Regular Surfaces

Definition 1. A subset S ⊂ R^{3} is a regular surface if, for each p ∈ S, there exists an open
neighborhood V in R^{3}, an open set U ⊂ R^{2} 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 ⊂ R^{3} → R^{2} defined on an open set W containing V ∩ S.

(3) (The regularity condition)For each q = (u, v) ∈ U, the linear map

dX_{q} =

∂x

∂u

∂x

∂v

∂y

∂u

∂y

∂v

∂z

∂u

∂z

∂v

: R^{2} → R^{3},

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 R^{3}, 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 dX_{q}
in the canonical bases e1 = (1, 0), e2 = (0, 1) of R^{2} with coordinates (u, v) and f1 = (1, 0, 0),
f_{2} = (0, 1, 0), f_{3} = (0, 0, 1) of R^{3}, with coordinates (x, y, z).

Let q = (u_{0}, v_{0}). The vector e_{1} is tangent to the curve u → (u, v_{0}) 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 = v_{0}) 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 (u_{0}, v_{0}) and a vector is indicated by its components
in the basis {f_{1}, f_{2}, f_{3}}. By the definition of differential

dX_{q}(e_{1}) = d

duX(u, v_{0})|_{u=u}_{0} = ∂x

∂u,∂y

∂u,∂z

∂u

= ∂X

∂u.

Similarly, using the coordinate curve u = u_{0} (image by X of the curve v → (u_{0}, v)), we
obtain

dX_{q}(e_{2}) = d

dvX(u_{0}, v)|_{v=v}_{0} = ∂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 S^{2} = {(x, y, z) ∈ R^{3}; x^{2}+ y^{2}+ z^{2} = 1} is a regular surface.

• Let U = {(x, y) ∈ R^{2}; x^{2} + y^{2} < 1}, R^{2} = {(x, y, z) ∈ R^{3}; z = 0} and the maps X_{1}, X_{2} :
U ⊂ R^{2} → R^{3} be defined by

X_{1}(x, y) = (x, y, +p

1 − x^{2}− y^{2}), X_{2}(x, y) = (x, y,p

1 − x^{2}− y^{2}), (x, y) ∈ U.

• Let U = {(x, z) ∈ R^{2}; x^{2}+ z^{2} < 1}, R^{2} = {(x, y, z) ∈ R^{3}; y = 0} and the maps X_{3}, X_{4} :
U ⊂ R^{2} → R^{3} be defined by

X_{3}(x, z) = (x, +√

1 − x^{2}− z^{2}, z), X_{4}(x, y) = (x, −√

1 − x^{2}− z^{2}, z), (x, z) ∈ U.

• Let U = {(y, z) ∈ R^{2}; y^{2} + z^{2} < 1}, R^{2} = {(x, y, z) ∈ R^{3}; x = 0} and the maps X_{5}, X_{6} :
U ⊂ R^{2} → R^{3} be defined by

X5(y, z) = (+√

1 − x^{2}− z^{2}, y, z), X6(y, z) = (−√

1 − x^{2} − z^{2}, y, z), (y, z) ∈ U.

Since {X_{i} : U → S^{2} | 1 ≤ i ≤ 6} are parametrizations covering S^{2} completely, S^{2} is a regular
surface.

For most applications, it is convenient to relate parametrizations to the geographical coordinates
on S^{2}. Let V = {(θ, φ); 0 < θ < π, 0 < φ < 2π} and let X : V → R^{3} be given by

X(θ, φ) = (sin θ cos φ, sin θ sin φ, cos θ).

Clearly, X(V ) ⊂ S^{2} and X is a parametrization of S^{2}, 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 S^{2} (including the two poles) and that S^{2} can be covered with the coordinate neighborhoods
of two parametrizations of this type.

Example shows that deciding whether a given subset of R^{3} 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 R^{2}, then the graph
of f over U , that is, the subset of R^{3} given by

S = the graph of a differentiable function f on an open subset U ⊂ R^{2}

= {(x, y, f (x, y)) | (x, y) ∈ U ⊂ R^{2}}
is a regular surface.

Proof It suffices to show that the map X : U → R^{3} 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 ⊂ R^{n} → R^{m} defined in an open set U of R^{n}, we
say that p ∈ U is a critical point of F if the differential dFp : R^{n} → R^{m} is not a surjective (or
onto) mapping. The image F (p) ∈ R^{m} of a critical point is called a critical value of F. A point
of R^{m} which is not a critical value is called aregular value of F.

