Math 162A - Introduction to Differential Geometry
Neil Donaldson
Winter 2024
Introduction
Classical Differential Geometry is the study of curves and surfaces in the plane and three-dimensional
space using multi-variable calculus, linear algebra & differential equations. At a more advanced level,
topology, analysis and abstract algebra become more important, but none of this is required for our
treatment.
Of particular interest is the notion of curvature: a measure of the ‘bendiness’ of a curve or surface.
Intuitively, a straight line should have zero curvature, while the curvature of a circle should vary
inversely as the radius: a very large circle should have very small curvature.
Zero curvature Small curvature Larger curvature Variable curvature
Understanding and quantifying this concept for more complicated curves is our first important goal.
The rough idea is to imagine a curve as a roller-coaster along which you travel at a constant speed;
the curvature is then the force necessary to keep you travelling along the curve.
Curvature is a more difficult concept for surfaces. In particular, we will hunt for quantities which
measure how much a surface appears to be dome- or saddle-shaped.
Dome-shaped Saddle-shaped More complicated
The third surface is saddle-shaped near the narrow neck and dome-shaped away from it.
1 Curves in Euclidean Space
1.1 Euclidean Space, Tangent Vectors & Regular Curves
We begin by refreshing and developing a little notation.
Definition 1.1. The set of n-tuples of real numbers is denoted R
n
.
An element can be thought of either as a point P or as its position vector
p =
−→
OP connecting the origin O = (0, . . . , 0) to P.
In co-ordinates, points are typically written as row vectors
P = (p
1
, . . . , p
n
) where each p
i
∈ R
P
p
O
P has position vector p
For vectors, either row or column vector notation is acceptable.
For each i, the co-ordinate function x
i
: R
n
→ R returns the i
th
co-ordinate of a point: x
i
(P) = p
i
.
Since the focus of the course is curves and surfaces in 2- and 3-dimensions, we’ll mostly restrict to
n ≤ 3 and quote theorems in this context.
1
We typically use x, y, z for the standard (rectangular)
co-ordinate functions
x(P) = p
1
, y(P) = p
2
, z(P) = p
3
You should be comfortable with this notation from previous classes and, in particular, with partial
derivatives of functions defined in terms of the co-ordinate functions x, y, z.
Examples 1.2. 1. If P = (3, 1, 5) ∈ R
3
, then y(P) = 1.
2. The function f : R
3
→ R defined by f = x
3
sin(yz) has partial derivatives
∂ f
∂x
= 3x
2
sin(yz)
∂ f
∂y
= x
3
z cos(yz)
∂ f
∂z
= x
3
y cos(yz)
A vector is a directed line segment joining two points. We’ve already seen the position vector of a point
P, namely
−→
OP. In differential geometry it is crucial to distinguish the vectors based at a given point.
Definition 1.3. A tangent vector v
p
is a pair of elements of R
3
: a base point p and
a direction v. It is the directed line segment from the point with position vector
p to the point with position vector p + v.
The tangent space at p is the set T
p
R
3
of all tangent vectors based at p. At each
point, R
3
has a different tangent space!
p
p
v
p
p + v
v
O
Be aware that v
p
= w
q
⇐⇒ p = q and v = w: the same direction at different base points means a
different tangent vector!
1
For simplicity’s sake, we’ll almost always state theorems in R
3
. The majority are valid in R
n
with a simple notational
modification {x, y, z} ⇝ {x
1
, . . . , x
n
}. For R
2
just delete z = x
3
; many results even make sense in R = R
1
!
2
The tangent space at p is suitably named, for it is indeed a
vector space: to add tangent vectors v
p
, w
p
∈ T
p
R
3
, simply
sum the direction vectors
v
p
+ w
p
:= (v + w)
p
(∗)
Scalar multiplication is similar: λ v
p
:= (λv)
p
.
p
p
v
p
w
w
p
v
p
+ w
p
v
O
In chapter 2 we will see a more abstract discussion of tangent vectors, vector fields, and their appli-
cation.
Euclidean Space: E
n
versus R
n
To describe curves and surfaces in differential geometry, we parametrize using functions.
Example 1.4. There are multiple ways to do this for a given curve: for instance
x : (−π, π] → R
2
: t 7→
cos t, sin t
and y : R → R
2
: s 7→
1 − s
2
1 + s
2
,
2s
1 + s
2
both parametrize (most of) the unit circle in the plane (y ignores the point (−1, 0)).
Plainly the codomain R
2
is where the geometric action is: in the above we have the same circle, and
concepts such as length and angle can be measured. This extra structure motivates us to distinguish
the codomain with new notation.
Definition 1.5. Euclidean space E
n
is R
n
equipped with the usual dot product. Specifically in E
3
:
The dot product of p and q is p ·q = p
T
q = (p
1
p
2
p
3
)
q
1
q
2
q
3
= p
1
q
1
+ p
2
q
2
+ p
3
q
3
The length of p is
||
p
||
=
√
p ·p =
q
p
2
1
+ p
2
2
+ p
2
3
The angle θ between p and q satisfies cos θ =
p ·q
||
p
||||
q
||
Vectors are orthogonal/perpendicular if p ·q = 0, equivalently θ =
π
2
; we write p ⊥ q.
Curves in E
2
and E
3
This course is primarily concerned with functions x : U ⊆ R
m
→ E
n
. In particular:
Plane curves: m = 1 and n = 2; for example the above circle.
Spacecurves: m = 1 and n = 3; we’ll see several momentarily.
Surfaces: m = 2 and n = 3. For instance, the parametrization x : R
2
7→ E
3
: (u, v) 7→ (u, v, u
2
+ v
2
) of
a paraboloid should be familiar.
Surfaces are the focus of Chapter 3. It is now time for the formal definition of a curve.
3
Definition 1.6. A (smooth parametrized) curve is a function, x : I → E
3
, x( t) =
x(t), y(t), z(t)
,
defined on an interval I and whose components x, y, z are infinitely differentiable
2
everywhere on I.
Its derivative (also velocity or tangent vector) is denoted
x
′
( t) =
dx
dt
=
x
′
( t), y
′
( t), z
′
( t)
The curve’s speed is the continuous scalar function
v(t) =
x
′
( t)
=
q
x
′
( t)
2
+ y
′
( t)
2
+ z
′
( t)
2
A curve is regular if its tangent vector x
′
( t) is everywhere non-zero.
O
x(t)
x
′
(t)
p
In the context of Definitions 1.1 and 1.3, note that for each t ∈ I:
x(t) is a position vector whose nose describes the location of a point on the curve.
x
′
( t) ∈ T
p
E
3
is a tangent vector based at the point p with position vector x(t).
A parametrized curve has an orientation (indicated by the blue arrow): as t increases along the interval
I, the point x(t) moves in a particular direction along the curve.
Examples 1.7. Straight line: The line through points with position
vectors a, b may be parametrized by
x(t) = a + t( b −a) = (1 − t)a + tb
The tangent vector at x(t) is the constant x
′
( t) = b −a and the
parametrization has constant speed
||
b −a
||
. For instance,
x(t) =
2 + t, 3 −2t
has constant velocity x
′
( t) = (1, −2) and speed v(t) =
√
5.
Circle (Example 1.4) The parametrization x(t) =
cos t, sin t
has ve-
locity x
′
( t) =
−sin t, cos t
and constant speed v(t) = 1.
By contrast, y(s) =
1
1+s
2
1 − s
2
, 2s
has non-constant speed
y
′
( s) =
2
(1 + s
2
)
2
−2s, 1 − s
2
v(s) =
2
1 + s
2
A particle moves quickest at s = 0 when v(0) = 2 and the speed
tends to zero as s → ±∞ (see the linked animation).
O
x(0) = a
x(1) = b
x
y
y(−2)
y
′
(− 2)
y(0)
y
′
(0)
y(0.4)
y
′
(0.4)
y(3)
y
′
(3)
2
The meaning of smooth depends on the author: at a minimum it means that x, y, z must be differentiable with continuous
derivative. We take the maximal approach for simplicity.
4
Helix x(t) =
cos t, sin t, t
parametrizes a helix (ascending spiral).
To help visualize this, imagine sitting on top of the z-axis and look-
ing down; you’d see its horizontal projection t 7→
cos t, sin t
(a
counter-clockwise circle). Since z(t) = t, the curve moves upwards
at constant speed. One can similarly project onto the xz- and yz-
planes.
−1
1
y
−1 1
x
z
−1 1
x
π
−π
2π
3π
z
−1 1
y
π
−π
2π
3π
The tangent vector at x(t) is x
′
( t) =
−sin t, cos t, 1
and the speed is constant v(t) =
√
2.
Tangent Line Let x : I → E
3
be regular and t
0
∈ I be fixed. The tangent line
at x(t
0
) is simply the straight line through the point with position
vector x(t
0
) oriented in the direction of the tangent vector x
′
( t
0
). It
is itself a parametrized curve, y : R → E
3
:
y(s) = x(t
0
) + sx
′
( t
0
)
For example, the tangent line to the above helix at t
0
=
7π
3
is
y(s) =
1
2
,
√
3
2
,
7π
3
!
+
−
√
3
2
,
1
2
, 1
!
s
The tangent line has the same speed as the helix
√
2.
