Regular Surfaces
Definition 1. A subset S ⊂ R3 is a regular surface if, for each p ∈ S, there exists an open neighborhood V in R3, an open set U ⊂ R2 and a map
X : U → V ∩ S
such that
(1) X is smooth, meaning that if we write
X(u, v) = (x(u, v), y(u, v), z(u, v)), (u, v) ∈ U,
then the real-valued functions x(u, v), y(u, v) and z(u, v) have continuous partial derivatives of all orders in U.
(2) X is a homeomorphism, meaning that it is a one-to-one correspondence between the points of U and V ∩ S which is continuous in both directions; that is, X−1 is the restriction of a continuous map F : W ⊂ R3 → R2 defined on an open set W containing V ∩ S.
(3) (The regularity condition)For each q = (u, v) ∈ U, the linear map
dXq =
∂x
∂u
∂x
∂v
∂y
∂u
∂y
∂v
∂z
∂u
∂z
∂v
: R2 → R3,
called the differential of X at q, is one-to-one.
The mapping X : U → V ∩ S is called a parametrizationor asystem of local coordinatesfor the surface S in the coordinate neighborhood V ∩ S of p.
Remarks Note that a surface is defined as a subset S of R3, not as a map as in the curve case. This is achieved by covering S with the traces of parameterization which satisfy the three conditions. Also note that
• Condition (1) is natural if we need to do differential calculus on S.
• Condition (2) has the purpose of preventing self-intersection in regular surfaces. It is also essential to prove that certain objects defined in terms of a parameterization do not depend on this parameterization but only on S itself.
• To give condition 3 a more familiar form, let us compute the matrix of the linear map dXq in the canonical bases e1 = (1, 0), e2 = (0, 1) of R2 with coordinates (u, v) and f1 = (1, 0, 0), f2 = (0, 1, 0), f3 = (0, 0, 1) of R3, with coordinates (x, y, z).
Let q = (u0, v0). The vector e1 is tangent to the curve u → (u, v0) whose image under X is the curve
u → (x(u, v0), y(u, v0), z(u, v0)).
This image curve (called the coordinate curve v = v0) lies on S and has at X(q) the tangent vector
∂x
∂u,∂y
∂u,∂z
∂u
= ∂X
∂u,
where the derivatives are computed at (u0, v0) and a vector is indicated by its components in the basis {f1, f2, f3}. By the definition of differential
dXq(e1) = d
duX(u, v0)|u=u0 = ∂x
∂u,∂y
∂u,∂z
∂u
= ∂X
∂u.
Similarly, using the coordinate curve u = u0 (image by X of the curve v → (u0, v)), we obtain
dXq(e2) = d
dvX(u0, v)|v=v0 = ∂x
∂v,∂y
∂v,∂z
∂v
= ∂X
∂v .
Condition (3) may now be expressed by requiring the two column vectors of this matrix to be linearly independent; or, equivalently, that the vector product ∂X
∂u ∧ ∂X
∂v 6= 0; or, in still another way, that one of the minors of order 2 of the matrix of dXq, that is, one of the Jacobian determinants
∂(x, y)
∂(u, v) =
∂x
∂u
∂x
∂v
∂y
∂u
∂y
∂v
, ∂(y, z)
∂(u, v), ∂(x, z)
∂(u, v),
is different from zero at q. So, Condition (3) will gurantee the existence of a tangent plane at all points of S.
Example Show that the unit sphere S2 = {(x, y, z) ∈ R3; x2+ y2+ z2 = 1} is a regular surface.
• Let U = {(x, y) ∈ R2; x2 + y2 < 1}, R2 = {(x, y, z) ∈ R3; z = 0} and the maps X1, X2 : U ⊂ R2 → R3 be defined by
X1(x, y) = (x, y, +p
1 − x2− y2), X2(x, y) = (x, y,p
1 − x2− y2), (x, y) ∈ U.
• Let U = {(x, z) ∈ R2; x2+ z2 < 1}, R2 = {(x, y, z) ∈ R3; y = 0} and the maps X3, X4 : U ⊂ R2 → R3 be defined by
X3(x, z) = (x, +√
1 − x2− z2, z), X4(x, y) = (x, −√
1 − x2− z2, z), (x, z) ∈ U.
• Let U = {(y, z) ∈ R2; y2 + z2 < 1}, R2 = {(x, y, z) ∈ R3; x = 0} and the maps X5, X6 : U ⊂ R2 → R3 be defined by
X5(y, z) = (+√
1 − x2− z2, y, z), X6(y, z) = (−√
1 − x2 − z2, y, z), (y, z) ∈ U.
Since {Xi : U → S2 | 1 ≤ i ≤ 6} are parametrizations covering S2 completely, S2 is a regular surface.
For most applications, it is convenient to relate parametrizations to the geographical coordinates on S2. Let V = {(θ, φ); 0 < θ < π, 0 < φ < 2π} and let X : V → R3 be given by
X(θ, φ) = (sin θ cos φ, sin θ sin φ, cos θ).
Clearly, X(V ) ⊂ S2 and X is a parametrization of S2, where θ is usually called the colatitude (the complement of the latitude) and φ the longitude. Note that X(V ) only omits a semicircle of S2 (including the two poles) and that S2 can be covered with the coordinate neighborhoods of two parametrizations of this type.
Example shows that deciding whether a given subset of R3 is a regular surface directly from the definition may be quite tiresome. Before going into further examples, we shall present two propositions which will simplify this task. Proposition 1 shows the relation which exists between the definition of a regular surface and the graph of a function z = f (x, y). Proposition 2 uses the inverse function theorem and relates the definition of a regular surface with the subsets of the form f (x, y, z) = constant.
Proposition 1. If f : U → R is a differentiable function in an open set U of R2, then the graph of f over U , that is, the subset of R3 given by
S = the graph of a differentiable function f on an open subset U ⊂ R2
= {(x, y, f (x, y)) | (x, y) ∈ U ⊂ R2} is a regular surface.
Proof It suffices to show that the map X : U → R3 given by X(u, v) = (u, v, f (u, v)) for (u, v) ∈ U
is a parametrization of the graph whose coordinate neighborhood covers every point of the graph.
