LECTURE 7: MINIMALITY AND VARIATIONAL CALCULATIONS

1. Minimal surfaces in 𝔼^{3}

We will say that a regular surface in 𝔼^{3} is minimal if it is locally area minimizing. More precisely, Σ ⊂ 𝔼^{3} is minimal if for any sufficiently small open set U ⊂ Σ, U has the minimum area of all surfaces in 𝔼^{3} with the same boundary as U. Classical examples are the plane, catenoid, and helicoid.

How would we go about finding minimal surfaces? If we define the area functional of a surface Σ to be

A(Σ) = ∫_{Σ} dA

then minimal surfaces should be critical points of this functional. But the space of surfaces in 𝔼^{3} is infinite-dimensional, so finding critical points of the functional A is somewhat complicated. The idea goes something like this: if Σ is a critical point of A, then for any smooth curve t → Σ_{t} in the space of surfaces in 𝔼^{3} with Σ_{0} = Σ we should have

\frac{d}{dt} |_{t=0} A(Σ_{t}) = 0.

Conversely, if Σ is not a critical point of A then there must exist a smooth curve t → Σ_{t} with Σ_{0} = Σ and \frac{d}{dt} |_{t=0} A(Σ_{t}) ≠ 0.

In order to make use of this idea we have to define what we mean by a smooth curve in the space of surfaces in 𝔼^{3}. This leads us to the notion of a variation of a surface Σ. Given a regular surface x : Σ → 𝔼^{3}, consider a smooth map

X : Σ × (−ε, ε) → 𝔼^{3}

where Σ_{t} = X(Σ, t) ⊂ 𝔼^{3} is a regular surface for all t ∈ (−ε, ε) and Σ_{0} = Σ. Such a map is called a variation of Σ. The variation is said to be compactly supported if there is a compact set U in the interior of Σ such that for every t ∈ (−ε, ε),

X(x, t) = X(x, 0)

for all x ∈ Σ \ U. If X is a compactly supported variation of Σ and Σ is a critical point of A, then

\frac{d}{dt} |_{t=0} A(Σ_{t}) = 0.

Conversely, if \frac{d}{dt} |_{t=0} A(Σ_{t}) = 0 for every compactly supported variation of Σ, then Σ is a critical point of the functional A. We will use this fact to investigate minimal surfaces.

Let {e_{1}, e_{2}, e_{3}} be an orthonormal frame on Σ with e_{3} normal to the tangent plane T_{x}Σ at each point x ∈ Σ. Recall that when the Maurer-Cartan forms of 𝔼^{3} are restricted to this frame, we have ω^{3} = 0 and

[\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}] .

Rotating the frame {e_{1}, e_{2}} changes the matrix [h_{ij}], but its determinant K = h_{11}h_{22} − h_{12}^{2} and its trace 2H = h_{11} + h_{22} are invariant under such changes of frame. K is the Gauss curvature of Σ at x, and H is called the mean curvature of Σ at x.

Now consider a compactly supported variation

X : Σ × (−ε, ε) → 𝔼^{3}.

Since reparametrizing the surface does not affect the area functional, we can assume that X is a normal variation of Σ. This means that the vector \frac{∂X}{∂t} is parallel to the unit normal of the surface Σ_{t} at each point. In

order to compute \frac{d}{dt} |_{t=0} A(Σ_{t}), we will define a frame on the variation X (i.e., a lifting X̃ : Σ × (−ε, ε) → E(3)) and consider the restriction of the Maurer-Cartan forms on 𝔼^{3} to this frame. For each (x, t) ∈ Σ × (−ε, ε), let {e_{1}(x, t), e_{2}(x, t), e_{3}(x, t)} be an orthonormal frame for the surface Σ_{t} at x with e_{3} normal to the tangent plane T_{x}Σ_{t}. The restrictions of the forms ω^{1}, ω^{2}, ω^{3} to this frame are defined by the equation

dX = ∑_{i=1}^{3} e_{i} ω^{i}.

Because {e_{1}, e_{2}, e_{3}} is adapted to the surface Σ_{t}, the forms

ω^{1} = ⟨dX, e_{1}⟩
ω^{2} = ⟨dX, e_{2}⟩

are the usual dual forms on the surface Σ_{t}. But instead of having ω^{3} = 0, we have

ω^{3} = ⟨dX, e_{3}⟩ = |\frac{∂X}{∂t} | dt.

Set f(x, t) = |\frac{∂X}{∂t} |. Then ω^{3} = f dt. Differentiating this equation yields

−ω_{1}^{3} ∧ ω^{1} − ω_{2}^{3} ∧ ω^{2} = df ∧ dt,

and therefore

ω_{1}^{3} ∧ ω^{1} + ω_{2}^{3} ∧ ω^{2} + df ∧ dt = 0.