Remark Recall that

Implicit Function Theorem

Let U ⊆ R^{n+m}≡ R^{n}× R^{m} be an open set,

f = (f_{1}, . . . , f_{m}) : U → R^{m} belong to Class C^{1}(U ),

(a, b) = (a_{1}, . . . , a_{n}, b_{1}, . . . , b_{m}) ∈ U be a point at which f (a, b) = 0 ∈ R^{m},
and the m × m matrix D_{y}f |_{(a,b)}= ∂f_{i}

∂y_{j}(a, b)

1≤i,j≤m

be invertible.

Then there exists

an open neighborhood A of a in R^{n},
an open neighborhood B of b in R^{m},

and a unique g : A → B belonging to Class C^{1}(A)
such that

b = g(a) and f (x, g(x)) = 0 ∈ R^{m} for all x ∈ A
and thus

f^{−1}(0) ∩ A × B = {(x, y) ∈ A × B ⊂ U ⊆ R^{n}× R^{m} | f (x, y) = 0 ∈ R^{m}}

= {(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 ⊂ R^{2} → 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 ⊂ R^{2} | f (x, y) = r} ⊂ R^{2},

since dfp = (fx(p), fy(p)) : R^{2} → R is surjective, (f^{x}(p), fy(p)) 6= (0, 0). By interchanging x
and y if necessary, we may assume that f_{y}(p) 6= 0, thus, by the Implicit Function Theorem,
there exists an open set

I_{1}× I_{2} = (a_{1}, b_{1}) × (a_{2}, b_{2}) containg p = (p_{1}, p_{2})
and a unique continuously differentiable function g : I_{1} → I_{2} such that

f (x, g(x)) = r for all x ∈ I_{1} =⇒ f^{−1}(r) ∩ I_{1}× I_{2} = {(x, g(x)) | x ∈ I_{1}} ⊂ R^{2}

is a regular plane over I1 = (a1, b1) since g is differentiable and the tangent vector (1, g^{0}(x)) 6=

(0, 0) at each (x, g(x)), x ∈ I_{1}. 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 = (F_{1}, F_{2}) : U ⊂ R^{3} → R^{2}
is a regular curve in R^{3}.

Proof Let r = (r_{1}, r_{2}) be a regular value of F = (F_{1}, F_{2}). For each

p = (p_{1}, p_{2}, p_{3}) ∈ F^{−1}(r) ∩ U = {(x, y, z) ∈ U ⊂ R^{3} | F (x, y, z) = r} ⊂ R^{3},
since

dF_{p} =(F_{1})_{x}(p) (F_{1})_{y}(p) (F_{1})_{z}(p)
(F_{2})_{x}(p) (F_{2})_{y}(p) (F_{2})_{z}(p)

: R^{3} → R^{2} is surjective,

the matrix (F_{1})_{x}(p) (F_{1})_{y}(p) (F_{1})_{z}(p)
(F_{2})_{x}(p) (F_{2})_{y}(p) (F_{2})_{z}(p)

is of rank 2. By interchanging x, y or z if nec- essary, we may assume that

(F_{1})_{y}(p) (F_{1})_{z}(p)
(F_{2})_{y}(p) (F_{2})_{z}(p)

is nonsingular, (F_{1})_{z}(p) 6= 0, (F_{2})_{y}(p) 6= 0,

thus, by the Implicit Function Theorem, there exist an open neighborhood I_{1} × I_{2}× I_{3} =
(a_{1}, b_{1}) × (a_{2}, b_{2}) × (a_{3}, b_{3}) of (p_{1}, p_{2}, p_{3}) and unique continuously differentiable functions
g = (g_{1}, g_{2}) : I_{1} → I_{2}× I_{2}, h : I_{1}× I_{2} → I_{3}, k : I_{1}× I_{3} → I_{2} such that

F (x, g(x)) = r for all x ∈ I_{1} =⇒ F^{−1}(r) ∩ I_{1}× I_{2}× I_{3} = {(x, g_{1}(x), g_{2}(x)) | x ∈ I_{1}} ⊂ R^{3}
is a regular curve in R^{3}

F_{1}(x, y, h(x, y)) = r_{1} for all (x, y) ∈ I_{1}× I_{2}

=⇒ (F_{1})^{−1}(r_{1}) ∩ I_{1}× I_{2}× I_{3} = {(x, y, h(x, y)) | (x, y) ∈ I_{1}× I_{2}} ⊂ R^{3}
is a graph of a differentiable function h over I_{1}× I_{2}

