The geometry of lightlike surfaces in
Minkowski space
New Directions in Exterior Differential Systems
A conference in honor of Robert Bryant’s 60th birthday
July 18, 2013
Jeanne Clelland
University of Colorado, Boulder
(Joint with Brian Carlsen)
Definition 1: Three-dimensional Minkowski space is the
manifold R2,1 defined by


 0
x




 1 0 1 2
2,1
R = x = x  : x , x , x ∈ R ,




x2
with the indefinite inner product h·, ·i defined on each tangent
space Tx R2,1 by
hv, wi = −v 0 w0 + v 1 w1 + v 2 w2 .
Definition 2: A nonzero vector v ∈ Tx R2,1 is called:
• spacelike if hv, vi > 0;
• timelike if hv, vi < 0;
• lightlike or null if hv, vi = 0.
Definition 2: A nonzero vector v ∈ Tx R2,1 is called:
• spacelike if hv, vi > 0;
• timelike if hv, vi < 0;
• lightlike or null if hv, vi = 0.
The set of all null vectors v ∈ Tx R2,1 forms a cone in the
tangent space Tx R2,1 , called the light cone or null cone at x.
Definition 2: A nonzero vector v ∈ Tx R2,1 is called:
• spacelike if hv, vi > 0;
• timelike if hv, vi < 0;
• lightlike or null if hv, vi = 0.
The set of all null vectors v ∈ Tx R2,1 forms a cone in the
tangent space Tx R2,1 , called the light cone or null cone at x.
Figure: Spacelike and timelike “spheres” and null cone
Definition 3: A regular surface Σ ⊂ R2,1 is called:
• spacelike if the restriction of h·, ·i to each tangent plane
Tx Σ is positive definite;
• timelike if the restriction of h·, ·i to each tangent plane Tx Σ
is indefinite;
• lightlike if the restriction of h·, ·i to each tangent plane Tx Σ
is degenerate.
Definition 3: A regular surface Σ ⊂ R2,1 is called:
• spacelike if the restriction of h·, ·i to each tangent plane
Tx Σ is positive definite;
• timelike if the restriction of h·, ·i to each tangent plane Tx Σ
is indefinite;
• lightlike if the restriction of h·, ·i to each tangent plane Tx Σ
is degenerate.
Figure: Spacelike, timelike, and lightlike planes
Goal for this talk: Use Cartan’s method of moving frames to
give a local characterization of the lightlike surfaces in R2,1 .
Goal for this talk: Use Cartan’s method of moving frames to
give a local characterization of the lightlike surfaces in R2,1 .
Examples: lightlike planes, null cones
Let Σ ⊂ R2,1 be a lightlike surface. The method of moving
frames starts by associating a collection of “adapted” frames
(e0 , e1 , e2 ) to each point x ∈ Σ.
Let Σ ⊂ R2,1 be a lightlike surface. The method of moving
frames starts by associating a collection of “adapted” frames
(e0 , e1 , e2 ) to each point x ∈ Σ.
Question: What should “adapted” mean?
Let Σ ⊂ R2,1 be a lightlike surface. The method of moving
frames starts by associating a collection of “adapted” frames
(e0 , e1 , e2 ) to each point x ∈ Σ.
Question: What should “adapted” mean?
For surfaces in Euclidean space R3 , we typically choose
orthonormal frames (e1 , e2 , e3 ), and “adapted” frames at a
point x ∈ Σ are those for which (e1 , e2 ) span Tx Σ and e3 is
normal to Tx Σ.
In R2,1 , the orthonormal frames (e0 , e1 , e2 ) are defined by the
conditions
he0 , e0 i = −1,
he1 , e1 i = he2 , e2 i = 1,
heα , eβ i = 0 if α 6= β;
i.e., e0 is a unit timelike vector, e1 , e2 are unit spacelike vectors,
and e0 , e1 , e2 are mutually orthogonal.
Orthonormal frames are easily adaptable to the geometry of
timelike or spacelike surfaces: a timelike surface has a spacelike
normal vector field, and a spacelike surface has a timelike
normal vector field.
Orthonormal frames are easily adaptable to the geometry of
timelike or spacelike surfaces: a timelike surface has a spacelike
normal vector field, and a spacelike surface has a timelike
normal vector field.
Figure: Normal vector fields on timelike and spacelike surfaces
But the normal vector field to a lightlike surface is a null vector
field, contained within the tangent plane at each point.
But the normal vector field to a lightlike surface is a null vector
field, contained within the tangent plane at each point.
Figure: Normal vector field on lightlike surface
But the normal vector field to a lightlike surface is a null vector
field, contained within the tangent plane at each point.
Figure: Normal vector field on lightlike surface
Upshot: orthonormal frames are not easily adaptable to the
