Harmonic Morphisms - Basics › ... › personal › sigma › Basic-talk.pdf · 2014-03-11 ·...

Harmonic Morphisms - Basics

Sigmundur Gudmundsson

Department of MathematicsFaculty of ScienceLund University

[email protected]

March 11, 2014

Harmonic Maps in Gaussian GeometryHarmonic Maps in Riemannian Geometry

Outline

1 Harmonic Maps in Gaussian GeometryHolomorphic Functions in One VariableMinimal Surfaces in R3

Harmonic Morphisms in R3

2 Harmonic Maps in Riemannian GeometryMinimal SurfacesHolomorphic Functions in Several VariablesHarmonic Morphisms

Sigmundur Gudmundsson Harmonic Morphisms - Basics

Outline

1 Harmonic Maps in Gaussian GeometryHolomorphic Functions in One VariableMinimal Surfaces in R3

Harmonic Morphisms in R3

2 Harmonic Maps in Riemannian GeometryMinimal SurfacesHolomorphic Functions in Several VariablesHarmonic Morphisms

Holomorphic Functions in One VariableMinimal Surfaces in R3Harmonic Morphisms in R3

Let W be an open subset of the complex plane C. A function

φ = u+ iv : W → C

is said to be holomorphic if it satisfies the Cauchy-Riemann equations

∂x=∂v

∂yand

∂y= −∂v

As a direct consequence of these equations we see that

∂x2=

∂x∂y= −∂

∂y2and

∂x2= − ∂2u

∂x∂y= −∂

∂y2.

This implies that the functions u, v : W → R are harmonic, hence

∆φ = ∆(u+ iv) = ∆u+ i∆v = 0.

Let W be an open subset of the complex plane C. A function

φ = u+ iv : W → C

is said to be holomorphic if it satisfies the Cauchy-Riemann equations

∂x=∂v

∂yand

∂y= −∂v

As a direct consequence of these equations we see that

∂x2=

∂x∂y= −∂

∂y2and

∂x2= − ∂2u

∂x∂y= −∂

∂y2.

This implies that the functions u, v : W → R are harmonic, hence

∆φ = ∆(u+ iv) = ∆u+ i∆v = 0.

Theorem 1.1

Let u : W → R be a real-valued C2-function defined on an open simplyconnected subset W of C. Then the following are equivalent,

i) u is the real part of a holomorphic φ = u+ iv : W → C,

ii) u is harmonic i.e. ∆u = 0.

For the gradients ∇u = (∂u/∂x, ∂u/∂y), ∇v = (∂v/∂x, ∂v/∂y) we have

〈∇u,∇v〉 =∂u

∂x+∂u

∂y= 0,

|∇u|2 = (∂u

∂x)2 + (

∂y)2 = (

∂x)2 + (

∂y)2 = |∇v|2.

This means that the two gradients are orthogonal and of the same length ateach point in W i.e. the map φ : W → C is conformal, or equivalently,

〈∇φ,∇φ〉 = 〈∇(u+ iv),∇(u+ iv)〉 = 0.

Theorem 1.1

〈∇u,∇v〉 =∂u

∂x+∂u

∂y= 0,

|∇u|2 = (∂u

∂x)2 + (

∂y)2 = (

∂x)2 + (

∂y)2 = |∇v|2.

〈∇φ,∇φ〉 = 〈∇(u+ iv),∇(u+ iv)〉 = 0.

Theorem 1.1

〈∇u,∇v〉 =∂u

∂x+∂u

∂y= 0,

|∇u|2 = (∂u

∂x)2 + (

∂y)2 = (

∂x)2 + (

∂y)2 = |∇v|2.

〈∇φ,∇φ〉 = 〈∇(u+ iv),∇(u+ iv)〉 = 0.

Definition 1.2 (minimal surface)

Let Σt with t ∈ (−ε, ε) be a family of surfaces with a common boundarycurve i.e. ∂Σ0 = ∂Σt for all t ∈ (−ε, ε). Then Σ0 is said to be a minimalsurface if it is a critical point for the area functional i.e.

∫Σt

dA)|t=0 = 0.

Theorem 1.3 (zero mean curvature)

A surface Σ in R3 is minimal if and only if its mean curvature vanishes i.e.

H =k1 + k2

2≡ 0.

Definition 1.2 (minimal surface)

Let Σt with t ∈ (−ε, ε) be a family of surfaces with a common boundarycurve i.e. ∂Σ0 = ∂Σt for all t ∈ (−ε, ε). Then Σ0 is said to be a minimalsurface if it is a critical point for the area functional i.e.

∫Σt

dA)|t=0 = 0.

Theorem 1.3 (zero mean curvature)

A surface Σ in R3 is minimal if and only if its mean curvature vanishes i.e.

H =k1 + k2

2≡ 0.

There is an interesting connection between the theory of minimalsurfaces, harmonic functions and hence complex analysis.

Theorem 1.4 (the Weierstrass representation - 1866)

Every minimal surface Σ in R3 can locally be parametrized by aconformal and harmonic map φ : W ⊂ C→ R3 of the form

φ : z 7→ Re

f(w)[1− g2(w), i(1 + g2(w)), 2g(w)

where f, g : W ⊂ C→ C are two holomorphic functions.

