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 2 w:Loops(R 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 •

Vin de Silva Pomona College 1. Winding Numbers Argument function

First attempt

2 arg : R 0 R { } ! This measures the angle of a point a = 0 from the positive x-axis. 6 [Example on board]

Corrected version

2 arg : R 0 R/2⇡Z { } ! Write [✓] for the equivalence class of ✓ modulo 2⇡.

Explicit formulas

[arctan(y/x)] if x > 0 8[arctan(y/x)+⇡]ifx < 0 arg((x, y)) = ⇡ >[ arctan(x/y)+ ]ify > 0 <> 2 [ arctan(x/y) ⇡ ]ify < 0 2 > These are continuous and agree:> where their domains overlap.

Vin de Silva Pomona College 1. Winding Numbers Lifts of the argument function

Lifts It is useful to consider functions

arg : R U ! that satisfy arg(a) =arg(a) for all a . 2 U At most one⇥ of the⇤ following can be satisﬁed: 2 = R 0 • U { } arg is continuous • The lift from a ray Let 2 arg : R 0 R { } ! be the unique lift taking values in [, +2⇡).

The lift arg is continuous except on the ray

1 R =arg ([]).

Vin de Silva Pomona College 1. Winding Numbers Angle cocycle

Line segments 2 Let a, b R . 2 [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 2 Let a, b R such that 0 [a, b] . 2 62 | | ⇤([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) •

Vin de Silva Pomona College 1. Winding Numbers Angle cocycle

Line segments 2 Let a, b R . 2 [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 2 Let a, b, p R such that p [a, b] . 2 62 | | ⇤([a, b], p)=‘anglesubtendedatp by [a, b]’

[Example on board]

Formally Let ⇤([a, b], p)betheuniquerealnumber✓ such that: ⇡ < ✓ < ⇡ • [✓]=arg(b p) arg(a p) •

Vin de Silva Pomona College 1. Winding Numbers 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 b p ⇤([a, b], p)=arg ⇡ . a p ✓ ◆ This is a composite of continuous functions, since

y arg ⇡(x + iy)=2arctan . x + x 2 + y 2 ! p

Vin de Silva Pomona College 1. Winding Numbers Closed polygons

Closed polygons

Let = P(a0, a1,...,an)denotethe‘directedpolygonwithverticesa0, a1,...,an and edges [a0, a1], [a1, a2],...,[an 1, an]’.

It is closed if a0 = an. Its support is

= [aj 1, aj ] | | | | j=1 [ We often write n

=[a0, a1]+[a1, a2]+ +[an 1, an]= [aj 1, aj ] ··· j=1 X

Vin de Silva Pomona College 1. Winding Numbers The winding number of a closed polygon

Winding number

Let = P(a0, a1,...,an)beclosed,andletp .Wedeﬁne: 62 | | n 1 w(, p)= ⇤([aj 1, aj ], p) 2⇡ j=1 X [Examples on board]

Vin de Silva Pomona College 1. Winding Numbers Properties of the winding number

⇤([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

n 1 w(, 0) = ⇤([aj 1, aj ], 0) = m1 + m2 + + mn. 2⇡ ··· j=1 X

Vin de Silva Pomona College 1. Winding Numbers Properties of the winding number

Properties of the winding number w(, p)isaninteger. • 2 w(, p)isconstantoneachcomponentofR . • | | w(, p)isequaltothenumberoftimes crosses a generic ray from p. • w(, p)=0ifminj aj p > maxj,k aj ak . • | | | | Proof 2 2 Each term ⇤([aj 1, aj ], p)iscontinuousonR [aj 1, aj ] and hence on R . | | | | Thus is continuous. Since it is integer-valued, it must be constant on each component.

Vin de Silva Pomona College 1. Winding Numbers Properties of the winding number

Properties of the winding number w(, p)isaninteger. • 2 w(, p)isconstantoneachcomponentofR . • | | w(, p)isequaltothenumberoftimes crosses a generic ray from p. • w(, p)=0ifminj aj p > maxj,k aj ak . • | | | | [Example on board] It is easiest to assume that the ray avoids the vertices of . Proof

We may assume p =0.Letarg be the lift from the ray R.Then

⇤([aj 1, aj ], 0) = arg (aj ) arg (aj 1)+2⇡⇠j

where ⇠j 0, 1 is the crossing number of the edge across the ray. Then 2 { ± } n 1 w(, p)= ⇤([aj 1, aj ], 0) = ⇠1 + ⇠2 + + ⇠n. 2⇡ ··· j=1 X

Vin de Silva Pomona College 1. Winding Numbers Properties of the winding number

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.