Self-intersections These are no problem for our formulation! The curve
x(t) =
sin
2t
3
, cos t
, t ∈ [0, 6π)
passes through the origin at both t
1
=
3π
2
and t
2
=
9π
2
, with
corresponding tangent vectors
x
′
(
3π
2
) =
−
2
3
, 1
, x
′
(
9π
2
) =
−
2
3
, −1
In this example, we shouldn’t talk about the tangent vector to the
curve at the origin, since it is non-unique. Rather we should refer
to the co-ordinates
3π
2
or
9π
2
.
−1
1
y
−1 1
x
The linked animation shows the variable speed v(t) =
q
4
9
cos
2
2t
3
+ sin
2
t of this curve.
5
Corners and Cusps To ensure that a tangent direction exists, a regular curve has everywhere non-
zero derivative. Here are a couple of examples of curves with non-regular points.
Examples 1.8. Corner A curve might enter and leave a point in dif-
ferent directions. For example, x(t) =
t, 1 −
|
t
|
has deriva-
tive
x
′
( t) =
(
(1, 1) if t < 0
(1, −1) if t > 0
At x(0) = (0, 1) the curve is non-differentiable and thus non-
smooth and non-regular.
Cusp The curve x(t) =
t
3
, t
2
has derivative
x
′
( t) =
3t
2
, 2t
The origin is a cusp, a special type of corner where the curve
leaves the point in the opposite direction to how it entered.
In this case the curve is differentiable at the origin, but is non-
regular since its speed v(0) is zero.
1
y
−1 0 1
x
1
y
−1 0 1
x
Exercises 1.1. 1. A twice-differentiable curve x(t) has the property that its second derivative x
′′
( t)
is identically zero. What can be said about x?
2. Find the unique curve such that x(0) = ( 1, 0, 5) and x
′
( t) = (t
2
, t, e
t
).
3. An ellipse in the plane has equation
x
2
4
+
y
2
9
= 1. By modifying the standard parametrization
of the circle, find a regular parametrization of this ellipse. What is its speed?
4. Show that x(t) =
e
t
+e
−t
2
,
e
t
−e
−t
2
) parametrizes half of the hyperbola x
2
− y
2
= 1. How would
you parametrize the other half?
5. (a) Find the speed of the re-parametrized standard helix y(s) = x(s
3
) =
cos s
3
, sin s
3
, s
3
.
(b) More generally, if x(t) is a regular curve, show that y(s) := x(s
3
) is non-regular.
6. Verify that our cusp example (above) may instead be parametrized y(u) =
u, u
2/3
. Is the new
parametrization still non-regular at the origin? Explain.
7. Show that the tangent vectors to the regular curve x( t) =
3t, 3t
2
, 2t
3
make a constant angle
with the vector ( 1, 0, 1).
8. Consider the plane curve x(t) =
t −1 + e
−t
, e
−t
. Find the equation of its tangent line at t = t
0
and find where the tangent line intersects the x-axis.
9. Let f : R → R be a smooth function. Find a parametrization for the graph of y = f (x) and find
its tangent line when x = x
0
.
10. Find a parametrization of the straight line through the points (1, −3, −1) and (6, 2, 1). Does this
line meet the line through the points (−1, 1, 0) and (−5, −1, −1)?
6
1.2 The Arc-length Parametrization and Curvature
As we’ve seen, the same ‘curve’ (viewed as a subset of E
3
) may be parametrized in different ways.
For instance, in Exercise 1.1.5, the standard helix x(t) = (cos t, sin t, t) was re-parametrized to obtain
y(s) = (cos s
3
, sin s
3
, s
3
) (∗)
This new parametrization is non-regular at s = 0; it slows down and stops before resuming its journey
up the helix! Regularity is not, therefore, an intrinsic property of a curve viewed as a set (range(x)),
rather it is a property of the parametrization.
Thankfully it is easy to create new parametrizations that remain regular.
Lemma 1.9. If x : I → E
3
is regular and α : J → I is smooth with nowhere-zero derivative, then we
obtain a new regular parametrization
y : J → E
3
, y(s) := x
α(s)
Proof. By the chain rule,
dy
ds
= α
′
( s)
dx
dt
, which is non-zero by assumption.
By contrast, if x(t) is non-regular, no smooth reparametrization can possible regularize it.
Since α
′
( s) is continuous and non-zero, there are two distinct cases:
3
α(s) increasing We call this an orientation-preserving re-parametrization, since a ‘particle’ travels along
the curve in the same direction.
α(s) decreasing The re-parametrization is orientation-reversing.
In the language of the Lemma, (∗) turned a regular parametrization into a non-regular one because
α(s) = s
3
has α
′
( s) = 3s
2
which is zero at s = 0.
Our next goal is to develop a special parametrization for regular curves. First we recall a concept
from multi-variable calculus.
Definition 1.10. The (signed) arc-length of a curve x : I → E
3
measured from x(t
0
) to x(t) is the
integral of the speed
s(t) =
Z
t
t
0
x
′
(T)
dT =
Z
t
t
0
v(T) dT
The arc-length is signed because it is negative if t < t
0
: we are measuring length against the orienta-
tion of the curve. Of course if x : [a, b] → E
3
has domain a closed bounded interval, then it is most
sensible to measure arc-length from t
0
= a so that s(t) ≥ 0 everywhere on the curve.
Example 1.11. The standard helix x(t) =
cos t, sin t, t
has constant speed
√
2, whence the arc-
length measured from x(0) is simply s(t) =
√
2t.
3
Observe that α
′
(s) is always positive or always negative. In particular, α(s) is 1–1. If, in addition, α is onto, then x and y
parametrize precisely the same subset of E
3
.
7
Recall the Fundamental Theorem of Calculus: is s(t) is the arc-length of a regular curve, then
ds
dt
=
d
dt
Z
t
t
0
x
′
(T)
dT =
x
′
( t)
= v(t)
is the curve’s speed, which is positive and continuous. The same is therefore true for its inverse function
dt
ds
=
1
s
′
( t)
=
1
v(t)
> 0
Definition 1.12. An arc-length parameter for a regular curve x(t) is the inverse α( s) = t(s) of an
arc-length function s(t).
Lemma 1.9 tells us that y(s) = x
α(s)
is a regular re-parametrization of our original curve. Indeed
it is a re-parametrization with a very special property:
y
′
( s)
= α
′
( s)
x
′
α(s)
=
1
v(t)
v(t) = 1 (†)
The curve y(s) has unit-speed. We have therefore proved a key result.
Theorem 1.13. Every regular curve has a unit-speed parametrization, namely by an arc-length pa-
rameter (measured from wherever you like).
The usefulness of the Theorem is abstract; by assuming that we have a unit-speed parametriza-
tion, certain analyses become much simpler. As a practical matter, explicitly finding an arc-length
parametrization might be essentially impossible (evaluate an integral then invert a function. . . ).
Examples 1.14. 1. Since the standard helix has arc-length parameter s(t) =
√
2t, it is trivial to
observe that the re-parametrization
y(s) =
cos
s
√
2
, sin
s
√
2
,
s
√
2
has unit speed.
2. More generally, if x( t) has constant speed v, then s(t) = vt is an arc-length parameter and
y(s) = x(
s
v
) a unit-speed re-parametrization.
3. The graph of y =
2
3
x
3/2
(t ≥ 0) may be parametrized by x(t) = (t,
2
3
t
3/2
). The arc-length
measured from the origin is then
s(t) =
Z
t
0
√
1 + T dT =
2
3
h
(1 + t)
3/2
−1
i
=⇒ α(s) = t(s) =
1 +
3
2
s
2/3
−1
We’ve obtained an explicit unit-speed parametrization
y(s) = x
α(s)
=
1 +
3
2
s
2/3
−1,
2
3
h
1 +
3
2
s
2/3
−1
i
3/2
though is it really something you ever want to compute with?!
8
Armed with unit-speed curves, we can now define our principal notion of bendiness.
Definition 1.15. The curvature of a unit-speed curve x : I → E
3
is
κ(s) =
x
′′
( s)
We modify this slightly for curves in the plane: κ(s) is positive/negative if the tangent vector rotates
counter-clockwise/clockwise as we traverse the curve. This corresponds to the usual right hand rule.
By Newton’s second law, a unit mass travelling along the curve at unit speed experiences a transverse
force of magnitude κ(s).
Examples 1.16. 1. A straight line has curvature zero. For example, the line joining (1, 4) and (−3, 1)
has unit-speed parametrization x(s) =
−3 +
4
5
s, 1 +
3
5
s
, whence x
′′
( s) = 0 =⇒ κ(s) = 0.
2. The circle of radius r has unit-speed parametrization x(s) = r
cos
s
r
, sin
s
r
, whence
x
′′
( s) = −
1
r
cos
s
r
sin
s
r
=⇒ κ(s) =
1
r
This is positive since the tangent vector rotates counter-clockwise. Observe that κ =
1
r
is in-
versely proportional to the radius: smaller circles have larger curvature.