Definition 1.5 (harmonic morphism)

The map φ = u+ iv : U ⊂ R3 → C is said to be a harmonic morphism ifthe composition f ◦ φ with any holomorphic function f : W ⊂ C→ C isharmonic.

Theorem 1.6 (Jacobi 1848)

The map φ = u+ iv : U ⊂ R3 → C is a harmonic morphism if and only if itis harmonic and horizontally (weakly) conformal i.e.

∆u = ∆v = 0, 〈∇u,∇v〉 = 0 and |∇u|2 = |∇v|2.

Proof.

∆(f ◦ φ) =∂f

∂z·∆φ+

∂z2· 〈∇φ,∇φ〉 = 0.

∆u = ∆v = 0, 〈∇u,∇v〉 = 0 and |∇u|2 = |∇v|2.

Proof.

∆(f ◦ φ) =∂f

∂z·∆φ+

∂z2· 〈∇φ,∇φ〉 = 0.

∆u = ∆v = 0, 〈∇u,∇v〉 = 0 and |∇u|2 = |∇v|2.

Proof.

∆(f ◦ φ) =∂f

∂z·∆φ+

∂z2· 〈∇φ,∇φ〉 = 0.

Theorem 1.7 (the Jacobi representation - 1848)

Let f, g : W ⊂ C→ C be holomorphic functions, then every local solutionz : U ⊂ R3 → C to the equation

〈f(z(x))[1− g2(z(x)), i(1 + g2(z(x))), 2g(z(x))

], x〉 = 1

is a harmonic morphism.

Theorem 1.8 (Baird, Wood 1988)

Every local harmonic morphism z : U → C in the Euclidean R3 is obtainedthis way.

Theorem 1.7 (the Jacobi representation - 1848)

Let f, g : W ⊂ C→ C be holomorphic functions, then every local solutionz : U ⊂ R3 → C to the equation

〈f(z(x))[1− g2(z(x)), i(1 + g2(z(x))), 2g(z(x))

], x〉 = 1

is a harmonic morphism.

Theorem 1.8 (Baird, Wood 1988)

Every local harmonic morphism z : U → C in the Euclidean R3 is obtainedthis way.

Example 1.9 (the outer disc example)

Let r ∈ R+ and choose g(z) = z, f(z) = −1/2irz then we obtain

(x1 − ix2)z2 − 2(x3 + ir)z − (x1 + ix2) = 0

with the two solutions

z±r =−(x3 + ir)±

1 + x22 + x2

3 − r2 + 2irx3

x1 − ix2.

Minimal SurfacesHolomorphic Functions in Several VariablesHarmonic Morphisms

Definition 2.1 (harmonic)

A map φ : (M, g)→ (N,h) between Riemannian manifolds is said to beharmonic if it is a critial point for the energy functional

E(φ) =1

|dφ|2dvolM .

The Euler-Lagrange equation for harmonic maps can be expressed, in localcoordinates x on M and y on N , as the following non-linear system ofPDEs

τ(φ) =

m∑i,j=1

∂2φγ

∂xi∂xj−

m∑k=1

Γkij∂φγ

∂xk+

n∑α,β=1

Γγαβ ◦ φ∂φα

∂φβ

where φα = yα ◦ φ and Γ and Γ the Christoffel symbols on M and N ,respectively.

Example 2.2 (geodesic)

A curve γ : I → (N,h) is harmonic if and only if it is a geodesic.

E(φ) =1

|dφ|2dvolM .

τ(φ) =m∑

∂2φγ

∂xi∂xj−

m∑k=1

Γkij∂φγ

∂xk+

n∑α,β=1

∂φβ

E(φ) =1

|dφ|2dvolM .

τ(φ) =m∑

∂2φγ

∂xi∂xj−

m∑k=1

Γkij∂φγ

∂xk+

n∑α,β=1

∂φβ

Theorem 2.3 (Eells, Sampson 1964)

A conformal immersion φ : (M2, g)→ (N,h) of a surface is harmonic ifand only if the image φ(M) is an immersed minimal surface in N .

Definition 2.4 (constant mean curvature)

A surface Σ in R3 is said to be of constant mean curvature if thereexists a constant c ∈ R such that the mean curvature H of Σ satisfies

H =k1 + k2

2≡ c.

Theorem 2.5 (Ruh, Vilms 1970)

A surface Σ in R3 is of constant mean curvature if and only if itsGauss map N : Σ→ S2 is harmonic.

H =k1 + k2

2≡ c.

H =k1 + k2

2≡ c.

Let φ = u+ iv : U → C be a holomorphic function defined on an opensubset U of Cm. Then it satisfies the Cauchy-Riemann equations i.e. if foreach k = 1, 2, . . . ,m

∂xk=

∂ykand

∂yk= − ∂v

∂xk.

As a direct consequence of these equations we get

∂x2k

=∂2v

∂xk∂yk= −∂

∂y2k

and∂2v

∂x2k

= − ∂2u

∂xk∂yk= −∂

∂y2k

This implies that the functions u, v : U → R are harmonic, hence

∆φ = ∆(u+ iv) = ∆u+ i∆v = 0.

