19. Curves and surfaces¶

How curved is a road at a given bend? How curved is a hill, or a saddle, or a pizza slice? Euclid's geometry measured straight things and circles; to measure bending in general you need calculus. This chapter starts differential geometry: you will measure the length of any curve, define its curvature (how fast it turns) and torsion (how fast it twists out of a plane), and then do the same for surfaces. There the story takes a famous turn. Gauss found in 1827 that one measure of a surface's curvature can be computed by an ant living on the surface, without ever leaving it. That fact explains why maps of the Earth must distort, why you fold a pizza slice to stop it drooping, and it leads to the Gauss–Bonnet theorem, which ties curvature to the shape of a whole surface.

In this chapter

You will learn:

parametrised curves, regular curves, arc length, and the arc-length parametrisation
curvature and torsion, the Frenet frame and the Frenet–Serret equations, and why they determine a curve
parametrised surfaces, tangent planes and the first fundamental form (lengths, angles and areas on a surface)
the second fundamental form, principal curvatures, and Gaussian and mean curvature
Gauss's Theorema Egregium: what it says, why it is surprising, and what it does to pizza and maps
geodesics: the straightest, locally shortest curves on a surface
the Gauss–Bonnet theorem, from geodesic triangles to the Euler characteristic

You need: Chapter 9 (dot and cross products), Chapter 11 (Euler's formula), Chapter 16 (Girard's theorem) and Chapter 17. From calculus: derivatives, integrals, partial derivatives and double integrals. The U of T guides on the definition of the integral and the derivative as a linear map cover the background.

Levels: 19.1 is intermediate; 19.2 to 19.4 are advanced (first university course in differential geometry); 19.5 to 19.7 are advanced to expert, with proofs sketched where they are long.

19.1 Parametrised curves and arc length¶

Level: intermediate. How long is a curve? For a straight segment you measure; for a circle, Archimedes trapped the answer between polygons (Chapter 6). The same idea works for any reasonable curve: mark points along it, join them by straight pieces, add up, and let the pieces get shorter.

To make that precise we describe a curve by a moving point. A parametrised curve is a function

\[ \gamma : I \to \mathbb{R}^n, \qquad t \mapsto \gamma(t) = (x_1(t), \dots, x_n(t)), \]

from an interval \(I\) of real numbers, with each coordinate a smooth (infinitely differentiable) function. Think of \(t\) as time and \(\gamma(t)\) as the position of a particle. Its velocity is \(\gamma'(t) = (x_1'(t), \dots, x_n'(t))\), a vector tangent to the curve, and its speed is the length \(|\gamma'(t)|\). The curve is regular if \(\gamma'(t) \neq 0\) for every \(t\): the particle never stops, so the curve has a well defined direction at every point. (A curve that stops can turn a sharp corner: \(\gamma(t) = (t^3, |t|^3)\) is smooth enough but traces the corner of \(y = |x|\).)

The same picture can be traced by many parametrisations: \(\gamma(t) = (\cos t, \sin t)\) and \(\delta(t) = (\cos 2t, \sin 2t)\) both run around the unit circle, the second one twice as fast. Geometry should not depend on the speed, so we look for quantities that do not change under reparametrisation \(t = \phi(s)\), with \(\phi\) smooth and \(\phi' > 0\).

Definition. The arc length of \(\gamma\) from \(t = a\) to \(t = b\) is

\[ L = \int_a^b |\gamma'(t)| \, dt . \]

Why this is the polygon limit. A small piece of the polygon from \(\gamma(t_i)\) to \(\gamma(t_{i+1})\) has length \(|\gamma(t_{i+1}) - \gamma(t_i)| \approx |\gamma'(t_i)| \, \Delta t\) by the definition of the derivative, and summing gives a Riemann sum for the integral. (Making "\(\approx\)" precise uses the uniform continuity of \(\gamma'\) on \([a, b]\).) And it does not depend on the parametrisation: if \(t = \phi(s)\) then \(|(\gamma \circ \phi)'(s)| = |\gamma'(\phi(s))| \, \phi'(s)\), and the substitution rule turns one integral into the other. Speed times time is distance, whatever the speed.

Worked example 1: the circle. \(\gamma(t) = (r\cos t, r\sin t)\) has \(\gamma'(t) = (-r\sin t, r\cos t)\) and speed \(r\). One turn: \(L = \int_0^{2\pi} r \, dt = 2\pi r\). Check: the familiar circumference.

Worked example 2: the helix. \(\gamma(t) = (a\cos t, a\sin t, bt)\) winds around a cylinder of radius \(a\), rising \(2\pi b\) per turn. Its speed is \(\sqrt{a^2 \sin^2 t + a^2 \cos^2 t + b^2} = \sqrt{a^2 + b^2}\), constant, so one turn has length \(2\pi\sqrt{a^2 + b^2}\). Check: cut the cylinder along a vertical line and unroll it: one turn becomes the diagonal of a rectangle \(2\pi a\) wide and \(2\pi b\) high, of length \(\sqrt{(2\pi a)^2 + (2\pi b)^2}\).

Worked example 3: the cycloid. A wheel of radius \(r\) rolls along the \(x\)-axis; a point on its rim traces \(\gamma(t) = (r(t - \sin t), r(1 - \cos t))\). Then \(\gamma'(t) = (r(1 - \cos t), r\sin t)\) and

\[ |\gamma'(t)|^2 = r^2\bigl((1 - \cos t)^2 + \sin^2 t\bigr) = 2r^2(1 - \cos t) = 4r^2 \sin^2 \tfrac{t}{2} . \]

One arch (\(0 \le t \le 2\pi\)) has length \(\int_0^{2\pi} 2r \sin\tfrac{t}{2}\, dt = \bigl[-4r\cos\tfrac{t}{2}\bigr]_0^{2\pi} = 8r\). Christopher Wren found this result in 1658: the arch is exactly four times the wheel's diameter, with no \(\pi\) in sight.

