If a time-like surface in R-j(n) is isothermic then there is another, uniquely determined surface which is either anti-conformal or conformal and sign-reversing with the same Gauss map. The choice of mapping depends on whether the shape operator is diagonalizable over R or C

Magid, Martin A

English

Wellesley College

Wellesley College
Wellesley College Digital Scholarship and Archive
Mathematics Faculty Scholarship Mathematics
2005
Lorentzian Isothermic Surfaces in R-j(n)
Martin A. Magid
Wellesley College
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Recommended Citation
Magid, M. A. (2005). Lorentzian isothermic surfaces in R-j(n). Rocky Mountain Journal of Mathematics, 35(2), 627–640.
doi:10.1216/rmjm/1181069750
ROCKY MOUNTAIN
JOURNAL OF MATHEMATICS
LORENTZIAN ISOTHERMIC SURFACES IN Rnj
M.A. MAGID
ABSTRACT. If a time-like surface in Rnj is isothermic then
there is another, uniquely determined surface which is either
anti-conformal or conformal and sign-reversing with the same
Gauss map. The choice of mapping depends on whether the
shape operator is diagonalizable over R or C.
1. Introduction. There has been much recent work on isothermic
surfaces in Euclidean spaces, see, for example, [1, 2, 4] and this
paper begins a companion study for the timelike case. We will prove,
for timelike surfaces in Rnj , the analogue of Palmer’s result in [9].
In the positive deﬁnite setting, of course, all shape operators are
diagonalizable over R, and the existence of shape operators which are
not diagonalizable over R makes the timelike case diﬀerent. Here
we have two cases to consider and the possibility of sign-reversing
mappings arise for the ﬁrst time.
2. Preliminaries.
2.1 Deﬁnitions and notations for timelike surfaces in Rnj . The
metric in Rnj , for n ≥ 3 and n− 1 ≥ j ≥ 1 is denoted by
〈v, w〉 = −v1w1 − v2w2 + · · · − vjwj + vj+1wj+1 + · · ·+ vnwn
for v = (v1, v2, . . . , vn) and w = (w1, w2, . . . , wn). We begin with
isothermal coordinates (t, s) on X, by which we mean that, for the
induced metric h on X and for the associated coordinate vectors ∂t
and ∂s, there is a nonzero function λ so that
(2.1)
h(∂t, ∂t) = −λ2
h(∂t, ∂s) = 0
h(∂s, ∂s) = λ2.
AMS Mathematics subject classiﬁcation. 53A15.
Received by the editors on August 7, 2001.
Copyright c©2004 Rocky Mountain Mathematics Consortium
1
2 M.A. MAGID
Let Ni, i = 1, 2, . . . , n − 2 denote a basis of normal vectors to the
image of X which are orthogonal and have length ±1. The fundamental
equations are:
(2.2)
Xtt =
λt
λ
Xt +
λs
λ
Xs +
n−2∑
i=1
eiNi
Xts =
λs
λ
Xt +
λt
λ
Xs +
n−2∑
i=1
fiNi
Xss =
λt
λ
Xt +
λs
λ
Xs +
n−2∑
i=1
giNi
(Nit)
T = 〈Ni, Ni〉
(
ei
λ2
Xt − fi
λ2
Xs
)
(Nis)
T = 〈Ni, Ni〉
(
fi
λ2
Xt − gi
λ2
Xs
)
,
where, for example, (Nit)T is the tangential component of Nit. Thus,
the shape operator corresponding to Ni is
(2.3) Ai = 〈Ni, Ni〉
[−ei/λ2 −fi/λ2
fi/λ
2 gi/λ
2
]
.
The mean curvature vector, η, is
(2.4) η =
n−2∑
i=1
(gi − ei)
2λ2
Ni.
Recall that, at each point on a timelike surface, there are three possi-
bilities for a shape operator. The shape operator is either a) diagonaliz-
able overR, b) diagonalizable overC but notR or c) not diagonalizable
over C, and has a single real eigenvalue [5, 8]. Case a) occurs if and
only if (
ei + gi
2
)2
− f2i > 0,
while case b) occurs if and only if the same quantity is negative.
LORENTZIAN ISOTHERMIC SURFACES 3
A Riemannian surface is called isothermic if there is an isothermal
coordinate system for which every shape operator is diagonalized. For
a timelike surface we need to adjust the deﬁnition. Essentially we will
call a surface isothermic if there is an isothermal coordinate system for
which each shape operator has one of the three canonical forms.
