A global property of plane curves: rotation index
Definition. A subset C ⊆ E2 is called immersed curve of the class Cr
, r ≥ 1 if there
exists a regular motion f : I → E2 tdy Cr
such that C = f(I) for some open interval I ⊆ R.
In immersed curve C ⊆ E2 is called immersed curve of the class Cr
if there exists
a parametrization f : [a, b] → E2, a, b ∈ R such that f([a, b]) = C, f(a) = f(b) and
f|(a,b) → E2 is a regular motion of the class Cr
such that f
(i)
+ (a) = f
(i)
− (b), i ≤ r.
If moreover maps f|[a,b) and f|(a,b] are injective then C is called simple closed immersed
curve.
For simplicity, we shall just talk about closed and simple closed curves (which will be
implicitly assumed to be immeresed). In this setting we shall introduce a new definition of
the curvature for which we shall consider E2 as oriented Euclidean space:
Definition. At each point of the curve f(t) we define oriented Frenet frame (f(t); e1(t), ¯e2(t))
as follows: e1(t) = f (t)
||f (t)||
and (e1(t), ¯e2(t)) is a positive orthonormal basis. Assuming f(s)
is the arc-length parametrization, we call the number ¯κ(s) ∈ R satisfying e1(s) = ¯κ(s)¯e2(s)
oriented curvature at the point f(s).
Frenet formulae are similar to the non-oriented version: e1(s) = ¯κ(s)¯e2(s) and e2(s) =
−¯κ(s)¯e1(s). Note the oriented curvature could be negative (think about examples).
Proposition. Let f : [a, b] → E2 be closed curve Cof the class Cr
. Then there is a
function θ : [a, b] → R of the class Cr
such that e1(t) = (cos θ(t), sin θ(t)) which satisfies
θ (t) = ¯κ(t)||f (t)||. Moreover, the difference θ(b) − θ(a) is independent on the choice of θ.
Proof. Existence of θ is obvious: choose θ(a) such that e1(a) = cos θ(a), sin θ(a) and then
extend θ continuously to the interval [a, b]. More precisly, using the arc-length parametrization
we have cos θ(s) = (e1(s), ε1) and sin θ(s) = −(¯e2(s), ε1) where ε1 is the first vector
of the standard basis. Then θ(s) is of tha class Cr
. (Do we need both previous relations?
Why?) By differentiating we obtain θ (s) = ¯κ(s). Reparametrization s = s(t), ds
dt
> 0 the
yields θ (t) = ¯κ(t)||f (t)||.
To show independence of the difference θ(b) − θ(a) on the choice of θ, assume ϕ(t) is
another function satisfying the proposition. Then θ(t) − ϕ(t) = 2k(t)π for some continuous
function k(t) ∈ Z. Thus k(t) is a constant.
Thus the difference θ(b) − θ(a) is obtained using a parametrization f : [a, b] → E2 of an
immersed closed curve (but is independent on a reparametrization). Consider e.g. various
parametrizations of the circle (cos t, sin t) where either t ∈ [0, 2π] or t ∈ [0, 4π] etc.
Definition. The number nC := 1
2π
θ(b) − θ(a) is called rotation index of the closed curve
uzavˇren´e kˇrivky C from the proposition.
Example. The curve f(t) = (cos 2πt, sin 2πt) for t ∈ [0, m], m ∈ N has the rotation index
m. (This curve is of course a circle.) What are examples of closed curves with negative
rotation index?
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Theorem. It holds nC = 1
2π
b
a
¯κ(t)||f (t)||dt.
Moreover, nC is independent on a reparametrization preserving the orientation. Reparametrizations
changing orientation change the sign of nC.
Proof. The first part of the proof follows from the relation θ (t) = ¯κ(t)||f (t)||. (In)dependence
on the reparametrization t = t(τ) follows from the form of the integral on the right hand
side after substituion t = t(τ).
Recall convex subset T ⊆ R2
satisfies pak x1x2 ⊆ T for each x1, x2 ∈ T. Here x1x2
denotes the segment with endpoints x1 and x2.
Lemma. Let T ⊆ R be a convex set and e : T → S1
be a function of the class Cr
. Then
there exists a function θ : T → R of the class Cr
satisfying e(x) = (cos θ(x), sin θ(x)), x ∈ T.
Moreover, if θ(x) and ϕ(x) are two such functions then they differ by 2kπ for some k ∈ Z.
Here we denote by S1
a circle. Note this technical lemma (stated without proof) is a
two-dimensional proof of the proposition
The following theorem is the main result of this section:
Theorem (Hopf’s Umlaufsatz). If f : [a, b] → E2 is a simple closed curve C then nC = ±1.
The opposite implication does not hold. Why?
Proof. We can assume a = 0 and that f is arc-length parametrization. Put ∆ = {(s, t) |
0 ≤ s ≤ b} ⊆ R2
and define the function h : ∆ → S1
as follows:
h(s, t) =



e1(s) s = t
−e1(0) (s, t) = (0, b)
f(t)−f(s)
||f(t)−f(s)||
otherwise.
The set ∆ is convex and function h(s, t) is continuous. Further we can assume that f(0) =
(0, 0) and that this is the “lowest” point on the curve, i.e. that this point has minimal ycoordinate.
Then e1(0) is (up to the sign) first vector of the standard basis, i.e. e1(0) = ±ε1.
Further we shall assume e1(0) = ε1 (which might change the orientation of C).
It follows from the lemma that h(s, t) = cos ˜θ(s, t), sin ˜θ(s, t) for continuous function
˜θ : ∆ → R. Using θ(s) from the proposition, then the theorem sayes that
nC =
1
2π
b
a
¯κ(s)ds =
1
2π
θ(b) − θ(0) =
1
2π
˜θ(b, b) − ˜θ(0, 0) =
=
1
2π
˜θ(0, b) − ˜θ(0, 0) + ˜θ(b, b) − ˜θ(0, b) .
Here N1 := ˜θ(0, b) − ˜θ(0, 0) is the angle which measures the change of the radious vector.
Hence N1 = π since the curve lies in the upper half-plane. Similarly, the angle N2 =
˜θ(b, b) − ˜θ(0, b) measures the change of the vector opposite to the radius vector, i.e. N2 = π.
Thus nC = 1.
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