LECTURE 5: SURFACES IN PROJECTIVE SPACE

1. Projective space

Definition: The n-dimensional projective space ℙ^{n} is the set of lines through the origin in the vector space ℝ^{n+1}.

ℙ^{n} may be thought of as the quotient space (ℝ^{n+1} \ {0})/ ∼ where ∼ represents the equivalence relation

(x^{0}, …, x^{n}) ∼ (λx^{0}, …, λx^{n}), λ ∈ ℝ^{∗}.

The equivalence class of the point (x^{0}, …, x^{n}) is denoted [x^{0}, …, x^{n}].

In order to describe ℙ^{n} as a homogenous space, we need to find its group of symmetries. Since the only structure on ℙ^{n} is that of lines through the origin in ℝ^{n+1}, we should begin by finding those symmetries of ℝ^{n+1} that preserve the set of lines through the origin. This is simply the matrix group GL(n + 1), so we might suppose that the group of symmetries of ℙ^{n} is also GL(n + 1).

However, there is a subtle point to consider here. While it is true that all elements of GL(n + 1) are symmetries of ℙ^{n}, some of them act trivially on ℙ^{n}. A matrix g ∈ GL(n + 1) fixes every line in ℝ^{n+1} if and only if g = λI for some λ ≠ 0. Thus the most natural choice for the symmetry group of ℙ^{n} is GL(n + 1)/ℝ^{∗}I. This group is isomorphic to SL(n + 1) if n is even and SL(n + 1)/{±I} if n is odd. In order to avoid the difficulties associated with working with a quotient group, we will take the symmetry group of ℙ^{n} to be SL(n + 1) in either case.

Now given a point [x] = [x^{0}, …, x^{n}] ∈ ℙ^{n}, we need to find its isotropy group H_{[x]}. First take [x_{0}] = [1, 0, …, 0]. It is straightforward to show that for g ∈ SL(n + 1), g · [x_{0}] = [x_{0}] if and only if

g = [\begin{matrix}(det A)^{−1} & r_{1} … r_{n} \\ 0 ⋮ 0 & A \end{matrix}]

where A ∈ GL(n). Thus

H_{[x_{0}]} = {[e_{0} … e_{n}] : e_{0} = (λ, 0, …, 0) for some λ ∈ ℝ^{∗}}.

Denote this group by H. For any other point [x] ∈ ℙ^{n}, H_{[x]} is conjugate to H, and ℙ^{n} is isomorphic to the set of left cosets of H in SL(n + 1). Thus ℙ^{n} may be thought of as the homogenous space ℙ^{n} ≅ SL(n + 1)/H.

A frame on ℙ^{n} is a set of vectors (e_{0}, …, e_{n}), e_{i} ∈ ℝ^{n+1}, with det[e_{0} … e_{n}] = 1. We can regard SL(n + 1) as the frame bundle of ℙ^{n}; it is a principal bundle with fibers isomorphic to H. We can define a projection map π : SL(n + 1) → ℙ^{n} by

π([e_{0} … e_{n}]) = [e_{0}].

The Maurer-Cartan forms {ω_{β}^{α}, 0 ≤ α, β ≤ n} on SL(n + 1) are defined by the equations

de_{α} = ∑_{β=0}^{n} e_{β} ω_{α}^{β}.

These forms satisfy the structure equations

dω_{β}^{α} = − ∑_{γ=0}^{n} ω_{γ}^{α} ∧ ω_{β}^{γ}

and the single relation

∑_{α=0}^{n} ω_{α}^{α} = 0.

The forms ω_{0}^{1}, …, ω_{0}^{n} are semi-basic for the projection π : SL(n + 1) → ℙ^{n}, while the remaining ω_{β}^{α}’s form a basis for the 1-forms on each fiber of π and so may be thought of as connection forms on the frame bundle.

2. Surfaces in ℙ^{3}

Consider a smooth, embedded surface [x] : Σ → ℙ^{3}, where Σ is an open set in ℝ^{2}. Because ℙ^{3} = ℝ^{4}/ ∼ is a quotient space, it is generally easier to work with the 3-dimensional submanifold Σ̃ ⊂ ℝ^{4} \ {0} defined by the property that x ∈ Σ̃ if and only if [x] ∈ Σ. Clearly Σ̃ consists of a 2-parameter family of lines through the origin of ℝ^{4} and so may be thought of as a cone over a 2-dimensional submanifold of ℝ^{4} \ {0}. We will use the geometry of the surface to construct an adapted frame {e_{0}(x), e_{1}(x), e_{2}(x), e_{3}(x)} ∈ SL(4) at each point x ∈ Σ̃.

