Intrinsic Geometry

Chapter 3

Surfaces: intrinsic geometry

It is the geometry of objects associated to the surface which depend only on the first fundamental form. Obvious examples are lengths of curves, angles between tangent vectors and areas of regions in the the surface. Less obvious examples are geodesics (yet to be defined) and the Gaussian curvature (Theo- rema Egregium). We will first discuss local questions.

3.1 Directional derivative

We start by recalling the concept of directional derivative of vector fields on Rn. Let Y : W Rn be a smooth vector field defined on an open subset W of n n → R , let v R be a fixed vector and p W . Then the directional derivative of Y ∈ ∈ along v at p is Y (p + tv) Y (p) DvY p = dYp(v) = lim − . t 0 | → t n Note that if γ : ( ǫ,ǫ) R is a smooth curve with γ(0) = p and γ′(0) = v, − → then Y (γ(t)) Y (γ(0)) d lim − = Y (γ(t)) t 0 t dt t=0 → = dY γ(0)(γ′(0)) (by the chain rule)

= dYp(v)

= DvY p, | in other words, DvY p depends only on the values of Y along a smooth curve | through p with velocity v. n As a particular case, consider the canonical basis e1,...,en of R and n { } denote by (x1,...,xn) the standard coordinates in R . Then

∂Y De Y p = i (p) i | ∂x 31 32 CHAPTER 3. SURFACES: INTRINSIC GEOMETRY

i and, if v = i v ei, then i ∂Y DvY = v . P ∂xi i X A word about notation: If X : W Rn is another smooth vector ﬁeld, then n → X(p) is a vector in R and we write

DX Y p = D Y p. | X(p) | 3.2 Vector ﬁelds on surfaces

Let S be a surface in R3, and let V S be an open subset. A (smooth) vector ⊂ 3 ﬁeld on V is a (smooth) map X : V R . We say that X is tangent to S (resp. → normal to S) if X(p) TpS (resp. X(p) TpS) for p V . 2 ∈ ⊥ ∈ Let ϕ : U R ϕ(U)= V S be a parametrization. In the following, we ⊂ → ⊂ 2 use (u1,u2) to denote coordinates on the parameter plane R and (x1, x2, x3) to denote coordinates in the ambient R3. According to the above deﬁnition,

∂ϕ 1 ∂ϕ 1 3 1 ϕ− , 2 ϕ− : V R ∂u ◦ ∂u ◦ → are vector fields on V tangent to S. For the sake of convenience, henceforth ∂ϕ ∂ϕ we will abuse terminology and say that ∂u1 , ∂u2 are vector fields tangent to S. Note then that any vector field X on V tangent to S can be written as a linear combination 1 ∂ϕ 2 ∂ϕ X = a ∂u1 + a ∂u2 where a1, a2 : V R. It is an easy exercise to check that X is smooth if and → only if a1, a2 are smooth functions (do it!). Similarly, any smooth vector field on V normal to S is of the form ∂ϕ ∂ϕ X = b 1 2 ∂u × ∂u for a smooth function b : V R. → 3.3 Covariant derivative

In this section we explain how to differentiate a vector field tangent to a surface along another tangent vector field to obtain a third tangent vector field. Let S be a surface in R3, V S an open subset, p V . Consider vector fields X, Y on ⊂ ∈ V such that X is tangent to S. Our previous discussion about the directional d derivative shows that DX Y p is well defined, namely, it equals t Y (γ(t)) | dt | =0 where γ : ( ǫ,ǫ) S is smooth and γ(0) = p, γ′(0) = v. The association − → p DX Y p 7→ | is a vector field on V . As a special case, if ϕ : U V is a parametrization of S → and ϕ(u)= p for some u U, ∈ ∂(Y ϕ) D ∂ϕ Y p = ∂u◦i (u). ∂ui | 3.3. COVARIANT DERIVATIVE 33

∂ϕ Further, if Y = ∂uj , ∂ϕ ∂2ϕ D ∂ϕ i p = i j (u). ∂ui ∂u | ∂u u Back to the general case, we now assume that both X and Y are tangent to S. The covariant derivative of Y along X at p is

X Y p = (DX Y p)⊤ ∇ | | where ( )⊤ denotes the orthogonal projection onto TpS. The symbol “ ” is · ∇ read “nabla”. In this way, the association

p X Y p 7→ ∇ | is a vector ﬁeld on V tangent to S.

Lemma 3.1 (Properties of D and ) Let X, X˜, Y , Y˜ be vector ﬁelds tangent to S ∇ and let f be a smooth function on S. Then:

1. ˜ Y = f X Y + ˜ Y ; ∇fX+X ∇ ∇X

2. X (Y + Y˜ )= X Y + X Y˜ ; ∇ ∇ ∇ 3. X (fY )= df(X)Y + f X Y ; ∇ ∇ 4. X Y, Y˜ = X Y, Y˜ + Y, X Y˜ ; h i h∇ i h ∇ i and the same identities hold for replaced by D. ∇ In this statement, fX denotes the tangent vector ﬁeld p f(p)X(p), and 7→ Y, Y˜ denotes the scalar function p Y (p), Y˜ (p) so that X Y, Y˜ (p) denotes h i 7→ h i h i the the directional derivative in the direction of X(p). Proof. We prove (3)and (4)and leave the restasan exercise. Let γ : ( ǫ,ǫ) − → S be a smooth curve with γ(0) = p, γ′(0) = X(p). Then d DX (fY ) p = t (fY )(γ(t)) | dt| =0 d = t f(γ(t))Y (γ(t)) dt| =0 d d = t f(γ(t)) Y (γ(0)) + f(γ(0)) t Y (γ(t)) dt| =0 · dt| =0 = dfp(X(p))Y (p)+ f(p)DX Y p. | Hence

X (fY ) p = (DX (fY ))⊤ ∇ | = (df(X)Y + fDX Y )⊤

= df(X)Y ⊤ + f ( X Y )⊤ (orthogonal projection is linear) ∇ = df(X)Y + f X Y (Y is tangent). ∇ 34 CHAPTER 3. SURFACES: INTRINSIC GEOMETRY

Next,

Let X, Y be tangent vector ﬁelds. Of course, ν is a normal vector ﬁeld, so Y, ν =0 and Lemma 3.1(4) says that h i

DX Y, ν + Y,DX ν =0. h i h i

Le A = dν be the Weingarten operator. Then A(X)= dν(X)= DX ν and − − − we have

X Y = (DX Y )⊤ ∇ = DX Y DX Y, ν ν − h i normal component

= DX Y + Y,D|X ν {zν } h i = DX Y Y,AX ν − h i = DX Y Π(X, Y )ν. − Hence we arrive at the Gauss formula

DX Y = X Y + Π(X, Y )ν. ∇ In the remaining of this section, we show that the covariant derivative ∇ of a surface S is an intrinsic object, namely, it is completely determined by the first fundamental form I; in particular, locally isometric surfaces have the same covariant derivative. This is not an obvious assertion in view of the fact that X Y was defined as the orthogonal projection onto the surface of the direc- ∇ 3 tional derivative DX Y , and so the ambient space R was used in this defini- tion. Let X, Y be vector fields on S. Since we are dealing with a local assertion, we can work in the image of a parametrization ϕ of S and write

2 2 i ∂ϕ j ∂ϕ X = a ∂ui , Y = b ∂uj . i=1 i=1 X X 3.3. COVARIANT DERIVATIVE 35

By using Lemma 3.1, we ﬁrst obtain a local expression for X Y : ∇

i j ∂ϕ X Y = a ∂ϕ b j ∇ ∇ ∂ui ∂u i,j X j i ∂b ∂ϕ j ∂ϕ = a i j + b ∂ϕ j . (3.2) ∂u ∂u ∇ ∂ui ∂u i,j X ∂ϕ Since ∂ϕ ∂uj is tangent to S, we can write ∇ ∂ui

