Rudiments of Surface Geometry in 
Structures of
is the Euclidean 3-dimensional space consisting of points represented by triples of real numbers
)
. The distinguished point )
is called the origin. In fact,
can also be regarded as a set of vectors, which are quantities that have both length and direction. In this way,
can be thought of as an arrow emanating from the origin, with the
tip at the point , or any other arrow having the same length and
direction.
There are various operations in . Addition of two vectors is
done componentwise:
+(x_2, y_2, z_2)=(x_1+x_2, y_1+y_2, z_1+z_2))
Similarly for subtraction:
-(x_2, y_2, z_2)=(x_1-x_2, y_1-y_2, z_1-z_2))
There are three kinds of multiplications
- scalar multiplication involves a real number and a vector, and the product is a vector.
=(\lambda x, \lambda y,
\lambda z))
- Inner product of two vectors is a real number(a scalar).
, (x_2, y_2, z_2)\rangle=
x_1x_2+y_1y_2+z_1z_2)
By Pythagoras' theorem, the length of is given by
, (x, y, z)\rangle})
We denote the length of a vector 
by
. So 
. In fact, for two vectors 
,
where 
is the included angle between
and 
.
Exercise 2.1 If 
and
, what can you say about
and ?
Solution
-
Vector product of two vectors is another vector which is perpendicular to that two vectors,
and points in a direction determined by the right-hand rule: the direction given by the thumb of your right hand
when the fingers curl from the direction of the first vector to that of the second vector. In coordinates,
\times (x_2, y_2, z_2)=
(y_1z_2-y_2z_1, x_2z_1-x_1z_2, x_1y_2-x_2y_1))
Exercise 2.2 Prove that 
, where
is the included angle between and .
Hint: prove that 
.
Solution
Exercise 2.3 Show that 
is the area of the parallelogram spanned by 
and .
Regular parametrized curves and surfaces
Note: In this subsection, we will deal with some differential calculus of functions of two variables. Given
)
, the partial differential of 
with respect to 
, denoted by 
,
can be obtained by differentiating with respect to as if the variable
were constant. Similarly for 
.
For instance, if )
, then )
, 
Let 
be given by
=(x(t), y(t), z(t)))
where 
are continuous functions mapping from an interval 
to
. If
can be differentiated infinitely many times, then we call 
a
smooth parametrized curve. If furthermore not all 
are 0 for all 
, then we call
a regular parametrized curve. Define =(x'(t), y'(t), z'(t)))
.
Example 2.1 Let be given by
=(R\cos t, R\sin t, 0), t\in[0, 2\pi])
is the circle of radius 
centered at the origin in the 
-plane. Since
=(-R\sin t, R\cos t, 0)\neq\vec{0},
t\in[0, 2\pi])
is a regular parametrized curve.
Example 2.2 Let be given by
=(R\cos t, R\sin t, bt),
t\in(-\infty, +\infty))
Then is a helix winding along the surface of an infinite cylinder of
radius with pitch 
.
Since
=(-R\sin t, R\cos t, b)\neq\vec{0},
t\in(-\infty, +\infty))
is a regular parametrized curve.
The helix
Exercise 2.4 Show that, if and
are smooth parametrized curves,
, \beta(t)\rangle=\langle\alpha'(t), \beta(t)\rangle+\langle\alpha(t), \beta'(t)\rangle)
Deduce that, if is a smooth parametrized curve lying on a sphere
centered at the origin in , then
, \alpha(t)\rangle=0)
.
Solution
Remark 2.2A In fact, )
is the tangent vector of the curve at
--it is the instantaneous direction at
of a particle moving along the curve in such a way that it is at
)
at time 
.
Put in another way, the assertion of the above exercise just says that any tangent vector of a smooth parametrized curve at
a point is perpendicular to the position vector of that point, if the curve lies on a sphere centered at the origin.
Definition 2.2B Define the curvature )
of a regular parametrized curve by
=\frac{\|\alpha'(t)\times\alpha''(t)\|}
{\|\alpha'(t)\|^3})
Exercise 2.5 If \|=1)
for all , then =
\|\alpha''(t)\|)
.
Solution
Example 2.3 The curvature of the circle of radius ,
as in Example 2.1, is
&=\frac{\|(-R\sin t, R\cos t, 0)\times(-R\cos t, -R\sin t, 0)\|}{\|(-R\sin t, R\cos t, 0)\|^3}\\
&=\frac{\|(0, 0, R^2)\|}{R^3}\\
&=\frac{1}{R}
\end{align*})
Exercise 2.6 Prove that the curvature of the helix as in Example 2.2 is
.
