There is an important notion called orientation. For simple rectifiable curves, you can think of
it as a direction of motion over the curve. Of course an orientation is imposed if you have
γ :
[a,b]
→ ℝ^{p} by simply letting t go from a to b. This does indeed give a direction of motion along the
curve. It is also reasonable to believe that there are exactly two directions of motion possible
depending on which end point one starts at and ends at. These things are discussed below. First
is a fundamental observation which says that one to one functions are either increasing or
decreasing.
Lemma 4.4.1Let ϕ :
[a,b]
→ ℝ be a continuous function and suppose ϕ is 1 − 1 on
(a,b)
. Thenϕ is either strictly increasing or strictly decreasing on
[a,b]
. Furthermore, ϕ^{−1}is continuous.
Proof:First it is shown that ϕ is either strictly increasing or strictly decreasing on
(a,b)
.
If ϕ is not strictly decreasing on
(a,b)
, then there exists x_{1}< y_{1}, x_{1},y_{1}∈
(a,b)
such that
(ϕ (y1)− ϕ (x1))(y1 − x1) > 0.
If for some other pair of points, x_{2}< y_{2} with x_{2},y_{2}∈
(a,b)
, the above inequality does not hold, then since
ϕ is 1 − 1,
(ϕ (y2)− ϕ (x2))(y2 − x2) < 0.
Let x_{t}≡ tx_{1} +
(1− t)
x_{2} and y_{t}≡ ty_{1} +
(1 − t)
y_{2}. Then x_{t}< y_{t} for all t ∈
[0,1]
because
tx1 ≤ ty1 and (1− t)x2 ≤ (1 − t)y2
with strict inequality holding for at least one of these inequalities since not both t and
(1− t)
can equal
zero. Now define
h (t) ≡ (ϕ (yt)− ϕ (xt))(yt − xt).
Since h is continuous and h
(0)
< 0, while h
(1)
> 0, there exists t ∈
(0,1)
such that h
(t)
= 0. Therefore,
both x_{t} and y_{t} are points of
(a,b)
and ϕ
(yt)
−ϕ
(xt)
= 0 contradicting the assumption that ϕ is one to one.
It follows ϕ is either strictly increasing or strictly decreasing on
(a,b)
.
This property of being either strictly increasing or strictly decreasing on
(a,b)
carries over to
[a,b]
by
the continuity of ϕ. Suppose ϕ is strictly increasing on
(a,b)
, a similar argument holding for ϕ strictly
decreasing on
(a,b)
. If x > a, then pick y ∈
(a,x)
and from the above, ϕ
(y)
< ϕ
(x)
. Now by continuity of
ϕ at a,
ϕ(a) = xli→ma+ ϕ(z) ≤ ϕ (y) < ϕ(x).
Therefore, ϕ
(a)
< ϕ
(x)
whenever x ∈
(a,b)
. Similarly ϕ
(b)
> ϕ
(x)
for all x ∈
(a,b)
.
It only remains to verify ϕ^{−1} is continuous. Suppose then that s_{n}→ s where s_{n} and s are points of
ϕ
([a,b])
. It is desired to verify that ϕ^{−1}
(sn)
→ ϕ^{−1}
(s)
. If this does not happen, there exists ε > 0 and a
subsequence, still denoted by s_{n} such that
||ϕ−1(sn)− ϕ−1 (s)||
≥ ε. Using the sequential compactness of
[a,b]
there exists a further subsequence, still denoted by n, such that ϕ^{−1}
(sn)
→ t_{1}∈
[a,b]
,t_{1}≠ϕ^{−1}
(s)
.
Then by continuity of ϕ, it follows s_{n}→ ϕ
(t1)
and so s = ϕ
(t1)
. Therefore, t_{1} = ϕ^{−1}
(s)
after all.
■
Definition 4.4.2Letη,γbe continuous one to one parametrizations for asimple curve. That is,γ
([a,b])
= η
([c,d])
andγ,ηare one to one on
[a,b]
and
[c,d]
respectively. Thus a simple curveis the one to one onto image of a closed interval. Earlier this was called a Jordan arc. Ifη^{−1}∘γis increasing, thenγandηare said to be equivalentparametrizations and this is written asγ∼η.It is also said that the two parametrizations give the same orientationfor the curve whenγ∼η.Asimple closedcurve is one which is the continuous one to one image of S^{1}, the unit circle.
First is a discussion of orientation of simple curves and after that, orientation of a simple closed curve is
considered. It is always about specifing a direction of motion along the curve.
In the case of a simple curve, there will be exactly two orientations because there is a single point in
γ^{−1}
(p)
for p a point on the curve. Thus γ^{−1} is well defined and continuous by Theorem 2.2.29. By Lemma
4.4.1, either η^{−1}∘γ is increasing or it is decreasing. In simple language, the message is that there are
exactly two directions of motion along a simple curve.