For our first frame adaptation we will choose a frame at each point x ∈ Σ̃ such that e_{0} = x and T_{x}Σ̃ is spanned by the vectors e_{0}, e_{1}, e_{2}. These conditions are clearly invariant under the action of SL(4) on ℝ^{4}, and any other frame {ẽ_{0}, ẽ_{1}, ẽ_{2}, ẽ_{3}} has the form

[ẽ_{0} ẽ_{1} ẽ_{2} ẽ_{3}] = [e_{0} e_{1} e_{2} e_{3}] [\begin{matrix}1 & s_{1} s_{2} & s_{3} \\ {0 \atop 0} & B & \begin{matrix}s_{4} \\ s_{5} \end{matrix} \\ 0 & 0 0 & (det B)^{−1} \end{matrix}]

where B ∈ GL(2). For such a frame, dx must be a linear combination of e_{0}, e_{1}, e_{2}. Therefore the structure equation

dx = de_{0} = ∑_{β=0}^{3} e_{β} ω_{0}^{β}

implies that ω_{0}^{3} = 0, while the 1-forms ω_{0}^{0}, ω_{0}^{1}, ω_{0}^{2} form a basis for the 1-forms on Σ̃. Thus we have dω_{0}^{3} = 0, and so

0 = dω_{0}^{3} = −ω_{1}^{3} ∧ ω_{0}^{1} − ω_{2}^{3} ∧ ω_{0}^{2}.

By Cartan’s Lemma, there exist functions h_{11}, h_{12}, h_{22} such that

[\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_{0}^{1} \\ ω_{0}^{2} \end{matrix}] .

In order to make our next frame adaptation we will compute how the matrix [h_{ij}] varies if we choose a different frame. Suppose that {ẽ_{0}, ẽ_{1}, ẽ_{2}, ẽ_{3}} is defined as above. Computing the Maurer-Cartan form of the new frame shows that

[\begin{matrix} & ω̃_{0}^{1} \\ & ω̃_{0}^{2} \end{matrix}] = B^{−1} [\begin{matrix}ω_{0}^{1} \\ ω_{0}^{2} \end{matrix}] , [\begin{matrix} & ω̃_{1}^{3} \\ & ω̃_{2}^{3} \end{matrix}] = (det B)B^{t} [\begin{matrix}ω_{1}^{3} \\ ω_{2}^{3} \end{matrix}]

and therefore

[\begin{matrix}h̃_{11} & h̃_{12} \\ h̃_{12} & h̃_{22} \end{matrix}] = (det B)B^{t} [\begin{matrix}h_{11} & h_{12} \\ h_{12} & h_{22} \end{matrix}] B.

This transformation has the property that det[h̃_{ij}] = (det B)^{4} det[h_{ij}], so the sign of the determinant is fixed. We will assume that det[h_{ij}] > 0; in this case the surface is said to be elliptic. Then we can choose the matrix B so that [h_{ij}] is the identity matrix. This determines the frame up to a transformation of the form

[ẽ_{0} ẽ_{1} ẽ_{2} ẽ_{3}] = [e_{0} e_{1} e_{2} e_{3}] [\begin{matrix}1 & s_{1} s_{2} & s_{3} \\ {0 \atop 0} & B & \begin{matrix}s_{4} \\ s_{5} \end{matrix} \\ 0 & 0 0 & 1 \end{matrix}]

with B ∈ SO(2).

The quadratic form

I = ω_{1}^{3} ω_{0}^{1} + ω_{2}^{3} ω_{0}^{2} = (ω_{0}^{1})^{2} + (ω_{0}^{2})^{2}

is now well-defined on Σ̃, but it is not well-defined on Σ; it varies by a constant multiple as we move along the fibers of the projection Σ̃ → Σ. Thus I determines a conformal structure on Σ which is invariant under the action of SL(4).

The restricted Maurer-Cartan forms on our frame now have the property that ω_{1}^{3} = ω_{0}^{1}, ω_{2}^{3} = ω_{0}^{2}. Differentiating these equations yields

(2ω_{1}^{1} − ω_{0}^{0} − ω_{3}^{3}) ∧ ω_{0}^{1} + (ω_{2}^{1} + ω_{1}^{2}) ∧ ω_{0}^{2} = 0
(ω_{2}^{1} + ω_{1}^{2}) ∧ ω_{0}^{1} + (2ω_{2}^{2} − ω_{0}^{0} − ω_{3}^{3}) ∧ ω_{0}^{2} = 0.