Solution
Exercise 2.7 Consider the smooth parametrized curve =(t^3, t^6, t^9))
,
)
. Is it a regular parametrized curve?
Can you compute the curvature at 
?
Solution
Remark 2.4 Curvature is a measure of how curved a curve is. Example 2.3
shows that the greater the radius is, the smaller the curvature of the circle becomes. This coincides with our geometric
intuition.
Let 
be given by
=(x(u, v), y(u, v), z(u, v)))
where 
and 
are intervals and , and
are functions in two variables )
.
Suppose that , and
can be differentiated with respect to 
and
infinitely many times. If we fix 
and let
vary in , we get a
curve )
, called a -coordinate curve.
Similarly, )
is a -coordinate curve,
with 
a constant. Put
)
)
They are the tangents of the - and -coordinate
curves respectively.
Definition 2.5 If 
and
are not multiple of each other for all
\in I_1\times I_2)
, then 
is a regular parametrized surface.
Remark 2.6 The above definition is different from the usual definition found in the literature,
which requires 
to be a homeomorphism, i.e.
can be morphed into the image of
without cutting and pasting. For the ease of presentation we omit
this requirement in our definition.
Exercise 2.8
- Show that and
are not multiple of each other iff
.
Solution
- and
are not multiple of each other iff there is a unique plane containing the two vectors passing through any particular point.
The above exercise shows that there exists a unique plane containing vectors
)
and )
passing through
)
. We call this plane the tangent plane of
at ,
and use 
to denote the tangent plane of a surface
at 
. Moreover we know
that \times \vec{x}_v(u_0, v_0)\neq \vec{0})
for all
\in I_1\times I_2)
. Define
the unit normal vector of . By the definition of vector
product, )
is perpendicular to the tangent plane at
.
Example 2.7 The unit sphere Let =(\sin u\cos v, \sin u\sin v, \cos u))
,
\in (0, \pi)\times[0, 2\pi))
. Then
is the unit sphere minus the two poles. The
-coordinate curves are the longitudes and the
-coordinate curves are the latitudes. Note that
\\
\vec{x}_v&=(-\sin u\sin v, \sin u\cos v, 0)\\
\|\vec{x}_u\times\vec{x}_v\|&=\|(\sin^2u\cos v, \sin^2u\sin v, \sin u\cos u)\|\\
&=\sin u\\
&\neq 0
\end{align*})
So is a regular parametrized surface.
The unit sphere
Example 2.8 Torus Let =((R+r\cos u)\cos v, (R+r\cos u)\sin v, r\sin u))
,
, \in[0, 2\pi)\times[0, 2\pi))
.
is a torus obtained by revolving a circle of radius
centered at )
around the -axis. The -coordinate
curves are vertical circles on the torus, whereas the -coordinate curves
are horizontal circles on the torus. We have
\\
\vec{x}_v&=(-(R+r\cos u)\sin v, (R+r\cos u)\cos v, 0)\\
\vec{x}_u\times\vec{x}_v&=(-r(R+r\cos u)\cos u\cos v, -r(R+r\cos u)\cos u\sin v, -r(R+r\cos u)\sin u)\\
\|\vec{x}_u\times\vec{x}_v\|&=r(R+r\cos u)\neq 0\text{ since }R>r
\end{align*})
So is a regular parametrized surface.
The torus
Example 2.9 Surface of revolution In general, if
=(p(u), q(u)))
is a regular parametrized curve in
with \neq 0)
,
then the equation of the surface of revolution obtained by rotating
around the -axis is
=(p(u)\cos v, p(u)\sin v, q(u)))
We have
\\
\vec{x}_v&=(-p\sin v, p\cos v, 0)\\
\vec{x}_u\times\vec{x}_v&=(-pq'\cos v, -pq'\sin v, pp')\\
\|\vec{x}_u\times\vec{x}_v\|&=p\sqrt{p'^2+q'^2}\neq 0\text{ since }p\neq 0\text{ and }\alpha\text{ is regular}.
\end{align*})
So is a regular parametrized surface.
Example 2.10 The graph of an infinitely differentiable function of two variables,
, is the parametrized surface
=(u, v, f(u, v)))
We have
\\
\vec{x}_v&=(0, 1, \frac{\partial f}{\partial v})\\
\vec{x}_u\times\vec{x}_v&=(-\frac{\partial f}{\partial u},
\frac{\partial f}{\partial v}, 1)\neq\vec{0}
\end{align*})
So the graph of is a regular parametrized surface.
Exercise 2.9 Show that the following are regular parametrized surfaces, and compute their normal vectors.
- The helicoid.