∂ϕ k ∂ϕ ∂ϕ j = Γij k , (3.3) ∇ ∂ui ∂u ∂u k X k for some smooth functions Γij , the so called Christofell symbols (of the second kind). Substituting into (3.2),

k i ∂b j k ∂ϕ X Y = a + b Γ , (3.4) ∇ ∂ui ij  ∂uk i j X X   which shows that depends only on Γk . ∇ { ij } We can also deﬁne the Christofell symbols of the ﬁrst kind by putting

∂ϕ ∂ϕ Γij,k = I ∂ϕ j , k . ∇ ∂ui ∂u ∂u By using (3.3), we have

∂ϕ ∂ϕ Γ = Γℓ I , ij,k ij ∂uℓ ∂uk ℓ X ℓ = Γij gℓk. ℓ X Multiplying through by gkm, where (gkm) denotes the inverse matrix of I = (gij ), we get m km Γij = Γij,kg . k X k Hence Γ depends only on Γij,k and I. { ij } { } In order to complete our argument, we need to show that Γij,k depends k { } only on I. It is important to notice that Γij (and Γij,k) is symmetric with respect to the indices i, j, namely, k k Γij =Γji. This is immediate from

∂ϕ ∂2ϕ ⊤ ∂ϕ j = i j . ∇ ∂ui ∂u ∂u ∂u 36 CHAPTER 3. SURFACES: INTRINSIC GEOMETRY

Next, by using Lemma 3.1(4), we write

∂gij ∂ϕ ∂ϕ ∂ϕ ∂ϕ k = ∂ϕ i , j + i ∂ϕ j , ∂u ∇ ∂uk ∂u ∂u ∂u ∇ ∂uk ∂u so ∂g ij =Γ +Γ . ∂uk ki,j kj,i Doing cyclic permutations on (i, j, k), we also obtain ∂g jk =Γ +Γ ∂ui ij,k ik,j and ∂g ki =Γ +Γ . ∂uj jk,i ji,k Summing the last two equations and subtracting the ﬁrst one ﬁnally yields that

∂gjk ∂gij ∂gki 2Γij,k = + , (3.5) ∂ui − ∂uk ∂uj which completes the proof that depends only on I = (gij ). ∇ Example 3.6 Consider a surface of revolution parametrized as in section 2.5. Since F =0, we can write

ϕu ϕv ϕ ϕu = (Dϕ ϕu)⊤ = (ϕuu)⊤ = ϕuu, ϕu + ϕuu, ϕv . ∇ u u h i E h i G Using the formulas from section 2.5,

1 2 2 ϕuu, ϕu = f ′f ′′ + g′g′′ = (f ′ + g′ )′ =0 h i 2 and ϕuu, ϕv =0. h i Hence ϕ ϕu =0. ∇ u Similarly, one computes that

f ′ ϕ ϕv = ϕ ϕu = ϕv ∇ u ∇ v f and ϕ ϕv = ff ′ϕu. ∇ v − We thus obtain

1 2 1 2 f ′ 1 2 Γ =Γ =0, Γ =0, Γ = , Γ = ff ′, Γ =0. 11 11 12 12 f 22 − 22 In particular, all Christofell symbols vanish along the parallels u = constant corresponding to critical points of f. 3.4. THE LIE BRACKET 37

3.4 The Lie bracket

n If X, Y are vector ﬁelds on R or on a surface, then, in general, DX Y = DY X. 6 For instance, take X = x2 e , Y = e . Then · 1 2 2 2 DX Y = Dx e1 e2 = x De1 e2 =0,

where De1 e2 =0 because e2 is constant, and

2 2 2 DY X = De (x e )= dx (e )e + x De e = e , 2 · 1 2 1 2 1 1 dx (e ) = ∂x2 = 1 where 2 2 ∂x2 . In general, the lack of comutativity is measured by the Lie bracket [X, Y ] := DX Y DY X. − If X, Y are tangent to S, then

X Y Y X = (DX Y II(X, Y )ν) (DY X II(Y,X)ν) ∇ − ∇ − − − = DX Y DY X − = [X, Y ].

As another example, for a parametrization ϕ of S we have

2 2 ∂ϕ ∂ϕ ∂ϕ ∂ϕ ∂ ϕ ∂ ϕ [ ∂ui , ∂uj ]= D ∂ϕ j D ∂ϕ i = i j j i =0. ∂ui ∂u − ∂uj ∂u ∂u ∂u − ∂u ∂u

This example shows that given a tangent frame X1,X2 to a surface, a neces- ∂ϕ ∂ϕ { } sary condition for it to be of the form 1 , 2 for some parametrization ϕ is { ∂u ∂u } that [X1,X2]=0. Let X, Y be vector ﬁelds on S. We obtain a local expression for their Lie bracket. In the image of a parametrization, we can write

2 2 i ∂ϕ j ∂ϕ X = a ∂ui , Y = b ∂uj . i=1 i=1 X X k Using (3.4) and the symmetry of Γij with respect to i, j, we have, on the image of ϕ:

[X, Y ] = X Y Y X ∇ − ∇ ∂bj ∂aj ∂ϕ = ai bi . ∂ui − ∂ui ∂uj i,j X 3.5 Parallel transport

Let S R3 be a surface and let γ : I S be a smooth parametrized curve ⊂ → (nonnecessarily regular). A (smooth) tangent vector ﬁeld on S along γ is a (smooth) map X : I R3 such that X(t) T S for all t I. → ∈ γ(t) ∈ 38 CHAPTER 3. SURFACES: INTRINSIC GEOMETRY

Examples 3.7 1.If Y is a tangent vector field on S and γ : I S is a parametrized → curve, then X(t)= Y (γ(t)) defines a tangent vector field on S along γ. 2. If γ : I S is a smooth parametrized curve, then X = γ′ defines a → tangent vector field on S along γ.

The covariant derivative of a tangent vector ﬁeld X on S along γ is deﬁned to be X d ⊤ ∇ = X(t) . dt t dt

In example (1) above,

X d ⊤ ∇ = Y (γ(t)) dt t dt = D ′ Y . γ (t) |γ(t) In example (2) above, X ∇ = (γ′′(t))⊤ . dt t X A tangent vector ﬁeld X on S along γ is said to be parallel if ∇ 0.A dt ≡ smooth parametrized curve γ : I S is said to be a geodesic if γ′ is parallel → along γ. X In the sequel, we write a local expression of the equation ∇ 0. In the dt ≡ image of a parametrization ϕ of S, we can write

2 ∂ϕ γ(t)= ϕ(u1(t),u2(t)), X(t)= ai(t) ∂ui (u1(t),u2(t)) i=1 X where t I. Then ∈ 2 2 2 i ∂ϕ i ∂ ϕ j X′(t)= (a )′ + a (u )′ ∂ui ∂ui∂uj i=1 i,j=1 X X and X ∇ = (X′(t))⊤ dt 2 2 i ∂ϕ i j ∂ϕ = (a )′ i + a (u )′ ∂ϕ j ∂u ∇ ∂ui ∂u i k i,j=1 X= X 2 2 k ∂ϕ i j k ∂ϕ = (a )′ + a (u )′Γ ∂uk ij ∂uk i k i,j,k X= X=1 2 2 k k i j ∂ϕ = (a )′ + Γ (u )′a .  ij  ∂uk k i,j=1 X=1 X   3.5. PARALLEL TRANSPORT 39

X Hence ∇ 0 is the following system of ﬁrst order linear ordinary differential dt ≡ equations in a1(t), a2(t):

2 k k 1 2 i j (a )′(t)+ Γij (u (t),u (t)) (u )′(t) a (t)=0 (k =1, 2). (3.8) i,j=1 X The theorem on existence and uniqueness of solution of linear ordinary differ- 1 1 2 2 ential equations says that, given the initial values a (t0) = a0, a (t0) = a0, for some t I, there exists a unique solution (a1(t),a2(t)) defined on I and satis- 0 ∈ fying the given initial values. Geometrically this means that, given v Tγ(t0)S 1 ∂ϕ ∈ 2 ∂ϕ and taking a parametrization ϕ : U S around γ(t0) with v = a0 ∂u1 + a0 ∂u2 , 1 ∂ϕ → 2 ∂ϕ the vector field X(t) = a (t) ∂u1 + a (t) ∂u2 is the only parallel vector field along γ such that X(t0)= v, defined on an interval centered at t0 and lying in 1 γ− (ϕ(U)). By covering the image of γ by finitely many opens sets V1,...,Vn, each of which the image of a parametrization, such that Vi Vi = ∅, and ∩ +1 6 applying this result to each one of V1,...,Vn in order, we deduce

Proposition 3.9 Given a smooth parametrized curve γ : [a,b] S and a tangent → 3 vector v T S, there exists a unique tangent vector ﬁeld X : [a,b] R on S ∈ γ(a) → along γ which is parallel and satisﬁes X(a)= v.

