1. winding numbers...[example on board] proof the inequalities confine to a disk centered at a0...
TRANSCRIPT
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
Vin de Silva Pomona College 1. Winding Numbers
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.
Vin de Silva Pomona College 1. Winding Numbers
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
([�]).
Vin de Silva Pomona College 1. Winding Numbers
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)
Vin de Silva Pomona College 1. Winding Numbers
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)
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
⇤([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
!.
Vin de Silva Pomona College 1. Winding Numbers
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 ]
Vin de Silva Pomona College 1. Winding Numbers
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 |.
Vin de Silva Pomona College 1. Winding Numbers
Properties of the winding number
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.
Vin de Silva Pomona College 1. Winding Numbers
Properties of the winding number
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
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.
Vin de Silva Pomona College 1. Winding Numbers
Properties of the winding number
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 |.
[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.
Vin de Silva Pomona College 1. Winding Numbers
Properties of the winding number
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 |.
[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.
Vin de Silva Pomona College 1. Winding Numbers
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)
Properties of the winding number
• w(�, p) is additive in �.
• 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 |.
Vin de Silva Pomona College 1. Winding Numbers
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)
Properties of the winding number
• 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.
• w(�, p) = 0 if minj |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
Let U ✓ R2
. Then Loops(U) is the set of continuous maps
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 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.
Vin de Silva Pomona College 1. Winding Numbers
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]
Vin de Silva Pomona College 1. Winding Numbers
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}?
Vin de Silva Pomona College 1. Winding Numbers
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
.
Vin de Silva Pomona College 1. Winding Numbers
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.
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
fT =
Pnk=1
⇥f (tk�1
), f (tk)⇤
Define
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 � > 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.
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 � > 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.
Vin de Silva Pomona College 1. Winding Numbers
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.
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) 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
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) 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
Vin de Silva Pomona College 1. Winding Numbers
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
z z3 + (2-i)z2 +iz + 2
Vin de Silva Pomona College 1. Winding Numbers
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
z z3 + (2-i)z2 +iz + 2
Vin de Silva Pomona College 1. Winding Numbers
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
z z3 + (2-i)z2 +iz + 2
Vin de Silva Pomona College 1. Winding Numbers
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
z z3 + (2-i)z2 +iz + 2
Vin de Silva Pomona College 1. Winding Numbers
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
z z3 + (2-i)z2 +iz + 2
Vin de Silva Pomona College 1. Winding Numbers
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
z z3 + (2-i)z2 +iz + 2
Vin de Silva Pomona College 1. Winding Numbers
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
z z3 + (2-i)z2 +iz + 2
Vin de Silva Pomona College 1. Winding Numbers
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
z z3 + (2-i)z2 +iz + 2
Vin de Silva Pomona College 1. Winding Numbers
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
z z3 + (2-i)z2 +iz + 2
Vin de Silva Pomona College 1. Winding Numbers
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
z z3 + (2-i)z2 +iz + 2
Vin de Silva Pomona College 1. Winding Numbers
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
z z3 + (2-i)z2 +iz + 2
Vin de Silva Pomona College 1. Winding Numbers
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
z z3 + (2-i)z2 +iz + 2
Vin de Silva Pomona College 1. Winding Numbers
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
z z3 + (2-i)z2 +iz + 2
Vin de Silva Pomona College 1. Winding Numbers
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
z z3 + (2-i)z2 +iz + 2
Vin de Silva Pomona College 1. Winding Numbers
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
z z3 + (2-i)z2 +iz + 2
Vin de Silva Pomona College 1. Winding Numbers
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
z z3 + (2-i)z2 +iz + 2
Vin de Silva Pomona College 1. Winding Numbers