(1)The Geometry of the Gauss Map Throughout this chapter, S will denote a regular orientable surface in which an orientation (i.e., a differentiable field of unit normal vectors N ) has been chosen

The Geometry of the Gauss Map

Throughout this chapter, S will denote a regular orientable surface in which an orientation (i.e., a differentiable field of unit normal vectors N ) has been chosen; this will be simply called a surface S with an orientation N.

Definition Let S ⊂ R³ be a surface with an orientation N. The map N : S → R³ takes its values in the unit sphere

S² = {(x, y, z) ∈ R³ | x²+ y²+ z² = 1}

The map N : S → S², thus defined, is called theGauss map of S.

It is straightforward to verify that the Gauss map is differentiable. Thus the differential dN_p is a linear map from T_pS to T_{N (p)}S². Since T_pS and T_{N (p)}S² are the same vector spaces, dN_p can be viewed as a linear mapdNp : TpS → TpS from TpS to itself defined as follows.

For each parametrized curve α(t) in S with α(0) = p, we consider the parametrized curve N ◦ α(t) = N (t) in the sphere S²; thhis amounts to restricting the normal vector N to the curve α(t). The tangent vector N⁰(0) = dNp(α⁰(0)) is a vector in TpS. It measures the rate of change of the normal vector N, restricted to the curve α(t), at t = 0. Thus, dN_p measures how N pulls away from N (p) in a neighborhood of p. In the case of curves, this measure is given by a number, the curvature. In the case of surfaces, this measure is characterized by a linear map.

Examples

1. Let S = {(x, y, z) ∈ R³ | ax + by + cz + d = 0} be a plane in R³. Then the unit normal vector N = (a, b, c)/√

a²+ b²+ c² is a constant, and therefore dN ≡ 0, i.e. dN_p(v) = 0v = 0 ∈ T_{N (p)}S = T_pS for all p ∈ S and all v ∈ T_pS.

2. Consider the unit sphere

S² = {(x, y, z) ∈ R³ | x²+ y²+ z² = 1}.

If α(t) = (x(t), y(t), z(t)) is a parametrized curve in S², then

2x(t)x⁰(t) + 2y(t)y⁰(t) + 2z(t)z⁰(t) = 0 ⇐⇒ hα(t), α⁰(t)i = 0,

which shows that the vector (x, y, z) is normal to the sphere at the point (x, y, z). Thus, N = (x, y, z) and N = (−x, −y, −z) are fields of unit normal vectors in S¯ ². Restricted to the curve α(t), the normal vectors

N (t) = (−x(t), −y(t), −z(t)) =⇒ dN (x⁰(t), y⁰(t), z⁰(t)) = N⁰(t) = (−x⁰(t), −y⁰(t), −z⁰(t)), N (t)¯ = (x(t), y(t), z(t)) =⇒ d ¯N (x⁰(t), y⁰(t), z⁰(t)) = ¯N⁰(t) = (x⁰(t), y⁰(t), z⁰(t)) that is, dN_p(v) = −v and d ¯N_p(v) = v for all p ∈ S² and all v ∈ T_pS².

3. Consider the cylinder S = {(x, y, z) ∈ R³ | x² + y² = 1}. By an argument similar to that of the previous example, we see that ¯N = (x, y, 0) and N = (−x, −y, 0) are unit normal vectors at (x, y, z). If (x(t), y(t), z(t)) is a parametrized curve in the cylinder, since

(x(t))²+ (y(t))² = 1 =⇒ 2x(t)x⁰(t) + 2y(t)y⁰(t) = 0, we are able to see that, along this curve,

N (t) = (−x(t), −y(t), 0) =⇒ dN (x⁰(t), y⁰(t), z⁰(t)) = N⁰(t) = (−x⁰(t), −y⁰(t), 0).

This implies that if v is a vector tangent to the cylinder and parallel to the z axis and if w is a vector tangent to the cylinder and parallel to the xy plane, then

dN (v) = 0 = 0v; dN (w) = −w.

It follows that v and w are eigenvectors of dN with eigenvalues 0 and −1, respectively.

4. Let us analyze the point p = (0, 0, 0) of the hyperbolic paraboloid z = y²− x². For this, we consider a parametrization X(u, v) given by

X(u, v) = (u, v, v²− u²),

and compute the normal vectorN (u, v). We obtain X_u = (1, 0, −2u), X_v = (0, 1, 2v) and

N = X_u∧ X_v =









u r

u²+ v²+1 4

, −v

r

u²+ v²+ 1 4

, 1

2 r

u²+ v²+ 1 4







 .

If α(t) = X(u(t), v(t)) is a curve with α(0) = p then the tangent vector α⁰(0) has coordinates (u⁰(0), v⁰(0), 0). Restricting N (u, v) to this curve and computing N⁰(0), we obtain

N⁰(0) = (2u⁰(0), −2v⁰(0), 0), and therefore, at p,

dN_p(u⁰(0), v⁰(0), 0) = (2u⁰(0), −2v⁰(0), 0).