geometry of lightlike surfaces.
But the normal vector field to a lightlike surface is a null vector
field, contained within the tangent plane at each point.
Figure: Normal vector field on lightlike surface
Upshot: orthonormal frames are not easily adaptable to the
geometry of lightlike surfaces.
So instead, we will choose frames as follows:
• Each tangent plane Tx Σ contains a unique null line.
Choose e0 parallel to this line.
• Choose e1 to be any vector in Tx Σ linearly independent
from e0 (and hence spacelike) with he1 , e1 i = 1.
• The plane orthogonal to e1 contains two linearly
independent null directions, one of which is parallel to e0 .
Choose e2 parallel to the other null direction, scaled so
that he0 , e2 i = 1.
• If necessary, reverse the sign of e1 so that (e0 , e1 , e2 ) is
positively oriented.
• Each tangent plane Tx Σ contains a unique null line.
Choose e0 parallel to this line.
• Choose e1 to be any vector in Tx Σ linearly independent
from e0 (and hence spacelike) with he1 , e1 i = 1.
• The plane orthogonal to e1 contains two linearly
independent null directions, one of which is parallel to e0 .
Choose e2 parallel to the other null direction, scaled so
that he0 , e2 i = 1.
• If necessary, reverse the sign of e1 so that (e0 , e1 , e2 ) is
positively oriented.
• Each tangent plane Tx Σ contains a unique null line.
Choose e0 parallel to this line.
• Choose e1 to be any vector in Tx Σ linearly independent
from e0 (and hence spacelike) with he1 , e1 i = 1.
• The plane orthogonal to e1 contains two linearly
independent null directions, one of which is parallel to e0 .
Choose e2 parallel to the other null direction, scaled so
that he0 , e2 i = 1.
• If necessary, reverse the sign of e1 so that (e0 , e1 , e2 ) is
positively oriented.
• Each tangent plane Tx Σ contains a unique null line.
Choose e0 parallel to this line.
• Choose e1 to be any vector in Tx Σ linearly independent
from e0 (and hence spacelike) with he1 , e1 i = 1.
• The plane orthogonal to e1 contains two linearly
independent null directions, one of which is parallel to e0 .
Choose e2 parallel to the other null direction, scaled so
that he0 , e2 i = 1.
• If necessary, reverse the sign of e1 so that (e0 , e1 , e2 ) is
positively oriented.
In other words, the vectors (e0 , e1 ) span the tangent plane Tx Σ
at each point x ∈ Σ, the frame (e0 , e1 , e2 ) is positively oriented,
and the frame vectors (e0 , e1 , e2 ) satisfy the inner product
relations:
he0 , e0 i = he0 , e1 i = he1 , e2 i = he2 , e2 i = 0,
he0 , e2 i = he1 , e1 i = 1.
We will call a frame satisfying these conditions 0-adapted.
(1)
The Maurer-Cartan forms (ω α , ωβα ) associated to a 0-adapted
frame field (e0 , e1 , e2 ) on Σ are the 1-forms on Σ defined by the
equations:
dx = eα ω α ,
deβ =
eα ωβα .
(2)
The Maurer-Cartan forms (ω α , ωβα ) associated to a 0-adapted
frame field (e0 , e1 , e2 ) on Σ are the 1-forms on Σ defined by the
equations:
dx = eα ω α ,
deβ =
(2)
eα ωβα .
The ω α are the dual forms (or solder forms), and the ωβα are the
connection forms.
The Maurer-Cartan forms (ω α , ωβα ) associated to a 0-adapted
frame field (e0 , e1 , e2 ) on Σ are the 1-forms on Σ defined by the
equations:
dx = eα ω α ,
deβ =
(2)
eα ωβα .
The ω α are the dual forms (or solder forms), and the ωβα are the
connection forms.
They satisfy the Maurer-Cartan structure equations:
dω α = −ωβα ∧ ω β ,
dωβα
=
−ωγα
∧
ωβγ .
(3)
Relations among the Maurer-Cartan forms, Part 1:
Relations among the Maurer-Cartan forms, Part 1:
Since the tangent plane Tx Σ at each point x ∈ Σ is spanned by
(e0 , e1 ), it follows from the equation
dx = e0 ω 0 + e1 ω 1 + e2 ω 2
that ω 2 = 0, and that ω 0 , ω 1 are linearly independent 1-forms
which span the cotangent space Tx∗ Σ at each point x ∈ Σ.
Differentiating the inner product relations (1) shows that the
connection forms satisfy the following relations:
ω02 = ω11 = ω20 = 0,
ω12 = −ω01 ,
ω22 = −ω00 ,
(4)
ω21 = −ω10 .
Differentiating the inner product relations (1) shows that the
connection forms satisfy the following relations:
ω02 = ω11 = ω20 = 0,
ω12 = −ω01 ,
ω22 = −ω00 ,
(4)
ω21 = −ω10 .
So the matrix Ω = [ωβα ] of connection forms may be written as
 0
  0

