LECTURE 8: WEIERSTRASS FORMULAE

1. The Weierstrass formula for minimal surfaces in 𝔼^{3}

Weierstrass showed that any minimal surface in 𝔼^{3} could be described in terms of holomorphic functions of a complex variable. We will use adapted frames on the surface to derive this result.

Let {e_{1}, e_{2}, e_{3}} be an orthonormal frame on a surface Σ ⊂ 𝔼^{3} with e_{3} normal to the surface at each point. Recall that for such a frame we have

[\begin{matrix}ω_{1}^{3} \\ ω_{2}^{3} \end{matrix}] = [\begin{matrix}h_{11} & h_{12} \\ h_{12} & h_{22} \end{matrix}] [\begin{matrix}ω^{1} \\ ω^{2} \end{matrix}]

for some functions h_{11}, h_{12}, h_{22} and that Σ is minimal if and only if h_{11} + h_{22} = 0.

Consider the complex vector-valued 1-form

ξ = (e_{1} − ie_{2})(ω^{1} + iω^{2})
= (e_{1} ω^{1} + e_{2} ω^{2}) + i(e_{1} ω^{2} − e_{2} ω^{1}).

ξ is well-defined independent of the choice of adapted frame, and its exterior derivative is

dξ = i(de_{1} ∧ ω^{2} + e_{1} dω^{2} − de_{2} ∧ ω^{1} − e_{2} dω^{1})
= i(h_{11} + h_{22})e_{3} ω^{1} ∧ ω^{2}.

Therefore dξ = 0 if and only if Σ is minimal.

Suppose that Σ is minimal and let z, f be local complex-valued functions on Σ such that

ω^{1} + iω^{2} = f dz.

(Such functions always exist for any 1-form on a surface, although this is not true on higher-dimensional manifolds.) The function f must be everywhere nonzero, and since

dz ∧ dz̄ = −\frac{2i}{|f|^{2}} ω^{1} ∧ ω^{2} ≠ 0,

we can regard z as a local complex coordinate on Σ; this defines a complex structure on Σ. If we define F (z, z̄) to be the vector valued function

F = (e_{1} − ie_{2})f

then

ξ = F dz.

The fact that Σ is minimal, and hence that dξ = 0, implies that F is a function of z alone and so is a holomorphic function on Σ. Moreover,

⟨F, F ⟩ = f^{2}⟨e_{1} − ie_{2}, e_{1} − ie_{2}⟩ = 0.

Since ξ is a closed (1,0)-form, locally there exists a holomorphic function X(z) such that ξ = dX (and so X′(z) = F (z)), and

Re(dX) = Re(ξ) = e_{1} ω^{1} + e_{2} ω^{2} = dx

where x is the position vector on the surface Σ. Therefore, up to a translation in 𝔼^{3} we have

X(z) = x(z) + iy(z)

for some real vector-valued function y on Σ. Conversely, if X(z) is any holomorphic ℂ^{3}-valued function with ⟨X′, X′⟩ = 0, then the surface x = Re(X) is a minimal surface in 𝔼^{3}.

This gives rise to the Weierstrass representation for minimal surfaces. Let U ⊂ ℂ be open, g : U → ℂ a meromorphic function, and f : U → ℂ a holomorphic function with the property that if g has a pole of order k at z_{0} ∈ U then f has a zero of order 2k at z_{0}. Choose z_{0} ∈ U and define X : U → ℂ^{3} by

X(z) = ∫_{z_{0}}^{z} [\begin{matrix}\frac{1}{2}f(ζ)(1 − g(ζ)^{2}) \\ \frac{i}{2}f(ζ)(1 + g(ζ)^{2}) \\ f(ζ)g(ζ) \end{matrix}] dζ.

Then ⟨X′, X′⟩ = 0 and so x = Re(X) is the position vector of a minimal surface Σ ⊂ 𝔼^{3}. Conversely, any minimal surface has a local representation of this form (up to translation) in a neighborhood of any point.