It follows that the vectors (1, 0, 0) and (0, 1, 0) are eigenvectors of dN_p with eigenvalues 2 and −2, respectively.

5. The method of the previous example, applied to the point p = (0, 0, 0) of the paraboloid x = x²+ ky², k > 0, shows that the unit vectors of the x axis and the y axis are eigenvectors of dN_p, with eigenvalues 2 and 2k, respectively (assuming that N is pointing outwards from the region bounded by the paraboloid).

Proposition The differential dN_p : T_pS → T_pS of the Gauss map is a self-adjoint linear map.

Proof Since dN_p : T_pS → T_pS is linear, it suffices to verify that

hdN_p(w₁), w₂i = hw₁, dN_p(w₂)i for a basis {w₁, w₂} of T_pS.

Let X(u, v) be a parametrization of S at p and {X_u, X_v} the associated basis of T_pS. If α(t) = X(u(t), v(t)) is a parametrized curve in S, with α(0) = p, we have

dNp(α⁰(0)) = dNp(Xuu⁰(0) + Xvv⁰(0))

= d

dtN (u(t), v(t))|_t=0

= Nuu⁰(0) + Nvv⁰(0);

in particular, dN_p(X_u) = N_u and dN_p(X_v) = N_v. Therefore, to prove that dN_p : T_pS → T_pS is self-adjoint, it suffices to show that

hN_u, X_vi = hdN_p(X_u), X_vi = hX_u, dN_p(X_v)i = hX_u, N_vi.

Differentiating the equations hN, X_ui = 0 and hN, X_vi = 0 with respect to v and u, respectively, we get

hN_v, X_ui + hN, X_uvi = 0 =⇒ hN_v, X_ui = −hN, X_uvi hN_u, X_vi + hN, X_vui = 0 =⇒ hN_u, X_vi = −hN, X_vui

=⇒ hNv, Xui = −hN, Xuvi = hNu, Xvi.

This proves that dN_p : T_pS → T_pS is a self-adjoint linear map.

Remark Let V be a vector space of dimension 2, endowed with an inner product h , i. We say that a linearmap A : V → V is self-adjoint if hAv, wi = hv, Awi for all v, w ∈ V.

If {e₁, e₂} is an orthonormal basis for V and (α_ij), i, j = 1, 2, is the matrix of A relative to that basis, then

hAe_j, e_ii = α_ij = he_j, Ae_ii = α_ji, that is, the matrix (α_ij) is symmetric.

To each self-adjoint linear map we associate a map B : V × V → R defined by B(v, w) = hAv, wi ∀ v, w ∈ V.

B is clearly bilinear; that is, it is linear in both v and w. Moreover, the fact that A is self-adjoint implies that B(v, w) = B(w, v); that is, B is a bilinear symmetric form in V.

Conversely, if B is a bilinear symmetric form in V, we can define a linear map A : V → V by hAv, wi = B(v, w) and the symmetry of B implies that A is self-adjoint.

On the other hand, to each symmetric, bilinear form B in V, there corresponds a quadratic form Q in V given by

Q(v) = B(v, v), v ∈ V,

and the knowledge of Q determines B completely, since B(v, w) = 1

2[Q(v + w) − Q(v) − Q(w)] .

Thus, a one-to-one correspondence is established between quadratic forms in V and self-adjoint linear maps of V.

The fact that dN_p : T_pS → T_pS is a self-adjoint linear map allows us to associate to dN_p a quadratic form in T_pS defined as follows.

Definition The quadratic form II_p : T_pS → R, defined by II_p(v) = −hdN_p(v), vi, v ∈ T_pS, is calledthe second fundamental form of S at p.

Definition Let C be a regular curve in S passing through p ∈ S, k the curvature of C at p, and cos θ = hn, N i, where n is the normal vector to C and N is the normal vector to S at p. The number k_n= k cos θ is then called thenormal curvature of C ⊂ S at p.

In other words, kn is the length of the projection of the vector kn over the normal to the surface at p, with a sign given by the orientation N of S at p.

Remarks

• The normal curvature of C does not depend on the orientation of C but changes sign with a change of orientation for the surface.