ω0 ω10 ω20
ω0 ω10
0

 

0
−ω10  .
Ω = ω01 ω11 ω21  = ω01
ω02 ω12 ω22
0 −ω01 −ω00
How many 0-adapted frames are there?
How many 0-adapted frames are there?
Any two 0-adapted frames based at a point x ∈ Σ must differ
by a transformation of the form


λ2
µ λ − 2µ


λ 
ẽ0 ẽ1 ẽ2 = e0 e1 e2 
(5)
0 1 − µ  ,
1
0 0
µ
with µ 6= 0 and λ ∈ R. Furthermore, if (e0 , e1 , e2 ) is any
0-adapted frame based at a point x ∈ Σ, then any frame
(ẽ0 , ẽ1 , ẽ2 ) at x given by (5) is also 0-adapted.
How many 0-adapted frames are there?
Any two 0-adapted frames based at a point x ∈ Σ must differ
by a transformation of the form


λ2
µ λ − 2µ


λ 
ẽ0 ẽ1 ẽ2 = e0 e1 e2 
(5)
0 1 − µ  ,
1
0 0
µ
with µ 6= 0 and λ ∈ R. Furthermore, if (e0 , e1 , e2 ) is any
0-adapted frame based at a point x ∈ Σ, then any frame
(ẽ0 , ẽ1 , ẽ2 ) at x given by (5) is also 0-adapted.
The 0-adapted frame fields on Σ are the local sections of a
principal G0 -bundle π : B0 → Σ, where G0 ⊂ GL(3) is the
subgroup of all matrices of the form (5).
Under a transformation of the form (5), The Maurer-Cartan
forms (ω̃ α , ω̃βα ) associated to the frame (ẽ0 , ẽ1 , ẽ2 ) are related to
those of the original frame by the equations:
1 0 λ 1
ω − ω ,
µ
µ
ω̃ 1 = ω 1 ,
ω̃ 0 =
ω̃00 = ω00 − λω01 +
1
dµ,
µ
ω̃01 = µω01 ,
1 0
ω̃10 =
ω1 + λω00 − 12 λ2 ω01 + dλ .
µ
(6)
General strategy:
General strategy:
• Find relations among the Maurer-Cartan forms associated
to a 0-adapted frame field on Σ. Typically these relations
contain functional coefficients.
General strategy:
• Find relations among the Maurer-Cartan forms associated
to a 0-adapted frame field on Σ. Typically these relations
contain functional coefficients.
• Determine how these functional coefficients transform if we
change from one 0-adapted frame field to another.
General strategy:
• Find relations among the Maurer-Cartan forms associated
to a 0-adapted frame field on Σ. Typically these relations
contain functional coefficients.
• Determine how these functional coefficients transform if we
change from one 0-adapted frame field to another.
• If possible, restrict to the subset of 0-adapted frames for
which these coefficients have some particularly simple form.
Call this subset of frames “1-adapted.”
General strategy:
• Find relations among the Maurer-Cartan forms associated
to a 0-adapted frame field on Σ. Typically these relations
contain functional coefficients.
• Determine how these functional coefficients transform if we
change from one 0-adapted frame field to another.
• If possible, restrict to the subset of 0-adapted frames for
which these coefficients have some particularly simple form.
Call this subset of frames “1-adapted.”
• Repeat this process until (if we’re lucky!) we have
determined a unique adapted frame at each point satisfying
the desired conditions.
Relations among the Maurer-Cartan forms, Part 2:
Relations among the Maurer-Cartan forms, Part 2:
When in doubt, differentiate!
Relations among the Maurer-Cartan forms, Part 2:
When in doubt, differentiate!
Differentiating the equation ω 2 = 0 yields
0 = dω 2 = −(ω02 ∧ ω 0 − ω12 ∧ ω 1 ),
and applying the relations (4) (in particular, ω02 = 0, ω12 = −ω01 )
yields
ω01 ∧ ω 1 = 0.
Relations among the Maurer-Cartan forms, Part 2:
When in doubt, differentiate!
Differentiating the equation ω 2 = 0 yields
0 = dω 2 = −(ω02 ∧ ω 0 − ω12 ∧ ω 1 ),
and applying the relations (4) (in particular, ω02 = 0, ω12 = −ω01 )
yields
ω01 ∧ ω 1 = 0.
By Cartan’s lemma, we have
ω01 = a1 ω 1
for some function a1 on Σ.
Now, if we change from one 0-adapted frame to another, how
does the function a1 transform?
Now, if we change from one 0-adapted frame to another, how
does the function a1 transform?
From the transformation rules (6) for the Maurer-Cartan forms,
we have
ω̃ 1 = ω 1 ,
ω̃01 = µω01 .
Now, if we change from one 0-adapted frame to another, how
does the function a1 transform?
From the transformation rules (6) for the Maurer-Cartan forms,
we have
ω̃ 1 = ω 1 ,
ω̃01 = µω01 .
It follows that
ã1 = µa1 .
Now, if we change from one 0-adapted frame to another, how
does the function a1 transform?
From the transformation rules (6) for the Maurer-Cartan forms,
we have
ω̃ 1 = ω 1 ,
ω̃01 = µω01 .
It follows that
ã1 = µa1 .
So a1 is a relative invariant: at each point x ∈ Σ, a1 is either
equal to zero for every 0-adapted frame based at x or nonzero
for every 0-adapted frame based at x.
Now, if we change from one 0-adapted frame to another, how
does the function a1 transform?
From the transformation rules (6) for the Maurer-Cartan forms,
we have
ω̃ 1 = ω 1 ,
ω̃01 = µω01 .
It follows that
ã1 = µa1 .
So a1 is a relative invariant: at each point x ∈ Σ, a1 is either
equal to zero for every 0-adapted frame based at x or nonzero
for every 0-adapted frame based at x.
We will assume that Σ is of constant type with respect to a1 ;
i.e., that a1 is either identically zero on Σ or never equal to zero
at any point of Σ.
First we consider the case where a1 ≡ 0.
First we consider the case where a1 ≡ 0.
Proposition 1: Let Σ be a connected, regular lightlike surface
in R2,1 , and suppose that a1 ≡ 0 for every 0-adapted frame on
Σ. Then Σ is contained in a lightlike plane in R2,1 .
First we consider the case where a1 ≡ 0.
Proposition 1: Let Σ be a connected, regular lightlike surface
in R2,1 , and suppose that a1 ≡ 0 for every 0-adapted frame on
Σ. Then Σ is contained in a lightlike plane in R2,1 .
Proof. If a1 ≡ 0 on Σ, then we have ω01 = 0, and therefore
de0 = e0 ω00 .
In particular, the line spanned by e0 is constant on Σ.
Choose a point x0 ∈ Σ, and consider the function
f (x) = hx − x0 , e0 i
on Σ.
Choose a point x0 ∈ Σ, and consider the function
f (x) = hx − x0 , e0 i
on Σ.
Differentiate:
df = hdx, e0 i + hx − x0 , de0 i
Choose a point x0 ∈ Σ, and consider the function
f (x) = hx − x0 , e0 i
on Σ.
Differentiate:
df = hdx, e0 i + hx − x0 , de0 i
= he0 ω 0 + e1 ω 1 , e0 i + hx − x0 , e0 ω00 i
Choose a point x0 ∈ Σ, and consider the function
f (x) = hx − x0 , e0 i
on Σ.
Differentiate:
df = hdx, e0 i + hx − x0 , de0 i
= he0 ω 0 + e1 ω 1 , e0 i + hx − x0 , e0 ω00 i
= f ω00 .
Choose a point x0 ∈ Σ, and consider the function
f (x) = hx − x0 , e0 i
on Σ.
Differentiate:
df = hdx, e0 i + hx − x0 , de0 i
= he0 ω 0 + e1 ω 1 , e0 i + hx − x0 , e0 ω00 i
= f ω00 .
Since f (x0 ) = 0, the existence/uniqueness theorem for ordinary
differential equations guarantees that the only solution of this
equation on Σ is f ≡ 0.
Choose a point x0 ∈ Σ, and consider the function
f (x) = hx − x0 , e0 i
on Σ.
Differentiate:
df = hdx, e0 i + hx − x0 , de0 i
= he0 ω 0 + e1 ω 1 , e0 i + hx − x0 , e0 ω00 i
= f ω00 .
Since f (x0 ) = 0, the existence/uniqueness theorem for ordinary
differential equations guarantees that the only solution of this
equation on Σ is f ≡ 0.
Therefore, Σ is contained in the lightlike plane passing through
the point x0 and orthogonal to the (constant) null direction e0 .
2
Now suppose that a1 6= 0 at every point of Σ.
Now suppose that a1 6= 0 at every point of Σ.
According to the transformation rule
ã1 = µa1 ,
we can adapt frames to arrange that a1 ≡ 1. We will call such a
frame 1-adapted.
Now suppose that a1 6= 0 at every point of Σ.
According to the transformation rule
ã1 = µa1 ,
we can adapt frames to arrange that a1 ≡ 1. We will call such a
frame 1-adapted.
Any two 1-adapted frames differ by a transformation of the
form (5) with µ = 1; i.e.,