= t so it is clearly an increasing function. If γ∼η
then γ^{−1}∘η is increasing. Now η^{−1}∘γ must also be increasing because it is the inverse of
γ^{−1}∘η. This verifies 4.22. To see 4.23, γ^{−1}∘θ =
(γ−1 ∘η)
∘
(η−1 ∘θ)
and so since both of
these functions are increasing, it follows γ^{−1}∘θ is also increasing. This proves the lemma.
■
Definition 4.4.4Let Γ be a simple curve and letγbe a parametrization for Γ. Denoting by
[γ ]
theequivalence class of parameterizations determined by the above equivalence relation, the following pair willbe called an oriented curve.
(Γ ,[γ ])
In simple language, an oriented curve is one which has a direction of motion specified.
Actually, people usually just write Γ and there is understood a direction of motion or orientation on Γ.
How can you identify which orientation is being considered? Is there a simple condition which shows that
the orientation defined by two parametrizations is the same for a curve? For a simple curve, we have the
following lemma.
Lemma 4.4.5Supposeγ^{∗} = η^{∗},γ :
[a,b]
→ ℝ^{p},η :
[c,d]
→ ℝ^{p}are both one to one. Thenη∼γif and only if there is a strictly increasing continuous function ϕ :
[a,b]
→
[c,d]
such thatη∘ϕ = γ.
Proof:⇒ In this case, it is given that η^{−1}∘γ is increasing. Let this function be ϕ. Then
γ = η∘ ϕ.
⇐ In this case, the increasing function ϕ exists such that η∘ ϕ = γ. Thus η^{−1}∘γ is increasing.
■
Note that, the end points of this curve consist of
{γ (a),γ(b)}
=
{η (c),η (d)}
. Otherwise, if an end
point were γ
(c)
not equal to either
{γ (a),γ(b)}
, you could remove γ
(c)
from γ^{∗} and the result would be
connected while [a,c) ∪ (c,b] is not connected. Now γ^{−1} : γ^{∗}→
[a,b]
is continuous so the image of
γ^{∗}∖
{γ (c)}
would need to be connected which would not be so. With a simple closed curve, the image of
[a,b) this may not be so. You could get the curve from two different parametrizations and the two could
yield different beginning and ending points. You remove a point from one of these and the
result is still connected. This is the difficulty of trying to use this approach on a simple closed
curve.
When the parametrizations are equivalent, this says they preserve the direction of motion
along the curve. Recall Theorem 4.1.3. Thus Lemma 4.4.5 shows that the line integral depends
only on the set of points in the curve and the orientation or direction of motion along the
curve.
The orientation of a simple curve is determined by two points on the curve. This is the idea of the
following proposition.
Proposition 4.4.6Let
(Γ ,[γ])
be an oriented simple curve,γ
([a,b])
= γ^{∗},γone to one on
[a,b]
and letp,qbe any two distinct points of Γ. Then
[γ]
is determined by the order ofγ^{−1}
(p)
andγ^{−1}
(q)
. This means thatη∈
[γ ]
if and only ifη^{−1}
(p )
andη^{−1}
(q)
occur in the same order asγ^{−1}
(p)
andγ^{−1}
(q)
. In other words, if and only ifp,qare encountered in the same order with bothparametrizations as the parameter increases.
Proof: Suppose γ^{−1}
(p)
<γ^{−1}
(q )
and let η∈
[γ]
. In this case, there is a strictly increasing function ϕ
such that γ = η∘ ϕ and so
is given. Therefore, γ^{−1}∘η can only be increasing. Thus η∼γ.
■
This shows that the direction of motion on the simple curve is determined by any two points and the
determination of which is encountered first by any parametrization in the equivalence class of
parameterizations which determines the orientation. Sometimes people indicate this direction of motion by
drawing an arrow.
Definition 4.4.7Let Γ be a simpleclosed curve. This means there isγ : S^{1}→ Γ which is one to one andonto. Recall also from Lemma 2.9.12this is equivalent to a parametrizationγ :
[a,b]
→ Γ such thatγis continuous and one to one on [a,b) butγ
(a)
= γ
(b)
. Here S^{1}isthe unit circle in theplane
2 2
x + y = 1
Simple closed curves are the continuous one to one image of the unit circle. However, one can take any
point on the simple closed curve and regard that point as the beginning and ending point. This differs from
a simple closed curve in which you have only two points which can be considered the beginning or the
end.
Proposition 4.4.8Let Γ be a simple closed curve. Then for any point p on Γ, there is aclosed interval
[a,b]
and a continuous mapγ :
[a,b]
→ Γ andγone to one on [a,b) such thatγ
(a)
= γ
(b)
= p. That is, p will be the beginning and ending point.
