Download isometric immersions of lorentz space with parallel second

TSUKUBA
J. MATH.
Vol. 8 No. 1 (1984). 31-54
ISOMETRIC
WITH
IMMERSIONS
PARALLEL
OF
SECOND
LORENTZ
SPACE
FUNDAMENTAL
FORMS
By
Martin
A. Magid
Introduction
In a series of papers, [Fl], [F2], [F3], [F4], D. Ferus
of euclidean space
form
many
with parallel second fundamental
an important
class of examples.
classified submanifolds
forms.
In [F4], Ferus
shows
These
submanifolds
that they appear in
topics in differentialgeometry.
There
has been much
spaces―notably
the work
recent work
on parallel submanifolds
of other ambient
of H. Naitoh, [Nal]-[Na4] and M. Takeuchi
interesting problem is to classify the parallel submanifolds
[T].
An
of euclidean spaces
equipped with indefinite metrics.
This paper studies special classes of parallelsubmanifolds in /2f, Lorentz space
of signature (1, m―1)
and
in Rf,
euclidean space of signature (2,m―2).
umbilical submanifolds
are classified,as well as isometric immersions
R" ->-J??+2and jR? -≫J?2"+2
with parallelsecond fundamental
indicate some
of the modifications which
forms.
All
Rn -> R"+k,
These
theorems
will be necessary in order to obtain a
complete classification.
The preliminary section(0) gives some basic results about indefinite Riemannian
geometry.
These
include
an indefinite version of the result of Allendoerfer and
Erbacher for reducing the codimension
second fundamental
forms for symmetric
of an isometric immersion
form (F*a = 0), and an improved
Q.
These
maps
Section 2 contains the main
R"-*RlHZ
product of the identity map
Received
umbilic with lightlike mean
classificationresults. These
state that isometric
are either quadratic in nature,
with lightlike mean
curvature
vector, or the
and previously determined low dimensional maps.
of the results in this paper are taken from the author's Brown
March
28, 1983.
with
circleof radius k￡C (1.12).
and /2f->/2 +2 with F*a―0
like the flat umbilical immersion
Most
version of Petrov's canonical
R＼->R＼, R＼-+R＼ and. Rn-+R"+Ic
include a flat ^-dimensional
curvature vector (1.4) and the complex
immersions
parallel
transformations of Lorentz space.
Section 1 classifiesi sometric immersions
F*a
with
University
32
Martin
doctoral
dissertation.
Nomizu's
constant
The
author
A. Mag
id
gratefully
encouragement
and
acknowledges
Professor
Katsumi
patience.
Notation and Terminology
Rfj denotes the ^-dimensional affine space with the metric ( , ) whose
canon-
ical form is
-h
Ip-i-j
o,
where
Ik is the kxk
non-degenerate
identity
In general, if M
metric
of signature
The
matrix
iff i = 0, in which
is a
and
case
/^-dimensional
(i,p―i―j,f)
indefinite sphere,
we
Oj is the jxj
we
write
manifold
write
hyperbolic
whose
The
isometric
The
tangent
and lightcone
S?-1(r)= {a;e/2f: (x,x) = r2},
O, x)=
LCff^ixeRf:
matrix.
metric
is
spaces
have
a
Mf4,
space
Hf:1＼r)={xeRf:
O
Rf.
r>0
-r2},
(x,x)=Q
and
are defined as follows:
r<0
xi-0)
immersions
t -≫(a cos (//≪),a sin (￡/#))
* -> (h sinh (*/&), & cosh (tlb))
t -> (6 cosh (f/&), 6 sinh (#/*))
are called circle maps.
S{cR2i
and
The
the third onto
Hlc:R＼.
are called one-dimensional
A
vector
timelike
product.
X
of its tangent
A
is
･■-, Xn
called
so that the inner
product
(Xi,Xj)=dij,
product
H＼dRl,
along
the
with
second
onto
the map
space is called spacelike
null or lightlike if (X,X)=O,
of an indefinite inner
orthonormal
and
{/,/, Xu
circle maps,
is called spacelike, timelike,
are unit timelike
basis
or
where
if (X,X)>0,
( , ) is
or null depending
on
the inner
the character
vectors.
basis Xh
(p, n―p)
SlczR2
maps.
and
curve
onto
The
in an indefinite inner
if (X,X)<0
A
first maps
if the vectors
the last n―p
■■-,Xn-i}.
l<z", j<,n
space
In such
2.
A
product
are pairwise
are unit spacelike.
is Lorentz,
we
space
can
with
the
signature
orthogonal,
the first p
If the signature
is (1, n―1),
define
a
pseudo-orthonormal
a basis (/, /)=()=(/, /)= (/, Xt)=(/, Xt＼ (I,!)= l and
a over
a
vector
will always
have
inner
product
1
Isometric Immersions
of Lorentz Space
33
with the hatless vector.
