1. Winding Numbers...[Example on board] Proof The inequalities conﬁne to a disk centered at a0 that avoids p.Pickaraythat avoids this disk. Let’s call this the Distant Cycle Lemma

1. Winding Numbers

Vin de Silva

Pomona College

CBMS Lecture Series, Macalester College, June 2017

Vin de Silva Pomona College 1. Winding Numbers

Goal

Continuous winding number

We seek a function

w : Loops(R2 � {0}) �! Z

that counts the number of times the loop winds around 0.

[Examples on board]

• Loops are oriented.

• Counterclockwise = positive.

Approaches

• Covering spaces, fundamental group, homotopy theory

• Contour integration

• Discrete approximations, homology theory


Argument function

First attempt

arg : R2 � {0} �! R

This measures the angle of a point a 6= 0 from the positive x-axis.

[Example on board]

Corrected version

arg : R2 � {0} �! R/2⇡Z

Write [✓] for the equivalence class of ✓ modulo 2⇡.

Explicit formulas

arg((x , y)) =

8>>><

>>>:

[arctan(y/x)] if x > 0

[arctan(y/x) + ⇡] if x < 0

[� arctan(x/y) + ⇡2

] if y > 0

[� arctan(x/y)� ⇡2

] if y < 0

These are continuous and agree where their domains overlap.


Lifts of the argument function

Lifts

It is useful to consider functions

arg : U �! R

that satisfy

⇥arg(a)

⇤= arg(a) for all a 2 U .

At most one of the following can be satisfied:

• U = R2 � {0}• arg is continuous

The lift from a ray

Let

arg� : R2 � {0} �! R

be the unique lift taking values in [�,�+ 2⇡).

The lift arg� is continuous except on the ray

R� = arg

�1

([�]).


Angle cocycle

Line segments

Let a, b 2 R2

.

[a, b] = ‘directed line segment from a to b’

|[a, b]| = {ta+ ub | t, u � 0, t + u = 1}= set of points which lie on [a, b]

Angle cocycle

Let a, b 2 R2

such that 0 62 |[a, b]|.

⇤([a, b], 0) = ‘angle subtended at 0 by [a, b]’

[Example on board]

Formally

Let ⇤([a, b], 0) be the unique real number ✓ such that:

• �⇡ < ✓ < ⇡

• [✓] = arg(b)� arg(a)


Angle cocycle

Line segments

Let a, b 2 R2

.

[a, b] = ‘directed line segment from a to b’

|[a, b]| = {ta+ ub | t, u � 0, t + u = 1}= set of points which lie on [a, b]

Angle cocycle

Let a, b, p 2 R2

such that p 62 |[a, b]|.

⇤([a, b], p) = ‘angle subtended at p by [a, b]’

[Example on board]

Formally

Let ⇤([a, b], p) be the unique real number ✓ such that:

• �⇡ < ✓ < ⇡

• [✓] = arg(b�p)� arg(a�p)


Continuity

Theorem

⇤(a, b, p) is continuous as a function of a, b, p (on its domain of definition).

Proof

Regarding a, b, p as complex numbers, one can show that

⇤([a, b], p) = arg�⇡

✓b � p

a� p

◆.

This is a composite of continuous functions, since

arg�⇡(x + iy) = 2 arctan

y

x +

px2

+ y 2

!.


Closed polygons

Closed polygons

Let � = P(a0

, a1

, . . . , an) denote the ‘directed polygon with vertices a0

, a1

, . . . , anand edges [a

0

, a1

], [a1

, a2

], . . . , [an�1

, an]’.

It is closed if a0

= an. Its support is

|�| =n[

j=1

|[aj�1

, aj ]|

We often write

� = [a0

, a1

] + [a1

, a2

] + · · ·+ [an�1

, an] =nX

j=1

[aj�1

, aj ]


The winding number of a closed polygon

Winding number

Let � = P(a0

, a1

, . . . , an) be closed, and let p 62 |�|. We define:

w(�, p) =1

2⇡

nX

j=1

⇤([aj�1

, aj ], p)

[Examples on board]

Properties of the winding number

• w(�, p) is an integer.

• w(�, p) is constant on each component of R2 � |�|.• w(�, p) is equal to the number of times � crosses a generic ray from p.

• w(�, p) = 0 if minj |aj � p| > maxj,k |aj � ak |.







Proof

We may assume p = 0. Let arg be any lift of arg. Then:

⇥⇤([aj�1

, aj ], 0)⇤=

⇥arg(aj)� arg(aj�1

)

⇤

in R/2⇡Z, so

⇤([aj�1

, aj ], 0) = arg(aj)� arg(aj�1

) + 2⇡mj

for some integers mj . In the sum, the arg terms cancel and we get

w(�, 0) =1

2⇡

nX

j=1

⇤([aj�1

, aj ], 0) = m1

+m2

+ · · ·+mn.







Proof

Each term ⇤([aj�1

, aj ], p) is continuous on R2� |[aj�1

, aj ]| and hence on R2� |�|.Thus � is continuous. Since it is integer-valued, it must be constant on each

component.