Definition 2. Given a differentiable map F : U ⊂ Rn → Rm defined in an open set U of Rn, we say that p ∈ U is a critical point of F if the differential dFp : Rn → Rm is not a surjective (or onto) mapping. The image F (p) ∈ Rm of a critical point is called a critical value of F. A point of Rm which is not a critical value is called aregular value of F.
Remark Recall that
Implicit Function Theorem
Let U ⊆ Rn+m≡ Rn× Rm be an open set,
f = (f1, . . . , fm) : U → Rm belong to Class C1(U ),
(a, b) = (a1, . . . , an, b1, . . . , bm) ∈ U be a point at which f (a, b) = 0 ∈ Rm, and the m × m matrix Dyf |(a,b)= ∂fi
∂yj(a, b)
1≤i,j≤m
be invertible.
Then there exists
an open neighborhood A of a in Rn, an open neighborhood B of b in Rm,
and a unique g : A → B belonging to Class C1(A) such that
b = g(a) and f (x, g(x)) = 0 ∈ Rm for all x ∈ A and thus
f−1(0) ∩ A × B = {(x, y) ∈ A × B ⊂ U ⊆ Rn× Rm | f (x, y) = 0 ∈ Rm}
= {(x, g(x)) | x ∈ A}
= the graph of g over A.
Example (2-2 Exercises #17) Prove that
(a) The inverse image of a regular value of a differentiable function f : U ⊂ R2 → R
is a regular plane curve. Give an example of such a curve which is not connected.
Proof Let r be a regular value of f. For each
p = (p1, p2) ∈ f−1(r) ∩ U = {(x, y) ∈ U ⊂ R2 | f (x, y) = r} ⊂ R2,
since dfp = (fx(p), fy(p)) : R2 → R is surjective, (fx(p), fy(p)) 6= (0, 0). By interchanging x and y if necessary, we may assume that fy(p) 6= 0, thus, by the Implicit Function Theorem, there exists an open set
I1× I2 = (a1, b1) × (a2, b2) containg p = (p1, p2) and a unique continuously differentiable function g : I1 → I2 such that
f (x, g(x)) = r for all x ∈ I1 =⇒ f−1(r) ∩ I1× I2 = {(x, g(x)) | x ∈ I1} ⊂ R2
is a regular plane over I1 = (a1, b1) since g is differentiable and the tangent vector (1, g0(x)) 6=
(0, 0) at each (x, g(x)), x ∈ I1. Since p is an arbitrary point in f−1(r) ∩ U, f−1(r) is a regular plane curve.
Example Let U = {(x, y) | x 6= 0} and f (x, y) = xy for (x, y) ∈ U. Then f−1(1) = {(x,1 x) | x > 0} ∪ {(x,1
x) | x < 0}= disjoint union of two regular plane curves.
(b) The inverse image of a regular value of a differentiable map F = (F1, F2) : U ⊂ R3 → R2 is a regular curve in R3.
Proof Let r = (r1, r2) be a regular value of F = (F1, F2). For each
p = (p1, p2, p3) ∈ F−1(r) ∩ U = {(x, y, z) ∈ U ⊂ R3 | F (x, y, z) = r} ⊂ R3, since
dFp =(F1)x(p) (F1)y(p) (F1)z(p) (F2)x(p) (F2)y(p) (F2)z(p)
: R3 → R2 is surjective,
the matrix (F1)x(p) (F1)y(p) (F1)z(p) (F2)x(p) (F2)y(p) (F2)z(p)
is of rank 2. By interchanging x, y or z if nec- essary, we may assume that
(F1)y(p) (F1)z(p) (F2)y(p) (F2)z(p)
is nonsingular, (F1)z(p) 6= 0, (F2)y(p) 6= 0,
thus, by the Implicit Function Theorem, there exist an open neighborhood I1 × I2× I3 = (a1, b1) × (a2, b2) × (a3, b3) of (p1, p2, p3) and unique continuously differentiable functions g = (g1, g2) : I1 → I2× I2, h : I1× I2 → I3, k : I1× I3 → I2 such that
F (x, g(x)) = r for all x ∈ I1 =⇒ F−1(r) ∩ I1× I2× I3 = {(x, g1(x), g2(x)) | x ∈ I1} ⊂ R3 is a regular curve in R3
F1(x, y, h(x, y)) = r1 for all (x, y) ∈ I1× I2
=⇒ (F1)−1(r1) ∩ I1× I2× I3 = {(x, y, h(x, y)) | (x, y) ∈ I1× I2} ⊂ R3 is a graph of a differentiable function h over I1× I2
=⇒ (F1)−1(r1) ∩ I1× I2× I3 is a regular surface in R3
F2(x, k(x, z), z) = r2 for all (x, z) ∈ I1× I3
=⇒ (F2)−1(r2) ∩ I1× I2× I3 = {(x, k(x, z).z) | (x, z) ∈ I1× I3} ⊂ R3 is a graph of a differentiable function k over I1× I3
=⇒ (F2)−1(r2) ∩ I1× I2× I3 is a regular surface in R3
Since p is an arbitrary point in F−1(r) ∩ U, these imply that F−1(r) is a regular curve, (F1)−1(r1) and (F2)−1(r2) are regular surfaces in R3 and F−1(r) = (F1)−1(r1) ∩ (F2)−1(r2) Proposition 2. If f : U ⊂ R3 → R is a differentiable function and a ∈ f(U) is a regular value of f, then the level set of a regular value f−1(a) = {(x, y, z) ∈ U | f (x, y, z) = a} is a regular surface in R3.
Proof Let p = (x0, y0, z0) be a point of f−1(a). Since a is a regular value of f, it is possible to assume, by renaming the axes if necessary, that fz(p) 6= 0. We define a mapping F : U ⊂ R3 → R3 byF (x, y, z) = (x, y, f (x, y, z)) for (x, y, z) ∈ U. Since
det(dFp) =
1 0 0
0 1 0
fx(p) fy(p) fz(p)
= fz(p) 6= 0,
and by the inverse funtion theorem, there exist open neighborhoods V of p and W of F (p) such that F : V → W is invertible and the inverse F−1 : W → V is differentiable.