The vector X(b) T S is called the parallel transport of v = X(a) along ∈ γ(b) γ. The parallel transport along γ deﬁnes a map P γ : T S T S which γ(a) → γ(b) is obviously linear, since the solutions to a linear ODE depend linearly on the initial values.

Proposition 3.10 If X, Y are parallel vector ﬁelds along γ, then X(t), Y (t) , X(t) h i || || and the angle between X(t) and Y (t) are constant functions.

Proof. We compute

d X(t), Y (t) = X′(t), Y (t) + X(t), Y ′(t) dth i h i h i X Y = ∇ , Y (t) + X(t), ∇ (since X, Y are tangent) h dt i h dt i = 0.

Hence X(t), Y (t) is constant, and the other assertions follow. h i Corollary 3.11 P γ : T S T S is a linear isometry. γ(a) → γ(b) Examples 3.12 1. For the plane, ϕ(u, v) = (u,v, 0) is a parametrization and I = du2 + dv2. Since the coefficients of I are constant, eqn. (3.5) yields that k k Γij = 0 for all i, j, k. The equations of parallel transport are thus (a )′ = 0, k =1, 2. Hence the parallel vector fields along γ are the constant vector fields. In particular, the parallel transport along γ depends on the endpoints of γ, but not on the curve itself. 40 CHAPTER 3. SURFACES: INTRINSIC GEOMETRY

2. We consider the cone C of equation z = k x2 + y2 for k > 0 and (x, y) = 0. It is obviously a graph of a smooth function, so it is a surface. 6 p We can also parametrize it as a surface of revolution by taking the generating γ(s) = (f(s), 0,g(s)) f(s) = 1 s g(s) = k s curve to be , where √k2+1 , √k2+1 . Then the Gaussian curvature K = f ′′/f =0. − In the sequel, we show that the cone is locally isometric to the plane. Note that the angle at the vertex of the cone is ψ = arccot k (0,π/2). Consider the ∈ open sector of the plane V given in polar coordinates by r> 0, 0 <θ< 2π sin ψ. We deﬁne a map θ θ Φ : V C, F (r, θ)= r cos sin ψ, r sin sin ψ, r cos ψ . → sin ψ sin ψ Then Φ is smooth and its image Φ(V ) is the cone minus the geratrix y = 0, x> 0, z = kx. The inverse map

1 1 2 2 x Φ− : Φ(V ) V, Φ(x,y,z)= x + y , sin ψ arccot → sin ψ y p is also smooth, since it is smooth as a function of (x, y) for y = 0. Hence 6 Φ : V Φ(V ) is a diffeomorphism. We finally show that Φ is an isome- → try, namely, the first fundamental forms of the plane and the cone coincide on points corresponding under Φ. The open set V is a regular surface parametrized by ϕ(u, v) = (u cos v,u sin v, 0), where (u, v) U = (0, + ) (0, 2π sin ψ), and the corresponding coefficients ∈ ∞ × of the first fundamental form are then E = 1, F = 0, G = u2. Since Φ is a diffeomorphism, ϕ˜ =Φ ϕ can be taken as a parametrization of Φ(V ) and then ◦ the corresponding coeffcients of the first fundamental form are E˜ = 1, F˜ = 0, G˜ = u2. Since E = E˜, F = F˜, G = G˜, Φ is an isometry. From the local expression (3.8), we see that parallel transport is an intrinsic object. Hence the parallel transport along a curve in the cone can be read off the parallel transport along the corresponding curve in the plane. Consider a parallel curve γ in the cone given by z = z0; we compute the parallel transport of its initial tangent vector v after one turn around the cone. The corresponding curve γ˜ in the plane is an arc of a circle of angle 2π sin ψ. The tangent vector to γ˜ rotates by an angle of measure 2π sin ψ after one turn, whereas the parallel transport is the identity. It follows that the parallel transport of v along γ is rotation by 2π sin ψ. − 3 3. Consider the unit sphere x2 + y2 + z2 = 1 in R . We describe parallel transport around a small circle γ of colatitude ϕ. There exists a cone which is tangent to the sphere along γ; the angle at the vertex of this cone is ψ = π ϕ. 2 − Since the tangent spaces of the sphere and the cone coincide along γ, also parallel transport along γ is the same whether we view it as a curve in the sphere or in the cone. Therefore parallel transport along the small circle of colatitude ϕ is rotation of angle 2π cos ϕ (with respect to a suitable orientation). Taking − ϕ 0, by continuity we see that parallel transport along the equator (after one → complete turn) is the identity. 3.6. GEODESICS 41

3.6 Geodesics

As we have already mentioned, a smooth parametrized curve γ : I S is a ′ (γ ) → geodesic if γ′ is parallel along γ. This means that 0 = ∇dt = (γ′′)⊤, so the 3 acceleration γ′′ in R is everywhere normal to S. In other words, γ does not accelerate viewed from S, so that “geodsics are the straightest curves in S”. If γ is geodesic, then γ′ is constant by Proposition 3.10. There are two || || cases: either γ′ = 0 and γ is a constant curve; or γ′ is a nonzero constant || || || || and γ is regular and parametrized proportionally to arc-length. Equations for geodesics contained in the image of a parametrization ϕ are immediately deduced from the equations for parallel vector ﬁelds (3.8); if γ(t)= 1 2 i i ϕ(u (t),u (t)), we take (a )′ = (u )′ and then

2 k k i j (u )′′ + Γij (u )′(u )′ =0, (k =1, 2). (3.13) i,j=1 X This a system of second order non-linear ordinary differential equations. Those equations show that geodesics are intrinsic objects. On the other hand, the nonlinearity implies that in general geodesics are only deﬁned locally. Namely, the theorem of existence and uniqueness for such equations is local in nature, so we have

Proposition 3.14 Given p S and v TpS, there exist ǫ> 0 and a unique geodesic ∈ ∈ γ : ( ǫ,ǫ) S such that γ(0) = p and γ′(0) = v. − →

k 1 Examples 3.15 1. For the plane, Γij 0, so the geodesic equations are (u )′′ = 2 ≡ (u )′′ =0. Hence the geodesics are the straight lines. 2 2 2 2. In the sphere S , let p S and v TpS = (Rv)⊤, v = 0, and consider ∈ ∈ 6 the great circle v γ(t) = cos(t v ) + sin(t v ) . || || || || v || || 2 2 Then γ(0) = p, γ′(0) = v and γ′′(t) = v γ(t) T S , so great circles are −|| || ⊥ γ(t) geodesics. Since there exist great circles thorugh any point with any speed, by the uniqueness part of Proposition 3.14, the great circles are all the geodesics. 3. The cilinder x2 + y2 =1 is locally isometric to the plane. In fact, ϕ(u, v)= (cos v, sin v,u) is a local isometry, since by restricting to (u, v) (u ,u +2π) R ∈ 0 0 × it becomes a parametrization with E = 1, F = 0, G = 1. It follows that the geodesics of the cilinder are images of the geodesics of the plane under ϕ. In particular, the geodesics through ϕ(0, 0) are of the form

t (cos(at), sin(at),bt), 7→ where a, b R. Note that we get horizontal circles (a =0, b =0), vertical lines ∈ 6 (a =0, b =0) and helices (otherwise). 6 42 CHAPTER 3. SURFACES: INTRINSIC GEOMETRY