The arc-length parametrisation. Fix a starting time \(a\) and let \(s(t) = \int_a^t |\gamma'(u)|\, du\), the distance travelled. For a regular curve \(s'(t) = |\gamma'(t)| > 0\), so \(s\) is increasing and has an inverse; reparametrising by \(s\) gives a curve with speed exactly 1. Every regular curve can be parametrised by arc length. It makes formulas cleaner, and we use \(s\) for this parameter from now on, with a dot or a prime for \(d/ds\). For the helix, \(s = \sqrt{a^2 + b^2}\, t\), so \(\gamma(s) = (a\cos(s/c), a\sin(s/c), bs/c)\) with \(c = \sqrt{a^2 + b^2}\).

19.2 Curvature and torsion; the Frenet frame¶

Level: advanced. A unit-speed curve has unit tangent vector \(T(s) = \gamma'(s)\). Along a straight line \(T\) is constant; along a bend it turns. How fast it turns is the curvature.

Definition. For a curve parametrised by arc length, the curvature is \(\kappa(s) = |T'(s)| = |\gamma''(s)|\).

Because \(T \cdot T = 1\), differentiating gives \(2T \cdot T' = 0\): the change \(T'\) is perpendicular to \(T\). Where \(\kappa \neq 0\) define the principal normal \(N = T' / \kappa\), a unit vector pointing to the inside of the bend. The circle that touches the curve at \(\gamma(s)\) with the same tangent and curvature, centred at \(\gamma(s) + N/\kappa\), is the osculating circle ("kissing circle"), and \(1/\kappa\) is the radius of curvature.

Example: the circle. \(\gamma(s) = (r\cos(s/r), r\sin(s/r))\) has \(T = (-\sin(s/r), \cos(s/r))\) and \(T' = -\tfrac1r(\cos(s/r), \sin(s/r))\), so \(\kappa = 1/r\) and \(N\) points to the centre. A small circle is very curved, a large one barely: this matches the everyday sense of "a sharp bend".

Formulas for any parametrisation. Reparametrising by arc length is often impossible to do in closed form, so it is useful to have formulas in terms of any parameter \(t\). For a plane curve \((x(t), y(t))\) the signed curvature (positive when the curve turns left) is

\[ \kappa = \frac{x'y'' - y'x''}{(x'^2 + y'^2)^{3/2}}, \]

and for a space curve

\[ \kappa = \frac{|\gamma' \times \gamma''|}{|\gamma'|^3} . \]

Proof of the space formula. Write \(v = |\gamma'|\), so \(\gamma' = vT\). Then \(\gamma'' = v'T + vT_t\), where \(T_t = dT/dt = (dT/ds)(ds/dt) = \kappa N v\). So \(\gamma'' = v'T + \kappa v^2 N\), and \(\gamma' \times \gamma'' = vT \times (v'T + \kappa v^2 N) = \kappa v^3\, T \times N\). Since \(T \perp N\) are unit vectors, \(|T \times N| = 1\), giving \(|\gamma' \times \gamma''| = \kappa v^3\). (A plane curve is a space curve with \(z = 0\); the third component of the cross product is \(x'y'' - y'x''\), which also gives the sign.) \(\square\)

Worked example: the parabola \(y = x^2\), parametrised as \((t, t^2)\): \(x' = 1\), \(y' = 2t\), \(x'' = 0\), \(y'' = 2\), so \(\kappa = 2/(1 + 4t^2)^{3/2}\). At the vertex \(\kappa = 2\), radius of curvature \(\tfrac12\); at \((1, 1)\), \(\kappa = 2/5^{3/2} \approx 0.179\), radius \(\approx 5.59\). Check: the bend is sharpest at the bottom, as the picture of a parabola suggests.

Worked example: the ellipse \((a\cos t, b\sin t)\): the plane formula gives

\[ \kappa(t) = \frac{ab}{(a^2\sin^2 t + b^2\cos^2 t)^{3/2}} . \]

At the end of the long axis (\(t = 0\)), \(\kappa = a/b^2\); at the end of the short axis, \(\kappa = b/a^2\). For \(a = 2\), \(b = 1\): curvature \(2\) (radius \(\tfrac12\)) at \((2, 0)\) and \(\tfrac14\) (radius 4) at \((0, 1)\).

Torsion and the Frenet frame. A curve in space can also twist out of its plane. Where \(\kappa \neq 0\) define the binormal \(B = T \times N\). The three vectors \(T, N, B\) are perpendicular unit vectors at each point, a moving frame attached to the curve: the Frenet frame. The plane of \(T\) and \(N\) is the osculating plane, the plane the curve "almost lies in" near the point. How fast that plane tilts is measured by \(B'\).

Theorem 19.1 (Frenet–Serret equations). For a unit-speed curve with \(\kappa > 0\) there is a function \(\tau(s)\), the torsion, such that

\[ T' = \kappa N, \qquad N' = -\kappa T + \tau B, \qquad B' = -\tau N . \]

Proof. The first equation is the definition of \(N\). For the others, express each derivative in the orthonormal basis \(T, N, B\); the coefficient of a vector \(X\) in \(Y'\) is \(Y' \cdot X\). Differentiating \(X \cdot Y = \text{const}\) (0 or 1) for any two frame vectors gives \(X' \cdot Y = -X \cdot Y'\), so the matrix of coefficients is skew-symmetric, with zero diagonal. We know \(T' \cdot N = \kappa\) and \(T' \cdot B = 0\). So \(N' \cdot T = -\kappa\), \(N' \cdot N = 0\), and we define \(\tau = N' \cdot B\). Then \(B' \cdot T = -T' \cdot B = 0\), \(B' \cdot N = -\tau\), \(B' \cdot B = 0\). \(\square\)

A plane curve has \(B\) constant (perpendicular to the plane) and \(\tau = 0\); conversely if \(\tau = 0\) then \(B\) is constant and \((\gamma \cdot B)' = T \cdot B = 0\), so the curve lies in a plane. Torsion is exactly the failure to be planar. For a general parameter,

\[ \tau = \frac{(\gamma' \times \gamma'') \cdot \gamma'''}{|\gamma' \times \gamma''|^2} . \]