Let F−1 be defined by F−1(u, v, t) = (g1(u, v, t), g2(u, v, t), g2(u, v, t)) for (u, v, t) ∈ W. Since (u, v, t) = F ◦ F−1(u, v, t) = F (g1, g2, g3) = (g1, g2, f (g1, g2, g3)) for (u, v, t) ∈ W, we have g1(u, v, t) = u, g2(u, v, t) = v and
(x, y, z) = F−1(u, v, t) = (u, v, g3(u, v, t)) for (u, v, t) ∈ W, (x, y, z) ∈ V.
This implies that π(V ) ∼= π(W ) and z = g3(u, v, a) = h(u, v) is a differentiable function for (u, v) ∈ π(W ), where π : R3 → R2 is the projection map defined by π(u, v, t) = (u, v) for (u, v, t) ∈ R3. Since
F (f−1(a) ∩ V ) = W ∩ {(u, v, t) | t = a}
=⇒ f−1(a) ∩ V = F−1(W ∩ {(u, v, t) | t = a}) = {(u, v, g3(u, v, a)) | (u, v) ∈ π(W )}
π(W )∼=π(V )
=⇒
g3(x,y,a)=h(x,y) f−1(a) ∩ V = {(x, y, h(x, y)) | (x, y) ∈ π(V )},
we conclude thatf−1(a) ∩ V is the graph of h over π(V ). By Prop. 1,f−1(a) ∩ V is a coordinate neighborhood of p. Therefore, every p ∈ f−1(a) can be covered by a coordinate neighborhood, and so f−1(a) is a regular surface.
Example The ellipsoid
x2 a2 +y2
b2 + z2 c2 = 1 is a regular surface. In fact, it is the set f−1(0) where
f (x, y, z) = x2 a2 + y2
b2 +z2 c2 − 1
is a differentiable function and 0 is a regular value of f. This follows from the fact that the partial derivatives fx = 2x/a2, fy = 2y/b2, fz = 2z/c2 vanish simultaneously only at the point (0, 0, 0), which does not belong to f−1(0). This example includes the sphere as a particular case (a = b = c = 1).
Example The hyperboloid of two sheets −x2− y2+ z2 = 1 is a regular surface, since it is given by S = f−1(0), where 0 is a regular value of f (x, y, z) = −x2− y2+ z2− 1. Note that the surface S is not connected; that is, given two points in two distinct sheets (z > 0 and z < 0) it is not possible to join them by a continuous curve α(t) = (x(t), y(t), z(t)) contained in the surface;
otherwise, z changes sign and, for some t0, we have z(t0) = 0, which means that α(t0) ∈ S.
Example Let a > r > 0, S1 = {(y, z) | (y − a)2+ z2 = r2} and T be the surface, called torus, generated by rotating S1 about z-axis. Hence the points (x, y, z) of T satisfy the equation
z2 = r2− (p
x2 + y2 − a)2. Therefore, T is the inverse image of r2 by the function
f (x, y, z) = z2+ (p
x2+ y2− a)2.
Note that f is differentiable for (x, y) 6= (0, 0), and r2 is a regular value of f. It follows that the torus T is a regular surface.
Proposition 3. Let S ⊂ R3 be a regular surface and p ∈ S. Then there exists an open neighborhood V of p in S such that V is the graph of a differentiable function which has one of the following three forms:
z = f (x, y), y = g(x, z), x = h(y, z).
That is,regular surface is locally a graph of a differentiable function.
Proof Let X : U ⊂ R2 → R3be a parametrization of S at p, and write X(u, v) = (x(u, v), y(u, v), z(u, v)), (u, v) ∈ U. By condition 3 of Def. 1, one of the Jacobian determinants
∂(x, y)
∂(u, v)
∂(y, z)
∂(u, v)
∂(z, x)
∂(u, v)
is not zero at X−1(p) = q.
Suppose first that ∂(x, y)
∂(u, v)(q) 6= 0, and consider the map π ◦ X : U → R2, where π(x, y, z) = (x, y) for (x, y, z) ∈ R3, defined by
π ◦ X(u, v) = (x(u, v), y(u, v)), for (u, v) ∈ U.
Since det d(π ◦ X)(q) = ∂(x, y)
∂(u, v)(q) 6= 0, by the inverse function theorem, there exist open neigh- borhoods V1 of q, V2 of π ◦ X(q) such that
• π ◦ X : V1 → V2 is one-to-one, onto and has a differentiable inverse (π ◦ X)−1 : V2 → V1 defined by
(π ◦ X)−1(x, y) = (u(x, y), v(x, y)) for (x, y) ∈ V2. It follows that
• the projection map π : X(V1) = V ⊂ S → V2 is one-to-one on V,
=⇒ V is a graph of z = z ◦ (π ◦ X)−1(x, y) on V2.
In fact, since X is a homeomorphism, V = X(V1) = (X−1)−1(V1) is an open neighborhood of p in S and since
• z = z(u, v) is differentiable for (u, v) ∈ V1,
• (u(x, y), v(x, y)) = (π ◦ X)−1(x, y) is differentiable for (x, y) ∈ V2,
=⇒ z = z(u(x, y), v(x, y)) = z ◦ (π ◦ X)−1(x, y) is differentiable in ∈ V2, and
V = {X(u, v) = (x(u, v), y(u, v), z(u, v)) | (u, v) ∈ V1}
= {(x, y, z) | z = z(u(x, y), v(x, y)) = z ◦ (π ◦ X)−1(x, y), (x, y) ∈ V2},
V is the graph of the differentiable function z = z(u(x, y), v(x, y)) = f (x, y) over V2, and this settles the first case.
The remaining cases can be treated in the same way, yielding x = h(y, z) and y = g(x, z).
Proposition 4. Let p ∈ S be a point of a regular surface S and let X : U ⊂ R2 → R3 be a map with p ∈ X(U ) ⊂ S such that conditions (1) and (3) of Def. 1 hold. Assume that X is one-to-one. Then X−1 is continuous.
Proof Let X : U ⊂ R2 → S ⊂ R3 be defined by
X(u, v) = (x(u, v), y(u, v), z(u, v)), for (u, v) ∈ U, and let p = X(q) for some q ∈ U.
By conditions 1 and 3, we can assume, by interchanging the coordinate axis if necessary, that
∂(x, y)
∂(u, v)(q) 6= 0.