=⇒ (F_{1})^{−1}(r_{1}) ∩ I_{1}× I_{2}× I_{3} is a regular surface in R^{3}

F_{2}(x, k(x, z), z) = r_{2} for all (x, z) ∈ I_{1}× I_{3}

=⇒ (F2)^{−1}(r2) ∩ I1× I2× I3 = {(x, k(x, z).z) | (x, z) ∈ I1× I3} ⊂ R^{3}
is a graph of a differentiable function k over I_{1}× I_{3}

=⇒ (F_{2})^{−1}(r_{2}) ∩ I_{1}× I_{2}× I_{3} is a regular surface in R^{3}

Since p is an arbitrary point in F^{−1}(r) ∩ U, these imply that F^{−1}(r) is a regular curve,
(F_{1})^{−1}(r_{1}) and (F_{2})^{−1}(r_{2}) are regular surfaces in R^{3} and F^{−1}(r) = (F_{1})^{−1}(r_{1}) ∩ (F_{2})^{−1}(r_{2})
Proposition 2. If f : U ⊂ R^{3} → 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 R^{3}.

Proof Let p = (x_{0}, y_{0}, z_{0}) 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 f_{z}(p) 6= 0. We define a mapping F : U ⊂ R^{3} → R^{3}
byF (x, y, z) = (x, y, f (x, y, z)) for (x, y, z) ∈ U. Since

det(dF_{p}) =

1 0 0

0 1 0

fx(p) fy(p) fz(p)

= f_{z}(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) = (g_{1}(u, v, t), g_{2}(u, v, t), g_{2}(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 g_{1}(u, v, t) = u, g_{2}(u, v, t) = v and

(x, y, z) = F^{−1}(u, v, t) = (u, v, g_{3}(u, v, t)) for (u, v, t) ∈ W, (x, y, z) ∈ V.

This implies that π(V ) ∼= π(W ) and z = g_{3}(u, v, a) = h(u, v) is a differentiable function for
(u, v) ∈ π(W ), where π : R^{3} → R^{2} is the projection map defined by π(u, v, t) = (u, v) for
(u, v, t) ∈ R^{3}. 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, g_{3}(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

x^{2}
a^{2} +y^{2}

b^{2} + z^{2}
c^{2} = 1
is a regular surface. In fact, it is the set f^{−1}(0) where

f (x, y, z) = x^{2}
a^{2} + y^{2}

b^{2} +z^{2}
c^{2} − 1

is a differentiable function and 0 is a regular value of f. This follows from the fact that the
partial derivatives f_{x} = 2x/a^{2}, f_{y} = 2y/b^{2}, f_{z} = 2z/c^{2} 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 −x^{2}− y^{2}+ z^{2} = 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) = −x^{2}− y^{2}+ z^{2}− 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 t_{0}, we have z(t_{0}) = 0, which means that α(t_{0}) ∈ S.

Example Let a > r > 0, S^{1} = {(y, z) | (y − a)^{2}+ z^{2} = r^{2}} and T be the surface, called torus,
generated by rotating S^{1} about z-axis. Hence the points (x, y, z) of T satisfy the equation

z^{2} = r^{2}− (p

x^{2} + y^{2} − a)^{2}.
Therefore, T is the inverse image of r^{2} by the function

f (x, y, z) = z^{2}+ (p

x^{2}+ y^{2}− a)^{2}.

Note that f is differentiable for (x, y) 6= (0, 0), and r^{2} is a regular value of f. It follows that the
torus T is a regular surface.

Proposition 3. Let S ⊂ R^{3} 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 ⊂ R^{2} → R^{3}be 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 → R^{2}, where π(x, y, z) = (x, y)
for (x, y, z) ∈ R^{3}, 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 V_{1} of q, V_{2} of π ◦ X(q) such that

• π ◦ X : V_{1} → V_{2} is one-to-one, onto and has a differentiable inverse (π ◦ X)^{−1} : V_{2} → V_{1}
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(V_{1}) = V ⊂ S → V_{2} is one-to-one on V,

=⇒ V is a graph of z = z ◦ (π ◦ X)^{−1}(x, y) on V_{2}.