• Let v ∈ T_pS be a unit tangent vector to S at p, and let C ⊂ S be a regular curve parametrized by α(s) : I → S, where s is the arc length of C, and with α(0) = p, α⁰(0) = v. Let N (s) = N ◦ α(s) be the restriction of the normal vector N to the curve α(s). For all s ∈ I, since

hN (s), α⁰(s)i = 0 =⇒ hN⁰(s), α⁰(s)i+hN (s), α⁰⁰(s)i = 0 =⇒ −hN⁰(s), α⁰(s)i = hN (s), α⁰⁰(s)i, we have

IIp(α⁰(0)) = −hdNp(α⁰(0)), α⁰(0)i = −hdNp(−α⁰(0)), −α⁰(0)i = IIp(−α⁰(0))

= −hN⁰(0), α⁰(0)i = hN (0), α⁰⁰(0)i

= hN (p), kn(p)i^(∗)= k_n(p) =⇒ k_n(p) depends only on v = α⁰(0) and II_p

that is, the valueof the second fundamental form II_p for a unit vector v ∈ T_pS is equal to the normal curvature of a regular curve passing through p and tangent to v. In particular, we obtained the following result.

Proposition (Meusnier) All curves lying on a surface S and having at a given point p ∈ S the same tangent line have at this point the same normal curvatures.

Remark Given a unit vector v ∈ T_pS, the intersection of S with the plane containing v and N (p) is called the normal section of S at p along v. In a neighborhood of p, a normal section of S at p isa regular plane curve on S whose normal vector

n(p) = ±N (p) or 0 (a zero vector when k = 0);

and, by(∗), k(p) = |k_n(p)|. With this terminology, the above proposition says that the absolute value of the normal curvature at p of a curve α(s) is equal to the curvature of the normal section of S at p along α⁰(0).

Examples

1. Let S be the surface of revolution obtained by rotating the curve z = y⁴, parametrized by α(t) = (0, t, t⁴), t ∈ R, about the z axis and let p = α(0) = (0, 0, 0) ∈ S. Since

• k(p) = |α⁰(0) × α⁰⁰(0)|

|α⁰(0)|³ = 0,

• T_pS = {(x, y, 0) | (x, y) ∈ R²}, the xy plane,

=⇒ N (p) // (0, 0, 1), i.e. the normal vector N (p) is parallel to the z axis,

and any normal section at p is obtained from the curve z = y⁴ by rotation; hence, it has curvature zero. It follows that all normal curvatures are zero at p, and thus dNp = 0.

2. • Let S = {(x, y, z) ∈ R³ | ax+by +cz +d = 0} be a plane in R³. Since all normal sections are straight lines, all normal curvatures are zero. Thus, the second fundamental form is identically zero at all points. This agrees with the fact that dN_p = 0 for all p ∈ S.

• Let S² = {(x, y, z) ∈ R³ | x²+ y²+ z² = 1} with N as orientation, the normal sections through a point p ∈ S² are circles with radius 1. Thus, all normal curvatures are equal to 1, and the second fundamental form is II_p(v) = 1 for all p ∈ S² and all v ∈ T_pS² with |v| = 1.

• Let S = {(x, y, z) ∈ R³ | x² + y² = 1} be a cylinder in R³. Since the normal sections at a point p vary from a circle perpendicular to the axis of the cylinder to a straight line parallel to the axis of the cylinder, passing through a family of ellipses, the normal curvatures varies from 1 to 0. It is not hard to see geometrically that 1 is the maximum and 0 is the minimum of the normal curvature at p.

Lemma If the function Q(x, y) = ax²+ 2bxy + cy², restricted to the unit circle x²+ y² = 1, has a maximum at the point (1, 0), then b = 0.

Proof Parametrize the circl x²+ y² = 1 by x = cos t, y = sin t, t ∈ (−ε, 2π + ε). Thus, t = 0 is an interior point of (−ε, 2π + ε), and Q, restricted to that circle, becomes a function of t:

Q(t) = a cos²t + 2b cos t sin t + c sin²t, t ∈ (−ε, 2π + ε).

Since Q has a maximum at the interior point (1, 0) we have

 dQ dt



t=0

= 2b = 0.

Hence, b = 0 as we wished.

Proposition Given a quadratic form Q in V, there exists an orthonormal basis {e1, e2} of V such that if v ∈ V is given by v = xe₁+ ye₂, then

Q(v) = λ1x²+ λ2y²,

where λ₁ and λ₂ are the maximum and minimum, repectively, of Q on the unit circle |v| = 1.

Proof Let λ₁ be the maximum of Q on the unit circle |v| = 1, and let e₁ be a unit vector with max|v|=1Q(v) = Q(e₁) = λ₁. Such an e₁ exists by continuity of Q on the compact set |v| = 1. Let e₂ be a unit vector that is orthogonal to e₁, and set λ₂ = Q(e₂). We shall show that the basis {e₁, e₂} satisfies the conditions of the proposition.

Let B be the symmetric bilinear form that is associated to Q and set v = xe₁+ ye₂. Then Q(v) = B(v, v) = B(xe₁+ ye₂, xe₁+ ye₂) = λ₁x²+ 2bxy + λ₂y², where b = B(e₁, e₂).

By the lemma, b = B(e₁, e₂) = 0, and thus Q(v) = λ₁x²+ λ₂y², for v = xe₁+ ye₂ ∈ V.