Proof:Let ξ : S^{1}→ Γ be one to one, onto and continuous. Then ξ^{−1}
(p)
≡
(cosα,sinα )
for
some α ∈ [0,2π). Then let θ : [α,α + 2π] → S^{1} be defined by θ
(t)
≡
(cost,sint)
. Then θ
is one to one on [α,α + 2π) and continuous on
[α,α + 2π]
and θ :
[α,α + 2π]
→ S^{1} is onto
and θ
(α)
= θ
(α + 2π)
. Consider ξ∘θ≡γ. γ
(α)
≡ξ
( )
ξ−1(p)
= p,γ is also continuous on
[α,α + 2π]
and is one to one on [α,α + 2π) and maps onto Γ. γ
(α + 2π)
= ξ
(θ (α + 2π))
= p.
■
If Γ = γ
([a,b])
where γ is one to one on [a,b), continuous on
[a,b]
and γ
(a)
= γ
(b)
, then if
c ∈
(a,b)
, you cannot have γ
([a,c])
= Γ or γ
([c,b])
= Γ because this would violate γ being
one to one on [a,b). Thus γ
(c)
is neither a beginning nor an ending point for this particular
parametrization.
Lemma 4.4.9Suppose Γ_{i},i = 1,2 are two simple curves which intersect only at the pointsp,qand thatp,qare also the endpoints of each Γ_{i}. Then the union of these two curves, denoted by Γ isa simple closed curve. Conversely, if Γ is a simple closed curve, it can be expressed as the union oftwo such simple curves.
PICT
Proof:⇒By assumption there exists γ_{i} :
[ai,bi]
→ Γ_{i} such that the end points γ_{i}
(a)
,γ_{i}
(b)
consist
of the points p,q. Without loss of generality, one can change the parameter if necessary to
have
γ (a ) = p,γ (b) = q,γ (a ) = q,γ (b ) = p
1 1 1 1 2 2 2 2
Now change the parameter again to have
[a1,b1]
=
[0,1]
and
[a2,b2]
=
[1,2]
. Let
{
γ1(t) if t ∈ [0,1],γ1(1) = q
γ (t) ≡ γ (t) if t ∈ [1,2],γ (2) = p
2 2
Then γ is one to one on [0,2) and has γ
(0)
= γ
(2)
. Thus if Γ ≡γ
([0,2])
, then Γ is a simple closed
curve.
⇐If Γ is a simple closed curve, pick p≠q. By Proposition 4.4.8, there is γ :
[a,b]
→ Γ which is onto, one
to one on [a,b) continuous, and γ
(a)
= γ
(b)
= p. Thus there is c ∈
(a,b)
such that γ
(c)
= q. Then the
two simple curves are γ
([a,c])
and γ
([c,b])
. ■
What is meant by an oriented simple closed curve? As in the case of a simple curve, it needs to be
defined in such a way as to specify a direction around the curve. What is easily understood is
that for the unit circle, there are exactly two ways to go around it, counter clockwise and
clockwise.
Definition 4.4.10Let Γ be a simple closed curve. Two parametrizationsγ : S^{1}→ Γ andη : S^{1}→ Γ are equivalent if and only ifη^{−1}∘γpreserves the direction of motion around S^{1}. Thatis, if for increasing t ∈ ℝ, x
(t)
is a point on S^{1}, then if x
(t)
moves clockwise for increasing t, sodoesη^{−1}∘γ
(x (t))
and if x
(t)
moves counter clockwise for increasing t, then so doesη^{−1}∘γ
(x(t))
.
Lemma 4.4.11The above definition yields an equivalence relation.
Proof: The proof is just like it was earlier in case of simple curves. It is obvious that γ∼γ. If γ∼η so
η^{−1}∘γ preserves direction of motion. How about γ^{−1}∘η? If x
(t)
is moving counter clockwise then
x
(t)
=
(η−1 ∘ γ)
(γ −1 ∘η (x(t)))
. This cannot be true unless t →γ^{−1}∘η
(x(t))
moves counter clockwise
as well since you cannot have equality in t of two points on S^{1} which move in opposite directions. Similarly,
clockwise motion must also be preserved. The transitive law is fairly obvious also. Say γ∼η and η∼ζ.
Then
( )( )
ζ−1∘γ = ζ− 1∘η η−1∘γ
and the two mappings on the right preserve motion on S^{1}. ■
Since there are exactly two directions of motion on S^{1}, this shows there are exactly two equivalence
classes of parametrizations of a simple closed curve. However, we will mainly use orientations on simple
curves to specify the direction of motion on a simple closed curve as in the following Proposition. The
situation is illustrated by the following picture in which there are two simple closed curves which
share a simple curve denoted by l. The case we have in mind is in the plane but it would work
as well if it were only the case that the two simple closed curves have the two points p,q in
common.