[Example on board]

It is easiest to assume that the ray avoids the vertices of �.

Proof

We may assume p = 0. Let arg� be the lift from the ray R�. Then

⇤([aj�1

, aj ], 0) = arg�(aj)� arg�(aj�1

) + 2⇡⇠j

where ⇠j 2 {0,±1} is the crossing number of the edge across the ray. Then

w(�, p) =1

2⇡

nX

j=1

⇤([aj�1

, aj ], 0) = ⇠1

+ ⇠2

+ · · ·+ ⇠n.







[Example on board]

Proof

The inequalities confine � to a disk centered at a0

that avoids p. Pick a ray that

avoids this disk.

Let’s call this the Distant Cycle Lemma.


Generalizations

Integer 1-cycles

An integer 1-cycle is a formal sum of directed edges

� =

nX

k=1

[ak , bk ]

such that each p 2 R2

occurs equally often in the lists (ak) and (bk).

[Example on board]

w(�, p) = 1

2⇡

Pnj=1

⇤([aj�1

, aj ], p)


• w(�, p) is additive in �.





Generalizations

1-cycles in a ring AA 1-cycle in A is a formal linear combination of directed edges

� =

nX

k=1

�k [ak , bk ]

such that

P��k | ak = p) =

P��k | bk = p) for each p 2 R2

.

[Example on board]

w(�, p) =Pn

j=1

�k ⇠([aj�1

, aj ],R�,p)


• w(�, p) is linear in �.

• w(�, p) is an element of A.• w(�, p) is constant on each component of R2 � |�|.• w(�, p) is independent of the choice of generic ray from p.




Continuous loops

The space of loops

Let U ✓ R2

. Then Loops(U) is the set of continuous maps

f : [0, 1] �! U

such that f (0) = f (1).


Continuous loops

Homotopy of loops

Two loops f , g 2 Loops(U) are homotopic if there exists a continuous map

H : [0, 1]⇥ [0, 1] �! U

on the unit square, such that

• H(0, t) = f (t) for all t 2 [0, 1],

• H(1, t) = g(t) for all t 2 [0, 1],

• H(s, 0) = H(s, 1) for all s 2 [0, 1].

H is called a homotopy between f and g . We write f ' g or f 'H g .

gf H

Homotopy ‘'’ is an equivalence relation.


Continuous loops

Theorem

Let U ✓ R2

be convex. Then any two loops f , g 2 Loops(U) are homotopic.

Proof

We have f 'H g via the straight-line homotopy

H(s, t) = (1� s)f (t) + sg(t).

Convexity implies that this is entirely contained in U .

[Example on board]


Continuous loops

Theorem

There are loops in R2 � {0} which are not homotopic to each other.

How do we prove that they are not homotopic in R2 � {0}?


The continuous winding number

Theorem

There exists a function Loops(R2 � {0}) ! Z, expressed by the notation

f 7! w(f , 0)

and called ‘the winding number of f around 0’, such that

• if f ' g in Loops(R2 � {0}) then w(f , 0) = w(g , 0); and

• if f = e2⇡idtthen w(f , 0) = d .

Define w(f , p) by translation, for any p 2 R2

.


The continuous winding number

Theorem

There are loops in R2 � {0} which are not homotopic to each other.

How do we prove that they are not homotopic in R2 � {0}?• The blue loop is homotopic to e2⇡i0t

and therefore has winding number 0.

• The red loop is equal to e2⇡i1tand therefore has winding number 1.


The continuous winding number: construction

Construction

Subdivide the interval by T = (t0

, t1

, . . . , tn), where

0 = t0

< t1

< · · · < tn = 1.

Consider the polygonal approximation

fT =

Pnk=1

⇥f (tk�1

), f (tk)⇤

Define

w(f , 0) = w(fT , 0).


The continuous winding number: well-defined

Is it well-defined?

Let M = mint |f (t)|. Let � > 0 such that |t� t0| < � implies |f (t)� f (t0)| < M.

Use subdivisions T such that maxk |tk � tk�1

| < �.

• It follows that the edges of T avoid 0.

• Adding further points does not change the winding number.

We used:

• Additivity of the winding number for 1-cycles.

• Distant Cycle Lemma.


The continuous winding number: homotopy invariance

What happens when f 'H g?

Let M = mins,t |H(s, t)|.Let � > 0 such that |s � s 0|, |t � t0| < � implies |H(s, t)� H(s 0, t0)| < M.

Divide the square into small squares of side less than �.

gf H

H maps each small square to a quadrilateral with winding number 0.

It follows that 0 = w(gT , 0)� w(fT , 0) = w(g , 0)� w(f , 0).

We used:

• Additivity of the winding number for 1-cycles.

• Distant Cycle Lemma.


The continuous winding number: normalization

What happens when f = e2⇡idt?

