Just like we defined a special type of curve that was free of singular points (regular curves), we shall now define a set of surfaces that are similiarly easy to work with. Yet due to the complexity of surfaces, when compared to curves, defining a regular surface will be much more difficult.
The goal in creating a definition of regular surface is twofold.
One, in order to apply calculus to surfaces, we need a tangent plane defined at every point. In other words, the map must not compress the tangent plane to a line for instance.
Two, we would like the definition to be independent of parameterization. So, if the surface is regular given one parameterization, we would like it to be regular for any parameterization.

I will now give the definition of a regular surface, and explain each part afterwards:

A subset S R3 is a regular surface if, for each p S, there exists a neighborhood V in R3 and a map x : U V S of an open set U R2 onto U S R3 such that:

  1. x is differentiable.
  2. x is a homeomorphism
  3. each map x: U S is a regular patch.

x is differentiable means simply that each component is differentiable. (ie. if g(u,v) = (x(u,v),y(u,v),z(u,v)), (u,v) U is differentiable, then each component, x(u,v), y(u,v), and z(u,v), is itself differentiable. It is a sufficient condition for differentiability that a function has continous partial derivatives. Although not necessary, this is probably what we will use throughout the text.

Example The map h(u,v) = (vsin(u),vcos(u),u) is a differentiable map, while p(u,v) = (cos(u)sin(v),sin(u)sin(v),cos(v) + ln(tan(v/2)) + u) is not.

 

x is a homeomorphism Translates into the following, x is a continous map and that x has a well defined continous inverse, x-1. Since this concept is very difficult to explain completely with using topology, see the following examples. These are all homeomorphisms, notice that all of them have smooth transitions, no jerks or starts, nor do they have any asymptotes.

 

Fig. 1
Fig. 2
Fig. 3
Fig. 4
 
each map x: U S is a regular patch This essentially means that the tangent plane is still two dimensional after the map. We don't want it to flatten into a line or a point. To aid the analysis, we shall now define the linear differential map, dxq. This is a map that transforms the basis vectors in the (u,v) space into 2 tangent vectors on the surface.
At some point q in U, we define two curves,
q = (u0,v0)
u (u, v0)
v (u0, v)

The tangent vectors of these curves in R2 is,

d(u,v0)
du
= (1,0) = e1
d(u0,v)
dv
= (0,1) = e2

The image of these curves under x is,

u x(u, v0) =
x(u,v0),y(u,v0),z(u,v0)
v x(u0, v) =
x(u0,v),y(u0,v),z(u0,v)

To find the tangent vectors of these cuves in R3 we simply take the derivative with respect to the free variable. In this case it is equilivent to taking partials,

d
x(u,v0),y(u,v0),z(u,v0)

du
=

x
u
, y
u
, z
u


= x
u
d
x(u0,v),y(u0,v),z(u0,v)

dv
=

x
v
, y
v
, z
v


= x
v

This shows that, under x,

e1 x
u
  and,
e2 x
v

 

Using e1 and e2 as basis as a linear map from those to the tangent vectors. The matrix representation is,

dxq =











x
u
x
v
 
y
u
y
v
 
z
u
z
v












We can easily verify that this matrix transform represents the linear map of the tangent vectors.

dxq(e1) =

x
u
, y
u
, z
u


dxq(e2) =

x
v
, y
v
, z
v


 

Looking back on our definition of Regular Surface, for x to be a regular map, its differential, dxq must be 1-1. In other words, the two column vectors in the matrix representation must be linearly independent. If they weren't, then the two basis vectors would map to a two linear dependent vectors that could at most span a line. This makes the tangent plane vanish, which is exactly what we want to avoid.


A couple quick ways of determining linear independence are,

  1. the vector product [(x)/( u)] [(x)/( v)] 0
  2. one of the minors of order 2 of the matrix dxq ie,
    (x,y)
    (u,v)
    =






    x
    u
    x
    v
     
    y
    u
    y
    v







    ,        (y,z)
    (u,v)
    ,        (x,z)
    (u,v)
    ,
    be different from zero at q.

 

Now, verifying each one of those requirements for each surface we discuss in this class would be time consuming. What follows are a couple short-cuts to determining whether a surface is regular.

  1. if f: U R is a differentiable function in an open set U R2, then the graph of f, the subset of R3 given by (u,v,f(u,v)), for u, v U, is a regular surface.

    Proof.
    we need only to show that the map x(u,v) = (u,v, f(u,v)) is regular.
    Condition 1. x is clearly differentiable, since we require f to be differentiable
    Condition 2. Since for each (u,v) there is a unique (x,y) in R3, x is 1-1, and this guarantees an inverse x-1. We can see that x-1 is a continous mapping if we just think of the the graph projecting down to the xy plane.
    Condition 3. [((x,y))/( (u,v))] 1

    Example   The graph of the function f(x,y) = sin(xy), as is (u,v) (u,v,usin(v)) by shortcut #1.
    (u,v,sin(u v))
    (u,v,usin(v))

  2. if f: U R3 R is a differentiable function and a f(U) is a regular value of f, then f-1(a) is a regular surface in R3

    Proof.
    Due to the complexity of this proof, only a sembelence of the proof will be given. A more rigorous proof can be found in the text.
    A regular value is a value a such that the partials of f do not vanish simultaneously. ie. for the sphere, x2 + y2 + z2 = f(x,y,z) : f(x,y,z) = a the only value of a which is not a regular value is a = 0, because that would require all of the partials, fx, fy, and fz to be 0.
    The proof basically goes as follows, first we define a mapping F:U Ã R3 Æ R3 by F(x,y,z) = (x,y,f(x,y,z))
    dFp =



    1
    0
    0
    0
    1
    0
    fx
    fy
    fz




    ,        det(dFp) = fz 0
    fz does not equal 0 because we required a to be a regular value (at least ONE of the partials must be apart from 0, relabeling of axes allows fz to always be not equal to 0).
    Therefor, by the inverse function theorm, there exists and inverse of F, F-1: (u,v,t) F-1(x,y,z) V. Now we define our surface (the level set of f, the one we're showing is regular) as {z = g(u,v,a) = h(x,y) : g(u,v,t) = F-1(x,y,z)} and by shortcut #1, this is a regular surface.

    Example   Apart from the parametric definition, it is possible to define a sphere very easily by using the implicit defnition x2 + y2 + z2 = r2 and by shortcut # 2, these are regular for all r > 0. In this case r = 0 is not a regular value.
    Even with assymptotes, the surface defined by x2y2 + x2z2 + y2z2 = 1 is regular. See picture below.


    Fig. 7
    Fig. 8