Furthermore, for any v = xe₁ + ye₂ with x² + y² = 1, since λ₁ = max

|v|=1Q(v) ≥ Q(e₂) = λ₂, Q(v) = λ₁x²+ λ₂y² ≥ λ₂(x²+ y²) = λ₂ = Q(e₂) =⇒ λ₂ = min

|v|=1Q(v).

Since

Theorem Let A : V → V be a self-adjoint linear map. Then there exists an orthonormal basis {e1, e2} of V such that A(e1) = λ1e1, A(e2) = λ2e2 (that is, e1 and e2 are eigenvectors and λ₁, λ₂ are eigenvalues of A). In the basis {e₁, e₂}, the matrix of A is clearly diagonal and the elements λ₁, λ₂, λ₁ ≥ λ₂, on the diagonal are the maximum and the minimum, respectively, of the quadratic form Q(v) = hAv, vi, on the unit circle of V.

Proof Consider the quadratic form Q(v) = hAv, vi. By proposition above, there exists an or- thonormal basis {e₁, e₂} of V such that

Q(e₁) = λ₁ = max

|v|=1|Q(v) ≥ min

|v|=1|Q(v) = λ₂ = Q(e₂).

By setting Ae₁ = α₁₁e₁+ α₂₁e₂ and Ae₂ = α₁₂e₁+ α₂₂e₂, since

α₁₁= hAe₁, e₁i = Q(e₁) = λ₁ and α₂₁ = hAe₁, e₂i = B(e₁, e₂) = 0 =⇒ Ae₁ = λ₁e₁, and since

α₁₂= hAe₂, e₁i = B(e₂, e₁) = 0 and α₂₂ = hAe₂, e₂i = Q(e₂) = λ₂ =⇒ Ae₂ = λ₂e₂, and in the basis {e1, e2}, the matrix of A is λ₁ 0

0 λ₂

 .

Remark For each p ∈ S there exists an orthonormal basis {e₁, e₂} of T_pS such that dNp(e1) = −k1e1 and dNp(e2) = −k2e2, where k1 = max

v∈TpS, |v|=1IIp(v) and k2 = min

v∈TpS, |v|=1IIp(v).

Definition The maximum normal curvature k₁ and the minimum normal curvature k₂ are called the principal curvatures at p; the corresponding directions, that is, the directions given by the eigenvectors e₁, e₂, are called principal directions at p.

Definition If a regular connected curve C on S is such that for all p ∈ C the tangent line of C is a principal direction at p, then C is called aline of curvature of S.

Proposition (Olinde Rodrigues) A necessary and sufficient condition for a connected regular curve C on S to be a line of curvature of S is that

N⁰(t) = λ(t)α⁰(t),

for any parametrization α(t) of C, where N (t) = N ◦ α(t) and λ(t) is a differentiable function of t. In this case, −λ(t) is the (principal) curvature along α⁰(t).

Proof It suffices to observe that if α⁰(t) is contained in a principal direction, then α⁰(t) is an eigenvector of dN and

N⁰(t) = dN (α⁰(t)) = λ(t)α⁰(t).

The converse is immediate (since dN (α⁰(t)) = N⁰(t) = λ(t)α⁰(t)).

Remark For p ∈ S, let {e₁, e₂} be the principal directions at p such that dN_p(e₁) = −k₁e₁ and dN_p(e₂) = −k₂e₂, where k₁ = max

v∈TpS, |v|=1II_p(v), k₂ = min

v∈TpS, |v|=1II_p(v) are the principal curvatures at p. For each unit vector v ∈ T_pS, since {e₁, e₂} forms an orthonormal basis of T_pS, we have

v = e₁cos θ + e₂sin θ,

where θ is the angle from e₁ to v in the orientation of T_pS. The normal curvature k_n along v is given by

k_n = II_p(v) = −hdN_p(v), vi

= −hdN_p(e₁cos θ + e₂sin θ), e₁cos θ + e₂sin θi

= he₁k₁cos θ + e₂k₂sin θ, e₁cos θ + e₂sin θi

= k₁cos²θ + k₂sin²θ.

The last expression is known classically as the Euler formula; actually, it is just the expression of the second fundamental form in the basis {e₁, e₂}.

Given a linear map A : V → V of a vector space of dimension 2 and given a basis {v₁, v₂} of V, we recall that

determinant of A = a₁₁a₂₂− a₁₂a₂₂, trace of A = a₁₁+ a₂₂

2 ,

where (a_ij) is the matrix of A in the basis {v₁, v₂}. It is known that these numbers do not depend on the choice of the basis {v₁, v₂} and are, therefore, attached to the linear map A.

Definition Let p ∈ S and let dN_p : T_pS → T_pS be the differential of the Gauss map. Then the determinant of dN_p is called the Gaussian curvatureK of S at p, and the negative of half of the trace of dN_p is called the mean curvatureH of S at p.