Let

T =

�0, 1

N, 2

N, . . . , N�1

N, 1�

where N > 2|d |. Then

w(e2⇡idt , 0) = w(

⇥e2⇡idt⇤

T, 0) = 1

2⇡

NX

k=1

2⇡dN

= d

as required.



Application: Brouwer’s Fixed Point Theorem

Theorem

Every continuous map p : D2 ! D2

has a fixed point.

(Here D2

is the closed unit disk in the plane.)

More generally

We may replace D2

by any space that is topologically equivalent.



Theorem


has a fixed point.

(Here D2


More generally

The theorem fails for the open disk.



Theorem


has a fixed point.

(Here D2


More generally

The theorem fails for the annulus.



Theorem


has a fixed point.

(Here D2


Proof

Suppose p has no fixed points. Then q(x) = x � p(x) is never zero.

• Let f (t) = e2⇡it. w(f , 0) = 1

• Let J(s, t) = e2⇡it � sp(e2⇡it). w(f , 0) = w(g , 0)

• Let g(t) = q(e2⇡it). w(g , 0) =?

• Let K(s, t) = q(se2⇡it). w(g , 0) = w(h, 0)

• Let h(t) = q(0). w(h, 0) = 0



Application: Fundamental Theorem of Algebra

Theorem

Every complex polynomial p(z) of degree d � 1 has at least one complex root.

(By repeated factorization, it therefore has d roots.)

Proof

Suppose p has no roots. Let R > 0 be very large.

• Let f (t) = p(Re2⇡it). w(f , 0) = d

• Let H(s, t) = p(sRe2⇡it). w(f , 0) = w(g , 0)

• Let g(t) = p(0). w(g , 0) = 0



Theorem



Proof


• Let f (t) = p(Re2⇡it). w(f , 0) = d


• Let g(t) = p(0). w(g , 0) = 0

z z3 + (2-i)z2 +iz + 2



Theorem



Proof


• Let f (t) = p(Re2⇡it). w(f , 0) = d


• Let g(t) = p(0). w(g , 0) = 0

z z3 + (2-i)z2 +iz + 2



Theorem



Proof


• Let f (t) = p(Re2⇡it). w(f , 0) = d


• Let g(t) = p(0). w(g , 0) = 0

z z3 + (2-i)z2 +iz + 2



Theorem



Proof


• Let f (t) = p(Re2⇡it). w(f , 0) = d


• Let g(t) = p(0). w(g , 0) = 0

z z3 + (2-i)z2 +iz + 2



Theorem



Proof


• Let f (t) = p(Re2⇡it). w(f , 0) = d


• Let g(t) = p(0). w(g , 0) = 0

z z3 + (2-i)z2 +iz + 2



Theorem



Proof


• Let f (t) = p(Re2⇡it). w(f , 0) = d


• Let g(t) = p(0). w(g , 0) = 0

z z3 + (2-i)z2 +iz + 2



Theorem



Proof


• Let f (t) = p(Re2⇡it). w(f , 0) = d


• Let g(t) = p(0). w(g , 0) = 0

z z3 + (2-i)z2 +iz + 2



Theorem



Proof


• Let f (t) = p(Re2⇡it). w(f , 0) = d


• Let g(t) = p(0). w(g , 0) = 0

z z3 + (2-i)z2 +iz + 2



Theorem



Proof


• Let f (t) = p(Re2⇡it). w(f , 0) = d


• Let g(t) = p(0). w(g , 0) = 0

z z3 + (2-i)z2 +iz + 2



Theorem



Proof


• Let f (t) = p(Re2⇡it). w(f , 0) = d


• Let g(t) = p(0). w(g , 0) = 0

z z3 + (2-i)z2 +iz + 2



Theorem



Proof


• Let f (t) = p(Re2⇡it). w(f , 0) = d


• Let g(t) = p(0). w(g , 0) = 0

z z3 + (2-i)z2 +iz + 2



Theorem



Proof


• Let f (t) = p(Re2⇡it). w(f , 0) = d


• Let g(t) = p(0). w(g , 0) = 0

z z3 + (2-i)z2 +iz + 2



Theorem



Proof


• Let f (t) = p(Re2⇡it). w(f , 0) = d


• Let g(t) = p(0). w(g , 0) = 0

z z3 + (2-i)z2 +iz + 2



Theorem



Proof


• Let f (t) = p(Re2⇡it). w(f , 0) = d


• Let g(t) = p(0). w(g , 0) = 0

z z3 + (2-i)z2 +iz + 2



Theorem



Proof


• Let f (t) = p(Re2⇡it). w(f , 0) = d


• Let g(t) = p(0). w(g , 0) = 0

z z3 + (2-i)z2 +iz + 2



Theorem



Proof


• Let f (t) = p(Re2⇡it). w(f , 0) = d


• Let g(t) = p(0). w(g , 0) = 0

z z3 + (2-i)z2 +iz + 2


Documents

1. Winding Numbers...[Example on board] Proof The inequalities conﬁne to a disk centered at a0 that avoids p.Pickaraythat avoids this disk. Let’s call this the Distant Cycle Lemma