By the inverse function theorem, there exist open neighborhoods V1 of q, V2 of π ◦ X(q) such that π ◦ X maps V1 diffeomorphically onto V2, where π : R3 → R2 is the projection defined by π(x, y, z) = (x, y) for (x, y, z) ∈ R3. It follows that (π ◦ X)−1 : V2 → V1 and π : R3 → R2 are continuous maps.
Assume now that X is one-to-one. Then X : V1 → V = X(V1) ⊂ S has an inverse X−1 : V → V1
and, sinceX−1 = (π ◦ X)−1◦ π : V → V1 is a composition of continuous maps, X−1 is continuous.
Examples
1. The one-sheetd cone C, given by z = +p
x2+ y2, (x, y) ∈ R2, is not a regular surface.
If C were a regular surface, it would be, in a neighborhood of (0, 0, 0) ∈ C, the graph of a differentiable function having one of three forms: y = h(x, z), x = g(y, z), z = f (x, y).
The two first forms can be discarded by the simple fact that the projections of C over the xz and yz planes are not one-to-one. The last form would have to agree, in a neighborhood of (0, 0, 0), with z = +p
x2+ y2. Since z = +p
x2+ y2 is not differentiable at (0, 0), this is impossible.
2. Let a > r > 0, S1 = {(y, z) | (y − a)2 + z2 = r2} and T be the torus generated by rotating S1 about z-axis. Then
X(u, v) = ((r cos u + a) cos v, (r cos u + a) sin v, r sin u) where 0 < u < 2π, 0 < v < 2π is a parametrization for the torus T.
Condition 1 of Def. 1 is easily checked, and condition 3 reduces to a straightforward computation, which is left as an exercise. Since we know that T is a regular surface, condition 2 is equivalent, by Prop. 4, to the fact that X is one-to-one.
To prove that X is one-to-one, we first observe that sin u = z/r also, ifp
x2+ y2 ≤ a, then π/2 ≤ u ≤ 3π/2, and if p
x2+ y2 ≥ a, then either 0 < u ≤ π/2 or 3π/2 ≤ u < 2π. Thus, given (x, y, z), this determines u, 0 < u < 2π, uniquely. By knowing u, x, and y we find cos v and sin v. This determines v uniquely, 0 < v < 2π. Thus, X is one-to-one.
It is easy to see that the torus can be covered by three such coordinate neighborhoods.
Change of Parameters; Differentiable Functions on Surface
Proposition 1 (Change of Parameters). Let p be a point of a regular surface S, and let X : U ⊂ R2 → S, Y : V ⊂ R2 → S be two parametrizations of S such that p ∈ X(U )∩Y (V ) = W.
Then the “change of coordinates” h = X−1◦ Y : Y−1(W ) → X−1(W ) is a diffeomorphism; that is, h is differentiable and has a differentiable inverse h−1.
In other words, if X and Y are given by
X(u, v) = (x(u, v), y(u, v), z(u, v)), (u, v) ∈ U Y (ξ, η) = (x(ξ, η), y(ξ, η), z(ξ, η)), (ξ, η) ∈ V then the change of coordinates h, given by
u = u(ξ, η), v = v(ξ, η), (ξ, η) ∈ Y−1(W ),
has the property that the functions u and v have continuous partial derivatives of all orders, and the map h can be inverted, yielding
ξ = ξ(u, v), η = η(u, v) (u, v) ∈ X−1(W ), where the functions ξ and η also have partial derivatives of all orders. Since
∂(u, v)
∂(ξ, η) · ∂(ξ, η)
∂(u, v) = 1,
this implies that the Jacobian determinants of both h and h−1 are nonzero everywhere.
Proof h = X−1 ◦ Y is a homeomorphism, since it is composed of homeomorphisms. It is not possible to conclude, by an analogous argument, that h is differentiable, since X−1 is defined in an open subset of S, and we do not yet know what is meant by a differentiable function on S.
We proceed in the following way. Let r ∈ Y−1(W ) and set q = h(r) ∈ X−1(W ).
Since X(u, v) = (x(u, v), y(u, v), z(u, v)) is a parametrization, we can assume, by renaming the axes if necessary, that
∂(x, y)
∂(u, v)(q) 6= 0.
We extend X to a map F : U × R → R3 defined by
F (u, v, t) = (x(u, v), y(u, v), z(u, v) + t), (u, v) ∈ U, t ∈ R.
Geometrically, F maps a vertical cylinder C over U into a “vertical cylinder” over X(U ) by mapping each section of C with height t into the surface X(u, v) + te3, where e3 is the unit vector of the z axis.
It is clear that F is differentiableand that the restriction F |U ×{0}= X. Since
det dFq =
∂x
∂u
∂x
∂v 0
∂y
∂u
∂y
∂v 0
∂z
∂u
∂z
∂v 1
= ∂(x, y)
∂(u, v)(q) 6= 0,
we can apply the inverse function theorem to find an open neighborhood M of p = X(q) in R3 such that F−1 exists and is differentiable in M.
By the continuity of Y, there exists an open neighborhood N = Y−1(M ∩ Y (V )) ⊂ V of r in V such that Y (N ) ⊂ M ∩ S. SinceX−1|Y (N ) = F−1|Y (N ), and restricted to N,
h|N = X−1◦ Y |N = F−1◦ Y |N is a composition of differentiable maps
=⇒ h is differentiable at r ∈ Y−1(W ).
=⇒ Since r is arbitrary, h is differentiable on Y−1(W ).
Exactly the same argument can be applied to show that the map h−1 is differentiable, and so h is a diffeomorphism.
Definition Let f : V ⊂ S → R be a function defined in an open subset V of a regular surface S. Then f is said to be differentiable at p ∈ V if, for some parametrization X : U ⊂ R2 → S with p ∈ X(U ) ⊂ V, the composition f ◦ X : U ⊂ R2 → R is differentiable at X−1(p). f is differentiable in V if it is differentiable at all points of V.
It follows immediately from the last proposition that the definition given does not depend on the choice of the parametrizationX. In fact, if Y : V ⊂ R2 → S is another parametrization with p ∈ Y (V ), and if h = X−1◦ Y, thenf ◦ Y = f ◦ X ◦ h is also differentiable, whence the asserted independence.
