LECTURE 6: PSEUDOSPHERICAL SURFACES AND BÄCKLUND’S THEOREM

1. Line congruences

Let G_{1}(𝔼^{3}) denote the Grassmanian of lines in 𝔼^{3}. A line congruence in 𝔼^{3} is an immersed surface L : U → G_{1}(𝔼^{3}), where U ⊂ ℝ^{2} is open. The points of the line L(u, v) are given by

L(u, v) = {y(u, v) + λw(u, v) : λ ∈ ℝ}

for some y(u, v), w(u, v) ∈ 𝔼^{3} with |w(u, v)| = 1.

A parametric curve u = u(t), v = v(t) in U defines a ruled surface

X(t, λ) = y(u(t), v(t)) + λw(u(t), v(t)) = y(t) + λw(t)

belonging to the congruence. The surface is called developable if

det [w(t) w′(t) y′(t)] = 0.

This is a quadratic equation for u′(t), v′(t). If it has distinct real roots, then the solutions of this equation define two distinct families of developable surfaces X. In the generic case each family consists of the tangent lines to a surface, and these two surfaces Σ, Σ̄ are called the focal surfaces of the congruence. The congruence gives a mapping f : Σ → Σ̄ with the property that the congruence consists of lines which are tangent to both Σ and Σ̄ and join x ∈ Σ to x̄ = f(x) ∈ Σ̄.

2. Bäcklund’s Theorem

Definition: let L be a line congruence in 𝔼^{3} with focal surfaces Σ, Σ̄, and let f : Σ → Σ̄ be the function defined above. The congruence is called pseudospherical if

1. The distance r = |x̄ − x| is a constant independent of x.

2. The angle α between the surface normals N(x), N(x̄) is a constant independent of x.

Bäcklund’s Theorem: Suppose that L is a pseudospherical line congruence in 𝔼^{3} with focal surfaces Σ, Σ̄. Then both Σ and Σ̄ have constant

negative Gauss curvature K = −\frac{sin^{2} α}{r^{2}}. (Such surfaces are called pseudospherical surfaces.)

This theorem can be proved using local coordinates on the surfaces Σ, Σ̄, but it is a computational mess. The proof can be greatly simplified by using the method of moving frames, because the frames can be adapted to the geometry of the problem in a way that local coordinates cannot. Whereas in previous lectures we have adapted our frames according to the geometry of a single surface, here we have to consider two surfaces and geometrical conditions relating them. We will use these considerations to choose frames on the surfaces Σ, Σ̄.

Proof: Let {e_{1}, e_{2}, e_{3}} be an orthonormal frame of T_{x}𝔼^{3} at x ∈ Σ such that e_{3} is the unit normal to Σ at x (and hence e_{1}, e_{2} span T_{x}Σ) and e_{1} is the unit vector in the direction of x̄ − x. We can then define an orthonormal frame {ē_{1}, ē_{2}, ē_{3}} of Tx̄𝔼^{3} by

ē_{1} = e_{1}
ē_{2} = (cos α)e_{2} + (sin α)e_{3}
ē_{3} = (− sin α)e_{2} + (cos α)e_{3}

Note that ē_{3} is the unit normal to Σ̄ at x̄.

A comment about the domain of definition of these frames may be in order. Since the line congruence gives a map f : Σ → Σ̄, we can think of the immersions x : Σ → 𝔼^{3} and x̄ = f ∘x : Σ → 𝔼^{3} as being defined on the same abstract surface Σ. Thus the pullback bundles x^{−1}(T 𝔼^{3}) and x̄^{−1}(T 𝔼^{3}) are naturally isomorphic as vector bundles over Σ, and this is the setting where it makes sense to say ē_{1} = e_{1}, etc. We will shortly have similar relations between the Maurer-Cartan forms restricted to Σ and Σ̄. These will make sense because all the forms in question are really pullbacks via x or x̄ and so are forms on the abstract surface Σ.

With frames chosen as above, the immersions x : Σ → 𝔼^{3} and x̄ : Σ̄ → 𝔼^{3} are related by the equation

x̄ = x + re_{1}.

