0$ and $|v(z)| = |z| = 1$ .

Conversely, suppose $v$ is a nowhere-zero tangent field on $S^n$ . We may assume it is smooth (approximate it by a polynomial field and project onto the tangent spaces; a small perturbation stays non-zero) and of length $1$ (divide by $|v|$ ). Then

F(x, t) = \cos(\pi t)\, x + \sin(\pi t)\, v(x)

is a unit vector, since $x \perp v(x)$ , so $F$ is a homotopy on $S^n$ from $F_0 = \mathrm{id}$ to $F_1 = a$ , the antipodal map. By Theorem 3, $1 = \deg \mathrm{id} = \deg a = (-1)^{n+1}$ , so $n$ is odd. ∎

∎

On the round Earth ( $n = 2$ ) every continuous horizontal wind field has a calm point — a statement about an idealised continuous model, but a theorem about that model.

ProofH

The Poincaré–Hopf theorem and Morse theory

Theorem— Theorem 8 (Poincaré–Hopf)

Let $M$ be a closed manifold and $v$ a smooth vector field on $M$ with only isolated zeros $p_1, \dots, p_k$ . Then

\sum_{i=1}^k \operatorname{ind}_{p_i}(v) = \chi(M),

the Euler characteristic of $M$ (the alternating sum of the numbers of cells in any cell decomposition, or of the Betti numbers).

Proof sketch (Milnor, Topology from the Differentiable Viewpoint, §6). Step 1: the sum does not depend on $v$ . Embed $M$ in $\R^N$ and take a closed tubular neighbourhood $T$ of $M$ . Extend $v$ to a vector field $w$ on $T$ that points outward along $\partial T$ and has the same zeros, with the same indices. On the compact manifold with boundary $T$ minus small balls round the zeros, $w/|w|$ is defined; by Corollary 4, the degree of $w/|w|$ on $\partial T$ equals the sum of the degrees on the small spheres, which is $\sum \operatorname{ind}(v)$ . But on $\partial T$ , $w/|w|$ is homotopic to the outward unit normal (the Gauss map of $\partial T$ ), which does not involve $v$ . Step 2: compute it for one field. Take a triangulation of $M$ and a field with one zero at the barycentre of each simplex, flowing from the barycentre of each simplex towards the barycentres of the higher-dimensional simplices containing it (so vertices are sources). At the barycentre of a $k$ -simplex the flow is attracting in the $k$ directions of the simplex and repelling in the $n - k$ others, so the index is $(-1)^k$ , so the sum is $\sum_k (-1)^k \#\{k\text{-simplices}\} = \chi(M)$ . Alternatively, use the gradient of a Morse function, as below. $\square$

Consequences:

The hairy ball theorem again: $\chi(S^2) = 2 \ne 0$ , so every vector field on $S^2$ vanishes somewhere. The torus and the Klein bottle ( $\chi = 0$ ) carry nowhere-zero fields; a surface of genus $g \ge 2$ does not.
Hopf's converse (stated): a closed connected manifold has a nowhere-zero vector field if and only if $\chi(M) = 0$ . Every closed odd-dimensional manifold has $\chi = 0$ (by Poincaré duality), so it has one.
Morse theory. If $h$ is a Morse function, its gradient (for any metric) has a zero of index $(-1)^\lambda$ at each critical point of Morse index $\lambda$ (from the normal form $h = h(p) - x_1^2 - \dots - x_\lambda^2 + x_{\lambda+1}^2 + \dots + x_n^2$ , the Morse lemma). So $\sum_\lambda (-1)^\lambda c_\lambda = \chi(M)$ , where $c_\lambda$ is the number of critical points of index $\lambda$ . For the upright torus, $1 - 2 + 1 = 0$ . Morse proved more: $M$ has a cell decomposition with one $\lambda$ -cell for each critical point of index $\lambda$ , which gives the Morse inequalities $c_\lambda \ge b_\lambda$ (the Betti numbers).

Transversality. Regular values generalise to transversality, and Sard's theorem to Thom's transversality theorem: if $F : X \times S \to Y$ is smooth and transverse to $Z$ , then $F_s = F(\cdot, s)$ is transverse to $Z$ for almost every $s$ . Hence any map can be perturbed to be transverse to a given submanifold. For $X$ compact with $\dim X + \dim Z = \dim Y$ , the number of points of $F_s^{-1}(Z)$ mod 2 is then a homotopy invariant, the mod 2 intersection number. A meridian and a longitude on the torus meet once, transversally, so no homotopy can pull them apart; any two closed curves on the sphere have intersection number $0$ .

Formal mathematicsF

Differential topology in Lean and Mathlib

Mathlib has a strong form of Sard's theorem in equal dimensions: the image of the set where the derivative is singular is null for any Haar measure, assuming only differentiability. We cite it and deduce that critical values are negligible:

sard_equidimensionalLean 4Applies a library theoremFormally verified

/-- **Sard's theorem (equal dimensions).** If `f : E → E` has derivative `f' x` at each point of
`s` and `f' x` is singular (`det = 0`) on `s`, then `f(s)` has measure zero for any Haar measure
(for instance Lebesgue measure on `ℝⁿ`). This is Mathlib's
`MeasureTheory.addHaar_image_eq_zero_of_det_fderivWithin_eq_zero`; note that it needs only
differentiability, not `C¹`. -/
theorem sard_equidimensional {E : Type*} [NormedAddCommGroup E] [NormedSpace ℝ E]
    [FiniteDimensional ℝ E] [MeasurableSpace E] [BorelSpace E]
    (μ : MeasureTheory.Measure E) [μ.IsAddHaarMeasure] {f : E → E} {s : Set E}
    {f' : E → E →L[ℝ] E} (hf' : ∀ x ∈ s, HasFDerivWithinAt f (f' x) s x)
    (hsing : ∀ x ∈ s, (f' x).det = 0) : μ (f '' s) = 0 :=
  MeasureTheory.addHaar_image_eq_zero_of_det_fderivWithin_eq_zero μ hf' hsing