Worked example: the helix \((a\cos t, a\sin t, bt)\). Here \(\gamma' = (-a\sin t, a\cos t, b)\), \(\gamma'' = (-a\cos t, -a\sin t, 0)\), \(\gamma''' = (a\sin t, -a\cos t, 0)\), and

\[ \gamma' \times \gamma'' = (ab\sin t, -ab\cos t, a^2), \qquad |\gamma' \times \gamma''| = a\sqrt{a^2 + b^2}. \]

So \(\kappa = a\sqrt{a^2 + b^2}/(a^2 + b^2)^{3/2} = a/(a^2 + b^2)\), and \((\gamma' \times \gamma'') \cdot \gamma''' = a^2 b\sin^2 t + a^2 b\cos^2 t = a^2 b\), so \(\tau = a^2 b / (a^2(a^2 + b^2)) = b/(a^2 + b^2)\). Both are constant: a helix bends and twists at the same rate everywhere. Check: \(b = 0\) gives a circle, \(\kappa = 1/a\), \(\tau = 0\).

Theorem 19.2 (fundamental theorem of curves). Given smooth functions \(\kappa(s) > 0\) and \(\tau(s)\) on an interval, there is a unit-speed curve with exactly this curvature and torsion, and any two such curves differ by a rigid motion (a rotation followed by a translation).

Idea of proof. The Frenet–Serret equations are a linear system of ordinary differential equations for the nine components of \(T, N, B\). By the existence and uniqueness theorem for such systems, a choice of initial frame determines the solution; the skew-symmetry keeps the frame orthonormal; and \(\gamma = \int T \, ds\). Two choices of initial point and frame differ by a rigid motion, and so do the resulting curves. \(\square\)

So curvature and torsion are a complete description of a space curve's shape: a curve with constant \(\kappa\) and \(\tau\) is a helix (or a circle). DNA, springs and the tendrils of climbing plants are all close to helices.

Global facts. Two classical theorems connect curvature to the whole curve. For a closed plane curve, the total signed curvature \(\int \kappa \, ds\) is \(2\pi\) times the number of times the tangent turns around (the turning number); for a simple closed curve (one that does not cross itself) this is \(\pm 2\pi\) (Hopf's Umlaufsatz, 1935). And for any closed space curve, \(\int \kappa\, ds \ge 2\pi\), with equality only for plane convex curves (Fenchel, 1929). These are first examples of the pattern that ends the chapter: integrate curvature, get a number that depends only on the shape's topology.

19.3 Surfaces and the first fundamental form¶

Level: advanced. Now let the moving point depend on two parameters. A parametrised surface (a chart of a surface) is a smooth map

\[ X : U \to \mathbb{R}^3, \qquad (u, v) \mapsto X(u, v), \]

from an open set \(U\) of the plane, such that the partial derivatives \(X_u = \partial X/\partial u\) and \(X_v = \partial X/\partial v\) are linearly independent at every point. Holding \(v\) fixed and moving \(u\) traces a curve on the surface with velocity \(X_u\); likewise for \(X_v\). The two vectors span the tangent plane \(T_pS\) at \(p = X(u, v)\), and

\[ n = \frac{X_u \times X_v}{|X_u \times X_v|} \]

is a unit normal. A whole surface, like a sphere, usually needs several charts; Chapter 20 makes this precise.

Examples.

Sphere of radius \(R\), by latitude-longitude: \(X(\varphi, \theta) = R(\sin\varphi\cos\theta, \sin\varphi\sin\theta, \cos\varphi)\), with \(\varphi \in (0, \pi)\) the angle from the north pole and \(\theta\) the longitude.
Cylinder of radius \(a\): \(X(\theta, z) = (a\cos\theta, a\sin\theta, z)\).
Graph of a function: \(X(x, y) = (x, y, f(x, y))\).
Torus, made by spinning a circle of radius \(r\) about an axis at distance \(R > r\) from its centre: \(X(u, v) = ((R + r\cos v)\cos u, (R + r\cos v)\sin u, r\sin v)\).

Measuring on the surface. An ant on the surface walks along a curve \(\alpha(t) = X(u(t), v(t))\). Its velocity, by the chain rule, is \(\alpha' = X_u u' + X_v v'\), and its speed squared is

\[ |\alpha'|^2 = E\,u'^2 + 2F\,u'v' + G\,v'^2, \qquad E = X_u \cdot X_u, \quad F = X_u \cdot X_v, \quad G = X_v \cdot X_v . \]

The quadratic form \(I = E\,du^2 + 2F\,du\,dv + G\,dv^2\) is the first fundamental form, also written \(ds^2\). It is the dot product of the tangent plane, written in the coordinates \(u, v\). With it the ant can compute, without ever looking outside the surface:

the length of a curve: \(\int \sqrt{E u'^2 + 2F u'v' + G v'^2}\, dt\);
the angle between two curves, from the dot product of their velocities;
the area of a region: \(\iint \sqrt{EG - F^2}\, du\, dv\), because \(|X_u \times X_v|^2 = |X_u|^2|X_v|^2 - (X_u \cdot X_v)^2 = EG - F^2\) is the squared area of the small parallelogram spanned by \(X_u\,du\) and \(X_v\,dv\).

These are intrinsic quantities: they depend only on \(E, F, G\), the measurements made inside the surface.

Worked example: the sphere. \(X_\varphi = R(\cos\varphi\cos\theta, \cos\varphi\sin\theta, -\sin\varphi)\) and \(X_\theta = R(-\sin\varphi\sin\theta, \sin\varphi\cos\theta, 0)\). So \(E = R^2\), \(F = 0\), \(G = R^2\sin^2\varphi\):

\[ ds^2 = R^2\, d\varphi^2 + R^2 \sin^2\varphi \, d\theta^2 . \]

Area: \(\int_0^{2\pi}\int_0^\pi R^2 \sin\varphi \, d\varphi\, d\theta = 4\pi R^2\). Check: Archimedes' result from Chapter 11. A meridian (\(\theta\) fixed) from pole to pole has length \(\int_0^\pi R\,d\varphi = \pi R\); a circle of latitude \(\varphi = \varphi_0\) has length \(2\pi R \sin\varphi_0\), shrinking toward the poles.