Taking the exterior derivative of this equation yields

dx̄ = dx + r de_{1}
= e_{1} ω^{1} + e_{2} ω^{2} + r(e_{2} ω_{1}^{2} + e_{3} ω_{1}^{3})
= e_{1} ω^{1} + e_{2}(ω^{2} + r ω_{1}^{2}) + e_{3}(r ω_{1}^{3}).

On the other hand, we also have

dx̄ = ē_{1} ω̄^{1} + ē_{2} ω̄^{2}
= e_{1} ω̄^{1} + e_{2}(cos α ω̄^{2}) + e_{3}(sin α ω̄^{2}).

Comparing these equations yields

ω̄^{1} = ω^{1}
(2.1)  cos α ω̄^{2} = ω^{2} + r ω_{1}^{2}
sin α ω̄^{2} = r ω_{1}^{3}.

The last two of these equations imply that

ω^{2} + rω_{1}^{2} = r cot α ω_{1}^{3}. (2.2)

We will use the fact that the Gauss curvature K̄ of Σ̄ satisfies the equation

ω̄_{1}^{3} ∧ ω̄_{2}^{3} = K̄ ω̄^{1} ∧ ω̄^{2}

to compute K̄. Recall that for an adapted frame with e_{3} = 0 on a surface Σ ⊂ 𝔼^{3} we have

ω_{1}^{3} = h_{11} ω^{1} + h_{12} ω^{2}
ω_{2}^{3} = h_{12} ω^{1} + h_{22} ω^{2}

for some functions h_{11}, h_{12}, h_{22}. (We will denote the corresponding functions for Σ̄ by h̄_{11}, h̄_{12}, h̄_{22}.) Since ω̄^{1}, ω̄^{2} are linearly independent forms, the first and third equations of (2.1) imply that h_{12} ≠ 0. Using equation (2.2) we compute that

ω̄_{1}^{3} = ⟨dē_{1}, ē_{3}⟩

= ⟨de_{1}, (− sin α)e_{2} + (cos α)e_{3}⟩

= (cos α) ω_{1}^{3} − (sin α) ω_{1}^{2}

= \frac{sin α}{r}ω^{2}

ω̄_{2}^{3} = ⟨dē_{2}, ē_{3}⟩

= ⟨(cos α) de_{2} + (sin α) de_{3}, (− sin α) e_{2} + (cos α) e_{3}⟩

= (cos^{2} α) ω_{2}^{3} − (sin^{2} α) ω_{3}^{2}

= ω_{2}^{3}

Therefore

ω̄_{1}^{3} ∧ ω̄_{2}^{3} = \frac{sin α}{r}ω^{2} ∧ ω_{2}^{3}

= −\frac{sin α}{r}h_{12} ω^{1} ∧ ω^{2}.

But by the last equation in (2.1) we also have

ω̄_{1}^{3} ∧ ω̄_{2}^{3} = K̄ ω̄^{1} ∧ ω̄^{2}

= K̄ ω^{1} ∧ (\frac{r}{sin α}ω_{1}^{3})

= K̄ \frac{r}{sin α}h_{12} ω^{1} ∧ ω^{2}.

Since h_{12} ≠ 0, comparing coefficients yields

K̄ = −\frac{sin^{2} α}{r^{2}}.

An analogous argument shows that K = −\frac{sin^{2} α}{r^{2}} as well. □

The map f : Σ → Σ̄ given by the line congruence is called a Bäcklund transformation of the surface Σ.

3. Pseudospherical surfaces and the sine-Gordon equation

Let Σ be a pseudospherical surface, and for simplicity assume that its Gauss curvature is K = −1. Since the Gauss curvature of Σ is negative, Σ must have no umbilic points. Therefore every point x ∈ Σ has a neighborhood on which there exists a local coordinate chart whose coordinate curves are principal curves in Σ. In such a coordinate system (u^{1}, u^{2}) we can choose an orthonormal frame \{\underline{e}_{1}, \underline{e}_{2}\} with

\underline{e}_{1} = \frac{1}{a_{1}} \frac{∂}{∂u^{1}} \underline{e}_{2} = \frac{1}{a_{2}} \frac{∂}{∂u^{2}}

for some nonvanishing functions a_{1}, a_{2} on Σ. Then we have

\underline{ω}^{1} = a_{1} du^{1},  \underline{ω}^{2} = a_{2} du^{2},  \underline{ω}_{1}^{3} = κ_{1}a_{1} du^{1},  \underline{ω}_{2}^{3} = κ_{2}a_{2} du^{2}

where κ_{1}, κ_{2} are the principal curvatures of Σ. The first and second fundamental forms of Σ are