Remark 1 We shall frequently make the notational abuse of indicating f and f ◦ X by the same symbol f (u, v), and say that f (u, v) is the expression of f in the system of coordinates X. This is equivalent to identifying X(U ) with U and thinking of (u, v), indifferently, as a point of U and as a point of X(U ) with coordinates (u, v). From now on, abuses of language of this type will be used without further comment.
Example 1 Let S be a regular surface and V ⊂ R3 be an open set such that S ⊂ V . Let f : V ⊂ R3 → R be a differentiable function. Then the restriction of f to S is a differentiable function on S. In fact, for any p ∈ S and any parametrization X : U ⊂ R2 → S in p, the function f ◦ X : U → R is differentiable. In particular, the following are differentiable functions:
1. The height function relative to a unit vector v ∈ R3, h : S → R, given by h(p) = p · v, p ∈ S, where the dot denotes the usual inner product in R3. h(p) is the height of p ∈ S relative to a plane normal to v and passing through the origin of R3.
2. The square of the distance from a fixed point p0 ∈ R3, f (p) = |p − p0|2, p ∈ S. The need for taking the square comes from the fact that the distance |p − p0| is not differentiable at p = p0.
Remark 2 The proof of Prop. 1 makes essential use of the fact that the inverse of a parametriza- tion is continuous. Since we need Prop. 1 to be able to define differentiable functions on surfaces (a vital concept), we cannot dispose of this condition in the definition of a regular surface.
The definition of differentiability can be easily extended to mappings between surfaces.
Definition A continuous map ϕ : V1 ⊂ S1 → S2 of an open set V1 of a regular surface S1 to a regular surface S2 is said to be differentiable at p ∈ V1 if, given parametrizations
X1 : U1 ⊂ R2 → S1, X2 : U2 ⊂ R2 → S2 with p ∈ X1(U1) and ϕ(X1(U1)) ⊂ X2(U2), the map
X2−1◦ ϕ ◦ X1 : U1 → U2 is differentiable at q = X1−1(p).
In other words, ϕ is differentiableif when expressed in local coordinates as ϕ(u1, v1) = (ϕ1(u1, v1), ϕ2(u1, v1))
the functions ϕ1 and ϕ2 have continuous partial derivatives of all orders.
Note that one can use Prop. 1 to show that this definition of differentiability of ϕ : S1 → S2 does not depend on the choice of parametrizations.
We should mention that the natural notion of equivalence associated with differentiability is the notion of diffeomorphism.
Definition Two regular surfaces S1 and S2 are diffeomorphicif there exists a differentiable map ϕ : S1 → S2 with a differentiable inverse ϕ−1 : S2 → S1. Such a ϕ is called a diffeomorphism from S1 to S2.
Note that, if ϕ : S1 → S2 is a diffeomorphism from S1 to S2, then f : S2 → R is differentiable on S2 if and only if f ◦ ϕ : S1 → R is differentiable on S1, i.e. two diffeomorphic surfaces are indistinguishable from the point of view of differentiability.
Example 2 If X : U ⊂ R2 → S is a parametrization, then X−1 : X(U ) → U ⊂ R2 is differentiable.
In fact, for any p ∈ X(U ) and any parametrization Y : V ⊂ R2 → S in p, we have that X−1◦ Y : Y−1(X(U ) ∩ Y (V )) → X−1(X(U ) ∩ Y (V )) is differentiable.
This shows thatU and X(U ) are diffeomorphic(i.e., every regular surface is locally diffeomorphic to a plane) and justifies the identification made in Remark 1.
Example 3 Let S1 and S2 be regular surfaces. Assume that S1 ⊂ V ⊂ R3, where V is an open set of R3, and that ϕ : V → R3 is a differentiable map such that ϕ(S1) ⊂ S2. Then the restriction ϕ|S1 : S1 → S2 is a differentiable map.
In fact, given p ∈ S1 and parametrizations
X1 : U1 ⊂ R2 → S1, X2 : U2 ⊂ R2 → S2 with p ∈ X1(U1) and ϕ(X1(U1)) ⊂ X2(U2), we have that the map
X2−1◦ ϕ ◦ X1 : U1 → U2 is differentiable.
The following are particular cases of this general example:
1. Let S be symmetric relative to the xy plane; that is, if (x, y, z) ∈ S, then also (x, y, −z) ∈ S.
Then the (antipodal) map σ : S → S, which takes p ∈ S into its symmetrical point, is differentiable, since it is the restriction to S of the differentiable map σ : R3 → R3 defined by
σ(x, y, z) = (x, y, z) for (x, y, z) ∈ R3.
This, of course, generalizes to surfaces symmetric relative to any plane of R3.
2. Let Rz,θ : R3 → R3 be the rotation of angle θ about the z axis, and let S ⊂ R3 be a regular surface invariant by this rotation; i.e., {Rz,θ(p) | p ∈ S} ⊆ S. Then the restriction Rz,θ : S → S is a differentiable map.
3. Let ϕ : R3 → R3 be given by
ϕ(x, y, z) = (xa, yb, zc), where a, b and c are nonzero real numbers.
Then ϕ is clearly differentiable, and the restriction ϕ|S2 is a differentiable map from the sphere
S2 = {(x, y, z) ∈ R3 | x2+ y2+ z2 = 1}
into the ellipsoid
{(x, y, z) ∈ R3 | x2 a2 +y2
b2 + z2 c2 = 1}.
Remark 3 Proposition 1 implies (cf. Example 2) that a parametrization X : U ⊂ R2 → S is a diffeomorphism of U onto X(U ). Actually, we can now characterize the regular surfaces as those subsets S ⊂ R3 which are locally diffeomorphic to R2; that is, for each point p ∈ S, there exists a neighborhood V of p in S, an open set U ⊂ R2, and a map X : U → V, which is a
diffeomorphism. This pretty characterization could be taken as the starting point of a treatment of surfaces (see Exercise 13).
Definition A parametrized surface X : U ⊂ R2 → R3 is a differentiable map X from an open set U ⊂ R2 into R3. The set X(U ) ⊂ R3 is called the trace of X. X is regular if the differential dXq : R2 → R3 is one-to-one for all q ∈ U (i.e., the vectors ∂X/∂u, ∂X/∂v are linearly independent for all q ∈ U ). A point p ∈ U where dXp is not one-to-one is called a singular point of X.