By Cartan’s Lemma, there exist functions h_{111}, h_{112}, h_{122}, h_{222} such that

[\begin{matrix}2ω_{1}^{1} − ω_{0}^{0} − ω_{3}^{3} \\ ω_{2}^{1} + ω_{1}^{2} \\ 2ω_{2}^{2} − ω_{0}^{0} − ω_{3}^{3} \end{matrix}] = [\begin{matrix}h_{111} & h_{112} \\ h_{112} & h_{122} \\ h_{122} & h_{222} \end{matrix}] [\begin{matrix}ω_{0}^{1} \\ ω_{0}^{2} \end{matrix}] .

In order to make further adaptations we need to compute how the h_{ijk}’s vary under a change of frame. This computation gets rather complicated, but we can make it simpler by breaking it down into two steps. Any two adapted frames at this stage vary by a composition of transformations of the form

[ẽ_{0} ẽ_{1} ẽ_{2} ẽ_{3}] = [e_{0} e_{1} e_{2} e_{3}] [\begin{matrix}1 & 0 0 & 0 \\ {0 \atop 0} & B & {0 \atop 0} \\ 0 & 0 0 & 1 \end{matrix}] (2.1)

with B ∈ SO(2) and

[ẽ_{0} ẽ_{1} ẽ_{2} ẽ_{3}] = [e_{0} e_{1} e_{2} e_{3}] [\begin{matrix}1 & s_{1} s_{2} & s_{3} \\ {0 \atop 0} & I & \begin{matrix}s_{4} \\ s_{5} \end{matrix} \\ 0 & 0 0 & 1 \end{matrix}] . (2.2)

First consider a change of frame of the form (2.2). It is left as an exercise that under such a change of frame,

h̃_{111} = h_{111} + 3(s_{1} − s_{4})
h̃_{112} = h_{112} + (s_{2} − s_{5})
h̃_{122} = h_{122} + (s_{1} − s_{4})
h̃_{222} = h_{222} + 3(s_{2} − s_{5}).

Thus we can choose the s_{i} so that h_{122} = −h_{111}, h_{112} = −h_{222}. For such a frame we have ω_{0}^{0} + ω_{3}^{3} = ω_{1}^{1} + ω_{2}^{2} = 0. (Exercise: why?) This condition is preserved under transformations of the form (2.1) and transformations of the form (2.2) with s_{4} = s_{1}, s_{5} = s_{2}. Transformations of the latter form fix all the h_{ijk}’s, while under a transformation of the form (2.1) we have

[h̃_{111} ; h̃_{222}] = B^{3} [\begin{matrix}h_{111} \\ h_{222} \end{matrix}]

so the quantity h_{111}^{2} + h_{222}^{2} is invariant.

3. The case h_{ijk} = 0

Now suppose that h_{111}^{2} + h_{222}^{2} ≡ 0. Then we have

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

Differentiating these equations yields

(ω_{1}^{0} − ω_{3}^{1}) ∧ ω_{0}^{1} = 0
(ω_{2}^{0} − ω_{3}^{2}) ∧ ω_{0}^{2} = 0
(ω_{2}^{0} − ω_{3}^{2}) ∧ ω_{0}^{1} + (ω_{1}^{0} − ω_{3}^{1}) ∧ ω_{0}^{2} = 0
−(ω_{1}^{0} − ω_{3}^{1}) ∧ ω_{0}^{1} − (ω_{2}^{0} − ω_{3}^{2}) ∧ ω_{0}^{2} = 0.

The fourth equation is obviously a consequence of the first two. Applying Cartan’s lemma to the first three of these equations shows that there exists a function λ such that

ω_{1}^{0} − ω_{3}^{1} = λ ω_{0}^{1}
ω_{2}^{0} − ω_{3}^{2} = λ ω_{0}^{2}.

Now consider a change of frame of the form

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

It is left as an exercise that under this change of frame,

λ̃ = λ − 2s_{3}.

Thus we can choose a frame with λ = 0, and for such a frame we have

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

Differentiating these equations yields

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

By Cartan’s lemma we have ω_{3}^{0} = 0. Finally, differentiating this equation yields an identity.