3. The standard helix with unit-speed parametrization x(s) =
cos
s
√
2
, sin
s
√
2
,
s
√
2
has
x
′′
( s) = −
1
2
cos
s
√
2
, sin
s
√
2
, 0
=⇒ κ(s) =
1
2
Since finding a unit-speed parametrization is difficult, there few curves for which this approach is
sensible. What we want is a method that works for arbitrary parametrization. This is indeed possible,
though for spacecurves it will take a while. For curves in the plane however, things are fairly easy.
Curvature of Plane Curves If y : I → E
2
has unit-speed, we can write
y
′
( s) =
cos θ(s)
sin θ(s)
where θ(s) is the angle between the tangent line and the positive x-axis.
Now observe that
y
′′
( s) = θ
′
( s)
−sin θ(s)
cos θ(s)
Since
−sin θ, cos θ
points to the left of y
′
( s), we conclude:
(
cos θ
sin θ
)
(
−sin θ
cos θ
)
θ(s)
κ(s) = θ
′
(s) > 0
Theorem 1.17. The curvature of a unit-speed plane curve is the rate of change κ(s) = θ
′
( s) of the
angle of its tangent line.
This should be intuitive for constant curvature examples such as the straight line and the circle.
9
Now suppose x(t) =
x(t), y(t)
is any regular parametrization of the same curve; its speed satisfies
v(t) =
q
x
′
( t)
2
+ y
′
( t)
2
= s
′
( t)
where s(t) is an arc-length function for x(t). Moreover, the angle θ(s) plainly satisfies
θ(s) = tan
−1
y
′
( t)
x
′
( t)
Now differentiate and applying the chain rule:
κ(s) =
d
ds
tan
−1
y
′
( t)
x
′
( t)
=
dt
ds
d
dt
tan
−1
y
′
( t)
x
′
( t)
= ···
The result is a formula for the curvature as a function of an arbitrary regular parametrization.
Corollary 1.18. A regular curve x(t) =
x(t), y(t)
has curvature
κ(t) =
y
′′
x
′
− x
′′
y
′
[x
′2
+ y
′2
]
3/2
=
y
′′
x
′
− x
′′
y
′
v
3
=
x
′′
· Jx
′
v
3
where Jx
′
=
0 −1
1 0
x
′
y
′
=
−y
′
x
′
. In particular, the graph of a smooth function y = f (x) has
curvature
κ(x) =
f
′′
(x)
h
1 +
f
′
(x)
2
i
3/2
Examples 1.19. 1. The graph of y =
2
3
x
3/2
has curvature
κ(x) =
1
2
x
−1/2
(1 + x)
3/2
=
1
2
p
x(1 + x)
3
2. If f (x) = sin x, then κ(x) =
−sin x
(1 + cos
2
x)
3/2
3. The spiral x(t) =
t cos t, t sin t
has
x
′
( t) =
cos t − t sin t
sin t + t cos t
, x
′′
( t) =
−2 sin t −t cos t
2 cos t − t sin t
=⇒ κ(t) =
(2 cos t −t sin t)(cos t −t sin t) − (−2 sin t − t cos t)(sin t + t cos t)
[
(cos t −t sin t)
2
+ (sin t + t cos t)
2
]
3/2
=
2 + t
2
[
1 + t
2
]
3/2
10
Exercises 1.2. 1. Compute the arc-length of the following curves by parametrizing and evaluating
an integral:
(a) The straight line between points (3, 1, 2) and (1, 1, 0).
(b) The circle centered at ( 1, −2) with radius 5 measured clockwise from (6, −2) to (1, 3).
(c) The graph of the function y =
2
3
x
3/2
−
1
2
x
1/2
for 1 ≤ x ≤ 9.
2. Find the curvature of the following plane curves (use Corollary 1.18).
(a) The graph of y = x
2
.
(b) The catenary: the graph of y =
1
2
( e
x
+ e
−x
) = cosh x
(c) The figure-eight curve x(t) =
cos t, sin 2t
(d) The exponential spiral x(t) =
e
t
cos t, e
t
sin t
.
3. Find a unit-speed parametrization of the straight line between points with position vectors
a = b in E
3
and hence verify that its curvature is zero.
4. Suppose x : R → E
3
has unit speed. Verify that x is parametrized by an arc-length parameter.
5. Find the curvature of the spacecurve x(s) =
5
13
cos s, sin s,
12
13
cos s
. What is this curve?
6. (a) Find the arc-length of the standard helix x(t) =
(
cos t, sin t, t
)
between t = −π and t = 2π.
(b) Suppose a particle travels down the standard helix so that y(0) = (1, 0, 2π) and such that
its speed is v(t) = 2
√
2t. Find a parametrization which describes this motion.
(c) Let r, h be positive constants. Find the curvature of the general circular helix
x(t) =
r cos t, r sin t, ht
and interpret how it depends on r and h.
7. Check the evaluation of κ(t) and κ(x) in the proof of Corollary 1.18.
8. We find the curvature of the exponential spiral x(t) =
e
t
cos t, e
t
sin t
the hard way.
(a) Calculate the arc-length s(t) measured from x(0).
(b) Find a unit-speed parametrization y(s) where y(0) = (1, 0).
(c) Hence compute κ(s) and show that it equals your answer from Exercise 2d.
9. A circle of radius 1 rolls at constant speed with-
out slipping along the x-axis so that the angle
indicated in the picture is t at time t.
The curve described by a point on the circum-
ference of the rolling circle is a cycloid.
0
1
2
y
x
t
π 2π
3π
2
π
2
0
(a) Find a parametrization x : [0, 2π] → E
2
.
(b) Find the curvature of the cycloid as a function of t.
(c) Compute the arc-length of the cycloid over a complete rotation of the circle.
11
1.3 Orthogonality, Moving Frames & The Structure Equations
Our plan is to analyze a curve with respect to a family of moving orthonormal bases. Before embark-
ing on this, we summarize the relevant ideas from linear algebra. Hopefully most of the concepts
are familiar. Most proofs are omitted, but will be met in a standard linear algebra class. As usual,
definitions and results are stated in 3-dimensions, but are valid in others, particularly 2-dimensions.
In E
3
, points are typically denoted with reference to the standard basis {i, j, k}. For instance,
v =
3
4
6
= 3i + 4j + 6k
The numbers 3, 4, 6 are the co-ordinates of v with respect to the standard basis. Of course other bases
are available. . .
Definition 1.20. A set β = {e
1
, e
2
, e
3
} ⊆ E
3
is a basis if every vector v ∈ E
3
can be expressed
uniquely
4
as a linear combination of e
1
, e
2
, e
3
: that is
v = c
1
e
1
+ c
2
e
2
+ c
3
e
3
(∗)
for unique c
1
, c
2
, c
3
∈ R, the co-ordinates of v with respect to β.
A basis is orthonormal if e
j
·e
k
= δ
jk
=
(
1 if j = k
0 if j = k
Consider the (invertible) matrix E = (e
1
e
2
e
3
) whose columns are the elements of β viewed as
column vectors (with respect to the standard basis). A basis is positively oriented if det E > 0.
Examples 1.21. 1.
n
1
√
2
1
1
0
,
1
√
6
1
−1
−2
,
1
√
3
1
−1
1
o
is a negatively oriented orthonormal basis of E
3
(det E = −1 < 0).
2. Every orthonormal basis of E
2
has the form
cos θ
sin θ
,
−sin θ
cos θ
or
cos θ
sin θ
,
sin θ
−cos θ
for some angle θ. The first is positively oriented (det = 1 > 0)
and the second negatively (det = −1 < 0).
(
cos θ
sin θ
)
(
−sin θ
cos θ
)
(
sin θ
−cos θ
)
θ
A positively oriented orthonormal basis in E
3
satisfies the right-hand rule: e
3
= e
1
×e
2
. In E
2
, positive
orientation means that e
2
is obtained by rotating e
1
counter-clockwise by 90°: we can write this as
e
2
= Je
1
=
0 −1
1 0
e
1
4
In a linear algebra class this is usually broken into two definitions which imply, respectively, the existence and unique-
ness of the linear combination (∗).
Spanning Set Every v ∈ E
3
can be expressed as a linear combination v = c
1
e
1
+ c
2
e
2
+ c
3
e
3
for some c
1
, c
2
, c
3
∈ R.
Linear Independence The only linear combination summing to 0 is trivial: c
1
e
1
+ c
2
e
2
+ c
3
e
3
= 0 =⇒ c
1
= c
2
= c
3
= 0.
12
Finding the co-ordinates of a vector with respect to a basis (∗) is really a matrix problem
5
v = E
c
1
c
2
c
3
=⇒
c
1
c
2
c
3
= E
−1
v
Inverting a 3 ×3 matrix is tedious. Thankfully the co-ordinates can be found more easily if the basis
is orthonormal just by taking dot products!
v ·e
i
= (c
1
e
1
+ c
2
e
2
+ c
3
e
3
) ·e
i
= c
i
Lemma 1.22. If β = {e
1
, e
2
, e
3
} is an orthonormal basis, then for any vector v ∈ E
3
,
v = (v ·e
1
) e
1
+ ( v ·e
2
) e
2
+ ( v ·e
3
) e
3
Example 1.23. β = {e
1
, e
2
} =
1
5
4
3
,
1
5
−3
4
is a positively oriented orthonormal basis of E
2
.