Worked example: the cylinder is "flat". For \(X(\theta, z) = (a\cos\theta, a\sin\theta, z)\), \(E = a^2\), \(F = 0\), \(G = 1\). Change the parameter to \(u = a\theta\) (arc length around the cylinder): then \(ds^2 = du^2 + dz^2\), exactly the first fundamental form of the plane in ordinary coordinates. Two surfaces whose first fundamental forms agree in suitable coordinates are locally isometric: an ant cannot tell them apart by any measurement of lengths, angles or areas in a small region. Indeed you can roll a sheet of paper into a cylinder without stretching or tearing. Whether you can do the same with a sphere is the question of 19.5.

19.4 Gaussian curvature¶

Level: advanced. The first fundamental form sees only the inside. How the surface bends in space is measured by how the normal \(n\) tilts as you move. Take a unit tangent vector \(w\) at \(p\) and slice the surface with the plane through \(p\) spanned by \(w\) and \(n\). The slice is a plane curve, the normal section in direction \(w\), and its signed curvature at \(p\) (positive if it bends toward \(n\)) is the normal curvature \(k_n(w)\).

The second fundamental form. With \(L = X_{uu} \cdot n\), \(M = X_{uv} \cdot n\), \(N = X_{vv} \cdot n\), the normal curvature in the direction of \(w = X_u\,du + X_v\,dv\) is

\[ k_n = \frac{L\,du^2 + 2M\,du\,dv + N\,dv^2}{E\,du^2 + 2F\,du\,dv + G\,dv^2} . \]

Why. Let \(\alpha(s)\) be a unit-speed curve on the surface through \(p\) with tangent \(w\). Since \(\alpha' \cdot n = 0\) all along it, differentiating gives \(\alpha'' \cdot n = -\alpha' \cdot n'\), and \(\alpha'' = X_{uu}u'^2 + 2X_{uv}u'v' + X_{vv}v'^2 + X_u u'' + X_v v''\); the last two terms are tangent, so \(\alpha'' \cdot n = L u'^2 + 2M u'v' + N v'^2\). For the normal section, \(\alpha''\) is \(\kappa\) times its normal, which is \(\pm n\), so \(\alpha'' \cdot n = k_n\); and with unit speed the denominator is 1. \(\square\)

The numerator is the second fundamental form \(II\). The formula says the normal curvature depends only on the direction, and is the ratio of two quadratic forms. As \(w\) turns around the circle of directions, \(k_n\) takes a maximum \(k_1\) and a minimum \(k_2\), the principal curvatures, in two perpendicular principal directions, and in between (Euler's formula, from the 1760s)

\[ k_n(\theta) = k_1 \cos^2\theta + k_2 \sin^2\theta, \]

where \(\theta\) is the angle from the first principal direction. (This is the diagonalisation of a symmetric matrix: \(II\) relative to \(I\) is a self-adjoint map of the tangent plane, the shape operator, and \(k_1, k_2\) are its eigenvalues.)

Definition. The Gaussian curvature is \(K = k_1 k_2\) and the mean curvature is \(H = \tfrac12(k_1 + k_2)\). As determinants,

\[ K = \frac{LN - M^2}{EG - F^2} . \]

The sign of \(K\) classifies points:

\(K > 0\) (elliptic point): both principal curvatures have the same sign; the surface bends the same way in all directions and lies on one side of its tangent plane nearby, like a dome or an egg.
\(K < 0\) (hyperbolic point): the principal curvatures have opposite signs; the surface crosses its tangent plane, like a saddle or a mountain pass.
\(K = 0\): one principal curvature is zero, as on a cylinder or a cone (or both are, as on a plane).

Note that changing the normal \(n\) to \(-n\) changes the signs of \(k_1\), \(k_2\) and \(H\) but not of \(K\): Gaussian curvature does not care which side is "up".

Worked example: graphs at a flat point. For \(z = f(x, y)\) at a point where \(f_x = f_y = 0\), the tangent plane is horizontal, \(E = G = 1\), \(F = 0\), and \(L, M, N\) are the second derivatives \(f_{xx}, f_{xy}, f_{yy}\). So

\[ K = f_{xx} f_{yy} - f_{xy}^2, \]

the determinant of the Hessian. (In general, \(K = (f_{xx}f_{yy} - f_{xy}^2)/(1 + f_x^2 + f_y^2)^2\).)

Paraboloid \(z = x^2 + y^2\) at the origin: \(K = 2 \cdot 2 - 0 = 4\).
Saddle \(z = xy\) at the origin: \(K = 0 \cdot 0 - 1^2 = -1\).
Trough \(z = x^2\): \(K = 2 \cdot 0 - 0 = 0\) everywhere.

This is the second-derivative test of calculus seen geometrically: \(K > 0\) at a maximum or minimum, \(K < 0\) at a saddle point.

Worked example: the sphere. At every point of a sphere of radius \(R\), every normal section is a great circle of radius \(R\), so \(k_1 = k_2 = 1/R\) (with the inward normal) and \(K = 1/R^2\). The Earth, with mean radius about 6371 km, has \(K \approx 2.5 \times 10^{-8}\) per square kilometre; Mars, radius about 3389.5 km, about \(8.7 \times 10^{-8}\), about three and a half times as curved.

Worked example: the torus. A calculation with the parametrisation above gives \(E = (R + r\cos v)^2\), \(F = 0\), \(G = r^2\), \(L = (R + r\cos v)\cos v\), \(M = 0\), \(N = r\) (with the inward normal), so

\[ K = \frac{\cos v}{r(R + r\cos v)} . \]

On the outer half (\(\cos v > 0\)) the torus is dome-like, \(K > 0\); on the inner half, around the hole, it is saddle-like, \(K < 0\); on the top and bottom circles \(v = \pm\pi/2\), \(K = 0\).