In terms of the principal curvatures we can write K = det dN_p =

−k₁ 0 0 −k₂

= det(−dN_p) = k₁k₂, H = tr (−dN_p) = tr k₁ 0 0 k₂



= k₁+ k₂ 2 . Definition A point p of a surface S is called

1. elliptic if det dN_p > 0, e.g. points of a sphere (x − a)² + (y − b)² + (z − c)² = r² and the point p = (0, 0, 0) of the paraboloid z = x²+ ky², k > 0.

2. hyperbolicif det dN_p < 0, e.g. the point p = (0, 0, 0) of the hyperbolic paraboloid z = y²−x². 3. parabolicif det dN_p = 0, with dN_p 6= 0, e.g. points of a cylinder (x − a)²+ (y − b)² = r². 4. planarif dN_p = 0, e.g. points of a plane ax + by + cz + d = 0.

It is clear that this classification does not depend on the choice of the orientation.

Definition A point p ∈ S is called anumbilical point of S if k₁ = k₂.

Observe that all points of a plane (k₁ = k₂ = 0) are (planar) umbilical points, and all points of a sphere of radius r (k₁ = k₂ = 1

r) or the point p = (0, 0, 0) of the paraboloid z = x²+ y² (k₁ = k₂ = 2) are (nonplanar) umbilical points.

It is an interesting fact that the only surfaces made up entirely of umbilical points are essentially spheres and planes.

Proposition If all points of a connected surface S are umbilical points, then S is either contained in a sphere or in a plane.

Proof Let p ∈ S and let X(u, v) be a parametrization of S at p such that the coordinate V = X(U ) ⊂ S is connected. Since each q ∈ V is an umbilical point, there exists a differentiable function λ : V → R such that

dN_q(w) = λ(q)w ∀ q ∈ V, ∀ w = a₁X_u+ a₂X_v ∈ T_qS

⇐⇒ N_ua₁+ N_va₂ = λ(q)(a₁X_u+ a₂X_v) ∀ a₁, a₂ ∈ R

=⇒ N_u = λ(q)X_u and N_v = λ(q)X_v.

Differentiating the first equation in v and the second one in u and subtracting the resulting equations, we obtain

λ_v(q)X_u− λ_u(q)X_v = 0.

Since X_u and X_v are linearly independent, we have

λ_u(q) = λ_v(q) = 0 ∀ q ∈ V.

Also since V is connected, λ(q) = λ (a constant) for all q ∈ V = X(U ) ⊂ S.

If λ = 0, then Nu(q) = Nv(q) = 0 for all q ∈ V and therefore N (q) = N0 (a constant vector) for all q ∈ V. Thus

hX(u, v), N₀i_u = hX(u, v), N₀i_v = 0 ∀ (u, v) ∈ U.

Since U ⊂ R² is connected, we have

hX(u, v), N₀i = d (a constant) ∀ (u, v) ∈ U and all points X(u, v) of V belong to a plane.

If λ 6= 0, since Nu = λXu, Nv = λXv, we have



X(u, v) − 1

λN (u, v)



u

=



X(u, v) − 1

λN (u, v)



v

= 0 ∀ (u, v) ∈ U.

Thus there exists a fixed point Y ∈ R³ such that X(u, v) − 1

λN (u, v) = Y ∀ (u, v) ∈ U =⇒ |(X(u, v) − Y |² = 1

λ²|N |² = 1

λ² ∀ (u, v) ∈ U.

Hence all points of V = X(U ) are contained in a sphere of center Y and radius 1

|λ|.

Furthermore, observe that if V = X(U ) and W = ¯X( ¯U ) are connected coordinate neighborhoods of p = X(u₀, v₀) = ¯X(¯u₀, ¯v₀) ∈ S, then V = X(U ) and W = ¯X( ¯U ) are contained in the same plane or in the same sphere by the continuity. This proves that if all points of a connected surface S are umbilical points, then S is either contained in a sphere or in a plane.

The Gauss Map in Local Coordinates

Let X(u, v) be a parametrization at a point p ∈ S of a surface S, and let α(t) = X(u(t), v(t)) be a parametrized curve on S, with α(0) = p. To simplify the notation, we shall make the convention that all functions to appear below denote their values at the point p.

The tangent vector to α(t) at p α⁰ = X_uu⁰+ X_vv⁰ and dN (α⁰) = d

dtN (u(t), v(t)) = N_uu⁰+ N_vv⁰. Since N = X_u∧ X_v

|X_u∧ X_v|, N_u, N_v ∈ T_pS, in basis {X_u, X_v} we may write

(∗)

( N_u = a₁₁X_u+ a₂₁X_v, N_v = a₁₂X_u+ a₂₂X_v, and therefore,

dN (α⁰) = (a₁₁u⁰+ a₁₂v⁰)X_u+ (a₂₁u⁰+ a₂₂v⁰)X_v; hence,

dNu⁰ v⁰



=a₁₁ a₁₂ a₂₁ a₂₂

 u⁰ v⁰

 .