With respect to β, the vector v =
1
1
can be written
1
1
= (v ·e
1
) e
1
+ ( v ·e
2
) e
2
=
7
5
e
1
+
1
5
e
2
Orthogonal Matrices Recall Definition 1.5. Given β = {e
1
, e
2
, e
3
} and its associated matrix E =
( e
1
e
2
e
3
), observe that
E
T
E =
e
T
1
e
T
2
e
T
3
( e
1
e
2
e
3
) =
||
e
1
||
2
e
1
·e
2
e
1
·e
3
e
2
·e
1
||
e
2
||
2
e
2
·e
3
e
3
·e
1
e
3
·e
2
||
e
3
||
2
When β is an orthonormal basis, this matrix is very simple.
Definition 1.24. A 3 ×3 matrix A is orthogonal if A
T
A = I (equivalently AA
T
= I). The set of all
such is denoted O
3
(R ). In addition, if det A = 1, we write A ∈ SO
3
(R ) (special orthogonal matrices).
Lemma 1.25. 1. If A ∈ O
3
(R ), then it is invertible with inverse A
T
(also orthogonal).
2. The product of two orthogonal matrices is orthogonal.
3. A is orthogonal if and only if (Ax) ·(Ay) = x ·y for all vectors x, y ∈ E
3
.
4. Let β = {e
1
, e
2
, e
3
} and E = (e
1
e
2
e
3
) ∈ M
3
(R ):
(a) E ∈ O
3
(R ) ⇐⇒ β is an orthonormal basis.
(b) E ∈ SO
3
(R ) ⇐⇒ β is a positively oriented orthonormal basis.
Parts 1 and 2 together say that O
3
(R ) forms a group under matrix multiplication; it is therefore known
as the orthogonal group.
6
5
For obvious reasons, this is known as the change of co-ordinate matrix from β to the standard basis {i, j, k}.
6
Knowledge of group theory is not necessary for these notes. Lie Groups, of which O
3
(R) is an example, are critical to
more advanced differential geometry.
13
Examples (1.21 cont). 1. It is no fun to check E
T
E = I directly, but since we have an orthonormal
basis, the lemma tells us that
E =
1
√
2
1
√
6
1
√
3
1
√
2
−1
√
6
−1
√
3
0
−2
√
6
1
√
3
∈ O
3
(R )
2. Every 2 ×2 orthogonal matrix has one of two forms:
Rotations A
θ
=
cos θ −sin θ
sin θ cos θ
∈ SO
2
(R )
The effect of the map x 7→ A
θ
x is to rotate x
counter-clockwise by θ radians.
7
Reflections B
θ
=
cos θ sin θ
sin θ −cos θ
(det B
θ
= −1)
The effect of x 7→ B
θ
x is to reflect x across the
line making angle
θ
2
with the positive x-axis.
1
y
−1 1
x
cos θ
sin θ
−sin θ
cos θ
θ
1
y
−1 1
x
cos θ
sin θ
sin θ
−cos θ
θ
2
θ
π + θ
Motivated by the 2 ×2 case, it is common to refer to every orthogonal matrix in O
3
(R ) as a rotation
(det = 1) or a reflection (det = −1).
8
Part 3 of Lemma 1.25 says that multiplication by an orthogonal matrix preserves the dot product and
thus (Definition 1.5) the lengths of vectors and the angles between them. We use this to define a useful
family of transformations of E
3
.
Definition 1.26. An isometry is a function S : E
3
→ E
3
acting on points/position vectors by
S(x) = Ax + b
where b is a constant vector and A ∈ O
3
(R ). We call S a direct isometry or rigid motion if det A = 1
(A ∈ SO
3
(R )), and an indirect isometry otherwise.
Isometry literally means equal length; it can be seen that every function S : E
3
→ E
3
which pre-
serves distances between all pairs of points is an isometry. Congruent geometric objects (in standard
Euclidean geometry) are precisely those which are related by an isometry.
7
Recall that the matrix of a linear map is found by evaluating the map on the standard basis: the 1
st
column of A
θ
is
the column vector A
θ
i =
cos θ
sin θ
. The pictures should verify the remaining columns; for B
θ
you might find it helpful to
consider how the required reflections of the standard basis vectors i, j may be computed using rotations.
8
A full analysis is more complicated. For instance, the map x 7→ Ex in the first example is the composition of a reflection
across a plane in E
3
followed by a rotation in that plane.
14
Moving Frames
Thus far we have analyzed curves with reference to the standard orthonormal basis ϵ = {i, j, k}. We
now replace this static frame of reference with one that moves. The goal is eventually to describe a
special moving frame with respect to which the fundamental properties of the curve are clear.
Definition 1.27. Let x : I → E
3
be a smooth curve. Suppose that e
1
, e
2
, e
3
are smooth functions on I
such that, for each t ∈ I,
{e
1
( t), e
2
( t), e
3
( t) } is a positively oriented orthonormal basis of the tangent space T
x(t)
E
3
We call this family of functions a moving frame along x.
Equivalently, E(t) =
e
1
( t) e
2
( t) e
3
( t)
is a smooth function E : I → SO
3
(R ). We will often refer to
this matrix-valued function as a moving frame.
e
1
(t)
e
2
(t)
e
1
(t)
e
2
(t)
e
1
(t)
e
2
(t)
A moving frame in E
2
A moving frame in E
3
The smoothness criterion needs a little unpacking. At each point on the curve, the tangent space
T
x(t)
E
3
has a standard basis of tangent vectors {i
x(t)
, j
x(t)
, k
x(t)
}, with respect to which
e
j
( t) = a
j
( t) i
x(t)
+ b
j
( t) j
x(t)
+ c
j
( t) k
x(t)
=
a
j
( t)
b
j
( t)
c
j
( t)
We require that the functions a
j
, b
j
, c
j
: I → R be smooth. Strictly speaking, e
j
( t) is a smooth vector
field along the curve.
Example 1.28. We define a moving frame along the unit
circle x(t) =
cos t, sin t
via
e
1
( t) =
cos 2t
sin 2t
e
2
( t) =
−sin 2t
cos 2t
Click on the picture to see how the frame rotates twice as one
travels once round the circle!
In accordance with the definition, for each t,
E(t) =
cos 2t −sin 2t
sin 2t cos 2t
∈ SO
3
(R )
−1
1
−1 1
x
y
e
1
(t)
e
2
(t)
e
1
(t)
e
2
(t)
e
1
(t)
e
2
(t)
15
The obvious disadvantage of a moving frame is that we have to understand how such a frame moves!
Theorem 1.29 (Structure equations). Suppose {e
1
( t), e
2
( t), e
3
( t) } is a moving frame (orthonormal
positive orientation). For each j, express the derivative as a linear combination
e
′
j
= e
1
w
1j
+ e
2
w
2j
+ e
3
w
3j
where each w
ij
( t) = e
i
·e
′
j
is a scalar function. Then the matrix W = (w
ij
) is skew-symmetric:
9
e
′
1
e
′
2
e
′
3
=
e
1
e
2
e
3
0 w
12
w
13
−w
12
0 w
23
−w
13
−w
23
0
(matrix form E
′
= EW)
In E
2
, there is only a single function w
12
= e
1
· e
′
2
. As we’ve done already, we often drop the (t) to
make things more readable; just remember that everything is still a function!
Proof. Since e
i
·e
j
is constant (equals 0 or 1), the product rule says that
0 =
d
dt
e
i
·e
j
= e
′
i
·e
j
+ e
i
·e
′
j
=⇒ w
ji
+ w
ij
= 0
Examples 1.30. 1. Example 1.28 described a moving frame in E
2
:
w
12
( t) = e
1
( t) ·e
′
2
( t) =
cos 2t
sin 2t
·
−2 cos 2t
−2 sin 2t
= −2
The structure equations are therefore
( e
′
1
e
′
2
) = (e
1
e
2
)
0 −2
2 0
2. A moving frame can be described without mentioning a specific curve x(t):
e
1
( t) =
cos
2
t
cos t sin t
sin t
e
2
( t) =
sin t
−cos t
0
e
3
( t) =
sin t cos t
sin
2
t
−cos t
The structure equations are easily computed
w
12
( t) = e
1
·e
′
2
= cos
3
t + cos t sin
2
t = cos t,
w
13
( t) = e
1
·e
′
3
= cos
2
t(cos
2
t −sin
2
t) + 2(cos t sin t)
2
+ sin
2
t = 1
w
23
( t) = e
2
·e
′
3
= sin t(cos
2
t −sin
2
t) −cos t(2 cos t sin t) = −sin t
e
′
1
e
′
2
e
′
3
=
e
1
e
2
e
3
0 cos t 1
−cos t 0 −sin t
−1 sin t 0
9
The set of skew-symmetric matrices is sometimes denoted so
3
(R). The structure equations are an example of the
relationship between a Lie group and its Lie algebra, a foundation on which much advanced differential geometry rests.
16
Exercises 1.3. 1. Express v =
5
1
= a
1
e
1
+ a
2
e
2
as a linear combination with respect to the or-
thonormal basis β = {e
1
, e
2
} =
1
13
5
12
,
1
13
12
−5
of E
2
.