In fact, since X is a homeomorphism, V = X(V_{1}) = (X^{−1})^{−1}(V_{1}) is an open neighborhood of p
in S and since

• z = z(u, v) is differentiable for (u, v) ∈ V_{1},

• (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 ∈ V_{2},
and

V = {X(u, v) = (x(u, v), y(u, v), z(u, v)) | (u, v) ∈ V_{1}}

= {(x, y, z) | z = z(u(x, y), v(x, y)) = z ◦ (π ◦ X)^{−1}(x, y), (x, y) ∈ V_{2}},

V is the graph of the differentiable function z = z(u(x, y), v(x, y)) = f (x, y) over V_{2}, 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 ⊂ R^{2} → R^{3} 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 ⊂ R^{2} → S ⊂ R^{3} 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 V_{1} of q, V_{2} of π ◦ X(q) such
that π ◦ X maps V_{1} diffeomorphically onto V_{2}, where π : R^{3} → R^{2} is the projection defined by
π(x, y, z) = (x, y) for (x, y, z) ∈ R^{3}. It follows that (π ◦ X)^{−1} : V_{2} → V_{1} and π : R^{3} → R^{2} 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 → V_{1} is a composition of continuous maps, X^{−1} is continuous.

Examples

1. The one-sheetd cone C, given by z = +p

x^{2}+ y^{2}, (x, y) ∈ R^{2}, 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

x^{2}+ y^{2}. Since z = +p

x^{2}+ y^{2} is not differentiable at (0, 0), this is
impossible.

2. Let a > r > 0, S^{1} = {(y, z) | (y − a)^{2} + z^{2} = r^{2}} and T be the torus generated by rotating
S^{1} 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

x^{2}+ y^{2} ≤ a, then
π/2 ≤ u ≤ 3π/2, and if p

x^{2}+ y^{2} ≥ 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 ⊂ R^{2} → S, Y : V ⊂ R^{2} → 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 → R^{3} 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) + te_{3}, where e_{3} 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 R^{3}
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 ⊂ R^{2} → S
with p ∈ X(U ) ⊂ V, the composition f ◦ X : U ⊂ R^{2} → 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 ⊂ R^{2} → 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 ⊂ R^{3} be an open set such that S ⊂ V . Let
f : V ⊂ R^{3} → 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 ⊂ R^{2} → 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 ∈ R^{3}, h : S → R, given by h(p) = p · v, p ∈ S,
where the dot denotes the usual inner product in R^{3}. h(p) is the height of p ∈ S relative to
a plane normal to v and passing through the origin of R^{3}.

2. The square of the distance from a fixed point p_{0} ∈ R^{3}, f (p) = |p − p_{0}|^{2}, p ∈ S. The need
for taking the square comes from the fact that the distance |p − p_{0}| is not differentiable at
p = p_{0}.

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 : V_{1} ⊂ S_{1} → S_{2} of an open set V_{1} of a regular surface S_{1} to a
regular surface S_{2} is said to be differentiable at p ∈ V_{1} if, given parametrizations

X_{1} : U_{1} ⊂ R^{2} → S_{1}, X_{2} : U_{2} ⊂ R^{2} → S_{2}
with p ∈ X_{1}(U_{1}) and ϕ(X_{1}(U_{1})) ⊂ X_{2}(U_{2}), the map

X_{2}^{−1}◦ ϕ ◦ X_{1} : U_{1} → U_{2}
is differentiable at q = X_{1}^{−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 ϕ_{1} and ϕ_{2} have continuous partial derivatives of all orders.

Note that one can use Prop. 1 to show that this definition of differentiability of ϕ : S_{1} → S_{2}
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 S_{1} and S_{2} are diffeomorphicif there exists a differentiable map
ϕ : S_{1} → S_{2} with a differentiable inverse ϕ^{−1} : S_{2} → S_{1}. Such a ϕ is called a diffeomorphism
from S_{1} to S_{2}.

Note that, if ϕ : S_{1} → S_{2} is a diffeomorphism from S_{1} to S_{2}, then f : S_{2} → R is differentiable
on S_{2} if and only if f ◦ ϕ : S_{1} → R is differentiable on S1, i.e. two diffeomorphic surfaces are
indistinguishable from the point of view of differentiability.

Example 2 If X : U ⊂ R^{2} → S is a parametrization, then X^{−1} : X(U ) → U ⊂ R^{2} is
differentiable.

In fact, for any p ∈ X(U ) and any parametrization Y : V ⊂ R^{2} → 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 S_{1} and S_{2} be regular surfaces. Assume that S_{1} ⊂ V ⊂ R^{3}, where V is an open
set of R^{3}, and that ϕ : V → R^{3} is a differentiable map such that ϕ(S_{1}) ⊂ S_{2}. Then the restriction
ϕ|_{S}_{1} : S_{1} → S_{2} is a differentiable map.