Deﬁnition. A timelike immersion is called isothermic if there is some
isothermal coordinate system (t, s) such that each Ai with respect to
the basis {∂t, ∂s} has one of the following forms:
a)
[
λ1 0
0 λ2
]
b)
[
a b
−b a
]
c)
[
ν ± a/2 −a/2
a/2 ν ∓ a/2
]
.
We call the ﬁrst case real isothermic and the second complex isothermic.
In this paper we are working locally and assume that, in a neighbor-
hood of a point, the immersion is either real isothermic or complex
isothermic. Note that, in the case of complex isothermic, each shape
operator satisﬁes ei + gi = 0. Finally note that we allow umbilic points
and a totally umbilic surface satisﬁes all three conditions.
2.3 Isothermal coordinate systems. One of the key elements of the
proof is changing from one isothermal coordinate system to another.
We will need the following
Proposition 2.1. Let (t, s) be an isothermal coordinate system
on X. Any other isothermal coordinate system (u, v) on X with
J(u, v)/J(t, s) > 0 satisﬁes
(2.5)
ut = σρ cosh(θ) vt = ρ sinh(θ)
us = ρ sinh(θ) vs = σρ cosh(θ)
with
(2.6) ρt = σρθs and ρs = σρθt.
Conversely, given any ρ and θ satisfying (2.6) then there exist a new
coordinate system (u, v) satisfying (2.5).
Here g(Xv, Xv) = µ2 and g(Xs, Xs) = λ2, ρ2 = λ2/µ2, ρ > 0 and
σ = ±1.
4 M.A. MAGID
The proof of this proposition is a straightforward computation.
Note that (2.6) implies θtt = θss. In addition θtt = θss implies the
existence of the coordinate system as well. We can ﬁnd a function β so
that βt = θs and βs = θt. Setting ρ = eβ gives the desired θ and ρ.
Proposition 2.2. With the new variables {u, v} given in Proposi-
tion 2.1, the normal components of the fundamental equations are:
(2.7)
eiuv =
ch2(θ)
ρ2
ei − σ sh(2θ)
ρ2
fi +
sh2(θ)
ρ2
gi
fiuv = −σ
sh(2θ)
2ρ2
ei +
ch(2θ)
ρ2
fi − σ sh(2θ)2ρ2 gi
giuv =
sh2(θ)
ρ2
ei − σ sh(2θ)
ρ2
fi +
ch2(θ)
ρ2
gi
(Here, for example, eiuv is the coeﬃcient of Ni in the expansion of
Xuu.)
3. Main theorem.
Theorem. Let X : (M,h) → (Rnj , 〈 , 〉) be a conformal immersion
and let γ : X(M) → G(R21 ⊂ Rnj ) be the Gauss map into the
Grassmannian of Lorentzian two-planes in Rnj .
There exists an anti-conformal immersion X˜ : (M,h) → (Rnj , 〈 , 〉)
such that γ˜ ◦ X˜ = γ where γ˜ is the Gauss map on the image of X˜ if
and only if X(M) is real isothermic.
There exists a sign-reversing conformal immersion Xˆ : (M,h) →
(Rnj , 〈 , 〉) such that γˆ ◦ Xˆ = γ where γˆ is the Gauss map on the image
of Xˆ if and only if X(M) is complex isothermic.
X˜ and Xˆ is unique, up to similarity, if M has no umbilic points.
Note that Xˆ is sign-reversing and conformal if and only if, for some
LORENTZIAN ISOTHERMIC SURFACES 5
µ = 0,
(3.1)
〈Xˆt, Xˆt〉 = µ2
〈Xˆt, Xˆs〉 = 0
〈Xˆs, Xˆs〉 = −µ2.
Proof. First consider the case where the immersion X is real isother-
mic, so that we may assume fi = 0 for i = 1, 2, . . . , n− 2. We deﬁne a
new immersion, for σ = ±1
X˜t = − σ
λ2
Xt, X˜s =
σ
λ2
Xs.
It is easy to check that X˜ts = X˜st, so that X˜ does deﬁne an
immersion.
Conversely, suppose X and X˜ exist as in the ﬁrst part of the theorem.