1 λ − 12 λ2


ẽ0 ẽ1 ẽ2 = e0 e1 e2 0 1 −λ  .
(7)
0 0
1
Now suppose that a1 6= 0 at every point of Σ.
According to the transformation rule
ã1 = µa1 ,
we can adapt frames to arrange that a1 ≡ 1. We will call such a
frame 1-adapted.
Any two 1-adapted frames differ by a transformation of the
form (5) with µ = 1; i.e.,


1 λ − 12 λ2


ẽ0 ẽ1 ẽ2 = e0 e1 e2 0 1 −λ  .
(7)
0 0
1
Moreover, the Maurer-Cartan forms associated to any
1-adapted frame field on Σ satisfy the condition
ω01 = ω 1 .
Differentiate again!
Differentiate again!
0 = d(ω01 − ω 1 )
Differentiate again!
0 = d(ω01 − ω 1 )
= −(ω01 ∧ ω00 + ω11 ∧ ω01 + ω21 ∧ ω02 ) + (ω01 ∧ ω 0 + ω11 ∧ ω 1 )
Differentiate again!
0 = d(ω01 − ω 1 )
= −(ω01 ∧ ω00 + ω11 ∧ ω01 + ω21 ∧ ω02 ) + (ω01 ∧ ω 0 + ω11 ∧ ω 1 )
= (ω00 − ω 0 ) ∧ ω 1 .
Differentiate again!
0 = d(ω01 − ω 1 )
= −(ω01 ∧ ω00 + ω11 ∧ ω01 + ω21 ∧ ω02 ) + (ω01 ∧ ω 0 + ω11 ∧ ω 1 )
= (ω00 − ω 0 ) ∧ ω 1 .
By Cartan’s lemma, we have
ω00 = ω 0 + a2 ω 1
for some function a2 on Σ.
Now, if we change from one 1-adapted frame to another, how
does the function a2 transform?
Now, if we change from one 1-adapted frame to another, how
does the function a2 transform?
Substituting µ = 1, ω01 = ω 1 into the transformation rules (6)
for the Maurer-Cartan forms yields
ω̃ 0 = ω 0 − λω 1 ,
ω̃ 1 = ω 1 ,
ω̃00 = ω00 − λω 1 .
Now, if we change from one 1-adapted frame to another, how
does the function a2 transform?
Substituting µ = 1, ω01 = ω 1 into the transformation rules (6)
for the Maurer-Cartan forms yields
ω̃ 0 = ω 0 − λω 1 ,
ω̃ 1 = ω 1 ,
It follows that
ã2 = a2 .
ω̃00 = ω00 − λω 1 .
Now, if we change from one 1-adapted frame to another, how
does the function a2 transform?
Substituting µ = 1, ω01 = ω 1 into the transformation rules (6)
for the Maurer-Cartan forms yields
ω̃ 0 = ω 0 − λω 1 ,
ω̃ 1 = ω 1 ,
It follows that
ã2 = a2 .
So a2 is an invariant function on Σ.
ω̃00 = ω00 − λω 1 .
Since the value of a2 is the same for all 1-adapted frames based
at a point x ∈ Σ, we need to...
Since the value of a2 is the same for all 1-adapted frames based
at a point x ∈ Σ, we need to...
Differentiate AGAIN!
Since the value of a2 is the same for all 1-adapted frames based
at a point x ∈ Σ, we need to...
Differentiate AGAIN!
This time, differentiate the equation ω00 = ω 0 + a2 ω 1 to obtain:
(da2 + 2a2 ω 0 ) ∧ ω 1 = 0.
Since the value of a2 is the same for all 1-adapted frames based
at a point x ∈ Σ, we need to...
Differentiate AGAIN!
This time, differentiate the equation ω00 = ω 0 + a2 ω 1 to obtain:
(da2 + 2a2 ω 0 ) ∧ ω 1 = 0.
By Cartan’s lemma, we have
da2 = −2a2 ω 0 + a3 ω 1
for some function a3 on Σ.
Now, if we change from one 1-adapted frame to another, how
does the function a3 transform?
Now, if we change from one 1-adapted frame to another, how
does the function a3 transform?
The transformation rules (6) for the Maurer-Cartan forms yield
ω̃ 0 = ω 0 − λω 1 ,
ω̃ 1 = ω 1 .
Now, if we change from one 1-adapted frame to another, how
does the function a3 transform?
The transformation rules (6) for the Maurer-Cartan forms yield
ω̃ 0 = ω 0 − λω 1 ,
ω̃ 1 = ω 1 .
Since ã2 = a2 , we have dã2 = da2 , and so
Now, if we change from one 1-adapted frame to another, how
does the function a3 transform?
The transformation rules (6) for the Maurer-Cartan forms yield
ω̃ 0 = ω 0 − λω 1 ,
ω̃ 1 = ω 1 .
Since ã2 = a2 , we have dã2 = da2 , and so
−2ã2 ω̃ 0 + ã3 ω̃ 1 = −2a2 ω 0 + a3 ω 1
Now, if we change from one 1-adapted frame to another, how
does the function a3 transform?
The transformation rules (6) for the Maurer-Cartan forms yield
ω̃ 0 = ω 0 − λω 1 ,
ω̃ 1 = ω 1 .
Since ã2 = a2 , we have dã2 = da2 , and so
−2ã2 ω̃ 0 + ã3 ω̃ 1 = −2a2 ω 0 + a3 ω 1
⇒ −2a2 ω 0 + (ã3 + 2a2 λ)ω 1 = −2a2 ω 0 + a3 ω 1 .
Now, if we change from one 1-adapted frame to another, how
does the function a3 transform?
The transformation rules (6) for the Maurer-Cartan forms yield
ω̃ 0 = ω 0 − λω 1 ,
ω̃ 1 = ω 1 .
Since ã2 = a2 , we have dã2 = da2 , and so
−2ã2 ω̃ 0 + ã3 ω̃ 1 = −2a2 ω 0 + a3 ω 1
⇒ −2a2 ω 0 + (ã3 + 2a2 λ)ω 1 = −2a2 ω 0 + a3 ω 1 .
Therefore,
ã3 = a3 − 2a2 λ.
So the ability to normalize a3 depends on the value of a2 :
• if a2 =
6 0, then we can adapt frames to arrange that a3 = 0;
• if a2 = 0, then a3 is invariant.
So the ability to normalize a3 depends on the value of a2 :
• if a2 =
6 0, then we can adapt frames to arrange that a3 = 0;
• if a2 = 0, then a3 is invariant.
This leads to another constant type assumption: we will assume
that a2 is either identically zero on Σ or never equal to zero at
any point of Σ.
First we consider the case where a2 ≡ 0.
First we consider the case where a2 ≡ 0.