2. (a) Show that β =
1
3
2
2
1
,
1
√
2
1
−1
0
,
1
3
√
2
−1
−1
4
is an orthonormal basis of E
3
. Is it positively
oriented?
(b) Find the co-ordinates of
1
0
0
with respect to β.
3. (a) Explain why the product rule
d
dt
( x ·y) = x
′
·y + x ·y
′
holds for differentiable curves x, y.
(b) Let x, y be differentiable on an interval and use the product rule to answer the following:
i. Suppose x(t
0
) and x
′
( t) are orthogonal to a fixed vector v (the latter for all t). Show
that x(t) is always orthogonal to v.
ii. If y(t
0
) is a point on y which is closest to the origin, show that y(t
0
) ⊥ y
′
( t
0
).
4. Find the function w
12
for the moving frame {e
1
, e
2
} =
1
1 + t
2
2t
1 − t
2
,
1
1 + t
2
t
2
−1
2t
5. Find the structure equations for the moving frame {e
1
, e
2
, e
3
} =
n
cos t
0
sin t
,
0
1
0
,
−sin t
0
cos t
o
6. (a) Explain why every moving frame in E
2
has the form {e
1
, e
2
} =
n
cos θ(t)
sin θ(t)
,
−sin θ(t)
cos θ(t)
o
for some function θ.
(b) Find the structure equations for this frame: how does w
12
relate to θ?
(c) If x(t) is parametrized at unit speed such that e
1
( t) = x
′
( t), what is w
12
( t)?
7. (a) Let E(t) be a square matrix-valued function. Show that
d
dt
(E(t))
−1
= −E
−1
E
′
E
−1
.
(b) Suppose E : I → O
3
(R ) is differentiable and define W(t) := E
−1
( t)E
′
( t). Use part (a) to
prove that W(t) is skew-symmetric (W
T
= −W).
8. (a) Verify parts 2 and 3 of Lemma 1.25.
(b) Suppose f , g are rigid motions. Show that f ◦ g and f
−1
are also rigid motions.
9. Let i =
1
0
. Suppose p ∈ E
2
and a unit vector v are given. Prove that there is a unique rigid
motion S : x 7→ Ax + b such that
S(0) = p and S(i) = p + v
Write i
0
=
0, i
∈ T
0
E
2
and v
p
=
p, v
∈ T
p
E
2
as tangent vectors, explain why it is reasonable
to write v
p
= S(i
0
) = (Ai)
p
: i.e., only A affects the directional part of a tangent vector.
10. (Hard) Suppose that a moving frame has structure equations
e
′
1
= −
1
√
2
( e
2
+ e
3
), e
′
2
=
1
√
2
e
1
, e
′
3
=
1
√
2
e
1
(a) By considering e
′′
1
, show that the vector e
1
×e
′
1
is constant.
(b) Show that
||
e
′
1
||
is constant.
(c) Prove that there exists a constant positively oriented orthonormal basis {a, b, c} such that
e
1
( t) = cos ta + sin tb and compute e
2
, e
3
in terms of this basis.
17
1.4 The Frenet Frame for a Spacecurve
In this section we analyze spacecurves with respect to a moving frame adapted to the curve. To do
this, we need to restrict our class of curves slightly. For this section, we work exclusively in E
3
.
Definition 1.31. A regular spacecurve x : I → E
3
is biregular if it has non-zero curvature κ.
Every biregular curve is necessarily regular, but the converse is false. For instance, a straight line is
regular but not biregular. Indeed for a biregular curve, the vectors x
′
( t) and x
′′
( t) must be linearly
independent.
Definition 1.32. Let x : I → E
3
be a biregular unit-speed curve. The Frenet frame E(t) = (T N B) is
the moving frame defined as follows:
T := x
′
is the unit tangent vector field
N :=
1
κ
T
′
is the principal normal vector field
B := T ×N is the binormal vector field
We verify that the Frenet frame is indeed a moving frame:
1. Since x is unit-speed, T has unit length.
2. By the product rule, T ·T = 1 =⇒ 2T
′
· T = 0 =⇒ N · T = 0. Moreover, the definition of
curvature tells us tht
||
N
||
=
1
κ
T
′
=
1
κ
x
′′
=
κ
κ
= 1
so that N is a unit vector perpendicular to T.
3. Standard properties of the cross product finish things off:
• B has unit length since
||
B
||
=
||
T
||||
N
||
sin θ = 1 (θ = 90° is the angle between T, N).
•
T × N) · z = det
T N z
with z = T or N says that B is perpendicular to T, N. Finally, let
z = B to see that the Frenet frame is positively oriented.
Theorem 1.33. The Frenet frame is a moving frame. Its structure equations are
T
′
N
′
B
′
=
T N B
0 −κ 0
κ 0 −τ
0 τ 0
T
′
= κN
N
′
= −κT + τB
B
′
= −τN
where κ > 0 is the curvature and τ = N
′
·B = −N ·B
′
is called the torsion.
The structure equations for the Frenet frame are also known as the Frenet–Serret equations. The mov-
ing planes spanned by pairs of these vectors have special names:
Span{T, N}, Span{T, B} and Span{N, B} are the osculating, rectifying and normal planes.
At any point, the tangent line lies in the osculating plane.
18
Examples 1.34. 1. We compute the Frenet frame and its structure equations for the standard helix
x(s) =
cos
s
√
2
, sin
s
√
2
,
s
√
2
parametrized by arc-length (3D pic)(animation)
T(s) = x
′
( s) =
1
√
2
−sin
s
√
2
cos
s
√
2
1
=⇒ T
′
( s) = −
1
2
cos
t
√
2
sin
t
√
2
0
=⇒ N(s) = −
cos
t
√
2
sin
t
√
2
0
, κ(s) =
1
2
=⇒ B(s) = T(s) ×N(s) =
1
√
2
sin
s
√
2
−cos
s
√
2
1
τ(s) = N
′
( s) ·B( s) =
1
2
sin
s
√
2
−cos
s
√
2
0
·
sin
s
√
2
−cos
s
√
2
1
=
1
2
The Frenet–Serret equations for the helix are therefore
T
′
N
′
B
′
=
T N B
0 −
1
2
0
1
2
0 −
1
2
0
1
2
0
2. Let x(s) =
1
3
(1 + s)
3/2
,
1
√
2
s,
1
3
(1 −s)
3/2
for s ∈ (−1, 1). First we verify this is unit-speed
x
′
( s) =
1
2
√
1 + s
√
2
−
√
1 − s
=⇒ v(s) =
x
′
( s)
=
1
2
√
1 + s + 2 + 1 −s = 1
It follows that T = x
′
. Now compute the rest of the Frenet apparatus:
T
′
=
1
4
(1 + s)
−1/2
0
(1 −s)
−1/2
=⇒ κ =
T
′
=
1
4
r
1
1 + s
+
1
1 − s
=
1
2
√
2
√
1 − s
2
N =
1
κ
T
′
=
2
√
2
√
1 − s
2
4
(1 + s)
−1/2
0
(1 −s)
−1/2
=
1
√
2
√
1 − s
0
√
1 + s
B = T ×N =
1
2
√
1 + s
−
√
2
−
√
1 − s
=⇒ τ = N
′
·B =
−1
2
√
2
√
1 − s
2
T
′
N
′
B
′
=
T N B
2
√
2
√
1 − s
2
0 −1 0
1 0 1
0 −1 0
19
The Frenet Frame in arbitrary parametrization
Since there are relatively few curves for which an explicit unit-speed parametrization can be found,
we want to be able to compute the Frenet frame for any biregular curve, regardless of parametriza-
tion. This requires nothing more than the careful application of the chain rule. . .
Example 1.35. We compute for the exponential spiral x(t) =
e
t
cos t, e
t
sin t, e
t
.
x
′
( t) = e
t
cos t − sin t
sin t + cos t
1
=⇒ v(t) =
√
3e
t
=⇒ T(t) =
1
√
3
cos t − sin t
sin t + cos t
1
Since T(t) has unit length, T
′
⊥ T. But then
T
′
( t) =
1
√
3
−sin t −cos t
cos t − sin t
0
=⇒ N(t) =
1
√
2
−sin t −cos t
cos t − sin t
0
(unit length ∥ T
′
)
=⇒ B(t) = T ×N =
1
√
6
−cos t + sin t
−sin t −cos t
2
It is tempting to think that the curvature should be
||
T
′
( t)
||
=
q
2
3
, but this is not so. Since x is not
unit-speed, we need to use the chain rule:
κ =
d
ds
T(t)
=
dt
ds
d
dt
T(t)
=
1
v(t)
T
′
( t)
=
√
2
3
e
−t
The torsion may be computed similarly
τ =
dN
ds
·B =
1
v(t)
N
′
( t) ·B( t) =
1
3
e
−t
For the general result, simply(!) repeat the example in the abstract.