In fact, given p ∈ S_{1} and parametrizations

X_{1} : U_{1} ⊂ R^{2} → S_{1}, X_{2} : U_{2} ⊂ R^{2} → S_{2}
with p ∈ X_{1}(U_{1}) and ϕ(X_{1}(U_{1})) ⊂ X_{2}(U_{2}), we have that the map

X_{2}^{−1}◦ ϕ ◦ X_{1} : U_{1} → U_{2} 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 : R^{3} → R^{3} defined
by

σ(x, y, z) = (x, y, z) for (x, y, z) ∈ R^{3}.

This, of course, generalizes to surfaces symmetric relative to any plane of R^{3}.

2. Let R_{z,θ} : R^{3} → R^{3} be the rotation of angle θ about the z axis, and let S ⊂ R^{3} be a
regular surface invariant by this rotation; i.e., {R_{z,θ}(p) | p ∈ S} ⊆ S. Then the restriction
R_{z,θ} : S → S is a differentiable map.

3. Let ϕ : R^{3} → R^{3} 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|_{S}^{2} is a differentiable map from the
sphere

S^{2} = {(x, y, z) ∈ R^{3} | x^{2}+ y^{2}+ z^{2} = 1}

into the ellipsoid

{(x, y, z) ∈ R^{3} | x^{2}
a^{2} +y^{2}

b^{2} + z^{2}
c^{2} = 1}.

Remark 3 Proposition 1 implies (cf. Example 2) that a parametrization X : U ⊂ R^{2} → S
is a diffeomorphism of U onto X(U ). Actually, we can now characterize the regular surfaces as
those subsets S ⊂ R^{3} which are locally diffeomorphic to R^{2}; that is, for each point p ∈ S, there
exists a neighborhood V of p in S, an open set U ⊂ R^{2}, 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 ⊂ R^{2} → R^{3} is a differentiable map X from an
open set U ⊂ R^{2} into R^{3}. The set X(U ) ⊂ R^{3} is called the trace of X. X is regular if the
differential dX_{q} : R^{2} → R^{3} 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 dX_{p} 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 ⊂ R^{2} → R^{3} is not necessary a homeomorphism (or a global one-to-one map).

Example Let α : I → R^{3} be a nonplanar regular parametrized curve. Define
X(t, v) = α(t) + vα^{0}(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) = |α^{00}(t) ∧ α^{0}(t)|

|α^{0}(t)|^{3} 6= 0, ∂X

∂t = α^{0}(t) + vα^{00}(t) and ∂X

∂v = α^{0}(t), we have

∂X

∂t ∧ ∂X

∂v = vα^{00}(t) ∧ α^{0}(t) 6= 0 for all (t, v) ∈ U.

It follows that the restriction X : U → R^{3} 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 ⊂ R^{2} → R^{3} be a regular parametrized surface and let q ∈ U. Then there
exists a neighborhood V of q in R^{2} such that X(V ) ⊂ R^{3} is a regular surface.

Proof Let X : U ⊂ R^{2} → R^{3} 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 → R^{3} 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 W_{1} of (q, 0) and W_{2} of F (q, 0)
such that F : W_{1} → W_{2} is a diffeomorphism. Set V = W_{1} ∩ U ⊂ R^{2} 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 α^{0}(0) = w. The set of tangent vectors to S at p, denoted byT_{p}S, is called the
tangent plane to S at p.

Proposition 1. Let X : U ⊂ R^{2} → S be a parametrization of a regular surface S and let q ∈ U.

The vector subspace of dimension 2,

dXq(R^{2}) ⊂ R^{3},

coincides with the set of tangent vectors to X at X(q), i.e. dX_{q}(R^{2}) = T_{X(q)}S.

Proof Let w ∈ T_{X(q)}S be a tangent vector at X(q), that is, let w = α^{0}(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} =⇒ dX_{q}(β^{0}(0)) = dX_{β(0)}(β^{0}(0)) = α^{0}(0) = w.

Hence, w ∈ dX_{q}(R^{2}) =⇒ T_{X(q)}S ⊆ dX_{q}(R^{2}).

On the other hand, letw = dX_{q}(v),where v ∈ R^{2}. 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 = α^{0}(0), where α = X ◦ γ. This shows thatw ∈ T_{X(q)}S is
a tangent vector and dX_{q}(R^{2}) ⊆ T_{X(q)}S.