Proposition 2: Let Σ be a connected, regular lightlike surface
in R2,1 with a1 6= 0, and suppose that a2 ≡ 0 for every 1-adapted
frame on Σ. Then Σ is contained in a null cone in R2,1 .
First we consider the case where a2 ≡ 0.
Proposition 2: Let Σ be a connected, regular lightlike surface
in R2,1 with a1 6= 0, and suppose that a2 ≡ 0 for every 1-adapted
frame on Σ. Then Σ is contained in a null cone in R2,1 .
Proof. If a2 ≡ 0 on Σ, then we have ω00 = ω 0 , and so
de0 = e0 ω00 + e1 ω01
First we consider the case where a2 ≡ 0.
Proposition 2: Let Σ be a connected, regular lightlike surface
in R2,1 with a1 6= 0, and suppose that a2 ≡ 0 for every 1-adapted
frame on Σ. Then Σ is contained in a null cone in R2,1 .
Proof. If a2 ≡ 0 on Σ, then we have ω00 = ω 0 , and so
de0 = e0 ω00 + e1 ω01
= e0 ω 0 + e1 ω 1
First we consider the case where a2 ≡ 0.
Proposition 2: Let Σ be a connected, regular lightlike surface
in R2,1 with a1 6= 0, and suppose that a2 ≡ 0 for every 1-adapted
frame on Σ. Then Σ is contained in a null cone in R2,1 .
Proof. If a2 ≡ 0 on Σ, then we have ω00 = ω 0 , and so
de0 = e0 ω00 + e1 ω01
= e0 ω 0 + e1 ω 1
= dx.
Therefore,
d(x − e0 ) = 0,
and so there exists a point p ∈ R2,1 such that
x − e0 = p
for every point x ∈ Σ.
Therefore,
d(x − e0 ) = 0,
and so there exists a point p ∈ R2,1 such that
x − e0 = p
for every point x ∈ Σ.
But then we have
x − p = e0 ,
from which it follows that x − p is a null vector for every x ∈ Σ.
Therefore,
d(x − e0 ) = 0,
and so there exists a point p ∈ R2,1 such that
x − e0 = p
for every point x ∈ Σ.
But then we have
x − p = e0 ,
from which it follows that x − p is a null vector for every x ∈ Σ.
Thus Σ is contained in the lightlike cone based at p, defined by
the equation
hx − p, x − pi = 0.
2
Now suppose that a2 6= 0 at every point of Σ.
Now suppose that a2 6= 0 at every point of Σ.
According to the transformation rule
ã3 = a3 − 2a2 λ,
we can adapt frames to arrange that a3 ≡ 0. We will call such a
frame 2-adapted.
Now suppose that a2 6= 0 at every point of Σ.
According to the transformation rule
ã3 = a3 − 2a2 λ,
we can adapt frames to arrange that a3 ≡ 0. We will call such a
frame 2-adapted.
Any two 2-adapted frames differ by a transformation of the
form (7) with λ = 0 — which means that, in fact, there is a
unique 2-adapted frame at each point of Σ.
Now suppose that a2 6= 0 at every point of Σ.
According to the transformation rule
ã3 = a3 − 2a2 λ,
we can adapt frames to arrange that a3 ≡ 0. We will call such a
frame 2-adapted.
Any two 2-adapted frames differ by a transformation of the
form (7) with λ = 0 — which means that, in fact, there is a
unique 2-adapted frame at each point of Σ.
Compute the remaining invariants:
Compute the remaining invariants:
So far, we know that the Maurer-Cartan forms associated to a
2-adapted frame satisfy the relations:
ω 2 = 0,
ω01 = ω 1 ,
ω00 = ω 0 + a2 ω 1 .
Compute the remaining invariants:
So far, we know that the Maurer-Cartan forms associated to a
2-adapted frame satisfy the relations:
ω 2 = 0,
ω01 = ω 1 ,
ω00 = ω 0 + a2 ω 1 .
Moreover, since a3 = 0, we have
da2 = −2a2 ω 0 .
(8)
Still more differentiating!
Still more differentiating!
Differentiating equation (8) yields
0 = d(da2 ) = −2da2 ∧ ω 0 − 2a2 dω 0
Still more differentiating!
Differentiating equation (8) yields
0 = d(da2 ) = −2da2 ∧ ω 0 − 2a2 dω 0
= −2(−2a2 ω 0 ) ∧ ω 0 − 2a2 (−ω00 ∧ ω 0 − ω10 ∧ ω 1 )
Still more differentiating!
Differentiating equation (8) yields
0 = d(da2 ) = −2da2 ∧ ω 0 − 2a2 dω 0
= −2(−2a2 ω 0 ) ∧ ω 0 − 2a2 (−ω00 ∧ ω 0 − ω10 ∧ ω 1 )
= 2a2 (ω10 − a2 ω 0 ) ∧ ω 1 = 0.
Still more differentiating!
Differentiating equation (8) yields
0 = d(da2 ) = −2da2 ∧ ω 0 − 2a2 dω 0
= −2(−2a2 ω 0 ) ∧ ω 0 − 2a2 (−ω00 ∧ ω 0 − ω10 ∧ ω 1 )
= 2a2 (ω10 − a2 ω 0 ) ∧ ω 1 = 0.
By Cartan’s lemma, we have
ω10 = a2 ω 0 + a4 ω 1
for some function a4 on Σ.
Finally, this gives a complete set of relations among the
Maurer-Cartan forms for a 2-adapted frame field:
ω 2 = 0,
ω01 = ω 1 ,
ω00
ω10
0
(9)
1
= ω + a2 ω ,
= a2 ω 0 + a4 ω 1 .
Finally, this gives a complete set of relations among the
Maurer-Cartan forms for a 2-adapted frame field:
ω 2 = 0,
ω01 = ω 1 ,
ω00
ω10
0
(9)
1
= ω + a2 ω ,
= a2 ω 0 + a4 ω 1 .
We already know that the invariant function a2 satisfies the
differential equation
da2 = −2a2 ω 0 ,
and differentiating the last equation in (9) shows that a4
satisfies the differential equation
da4 = ((a2 )2 − 2a4 )ω 0 + a5 ω 1
for some function a5 on Σ.
This is just about all we can learn from differentiating.
So what do we do next?
This is just about all we can learn from differentiating.