=(av\cos u, av\sin u, bu)$, $(u, v)\in \mathbb{R}^2, a>0, b\neq 0)
. Note: One can obtain a helicoid by dipping
a wire in the shape of a helix into a soap solution and taking it out, as shown on the Homepage.
Solution can be found in Example 3.4
The helicoid
- The catenoid.
=(\cosh u\cos v,
\cosh u\sin v, u))
, \in\mathbb{R}^2)
, where
. This is a special function which can be
found in your calculator. Note: The curve which is revolved around the -axis
to get the catenoid is called the catenary. Its equation is 
.
It is in the shape of a freely hanging chain with each end attached to the top of a vertical pole. The catenoid can be
obtained by dipping two identical circular wires into a soap solution and slowly drawing them apart.
Solution can be found here.
The catenoid
Various curvatures of regular parametrized surfaces
There are several notions of curvatures of surfaces, all of which are related to a special
map called the Gauss map.
Definition 2.11 The Gauss map
maps
to the unit normal vector of
at .
As you may see, the Gauss map measures how `rough' a surface is. The image of the Gauss map
of a plane consists of only one vector, whereas the Gauss map of a surface with lots of
sharp bumps and troughs has a `large' image which may be thought of as a large area on the
unit sphere swept out by the unit normal vectors. Moreover, the local behaviour of the
Gauss map gives a measure of how curved a surface is around a certain point.
Definition 2.12 The Gauss curvature
at of a surface
is the limit of the following quotient
\text{ on }\mathbb{S}^2}
{\text{Area of a neighbourhood }\mathcal{O}_p\text{ around }p\text{ on }S})
as 
shrinks to
. The sign is determined as follows:
if is a small closed curve
around for
,
=\alpha(1))
and traced out
counterclockwise as ranges from 0 to 1
when you view the surface from the side to which )
is pointing, and so is ))
,
then the sign should be positive. Otherwise, negative sign should be attached.
Example 2.13 The Gauss map for the unit sphere
is the identity map. So the
Gauss curvature is 1. On the other hand, the Gauss curvature of any plane is 0 because the
image of the Gauss curvature consists of only one vector.
Example 2.14 The Gauss map for the cylinder
=(\cos v, \sin v, u))
has the equator of as its
image, which has no area. So the Gauss curvature of the cylinder is 0.
Though intuitive, the definition above will not be made used of to compute the Gauss
curvature of other surfaces. Instead, we will introduce the first and second fundamental
forms.
Definition 2.14A The first fundamental form 
at
maps a pair of tangent vectors of
at
to their inner product:
=
\langle \vec{v}, \vec{w}\rangle, \vec{v}, \vec{w}\in T_pS)
Remark 2.14B The first fundamental form reflects the intrinsic property of surfaces, meaning that
it gives the local information of length and area on the surface. For a regular
parametrized surface )
, put
,
F=I(\vec{x}_u, \vec{x}_v), G=I(\vec{x}_v, \vec{x}_v))
Definition 2.14C The second fundamental form
at
is defined to be
=
\left\langle-\frac{d}{dt}N(\alpha(t)), \vec{w}\right\rangle)
where and
are tangent vectors at
, and
is any curve on
such that
=p)
and
=\vec{v})
.
Remark 2.14D The second fundamental form reflects the
extrinsic property of surfaces, meaning that it shows how surfaces are embedded in
.
For a regular parametrized surface ,
put
,
f=II(\vec{x}_u, \vec{x}_v), g=II(\vec{x}_v, \vec{x}_v))
In fact, one can show that
Exercise 2.10 Show that
,
,
,
where )
, the normal vector at
,
and
, etc.
Solution
In fact, for which can be
differentiated infinitely many times, the order of taking partial differentiations is immaterial, i.e.
. By the above
exercise, =
II(\vec{x}_v, \vec{x}_u))
. If and
are tangent vectors at
, then
+
b\vec{x}_v(u_0, v_0))
and
+
b'\vec{x}_v(u_0, v_0))
for some
. It can be shown
that
=aa'e+(ab'+a'b)f+bb'g=II(\vec{w}, \vec{v}))
Theorem 2.15 The Gauss curvature is given by
Another way to compute the Gauss curvature, which may be more tedious but relevant to the
next section, is to use principal curvatures
and 
,
which are defined as in
Definition 2.15A
=\underset{\|v\|=1, v\in T_pS}{\max}II_p(v, v))
=\underset{\|v\|=1, v\in T_pS}{\min}II_p(v, v))
Theorem 2.16
Definition 2.16A Define the
mean curvature 
.
Theorem 2.17
})
Now with the various tools in our hands, we are able to compute
and 
for all the regular parametrized surfaces in the previous examples.