Corollary 1.36. Let x(t) be a biregular spacecurve with arbitrary parametrization. The speed, cur-
vature, torsion, Frenet frame, and structure equations are as follows.
v(t) =
x
′
( t)
κ(t) =
||
x
′
×x
′′
||
v
3
τ(t) =
( x
′
×x
′′
) ·x
′′′
v
6
κ
2
T(t) =
1
v
x
′
N(t) =
vx
′′
−v
′
x
′
v
3
κ
B(t) =
x
′
×x
′′
v
3
κ
T
′
N
′
B
′
=
T N B
0 −vκ 0
vκ 0 −vτ
0 vτ 0
The curvature formula also holds if x(t) is merely regular.
20
Exercises 1.4. 1. Compute the curvature and torsion of the spiral x(t) =
e
t
cos t, e
t
sin t, e
t
directly
using the expressions in Corollary 1.36.
2. A circular helix has the form x(t) =
r cos t, r sin t, ht
, where r > 0 and h are constants. Find
its Frenet frame and show that its curvature and torsion are given by
κ =
r
r
2
+ h
2
, τ =
h
r
2
+ h
2
3. Find the curvature and torsion of the curve x(t) =
t, t
2
, t
3
.
4. Given x(t) =
1
√
5
√
1 + t
2
, 2t, ln
t +
√
1 + t
2
, find the Frenet frame, curvature and torsion.
5. Let f (t) =
√
2
R
t
0
√
1 − e
−2u
du, and define the curve x(t) =
1
√
2
e
−t
cos t, e
−t
sin t, f (t)
, t > 0.
(a) Verify that x(t) has unit speed.
(b) Calculate the curvature of x and show that lim
t→∞
κ(t) = 0.
6. Let a, b be positive constants and x(t) =
4a cos
3
t, 4a sin
3
t, 3b cos 2t
where 0 < t <
π
2
. Find the
Frenet frame, curvature and torsion of x.
7. Let x : I → E
3
be a twice-differentiable regular curve.
(a) Prove the formula for κ in Corollary 1.36:
κ(t) =
||
x
′
×x
′′
||
v
3
Hence conclude that κ(t
0
) = 0 ⇐⇒ x
′
( t
0
) and x
′′
( t
0
) are parallel.
(Hint: let x(t) = y
s(t)
where y(s) has unit speed)
(b) Prove as much else as you can tolerate of Corollary 1.36.
8. Suppose x : I → E
3
is a curve lying on the surface of the unit sphere (
||
x
||
= 1).
(a) If x has unit speed, show that x
′′
·x = −1.
(b) Hence or otherwise, prove that the curvature of x is at least 1 everywhere.
(Hint: x and x
′
are orthonormal. . . )
(c) What happens if x lies on the surface of the sphere
||
x
||
= r of radius r > 0?
(d) (Hard) If a unit-speed curve lies on a sphere of radius r, show that
κ
2
τ
2
(r
2
κ
2
−1) = (κ
′
)
2
(Hint: compute the coefficients of x with respect to the Frenet frame)
9. (Hard) Let d(t) > 0. Suppose x(t) and y(t) = x(t) + dN(t) are unit-speed curves such that
the principal normal vector field N of x is the translate
a
of the binormal vector field
ˆ
B of y.
Prove that the distance d between corresponding points of the curves is constant. Prove also
that the curvature and torsion of x satisfy 2κ = d(κ
2
+ τ
2
).
(Hint: Compute
ˆ
T and take dot products with something useful. . . )
a
That is, the directional parts of N,
ˆ
B are identical: of course these are members of different tangent spaces.
21
1.5 The Fundamental Theorem of Biregular Spacecurves
Our goal for this section is to see that curvature and torsion determine a spacecurve uniquely up to
rigid motions. We do this by recognizing the Frenet–Serret equations satisfied by the Frenet frame as
a system of ordinary differential equations; provided sufficient initial conditions (starting point and
orientation), the usual existence and uniqueness theorem for initial value problems shows that there
is a unique curve with this data.
As a precursor, we consider how to interpret curvature and torsion, and how they change (or don’t!)
under rigid motions of a curve.
Theorem 1.37. 1. A regular spacecurve has κ ≡ 0 if and only if it is a straight line.
2. A biregular spacecurve has τ ≡ 0 if and only if it is contained in a fixed plane (the unmoving
osculating plane of the curve).
Proof. In both cases, we assume, without loss of generality, that x(s) is a unit-speed parametrization
of our spacecurve.
1. κ(s) =
||
x
′′
( s)
||
= 0 ⇐⇒ x
′′
( s) = 0. Thus x is a straight line.
2. (⇐) Suppose the curve lies in a fixed plane. Then x
′
and x
′′
are parallel to this plane, whence
T and N are also. But then B is a continuous unit vector orthogonal to the plane and is therefore
constant. From the Frenet equations, −τN = B
′
= 0 =⇒ τ ≡ 0.
(⇒) As above, if τ ≡ 0, then B is constant. But then
( x ·B)
′
= x
′
·B + x ·B
′
= T ·B = 0
from which x ·B is constant. The curve therefore lies in a fixed plane perpendicular to B.
Curvature measures the deviation of a curve from a straight line; its bending. Torsion measures how
badly a curve fails to be planar; its twisting.
To visualize the difference, the pictures below show a segment of a standard helix. In the first we
look down the binormal onto the osculating plane; the non-zero curvature is clearly visible. In the
second we look along the principal normal vector N and across the osculating plane; the positive
torsion (τ =
1
2
) indicates that the curve crosses the plane similarly to how the cubic function y = x
3
crosses the x-axis. The full 3D curve is linked via either picture.
22
Theorem 1.38. Under an isometry
ˆ
x := Ax + b (recall Definition 1.26), the curvature and torsion of
a biregular spacecurve transform as follows:
Direct isometry/rigid motion:
ˆ
κ = κ,
ˆ
τ = τ.
Indirect isometry:
ˆ
κ = κ,
ˆ
τ = −τ.
Proof. Suppose x(s) has unit-speed. We relate the Frenet frame (
ˆ
T
ˆ
N
ˆ
B) of
ˆ
x to the original.
10
Since orthogonal matrices preserve the dot product (Lemma 1.25),
ˆ
x has unit-speed also:
ˆ
x
′
( s) = Ax
′
( s) =⇒
ˆ
v(s) =
ˆ
x
′
( s)
=
x
′
( s)
= 1 =⇒
ˆ
T = AT
Moreover, since A is constant and both
ˆ
N and N have unit length,
1
ˆ
κ
ˆ
N =
ˆ
T
′
= AT
′
=
1
κ
AN =⇒
ˆ
κ = κ and
ˆ
N = AN
Curvature is therefore invariant under any isometry. Since A preserves angles, AB is perpendicular
to both
ˆ
T and
ˆ
N, and so AB = ±
ˆ
B. Since the Frenet frame E = (T N B) is a special orthogonal
matrix, AE is also orthogonal, and moreover
det AE = det A det E = det A
We conclude that det A = 1 ⇐⇒ AE = (AT AN AB) = (
ˆ
T
ˆ
N AB) is positively oriented, whence
ˆ
B = (det A)AB =
(
AB if the isometry is direct,
−AB if the isometry is indirect.
Finally, we compute the torsion:
ˆ
τ =
ˆ
N
′
·
ˆ
B = AN
′
·
(det A)(AB)
= (det A)(AN
′
) ·(AB) = (det A)N
′
·B = (det A)τ
Existence and Uniqueness of Solutions to ODEs
Our classification of spacecurves depends on the ‘usual’ existence and uniqueness result for ODEs.
Here is a version suitable for our needs.
Theorem 1.39 (Existence/Uniqueness for Linear ODE (Picard, Lindel¨of, etc.)).
Let t
0
∈ R and c ∈ R
n
be given, and let M(t) ∈ M
n
(R ) be a continuous matrix-valued function
defined on an interval
|
t −t
0
|
≤ T. Then the initial value problem
dE
dt
= M(t)E, E(t
0
) = c
has a unique solution E : [t
0
− T, t
0
+ T] → R
n
.
10
Recall Exercise 1.3.9: when we write
ˆ
T = AT we mean that the directional parts of the tangent vectors are thus related.
23
The rough idea of the proof is to define a sequence of functions
E
0
( t) := c, E
1
( t) := c +
Z
t
t
0
M(u)E
0
( u) du, E
2
( t) := c +
Z
t
t
0
M(u)E
1
( u) du, . . .
which are seen to converge to the required solution; this last step requires advanced ideas from
topology/analysis. A simple example should convince you of the approach.
Example 1.40. Given the initial value problem
dE
dt
= 2tE, E(0) = 1, we obtain
E
0
( t) = 1, E
1
( t) = 1 +
Z
t
0
2u du = 1 + t
2
, E
2
( t) = 1 +
Z
t
0
2u(1 + u
2
) du = 1 + t
2
+
1
2
t
4
, . . .
The Picard iteration builds up the correct solution as a power series
E(t) = e
t
2
=
∞
∑
n=0
t
2n
n!
= 1 + t
2
+
1
2
t
4
+
1
6
t
6
+ ···
Corollary 1.41. Let O be an orthogonal matrix, I = [t
0
− T, t
0
+ T] an interval, and W : I → M
3
(R )
a matrix-valued function such that each W(t) is skew-symmetric. Then:
1. There exists a unique solution E : I → O
3
(R ) to the initial value problem
dE
dt
= EW, E(t
0
) = O
2. If det O = 1, then E : I → SO
3
(R ).