For the two gradients ∇u,∇v we have the following

〈∇φ,∇φ〉 = 〈∇(u+ iv),∇(u+ iv)〉 = 0.

If ψ : U → C is another holomorphic function then the polar identity

4〈∇φ,∇ψ〉 = 〈∇(φ+ ψ),∇(φ+ ψ)〉 − 〈∇(φ− ψ),∇(φ− ψ)〉

gives〈∇φ,∇ψ〉 = 0.

Let φ = u+ iv : U → C be a holomorphic function defined on an opensubset U of Cm. Then it satisfies the Cauchy-Riemann equations i.e. if foreach k = 1, 2, . . . ,m

∂xk=

∂ykand

∂yk= − ∂v

∂xk.

As a direct consequence of these equations we get

∂x2k

=∂2v

∂xk∂yk= −∂

∂y2k

and∂2v

∂x2k

= − ∂2u

∂xk∂yk= −∂

∂y2k

This implies that the functions u, v : U → R are harmonic, hence

∆φ = ∆(u+ iv) = ∆u+ i∆v = 0.

For the two gradients ∇u,∇v we have the following

〈∇φ,∇φ〉 = 〈∇(u+ iv),∇(u+ iv)〉 = 0.

If ψ : U → C is another holomorphic function then the polar identity

4〈∇φ,∇ψ〉 = 〈∇(φ+ ψ),∇(φ+ ψ)〉 − 〈∇(φ− ψ),∇(φ− ψ)〉

gives〈∇φ,∇ψ〉 = 0.

A map φ : (M, g)→ (N,h) between Riemannian manifolds is said to be aharmonic morphism if, for any harmonic function f : U → R defined onan open subset U of N with φ−1(U) non-empty, f ◦ φ : φ−1(U)→ R is aharmonic function.

Theorem 2.7 (Fuglede 1978)

Let φ : (M, g)→ (Nn, h) be a non-constant harmonic morphism. Thenm ≥ n and

M∗ = {x ∈M | rank dφx = n}

is an open and dense subset of M .

Let φ : (M, g)→ (N,h) be a non-constant harmonic morphism. If M iscompact, then N is compact.

Theorem 2.9 (Fuglede 1978, Ishihara 1979)

A map φ : (M, g)→ (N,h) between Riemannian manifolds is a harmonicmorphism if and only if it is harmonic and horizontally (weakly)conformal.

Definition 2.10 (horizontally (weakly) conformal)

A map φ : (M, g)→ (N,h) between Riemannian manifolds is said to behorizontally (weakly) conformal at x ∈M if either dφx = 0, or dφxmaps the horizontal space Hx = (Ker(dφx))⊥ conformally onto Tφ(x)N i.e.there exist a function λ : M → R+

0 such that for all X,Y ∈ Hx we have

h(dφx(X), dφx(Y )) = λ2(x)g(X,Y ).

Theorem 2.9 (Fuglede 1978, Ishihara 1979)

A map φ : (M, g)→ (N,h) between Riemannian manifolds is a harmonicmorphism if and only if it is harmonic and horizontally (weakly)conformal.

Definition 2.10 (horizontally (weakly) conformal)

A map φ : (M, g)→ (N,h) between Riemannian manifolds is said to behorizontally (weakly) conformal at x ∈M if either dφx = 0, or dφxmaps the horizontal space Hx = (Ker(dφx))⊥ conformally onto Tφ(x)N i.e.there exist a function λ : M → R+

0 such that for all X,Y ∈ Hx we have

h(dφx(X), dφx(Y )) = λ2(x)g(X,Y ).

Example 2.11 (Fuglede 1978)

Let φ = u+ iv : U → C be a holomorphic function defined on an opensubset U of Cm. Then we know that φ is harmonic i.e.

∆φ = ∆(u+ iv) = ∆u+ i∆v = 0.

Furthermore〈∇φ,∇φ〉 = 〈∇(u+ iv),∇(u+ iv)〉 = 0.

This means that the two gradients are orthogonal and of the same length ateach point i.e. the map φ : U → C is horizontally conformal, hence aharmonic morphism.

Theorem 2.12 (Baird-Eells 1981)

Let φ : (M, g, J)→ C be a holomorphic function from a Kahler manifold.Then φ is a harmonic morphism.

Example 2.11 (Fuglede 1978)

Let φ = u+ iv : U → C be a holomorphic function defined on an opensubset U of Cm. Then we know that φ is harmonic i.e.

∆φ = ∆(u+ iv) = ∆u+ i∆v = 0.

Furthermore〈∇φ,∇φ〉 = 〈∇(u+ iv),∇(u+ iv)〉 = 0.

This means that the two gradients are orthogonal and of the same length ateach point i.e. the map φ : U → C is horizontally conformal, hence aharmonic morphism.

Theorem 2.12 (Baird-Eells 1981)

Let φ : (M, g, J)→ C be a holomorphic function from a Kahler manifold.Then φ is a harmonic morphism.

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