Observe that a parametrized surface, even when regular, may have self-intersections in its trace since X : U ⊂ R2 → R3 is not necessary a homeomorphism (or a global one-to-one map).
Example Let α : I → R3 be a nonplanar regular parametrized curve. Define X(t, v) = α(t) + vα0(t), (t, v) ∈ I × R.
X is a parametrized surface called the tangent surface of α.
Suppose that the curvature k(t) 6= 0, for all t ∈ I, and restrict the domain of X to U = {(t, v) | (t, v) ∈ I × R; v 6= 0}.
Since k(t) = |α00(t) ∧ α0(t)|
|α0(t)|3 6= 0, ∂X
∂t = α0(t) + vα00(t) and ∂X
∂v = α0(t), we have
∂X
∂t ∧ ∂X
∂v = vα00(t) ∧ α0(t) 6= 0 for all (t, v) ∈ U.
It follows that the restriction X : U → R3 is a regular parametrized surface, the trace of which consists of two connected pieces whose common boundary is the set α(I).
Proposition Let X : U ⊂ R2 → R3 be a regular parametrized surface and let q ∈ U. Then there exists a neighborhood V of q in R2 such that X(V ) ⊂ R3 is a regular surface.
Proof Let X : U ⊂ R2 → R3 be defined by
X(u, v) = (x(u, v), y(u, v), z(u, v)), (u, v) ∈ U.
By regularity, we can assume that ∂(x, y)
∂(u, v)(q) 6= 0. Define a map F : U × R → R3 by F (u, v, t) = X(u, v) + t(0, 0, 1) = (x(u, v), y(u, v), z(u, v) + t), (u, v) ∈ U, t ∈ R.
Then
det(dF(q,0)) = ∂(x, y)
∂(u, v)(q) 6= 0.
By the inverse function theorem, there exist open neighborhoods W1 of (q, 0) and W2 of F (q, 0) such that F : W1 → W2 is a diffeomorphism. Set V = W1 ∩ U ⊂ R2 and observe that the restriction F |V = X|V. Thus, X(V ) = F (V ) is diffeomorphic to V, and hence a regular surface.
The Tangent Plane; The Differential of a Map
Definition Let S be a regular surface and p be a point in S. Then a vector w is calleda tangent vector to S at the point pif there exists a differentiable parametrized curve α : (−ε, ε) → S such that α(0) = p and α0(0) = w. The set of tangent vectors to S at p, denoted byTpS, is called the tangent plane to S at p.
Proposition 1. Let X : U ⊂ R2 → S be a parametrization of a regular surface S and let q ∈ U.
The vector subspace of dimension 2,
dXq(R2) ⊂ R3,
coincides with the set of tangent vectors to X at X(q), i.e. dXq(R2) = TX(q)S.
Proof Let w ∈ TX(q)S be a tangent vector at X(q), that is, let w = α0(0), where α : (−ε, ε) → X(U ) ⊂ S is differentiable and α(0) = X(q). Since U and X(U ) are diffeomorphic, X−1 : X(U ) → U is differentiable and the curve β = X−1 ◦ α : (−ε, ε) → U is differentiable. This implies that X ◦ β(t) = α(t) for t ∈ (−ε, ε) and
d
dtX ◦ β(t)|t=0 = d
dtα(t)|t=0 =⇒ dXq(β0(0)) = dXβ(0)(β0(0)) = α0(0) = w.
Hence, w ∈ dXq(R2) =⇒ TX(q)S ⊆ dXq(R2).
On the other hand, letw = dXq(v),where v ∈ R2. It is clear that v is the velocity vector of the curve γ : (−ε, ε) → U given by
γ(t) = tv + q, t ∈ (−ε, ε).
By the definition of the differential, w = α0(0), where α = X ◦ γ. This shows thatw ∈ TX(q)S is a tangent vector and dXq(R2) ⊆ TX(q)S.
By the above proposition, the planedXq(R2),which passes through X(q) = p,does not depend on the parametrization X.This plane will be called the tangent plane to S at p and will be denoted by TpS. The choice of the parametrization X determines a basis {(∂X/∂u)(q), (∂X/∂v)(q)} of TpS, called the basis associated to X. Sometimes it is convenient to write ∂X/∂u = Xu and
∂X/∂v = Xv.
The coordinates of a vector w ∈ TpS in the basis associated to a parametrization X are deter- mined as follows. Let w = α0(0) for some α(t) = X(u(t), v(t)) with (u(0), v(0)) = q = X−1(p).
Thus,
w= α0(0) = d
dtX(u(t), v(t))|t=0= Xu(q)u0(0) + Xv(q)v0(0).
That is, in the basis {Xu(q), Xv(q)} of TpS, w has coordinates (u0(0), v0(0)), where (u(t), v(t)) is the expression, in the parametrization X, of a curve whose velocity vector at t = 0 is w.
With the notion of a tangent plane, we can talk about the differential of a (differentiable) map between surfaces. Let S1 and S2 be two regular surfaces and let ϕ : V ⊂ S1 → S2 be a differentiable mapping of an open set V of S1 into S2. For each p ∈ V, there is a map, called the differential of ϕ at p,
dϕp : TpS1 → Tϕ(p)S2 defined as follows.
For each w ∈ TpS1, let α : (−ε, ε) → V ⊂ S1 be a differentiable parametrized curve such that α(0) = p and α0(0) = w. Since β = ϕ ◦ α : (−ε, ε) → S2 is a curve in S2 such that β(0) = ϕ(p), this implies that β0(0) ∈ Tϕ(p)S2 and it is defined to be dϕp(w), i.e. β0(0) = dϕp(w).
Note that the coordinates of a vectorβ0(0) ∈ Tϕ(p)S2 in the basis associated to a parametrization X are determined as follows. Let β(t) = ϕ ◦ α(t) = ¯¯ X(¯u(t), ¯v(t)) with (¯u(0), ¯v(0)) = r = X¯−1(ϕ(p)). Then
β0(0) = d dt
X(¯¯ u(t), ¯v(t))|t=0= ¯Xu¯(r)¯u0(0) + ¯X¯v(r)¯v0(0).