By the above proposition, the planedX_{q}(R^{2}),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 T_{p}S. The choice of the parametrization X determines a basis {(∂X/∂u)(q), (∂X/∂v)(q)} of
T_{p}S, called the basis associated to X. Sometimes it is convenient to write ∂X/∂u = X_{u} and

∂X/∂v = Xv.

The coordinates of a vector w ∈ T_{p}S in the basis associated to a parametrization X are deter-
mined as follows. Let w = α^{0}(0) for some α(t) = X(u(t), v(t)) with (u(0), v(0)) = q = X^{−1}(p).

Thus,

w= α^{0}(0) = d

dtX(u(t), v(t))|_{t=0}= X_{u}(q)u^{0}(0) + X_{v}(q)v^{0}(0).

That is, in the basis {X_{u}(q), X_{v}(q)} of T_{p}S, w has coordinates (u^{0}(0), v^{0}(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 S_{1} and S_{2} be two regular surfaces and let ϕ : V ⊂ S_{1} → S_{2} be a
differentiable mapping of an open set V of S_{1} into S_{2}. For each p ∈ V, there is a map, called the
differential of ϕ at p,

dϕ_{p} : T_{p}S_{1} → T_{ϕ(p)}S_{2}
defined as follows.

For each w ∈ T_{p}S_{1}, let α : (−ε, ε) → V ⊂ S_{1} be a differentiable parametrized curve such that
α(0) = p and α^{0}(0) = w. Since β = ϕ ◦ α : (−ε, ε) → S_{2} is a curve in S_{2} such that β(0) = ϕ(p),
this implies that β^{0}(0) ∈ T_{ϕ(p)}S2 and it is defined to be dϕp(w), i.e. β^{0}(0) = dϕp(w).

Note that the coordinates of a vectorβ^{0}(0) ∈ T_{ϕ(p)}S_{2} 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

β^{0}(0) = d
dt

X(¯¯ u(t), ¯v(t))|_{t=0}= ¯X_{u}_{¯}(r)¯u^{0}(0) + ¯X_{¯}_{v}(r)¯v^{0}(0).

That is, in the basis { ¯X_{u}_{¯}(r), ¯X_{v}_{¯}(r)} of T_{ϕ(p)}S_{2}, β^{0}(0) has coordinates (¯u^{0}(0), ¯v^{0}(0)).

Proposition 2. In the discussion above, given w, the vector β^{0}(0) does not depend on the choice
of α. The map dϕ_{p} : T_{p}S_{1} → T_{ϕ(p)}S_{2} defined by dϕ_{p}(w) = β^{0}(0) is linear.

Proof Idea: Find a matrix representation of dϕ_{p} : T_{p}S_{1} → T_{ϕ(p)}S_{2} in the bases {X_{u}, X_{v}} of T_{p}S_{1}
and { ¯X_{u}_{¯}, ¯X_{v}_{¯}} of T_{ϕ(p)}S_{2}.

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) → S_{1}, 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) = (ϕ_{1}(u, v), ϕ_{2}(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)) = (ϕ_{1}(u(t), v(t)), ϕ_{2}(u(t), v(t))) for t ∈ (−ε, ε).

Since

¯
u^{0}(0)

¯
v^{0}(0)

=

∂ϕ_{1}

∂u u^{0}(0) + ∂ϕ_{1}

∂v v^{0}(0)

∂ϕ_{2}

∂u u^{0}(0) + ∂ϕ_{2}

∂v v^{0}(0)

=

∂ϕ_{1}

∂u

∂ϕ_{1}

∂v

∂ϕ_{2}

∂u

∂ϕ_{2}

∂v

u^{0}(0)
v^{0}(0)

,

and since dX_{q}(R^{2}) = T_{p}S_{1}, d ¯X_{p}^{−1}(T_{ϕ(p)}S_{2}) = R^{2}, and

• (u^{0}(0), v^{0}(0)) ∈ R^{2} is the coordinates of w in the basis {X_{u}, X_{v}} of T_{p}S_{1},

• (¯u^{0}(0), ¯v^{0}(0)) ∈ R^{2} is the coordinates of β^{0}(0) in the basis { ¯X_{u}_{¯}, ¯X_{¯}_{v}} of T_{ϕ(p)}S_{2},

• the matrix dΦ_{q} =∂ϕ_{1}/∂u ∂ϕ_{1}/∂v

∂ϕ_{2}/∂u ∂ϕ_{2}/∂v

depends only on Φ, i.e.