4. Consider a surface of revolution parametrized as in section 2.5. For a ∂ϕ curve v-constant, η(t) = ϕ(t, v0), we have η′ = ∂u , so by the computations in example 3.6, η′ ∇ = η′ η′ = ϕ ϕu =0. dt ∇ ∇ u Hence meridians are always geodesics. Similarly, for the curves u-constant, u = u , we have ϕ ϕv u u = f(u )f ′(u )ϕu and f > 0, so precisely the 0 ∇ v | = 0 − 0 0 parallels corresponding to critical points of f are geodesics. More generally, using the Christofell symbols computed in the quoted example, we can write the geodesic equations as

2 u′′ f(u)f ′(u)(v′) = 0 − f ′(u) v′′ +2 u′v′ = 0 f(u)

Note that

2 2 f (u)v′ ′ = 2f(u)f ′(u)u′v′ + f (u)v′′

2 f ′(u) = f (u) v′′ +2 u′v′ f(u) = 0

2 along a solution curve, so f (u)v′ is constant along a geodesic (this function is a ﬁrst integral of the system). As an application, we consider a geodesic γ(t) = (u(t), v(t)) and, for each t, the parallel ζ(r) = ϕ(u(t), r) which crosses γ(t) at r = v(t), and compute the inner product

∂ϕ ∂ϕ ∂ϕ 2 γ′, ζ′ = u′ + v′ , = u′F + v′G = f (u)v′ h i h ∂u ∂v ∂v i to be constant with respect to t. On the other hand,

γ′, ζ′ = γ′ ζ′ cos θ = γ′ f(u)cos θ, h i || |||| || || || ∂ϕ where γ′ is constant and θ(t) is the angle between γ′(t) and ζ′(v(t)) = (u(t), v(t)), || || ∂v and f(u(t)) is the radius of the parallel ζ. Hence the cosine of the angle at which γ meets a parallel multiplied by the radius of that parallel is constant; this is known as Clairaut’s relation. In fact, the ﬁrst integral allows to explicitly integrate the geodesic equations. For the sake of simplicity, assume the geodesic γ is parametrized by arc-length; together with Clairaut’s relation, this gives the system

2 2 2 (u′) + f (v′) = 1 2 f v′ = c where c is a constant; by changing the orientation of γ if necessary we may assume that c> 0. If γ is not a meridian, v′ is never zero and we can use v as a 3.7. THE INTEGRABILITY EQUATIONS AND THE THEOREMA EGREGIUM43 parameter along γ, so that u = u(v). We have

du 2 (u )2 = ′ dv (v )2 ′ 2 2 1 f (v′) = − 2 (v′) 1 2 = 2 f (v′ ) − f 4 = f 2 c2 − f 2 = (f 2 c2) c2 − It follows that f c ≥ and, if γ is not a parallel, du =0 and f >c so that dv 6 1 v = c du. f(u) f(u)2 c2 Z − p 3.7 The integrability equations and the Theorema Egregium

Our next goal is to investigate how true is the fact that the ﬁrst and second fundamental forms locally determine a surface. We will start by looking at necessary conditions for I, II to correspond to a surface. Suppose ϕ : U S is a regular parametrized surface deﬁned on an open 2 → set U R , and let I = (gij ), II = (hij ) be its fundamental forms. Recall the ⊂ Gauss formula DX Y = X Y + II(X, Y )ν (3.16) ∇ and the Weingarten equation

DX ν = A(X), (3.17) − where I(AX, Y )= II(X, Y ). Writing the Gauss formula in terms of the parametrization, we get

∂2ϕ ∂ϕ ϕ i j = D ∂ j ∂u ∂u ∂ui ∂u ∂ϕ ∂ϕ ∂ϕ ϕ = ∂ j + II( i j ) ν ∇ ∂ui ∂u ∂u ∂u ∂ϕ = Γk + h ν. (3.18) ij ∂uk ij k X 44 CHAPTER 3. SURFACES: INTRINSIC GEOMETRY

Doing the same with the Weingarten equation,

∂ν i = D ∂ϕ ν ∂u ∂ui ∂ϕ = A − ∂ui ∂ϕ = hj − i ∂uj j X kj ∂ϕ = hikg . (3.19) − ∂uj j,k X We would like to integrate the equations (3.18) and (3.19) in terms of ϕ. There are obvious neccesaary conditions for that, namely

∂3ϕ ∂3ϕ 0 = ∂uk∂ui∂uj − ∂uj∂ui∂uk ∂ ∂ϕ ∂ ∂ϕ = Γs + h ν Γr + h ν ∂uk ij ∂us ij ∂uj ik ∂ur ik s ! − r ! X X ∂Γs ∂ϕ ∂2ϕ ∂h ∂ν = ij +Γs + ij ν + h ∂uk ∂us ij ∂uk∂us ∂uk ij ∂uk s X ∂Γr ∂ϕ ∂2ϕ ∂h ∂ν ik +Γr ik ν + h ∂uj ∂ur ik ∂uj∂ur ∂uj ik ∂uj − r − X ∂Γs ∂Γs ∂ϕ = ij ik ∂uk ∂uj ∂us s − X ∂ϕ ∂ϕ + Γr Γs + h ν Γr Γs + h ν ij kr ∂us kr ik jr ∂us jr r s ! − r s ! X X X X ∂h ∂h ∂ϕ ∂ϕ + ij ik ν h h gms + h h gms . ∂uk ∂uj ij km ∂us ik jm ∂us − − m,s ms X X Considering separately the tangential and normal components, we respectively get the Gauss equation

∂Γs ∂Γs ij ik + Γr Γs Γr Γs ∂uk ∂uj ij rk ik rj − r − X ms = (hij hkm hikhjm)g for all i, j, k, s (3.20) m − X and the Codazzi-Mainard equation

∂hij ∂hik + Γr h Γr h =0 for all i, j, k. (3.21) ∂uk ∂uj ij rk ik rj − r − X 3.8. A VERY QUICK DIGRESSION ON SYSTEMSOF FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS45

Taken together, they are known as the integrability (or compatibility) equations. As a by-product of our computation, we derive the following consequence of the integrability equations. Put i = j = 1, k = 2 in the Gauss eqn. (3.20), multiply through by gs2, and sum over s to get

∂Γ2 ∂Γs 11 12 g + (Γr Γs Γr Γs ) g ∂u2 ∂u1 s2 11 r2 12 r1 s2 s − r,s − X X ms = (h11h2m h12h1m)g gs2 m,s − X = h h (h )2 11 22 − 12 = det(II).

k The left-hand side of this equation involves only the gij ’s and the Γij ’s, and we already know that the Christofell symbols are completely determined by the gij ’s. Hence det(II) depends only on I. Recalling that K = det(II)/ det(I), we ﬁnally get the Theorema Egregium.

Theorem 3.22 (Gauss, 1826) The Gaussian curvature of a surface is an intrinsic invariant of the surface.

In other words, locally isometric surfaces have the same Gaussian curvature at corresponding points.

Remark 3.23 The expression

∂Γ2 ∂Γs 11 12 + (Γr Γs Γr Γs ) =: Rs ∂u2 ∂u1 11 r2 12 r1 121 − r − X is a component of the so called Riemann curvature tensor, and

s R121gs2 = R1212, s X so the Theorema Egregium can be restated in the form

R K = 1212 . g g (g )2 11 22 − 12 3.8 A very quick digression on systems of ﬁrst order partial differential equations

In constrast to the case of ordinary differential equations, systems of PDE’s do not always have solutions. We start with a simple example. Consider a vector ﬁeld X~ = P~i + Q~j deﬁned on an open set U R2, where P , Q : U R2 are ⊂ → 46 CHAPTER 3. SURFACES: INTRINSIC GEOMETRY smooth functions. Finding a potencial for X~ , i.e. a smooth scalar function f on U such that grad f = X~ , is equivalent to solving the system ∂f = P (x, y) ∂x ∂f = Q(x, y). ∂y

∂2f ∂2f Since ∂y∂x = ∂x∂y , a necessary condition for the existence of solutions is that ∂P ∂Q = , ∂y ∂x namely, the rotational rot X~ = ~0. It is also known that if U is a rectangle, or star-shaped, or even simply-connected, then this condition is also sufﬁcient. A more general system has the form ∂z = P (x,y,z) ∂x ∂z = Q(x,y,z). ∂y

In this case, the necessary condition is easily seen to be ∂P ∂P ∂Q ∂Q + Q = + P. ∂y ∂z ∂x ∂z The general case is dealt with the following theorem, for whose proof the reader is referred to app. B in J. J. Stoker, Differential Geometry, Wiley Inter- science, 1969.