I = (\underline{ω}^{1})^{2} + (\underline{ω}^{2})^{2} = (a_{1})^{2} (du^{1})^{2} + (a_{2})^{2} (du^{2})^{2}
II = \underline{ω}_{1}^{3} \underline{ω}^{1} + \underline{ω}_{2}^{3} \underline{ω}^{2} = κ_{1}(a_{1})^{2} (du^{1})^{2} + κ_{2}(a_{2})^{2} (du^{2})^{2}.

The structure equations of the Maurer-Cartan forms imply that

\underline{ω}_{2}^{1} = \frac{1}{a_{2}} \frac{∂a_{1}}{∂u^{2}} du^{1} − \frac{1}{a_{1}} \frac{∂a_{2}}{∂u^{1}} du^{2}

and the Codazzi equations take the form

\frac{1}{κ_{1} − κ_{2}} \frac{∂κ_{1}}{∂u^{2}} = −\frac{∂(ln a_{1})}{∂u^{2}}
\frac{1}{κ_{2} − κ_{1}} \frac{∂κ_{2}}{∂u^{1}} = −\frac{∂(ln a_{2})}{∂u^{1}}.

Now since K = −1, we have κ_{1}κ_{2} = −1. Thus the Codazzi equations can be written as

\frac{κ_{1}}{κ_{1}(κ_{1} − κ_{2})} \frac{∂κ_{1}}{∂u^{2}} = −\frac{∂(ln a_{1})}{∂u^{2}}
\frac{κ_{2}}{κ_{2}(κ_{2} − κ_{1})} \frac{∂κ_{2}}{∂u^{1}} = −\frac{∂(ln a_{2})}{∂u^{1}},

or

\frac{1}{2} \frac{∂(ln(κ_{1}^{2} + 1))}{∂u^{2}} = −\frac{∂(ln a_{1})}{∂u^{2}}
\frac{1}{2} \frac{∂(ln(κ_{2}^{2} + 1))}{∂u^{1}} = −\frac{∂(ln a_{2})}{∂u^{1}}.

Therefore there must exist functions c_{1}(u^{1}), c_{2}(u^{2}) such that

κ_{i}^{2} + 1 = \frac{c_{i}(u^{i})}{a_{i}^{2}}, i = 1, 2.

Making a change of coordinates of the form ũ^{1} = ũ^{1}(u^{1}), ũ^{2} = ũ^{2}(u^{2}), we can assume that c_{i} ≡ 1. Then there exists a function ψ such that

κ_{1} = tan ψ,  κ_{2} = − cot ψ,  a_{1} = cos ψ,  a_{2} = sin ψ,

so the first and second fundamental forms of Σ are

I = cos^{2} ψ (du^{1})^{2} + sin^{2} ψ (du^{2})^{2}
II = sin ψ cos ψ ((du^{1})^{2} − (du^{2})^{2}).

From this we can compute that the angle between the asymptotic directions at any point is 2ψ. The connection form is

\underline{ω}_{2}^{1} = −\frac{∂ψ}{∂u^{2}} du^{1} − \frac{∂ψ}{∂u^{1}} du^{2}

and the Gauss equation is equivalent to

\frac{∂^{2}ψ}{∂(u^{1})^{2}} − \frac{∂^{2}ψ}{∂(u^{2})^{2}} = sin ψ cos ψ.

In other words, the angle ϕ = 2ψ between the asymptotic directions satisfies the sine-Gordon equation

\frac{∂^{2}ϕ}{∂(u^{1})^{2}} − \frac{∂^{2}ϕ}{∂(u^{2})^{2}} = sin ϕ. (3.1)

In fact, there is a one-to-one correspondence between local solutions ϕ of the sine-Gordon equation with 0 < ϕ < π and local surfaces of constant Gauss curvature K = −1 in 𝔼^{3} up to rigid motion.