Example 2.18 The torus.
^2)
\\
\vec{x}_{uu}&=(-r\cos u\cos v, -r\cos u\sin v, -r\sin u)\\
\vec{x}_{uv}&=(r\sin u\sin v, -r\sin u\cos v, 0)\\
\vec{x}_{vv}&=(-(R+r\cos u)\cos v, -(R+r\cos u)\sin v, 0)
\end{align*})
)
}\\
H&=\frac{Eg+Ge-2Ff}{2(EG-F^2)}=\frac{R+2r\cos u}{2r(R+r\cos u)}
\end{align*})
Let us also verify the Gauss and mean curvatures by computing the principal curvatures.
Let
&=\frac{\vec{x}_u(u, v)}{\|\vec{x}_u(u, v)\|}=(-\sin u\cos v, \sin u\sin v, \cos u)\\
\vec{v}(u, v)&=\frac{\vec{x}_v(u, v)}{\|\vec{x}_v(u, v)\|}=(-\sin v, \cos v, 0)
\end{align*})
They are unit tangent vectors which are perpendicular to each other. It is clear that any
unit tangent vector can be written as 
with 
. Hence
&=a^2II(\vec{u}, \vec{u})+2abII(\vec{u}, \vec{v})+b^2II(\vec{v}, \vec{v})\\
&=\frac{a^2e}{r^2}+\frac{b^2g}{(R+r\cos u)^2}\\
&=\frac{a^2}{r}+\frac{b^2\cos u}{R+r\cos u}
\end{align*})
It can be shown that the above expression attains maximum when
, and minimum when
. So
Indeed, 
,
.
Example 2.19 Surface of revolution
\\
\vec{x}_{uu}&=(p''\cos v, p''\sin v, q'')\\
\vec{x}_{uv}&=(-p'\sin v, p'\cos v, 0)\\
\vec{x}_{vv}&=(-p\cos v, -p\sin v, 0)
\end{align*})
q'}{p(p'^2+q'^2)}\\
H&=\frac{(p'^2+q'^2)q'+p(p'q''-p''q')}{p(p'^2+q'^2)^{\frac{3}{2}}}
\end{align*})
Example 2.20 The graph of an infinitely differentiable
function
^2, F=\frac{\partial f}{\partial u}\frac{\partial f}{\partial v},
G=1+\left(\frac{\partial f}{\partial v}\right)^2)
^2+\left(\frac{\partial f}{\partial v}\right)^2}}\left(-\frac{\partial f}{\partial u}, -\frac{\partial f}{\partial v}, 1\right)\\
\vec{x}_{uu}&=\left(0, 0, \frac{\partial^2 f}{\partial u^2}\right)\\
\vec{x}_{uv}&=\left(0, 0, \frac{\partial^2 f}{\partial u\partial v}\right)\\
\vec{x}_{vv}&=\left(0, 0, \frac{\partial^2 f}{\partial v^2}\right)
\end{align*})
^2+\left(\frac{\partial f}
{\partial v}\right)^2}}, f=\frac{\frac{\partial^2f}{\partial u\partial v}}
{\sqrt{1+\left(\frac{\partial f}{\partial u}\right)^2+\left(\frac{\partial f}{\partial v}
\right)^2}}, g=\frac{\frac{\partial^2f}{\partial v^2}}{\sqrt{1+\left(\frac{\partial f}
{\partial u}\right)^2+\left(\frac{\partial f}{\partial v}\right)^2}})
^2}{\left(1+\left(\frac{\partial f}
{\partial u}\right)^2+\left(\frac{\partial f}{\partial v}\right)^2\right)
\left(\left(1+\left(\frac{\partial f}{\partial u}\right)^2\right)
\left(1+\left(\frac{\partial f}{\partial v}\right)^2\right)-\left(\frac{\partial f}
{\partial u}\frac{\partial f}{\partial v}\right)^2\right)})
^2\right)\frac{\partial^2 f}{\partial v^2}+\left(1+\left(\frac{\partial f}
{\partial v}\right)^2\right)\frac{\partial^2 f}{\partial u^2}-2\frac{\partial f}
{\partial u}\frac{\partial f}{\partial v}\frac{\partial^2 f}{\partial u\partial v}}
{2\sqrt{1+\left(\frac{\partial f}{\partial u}\right)^2+\left(\frac{\partial f}
{\partial v}\right)^2}\left(\left(1+\left(\frac{\partial f}{\partial u}\right)^2\right)
\left(1+\left(\frac{\partial f}{\partial v}\right)^2\right)-\left(\frac{\partial f}
{\partial u}\frac{\partial f}{\partial v}\right)^2\right)})