This shows that in the basis {X_u, X_v}, dN is given by the matrix (a_ij), i, j = 1, 2. Notice that this matrix is not necessarily symmetric, unless {X_u, X_v} is an orthonormal basis.

On the other hand, the expression of the second fundamental form in the basis {X_u, X_v}is given by

II_p(α⁰) = −hdN (α⁰), α⁰i = −hN_uu⁰+ N_vv⁰, X_uu⁰+ X_vv⁰i

= e(u⁰)²+ 2f u⁰v⁰+ g(v⁰)², where, since hN, X_ui = hN, X_vi = 0,

e = −hN_u, X_ui = hN, X_uui,

f = −hN_v, X_ui = hN, X_uvi = hN, X_vui = −hN_u, X_vi, g = −hNv, Xvi = hN, Xvvi.

We shall now obtain the values of a_ij in terms of the coefficients e, f, g. We shall now obtain the values of aij in terms of the coefficients e, f, g. From equations(∗) for N_u, N_v, we have

−f = −hN_u, X_vi = a₁₁F + a₂₁G,

−f = −hNv, Xui = a12E + a22G,

−e = −hN_u, X_ui = a₁₁E + a₂₁F,

−g = −hN_v, X_vi = a₁₂F + a₂₂G.

where E, F and G are the coefficients of the first fundamental form in the basis {X_u, X_v}. In matrix form, we have

− e f f g



=a₁₁ a₂₁ a₁₂ a₂₂

 E F

F G



⇐⇒ a₁₁ a₂₁ a₁₂ a₂₂



= − e f f g

 E F

F G

⁻¹ (†),

and thus

a₁₁ a₂₁ a12 a22



= −1

EG − F²

 e f f g

  G −F

−F E



= −1

EG − F²

 eG − f F −eF + f E f G − gF −f F + gE

 . Note that the Equations(∗), with (aij) defined in (†), are nonlinear partial differential equations of 2^nd order for X = X(u, v), called the equations of Weingarten.

From Eq. (†), we immediately obtain the Gaussian curvature K(p) = det(−dNp) = det(aij) = eg − f²

EG − F².

To compute the mean curvature, we recall that −k₁, −k₂ are the eigenvalues of dN Therefore, k1 and k2 satisfy the equation

dN (v) = −kv = −kIv for some v ∈ TpS, v 6= 0,

where I is the identity map. It follows that the linear map dN + kI is not invertible; hence, it has zero determinant. Thus,

deta₁₁+ k a₁₂ a₂₁ a₂₂+ k



= 0 ⇐⇒ k²+ (a₁₁+ a₂₂)k + (a₁₁a₂₂− a₁₂a₂₁) = 0.

Since k1 and k2 are the roots of the above quadratic equation, we conclude that H = 1

2(k1+ k2) = −1

2(a11+ a22) = 1 2

eG − 2f F + gE EG − F² ;

hence,

k²− 2Hk + K = 0 ⇐⇒ k = H ±√

H²− K.

From this relation, it follows that if we choose k₁(q) ≥ k₂(q), q ∈ S, then the functions k₁ and k₂ are continuous in S. Moreover, k₁ and k₂ are differentiable in S, except perhaps at theumbilical points (H² = K)of S.

Examples

1. Let U be an open subset of R²and let S be the graph of a differentiable function z = h(x, y), (x, y) ∈ U. Then S is parametrized by

X(x, y) = (x, y, h(x, y)), (x, y) ∈ U.

A simple computation shows that K = h_xxh_yy− h²_xy

(1 + h²_x+ h²_y)², 2H = (1 + h²_x)h_yy− 2h_xh_yh_xy+ (1 + h²_y)h_xx (1 + h²_x+ h²_y)^3/2 . 2. Consider a surface of revolution parametrized by

X(u, v) = (ϕ(v) cos u, ϕ(v) sin u, ψ(v)) 0 < u < 2π, a < v < b, ϕ(v) 6= 0.

The coefficients of the first fundamental form are given by E = ϕ², F = 0, G = (ϕ⁰)²+ (ψ⁰)².

It is convenient to assume that the rotating curve is parametrized by arc length, that is, that

(ϕ⁰)²+ (ψ⁰)² = G = 1.