That is, in the basis { ¯Xu¯(r), ¯Xv¯(r)} of Tϕ(p)S2, β0(0) has coordinates (¯u0(0), ¯v0(0)).
Proposition 2. In the discussion above, given w, the vector β0(0) does not depend on the choice of α. The map dϕp : TpS1 → Tϕ(p)S2 defined by dϕp(w) = β0(0) is linear.
Proof Idea: Find a matrix representation of dϕp : TpS1 → Tϕ(p)S2 in the bases {Xu, Xv} of TpS1 and { ¯Xu¯, ¯Xv¯} of Tϕ(p)S2.
Let X(u, v), ¯X(¯u, ¯v), be parametrizations in neighborhoods of p = X(q) = ¯X(r) and ϕ(p), respectively, such that ϕ(X(u, v)) ∈ ¯X( ¯U ) for all (u, v) ∈ U. Let
α(t) = X(u(t), v(t)) : (−ε, ε) → S1, β(t) = ϕ ◦ α(t) = ¯X(¯u(t), ¯v(t)) such that α(0) = p and β(0) = ϕ(p).
Consider the map Φ = ¯X−1◦ ϕ ◦ X : U → ¯U given by
Φ(u, v) = (ϕ1(u, v), ϕ2(u, v)) for (u, v) ∈ U, and the curve ¯X−1◦ β = ¯X−1◦ ϕ ◦ α = ¯X−1◦ ϕ ◦ X(u(t), v(t)) defined by
(¯u(t), ¯v(t)) = Φ(u(t), v(t)) = (ϕ1(u(t), v(t)), ϕ2(u(t), v(t))) for t ∈ (−ε, ε).
Since
¯ u0(0)
¯ v0(0)
=
∂ϕ1
∂u u0(0) + ∂ϕ1
∂v v0(0)
∂ϕ2
∂u u0(0) + ∂ϕ2
∂v v0(0)
=
∂ϕ1
∂u
∂ϕ1
∂v
∂ϕ2
∂u
∂ϕ2
∂v
u0(0) v0(0)
,
and since dXq(R2) = TpS1, d ¯Xp−1(Tϕ(p)S2) = R2, and
• (u0(0), v0(0)) ∈ R2 is the coordinates of w in the basis {Xu, Xv} of TpS1,
• (¯u0(0), ¯v0(0)) ∈ R2 is the coordinates of β0(0) in the basis { ¯Xu¯, ¯X¯v} of Tϕ(p)S2,
• the matrix dΦq =∂ϕ1/∂u ∂ϕ1/∂v
∂ϕ2/∂u ∂ϕ2/∂v
depends only on Φ, i.e.
β0(0) = dϕp(w) ⇐⇒
¯ u0(0)
¯ v0(0)
coordinates of β0(0)
=
∂ϕ1
∂u
∂ϕ1
∂v
∂ϕ2
∂u
∂ϕ2
∂v
u0(0) v0(0)
coordinates of w
,
we conclude that
• β0(0) is independent of α,
• the matrix dΦq is a linear map from TpS1 in the basis {Xu, Xv} to Tϕ(p)S2 in the basis { ¯Xu¯, ¯X¯v},
• dΦq is the matrix representation of dϕp in the bases {Xu, Xv} of TpS1 and { ¯Xu¯, ¯X¯v} of Tϕ(p)S2.
Hence dϕp is a linear mapping from TpS1 into Tϕ(p)S2.
The linear mapdϕp defined by Prop. 2 is called thedifferential of ϕ at p ∈ S1. In a similar way we define the differential of a (differentiable) function f : U ⊂ S → R at p ∈ U as a linear map dfp : TpS → R.
Example Let v ∈ R3 be a unit vector and let h : S → R, h(p) = v · p, p ∈ S, be the height function defined in Example 1 of Sec. 2-3. To compute dhp(w), w ∈ TpS, choose a differentiable curve α : (−ε, ε) → S with α(0) = p, α0(0) = w. Since h(α(t)) = α(t) · v, we obtain
dhp(w) = d
dth(α(t))|t=0= α0(0) · v = w · v.
Example Let S2 ⊂ R3 be the unit sphere
S2 = {(x, y, z) ∈ R3 | x2+ y2+ z2 = 1}
and let Rz,θ : R3 → R3 be the rotation of angle θ about the z axis. Then Rz,θ restricted to S2 is a differentiable map of S2. We shall compute (dRz,θ)p(w), p ∈ S2, w ∈ TpS2. Let α : (−ε, ε) → S2 be a differentiable curve with α(0) = p, α0(0) = w. Then, since Rz,θ is linear,
(dRz,θ)p(w) = d
dt(Rz,θ◦ α(t)) |t=0 = Rz,θ(α0(0)) = Rz,θ(w).
Observe that Rz,θ leaves the north pole N = (0, 0, 1) fixed, and that (dRz,θ)N : TNS2 → TNS2 is just a rotation of angle θ in the plane TNS2.
We shall say that a mapping ϕ : U ⊂ S1 → S2 alocal diffeomorphismat p ∈ U if there exists an open neighborhood V ⊂ U of p such that ϕ restricted to V is a diffeomorphism onto an open set ϕ(V ) ⊂ S2. In these terms, the version of the inverse of function theorem for surfaces is expressed as follows.
Proposition 3. If S1 and S2 are regular surfaces and ϕ : U ⊂ S1 → S2 is a differentiable mapping of an open set U ⊂ S1 such that the differential dϕp of ϕ at p ∈ U is an isomorphism, then ϕ is a local diffeomorphism at p.
Remark By fixing a parametrization X : U ⊂ R2 → S at p ∈ S, we can make a definite choice of a unit normal vector at each point q ∈ X(U ) by the rule
N (q) = Xu∧ Xv
|Xu∧ Xv|(q).
Note that N : X(U ) → R3 is a differentiable map on X(U ) ⊂ S and it is not always possible to extend this map differentiably to the whole surface S.
The First Fundamental Form; Area
Definition Let S be a regular surface in R3. For each p ∈ S and tangent vectors w1, w2 ∈ TpS ⊂ R3, there is an (induced) inner product
h , ip : TpS × TpS → R defined by
hw1, w2ip = hw1, w2i = the inner product of w1and w2as vectors in R3.