By Cartan’s Lemma,

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

for some functions h_{11}, h_{12}, h_{22}, f_{1}, f_{2}, f_{3} on Σ × (−ε, ε). The h_{ij} are the coefficients of the second fundamental form of the surface Σ_{t}, while the f_{i} are the directional derivatives of f in the directions of the e_{i}.

Now the area form on the surface Σ_{t} is dA = ω^{1} ∧ ω^{2}, so the area functional is

A(Σ_{t}) = ∫_{Σ_{t}} ω^{1} ∧ ω^{2}.

Its derivative at t = 0 is

\frac{d}{dt} |_{t=0} A(Σ_{t}) = \frac{d}{dt} |_{t=0} ∫_{Σ_{t}} ω^{1} ∧ ω^{2}

= ∫_{Σ} L_{∂/∂t}(ω^{1} ∧ ω^{2})

= ∫_{Σ} \frac{∂}{∂t} d(ω^{1} ∧ ω^{2})

= ∫_{Σ} \frac{∂}{∂t} (−ω_{3}^{1} ∧ ω^{3} ∧ ω^{2} + ω^{1} ∧ ω_{3}^{2} ∧ ω^{3})

= ∫_{Σ} \frac{∂}{∂t} (−h_{11} − h_{22})f ω^{1} ∧ ω^{2} ∧ dt

= ∫_{Σ} −2Hf ω^{1} ∧ ω^{2}.

This computation shows that the derivative of the area functional at t = 0

is the integral over Σ of the function H|\frac{∂X}{∂t} |. This integral vanishes for all compactly supported normal variations of Σ if and only if the mean curvature H of Σ is identically zero.

We have proved the following theorem, which is often taken as a definition of minimal surfaces:

Theorem: A regular surface in 𝔼^{3} is minimal if and only if its mean curvature H is identically zero.

2. Minimal surfaces in 𝔸^{3}

Recall that for elliptic surfaces Σ ⊂ 𝔸^{3} we found adapted frames {e_{1}, e_{2}, e_{3}} for which the Maurer-Cartan forms satisfy the conditions

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

Such a frame is determined up to a transformation of the form

[ẽ_{1} ẽ_{2} ẽ_{3}] = [e_{1} e_{2} e_{3}] [\begin{matrix}B & {0 \atop 0} \\ 0 0 & 1 \end{matrix}]

with B ∈ SO(2), and we also 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}]

for some functions ℓ_{11}, ℓ_{12}, ℓ_{22}. The quadratic forms

I = ω_{1}^{3} ω^{1} + ω_{2}^{3} ω^{2} = (ω^{1})^{2} + (ω^{2})^{2}
II = ω_{3}^{1} ω^{1} + ω_{3}^{2} ω^{1} = ℓ_{11}(ω^{1})^{2} + 2ℓ_{12} ω^{1} ω^{2} + ℓ_{22}(ω^{2})^{2}

are well-defined; the affine first fundamental form I defines a metric on Σ, and the affine second fundamental form II is the analog of the second fundamental form for surfaces in 𝔼^{3}. Its trace

2L = ℓ_{11} + ℓ_{22}

is well-defined, and the quantity L is called the affine mean curvature of Σ.

The affine first fundamental form gives rise to a well-defined affine area form dA = ω^{1} ∧ ω^{2} on Σ, and we define the affine area of Σ to be

A(Σ) = ∫_{Σ} ω^{1} ∧ ω^{2}.

By analogy with the Euclidean case, we can ask what properties a surface Σ ⊂ 𝔸^{3} must satisfy in order to be a critical point of this area functional. Such surfaces are called affine minimal surfaces.

We proceed as in the Euclidean case by considering a compactly supported normal variation

X : Σ × (−ε, ε) → 𝔸^{3}.

Here “normal” means that the vector \frac{∂X}{∂t} is parallel to the affine normal of the surface Σ_{t} = X(Σ, t) at each point. For each (x, t) ∈ Σ × (−ε, ε) let {e_{1}(x, t), e_{2}(x, t), e_{3}(x, t)} be a frame for the surface Σ_{t} at x which is adapted as described above. As in the Euclidean case, the Maurer-Cartan forms ω^{1}, ω^{2} are the usual dual forms on the surface Σ_{t}, while ω^{3} = f dt

where f(x, t) = |\frac{∂X}{∂t} |. The derivative of the affine area functional at t = 0 is