At this point the Maurer-Cartan form for the reduced frame bundle is

ω = [\begin{matrix}ω_{0}^{0} & ω_{1}^{0} & ω_{2}^{0} & 0 \\ ω_{0}^{1} & 0 & ω_{2}^{1} & ω_{1}^{0} \\ ω_{0}^{2} & −ω_{2}^{1} & 0 & ω_{2}^{0} \\ 0 & ω_{0}^{1} & ω_{0}^{2} & −ω_{0}^{0} \end{matrix}] .

We have not found a unique frame over each point of Σ̃, but since differentiating the structure equations yields no further relations, this is as far as the frame bundle can be reduced. What this means is that Σ̃ is itself a homogenous space G/H where G is the Lie group whose Lie algebra 𝔤 is the set of matrices with the symmetries of the Maurer-Cartan form above. All that remains is to identify this group G and to describe Σ̃ as a homogenous space G/H. Because Σ̃ is a homogenous space, perhaps it will not come as a surprise that Σ̃ is, up to a projective transformation, the cone over the sphere S^{2}. The details will be left to the exercises.

Exercises

1. Suppose that instead of being elliptic, Σ̃ has h_{ij} = 0. Prove that Σ is a plane in ℙ^{3}. (Hint: Σ is a plane if and only if Σ̃ is a hyperplane in ℝ^{4}. Show that the plane spanned by the vectors e_{0}, e_{1}, e_{2} is constant, and that therefore Σ̃ must be contained in this plane.)

2. Suppose that Σ ⊂ ℙ^{3} is elliptic and that we have adapted our frames so that ω_{1}^{3} = ω_{0}^{1}, ω_{2}^{3} = ω_{0}^{2}. Show that

a) The quadratic form I is well-defined on Σ̃.

b) I_{λx} = λ^{2}I_{x}, where I_{x} denotes the quadratic form I based at the point x ∈ Σ̃. (Hint: moving from x to λx corresponds to making a change of frame with ẽ_{0} = λe_{0}. Since I is well-defined, the change of frame can otherwise be made as simple as possible for ease of computation. Set

[ẽ_{0} ẽ_{1} ẽ_{2} ẽ_{3}] = [e_{0} e_{1} e_{2} e_{3}] [\begin{matrix}λ & 0 & 0 & 0 \\ 0 & μ & 0 & 0 \\ 0 & 0 & μ & 0 \\ 0 & 0 & 0 & \frac{1}{λμ^{2}} \end{matrix} ]

and show that in order to preserve the condition that ω_{i}^{3} = ω_{0}^{i}, i = 1, 2, we must have μ = ±1. Now show that under this change of frame, ω̃_{0}^{i} = λω_{0}^{i}, i = 1, 2 so that I_{λx} = λ^{2}I_{x}.)

3. Fill in the details of the frame computations: a) Suppose that the surface Σ ⊂ ℙ^{2} is elliptic and that we have restricted to frames for which ω_{1}^{3} = ω_{0}^{1}, ω_{2}^{3} = ω_{0}^{2}. Show that differentiating these equations yields

(2ω_{1}^{1} − ω_{0}^{0} − ω_{3}^{3}) ∧ ω_{0}^{1} + (ω_{2}^{1} + ω_{1}^{2}) ∧ ω_{0}^{2} = 0
(ω_{2}^{1} + ω_{1}^{2}) ∧ ω_{0}^{1} + (2ω_{2}^{2} − ω_{0}^{0} − ω_{3}^{3}) ∧ ω_{0}^{2} = 0

and that Cartan’s Lemma implies that there exist functions h_{111}, h_{112}, h_{122}, h_{222} such that

[\begin{matrix}2ω_{1}^{1} − ω_{0}^{0} − ω_{3}^{3} \\ ω_{2}^{1} + ω_{1}^{2} \\ 2ω_{2}^{2} − ω_{0}^{0} − ω_{3}^{3} \end{matrix}] = [\begin{matrix}h_{111} & h_{112} \\ h_{112} & h_{122} \\ h_{122} & h_{222} \end{matrix}] [\begin{matrix}ω_{0}^{1} \\ ω_{0}^{2} \end{matrix}] .

b) Show that under a change of frame of the form (2.2),

h̃_{111} = h_{111} + 3(s_{1} − s_{4})
h̃_{112} = h_{112} + (s_{2} − s_{5})
h̃_{122} = h_{122} + (s_{1} − s_{4})
h̃_{222} = h_{222} + 3(s_{2} − s_{5}).