Theorem 3.24 (Frobenius, 1877) Consider the ﬁrst-order system of partial differential equations ∂yi = P i(x1,...,xm; y1,...,yn), ∂xj j i where 1 i n and 1 j m and Pj admit continuous derivatives of second order ≤ ≤ ≤ ≤ j in all arguments. Suppose that the Pi satisfy the integrability conditions

∂P i ∂P i ∂P i ∂P i j + j P k = k + k P j . ∂xk ∂yℓ ℓ ∂xj ∂yℓ ℓ ℓ ℓ X X Then there exist a unique solution satisfying the initial conditions

i 1 m i y (x0,...,x0 )= y0 for 1 i n. ≤ ≤ 3.9. THE FUNDAMENTALTHEOREM OF THE LOCAL THEORY OF SURFACES47

3.9 The fundamental theorem of the local theory of surfaces

The ﬁrst result asserts the invariance of the fundamental forms of a surface under orientation-preserving rigid motions of Euclidean space, and contains the uniqueness part of Theorem 3.28.

Lemma 3.25 Let ϕ : U S be a regular parametrized surface, and let T : R3 R3 → → be an orientation-preserving rigid motion, i.e. T (x) = A(x)+ b where A SO(3) 3 ∈ and b R . Put ϕ˜ = T ϕ. Then ϕ˜ : U S˜ is a regular parametrized surface ∈ ◦ → and the coefﬁcients of the fundamental forms g˜ij = gij , h˜ij = hij . Conversely, given two regular parametrized surfaces ϕ : U S and ϕ˜ : U S˜, where U is connected, → → satisfying gij =g ˜ij , hij = h˜ij , there exists an orientation-preserving rigid motion T of R3 such that ϕ˜ = T ϕ. ◦ Proof. Differentiate ϕ˜ = Aϕ + b; owing to the constancy of A, b, we get ∂ϕ˜ ∂ϕ ∂ui = A ∂ui . Using this and the fact that A is orientation-preserving, ∂ϕ˜ ∂ϕ˜ ∂ϕ˜ ∂ϕ˜ A 1 2 1 2 ∂u ∂u ν˜ = ∂u × ∂u = × = A(ν). ∂ϕ˜ ∂ϕ˜ ∂ϕ˜ ∂ϕ˜ ∂u1 ∂u2 A 1 2 || × || || ∂u × ∂u || Since A is ortogonal, we immediately see that g˜ij = gij , h˜ij = hij . For the second part, deﬁne A(u) : R3 R3 for u U by setting → ∈ ∂ϕ ∂ϕ˜ A(u) = (i =1, 2), A(u)(ν(u))=ν ˜(u). ∂ui u ∂ui u

Plainly, A(u) SO(3) ; we next show that A is constant. On one hand, ∈ ∂2ϕ˜ ∂ ∂ϕ ∂A ∂ϕ ∂2ϕ = A = + A . ∂ui∂uj ∂ui ∂uj ∂ui ∂uj ∂ui∂uj On the other hand, by the Gauss formula (3.18),

∂2ϕ˜ ∂ϕ˜ ∂ϕ ∂2ϕ = Γ˜k + h˜ ν˜ = Γk A + h A(ν)= A . ∂ui∂uj ij ∂uk ij ij ∂uk ij ∂ui∂uj k k X X Putting this together yields ∂A ∂ϕ =0. (3.26) ∂ui ∂uj Similarly, ∂ν˜ ∂A ∂ν = ν + A , ∂ui ∂ui ∂ui 48 CHAPTER 3. SURFACES: INTRINSIC GEOMETRY and the Weingarten equation

∂ν˜ kj ∂ϕ˜ kj ∂ϕ ∂ν = h˜ikg˜ = hikg A = A ∂ui − ∂uj − ∂uj ∂ui j,k j,k X X together imply that ∂A ν =0. (3.27) ∂ui ∂A From (3.26) and (3.27), we get that ∂ui = 0 on U and hence A is constant. ∂ ∂ϕ˜ ∂ϕ Finally, also ϕ˜ A ϕ is a constant b, for i (˜ϕ A ϕ) = i A( i )=0, − ◦ ∂u − ◦ ∂u − ∂u which ﬁnishes the proof.

Theorem 3.28 (Bonnet, 1867) Let be given smooth functions gij , hij deﬁned on an 2 open set U R (1 i, j 2) such that gij = gji, hij = hji and the matrix (gij ) ⊂ ≤ ≤ is positive-deﬁnite. Suppose that gij , hij satisfy the equations of Gauss (3.20) and Codazzi-Mainardi (3.21). Then, for given initial conditions

3 3 u U, p R , X , , X , , ν R 0 ∈ 0 ∈ 1 0 2 0 0 ∈ with ν0 a unit vector and X1,0,X2,0 = gij (u0), there exists an open neighborhood h i 3 V of u contained in U and a unique regular parametrized surface ϕ : V R whose 0 → Gauss map is ν with the following properties:

1. ϕ(u0)= p0;

∂ϕ 2. ∂ui (u0)= Xi,0;

3. ν(u0)= ν0;

4. (gij ) and (hij ) are the fundamental forms of ϕ.

Proof. We introduce new vector-valued variables X1, X2

∂ϕ = X (3.29) ∂uj j and write the Gauss formula and the Weingarten equation in the form of a ﬁrst-order system of PDE’s in X1, X2, ν:

∂X j = Γk X + h ν ∂ui ij k ij k X ∂ν jk = hij g Xk. (3.30) ∂ui − j X We first solve (3.30): the integrability conditions of Theorem 3.24 are exactly the equations of Gauss and Codazzi-Mainardi, which are satisfied by assumption, so there exists a unique solution (X1,X2, ν) defined on a neighborhood of u0 3.9. THE FUNDAMENTALTHEOREM OF THE LOCAL THEORY OF SURFACES49

and satisfying the initial conditions (X1,0,X2,0, ν0) at u0. Let us check that the solutions satisfy

ν, ν =1, ν,Xi =0, Xi,Xj = gij (3.31) h i h i h i on a neighborhood of u0. We differentiate the left-hand sides of these equations to get

∂ ∂ν kl ν, ν = 2 , ν = 2 hikg ν,Xl ∂ui h i h∂ui i − h i k,l X ∂ ∂ν ∂Xj ν,Xj = ,Xj + ν, ∂ui h i h∂ui i h ∂ui i kl k = hikg Xl,Xj + Γ ν,Xk + hij ν, ν − h i ij h i h i k,l k X X ∂ ∂Xi ∂Xj Xi,Xj = ,Xj + Xi, ∂uk h i h ∂ui i h ∂ui i r = Γik Xr,Xj + hik ν,Xj r h i h i X s + Γjk Xs,Xi + hjk ν,Xi . s h i h i X

These identities show that the functions ν, ν , ν,Xi , Xi,Xj satisfy a system h i h i h i of PDE’s in U. It is easy to check that the functions 1, 0, gij satisfy the same system of PDE’s. (Do it!) Since the values of the two triples of functions coincide at the point u0, by the uniqueness part of Theorem 3.24, the equations (3.31) are satisfied on a neighborhood of u0. ∂Xj ∂Xi The final step is to solve (3.29) in ϕ. The integrability conditions ∂ui = ∂uj k k are satisfied because Γij = Γji and hij = hji. Therefore there exists a smaller 3 neighborhood V of u0 and a unique solution ϕ : V R with ϕ(u0) = p0. ∂ϕ → Since ν, ν = 1, ν, i = ν,Xi = 0, ν is a unit normal vector field along ϕ. h i h ∂u i h i Moreover, the fundamental forms of ϕ are

