Visualizing the Schwarzian Derivative - University of Michigan

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Visualizing the Schwarzian Derivative Noah Luntzlara, Piriyakorn Piriyatamwong, Mengxi Wang, Sharon Ye Patrick Haggerty, Didac Martinez-Granado, Maxime Scott, Dylan Thurston Laboratory of Geometry at Michigan Introduction Goal Draw pictures to better understand the Schwarzian Derivative. Some Preliminaries We need to know about the stereographic projection and M¨ obius transformations. Definition. A(n extended) stereographic projection is a function π : S 2 b C that maps S 2 , i.e. the surface of the unit sphere in R 3 , to the extended complex plane b C = C ∪ {∞} (we identify C with the xy -plane in R 3 ). x y z N π (P ) P 1 Let N in S 2 denote the north pole. Each point P in S 2 -{N } is mapped to the intersection of the unique line through N and P with the xy -plane, denoted as π (P ). However, when P = N , this procedure fails, because there is no unique line through N and P . Notice that the points near N are mapped to points that have large absolute value, so it makes sense to map N to . So we set π (N )= . This produces a bijection from the sphere S 2 to the extended complex numbers b C, thereby equipping S 2 with the algebraic structure of b C; we call this the Riemann Sphere. Definition. AM¨ obius transformation is a complex function of the form M (z )= az + b cz + d where a, b, c, d are complex numbers and ad - bc 6=0. Being careful, we can extend the domain and range of M to the extended complex plane b C. One of the key properties of M¨ obius transformations is that they preserve circles on the Riemann Sphere. Here we give three examples of M¨ obius transformations. Every M¨ obius transformation can be decomposed as a composition of M ¨ obius transformations similar to these three. ( a ) Hyperbolic: z 72z ( b ) Parabolic: z 7z +1 ( c ) Elliptic: z 7iz What is the Schwarzian Derivative? Definition. Let f : C C be a holomorphic function. The Schwarzian derivative of f is the function S (f )= f 000 f 0 - 3 2 f 00 f 0 2 The Schwarzian measures how much a function deviates from a M¨ obius transformation; it vanishes for M¨ obius transformations but gives useful information about other maps. We noted in the previous section that M¨ obius transformations take circles in the Riemann sphere to other circles; but other functions may distort circles. Example 1. Images of concentric circles un- der the polynomial f (z )= z + z 3 /3; note the self-intersection. Definition. Let f : C C be a holomorphic function, and let s C. The osculating M ¨ obius transformation of f at s is the unique M¨ obius transformation M (f,s)(z ) that matches f in value, first, and second derivative at s: M (f,s)(s)= f (s),M 0 (f,s)(s)= f 0 (s),M 00 (f,s)(s)= f 00 (s). The Schwarzian derivative is determined by the rate of change of the osculating M¨ obius transformation as we change s. After some renormalization, ∂w M (f,w) is a quadratic in z - w. The leading coefficient of this quadratic is the Schwarzian derivative. Why is the Schwarzian derivative important? The Schwarzian is the unique operator with the following properties. Vanishes precisely for M¨ obius transformations: S (f )=0 ⇐⇒ f is a M ¨ obius transformation Invariant under composition with M¨ obius transformations; if μ is a M¨ obius transforma- tion, S (μ f )= S (f ) It appears in many branches of mathematics, including hypergeometric functions and Te- ichm ¨ uller theory. Relation to Curvature Motivation. While M¨ obius transformations are the right framework for doing computations, they can be difficult to visualize. On the other hand, it is relatively easy to see curvature–how ”bent” one-dimensional curves are. Luckily, curvature is related to the Schwarzian derivative! Definitions. Let γ : R C be a smooth parameterized curve. Let T be the unit tangent vector T (t)= γ 0 (t)/||γ 0 (t)||. The curvature of γ at t is defined to be κ(t)= d dt T (t). (Note that this is invariant under different parameterizations for the curve im(γ ).) The osculating circle of γ at t is the unique circle tangent to γ at t with the same curva- ture; it has radius r =1. The name osculating circle might remind you of osculating M¨ obius transformations. The two are related by the following theorem. Theorem 1. Let f : C C be a holomorphic function, let γ : R C be a parameterized circle in C, and let z 0 = γ (t 0 ) be a point on this circle. If μ = M (f,z 0 ) is the osculating M ¨ obius transformation of f at z 0 , then μ sends the circle γ to the osculating circle of the curve f γ at the point f (z 0 ). To prove this, we use the following lemma relating curvature before and after applying a holomorphic map. Lemma. Let γ : R C be a smooth parameterized curve, T be the unit tangent vector T (t)= γ 0 (t)/||γ 0 (t)||. Let p = γ (s 0 ) be a point on this curve. Let κ be the curvature of γ at p. Then the curvature of f γ at f (p) is given by ˜ κ = 1 |f 0 (p)| = h f 00 (p)T (s 0 ) f 0 (p) i + κ . The proof is a computation. Proof of Theorem 1. Note that since μ = M (f,z 0 ), we have μ(z 0 )= f (z 0 ), μ 0 (z 0 )= f 0 (z 0 ), μ 00 (z 0 )= f 00 (z 0 ). Since the osculating circle is unique, it suffices to show that μ γ is tangent to f γ at f (z 0 ) and has the same curvature. Tangent: We have μ(z 0 )= f (z 0 ), so the two curves intersect. Moreover, μ 0 (z 0 )= f 0 (z 0 ), so by the chain rule (μ γ ) 0 (t 0 )= μ 0 (z 0 )γ 0 (t 0 )= f 0 (z 0 )γ 0 (t 0 )=(f γ ) 0 (t 0 ). Thus the two curves are tangent. Equal curvature: by the lemma, we can write an expression in terms of derivatives for the curvature of μ γ at μ(z 0 )= f (z 0 ), and similarly for f . Since the first two derivatives are equal, 1 |μ 0 (p)| = h μ 00 (p)T (s 0 ) μ 0 (p) i + κ = 1 |f 0 (p)| = h f 00 (p)T (s 0 ) f 0 (p) i + κ . Thus the circle im(μ γ ) is the osculating circle of im(f γ ) at f (z 0 ). By drawing nearby osculating circles to the images of circles under the map f , we can see how the osculating obius transformation changes and glean information about the shape of the Schwarzian derivative. Example 2. Left: the images of concentric circles under f (z )= z + z 3 /3 (red), and oscu- lating circles at equidistant points around them (blue). Right: the images under the same map of circles with equal radii through a point (red), and osculating circles at that point (blue). Reference [1] Needham, Tristan. Visual Complex Analysis, Oxford University Press, (1997). [2] Osgood, Brad. (1998). Old and New on the Schwarzian Derivative. [3] Haggerty Patrick, Mart´ ınez Granado D´ ıdac, Scott Maxime. LOG(M): Visualiz- ing the Schwarzian derivative, online notes https://www.overleaf.com/read/ tvfpfcmqytdp#/41298056/ LOG(M) Poster Session 2018