Since the Gauss maps agree we must have:
X˜t = aXt + bXs, X˜s = cXt + dXs,
with ad− bc < 0. By the deﬁnition of anti-conformality we have:
〈X˜t, X˜t〉 = −µ2 = −a2λ2 + b2λ2
〈X˜s, X˜s〉 = µ2 = −c2λ2 + d2λ2
〈X˜t, X˜s〉 = 0 = −ac + bd.
Thus, if we set ν2 = µ2/λ2 we see that
(3.2)
a = ντch(θ), b = νsh(θ), c = −νsh(θ), and d = −ντch(θ),
for some function θ and τ = ±1.
From X˜ts = X˜st we arrive at three equations deﬁned in terms of the
λ, ei, fi, and gi associated to X(t, s):
as + bt + 2a
λs
λ
+ 2b
λt
λ
= 0,(I1)
bs + at + 2a
λt
λ
+ 2b
λs
λ
= 0,(I2)
2a
(∑
fiNi
)
+ b
(∑
eiNi
)
+ b
(∑
giNi
)
= 0.(I3)
6 M.A. MAGID
Note the last equation gives 2afi + b(ei + gi) = 0 for i = 1, . . . , n− 2.
We want to ﬁnd an angle ψ so that, for each i,
fiuv = 0 = −σ
sh(2ψ)
2ρ2
ei +
ch(2ψ)
ρ2
fi − σ sh(2ψ)2ρ2 gi.
In other words, we want ψ so that
tanh(2σψ) =
2fi
ei + gi
= − b
a
.
We need only set ψ = −τσθ/2. We have to check that ψtt − ψss = 0,
or, equivalently, that θtt − θss = 0. We note that θ is proportional to
the arctanh (b/a) and that
θt =
abt − bat
a2 − b2
θs =
abs − bas
a2 − b2 ,
so that the numerator of θtt − θss is
(3.3) (a2 − b2)(abtt − batt − abss + bass)− 2(abt − bat)(aat − bbt)
+ 2(abs − bas)(aas − bbs).
By diﬀerentiating (I1) and (I2) with respect to s and t and substi-
tuting for att, ass, btt and bss and then, again, using (I1) and (I2) to
substitute for at and bt the equation (3.3) can be seen to be zero. Thus
X˜ can be seen to be real isothermic. We postpone the proof of the
uniqueness of X˜ to the end of this proof.
Next suppose that X is complex isothermic, so that ei + gi = 0. We
deﬁne Xˆ by
Xˆt = − σ
λ2
Xs, Xˆs =
σ
λ2
Xt.
Again, Xˆts = Xˆst. Next, assume that X and Xˆ exist, so that
Xˆt = aXt + bXs, Xˆs = cXt + dXs,
LORENTZIAN ISOTHERMIC SURFACES 7
with ad − bc > 0. By the deﬁnition of a sign-reversing conformal
immersion we have:
〈Xˆt, Xˆt〉 = µ2 = −a2λ2 + b2λ2
〈Xˆs, Xˆs〉 = −µ2 = −c2λ2 + d2λ2
〈Xˆt, Xˆs〉 = 0 = −ac + bd.
Thus, if we set ν2 = µ2/λ2 we see that
a = νsh(θ), b = τνch(θ), c = −τνch(θ), and d = −νsh(θ),
for some function θ and τ = ±1.
From Xˆts = Xˆst we arrive at the same three equations as before:
(I1), (I2) and (I3).
We want to ﬁnd an angle ψ so that ei{uv}+ gi{uv} = 0, or, for each i,
ch(2ψ)
ρ2
ei − 2σ sh(2ψ)
ρ2
fi +
ch(2ψ)
ρ2
gi = 0.
Thus we set
tanh(2σψ) =
ei + gi
2fi
= −a
b
= −τ tanh(θ),
or ψ = −τσθ/2. The only diﬀerence here is that we have a/b instead
of b/a. But the equations (I1) and (I2) are interchanged if a and b are
interchanged. Thus we again see that ψtt − ψss = 0.
To see uniqueness, suppose we have two X˜ which are anti-conformal
to the original X. Then, for some choice of isothermal coordinates,
each fi = 0. By (I3) we get b(ei + gi) = 0. Since M is not umbilic at
any point, for some i, (ei + gi) = 0 at every point. Then b = 0 and, so,
θ = 0 . Finally, using, (I1) and (I2) we ﬁnd that a = −k/λ2 for some
constant k. Thus any two X˜ are similar. The same argument will work
for Xˆ.