So what do we do next?
When you can’t differentiate — integrate!
Canonical coordinates and local normal forms
Canonical coordinates and local normal forms
First, observe that
dω 0 = −ω00 ∧ ω 0 − ω10 ∧ ω 1
Canonical coordinates and local normal forms
First, observe that
dω 0 = −ω00 ∧ ω 0 − ω10 ∧ ω 1
= −(ω 0 + a2 ω 1 ) ∧ ω 0 − (a2 ω 0 + a4 ω 1 ) ∧ ω 1
Canonical coordinates and local normal forms
First, observe that
dω 0 = −ω00 ∧ ω 0 − ω10 ∧ ω 1
= −(ω 0 + a2 ω 1 ) ∧ ω 0 − (a2 ω 0 + a4 ω 1 ) ∧ ω 1
= 0.
Canonical coordinates and local normal forms
First, observe that
dω 0 = −ω00 ∧ ω 0 − ω10 ∧ ω 1
= −(ω 0 + a2 ω 1 ) ∧ ω 0 − (a2 ω 0 + a4 ω 1 ) ∧ ω 1
= 0.
By Poincaré’s lemma, locally there exists a function u on Σ,
unique up to an additive constant, such that
ω 0 = du.
Next, we have
dω 1 = −ω01 ∧ ω 0 − ω11 ∧ ω 1
Next, we have
dω 1 = −ω01 ∧ ω 0 − ω11 ∧ ω 1
= ω0 ∧ ω1.
Next, we have
dω 1 = −ω01 ∧ ω 0 − ω11 ∧ ω 1
= ω0 ∧ ω1.
By the Frobenius theorem, locally there exist functions v, σ on
Σ, with σ 6= 0, such that
ω 1 = σ dv.
Next, we have
dω 1 = −ω01 ∧ ω 0 − ω11 ∧ ω 1
= ω0 ∧ ω1.
By the Frobenius theorem, locally there exist functions v, σ on
Σ, with σ 6= 0, such that
ω 1 = σ dv.
The independence condition
ω 0 ∧ ω 1 6= 0
then implies that (u, v) form a local coordinate system on Σ,
and so σ may be regarded as a function σ(u, v).
Substituting ω 0 = du, ω 1 = σ dv into the equation
dω 1 = ω 0 ∧ ω 1
yields
σu du ∧ dv = σ du ∧ dv.
Substituting ω 0 = du, ω 1 = σ dv into the equation
dω 1 = ω 0 ∧ ω 1
yields
σu du ∧ dv = σ du ∧ dv.
Thus the function σ satisfies the partial differential equation
σu = σ,
and hence
σ(u, v) = eu σ0 (v)
for some nonvanishing function σ0 (v) on Σ.
Let ṽ be the function
Z
ṽ =
σ0 (v) dv;
then we can write
ω 1 = eu σ0 (v) dv = eu dṽ.
Let ṽ be the function
Z
ṽ =
σ0 (v) dv;
then we can write
ω 1 = eu σ0 (v) dv = eu dṽ.
Replacing v with ṽ (and dropping the tilde) produces local
coordinates (u, v) on Σ such that
ω 0 = du,
ω 1 = eu dv.
(10)
Let ṽ be the function
Z
ṽ =
σ0 (v) dv;
then we can write
ω 1 = eu σ0 (v) dv = eu dṽ.
Replacing v with ṽ (and dropping the tilde) produces local
coordinates (u, v) on Σ such that
ω 0 = du,
ω 1 = eu dv.
(10)
These coordinates are unique up to transformations of the form
u → u + r,
where r, s ∈ R are constants.
v → e−r v + s,
(11)
Now consider the functions a2 , a4 on Σ as functions of (u, v).
Now consider the functions a2 , a4 on Σ as functions of (u, v).
The equation
da2 = −2a2 ω 0
is equivalent to the PDE system
(a2 )u = −2a2 ,
(a2 )v = 0.
Now consider the functions a2 , a4 on Σ as functions of (u, v).
The equation
da2 = −2a2 ω 0
is equivalent to the PDE system
(a2 )u = −2a2 ,
(a2 )v = 0.
Therefore,
a2 = ce−2u
for some (nonzero) constant c ∈ R.
Now consider the functions a2 , a4 on Σ as functions of (u, v).
The equation
da2 = −2a2 ω 0
is equivalent to the PDE system
(a2 )u = −2a2 ,
(a2 )v = 0.
Therefore,
a2 = ce−2u
for some (nonzero) constant c ∈ R.
By a coordinate transformation as above, we can arrange that
c = ±1, and by reversing the orientation of Σ if necessary, we
can assume that c = 1, so that a2 = e−2u .
The equation
da4 = ((a2 )2 − 2a4 )ω 0 + a5 ω 1
is equivalent to the PDE
(a4 )u = e−4u − 2a4 .
The equation
da4 = ((a2 )2 − 2a4 )ω 0 + a5 ω 1
is equivalent to the PDE
(a4 )u = e−4u − 2a4 .
The general solution to this equation is
a4 = − 12 e−4u + f (v)e−2u ,
where f (v) is an arbitrary function of v.
(12)
It is straightforward to check that with ω 0 , ω 1 , a2 , a4 as above,
all the Maurer-Cartan structure equations are satisfied.
It is straightforward to check that with ω 0 , ω 1 , a2 , a4 as above,
all the Maurer-Cartan structure equations are satisfied.
Thus we have:
Theorem 1: Let Σ be a connected, regular lightlike surface of
constant type in R2,1 with a1 , a2 6= 0. Then there exists a local
coordinate system (u, v) on Σ (unique up to translation in v)
and a unique 2-adapted frame field (e0 , e1 , e2 ) on Σ with
associated Maurer-Cartan forms
w0 = du,
ω01 = eu dv,
ω 1 = eu dv,
ω00 = du + e−u dv,
ω10 = e−2u du + (− 12 e−3u + f (v)e−u )dv
for some smooth function f (v).
(13)
According to the general theory of moving frames, the converse
is true as well: for any smooth function f (v), locally there exist
smooth functions
x, e0 , e1 , e2 : U ⊂ R2 → R2,1
(unique up to an isometry of R2,1 ) such that Σ = x(U ) is a
lightlike surface in R2,1 and (e0 , e1 , e2 ) is a 2-adapted frame
field along Σ whose Maurer-Cartan forms are given by (13).
We can construct the surface associated to a given function
f (v) as follows:
We can construct the surface associated to a given function
f (v) as follows:
The equations
dx = eα ω α ,
deβ = eα ωβα
are equivalent to the compatible, overdetermined PDE system:
xu = e0 ,
xv = eu e1 ,
(e0 )u = e0 ,
(e0 )v = (e−u e0 + eu e1 ),
(e1 )u = e−2u e0 ,
(e1 )v = (− 21 e−3u + f (v)e−3u )e0 − eu e2 ,
(e2 )u = −(e−2u e1 + e2 ),
(e2 )v = ( 12 e−3u − f (v)e−3u )e1 − e−u e2 .
Start with the equations for the u-derivatives of the eα :
Start with the equations for the u-derivatives of the eα :
From
(e0 )u = e0 ,
we obtain
e0 = eu G0 (v)
for some R2,1 -valued function G0 (v).
Start with the equations for the u-derivatives of the eα :
From
(e0 )u = e0 ,
we obtain
e0 = eu G0 (v)
for some R2,1 -valued function G0 (v).
Substituting this expression into the equation for (e1 )u yields
(e1 )u = e−u G0 (v),
and therefore
e1 = −e−u G0 (v) + G1 (v)
for some R2,1 -valued function G1 (v).
Now the equation for (e2 )u becomes
(e2 )u = e−3u G0 (v) − e−2u G1 (v) − e2 ,
and therefore,
e2 = − 21 e−3u G0 (v) + e−2u G1 (v) + e−u G2 (v)
for some R2,1 -valued function G2 (v).
Next, we substitute these expressions into the equations for the
v-derivatives of the eα to obtain the following ODE system for
the functions Gα (v):
G00 (v) = G1 (v),
G01 (v) = f (v)G0 (v) − G2 (v),
G02 (v) = −f (v)G1 (v).
Next, we substitute these expressions into the equations for the
v-derivatives of the eα to obtain the following ODE system for
the functions Gα (v):
G00 (v) = G1 (v),
G01 (v) = f (v)G0 (v) − G2 (v),
G02 (v) = −f (v)G1 (v).
This system is equivalent to the third-order ODE
0
0
G000
0 (v) − 2f (v)G0 (v) − f (v)G0 (v) = 0
for the function G0 (v), together with the equations
G1 (v) = G00 (v),
G2 (v) = f (v)G0 (v) − G000 (v)
for G1 (v), G2 (v).
(14)
It turns out that solutions of (14) are related to solutions of a
second-order Sturm-Liouville equation associated to the
function f (v):
It turns out that solutions of (14) are related to solutions of a
second-order Sturm-Liouville equation associated to the
function f (v):
Proposition 3: Let h(v) be any real-valued solution of the
Sturm-Liouville equation
!
2
d
(15)
− 12 f (v) h(v) = 0.
dt
Then the function g(v) = (h(v))2 is a real-valued solution of
(14). Moreover, any real-valued solution g(v) of (14) is a linear
combination of solutions of this form.
Proof. Direct computation shows that for any real-valued
solution h(v) of (15), the function g(v) = (h(v))2 is a solution of
(14).
Proof. Direct computation shows that for any real-valued
solution h(v) of (15), the function g(v) = (h(v))2 is a solution of
(14).
Now, let h1 (v), h2 (v) be a basis for the real-valued solutions of
(15). Since the solution space of (14) must contain all linear
combinations of squares of solutions of (15), it must contain all
linear combinations of the the three independent functions
(h1 (v))2 ,
h1 (v)h2 (v),
(h2 (v))2 .
(16)
Proof. Direct computation shows that for any real-valued
solution h(v) of (15), the function g(v) = (h(v))2 is a solution of
(14).
Now, let h1 (v), h2 (v) be a basis for the real-valued solutions of
(15). Since the solution space of (14) must contain all linear
combinations of squares of solutions of (15), it must contain all
linear combinations of the the three independent functions
(h1 (v))2 ,
h1 (v)h2 (v),
(h2 (v))2 .
(16)
But the solution space of (14) is precisely 3-dimensional; hence
every real-valued solution of (14) is a linear combination of the
functions (16). 2
It follows from Proposition 3 that the entries of any R2,1 -valued
solution G0 (v) of (14) must be linear combinations of the
functions (16).
It follows from Proposition 3 that the entries of any R2,1 -valued
solution G0 (v) of (14) must be linear combinations of the
functions (16).
These entries are subject to the the inner product relations for
the frame (e0 , e1 , e2 ) defined by G0 (v). Fortunately, the
symmetries of the Maurer-Cartan connection forms guarantee
that if these relations hold at any point of Σ, then they hold
identically on Σ. We can arrange this by imposing appropriate
restrictions on the initial conditions for the ODE (14).
Finally, substituting the expressions for e0 , e1 into the
equations for xu , xv yields
xu = eu G0 (v),
xv = −G0 (v) + eu G00 (v).
Finally, substituting the expressions for e0 , e1 into the
equations for xu , xv yields
xu = eu G0 (v),
xv = −G0 (v) + eu G00 (v).
Integrating this equation (and translating so that x(0, 0) = 0)
gives the following parametrization for Σ:
Z v
u
x(u, v) = e G0 (v) −
G0 (τ ) dτ − G0 (0).
(17)
0
By an action of the Minkowski isometry group, we can arrange
that at (u, v) = (0, 0):
 