Transcript of Visualizing the Schwarzian Derivative - University of Michigan

Page 1: Visualizing the Schwarzian Derivative - University of Michigan

Visualizing the Schwarzian DerivativeNoah Luntzlara, Piriyakorn Piriyatamwong, Mengxi Wang, Sharon Ye

Patrick Haggerty, Didac Martinez-Granado, Maxime Scott, Dylan ThurstonLaboratory of Geometry at Michigan

Introduction

GoalDraw pictures to better understand the Schwarzian Derivative.

Some PreliminariesWe need to know about the stereographic projection and Mobius transformations.

Definition. A(n extended) stereographic projection is a function π : S2 → C that maps S2,i.e. the surface of the unit sphere in R3, to the extended complex plane C = C ∪ {∞} (weidentify C with the xy-plane in R3).

x

y

z

N

π(P )

P1

Let N in S2 denote the north pole. Each point P in S2−{N} is mapped to the intersectionof the unique line through N and P with the xy-plane, denoted as π(P ). However, whenP = N , this procedure fails, because there is no unique line through N and P . Notice thatthe points near N are mapped to points that have large absolute value, so it makes senseto map N to∞. So we set π(N) =∞.This produces a bijection from the sphere S2 to the extended complex numbers C, therebyequipping S2 with the algebraic structure of C; we call this the Riemann Sphere.

Definition. A Mobius transformation is a complex function of the form

M(z) =az + b

cz + d

where a, b, c, d are complex numbers and ad − bc 6= 0. Being careful, we can extend thedomain and range of M to the extended complex plane C.

One of the key properties of Mobius transformations is that they preserve circles on theRiemann Sphere.Here we give three examples of Mobius transformations. Every Mobius transformationcan be decomposed as a composition of Mobius transformations similar to these three.

( a ) Hyperbolic: z 7→ 2z ( b ) Parabolic: z 7→ z + 1 ( c ) Elliptic: z 7→ iz

What is the Schwarzian Derivative?

Definition. Let f : C → C be a holomorphic function. The Schwarzian derivative of fis the function

S(f ) = f ′′′

f ′− 3

2

(f ′′f ′

)2The Schwarzian measures how much a function deviates from a Mobius transformation;it vanishes for Mobius transformations but gives useful information about other maps.We noted in the previous section that Mobius transformations take circles in the Riemannsphere to other circles; but other functions may distort circles.