In the next proposition we record the basic properties of X˜ and Xˆ.
For simplicity we consider surfaces in Lorentz 3-space
8 M.A. MAGID
Proposition. Let X : (M,h) → (R31, 〈 , 〉) be a conformal immer-
sion.
1. If X is real isothermic then X˜ is real isothermic and, if X is
neither umbilic nor minimal, then ˜˜X = X.
2. If X is complex isothermic and if X is neither umbilic nor minimal
then Xˆ is complex isothermic and ˆˆX = X.
Proof. The statement follows from looking at the shape operators of
X˜ and Xˆ. Indeed, if the shape operator of X is
A =
[−e/λ2 0
0 g/λ2
]
,
then the shape operator for X˜ is
A˜ =
[
σe 0
0 σg
]
.
In the same way, if the shape operator of X is
A =
[−e/λ2 −f/λ2
f/λ2 −e/λ2
]
,
then
Aˆ =
[−σf σe
−σe −σf
]
.
Thus, if the immersion is neither umbilic nor minimal then X˜ or Xˆ
can be called the dual of X.
4. Examples.
4.1. Surfaces in Lorentz space with constant mean curvature, in-
cluding minimal surfaces, are isothermic. (We assume that the shape
operator is either diagonalizable over R or C.) We give a proof of this
fact, since it does not seem to appear in the literature.
LORENTZIAN ISOTHERMIC SURFACES 9
Suppose that X : M → R31 is an isometric immersion with constant
mean curvature. The Codazzi equations in this case become, from the
fundamental equations with e1 = e, f1 = f and g1 = g,
(4.1) (e + g)t = 2fs and (e + g)s = 2ft.
This implies that ftt = fss. Assume ﬁrst that the surface has a shape
operator which is diagonalizable over R. As before, we seek a function
θ(t, s) so that the normal component of Xuv is zero. Equivalently we
want
θ =
σ
2
arctanh
(
2f
e + g
)
.
As in the proof of the theorem, if we use (4.1) and ftt = fss, we can
see that θtt = θss and the surface is isothermic.
If the shape operator is diagonalizable over C then we want
θ =
σ
2
arctanh
(
e + g
2f
)
.
We see again that θtt = θss.
For an explicit complex isothermic example, consider a Lorentzian
helicoid:
X(t, s) = (cosh(s) sinh(t), sinh(t) sinh(s), s),
with normal vector
N(t, s) =
1
cosh(t)
(− sinh(s),− cosh(s), sinh(t)).
The surface looks like Figure 1.
It is easy to see that the dual surface to this surface is:
Xˆ(t, s) =
1
cosh(t)
(sinh(s), cosh(s),− sinh(t)),
which, as we know, is a parameterization of part of the standard
immersion of the Lorentzian sphere.
10 M.A. MAGID
FIGURE 1.
4.2. Lorentzian rotation surfaces in R31 are real isothermic. There
are several diﬀerent kinds of rotation surfaces in R31, depending on
length of the axis of rotation. Details are given in [7].
a) X(t, s) = (x(s) sinh(t), x(s) cosh(t), y(s)), with x(s) = 0. We can
see that
N(t, s) =
1√
x′2 + y′2
(y′ sinh(t), y′ cosh(t),−x′).
b) X(t, s) = (x(t) cosh(s), x(t) sinh(s), y(t)), with (x(t), y(t)) a time-
like curve. We can see that
N(t, s) =
1√
x′2 − y′2
(y′ cosh(s), y′ sinh(s), x′).
c) X(t, s) = (x(t), y(t) cos(s), y(t) sin(s)), with y(t) = 0 and (x(t), y(t))
a timelike curve.
N(t, s) =
1√
x′2 − y′2
(y′, cos(s)x′, sin(s)x′).
d)
X(t, s) =
(
a(t)− b(t)√
2
+
a(t)s2
2
√
2
,
a(t) + b(t)√
2
− a(t)s
2
2
√
2
, sa(t)
)
LORENTZIAN ISOTHERMIC SURFACES 11
with a′(t)b′(t) < 0 and
N(t, s) =
(
1√
−2a′(t)b′(t)
)(
a′(t)+b′(t)√
2
+a
′(t)s2
2
√
2
, a
′(t)−b′(t)√
2
−a′(t)s2
2
√
2
, sa(t)′
)
.
These surfaces are all real isothermic. The parameterizations above all
have diagonalized shape operators, but the coordinates are isothermal
only if the proﬁle curves are parameterized properly.