The computation of the coefficients of the second fundamental form is straightforward and yields

e = (X_u, X_v, X_uu)

√EG − F² = 1

√EG − F²      

−ϕ sin u ϕ cos u 0 ϕ⁰cos u ϕ⁰sin u ψ⁰

−ϕ cos u −ϕ sin u 0      

= −ϕψ⁰

f = (X_u, X_v, X_uv)

√EG − F² = 1

√EG − F²      

−ϕ sin u ϕ cos u 0 ϕ⁰cos u ϕ⁰sin u ψ⁰

−ϕ⁰sin u ϕ⁰cos u 0      

= 0

g = (X_u, X_v, X_vv)

√EG − F² = 1

√EG − F²      

−ϕ sin u ϕ cos u 0 ϕ⁰cos u ϕ⁰sin u ψ⁰ ϕ⁰⁰cos u ϕ⁰⁰sin u ψ⁰⁰

= ψ⁰ϕ⁰⁰− ψ⁰⁰ϕ⁰

Since F = f = 0, we conclude that the parallels (v =const.) and the meridians (u =const.) of a surface of revolution are lines of curvature of such a surface.

Because

K = eg − f²

EG − F² = −ψ⁰(ψ⁰ϕ⁰⁰− ψ⁰⁰ϕ⁰) ϕ

and ϕ is always positive, it follows that the parabolic points are given by either ψ⁰ = 0 (the tangent line to the generator curve is perpendicular to the axis of rotation) or ψ⁰ϕ⁰⁰−ψ⁰⁰ϕ⁰ = 0

(the curvature of the generator curve is zero). A point which satisfies both conditions is a planar point, since these conditions imply that e = f = g = 0.

It is convenient to put the Gaussian curvature in still another form. By differentiating (ϕ⁰)²+ (ψ⁰)² = 1 we obtain ϕ⁰ϕ⁰⁰= −ψ⁰ψ⁰⁰. Thus

K = −ψ⁰(ψ⁰ϕ⁰⁰− ψ⁰⁰ϕ⁰)

ϕ = −(ψ⁰)²ϕ⁰⁰+ (ϕ⁰)²ϕ⁰⁰

ϕ = −ϕ⁰⁰

ϕ. 3. Let a > r > 0, and consider the parametrization

X(u, v) = ((r cos u + a) cos v, (r cos u + a) sin v, r sin u) , 0 < u < 2π, 0 < v < 2π of the torus generated by rotating S¹ = {(y, z) | (y − a)²+ z² = r²} about z-axis. Since

Xu = (−r sin u cos v, −r sin u sin v, r cos u) , X_v = (−(r cos u + a) sin v, (r cos u + a) cos v, 0) , X_uu = (−r cos u cos v, −r cos u sin v, −r sin u) , X_uv = (r sin u sin v, −r sin u cos v, 0) ,

X_vv = (−(r cos u + a) cos v, −(r cos u + a) sin v, 0) ,

we obtain X_u∧X_v = (−r cos u (r cos u + a) cos v, −r cos u (r cos u + a) sin v, −r sin u (r cos u + a)) , E = hX_u, X_ui = r², F = hX_u, X_vi = 0, G = hX_v, X_vi = (r cos u + a)²,

|X_u∧ X_v| =√

EG − F² = r(r cos u + a), and e = hN, X_uui = hX_u∧ X_v

|xu ∧ Xv|, X_uui = hX_u∧ X_v, X_uui

√EG − F² = r²(r cos u + a) r(r cos u + a) = r, f = hN, Xuvi = hX_u∧ X_v

|x_u∧ X_v|, Xuvi = hX_u∧ X_v, X_uvi

√EG − F² = 0, g = hN, X_vvi = hXu∧ Xv

|x_u∧ X_v|, X_vvi = hXu∧ Xv, Xvvi

√EG − F² = r cos u (r cos u + a)²

r(r cos u + a) = cos u(r cos u + a), and K = eg − f²

EG − F² = cos u

r(r cos u + a). Note that K = 0 when u = π/2 or u = 3π/2, the points of such parallels are parabolic points; K < 0 when π/2 < u < 3π/2, the points in this region are hyperbolic points; and K > 0 when 0 < u < π/2, or 3π/2 < u < 2π, the points in this region are elliptic points.

Proposition Let p ∈ S be an elliptic point of a surface S. Then there exists a neighborhood V of p in S such that all points in V belong to the same side of the tangent plane T_pS. Let p ∈ S be a hyperbolic point. Then in each neighborhood of p there exist points of S in both sides of T_pS.

Proof Let X(u, v) be a parametrization of S at p, with X(0, 0) = p, and let d : X(U ) → R be a function defined by

d(q) = hX(u, v) − X(0, 0), N (p)i, for q = X(u, v) ∈ X(U ).

Since X(u, v) is differentiable and by the Taylor’s formula, we have X(u, v) = X(0, 0) + X_u(0, 0)u + X_v(0, 0)v + 1

2(X_uu(0, 0)u²+ 2X_uv(0, 0)uv + X_vv(0, 0)v²) + ¯R, where the remainder ¯R satisfies that

lim

(u,v)→(0,0)

R¯

u²+ v² = 0.