\frac{d}{dt} |_{t=0} A(Σ_{t}) = \frac{d}{dt} |_{t=0} ∫_{Σ_{t}} ω^{1} ∧ ω^{2}

= ∫_{Σ} L_{∂/∂t}(ω^{1} ∧ ω^{2})

= ∫_{Σ} \frac{∂}{∂t} d(ω^{1} ∧ ω^{2})

= ∫_{Σ} \frac{∂}{∂t} (−ω_{1}^{1} ∧ ω^{1} ∧ ω^{2} − ω_{3}^{1} ∧ ω^{3} ∧ ω^{2}

+ ω^{1} ∧ ω_{2}^{2} ∧ ω^{2} + ω^{1} ∧ ω_{3}^{2} ∧ ω^{3})

= ∫_{Σ} \frac{∂}{∂t} (ℓ_{11} + ℓ_{22})f ω^{1} ∧ ω^{2} ∧ dt

= ∫_{Σ} 2Lf ω^{1} ∧ ω^{2}.

This integral vanishes for all compactly supported normal variations of Σ if and only if the affine mean curvature L of Σ is identically zero. Thus we have proved the following theorem:

Theorem: A regular surface in 𝔸^{3} is affine minimal if and only if its affine mean curvature L is identically zero.

Exercises

1. The catenoid is the surface Σ ⊂ 𝔼^{3} obtained by rotating the curve x = cosh z about the z axis. It can be parametrized by

x(u, v) = (cos(u) cosh(v), sin(u) cosh(v), v).

a) Show that the frame

e_{1} = \frac{x_{u}}{|x_{u}|} = (− sin(u), cos(u), 0)
e_{2} = \frac{x_{v}}{|x_{v}|} = \frac{1}{cosh(v)}(cos(u) sinh(v), sin(u) sinh(v), 1)
e_{3} = e_{1} × e_{2} = \frac{1}{cosh(v)}(cos(u), sin(u), − sinh(v))

is orthonormal and that e_{1}, e_{2} span the tangent space to Σ at each point.

b) Show that the dual forms of this frame are

ω^{1} = cosh(v) du, ω^{2} = cosh(v) dv.

c) Compute de_{3} and show that

ω_{1}^{3} = −\frac{1}{cosh(v)} du, ω_{2}^{3} = \frac{1}{cosh(v)} dv.

Use this to compute the matrix [h_{ij}] and show that the mean curvature of Σ is H ≡ 0.

2. Repeat the computation of Exercise 1 for an arbitrary surface of revolution parametrized by

x(u, v) = (f(v) cos(u), f(v) sin(u), v)

(you may want to use a computer algebra package such as Maple to assist with part c) and show that the surface is minimal if and only if f satisfies the differential equation

ff′′ = (f′)^{2} + 1.

Show that the only solutions of this equation are

f(v) = \frac{1}{a} cosh(av + b)

where a, b are constants. Conclude that catenoids are the only non-planar minimal surfaces of revolution.

3. The helicoid is the ruled surface Σ ⊂ 𝔼^{3} parametrized by

x(u, v) = (v cos(u), v sin(u), u).

a) Show that the frame

e_{1} = \frac{x_{u}}{|x_{u}|} = \frac{1}{\sqrt{v^{2} + 1}}(−v sin(u), v cos(u), 1)
e_{2} = \frac{x_{v}}{|x_{v}|} = (cos(u), sin(u), 0)
e_{3} = e_{1} × e_{2} = \frac{1}{\sqrt{v^{2} + 1}}(− sin(u), cos(u), −v)

is orthonormal and that e_{1}, e_{2} span the tangent space to Σ at each point.

b) Show that the dual forms of this frame are

ω^{1} = \sqrt{v^{2} + 1} du, ω^{2} = dv.

c) Compute de_{3} and show that

ω_{1}^{3} = \frac{1}{v^{2} + 1} dv, ω_{2}^{3} = \frac{1}{\sqrt{v^{2} + 1}} du.

Use this to compute the matrix [h_{ij}] and show that the mean curvature of Σ is H ≡ 0.

4. a) Show that for any values of a, b, c with ac−b^{2} > 0 the elliptic paraboloid z = a x^{2} + b xy + c y^{2} is affinely equivalent to the paraboloid Σ ⊂ 𝔸^{3} given by z = \frac{1}{2}(x^{2} + y^{2}). b) Show that the affine frame

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

is adapted to Σ. Compute its dual and connection forms and show that they satisfy the conditions

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

c) Show that the affine mean curvature of Σ is identically zero. Conclude that any elliptic paraboloid is affine minimal.