∂ϕ ∂ϕ i , j = Xi,Xj = gij h ∂u ∂u i h i and ∂2ϕ ∂ϕ , ν = Γk + h ν, ν = h , ∂ui∂uj ij ∂uk ij ij * k + X as we wished. As a ﬁnal remark, note that

∂ϕ ∂ϕ 1 2 ν = ∂u × ∂u ± ∂ϕ ∂ϕ || ∂u1 × ∂u2 || where the sign is “+”or“ ” according to whether X1,0,X2,0, ν0 is a positive 3 − { } basis of R or not. 50 CHAPTER 3. SURFACES: INTRINSIC GEOMETRY

3.10 Differential forms

Let U be an open set of Rn.A differential 1-form (or a differential form of degree 1) on U is a map n ω : p U (R )∗; ∈ → ω is said to be smooth if, for all i, the function ω(ei) : U R given by → n ω(ei)(p)= ωp(ei) is smooth, where e ,...,en is the canonical basis of R . { 1 } Suppose X ,...,Xn is a smooth orthonormal frame on U, that is, the { 1 } Xi are smooth vector ﬁelds and X (p),...,Xn(p) is an orthonormal basis { 1 } for p U. Then we can consider the dual coframe ω1,...,ωn by specifying i ∈ i { } ω (Xj )= δj (Kronecker delta). For instance, the dual coframe to the canonical basis of Rn is usually denoted dx1,...,dxn . Of course, dxi is just the linear n{ } projection onto the xi-axis of R .

Example 3.32 On U = R2 (0, 0) , consider the orthonormal frame \{ } x x X = 1 e + 2 e , 1 2 2 1 2 2 2 x1 + x2 x1 + x2 x x X = p − 2 e + p 1 e . 2 2 2 1 2 2 2 x1 + x2 x1 + x2 Then the dual coframe has p p x x ω1 = 1 dx1 + 2 dx2, 2 2 2 2 x1 + x2 x1 + x2 x x ω2 = p − 2 dx1 + p 1 dx2. 2 2 2 2 x1 + x2 x1 + x2

We will also consider differentialp forms ofp degree 2.A differential 2-form on U Rn is a map Ω that takes p U to a skew-symmetric bilinear map ⊂ ∈ n n Ωp : R R R; × →

Ω is said to be smooth if Ω(ei,ej) : U R is a smooth function for all i, j. There → are two important ways of manufacturing 2-forms starting with 1-forms. The ﬁrst one is the exterior product. If ω, η are 1-forms on U, their exterior product is deﬁned to be the 2-form ω η given by ∧

(ω η)p(u, v)= ωp(u)ηp(v) ωp(v)ηp(u), ∧ − n where p U and u, v R . It is immediate to see that (ω η)p is skew- ∈ ∈ ∧ symmetric and bilinear. It is also easy to see that ω η is smooth if ω, η are ∧ smooth. For future use, we note the following properties:

1. ω η = η ω; ∧ − ∧ 2. ω ω =0. ∧ 3.10. DIFFERENTIAL FORMS 51

The second way of constructing 2-forms from 1-forms is exterior derivation. n i If ω is a 1-from on U, we can write ω = a dxi, where ai : U R are i=1 → smooth functions. The exterior derivative of ω is the 2-form P n i dω = da dxi ∧ i=1 Xn i = ∂a dxj dxi. ∂xj ∧ i,j=1 X Example 3.33 Referring to Example 3.32, we have 1 dω1 =0 and dω2 = dx dx . 2 2 1 2 x1 + x2 ∧ The next lemma shows how to computep dω without invoking coordinates. However, note that on left-hand side of (3.35), X and Y need to be vector ﬁelds, whereas (dω)p(X, Y ) makes sense even if X, Y are just vectors. Lemma 3.34 If ω is a smooth 1-form on an open set U of Rn, and X, Y are smooth vector ﬁelds on U, then

dω(X, Y )= DX (ω(Y )) DY (ω(X)) ω([X, Y ]). (3.35) − − i Proof. Write ω = i a dxi. Since both hand sides of (3.35) are linear in ω, we may assume that ω = a dxi, where a : U R is smooth. We write i Pi → X = i X ei, Y = i Y ei and compute

P DX (ω(Y ))P DY (ω(X)) ω([X, Y ]) − i − i i = DX (a dx (Y )) DY (a dx (X)) a dx ([X, Y ]) i − i i− i i = (DX a)Y + aDX Y (DY a)X aDY X a[X, Y ] i i− i− i − i = da(X)Y da(Y )X + a (DX Y DY X [X, Y ] ) − − − =0 = da dxi(X, Y ) ∧ | {z } = dω(X, Y ), as we wished. Examples 3.36 1. Every vector field X defines a 1-form via the equation ω(Y )= X, Y . h i 2. The line integral of a 1-form ω in U along a smooth curve γ : [a,b] U is → defined to be b ω = ω(γ ˙ (t)) dt. Zγ Za 3. The differential of a smooth function f : U R is the 1-form → ∂f i df = ∂xi dx . i X 52 CHAPTER 3. SURFACES: INTRINSIC GEOMETRY

Then

d2f = d(df) = d ∂f dxi ∂xi ∧ i X ∂2f = dxj dxi ∂xi∂xj ∧ i,j X ∂2f ∂2f = dxi dxj ∂xi∂xj − ∂xj ∂xi ∧ i

In particular, d(dxi)=0. 4. More generally, a 1-form ω is called exact if there exists a function f such that df = ω.A 1-form ω is called closed if dω = 0. We have shown that every exact 1-form is closed. The converse is true if the domain U simply-connected; this can be proven using Theorem 3.24.

3.11 Connection forms and the integrability equations

Our next goal is to express the Gauss and Codazzi-Mainardi equations in terms of differential forms. In order to express the covariant derivative of a surface in terms of differential forms, we first consider the directional derivative in the ambient space; fix fix a frame X ,X ,X in R3 with dual coframe { 1 2 3} ω1,ω2,ω3 and define 1-forms by setting { } i i ωj (Y )= ω (DY Xj) for i, j =1, 2, 3. Then 3 i DY Xj = ωj(Y )Xi. i=1 X We also have that i j ωj + ωi =0 for all i, j, because

i j ω (Y )+ ω (Y ) = DY Xj ,Xi + DY Xi,Xj j i h i h i = DY Xi,Xj h i = 0, since Xi,Xj is constant. h i 3 Henceforth we suppose that a surface S in R is given and we take an adapted orthonormal frame, namely, assume that X1, X2 are tangent to S and 3.11. CONNECTION FORMS AND THE INTEGRABILITY EQUATIONS 53

X = ν is normal to S. Let ω1,ω2,ω3 denote the dual coframe. Since the 3 { } covariant derivative is the tangential component of the directional derivative, for a tangent vector Y and i, j =1, 2, we have that i i i ω (Y )= ω (DY Xj )= ω ( Y Xj ). j ∇ i The restrictions of the ωj (i, j = 1, 2) to the tangent spaces of S are called connection forms of S. They determine (and are determined by) the covariant i derivative. Since ωj is skew-symmetric in i, j, there is in fact only one curvature 1 form ω2. If Y is tangent to S and j =1, 2, then

2 i Y Xj = (DY Xj )⊤ = ω (Y )Xj (3.37) ∇ j i=1 X and 3 ω (Y )= DY Xj ,X = Xj,DY ν = Xj , AY = II(Xj , Y ) (3.38) j h 3i −h i h i Equations (3.37), (3.38) are the Gauss formula (3.16) and Weingarten equation (3.17) written in the language of differential forms. In order to do the same with the integhrability equations (3.20) and (3.21), let us first prove the following lemma. Lemma 3.39 If X, Y , Z are smooth vector fields defined on an open set U of Rn, then

DX DY Z DY DX Z = D X, Y ]Z. − [ (Equivalently, [DX ,DY ]= D[X,Y ] as operators)

Remark 3.40 In case X = ei, Y = ej , the lemma follows form the equality ∂2Z of mixed second partial derivatives. Indeed, Dei Dej Z Dej Dei Z = ∂xi∂xj ∂2Z − − ∂xj ∂xi =0 and D[ei,ej ] =0 (since [ei,ej ]=0). i j Proof of lemma 3.39. Write X = i X ei and Y = j Y ej . Then P P i j j i DX DY Z DY DX Z = X De Y ej Y De X ei − i   − j i j j i ! X X X X i j   i ∂Y j = X Y Dei Dej Z + X ∂xi Dej Z i,j i,j X X j i j ∂Xi Y X De De Z + Y j De Z − j i ∂x i i,j i,j X X i ∂Y j i ∂Xj = X i Y i De Z ∂x − ∂x j i,j X j = [X, Y ] Dej Z j X = D[X,Y ]Z, 54 CHAPTER 3. SURFACES: INTRINSIC GEOMETRY as desired.