Example 1. Images of concentric circles un-der the polynomial f (z) = z + z3/3; note theself-intersection.

Definition. Let f : C → C be a holomorphic function, and let s ∈ C. Theosculating Mobius transformation of f at s is the unique Mobius transformation M(f, s)(z)that matches f in value, first, and second derivative at s:

M(f, s)(s) = f (s), M ′(f, s)(s) = f ′(s), M ′′(f, s)(s) = f ′′(s).

The Schwarzian derivative is determined by the rate of change of the osculating Mobiustransformation as we change s. After some renormalization, ∂

∂wM(f, w) is a quadratic inz − w. The leading coefficient of this quadratic is the Schwarzian derivative.

Why is the Schwarzian derivative important?The Schwarzian is the unique operator with the following properties.• Vanishes precisely for Mobius transformations: S(f ) = 0⇐⇒ f is a Mobius transformation• Invariant under composition with Mobius transformations; if µ is a Mobius transforma-

tion, S(µ ◦ f ) = S(f )It appears in many branches of mathematics, including hypergeometric functions and Te-ichmuller theory.

Relation to Curvature

Motivation.While Mobius transformations are the right framework for doing computations, they canbe difficult to visualize. On the other hand, it is relatively easy to see curvature–how ”bent”one-dimensional curves are. Luckily, curvature is related to the Schwarzian derivative!Definitions. Let γ : R → C be a smooth parameterized curve. Let T be the unit tangentvector T (t) = γ′(t)/||γ′(t)||.• The curvature of γ at t is defined to be

κ(t) =d

dtT (t).

(Note that this is invariant under different parameterizations for the curve im(γ).)• The osculating circle of γ at t is the unique circle tangent to γ at t with the same curva-

ture; it has radius r = 1/κ.The name osculating circle might remind you of osculating Mobius transformations. Thetwo are related by the following theorem.

Theorem 1. Let f : C→ C be a holomorphic function, let γ : R→ C be a parameterizedcircle in C, and let z0 = γ(t0) be a point on this circle.If µ = M(f, z0) is the osculating Mobius transformation of f at z0, then µ sends thecircle γ to the osculating circle of the curve f ◦ γ at the point f (z0).

To prove this, we use the following lemma relating curvature before and after applying aholomorphic map.Lemma. Let γ : R → C be a smooth parameterized curve, T be the unit tangent vectorT (t) = γ′(t)/||γ′(t)||. Let p = γ(s0) be a point on this curve. Let κ be the curvature of γ atp. Then the curvature of f ◦ γ at f (p) is given by

κ =1

|f ′(p)|

(=[f ′′(p)T (s0)

f ′(p)

]+ κ).

The proof is a computation.Proof of Theorem 1. Note that since µ = M(f, z0), we have µ(z0) = f (z0), µ′(z0) = f ′(z0),µ′′(z0) = f ′′(z0).Since the osculating circle is unique, it suffices to show that µ◦γ is tangent to f ◦γ at f (z0)and has the same curvature.• Tangent: We have µ(z0) = f (z0), so the two curves intersect. Moreover, µ′(z0) = f ′(z0),

so by the chain rule

(µ ◦ γ)′(t0) = µ′(z0)γ′(t0) = f ′(z0)γ

′(t0) = (f ◦ γ)′(t0).

Thus the two curves are tangent.• Equal curvature: by the lemma, we can write an expression in terms of derivatives for

the curvature of µ ◦ γ at µ(z0) = f (z0), and similarly for f . Since the first two derivativesare equal,

1

|µ′(p)|

(=[µ′′(p)T (s0)

µ′(p)

]+ κ)=

1

|f ′(p)|

(=[f ′′(p)T (s0)

f ′(p)

]+ κ).

Thus the circle im(µ ◦ γ) is the osculating circle of im(f ◦ γ) at f (z0).By drawing nearby osculating circles to the images of circles under themap f , we can see how the osculating Mobius transformation changesand glean information about the shape of the Schwarzian derivative.

Example 2. Left: the images of concentric circles under f (z) = z + z3/3 (red), and oscu-lating circles at equidistant points around them (blue). Right: the images under the samemap of circles with equal radii through a point (red), and osculating circles at that point(blue).

Reference

[1] Needham, Tristan. Visual Complex Analysis, Oxford University Press, (1997).[2] Osgood, Brad. (1998). Old and New on the Schwarzian Derivative.[3] Haggerty Patrick, Martınez Granado Dıdac, Scott Maxime. LOG(M): Visualiz-ing the Schwarzian derivative, online notes https://www.overleaf.com/read/tvfpfcmqytdp#/41298056/

LOG(M) Poster Session 2018