Exercises

1. Consider the change of coordinates

x = \frac{1}{2}(u^{1} + u^{2}), y = \frac{1}{2}(u^{1} − u^{2})

where u^{1}, u^{2} are the coordinates for which

I = cos^{2} ψ (du^{1})^{2} + sin^{2} ψ (du^{2})^{2}
II = sin ψ cos ψ ((du^{1})^{2} − (du^{2})^{2})

on the surface Σ with K = −1.

a) Show that x, y are asymptotic coordinates on Σ (this is equivalent to the statement that II = f dx dy for some function f on Σ) and that the first and second fundamental forms on Σ are

I = dx^{2} + 2 cos(2ψ) dx dy + dy^{2}
II = 2 sin(2ψ) dx dy

b) Show that the Maurer-Cartan forms corresponding to the principal frame \{\underline{e}_{1}, \underline{e}_{2}, \underline{e}_{3}\} are

\underline{ω}^{1} = cos ψ(dx + dy)
\underline{ω}^{2} = sin ψ(dx − dy)
\underline{ω}_{1}^{3} = sin ψ(dx + dy)
\underline{ω}_{2}^{3} = − cos ψ(dx − dy)
\underline{ω}_{2}^{1} = \frac{∂ψ}{∂y} dy − \frac{∂ψ}{∂x} dx.

c) Show that in these coordinates, the sine-Gordon equation takes the form

\frac{∂^{2}ϕ}{∂x∂y} = sin ϕ.

2. Suppose that we have a Bäcklund transformation between two pseudospherical surfaces Σ, Σ̄ with K = −1. Let {e_{1}, e_{2}, e_{3}} be the frame adapted to the transformation as in the lecture, and let η denote the angle between e_{1} and \underline{e}_{1}. Then we have

[e_{1} e_{2}] = [\underline{e}_{1} \underline{e}_{2}] [\begin{matrix}cos η & − sin η \\ sin η & cos η \end{matrix}] .

a) Show that

[\begin{matrix}ω^{1} \\ ω^{2} \end{matrix}] = [\begin{matrix}cos η & sin η \\ − sin η & cos η \end{matrix}] [\begin{matrix}\underline{ω} & 1 \\ \underline{ω} & 2 \end{matrix}]

[\begin{matrix}ω_{1}^{3} \\ ω_{2}^{3} \end{matrix}] = [\begin{matrix}cos η & sin η \\ − sin η & cos η \end{matrix}] [\begin{matrix}\underline{ω} & 3 1 \\ \underline{ω} & 3 2 \end{matrix}]

ω_{2}^{1} = \underline{ω}_{2}^{1} − dη.

b) Show that the Bäcklund equation (2.2) is equivalent to the first-order system of partial differential equations

ψ_{x} + η_{x} = λ sin(ψ − η)
ψ_{y} − η_{y} = \frac{1}{λ} sin(ψ + η)

where λ = cot α − csc α is constant.

3. Suppose that ψ(x, y), η(x, y) are any two solutions of the PDE system

(3.2)  ψ_{x} + η_{x} = λ sin(ψ − η)
ψ_{y} − η_{y} = \frac{1}{λ} sin(ψ + η)

where λ ≠ 0 is constant.

a) Show that the functions 2ψ, 2η must both be solutions of the sine-Gordon equation

\frac{∂^{2}ϕ}{∂x∂y} = sin ϕ.

b) If 2ψ is any known solution of the sine-Gordon equation, then the system (3.2) is a compatible, overdetermined system for the unknown function η. Therefore it can be solved using only techniques of ordinary differential equations. The system (3.2) is called a Bäcklund transformation for the sine-Gordon solution. Suppose that ψ is the trivial solution ψ(x, y) ≡ 0. Show that the corresponding solutions η are

η(x, y) = 2 tan^{−1}(Ce^{−(λx+ \frac{1}{λ} y)})

where C ≠ 0 is constant. (Hint: you may find the trig identity csc η+cot η = cot(\frac{1}{2}η) useful.) The functions

2η = 4 tan^{−1}(Ce^{−(λx+ \frac{1}{λ} y)})

are called the 1-soliton solutions of the sine-Gordon equation. Iterating this procedure gives the 2-solitons, etc. The trivial solution ψ = 0 corresponds to the degenerate “surface” consisting of a straight line in 𝔼^{3}, while the family of surfaces corresponding to the 1-solitons includes the classical pseudosphere.