Proposition 3.41 (Maurer-Cartan structural equations) For all i,, j, wehavethat

3 dωi + ωi ωk =0. j k ∧ j k X=1 In particular

dω1 + ω1 ω3 = 0 (Gauss eqn.) 2 3 ∧ 2 dω1 + ω1 ω2 = 0 (Codazzi-Mainardi eqn.) 3 2 ∧ 3

Proof. If X, Y are smooth vector ﬁelds on R3, then

3 3 dωi + ωi ωk (X, Y ) = dωi (X, Y )+ (ωi ωk)(X, Y ) j k ∧ j j k ∧ j k=1 ! k=1 X i X i i = DX (ω (Y )) DY (ω (X)) ω ([X, Y ]) j − j − j + (ωi (X)ωk(Y ) ωi (Y )ωk(X)) k j − k j k X k i k i = DX (ω (Y ))ω (Xk) DY (ω (Y ))ω (Xk) j − j k k X X k i ω ([X, Y ])ω (Xk) − j k X k i k i + ω (Y )ω (DX Xk) ω (X)ω (DY Xk) j − j k k X X i k k = ω DX ( ω (Y )Xk) DY ( ω (X)Xk) j − j k k X X k ω ([X, Y ])Xk − j k i X = ω (DX DY Xj DY DX Xj D Xj ) − − [X,Y ] = 0, using Lemma 3.39, as we wished. The curvature form of S is the 2-form deﬁned on S (meaning that it is restricted to the tangent spaces of S) 1 1 Ω2 = dω2. The relation to (Gaussian) curvature is expressed by the following proposition. Proposition 3.42 We have that Ω1 = Kω1 ω2, 2 ∧ where K is the Gaussian curvature of S. 3.12. THE GAUSS-BONNET THEOREM 55

Proof. By skew-symmetry, he only nontrivial ordered pair of vectors on which to evaluate a 2-form on S is (X1,X2). On one hand,

1 1 Ω2(X1,X2) = dω2(X1,X2) = ω1 ω3(X ,X ) (by Propostion 3.41) − 3 ∧ 2 1 2 = ω1(X )ω3(X )+ ω1(X )ω3(X ) − 3 1 2 2 3 2 2 1 = ω3(X )ω3(X ) ω3(X )ω3(X ) 1 1 2 2 − 1 2 2 1 = II(X ,X )II(X ,X ) II(X ,X )II(X ,X ) 1 1 2 2 − 1 2 2 1 = det(II) = K (since X ,X is orthonormal). { 1 2} On the other hand, ω1 ω2(X ,X )=1. ∧ 1 2 Finally, Ω1(X ,X )= Kω1 ω2(X ,X ) implies that Ω1 = Kω1 ω2. 2 1 2 ∧ 1 2 2 ∧ Remark 3.43 The reasoning in the proof of Proposition 3.42 shows in fact that any smooth 2-form Ω on S can be written Ω = fω1 ω2 for some smooth ∧ function f : S R. Suppose ϕ : U S is a parametrization and B U is → → ⊂ compact. The integral of the 2-form Ω on B˜ = ϕ(B) is deﬁned to be the surface integral of f on B˜:

Ω := f dA = f ϕ EG F 2 dudv, B˜ B˜ B ◦ − Z Z Z Z Z p and its value is known not to depend on the parametrization. On the other hand, by taking X ,X to be the result of the Gram-Schmidt orthonormal- { 1 2} ization of ∂ϕ , ∂ϕ , { ∂u ∂v } ω1 ω2( ∂ϕ , ∂ϕ ) = X , ∂ϕ X , ∂ϕ X , ∂ϕ X , ∂ϕ ∧ ∂u ∂v h 1 ∂u ih 2 ∂v i − h 1 ∂v ih 2 ∂u i E F = G 2 0 √E r − E − = EG F 2. − It is also easy to see that for anyp other positively oriented orthonormal frame 1 2 1 2 1 2 X′ ,X′ , the dual coframe ω ′,ω ′ satisﬁes ω ′ ω ′ = ω ω . For this { 1 2} { } ∧ ∧ reason, ω1 ω2 is called the area element of S and written ω1 ω2 = dA. ∧ ∧ 3.12 The Gauss-Bonnet theorem

The Gauss-Bonnet is one of the most important theorems in Differential Ge- ometry. The global version expresses the invariance the total (Gaussian) curvature of a closed orientable surface under deformations in the ambient space preserving the topology. For this reason, it is said that this theorem relates the geometry and topology of a closed surface. 56 CHAPTER 3. SURFACES: INTRINSIC GEOMETRY

We start our discussion with the notion of geodesic curvature. Let γ : I S be a curve parametrized by arc-length whose image lies in a regular → parametrized surface ϕ : U S. We want to consider the curvature of γ from → the point of view of an observer in S. We construct a frame along γ which is adapted to S: take e1 = γ′ and e2 = e1 ν, where the sign is chosen so that 3 ± × 3 e ,e , ν is a positive basis of R . The curvature κ of γ as a curve in R is of { 1 2 } course

κ = e′ = De e . || 1|| || 1 1|| Since e has constant length 1, De e ,e = 0. The normal component is the 1 h 1 1 1i normal curvature,

κν = De e , ν = e ,De ν = e , Ae = II(e ,e ), h 1 1 i −h 1 1 i h 1 1i 1 1 and the tanegential component is the geodesic curvature,

κg = De e ,e . h 1 1 2i In particular, 2 2 2 κ = κν + κg.

Note that κg = 0 if and only if e e ,e = 0 (since e is tangent) if and only h∇ 1 1 2i 2 if e e = 0 (since e e ,e =0), which is the same as γ′ γ′ = 0, namely, γ ∇ 1 1 h∇ 1 1 1i ∇ is a geodesic. This shows that the geodesic curvature is a measure of how far from being a geodesic the curve is. Plainly, we also have the equations

e e = κge ∇ 1 1 2 e e = κge . ∇ 1 2 − 1 We can now state the ﬁrst theorem.

Theorem 3.44 (Gauss-Bonnet, ﬁrst local version) Let ϕ : U S be a regular ⊂ parametrized surface, and consider a subset B U diffeomorphic to a closed disk, ⊂ where the boundary ∂B in oriented in the counter-clockwise sense. Then

K dA + κg ds =2π. Zϕ(B) Zϕ(∂B) Examples 3.45 1. For a disk of radius r in the plane R2 R3, we have K = 0 1 1 1 ⊂ and κg = r , so K dA + κg ds = r ds = r 2πr =2π. 2. For the closed hemisphere R R R S2 = (x,y,z) R3 : x2 + y2 + z2 =1, z 0 + { ∈ ≥ } we have K = 1, κg = 0 (the equator is a geodesic), so K dA + κg ds = dA = area(S2 ) = 1 4π =2π. + 2 R R R 3.12. THE GAUSS-BONNET THEOREM 57