When we restrict to those frames for which h_{122} = −h_{111}, h_{112} = −h_{222}, why do we have ω_{0}^{0} + ω_{3}^{3} = ω_{1}^{1} + ω_{2}^{2} = 0?

c) Suppose that the invariant h_{111}^{2} + h_{222}^{2} vanishes identically. Show that differentiating the equations

ω_{1}^{1} = ω_{2}^{2} = ω_{2}^{1} + ω_{1}^{2} = ω_{0}^{0} + ω_{3}^{3} = 0

yields

(ω_{1}^{0} − ω_{3}^{1}) ∧ ω_{0}^{1} = 0
(ω_{2}^{0} − ω_{3}^{2}) ∧ ω_{0}^{2} = 0
(ω_{2}^{0} − ω_{3}^{2}) ∧ ω_{0}^{1} + (ω_{1}^{0} − ω_{3}^{1}) ∧ ω_{0}^{2} = 0
−(ω_{1}^{0} − ω_{3}^{1}) ∧ ω_{0}^{1} − (ω_{2}^{0} − ω_{3}^{2}) ∧ ω_{0}^{2} = 0.

Use Cartan’s lemma to conclude that there exists a function λ such that

ω_{1}^{0} − ω_{3}^{1} = λ ω_{0}^{1}
ω_{2}^{0} − ω_{3}^{2} = λ ω_{0}^{2}.

d) Show that under a change of frame of the form

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

we have

λ̃ = λ − 2s_{3}

so we can choose a frame for which λ = 0.

e) Show that differentiating the equations ω_{3}^{1} = ω_{1}^{0}, ω_{3}^{2} = ω_{1}^{0} yields

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

Use Cartan’s lemma to conclude that ω_{3}^{0} = 0. Show that differentiating this equation yields no further relations among the ω_{β}^{α}’s.

4. In this exercise we will show that when h_{111}^{2} + h_{222}^{2} = 0, Σ is a sphere in ℙ^{2}.

Let Q be the matrix

Q = [\begin{matrix}0 & 0 & 0 & −1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ −1 & 0 & 0 & 0 \end{matrix}] .

Q represents the quadratic form

q = (x^{1})^{2} + (x^{2})^{2} − 2x^{0}x^{3}

which is a Lorentzian metric on ℝ^{4}. The Lie Group O(Q) is the group of matrices which preserves this metric; it is defined by

O(Q) = {A ∈ GL(4) : ^{t}AQA = Q}

and is isomorphic to the Lie group O(3, 1).

a) Differentiate the equation

tAQA = Q

to show that the Lie algebra 𝔬(Q) is defined by

𝔬(Q) = {a ∈ 𝔤𝔩(4) : ^{t}aQ + Qa = 0.

b) Show that 𝔬(Q) consists of the matrices of the form

[\begin{matrix}a_{0}^{0} & a_{1}^{0} & a_{2}^{0} & 0 \\ a_{0}^{1} & 0 & a_{2}^{1} & a_{1}^{0} \\ a_{0}^{2} & −a_{2}^{1} & 0 & a_{2}^{0} \\ 0 & a_{0}^{1} & a_{0}^{2} & −a_{0}^{0} \end{matrix}] .

Conclude that the reduced Maurer-Cartan form at the end of the lecture takes values in 𝔬(Q). c) Recall that for a given reduced frame

g(x) = [e_{0}(x) e_{1}(x) e_{2}(x) e_{3}(x)]

the Maurer-Cartan form is ω = g^{−1}dg. Show that any reduced frame has the form

g(x) = CA(x)

where C ∈ SL(4) is a constant matrix and A(x) ∈ O(Q). C may be thought of as a symmetry of ℙ^{3}, so the surface Σ̃ is equivalent to the surface C^{−1} · Σ̃. Thus we can assume that g(x) ∈ O(Q).

d) Define a projection map π : O(Q) → ℝ^{4} \ {0} by

π([e_{0} e_{1} e_{2} e_{3}]) = e_{0}.

Show that in the Lorentzian metric defined by Q, e_{0} is a null vector, i.e., ⟨e_{0}, e_{0}⟩ = 0. Since the set of null vectors in ℝ^{4} \{0} is 3-dimensional, Σ̃ must be an open subset of the hypersurface defined by the equation ⟨x, x⟩ = 0. This hypersurface is the cone over the unit sphere S^{2} of the hyperplane {x^{0} = 1} ⊂ ℝ^{4}; therefore Σ is an open subset of the sphere in ℙ^{3}.