Vin de Silva Pomona College 1. Winding Numbers Generalizations

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

= [ak , bk ] Xk=1 2 such that each p R occurs equally often in the lists (ak )and(bk ). 2 [Example on board]

1 n w(, p)= ⇤([aj 1, aj ], p) 2⇡ j=1 Properties of the winding number P w(, p)isadditivein. • w(, p)isaninteger. • 2 w(, p)isconstantoneachcomponentofR . • | | w(, p)isequaltothenumberoftimes crosses a generic ray from p. • w(, p)=0ifminj aj p > maxj,k aj ak . • | | | |

Vin de Silva Pomona College 1. Winding Numbers Generalizations

1-cycles in a ring A A1-cycleinA is a formal linear combination of directed edges

= k [ak , bk ] Xk=1 2 such that k ak = p)= k bk = p) for each p R . | | 2 [Example onP board] P

n w(, p)= k ⇠([aj 1, aj ], R,p) j=1 Properties of the winding numberP w(, p)islinear in . • w(, p)isanelement of A. • 2 w(, p)isconstantoneachcomponentofR . • | | w(, p)isindependent of the choice of generic ray from p. • w(, p)=0ifminj aj p > maxj,k aj ak . • | | | | Vin de Silva Pomona College 1. Winding Numbers Vin de Silva Pomona College 1. Winding Numbers Continuous loops

The space of loops 2 Let R .ThenLoops( )isthesetofcontinuousmaps U ✓ U f :[0, 1] ! U such that f (0) = f (1).

Vin de Silva Pomona College 1. Winding Numbers Continuous loops

Homotopy of loops Two loops f , g Loops( )arehomotopic if there exists a continuous map 2 U H :[0, 1] [0, 1] ⇥ ! U on the unit square, such that H(0, t)=f (t) for all t [0, 1], • 2 H(1, t)=g(t) for all t [0, 1], • 2 H(s, 0) = H(s, 1) for all s [0, 1]. • 2 H is called a homotopy between f and g.Wewritef g or f H g. ' '

f H g

Homotopy ‘ ’isanequivalencerelation. '

Vin de Silva Pomona College 1. Winding Numbers Continuous loops

Theorem 2 Let R be convex. Then any two loops f , g Loops( )arehomotopic. U ✓ 2 U 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]

Vin de Silva Pomona College 1. Winding Numbers Continuous loops

Theorem 2 There are loops in R 0 which are not homotopic to each other. { }

2 How do we prove that they are not homotopic in R 0 ? { }

Vin de Silva Pomona College 1. Winding Numbers The continuous winding number

Theorem 2 There exists a function Loops(R 0 ) Z,expressedbythenotation { } ! f w(f , 0) 7! and called ‘the winding number of f around 0’, such that 2 if f g in Loops(R 0 )thenw(f , 0) = w(g, 0); and • ' { } if f = e2⇡idt then w(f , 0) = d. • 2 Deﬁne w(f , p) by translation, for any p R . 2

Vin de Silva Pomona College 1. Winding Numbers The continuous winding number

Theorem 2 There are loops in R 0 which are not homotopic to each other. { }

2 How do we prove that they are not homotopic in R 0 ? { } The blue loop is homotopic to e2⇡i0t and therefore has winding number 0. • The red loop is equal to e2⇡i1t and therefore has winding number 1. •

Vin de Silva Pomona College 1. Winding Numbers The continuous winding number: construction

Construction

Subdivide the interval by T =(t0, t1,...,tn), where

0=t0 < t1 < < tn =1. ··· Consider the polygonal approximation

n fT = f (tk 1), f (tk ) k=1 Deﬁne P ⇥ ⇤ w(f , 0) = w(fT , 0).

Vin de Silva Pomona College 1. Winding Numbers The continuous winding number: well-deﬁned

Is it well-deﬁned?

Let M =mint f (t) .Let > 0suchthat 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. •

Vin de Silva Pomona College 1. Winding Numbers The continuous winding number: homotopy invariance

What happens when f H g? ' Let M =mins,t H(s, t) . | | Let > 0suchthat s s0 , t t0 < implies H(s, t) H(s0, t0) < M. | | | | | | Divide the square into small squares of side less than .

f H g

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. •

Vin de Silva Pomona College 1. Winding Numbers The continuous winding number: normalization

What happens when f = e2⇡idt ? Let 1 2 N 1 T = 0, N , N ,..., N , 1 where N > 2 d .Then | | N 2⇡idt 2⇡idt 1 2⇡d w(e , 0) = w( e T , 0) = 2⇡ N = d k=1 ⇥ ⇤ X as required.

Vin de Silva Pomona College 1. Winding Numbers Vin de Silva Pomona College 1. Winding Numbers 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.

Vin de Silva Pomona College 1. Winding Numbers 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 The theorem fails for the open disk.

Vin de Silva Pomona College 1. Winding Numbers 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 The theorem fails for the annulus.

Vin de Silva Pomona College 1. Winding Numbers 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.)

Proof Suppose p has no ﬁxed points. Then q(x)=x p(x)isneverzero. 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 •

Vin de Silva Pomona College 1. Winding Numbers Vin de Silva Pomona College 1. Winding Numbers Application: Fundamental Theorem of Algebra

Theorem Every complex polynomial p(z)ofdegreed 1hasatleastonecomplexroot. (By repeated factorization, it therefore has d roots.)