Proof. 1. The initial value problem is a system of nine linear first order ODEs in the entries of the
3 × 3 matrix E. We are therefore in the case of Picard’s theorem where E : I → R
9
. There
therefore exists a unique solution E : I → M
3
(R ). Now differentiate:
d
dt
(EE
T
) = E
′
E
T
+ E(E
′
)
T
= EWE
T
+ E(EW)
T
= EWE
T
+ EW
T
E
T
= EWE
T
+ E(−W)E
T
= 0 (W
T
= −W!)
Thus EE
T
is constant. However E(t
0
)E(t
0
)
T
= I since E(t
0
) = O is orthogonal. We conclude
that E(t) is always orthogonal.
2. Determinant is continuous (it is a polynomial!); E is differentiable, and so det E : I → R is
continuous on an interval. But det E = ±1 since E is orthogonal. It follows that det E is the
constant 1.
For simple W, we might be able to state the solution using the matrix exponential; for instance
W constant =⇒ E(t) = Oe
tW
This is of limited utility: the matrix exponential is rarely computable except as an infinite series, and
the approach fails for general W(t).
24
Corollary 1.42 (Fundamental theorem of biregular spacecurves).
Suppose we are given the following data:
• Smooth functions κ > 0 and τ on an interval I = [t
0
− T, t
0
+ T].
• A position vector c ∈ E
3
and a positively oriented orthonormal basis (T
0
N
0
B
0
) of T
c
E
3
.
Then there exists a unique unit-speed biregular spacecurve x : I → E
3
with curvature κ, torsion τ,
initial position x(t
0
) = c and Frenet frame E(t
0
) = (T
0
N
0
B
0
) at x(t
0
).
Proof. The structure equations E
′
= EW put us in the situation of Corollary 1.41; there exists a unique
solution E = (T N B) : I → SO
3
(R ). Integrate the unit tangent vector field to finish:
x(t) = c +
Z
t
t
0
T(u) du
This is plainly the unique curve with the required initial conditions, curvature and torsion.
Alternatively, a biregular curve is determined up to rigid motions by its curvature and torsion.
Corollary 1.43. Given two biregular spacecurves with the same curvature and torsion functions,
there exists a unique direct isometry transforming one to the other.
Proof. Suppose x
1
: I → E
3
and x
2
: I → E
3
have Frenet frames E
1
, E
2
, and the same curvature and
torsion functions. Choose some (any!) t
0
∈ I. The required rigid motion S : x 7→ Ax + b must satisfy
the conditions at t
0
, whence
11
S
x
1
( t
0
)
= x
2
( t
0
) and AE
1
( t
0
) = E
2
( t
0
)
Plainly A = E
2
( t
0
)
E
1
( t
0
)
−1
and b = x
2
( t
0
) − Ax
1
( t
0
) provide the unique isometry S. Moreover
det A = 1 since both E
1
and E
2
do so also.
By Theorem 1.38, x
3
:= S(x
1
) is a spacecurve with the same initial conditions (at t
0
), curvature and
torsion as x
2
. The Fundamental Theorem says that x
2
= x
3
= S(x
1
).
Compare what we’ve done to the standard acceleration/position kinematics problem, where three
scalar functions x
′′
( t) =
x
′′
( t), y
′′
( t), z
′′
( t)
and six scalar initial conditions x( t
0
) and x
′
( t
0
) recover the
motion by twice integrating.
The Fundamental Theorem says that a spacecurve is determined uniquely by three scalar functions
v(t), κ(t), τ(t) and the initial conditions x(t
0
), T(t
0
), N(t
0
), which also amount to six scalar constants.
12
One benefit of our result is that, by standardizing v( t) ≡ 1 and ignoring rigid motions, we see that
the physical shape of a curve depends only on two scalar functions κ(t) and τ(t).
11
As in Theorem 1.38, S acts on position vectors but Frenet frames consist of tangent vectors and thus only see A.
12
You don’t need explicitly to specify B(t
0
) = T(t
0
) × N(t
0
)! The position x(t
0
) requires three constants; T(t
0
) needs
two angles (spherical polar co-ordinates), and N(t
0
) a single angle in the plane (T(t
0
))
⊥
.
25
We finish this discussion with a quick application of the Fundamental Theorem.
Corollary 1.44. Every biregular curve with κ and τ constant is a circular helix (circle if τ ≡ 0).
Proof. By (Exercise 1.4.2), the circular helix x(t) =
r cos t, r sin t, ht
has constant curvature κ =
r
r
2
+h
2
and torsion τ =
h
r
2
+h
2
.
Given constant κ, τ, it is a simple exercise to find suitable r, h. By Corollary 1.43, this is the only such
curve up to direct isometry (and constant speed re-parametrization).
What changes in other dimensions?
For plane curves things are a little simpler. Here is a summary.
Assume: x : I → E
2
is regular; we don’t need biregularity.
Frenet frame: Define T :=
1
v
x
′
and N := JT, where J =
0 −1
1 0
is rotation by 90° counter-clockwise; no differentiation is re-
quired to compute N!
Curvature: κ =
1
v
T
′
·N is signed as we saw in Section 1.2:
Frenet–Serret equations: In arbitrary parametrization
T
′
N
′
=
T N
0 −vκ
vκ 0
Isometries: Direct preserves κ, indirect changes its sign.
Fundamental Theorem: Given κ(s) , x(s
0
) ∈ E
2
and T
0
∈ T
x(s
0
)
E
2
,
there exists a unique unit-speed curve with curvature κ(s)
and these initial data. Exercise 7 gives an elementary proof.
T(t)
N(t)
κ > 0: x bends towards N
T(t)
N(t)
κ < 0
κ < 0: x bends away from N
Example 1.45. Constant κ curves are circles, as we can see explicitly in a couple of ways. The unit-
speed structure equations E
′
= E
0 −κ
κ 0
become T
′′
= −κ
2
T which may be explicitly integrated.
Alternatively, θ
′
( t) = κ =⇒ θ(t) = κt + c, yields an explicit circle of radius
1
κ
:
T(t) =
cos(κt + c)
sin(κt + c)
=⇒ x(t) = x(t
0
) +
Z
t
t
0
T(u) du = x(t
0
) +
1
κ
−sin(κt + c)
cos(κt + c)
We can also play this game in higher dimensions. Given a unit-
speed curve x : I → E
n
whose first n −1 derivatives at each
point are linearly independent, apply Gram-Schmidt orthog-
onalization to obtain a moving frame E = (e
1
··· e
n
) and
functions κ
1
, . . . , κ
n−1
(the generalized curvatures) satisfying
the structure equations shown.
E
′
= E
0 −κ
1
0 ··· 0 0
κ
1
0 −κ
2
0 0
0 κ
2
0 0 0
.
.
.
.
.
.
.
.
.
0 0 0 0 −κ
n−1
0 0 0 ··· κ
n−1
0
Conversely, the n −1 generalized curvatures determine the curve up to rigid motions.
26
Exercises 1.5. 1. Find an explicitly parametrized curve with constant curvature κ and torsion τ.
2. Reflection in the xy-plane S(x) =
1 0 0
0 1 0
0 0 −1
x is an indirect isometry. Explicitly compare the
curvature and torsion of the standard helix x(t) =
cos t, sin t, t
with those of S(x).
3. In the manner of Example 1.40, compute the Picard iteration process up to E
3
( t) for the initial
value problem
dE
dt
=
0 −1
1 0
E, E(0) =
1
0
Verify that this comports with the correct solution E(t) =
(
cos t
sin t
)
to this system of ODEs.
4. Suppose f is a function such that x(t) =
cos t, sin t, f (t)
lies in a fixed plane. Show that f
satisfies the 3
rd
-order linear ODE f
′′′
( t) + f
′
( t) = 0 and thus find all possible functions f .
(Hints: What is the torsion of a plane curve?)
5. Assume that all principal normals of a biregular curve in E
3
pass through a fixed point: ∃α(t)
and a constant n such that x(t) + α(t)N(t) = n. Show that the curve is (part of) a circle.
6. Let x : I → E
2
be a regular curve and let y = S(x) = Ax + b be a new curve resulting from a
rigid motion. Prove that the curvatures of x and y are identical.
7. For regular curves in E
2
, the Fundamental Theorem is relatively simple to prove.
(a) Suppose you are given a smooth function κ : I → R on an interval I containing t
0
, an initial
position x(t
0
) = (a, b) and an initial direction θ(t
0
) = θ
0
(angle with positive x-axis).
Use the Fundamental Theorem of Calculus to describe the unique unit-speed curve x :
I → E
2
with curvature κ and given initial data.
(Hints: use θ(t) := θ
0
+
R
t
t
0
κ(u) du to define T(t) and integrate! Your answer will contain
definite integrals.)
(b) Suppose x : R → E
2
is unit-speed with κ( t) =
1
1+t
2
, x(0) = (0, 0), and x
′
(0) = (1, 0). Find
x(t).
8. (Hard) A cylindrical helix is a curve x(t) whose unit tangent field T( t) makes a constant angle
θ ∈ (0,
π
2
) with a fixed vector n.
(a) If x(t) =
cos t, sin t, t
is the standard circular helix, describe a suitable vector n.