 
 
 
0
0
1
−1
1  
1  
 
 
x = 0 , e0 = √ 1 , e1 = 0 , e2 = √  1  .
2
2
0
0
1
0
By an action of the Minkowski isometry group, we can arrange
that at (u, v) = (0, 0):
 
 
 
 
0
0
1
−1
1  
1  
 
 
x = 0 , e0 = √ 1 , e1 = 0 , e2 = √  1  .
2
2
0
0
1
0
This corresponds to choosing initial conditions
 
 1 
 1

√
√ (f (0) + 3 )
1
2
2
2
1  
 


G0 (0) = √ 1 , G00 (0) =  √12  , G000 (0) =  √12 (f (0) − 12 )
2
0
1
1
for the function G0 (v).
Together with Propositions 1 and 2, this proves the following
classification theorem:
Theorem 2: Let Σ ⊂ R2,1 be a regular lightlike surface of
constant type. Then up to a Minkowski isometry, Σ is either:
• contained in a lightlike plane in R2,1 ;
• contained in a null cone in R2,1 ; or
• locally parametrized by (17), where G0 (v) is the unique
solution of (14) satisfying the initial conditions (54), and
f (v) is an arbitrary function of one variable.
Together with Propositions 1 and 2, this proves the following
classification theorem:
Theorem 2: Let Σ ⊂ R2,1 be a regular lightlike surface of
constant type. Then up to a Minkowski isometry, Σ is either:
• contained in a lightlike plane in R2,1 ;
• contained in a null cone in R2,1 ; or
• locally parametrized by (17), where G0 (v) is the unique
solution of (14) satisfying the initial conditions (54), and
f (v) is an arbitrary function of one variable.
This theorem says that locally, a generic lightlike surface Σ in
R2,1 is completely determined up to isometry by the choice of
one arbitrary function f (v), and Σ can be constructed explicitly
from solutions to the Sturm-Liouville equation (15) determined
by f .
As a consequence of the formula (17), we obtain the following
corollary:
Corollary: Every lightlike surface Σ of constant type in R2,1 is
ruled by null lines.
As a consequence of the formula (17), we obtain the following
corollary:
Corollary: Every lightlike surface Σ of constant type in R2,1 is
ruled by null lines.
Proof. The statement is clearly true for lightlike planes and null
cones. For surfaces of the form (17), each u-parameter curve
with v = v0 is a line parallel to the vector G0 (v0 ), which is a
multiple of the null vector e0 at each point of Σ. Therefore, the
u-parameter curves are null lines. 2
Examples!
Examples!
The simplest examples are those for which f (v) is a constant
function, since then the Sturm-Liouville equation (15) can be
solved explicitly.
Example 1: f (v) = 0.
Example 1: f (v) = 0.
In this case, solving for G0 (v) and substituting into (17)
produces the parametrization:


1
√
eu (3v 2 + 4v + 4) − (v 3 + 2v 2 + 4v + 4)
4 2
 1

u
2
3
2

√
x(u, v) = 
 12 2 e (−3v + 12v + 12) + (v − 6v − 12v − 12)  .
1
u
2
3
2
6 e (3v + 6v) − (v + 3v )
Figure: Lightlike surface with f (v) = 0
Example 2: f (v) = 1.
Example 2: f (v) = 1.
In this case, solving for G0 (v) and substituting into (17)
produces the parametrization:
x(u, v) =

√

√
√
√
√
√
√
e 2v ((5 2+4)eu −5−2 2) + e− 2v ((5 2−4)eu +5−2 2) + 2 2(v−eu −2)
√

√
√
√
√
√
√
e 2v (( 2+4)eu −1−2 2) + e− 2v (( 2−4)eu +1−2 2) + 2 2(−3v+3eu −2) 

√
√
√
√
√
√
1
2v
2v
u
−
u
u
e
((2 2+2)e −2− 2) + e
((−2 2+2)e −2+ 2) + 4(v−e + 1)
8
1
16
1
 16

Figure: Lightlike surface with f (v) = 1
Example 3: f (v) = −1.
Example 3: f (v) = −1.
In this case, solving for G0 (v) and substituting into (17)
produces the parametrization:
x(u, v) =
√
√
√

1 √
2(2 − eu ) cos( 2v) + (1 + 4eu ) sin( 2v) − 2(5v − 5eu + 6)
8
1 √
√
√
√


2(2 + 3eu ) cos( 2v) + (−3 + 4eu ) sin( 2v) − 2(v − eu + 6) 
8

√
√
√
1
u
u
u
4 (2 − 2e ) cos( 2v) + 2(1 + 2e ) sin( 2v) − (2v − 2e + 2)
Figure: Lightlike surface with f (v) = −1
For nonconstant functions f (v), equation (14) generally cannot
be solved explicitly for G0 (v); however, we can still construct
the corresponding surfaces x(u, v) via numerical integration.
Figure: Lightlike surface with f (v) = v
Figure: Lightlike surface with f (v) = v = sin v
Happy Birthday, Robert!