It follows that hX_u(0, 0), N (p)i = hX_v(0, 0), N (p)i = 0 and d(q) = hX(u, v) − X(0, 0), N (p)i

= 1

2hX_uu(0, 0), N (p)iu²+ 2hX_uv(0, 0), N (p)iuv + hX_vv(0, 0), N (p)iv² + h ¯R, N (p)i

= 1

2eu² + 2f uv + gv² + h ¯R, N (p)i

= 1

2IIp(w) + h ¯R, N (p)i,

where w = X_u(0, 0)u + X_v(0, 0)v ∈ T_pS and lim

(u,v)→(0,0)

h ¯R, N (p)i u² + v² = 0.

For an elliptic point p, K(p) > 0, so the principal curvatures k₁, k₂ have the same sign and thus II_p(w) = k_n has a fixed sign for all w ∈ T_pS satisfying |w| = 1 (by the Euler formula). Therefore, for all (u, v) sufficiently near p, d has the same sign as IIp(w); that is, all such (u, v) belong to the same side of T_pS.

For a hyperbolic point p, K(p) < 0, so the principal curvatures k₁, k₂ have the opposite signs, and in each neighborhood of p there exist points (u, v) and (¯u, ¯v) such that II_p(w/|w|) = k₁

and II_p( ¯w/| ¯w|) = k₂ have opposite signs (here w = X_u(0, 0)u + X_v(0, 0)v and ¯w = X_u(0, 0)¯u + X_v(0, 0)¯v are principal directions); such points belong therefore to distinct sides of T_pS.

Proposition Let p be a point of a surface S such that the Gaussian curvature K(p) 6= 0, and let V be a connected neighborhood of p where K does not change sign. Then

K(p) = lim

A→0

A⁰ A, where

• A is the area of a a region B ⊂ V containing p,

• A⁰ is the area of the region N (B) in S²,

and the limit is taken through a sequence of regions B_n that converges to p, in the sense that any sphere around p conatins all B_n, for n sufficiently large.

Proof Suppose K > 0 in V. Let X : U → S be a parametrization of S at p such that V ⊂ X(U ) and let B = X(R). Since

A = Z Z

R

|X_u∧ X_v| du dv, and A⁰ = Z Z

R

|N_u ∧ N_v| du dv = Z Z

R

K|X_u∧ X_v| du dv, we have

A→0lim A⁰

A = lim

A→0

A⁰/A(R) A/A(R) =

lim

A(R)→0

1 A(R)

Z Z

R

K|X_u∧ X_v| du dv lim

A(R)→0

1 A(R)

Z Z

R

|X_u∧ X_v| du dv

= K|Xu∧ Xv|

|x_u∧ X_v| = K(p).

Remark In the proof, we have used the following Theorems from Advanced Calculus.

• Change of Variables Theorem Let F : U → V be a diffeomorphism between open subsets of U, V ⊂ Rⁿ, let D^∗ ⊂ U and D = F (D^∗) ⊂ V be bounded subsets, and let f : D → R be a bounded function. Then

Z

D

f (y1, . . . , yn) dy1· · · dyn = Z

D^∗

f (F (x1, . . . , xn)) | det DF (x1, . . . , xn)| dx1· · · dxn

= Z

D^∗

f (F (x₁, . . . , x_n))    

∂(y1, . . . , yn)

∂(x₁, . . . , x_n)    

dx₁· · · dx_n.

• Theorem Let f : B_r(p) → R be a function defined on the ball Br(p) ⊂ Rⁿ of radius r and center p. If f is continuous at p, then

ρ→0lim 1 V (B_ρ(p))

Z

Bρ(p)

f (x) dx = f (p), where V (B_ρ(p)) = Z

Bρ(p)

dx = the volume of B_ρ(p).

Proof Since

f (p) = f (p) · 1 V (B_ρ(p))

Z

Bρ(p)

dx = 1

V (B_ρ(p)) Z

Bρ(p)

f (p) dx and lim

x→pf (x) = f (p),

we have for any ε > 0, there is a δ > 0 such that if x ∈ B_δ(p) then |f (x) − f (p)| < ε, so for all 0 < ρ < δ, we have

ρ→0lim 1 V (B_ρ(p))

Z

Bρ(p)

f (x) dx − f (p)     

=     

ρ→0lim 1 V (B_ρ(p))

Z

Bρ(p)

[f (x) − f (p)] dx     

≤ lim

ρ→0

1 V (B_ρ(p))

Z

Bρ(p)

|f (x) − f (p)| dx

< lim

ρ→0

1 V (B_ρ(p))

Z

Bρ(p)

ε dx

= ε.

Since ε > 0 is arbitrary, we have

ρ→0lim 1 V (B_ρ(p))

Z

Bρ(p)

f (x) dx = f (p).