(b) Use the Frenet–Serret formulas to prove that a (unit-speed) non-planar curve is a cylindri-
cal helix if and only if κ/τ is constant.
9. (Very hard) Suppose a moving frame e
1
, e
2
, e
3
has structure equations where all three func-
tions w
12
, w
13
, w
23
are constant. Find the moving frame f
1
, f
2
, f
3
where f
1
= e
1
such that f
1
, f
2
, f
3
is the Frenet frame of a unit-speed circular helix. Calculate the curvature κ and torsion τ of this
helix in terms of w
12
, w
13
, w
23
. Can you find an orthogonal matrix A such that
A
−1
0 w
12
w
13
−w
12
0 w
23
−w
13
−w
23
0
A =
0 −κ 0
κ 0 −τ
0 τ 0
?
27
1.6 Radii of curvature
We have seen how curvature measures the deviation of a curve from a straight line and that the only
planar curves with constant curvature κ are circles of radius
1
κ
. We could have started with this as our
definition; at a given point, a curve has curvature κ if the circle which best approximates the curve
has radius
1
κ
. Of course, we have to define what is meant by best approximation.
Definition 1.46. Unit-speed curves x, y have n
th
order contact at an intersection point x(t
0
) = y(s
0
),
if their first n derivatives agree there: x
(j)
( t
0
) = y
(j)
( s
0
) for all 1 ≤ j ≤ n.
Let x(t) be a unit-speed curve in E
2
, fix r = 0 and consider the
unit-speed circle c(s) with (signed) radius r for which
c(0) = x(t
0
) and c
′
(0) = x
′
( t
0
)
We take r > 0 ⇐⇒ the circle lies on the same side of the curve
as the principal normal vector N.
The circle is straightforward to parametrize:
c(s) = x(t
0
) + rN(t
0
)
| {z }
center
+ r sin(s/r)T(t
0
) −r cos(s/r)N(t
0
)
| {z }
rotation
T( t
0
)
N( t
0
)
r > 0
r < 0
Certainly this circle has 1
st
-order contact with the curve: c(0) = x(t
0
) and
c
′
( s) = cos(s/r)T(t
0
) + sin(s/r)N(t
0
) =⇒ c
′
(0) = T(t
0
) = x
′
( t
0
)
Moreover,
c
′′
( s) = −
1
r
sin(s/r)T(t
0
) +
1
r
cos(s/r)N(t
0
) =⇒ c
′′
(0) =
1
r
N(t
0
)
The circle has second-order contact with the curve if and only if
c
′′
(0) = x
′′
( t
0
) ⇐⇒
1
r
= κ(t
0
)
There is nothing stopping us from finding this circle for an arbitrary speed regular curve, since all
we need is the curvature and the Frenet frame at the relevant point.
Definition 1.47. Let x(t) be a regular curve. At a point x(t
0
) with non-zero curvature:
• The radius of curvature is r =
1
κ(t
0
)
.
• The center of curvature is the point with position vector x(t
0
) + rN(t
0
).
• The osculating circle is the radius r circle centered at the center of curvature. It has unit-speed
parametrization
c(s) = x(t
0
) +
1
κ(t
0
)
sin(s/r)T(t
0
) + (1 −cos(s/r))N(t
0
)
Osculating means ‘kissing.’ If κ(t
0
) = 0, some consider the tangent line to be an osculating circle with
infinite radius!
28
Example 1.48. We find the osculating circles for the parabola y = x
2
parametrized in the obvious
manner x(t) =
t, t
2
. The relevant ingredients are
x
′
( t) =
1
2t
=⇒ T(t) =
1
√
1 + 4t
2
1
2t
N(t) =
1
√
1 + 4t
2
−2t
1
x
′′
( t) =
0
2
, κ(t) =
2
(1 + 4t
2
)
3/2
The center of curvature when t = t
0
has position vector
x(t
0
) +
1
κ(t
0
)
N(t
0
) =
−4t
3
0
1
2
+ 3t
2
0
Several osculating circles are drawn and their centers of
curvature indicated.
1
2
−2 −1 0 1 2
The centers of curvature describe a curve that is interesting in its own right.
Definition 1.49. Let x(t) be a regular plane curve with non-zero curvature. The curve e(t) defined
by the centers of curvature is the evolute of x(t):
e(t) = x(t) +
1
κ(t)
N(t)
Example (1.48 cont). The evolute of the parabola
x(t) = (t, t
2
) was found above:
e(t) = x(t) +
1
κ(t)
N(t) =
−4t
3
1
2
+ 3t
2
Alternatively, this is the graph y =
1
2
+ 3
x
4
2/3
:
notice that this isn’t regular at x = 0.
The picture now animates to show the osculating
circles and the construction of the evolute.
1
2
3
4
−4 −3 − 2 −1 0 1 2 3 4
The gray lines are the normal lines to the parabola, and are also tangent to the evolute.
e
′
= x
′
−
κ
′
κ
2
N +
1
κ
( −vκT) = −
κ
′
κ
2
N
This last means that the evolute is a focal curve for the family of normal lines. The same equation
shows that the evolute is regular precisely when κ
′
( t) = 0.
29
A related notion is the involute, which may be imagined by rolling a line along a curve and seeing
what curve the end of the line traces out.
Definition 1.50. Suppose x(t) has unit speed. Its involute is the curve
i(t) := x(t) − tx
′
( t) = x(t) − tT(t)
An involute depends crucially on its parametrization: it intersects its source curve when t = 0.
Examples 1.51. 1. The unit speed unit circle x(t) =
cos t, sin t
. Its involute is therefore
i(t) = x(t) − tT(t) =
cos t + t sin t
sin t − t cos t
2. The involute of the unit speed catenary x(t) =
sinh
−1
t,
√
1 + t
2
is the tractrix:
i(t) =
sinh
−1
t −t(1 + t
2
)
−1/2
(1 + t
2
)
−1/2
This is the curve obtained when an object starting at the point (0, 1) is dragged (subjected to
traction) by attaching a rope of length 1 to a vehicle moving along the x-axis.
−6
−4
−2
2
−4 −2 2
1
2
3
4
−2 −1 0 1 2
Circle and involute (spiral) Catenary and involute (tractrix)
Another way to visualize the involute of the catenary is to imagine attaching a weight at (0, 1)
to a long string wrapped tightly along the catenary and then releasing the weight. Similarly,
imagine a string is wound tightly around the circle and then uncoiled; the result is the involute.
Theorem 1.52. The evolute of any involute is the original curve, except where t = 0 or κ = 0.
We leave the argument as an exercise. The reverse process fails, as an observation of the parabola
example should convince you: remember that an involute intersects its source curve at t = 0. . .
30
Exercises 1.6. 1. Find the center of curvature for the curve x(t) = (1 − t
−1
, 1 + t) at t = 1.
2. Consider the ellipse x(t) =
a cos t, b sin t
where a > b > 0.
(a) Compute the curvature of the ellipse.
(b) Show that its evolute is the astroid e(t) = (a
2
−b
2
)
a
−1
cos
3
t
−b
−1
sin
3
t
(c) The four-vertex theorem states that a simple closed plane curve with differentiable curva-
ture has at least four points where κ
′
= 0. Show that the ellipse has precisely four.
3. Describe the involutes of a straight line.
(Hint: this is a trick question!)
4. In Example 1.51.2 we constructed the tractrix as the involute of the catenary.
(a) Use sinh
−1
t = ln
t +
√
1 + t
2
to verify that x(t) has unit speed and thus confirm the
derivation of i(t).
(b) Compute the tangent line to the tractrix when t > 0 and show that this line cuts the x-axis
a distance 1 from the curve, thus justifying the traction claim.
5. Suppose that the graph of a smooth function y = f (x) passes horizontally through the origin:
f (0) = 0 = f
′
(0). Show that its Maclaurin series is
f (x) ≈
1
2
κ(0)x
2
+ higher order terms
Use this to quickly state the curvature at x = 0 of the graph of y = x
2
(7x
2
−29).
6. Let x(t) be unit speed with non-zero curvature κ and Frenet frame {T, N}. Moreover, let
i(t) = x(t) − tT(t) be an involute and denote the speed and corresponding data for the in-
volute
ˆ
v,
ˆ
κ,
ˆ
T,
ˆ
N. For simplicity, suppose κ, t > 0.
(a) Compute the Frenet frame of i(t) in terms of T and N.
(b) Show that
ˆ
κ(t) =
1
t
.
(c) Show that the evolute of i(t) is the original curve x(t).
7. We see how an involute of the evolute fails to recover the original curve.
Let x(t) be regular with non-zero curvature, κ
′
( t) = 0, and evolute e(t) = x(t) +
1
κ(t)
N(t). Since
e(t) is regular, we may assume it is parametrized by arc-length.
(a) If κ
′
> 0, explain why κ
′
= κ
2
.
(b) Show that the natural involute of the evolute is
e(t) −te
′
( t) = x(t) +
1
κ(0)
N(t)
that is, the original curve shifted a constant distance
1
κ(0)
in its normal direction.
(Hint: the ODE in part (a) is separable)
(c) Find the involute of the evolute of the parabola x(t) = (t, t
2
).
31