To this inner product, which is a symmetric bilinear form, there corresponds aquadratic form Ip : TpS → R
defined by
Ip(w) = hw, wip = hw, wi = |w|2 ≥ 0 for w ∈ TpS,
and the quadratic form Ip on TpS is called the first fundamental form of the regular surface S ⊂ R3 at p ∈ S.
Remark Let U be an open set in the uv-plane and let X : U ⊂ R2 → S be a parametrization of the regular surface at p = X(u0, v0) ∈ S. For each w ∈ TpS, since there is a parametrized curve
α(t) = X(u(t), v(t)) ∈ X(U ), t ∈ (−ε, ε), such that p = α(0) = X(u0, v0) and w = α0(0), we have
Ip(w) = Ip(α0(0)) = hα0(0), α0(0)ip
= hXuu0+ Xvv0, Xuu0+ Xvv0ip
= hXu, Xuip(u0)2+ 2hXu, Xvipu0v0+ hXv, Xvip(v0)2
= E(u0)2 + 2F u0v0+ G(v0)2,
where the values of the functions involved are computed for t = 0, and E(u0, v0) = hXu, Xuip,
F (u0, v0) = hXu, Xvip, G(u0, v0) = hXv, Xvip
are the coefficients of the first fundamental form in the basis {Xu, Xv} of TpS. By letting p run in the coordinate neighborhood corresponding to X(u, v) we obtain functions E(u, v), F (u, v), G(u, v) which are differentiable in that neighborhood.
From now on we shall drop the subsript p in the indication of the inner product h , ip, or the quadratic form Ip when it is clear from the context which point we are referring to. It will also be convenient to denote the natural inner product of R3 by the same symbol h , i rather than the previous dot.
Examples
1. Let p0 = (x0, y0, z0) be a point in R3, w1 = (a1, a2, a3) and w2 = (b1, b2, b3) be orthonormal vectors in R3 and
P = {X(u, v) = p0+ uw1+ vw2 | (u, v) ∈ R2}.
Then P is a plane and E = 1, F = 0 and G = 1 at every point in P.
2. Let U = {(u, v) | 0 < u < 2π, −∞ < v < ∞} and
X(u, v) = (cos u, sin u, v), for (u, v) ∈ U.
Then X(U ) is an open subset of the cylinder C = {(x, y, z) ∈ R3 | x2+ y2 = 1} and E = 1, F = 0 and G = 1 at every point X(u, v) in C.
We remark that, although the cylinder and the plane are distinct surfaces, we obtain the same result in both cases.
3. Let a > 0 and α(u) = (cos u, sin u, au) denote a helix. Through each point of the helix, draw a line parallel to the xy plane and intersecting the z axis. The surface generated by these lines is called a helicoid and admits the following parametrization
X(u, v) = (v cos u, v sin u, au), (u, v) ∈ U = {(u, v) |0 < u < 2π, −∞ < v < ∞}.
Then E(u, v) = v2+ a2, F (u, v) = 0 and G(u, v) = 1.
Remarks
1. Let S be a regular surface S in R3 and let the arc length s = s(t) of a parametrized curve α : I → S be given by
s(t) = Z t
0
|α0(τ )| dτ = Z t
0
pI(α0(τ )) dτ.
In particular, if α(t) = X(u(t), v(t)) is contained in a coordinate neighborhood correspond- ing to the parametrization X(u, v), we can computethe arc length of αbetween, say, 0 and t by
s(t) = Z t
0
pE(u0)2+ 2F u0v0+ G(v0)2dτ
which implies that
ds dt
2
= E du dt
2
+ 2Fdu dt
dv
dt + G dv dt
2
and the “element” of arc length, dsof S, satisfies that ds2 = Edu2+ 2F dudv + Gdv2.
2. The angle θ under which two parametrized regular curves α : I → S, β : I → S intersect at t = t0 is given by
cos θ = hα0(t0), β0(t0)i
|α0(t0)||β0(t0)|.
In particular,the angle ϕ of the coordinate curves of a parametrization X(u, v) is cos ϕ = hXu, Xvi
|Xu||Xv| = F
√EG
it follows that the coordinate curves of a parametrization are orthogonal if and only if F (u, v) = 0 for all (u, v). Such a parametrization is called an orthogonal parametrization.
Definition Let R ⊂ S be a bounded region of a regular surface contained in the coordinate neighborhood of the parametrization X : U ⊂ R2 → S. Then the area A(R) of R = X(Q) is given by
A(R) = Z Z
Q
|Xu∧ Xv| du dv, where Q = X−1(R).
Claim The integral Z Z
Q|Xu∧ Xv| du dv does not depend on the parametrization X.
Proof of the Claim Suppose that ¯U is an open set in the ¯u¯v-plane, ¯X : ¯U ⊂ R2 → S is another parametrization such that R ⊂ ¯X( ¯U ), and ¯Q = ¯X−1(R), (¯u, ¯v) ∈ ¯Q, then
X¯u¯ = Xu∂u
∂ ¯u + Xv∂v
∂ ¯u, X¯v¯ = Xu∂u
∂ ¯v + Xv∂v
∂ ¯v =⇒ | ¯X¯u∧ ¯Xv¯| = |Xu∧ Xv|
∂(u, v)
∂(¯u, ¯v) ,
and Z Z
Q¯| ¯Xu¯∧ ¯X¯v| d¯u d¯v = Z Z
Q¯|Xu∧ Xv|
∂(u, v)
∂(¯u, ¯v)
d¯u d¯v = Z Z
Q|Xu∧ Xv| du dv.
Thus the definition of the area of R does not depend on the parametrization X.
Remark Since
|Xu∧ Xv|2+ hXu, Xvi2 = |Xu|2|Xv|2(sin2θ + cos2θ) = |Xu|2|Xv|2 =⇒ |Xu∧ Xv|2 = EG − F2, the area of R = X(Q) ⊂ S can be written as
A(R) = Z Z
Q|Xu∧ Xv| du dv = Z Z
Q
√EG − F2du dv.
Example Let a > r > 0, S1 = {(y, z) | (y − a)2 + z2 = r2} and T be the torus generated by rotating S1 about z-axis. Find the area of T.