5. (This exercise should be done with the aid of a computer algebra package such as Maple.) Suppose that a surface Σ ⊂ 𝔸^{3} is described by a graph z = f(x, y). Consider the affine frame

\underline{e}_{1} = (1, 0, f_{x})
\underline{e}_{2} = (0, 1, f_{y})
\underline{e}_{3} = (0, 0, 1).

a) Show that the dual forms of this frame are

\underline{ω}^{1} = dx,  \underline{ω}^{2} = dy,  \underline{ω}^{3} = 0

and that the only nonzero connection forms are

\underline{ω}_{1}^{3} = f_{xx} dx + f_{xy} dy
\underline{ω}_{2}^{3} = f_{xy} dx + f_{yy} dy.

Thus we have

[\begin{matrix}\underline{h}_{11} & \underline{h}_{12} \\ \underline{h}_{12} & \underline{h}_{22} \end{matrix}] = [\begin{matrix}f_{xx} & f_{xy} \\ f_{xy} & f_{yy} \end{matrix}] .

Assume that f_{xx}f_{yy} −f_{xy}^{2} > 0, so that Σ is elliptic, and for simplicity assume that f_{xx} > 0. In order to decide whether Σ is affine minimal, we must adapt frames so that

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

Consider an affine change of frame

[e_{1} e_{2} e_{3}] = [\underline{e}_{1} \underline{e}_{2} \underline{e}_{3}] [\begin{matrix}B & \begin{matrix}r_{1} \\ r_{2} \end{matrix} \\ 0 0 & (det B)^{−1} \end{matrix}]

with B ∈ GL(2).

b) Show that if we take

B = [\begin{matrix}\frac{(f_{xx}f_{yy} − f_{xy}^{2})^{1/8}}{\sqrt{f_{xx}}} & −\frac{f_{xy}}{\sqrt{f_{xx}}(f_{xx}f_{yy} − f_{xy}^{2})^{3/8}} \\ 0 & \frac{\sqrt{f_{xx}}}{(f_{xx}f_{yy} − f_{xy}^{2})^{3/8}} \end{matrix} ]

then

[\begin{matrix}h_{11} & h_{12} \\ h_{12} & h_{22} \end{matrix}] = [\begin{matrix}1 & 0 \\ 0 & 1 \end{matrix}]

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

c) Show that under this change of basis (with r_{1}, r_{2} still arbitrary),

ω_{3}^{3} = [\frac{f_{xx}r_{1}}{(f_{xx}f_{yy} − f_{xy}^{2})^{1/4}} + \frac{f_{xy}r_{2}}{(f_{xx}f_{yy} − f_{xy}^{2})^{1/4}} + \frac{f_{xx}f_{xyy} − 2f_{xy}f_{xxy} + f_{yy}f_{xxx}}{4(f_{xx}f_{yy} − f_{xy}^{2})} ] dx

+ [\frac{f_{xy}r_{1}}{(f_{xx}f_{yy} − f_{xy}^{2})^{1/4}} + \frac{f_{yy}r_{2}}{(f_{xx}f_{yy} − f_{xy}^{2})^{1/4}} + \frac{f_{xx}f_{yyy} − 2f_{xy}f_{xyy} + f_{yy}f_{xxy}}{4(f_{xx}f_{yy} − f_{xy}^{2})} ] dy.

Conclude that by choosing

[\begin{matrix}r_{1} \\ r_{2} \end{matrix}] = − [\begin{matrix}f_{xx} & f_{xy} \\ f_{xy} & f_{yy} \end{matrix}]^{−1} [\frac{f_{xx}f_{xyy} − 2f_{xy}f_{xxy} + f_{yy}f_{xxx}}{4(f_{xx}f_{yy} − f_{xy}^{2})^{3/4}} ; \frac{f_{xx}f_{yyy} − 2f_{xy}f_{xyy} + f_{yy}f_{xxy}}{4(f_{xx}f_{yy} − f_{xy}^{2})^{3/4}} ]

we can arrange that ω_{3}^{3} = 0.

d) Show that under this change of frame we have

ω^{1} = \frac{\sqrt{f_{xx}}}{(f_{xx}f_{yy} − f_{xy}^{2})^{1/8}} dx + \frac{f_{xy}}{\sqrt{f_{xx}}(f_{xx}f_{yy} − f_{xy}^{2})^{1/8}} dy

ω^{2} = \frac{(f_{xx}f_{yy} − f_{xy}^{2})^{3/8}}{\sqrt{f_{xx}}} dy

Also compute ω_{3}^{1}, ω_{3}^{2} and find the functions ℓ_{ij} such that

[\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}] .

LIE GROUPS AND THE METHOD OF THE MOVING FRAME 55

Show that the affine mean curvature equation

L = ℓ_{11} + ℓ_{22} = 0

is a fourth-order differential equation for f. Pretty messy, huh?