4.3. Every member of a Bonnet family is isothermic [3, 6].
APPENDIX
1. Null Coordinates. If we use null coordinates {x, y} on our
surface, then many of the proofs are simpliﬁed but lose some of the
geometric ﬂavor. In this appendix some of the proofs are given with
respect to null coordinates, deﬁned by
x =
t− s√
2
, y =
t + s√
2
t =
x + y√
2
, s =
y − x√
2
.
Thus 〈Xx, Xy〉 = −λ2 and the fundamental equations become:
(5.1)
Xxx = 2
λx
λ
Xx +
n−2∑
i=1
(ei + gi
2
− fi
)
Ni
Xxy =
n−2∑
i=1
ei − gi
2
Ni
Xyy = 2
λy
λ
Xy +
n−2∑
i=1
(ei + gi
2
+ fi
)
Ni.
We write the coeﬃcients of the normal components as
(5.2)
e¯i =
ei + gi
2
− fi
f¯i =
ei − gi
2
g¯i =
ei + gi
2
+ fi
12 M.A. MAGID
Proposition 5.1. If new null variables {x˜, y˜} are given so that
x = α(x˜) and y = β(y˜), then the normal components of fundamental
equations become:
e¯ix˜y˜ = α
′2e¯i
f¯ix˜y˜ = α
′β·f¯i
g¯ix˜y˜ = β
·2g¯i
The expressions for X˜ and Xˆ are
X˜x = − σ
λ2
Xy, X˜y = − σ
λ2
Xx
Xˆx = − σ
λ2
Xy, Xˆy =
σ
λ2
Xx.
In the proof of the main theorem we can, of course, use null coordi-
nates. For example, in the case of Xˆ, suppose that we are given
Xˆx = aXx + bXy
Xˆy = cXx + dXy
with ad − bc > 0 and 〈Xˆx, Xˆy〉 > 0. This yields that a = 0 = d and
bc < 0. Xˆxy = Xˆyx gives
(5.3)
by +
2bλy
λ
= 0
cx +
2cλx
λ
= 0
bg¯i = ce¯i.
We are looking for a change of null coordinates x˜ = α(x), y˜ = β(y) so
that e¯i + g¯i = 0. From (5.3) we ﬁnd
b =
γ(x)
λ2
, c =
δ(y)
λ2
.
We note that b and c have opposite signs; assume, without loss of
generality, that b < 0. We need to ﬁnd α(x) and β(y) so that
α′(x)2e¯i + β·(y)
2
g¯i = 0.
LORENTZIAN ISOTHERMIC SURFACES 13
We know that
γ(x)g¯i = δ(y)e¯i
with γ(x) < 0 and δ(y) > 0, or
g¯i
δ(y)
− e¯i
γ(x)
= 0.
Just set
α′(x)2 = − 1
γ(x)
β·(y)2 =
1
δ(y)
.
REFERENCES
1. F.E. Burstall, Isothermic surfaces: Conformal geometry, Cliﬀord algebras and
integrable systems, preprint arXiv:math.DG/003096 v2.
2. F. Burstall, U. Hertrich-Jeromin, F. Pedit and U. Pinkall, Curved ﬂats and
isothermic surfaces, Math. Z. 225 (1997), 199 209.
3. A. Fujioka and J.-I. Inoguchi, Timelike Bonnet surfaces in space forms,
preprint.
4. G. Kamberov, F. Pedit and U. Pinkall, Bonnet pairs and isothermic surfaces,
Duke Math. J. 92 (1998), 637 644.
5. Martin Magid, Isometric immersions of Lorentz space with parallel second
fundamental forms, Tsukuba J. Math. 8 (1984), 31 54.
6. , A theorem of Cartan for timelike Bonnet surfaces forms, preprint.
7. Louise McNertny, One-parameter families of surfaces with constant curvature
in Lorentz 3-space, Doctoral thesis, Brown University.
8. Barrett O’Neill, Semi-Riemannian geometry, Academic Press, New York, 1983.
9. Bennet Palmer, Isothermic surfaces and the Gauss map, Proc. Amer. Math.
Soc. 104 (1988), 876 884.
Department of Mathematics, Wellesley College, Wellesley, MA, 02181
Email address: mmagid@wellesley.edu


Lorentzian Isothermic Surfaces in R-j(n)

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