Proof of Theorem 3.44. Recall that S is oriented by the unit normal ν coming from ϕ. Consider a parametrization γ : [a,b] ∂B (in the counter-clockwise → sense) so that ϕ γ : [a,b] ϕ(∂B) is a parametrization by arc-length. To ◦ → compute the geodesic curvature, we need an adapted positive orthonormal frame e ,e along where e = (ϕ γ)′. We also consider X ,X , positive { 1 2} 1 ◦ { 1 2} orthonormal frame on S which is the result of the Gram-Schmidt process to ∂ϕ , ∂ϕ ; in particular, X = ∂ϕ / ∂ϕ . Then there exists a smooth function { ∂u ∂v } 1 ∂u || ∂u || θ : [a,b] R such that →

e cos θ sin θ X 1 = 1 . e sin θ cos θ X 2 − 2

Writing γ(s) = (x(s),y(s)), we have

I((ϕ γ) , ∂ϕ ) ∠ ′ ∂u x′g11 + y′g12 cos (e1,X1)= ◦ ∂ϕ = . (ϕ γ) √g11 || ◦ ′|||| ∂u ||

We deﬁne a one-parameter family of inner products

t g = (1 t)gij + tδij , ij −

continuous deformation of the ﬁrst fundamental form of S into the standard 2 inner product of R . The angle ∠(e1,X1)(s,t) is continuous as function of s and t. Since the difference ∠(e ,X )(b,t) ∠(e ,X )(a,t) is always an inte- 1 1 − 1 1 gral multiple of 2π, it must be a constant function of t. By the Umlaufsatz, ∠(e ,X )(b, 1) ∠(e ,X )(a, 1) = 2π. Hence θ(b) θ(a) = ∠(e ,X )(b, 0) 1 1 − 1 1 − 1 1 − ∠(e1,X1)(a, 0)=2π. 58 CHAPTER 3. SURFACES: INTRINSIC GEOMETRY

We compute

2π = θ(b) θ(a) − b dθ = ds ds Za b 1 = ( e e ,X + e , e X ) ds − sin θ h∇ 1 1 1i h 1 ∇ 1 1i Za b 1 = (cos θ e1 e1,e1 sin θ e1 e1,e2 − a sin θ h∇ i − h∇ i Z =0 +cos θ X , e X + sin θ X , e X ) ds h 1 |∇ 1 {z1i } h 2 ∇ 1 1i =0 b 1| {z } = (κg + ω2(e1)) ds Za 1 = κg ds + ω2 Zϕ(∂B) Zϕ∂B) 1 = κg ds + dω2 (by Stokes theorem) Zϕ(∂B) Zϕ(B)

= κg ds + K dA, Zϕ(∂B) Zϕ(B) as desired. In the second version of the local Gauss-Bonnet theorem, we alllow ∂B to have corners.

Theorem 3.46 (Gauss-Bonnet, second local version) Same assumptions as in 3.44, except that now B is only homeomorphic to a closed disk and ∂B is piecewise smooth. Let αi be the exterior angle at the ith vertex of ∂B. Then

K dA + κg ds + αi =2π. ϕ(B) ϕ(∂B) i Z Z X

We only make some remarks about what needs to be changed in the proof of this theorem in relation to Theorem 3.44. The ﬁrst ingredient is the Umalufsatz for piecewise smooth closed curves. Suppose γ : [a,b] ∂B is continuous, → closed (γ(a) = γ(b)), and piecewise smooth in the sense that there exists a partition a = s < s < ...sn = b such that γ is smooth for i = 0 1 +1 |[si,si+1] 0,...,n. For the sake of convenience, we also assume that γ(s0) = γ(sn) is not a vertex. By smoothing γ near its vertices, one shows that the Umlaufsatz 3.12. THE GAUSS-BONNET THEOREM 59 remains valid, te index of rotation of γ is 1. Now we can write

2π = θ(b) θ(a) n − n = (θ(si ) θ(si+)) + (θ(si+) θ(si )) +1− − − − i=0 i=1 Xn n X si+1 dθ = ds + α , ds i i=0 si i=1 X Z X and the rest of the proof goes as before.

Corollary 3.47 (Geodesic n-gon) If the sides of ∂B are geodesic segments, then

n K dA =2π αi. − ϕ(B) i=1 Z X Corollary 3.48 (Theorema Elegantissimum, Gauss) For a geodesic triangle (n = 3), we have that K dA = β + β + β π, 1 2 3 − Zϕ(B) where βi = π αi is the interior angle. − Corollary 3.49 The sum of the interior angles of a geodesic triangle in a surface S is > π K > 0 = π if K =0 , resp.    < π  K < 0  As an application of the second version of the local Gauss-Bonnt theorem, we have the following proposition.

Proposition 3.50 If the Gaussian curvature K 0 on a simply-connected surface ≥ S, then two geodesics that start at a point p S cannot meet again (i.e. there is no ∈ geodesic 2-gon in S).

Proof. Suppose the geodesics meet again. Then they bound a region R diffeomorphic to a disk, by simply-connectedness of S. By Theorem 3.46, K dA + α1 + α2 = 2π. Note that α1, α < π since two distinct geodesics cannotR be tangent at a point, and the integral term is nonpositive, so we get a R R contradiction. In particular (case α1 = α =2):

Corollary 3.51 There is no simple closed geodesic in a simply-connected nonposi- tively curved surface.

Of course, a cilinder is not simply-connected and violates the conclusion of the preceding corollary. 60 CHAPTER 3. SURFACES: INTRINSIC GEOMETRY

Theorem 3.52 (Gauss-Bonnet, global version) Let S be a compact orientable surface. Then

K dA =2πχ(S), Z ZS where χ(S) is the Euler characteristic of S.

Remark 3.53 (Digression on the Euler characteristic of a compact surface) A triangulation of a compact surface S is a decomposition S = ∆i into finitely T ∪ many triangles such that a non-empty intersection ∆i ∆j (i = j) consists of ∩ 6 one common side or one common vertex. Radóproved in 1925 that every compact surface admits a triangulation. The Euler characteristic of S with respect to is defined to be χ(S, ) = V E + F , where V , E , F denote re- T T T − T T T T T spectively the total numbers of vertices, edges, faces of triangles in . It is not T difficult to see that χ(S, ) = χ(S, ′) for two triangulations , ′ of S. This T T T T can be proved in two stages: first one checks that it is true in case ′ si a refine- T ment of ; in the general case, one sees that , ′ admit a commn refinement. T T T This being so, one can define the Euler characteristic of S as χ(S) = χ(S, ) T for any triangulation of S. In fact, Poincaréshowed that χ(S) is a topo- T logical invariant of S, that is, two homemorphic surfaces have the same Euler characteristic. Theorem 3.52 then says that the total curvature S K dA is a topological invariant. R R The sphere S2 has χ(S2)=2, as is easily seen. This relation is reminiscent of formulas by Descartes and Euler (Euler’s relation for convex polyhedra). Similarly, one sees that the torus T 2 has χ(T 2)=0. More generally, every compact orientable surface is homeomorphic to a sphere with g handles (cylinders) attached; the number is a topological invariant of the surface called genus. So g = 0 for the sphere and g = 1 for the torus. Also, the relation χ = 2 2g is − easily checked.

Proof of Theorem 3.52. It is possible to choose a triangulation of S such T that each triangle ∆i is contained in the image of a parametrization compatible with the orientation of S. By Theorem 3.46,

3 K dA + κg ds + αij =2π, ∆i ∂∆i j=1 Z Z Z X where αij are the external angles of ∆i. Summing over i =1,...,F , we get

F F 3 K dA + κg ds + αij =2π F. S i=1 ∂∆i i=1 j=1 Z Z X Z X X The second term vanishes, because each edge is counted twice, each time with a different orientation induced by the corresponding triangle. Moreover, if 3.12. THE GAUSS-BONNET THEOREM 61

βij = π αij is the internal angle, − F 3 F 3 F 3 αij = π βij − i=1 j=1 i=1 j=1 i=1 j=1 X X X X X X = 3F π 2π V, − since the sum of the internal angles around a vertex is 2π. The proof is com- pleted by noting that 3F =2E since each edge is shared by two faces. As an easy application, we have:

Corollary 3.54 A compact orientable surface in R3 with constant (Gaussian) curvature is homeomorphic to the sphere.

Proof. Indeed, a compact surface S in R3 admits a point p with K(p) > 0, so the constant curvature must be positive. By Gauss-Bonnet, 2 2g = χ(S) > 0 − implying g =0. Indeed Liebmann’s theorem (1899) says that S must be a round sphere of radius 1/√K.