Mean curvature and soap films. A soap film spanning a wire frame minimises area, and a calculation of the first variation of area shows that a surface is a critical point of area exactly when \(H = 0\) everywhere: a minimal surface. A minimal surface has \(k_2 = -k_1\), so \(K \le 0\): soap films are saddle-shaped at every point. The catenoid (the surface spun from a hanging chain, found by Euler in 1744) and the helicoid are the classical examples.

19.5 The Theorema Egregium¶

Level: advanced to expert. The formula \(K = (LN - M^2)/(EG - F^2)\) uses \(L, M, N\), which depend on the normal \(n\) and so on how the surface sits in space. Gauss discovered something no one expected.

Theorem 19.3 (Theorema Egregium, Gauss 1827). The Gaussian curvature \(K\) depends only on the first fundamental form \(E, F, G\) and their first and second derivatives. It is an intrinsic quantity, unchanged by any bending of the surface that does not stretch it (any local isometry).

Gauss called it the "remarkable theorem" (egregium) in his Disquisitiones generales circa superficies curvas (General investigations of curved surfaces), presented in 1827. He had spent years surveying the Kingdom of Hanover, and the question of what can be measured on the Earth's surface from inside it was a practical one for him.

For coordinates with \(F = 0\) (the coordinate curves meet at right angles, as for latitude and longitude) the formula is

\[ K = -\frac{1}{2\sqrt{EG}} \left[ \frac{\partial}{\partial u}\left( \frac{G_u}{\sqrt{EG}} \right) + \frac{\partial}{\partial v}\left( \frac{E_v}{\sqrt{EG}} \right) \right] . \]

Check on the sphere. \(E = R^2\), \(G = R^2\sin^2\varphi\) with \(u = \varphi\): \(\sqrt{EG} = R^2\sin\varphi\), \(E_v = 0\), and \(G_u/\sqrt{EG} = 2R^2\sin\varphi\cos\varphi / (R^2\sin\varphi) = 2\cos\varphi\). Its \(\varphi\)-derivative is \(-2\sin\varphi\), so

\[ K = -\frac{1}{2R^2\sin\varphi}(-2\sin\varphi) = \frac{1}{R^2} . \]

The same answer as before, now computed by an ant that knows only how to measure distances on the sphere.

Check on the plane in polar coordinates. \(ds^2 = dr^2 + r^2 d\theta^2\): \(E = 1\), \(G = r^2\), \(\sqrt{EG} = r\), \(G_r/\sqrt{EG} = 2\), whose derivative is 0. So \(K = 0\), correctly, even though the coordinate lines are curved.

Idea of the proof. Write the second derivatives \(X_{uu}, X_{uv}, X_{vv}\) in the basis \(X_u, X_v, n\). Their tangential parts have coefficients \(\Gamma^k_{ij}\), the Christoffel symbols, which can be solved for from the dot products \(X_{uu} \cdot X_u = \tfrac12 E_u\) and the like, so they depend only on \(E, F, G\) and their first derivatives. Their normal parts are \(L, M, N\). Now use that mixed third derivatives agree: \((X_{uu})_v = (X_{uv})_u\). Taking the component of this identity along \(X_u\) or \(X_v\) gives an equation (the Gauss equation) in which \(LN - M^2\) appears on one side and only Christoffel symbols and their derivatives on the other. Dividing by \(EG - F^2\) expresses \(K\) in terms of \(E, F, G\) alone. The algebra fills a page or two; it is done in full in do Carmo's book (section 4-3) and in most texts on curves and surfaces. \(\square\)

Consequences.

No perfect flat map. A map that kept all distances (up to one scale factor) between a region of the sphere and a region of the plane would be a local isometry, so it would carry \(K = 1/R^2\) to \(K = 0\). Impossible. Every map of the Earth or of Mars distorts something; Chapter 16 showed which projections keep angles (Mercator, stereographic) or areas, but never both.
Paper can roll but not wrap. A flat sheet has \(K = 0\). Without stretching it can become a cylinder, a cone or any other surface with \(K = 0\) (a developable surface), but never a piece of a sphere. Gift-wrapping a ball always crumples the paper.
The pizza. A slice of pizza is flat: \(K = 0\), and it stays 0 as long as the dough does not stretch. If you curl the crust into a U, the slice gets a nonzero principal curvature \(k_1\) across its width. Since \(k_1 k_2\) must remain 0, the other principal curvature \(k_2\), along the length of the slice, is forced to be 0: the slice stays straight from crust to tip, and the tip no longer droops. Corrugated roofing and the folds in a sheet of card use the same principle: curvature in one direction gives stiffness in the other.

19.6 Geodesics¶

Level: advanced. What is a "straight line" for an ant on a surface? Two answers, which turn out to agree locally: a curve that never turns within the surface, and a curve that is the shortest path between nearby points.

Let \(\alpha(s)\) be a unit-speed curve on a surface. Its acceleration \(\alpha''\) splits into a part along the normal \(n\), which the surface forces on it (you cannot follow a sphere without accelerating toward its centre), and a part in the tangent plane, which is the turning the ant feels. The length of the tangential part (with a sign) is the geodesic curvature \(k_g\), and

\[ \kappa^2 = k_n^2 + k_g^2 . \]

Definition. A geodesic is a curve on the surface with \(k_g = 0\) everywhere: its acceleration is purely normal to the surface.

Example: great circles. On the sphere \(|x| = R\), take the great circle \(\alpha(s) = R(\cos(s/R)\, e + \sin(s/R)\, f)\) with \(e, f\) perpendicular unit vectors. Then \(\alpha''(s) = -\alpha(s)/R^2\), which points along the radius, the normal of the sphere. So great circles are geodesics. A small circle of latitude is not (except the equator): its acceleration points to the centre of the small circle, which is not the centre of the sphere, so it has a tangential component pointing toward the pole. That is why a long-haul flight from Toronto to Tokyo goes far north: the great circle, not the line of latitude, is straight on the sphere (Chapter 16).