β^{0}(0) = dϕ_{p}(w) ⇐⇒

¯
u^{0}(0)

¯
v^{0}(0)

coordinates of β^{0}(0)

=

∂ϕ_{1}

∂u

∂ϕ_{1}

∂v

∂ϕ_{2}

∂u

∂ϕ_{2}

∂v

u^{0}(0)
v^{0}(0)

coordinates of w

,

we conclude that

• β^{0}(0) is independent of α,

• the matrix dΦ_{q} is a linear map from T_{p}S_{1} in the basis {X_{u}, X_{v}} to T_{ϕ(p)}S_{2} 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
T_{ϕ(p)}S_{2}.

Hence dϕ_{p} is a linear mapping from T_{p}S_{1} into T_{ϕ(p)}S_{2}.

The linear mapdϕ_{p} defined by Prop. 2 is called thedifferential of ϕ at p ∈ S_{1}. In a similar way
we define the differential of a (differentiable) function f : U ⊂ S → R at p ∈ U as a linear map
df_{p} : T_{p}S → R.

Example Let v ∈ R^{3} 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 dh_{p}(w), w ∈ T_{p}S, choose a differentiable
curve α : (−ε, ε) → S with α(0) = p, α^{0}(0) = w. Since h(α(t)) = α(t) · v, we obtain

dh_{p}(w) = d

dth(α(t))|_{t=0}= α^{0}(0) · v = w · v.

Example Let S^{2} ⊂ R^{3} be the unit sphere

S^{2} = {(x, y, z) ∈ R^{3} | x^{2}+ y^{2}+ z^{2} = 1}

and let R_{z,θ} : R^{3} → R^{3} be the rotation of angle θ about the z axis. Then R_{z,θ} restricted to S^{2} is a
differentiable map of S^{2}. We shall compute (dRz,θ)p(w), p ∈ S^{2}, w ∈ TpS^{2}. Let α : (−ε, ε) → S^{2}
be a differentiable curve with α(0) = p, α^{0}(0) = w. Then, since R_{z,θ} is linear,

(dR_{z,θ})_{p}(w) = d

dt(R_{z,θ}◦ α(t)) |_{t=0} = R_{z,θ}(α^{0}(0)) = R_{z,θ}(w).

Observe that Rz,θ leaves the north pole N = (0, 0, 1) fixed, and that (dRz,θ)N : TNS^{2} → TNS^{2} is
just a rotation of angle θ in the plane T_{N}S^{2}.

We shall say that a mapping ϕ : U ⊂ S_{1} → S_{2} 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 ) ⊂ S_{2}. In these terms, the version of the inverse of function theorem for surfaces is expressed
as follows.

Proposition 3. If S_{1} and S_{2} are regular surfaces and ϕ : U ⊂ S_{1} → S_{2} is a differentiable
mapping of an open set U ⊂ S_{1} such that the differential dϕ_{p} of ϕ at p ∈ U is an isomorphism,
then ϕ is a local diffeomorphism at p.

Remark By fixing a parametrization X : U ⊂ R^{2} → 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) = X_{u}∧ X_{v}

|X_{u}∧ X_{v}|(q).

Note that N : X(U ) → R^{3} 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 R^{3}. For each p ∈ S and tangent vectors w_{1}, w_{2} ∈ T_{p}S ⊂
R^{3}, there is an (induced) inner product

h , ip : TpS × TpS → R defined by

hw_{1}, w_{2}i_{p} = hw_{1}, w_{2}i = the inner product of w_{1}and w_{2}as vectors in R^{3}.

To this inner product, which is a symmetric bilinear form, there corresponds aquadratic form
I_{p} : T_{p}S → R

defined by

I_{p}(w) = hw, wi_{p} = hw, wi = |w|^{2} ≥ 0 for w ∈ T_{p}S,

and the quadratic form Ip on TpS is called the first fundamental form of the regular surface
S ⊂ R^{3} at p ∈ S.