The first fundamental form of Σ may be written as

I = \frac{1}{2}⟨ξ, ξ̄⟩
= \frac{1}{2}⟨dX, dX̄⟩
= (ω^{1} + iω^{2})(ω^{1} − iω^{2})
= (ω^{1})^{2} + (ω^{2})^{2}.

This leads to the following observation. Let t ∈ ℝ, and set

X_{t} = e^{it}X.

The family of minimal surfaces Σ_{t} with position vector x_{t} = Re(X_{t}) is called the associated family of Σ. All the surfaces in this family clearly have the same first fundamental form and so are isometric. In particular, the surface Σ_{3π/2} with position vector y = Im(X) is isometric to Σ; this surface is called the conjugate surface of Σ.

2. A Weierstrass-type formula for minimal surfaces in 𝔸^{3}

Now let Σ ⊂ 𝔸^{3} be an elliptic surface, and let {e_{1}, e_{2}, e_{3}} be an orthonormal frame on Σ for which the Maurer-Cartan forms satisfy the conditions

ω_{1}^{3} = ω^{1},  ω_{2}^{3} = ω^{2},  ω_{3}^{3} = 0.

Recall that for such a frame we have

[\begin{matrix}ω_{3}^{1} \\ ω_{3}^{2} \end{matrix}] = [\begin{matrix}ℓ_{11} & ℓ_{12} \\ ℓ_{12} & ℓ_{22} \end{matrix}] [\begin{matrix}ω^{1} \\ ω^{2} \end{matrix}]

[\begin{matrix}2ω_{1}^{1} \\ ω_{2}^{1} + ω_{1}^{2} \\ 2ω_{2}^{2} \end{matrix}] = [\begin{matrix}h_{1} & −h_{2} \\ −h_{2} & −h_{1} \\ −h_{1} & h_{2} \end{matrix}] [\begin{matrix}ω^{1} \\ ω^{2} \end{matrix}]

where h_{1} = h_{111} = −h_{122}, h_{2} = h_{222} = −h_{112}, and that Σ is affine minimal if and only if ℓ_{11} + ℓ_{22} = 0.

Let 𝔸_{ℂ}^{3} denote the complexified affine space 𝔸^{3} ⊗ ℂ, and consider the Λ^{2}𝔸_{ℂ}^{3}- valued 1-form

ξ = \frac{1}{2}e_{3} ∧ (e_{1} − ie_{2})(ω^{1} + iω^{2})
= \frac{1}{2}e_{3} ∧ [(e_{1} ω^{1} + e_{2} ω^{2}) + i(e_{1} ω^{2} − e_{2} ω^{1})].

ξ is well-defined independent of the choice of adapted frame, and a straightforward computation shows that its exterior derivative is

dξ = \frac{1}{2}(ℓ_{11} + ℓ_{22})(e_{1} ∧ e_{2}) ω^{1} ∧ ω^{2}.

Therefore dξ = 0 if and only if Σ is affine minimal.

Suppose that Σ is affine minimal and let z, f be complex-valued functions on Σ such that

ω^{1} + iω^{2} = f dz.

By the same reasoning as in the Euclidean case, z can be thought of as a local complex coordinate on Σ, and locally there exists a holomorphic Λ^{2}𝔸_{ℂ}^{3}-valued function X(z) on Σ such that ξ = dX.

For ease of notation, let

e = \frac{1}{2}(e_{1} − ie_{2})
ω = ω^{1} + iω^{2}.

Then dX = ξ = e_{3} ∧ e ω, and by conjugation dX̄ = ξ̄ = e_{3} ∧ ēω̄. A computation shows that

d(e ∧ ē) = \frac{1}{2}(e_{3} ∧ ēω̄ − e_{3} ∧ e ω) = \frac{1}{2}(dX̄ − dX).

It follows that

X̄ − X = 2e ∧ ē + 2ic

for some real-valued constant c ∈ Λ^{2}𝔸^{3}. By adding an imaginary constant to X, we can assume that c = 0.