The geodesic equations. In coordinates, \(\alpha(s) = X(u(s), v(s))\) is a geodesic exactly when

\[ u'' + \Gamma^u_{uu}u'^2 + 2\Gamma^u_{uv}u'v' + \Gamma^u_{vv}v'^2 = 0, \qquad v'' + \Gamma^v_{uu}u'^2 + 2\Gamma^v_{uv}u'v' + \Gamma^v_{vv}v'^2 = 0, \]

using the Christoffel symbols of 19.5. These depend only on \(E, F, G\), so geodesics are intrinsic: a local isometry carries geodesics to geodesics. Being second-order differential equations, they have a unique solution for each starting point and starting direction: from every point, in every direction, there is exactly one geodesic.

Theorem 19.4 (geodesics are locally shortest). Every point of a surface has a neighbourhood in which any two points are joined by a unique shortest curve, and that curve is a geodesic. Conversely a shortest curve between two points (when one exists) is a geodesic.

Idea of proof. The converse is a calculus-of-variations computation: if a curve is shortest, the first variation of length for every small deformation is zero, and integrating by parts shows this forces \(k_g = 0\). For the first part one builds geodesic polar coordinates around the point, using the geodesics from it as radial lines; in them \(ds^2 = dr^2 + G\,d\theta^2\) (Gauss's lemma: the circles \(r = \text{const}\) meet the radial geodesics at right angles), and then any curve from the centre to the circle of radius \(r\) has length at least \(\int dr = r\), with equality only along a radial geodesic. \(\square\)

"Locally" matters: on a sphere a great circle stops being shortest once it passes the antipodal point, and on a cylinder there are infinitely many geodesics between two points (helices winding different numbers of times).

Worked example: the cylinder by unrolling. The cylinder is locally isometric to the plane, and isometries carry geodesics to geodesics, so the geodesics on a cylinder are the images of straight lines on the unrolled sheet: vertical lines, horizontal circles and helices.

On a cylinder of radius 1, what is the shortest path from the point at angle 0 and height 0 to the point at angle \(\pi\) and height 2? Unroll: the second point is at horizontal distance \(\pi\) (going either way around) and height 2, so the shortest path is a half turn of a helix of length \(\sqrt{\pi^2 + 4} \approx 3.724\). Check: going straight up and then around the circle would cost \(2 + \pi \approx 5.14\).

Clairaut's relation. On a surface of revolution (spin a curve about the \(z\)-axis), let \(r\) be the distance from the axis and \(\psi\) the angle a geodesic makes with the circle of latitude. Then along every geodesic

\[ r \cos\psi = \text{constant} . \]

(It is a conservation law, of angular momentum about the axis, coming from the rotational symmetry; it follows from the geodesic equation for the angle coordinate.) On a sphere it says, for example, that a great circle that starts at the equator at angle \(\psi_0\) reaches a highest latitude where \(r = R\cos\psi_0\), the point where it runs due east.

19.7 The Gauss–Bonnet theorem¶

Level: advanced to expert. In Chapter 16, Girard's theorem said that a triangle on a sphere of radius \(R\) with angles \(\alpha, \beta, \gamma\) has area \(R^2(\alpha + \beta + \gamma - \pi)\). Write this as

\[ \iint_T K\, dA = \frac{1}{R^2} \cdot \text{area} = \alpha + \beta + \gamma - \pi . \]

In Chapter 17, a hyperbolic triangle (\(K = -1\)) has area \(\pi - (\alpha + \beta + \gamma)\), which is the same formula. And in the plane (\(K = 0\)), \(\alpha + \beta + \gamma = \pi\). Gauss saw that one theorem lies behind all three.

Theorem 19.5 (local Gauss–Bonnet for geodesic triangles; Gauss 1827). If \(T\) is a triangle on a surface whose sides are geodesics, contained in a region covered by one chart, with interior angles \(\alpha, \beta, \gamma\), then

\[ \iint_T K \, dA = \alpha + \beta + \gamma - \pi . \]

So on a surface with \(K > 0\) geodesic triangles have angle sum more than \(\pi\), with \(K < 0\) less than \(\pi\), and the excess measures the total curvature inside.

Proof sketch. Put the vertex \(A\) at the centre of geodesic polar coordinates \((r, \theta)\), so \(ds^2 = dr^2 + G\,d\theta^2\). With \(E = 1\), \(F = 0\) the Theorema Egregium formula simplifies to \(K = -(\sqrt{G})_{rr}/\sqrt{G}\), and the area element is \(\sqrt{G}\, dr\, d\theta\). The sides \(AB\) and \(AC\) are radial geodesics at angles \(\theta = 0\) and \(\theta = \alpha\), and the third side is \(r = \rho(\theta)\). So

\[ \iint_T K\, dA = \int_0^\alpha \int_0^{\rho(\theta)} -(\sqrt{G})_{rr}\, dr\, d\theta = \int_0^\alpha \Bigl(1 - (\sqrt{G})_r\bigl(\rho(\theta), \theta\bigr)\Bigr)\, d\theta, \]

using that \((\sqrt{G})_r \to 1\) as \(r \to 0\) (near the centre the coordinates look like ordinary polar coordinates). The first term integrates to \(\alpha\). For the second, let \(\psi(\theta)\) be the angle between the side \(BC\) and the radial geodesic through its point at angle \(\theta\). A computation with the geodesic equations (it is the formula for the geodesic curvature of \(BC\) in these coordinates) shows that because \(BC\) is a geodesic, \(d\psi = -(\sqrt{G})_r\, d\theta\) along it. So the second term is the total change of \(\psi\) from \(B\) to \(C\). At \(B\) the side \(BC\) makes angle \(\pi - \beta\) with the outgoing radial direction, and at \(C\) the angle is \(\gamma\), so the change is \(\gamma - (\pi - \beta)\). Altogether, \(\alpha + \gamma - \pi + \beta\). \(\square\)

Check on the sphere. The triangle in the figure, with vertices at the north pole and two points on the equator a quarter turn apart, is one eighth of the sphere: area \(4\pi R^2/8 = \pi R^2/2\), so \(\iint K\,dA = \pi/2\). Its three angles are right angles, and \(3 \cdot \tfrac{\pi}{2} - \pi = \tfrac{\pi}{2}\).