Remark Let U be an open set in the uv-plane and let X : U ⊂ R^{2} → S be a parametrization of
the regular surface at p = X(u_{0}, v_{0}) ∈ S. For each w ∈ T_{p}S, since there is a parametrized curve

α(t) = X(u(t), v(t)) ∈ X(U ), t ∈ (−ε, ε), such that p = α(0) = X(u_{0}, v_{0}) and w = α^{0}(0),
we have

I_{p}(w) = I_{p}(α^{0}(0)) = hα^{0}(0), α^{0}(0)i_{p}

= hX_{u}u^{0}+ X_{v}v^{0}, X_{u}u^{0}+ X_{v}v^{0}i_{p}

= hX_{u}, X_{u}i_{p}(u^{0})^{2}+ 2hX_{u}, X_{v}i_{p}u^{0}v^{0}+ hX_{v}, X_{v}i_{p}(v^{0})^{2}

= E(u^{0})^{2} + 2F u^{0}v^{0}+ G(v^{0})^{2},

where the values of the functions involved are computed for t = 0, and E(u0, v0) = hXu, Xuip,

F (u_{0}, v_{0}) = hX_{u}, X_{v}i_{p},
G(u_{0}, v_{0}) = hX_{v}, X_{v}i_{p}

are the coefficients of the first fundamental form in the basis {X_{u}, X_{v}} of T_{p}S. 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 , i_{p}, 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 R^{3} by the same symbol h , i rather than
the previous dot.

Examples

1. Let p_{0} = (x_{0}, y_{0}, z_{0}) be a point in R^{3}, w_{1} = (a_{1}, a_{2}, a_{3}) and w_{2} = (b_{1}, b_{2}, b_{3}) be orthonormal
vectors in R^{3} and

P = {X(u, v) = p_{0}+ uw_{1}+ vw_{2} | (u, v) ∈ R^{2}}.

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) ∈ R^{3} | x^{2}+ y^{2} = 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) = v^{2}+ a^{2}, F (u, v) = 0 and G(u, v) = 1.

Remarks

1. Let S be a regular surface S in R^{3} and let the arc length s = s(t) of a parametrized curve
α : I → S be given by

s(t) = Z t

0

|α^{0}(τ )| dτ =
Z t

0

pI(α^{0}(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(u^{0})^{2}+ 2F u^{0}v^{0}+ G(v^{0})^{2}dτ

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
ds^{2} = Edu^{2}+ 2F dudv + Gdv^{2}.

2. The angle θ under which two parametrized regular curves α : I → S, β : I → S intersect at
t = t_{0} is given by

cos θ = hα^{0}(t_{0}), β^{0}(t_{0})i

|α^{0}(t_{0})||β^{0}(t_{0})|.

In particular,the angle ϕ of the coordinate curves of a parametrization X(u, v) is
cos ϕ = hX_{u}, X_{v}i

|X_{u}||X_{v}| = 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 ⊂ R^{2} → S. Then the area A(R) of R = X(Q) is
given by

A(R) = Z Z

Q

|X_{u}∧ X_{v}| du dv, where Q = X^{−1}(R).

Claim The integral Z Z

Q|X_{u}∧ X_{v}| 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 ⊂ R^{2} → S is another
parametrization such that R ⊂ ¯X( ¯U ), and ¯Q = ¯X^{−1}(R), (¯u, ¯v) ∈ ¯Q, then

X¯_{u}_{¯} = X_{u}∂u

∂ ¯u + X_{v}∂v

∂ ¯u, X¯_{v}_{¯} = X_{u}∂u

∂ ¯v + X_{v}∂v

∂ ¯v =⇒ | ¯X_{¯}_{u}∧ ¯X_{v}_{¯}| = |X_{u}∧ X_{v}|

∂(u, v)

∂(¯u, ¯v) ,

and Z Z

Q¯| ¯X_{u}_{¯}∧ ¯X_{¯}_{v}| d¯u d¯v =
Z Z

Q¯|X_{u}∧ X_{v}|

∂(u, v)

∂(¯u, ¯v)

d¯u d¯v = Z Z

Q|X_{u}∧ X_{v}| du dv.

Thus the definition of the area of R does not depend on the parametrization X.

Remark Since

|X_{u}∧ X_{v}|^{2}+ hX_{u}, X_{v}i^{2} = |X_{u}|^{2}|X_{v}|^{2}(sin^{2}θ + cos^{2}θ) = |X_{u}|^{2}|X_{v}|^{2} =⇒ |X_{u}∧ X_{v}|^{2} = EG − F^{2},
the area of R = X(Q) ⊂ S can be written as

A(R) = Z Z

Q|X_{u}∧ X_{v}| du dv =
Z Z

Q

√EG − F^{2}du dv.

Example Let a > r > 0, S^{1} = {(y, z) | (y − a)^{2} + z^{2} = r^{2}} and T be the torus generated by
rotating S^{1} about z-axis. Find the area of T.