At this point we need the “special linear cross product”. This is the unique skew-symmetric bilinear map

× : Λ^{2}𝔸^{3} × Λ^{2}𝔸^{3} → 𝔸^{3}

that is SL(3)-equivariant and satisfies

(e_{1} ∧ e_{2}) × (e_{1} ∧ e_{3}) = e_{1}

for any unimodular basis {e_{1}, e_{2}, e_{3}} of 𝔸^{3}. Geometrically, we can think of v_{1} ∧ v_{2} ∈ Λ^{2}𝔸^{3} as the plane spanned by v_{1} and v_{2}. The cross product of v_{1} ∧ v_{2} and w_{1} ∧ w_{2} is a vector which spans the line of intersection of the two planes. It can be computed using the ordinary cross product formula in ℝ^{3} by

(v_{1} ∧ v_{2}) × (w_{1} ∧ w_{2}) = (v_{1} × v_{2}) × (w_{1} × w_{2}).

This cross product can be extended in the obvious way to Λ^{2}𝔸_{ℂ}^{3}.

Now, using this formula for the cross product we compute that

(X̄ − X) × d(X̄ + X) = −i(e ω + ēω̄) = −i dx

where x is the position vector of Σ. Therefore

dx = i[(X̄ − X) × d(X̄ + X)]
= i[X̄ × dX̄ − X × dX + d(X̄ × X)]

and so the position vector x of the surface Σ is given by

x(z) = x(z_{0}) + i[\overline{X(z)} × X(z) − \overline{X(z_{0})} × X(z_{0}) + ∫_{z_{0}}^{z}(X̄ × dX̄ − X × dX)].

for some z_{0} ∈ Σ.

Conversely, let U ⊂ ℂ be open, and let X : U → Λ^{2}𝔸_{ℂ}^{3} be a holomorphic function that satisfies the open conditions dX ≠ 0 and X̄ ≠ X. Then the formula above gives the position vector x of an affine minimal surface Σ. This Weierstrass-type representation for affine minimal surfaces is due to Blaschke.

Exercises

1. In the Weierstrass representation for surfaces in 𝔼^{3}, let f(z) = 2 and g(z) = z. Show that the resulting minimal surface is parametrized by

x(u, v) = [\begin{matrix}u − \frac{1}{3}u^{3} + uv^{2} \\ −v + \frac{1}{3}v^{3} − vu^{2} \\ u^{2} − v^{2} \end{matrix}]

where z = u + iv. This is called Enneper’s surface. If you have access to a software package such as Maple, try sketching the surface over various intervals in u and v.

2. Recall from Lecture 7, Exercise 1 that the catenoid is parametrized by

x(u, v) = [\begin{matrix}cos u cosh v \\ sin u cosh v \\ v \end{matrix}] .

Show that the Weierstrass representation of the catenoid is obtained by taking f(z) = −ie^{−iz}, g(z) = e^{iz}, and that its conjugate surface is the helicoid. (Hint: the formula you’ll find for the conjugate surface will require a change of parameters before it looks like the parametrization for the helicoid from Lecture 7, Exercise 2.)

3. Consider the affine elliptic paraboloid z = \frac{1}{2}(x^{2} + y^{2}) with its adapted frame

e_{1} = (1, 0, x)
e_{2} = (0, 1, y)
e_{3} = (0, 0, 1).

Show that its Weierstrass-type representation is obtained by taking

X(z) = −\frac{i}{2}ε_{1} ∧ ε_{2} + \frac{i}{2}z ε_{2} ∧ ε_{3} + \frac{1}{2}z ε_{3} ∧ ε_{1}

where {ε_{1}, ε_{2}, ε_{3}} represents the standard basis of 𝔸^{3}. (Hint: Write the e_{i} as

e_{1} = ε_{1} + xε_{3}
e_{2} = ε_{2} + yε_{3}
e_{3} = ε_{3}

and note that, since {ε_{1}, ε_{2}, ε_{3}} is a unimodular basis,

(ε_{1} ∧ ε_{2}) × (ε_{1} ∧ ε_{3}) = ε_{1}
(ε_{2} ∧ ε_{3}) × (ε_{2} ∧ ε_{1}) = ε_{2}
(ε_{3} ∧ ε_{1}) × (ε_{3} ∧ ε_{2}) = ε_{3}.)