The global theorem. Now cover a closed surface \(S\) (compact, without boundary, like a sphere or a torus) by geodesic triangles: a triangulation with \(V\) vertices, \(E\) edges and \(F\) triangles. Add up the local theorem over all the triangles:

\[ \iint_S K\, dA = \sum_{\text{triangles}} (\alpha + \beta + \gamma - \pi) = 2\pi V - \pi F, \]

because the angles around each vertex add up to \(2\pi\). Each triangle has 3 edges and each edge borders 2 triangles, so \(3F = 2E\), i.e. \(\pi F = 2\pi E - 2\pi F\). Substituting,

\[ \iint_S K \, dA = 2\pi V - 2\pi E + 2\pi F = 2\pi (V - E + F) . \]

Theorem 19.6 (Gauss–Bonnet). For every compact surface \(S\) without boundary,

\[ \iint_S K \, dA = 2\pi\, \chi(S), \]

where \(\chi(S) = V - E + F\) is the Euler characteristic of any triangulation of \(S\).

(The sketch above assumed the surface can be cut into small geodesic triangles, which is true but takes work; the full proof, also covering surfaces with boundary and curved edges, is in do Carmo, section 4-5.)

This is remarkable twice over. The left side is pure geometry: bend the surface, and \(K\) changes from point to point. The right side is pure topology: it does not change under any deformation at all. So however you dent a sphere, its total curvature stays \(2\pi \cdot 2 = 4\pi\). And Euler's formula \(V - E + F = 2\) for polyhedra (Chapter 11) appears as a fact about curvature: a convex polyhedron is a "sphere" whose curvature is concentrated at the corners, where the angle defect (\(2\pi\) minus the sum of face angles at the vertex) plays the role of \(\iint K\,dA\). Descartes's theorem says the defects of a convex polyhedron add up to \(4\pi\). Check on a cube: each corner has three right angles, defect \(2\pi - 3\pi/2 = \pi/2\), and eight corners give \(4\pi\).

Worked example: the torus. With the formulas from 19.4, \(K\, dA = \dfrac{\cos v}{r(R + r\cos v)} \cdot r(R + r\cos v)\, du\, dv = \cos v\, du\, dv\), and

\[ \iint K\, dA = \int_0^{2\pi}\int_0^{2\pi} \cos v \, du\, dv = 0 = 2\pi \chi(\text{torus}) . \]

The positive curvature outside exactly cancels the negative curvature around the hole, as it must, since the torus has \(\chi = 0\). A surface with \(g\) holes has \(\chi = 2 - 2g\); so a two-holed torus has total curvature \(-4\pi\), and every metric on it, however it is bent, has some negative curvature.

Consequences.

No surface shaped like a torus can have \(K > 0\) everywhere, or \(K < 0\) everywhere: the integral must be 0.
A closed surface with \(K > 0\) everywhere has \(\chi > 0\), so it is a sphere (among orientable surfaces).
In Chapter 20 the uniformization theorem goes further: every closed surface carries a metric of constant curvature, and Gauss–Bonnet fixes its sign by \(\chi\).

A short history¶

Curvature of plane curves was studied by Huygens (in his 1673 work on pendulum clocks), Newton and Leibniz; Euler (in the 1760s) found the principal curvatures of surfaces, and Monge's school in Paris developed the geometry of surfaces in space. Frenet (thesis 1847) and Serret (1851) found the moving-frame equations independently. Gauss's Disquisitiones generales circa superficies curvas (1827) introduced the first fundamental form, Gaussian curvature, the Theorema Egregium and the geodesic triangle theorem; Pierre Ossian Bonnet extended it to regions with curved boundary (1848), and the link to the Euler characteristic of a closed surface came later in the nineteenth century. The global theorem in all dimensions (Allendoerfer and Weil 1943, Chern 1944) became a cornerstone of modern geometry. Gauss's student Riemann took the next step, to curved spaces of any dimension; that is Chapter 20.

Check yourself¶

1. Find the length of one turn of the helix \((\cos t, \sin t, t)\), and of the catenary \(y = \cosh x\) from \(x = 0\) to \(x = a\).

For the helix, \(|\gamma'| = |(-\sin t, \cos t, 1)| = \sqrt2\), so one turn (\(0 \le t \le 2\pi\)) has length \(2\pi\sqrt2 \approx 8.886\). For the catenary \((x, \cosh x)\), \(|\gamma'| = \sqrt{1 + \sinh^2 x} = \cosh x\), so the length is \(\int_0^a \cosh x\, dx = \sinh a\).

2. Find the curvature and the radius of curvature of \(y = x^2\) at \((1, 1)\), and of \(y = \sin x\) at \(x = \pi/2\).

For \((t, t^2)\): \(\kappa = 2/(1 + 4t^2)^{3/2}\), at \(t = 1\): \(2/5^{3/2} \approx 0.179\), radius \(5^{3/2}/2 \approx 5.59\). For \((t, \sin t)\): \(x' = 1\), \(y' = \cos t\), \(y'' = -\sin t\), so \(\kappa = -\sin t/(1 + \cos^2 t)^{3/2}\); at \(t = \pi/2\): \(\kappa = -1\) (turning right, at the top of the hump), radius 1.

3. Compute the curvature and torsion of the twisted cubic \(\gamma(t) = (t, t^2, t^3)\) at \(t = 0\).

\(\gamma'(0) = (1, 0, 0)\), \(\gamma''(0) = (0, 2, 0)\), \(\gamma'''(0) = (0, 0, 6)\). Then \(\gamma' \times \gamma'' = (0, 0, 2)\), so \(\kappa = |(0, 0, 2)|/1^3 = 2\), and \(\tau = (0, 0, 2) \cdot (0, 0, 6)/2^2 = 12/4 = 3\).

4. Show that a unit-speed curve with constant curvature \(\kappa > 0\) and zero torsion is a circle of radius \(1/\kappa\).

With \(\tau = 0\), \(B\) is constant and the curve is planar. Let \(c(s) = \gamma(s) + N(s)/\kappa\). Then \(c' = T + N'/\kappa = T + (-\kappa T)/\kappa = 0\), so \(c\) is a fixed point, and \(|\gamma - c| = |N|/\kappa = 1/\kappa\). So the curve lies on the circle of radius \(1/\kappa\) about \(c\) in its plane.

5. Find the first fundamental form of the cone \(X(r, \theta) = (r\cos\theta, r\sin\theta, r)\) and show it is locally isometric to the plane.

\(X_r = (\cos\theta, \sin\theta, 1)\), \(X_\theta = (-r\sin\theta, r\cos\theta, 0)\): \(E = 2\), \(F = 0\), \(G = r^2\), so \(ds^2 = 2\,dr^2 + r^2\, d\theta^2\). Put \(\rho = \sqrt2\, r\) and \(\phi = \theta/\sqrt2\): then \(ds^2 = d\rho^2 + \rho^2 d\phi^2\), the plane in polar coordinates. As \(\theta\) runs through \(2\pi\), \(\phi\) runs through \(2\pi/\sqrt2 \approx 254.6°\): the cone unrolls to a sector of the plane with that angle. So \(K = 0\) on the cone (away from the tip).

6. Find the Gaussian curvature at the origin of \(z = x^2 - 3y^2\), of \(z = 2x^2 + xy + y^2\), and describe each point.

At the origin \(f_x = f_y = 0\), so \(K = f_{xx} f_{yy} - f_{xy}^2\). First: \(2 \cdot (-6) - 0 = -12\), a saddle (hyperbolic) point. Second: \(f_{xx} = 4\), \(f_{yy} = 2\), \(f_{xy} = 1\), so \(K = 8 - 1 = 7 > 0\), an elliptic point (a bowl).

7. Use the first fundamental form to find the area of the band of a sphere of radius \(R\) between latitudes given by \(\varphi = \varphi_1\) and \(\varphi = \varphi_2\). What do you notice?

\(\int_0^{2\pi}\int_{\varphi_1}^{\varphi_2} R^2\sin\varphi\, d\varphi\, d\theta = 2\pi R^2(\cos\varphi_1 - \cos\varphi_2)\). The heights of the two circles are \(R\cos\varphi_1\) and \(R\cos\varphi_2\), so the area is \(2\pi R \times\) (height of the band): equal to the area of the matching band on the circumscribed cylinder, as Archimedes found.

8. On a cylinder of radius 2, find the shortest path from the point at angle 0 and height 0 to the point at angle \(\pi/2\) and height 3. Then find a second geodesic between them.

Unrolled, the horizontal distance is \(2 \cdot \pi/2 = \pi\) one way, and the height is 3: shortest length \(\sqrt{\pi^2 + 9} \approx 4.344\). Going the other way round, the horizontal distance is \(2 \cdot 3\pi/2 = 3\pi\), giving a longer helix of length \(\sqrt{9\pi^2 + 9} \approx 9.891\). (Adding full turns gives infinitely many more.)

9. A geodesic triangle on a sphere of radius 2 has angles \(90°\), \(60°\) and \(60°\). Find its area.

The excess is \(90° + 60° + 60° - 180° = 30° = \pi/6\). By Gauss–Bonnet with \(K = 1/4\): area \(/4 = \pi/6\), so the area is \(2\pi/3 \approx 2.094\).

10. Check Descartes's theorem (the angle defects add up to \(4\pi\)) on a prism whose faces are two equilateral triangles and three squares.

At each of the 6 vertices: two right angles and one \(60°\) angle, total \(240°\), defect \(360° - 240° = 120° = 2\pi/3\). Total \(6 \times 2\pi/3 = 4\pi\). This is \(2\pi\chi\) with \(\chi = V - E + F = 6 - 9 + 5 = 2\).

11. Show that a compact surface in space with \(K < 0\) everywhere is impossible. (Hint: the point farthest from the origin.)

Let \(p\) be a point of the surface farthest from the origin, at distance \(d\). The surface lies inside the sphere of radius \(d\) and touches it at \(p\), so near \(p\) it bends at least as much as the sphere in every direction: every normal curvature at \(p\) is at least \(1/d\) in size, with the same sign. So \(K(p) \ge 1/d^2 > 0\). (So the saddle-shaped surfaces of hyperbolic geometry cannot be closed surfaces in 3-space; Chapter 17 met Hilbert's stronger theorem that the whole hyperbolic plane cannot sit in 3-space.)

12. A surface shaped like a sphere has been dented so that its curvature is negative on a region of total curvature \(-\pi\). What is the total curvature of the rest of the surface?

By Gauss–Bonnet the total is \(2\pi\chi = 4\pi\) whatever the shape. So the rest has total curvature \(4\pi - (-\pi) = 5\pi\): the dent forces more positive curvature elsewhere (around its rim).

What you learned¶

A regular curve has length \(\int |\gamma'|\,dt\), and can be parametrised by arc length.
Curvature measures turning, torsion measures twisting out of a plane; the Frenet–Serret equations show they determine a space curve up to rigid motion.
A surface's first fundamental form \(E, F, G\) gives lengths, angles and areas measured from inside the surface.
The second fundamental form gives normal and principal curvatures; Gaussian curvature \(K = k_1k_2\) classifies points as dome-like, flat or saddle-like.
The Theorema Egregium: \(K\) depends only on \(E, F, G\). So no flat map of a sphere is perfect, and folded pizza does not droop.
Geodesics are curves that do not turn within the surface; locally they are shortest paths.
Gauss–Bonnet: the total curvature of a closed surface is \(2\pi\) times its Euler characteristic.

Next: Chapter 20, where Riemann takes intrinsic geometry to any number of dimensions, and Einstein uses it to describe gravity.