We
recall the following notation from [K-N], vol.II,ch. 7. If /: (M, g)->(M, g)
is an isometric immersion
if the Riemannian
and Y on M
of one indefinite Riemannian
manifold into another, and
connections are denoted by F and F, then
we have
for vector fields X
the following decomposition,
I. Vxf*Y=f*VxY+a{X,Y).
Here/*FAY
is the tangential component
a is called the second fundamental
and
form
a{X,Y)
is the normal
component.
of /. For a normal fieldf to M
we have
II. PxS^-f+AtX+Vg.
At is called the shape operator associated to ￡ and F1 is called the normal
con-
nection. A and a are related by the equation
g(A(X,Y)
= g(a(X,Y),e.
a is said to be parallelif
(FtaYX,
Y): = Fha(X, Y)-a(FzX,
Y)-a(X,
FZY) = O.
0. Preliminaries
Defining the mean
curvature vector rjas in the Riemannian
case, we
have
the following.
0.1
Proposition,
Riemannian
If /:
M-+M
is an
(i)
the mean
(ii)
the firstnormal
spaces N^x),
xsM
where
(Indefinite version of a
Let /: M" -*･Rf
exists a complete n+k
n = $imM
with
parallel first normal
with parallelsecond fundamental
theorem
be an isometric immersion
with signature (i,n―i) into R
of Allendoerfer and
of an indefinite Riemannian
forms.
Erbacher).
manifold
. If the firstnormal spaces are parallel,then there
dimensional totally geodesic submanifold M*
of Rf (where
and k = A＼mNl) such that /(M)cAf*.
Note:
theorem
are P-parallel.
next theorem is about isometric immersions
Theorem
form, then
N°(x)= {￡eN(x): ^=0}.
spaces and so can be applied toimmersions
0.2
of one indefinite
curvature vector is parallel: F￡=Q.
Nl(x) : = {N＼x)}1
The
isometric immersion
manifold into another with parallel second fundamental
M* = R's,t*for some
shows
that t measures
s, / and ^ need
not be zero. The
the degenerate part of N＼
proof of the
34
Martin
Proof.
The
assumption
displacement
of normal
maps
onto
Nl(x)
a normal
X
vector
in TX{M).
vectors
Nl{y).
+ (￡>VxO)Let
f(x0) which
of E
in M
starting
at x0.
For
and
means
any
z from
x to y the parallel
to the
normal
of Nl is a constant,
for each
x in M,
then
the subspaces
In fact ￡gN＼
0&N°
consider
the
If ￡ is
F^eN^x)
for
all
N°(x) are also parallel
implies
n + k
0 = X($, 6) = (V^,
dimensional
to N°(x0). E― TX(S(M) 0
f0 in Na(xo)
be
subspace
Nl(x0).
that f(M)czE.
let &
^/..so that $t^N0(x,).
The
Let
the result
For the euclidean
―(/(■Xd-A-Vo),
E
0)
of
of jR
degenerate
xt be any
P-parallel
connection
that gtis parallelin Rf and so is a constant vector.
we
Now
curve
dishave
we have
￡,)= (/*(*,.), fo)
=
so that f(xt) lies in E.
connection
say k.
is zero, so is the second.
is perpendicular
of ￡<>along
respect
connection,
connection.
of M
with
is N°(xo)r＼Ni(xo). It will be shown
placement
which
that feiV(x)
the first term
x0 be a point
through
part
Since
r
the dimension
F1 is a metric
relative to the normal
id
that for any curve
along
Thus
field such
Since
means
A. Mag
0
Since xt is an arbitrary curve in M, f{M)cE.
Next is an indefinite version of Moore's lemma
that allows an isometric immersion
[Mo 1] which
gives a condition
of a product manifold to be decomposed
into
a product of immersions.
0.3
Theorem.
Let M
be an indefinite Riemannian
R] is an isometric immersion
then /=/iX/2:
The
M1xM2->R"11xRj*
decomposition theorems
manifolds.
A
shape
product M,xM2.
with a(X, F) = 0 whenever
XsTx(Mi)
If /: M-≫
and YsTx(M2)
and each /≫is an isometric immersion.
here involve the shape operators of Riemannian
operator of a Riemannian
manifold is always diagonalizable,
but this is not the case for a shape operator of a Lorentzian manifold. In order
to analyze the latter,a special case of Petrov's Principal axis theorem
for a tensor
is needed ([P] pp. 50-55).
0.4 Theorem.
Let V be a real ^-dimensional vector space equipped with a
Lorentzian metric and let A be linear transformation of V which is symmetric
with respect to this metric. Then
A can be put into one of the following four
forms with respect to bases whose inner products are given by G.
Isometric Immersions
of Lorentz Space
X
35
1
1
h
A=
/"≫
1
An t
X
1
0
1
0
X
1
0
X
A=
1
/"<_
An-2
A=
X
0
1
0
X
0
0
1^
0
1
1
0
c=
1
X
a
j8
B
a
+1
-1
^
A=
1
Cr ―
B^O
1
AM
These
is a
―2
will be referred to as the case where
there: are simple real eigenvalues,
non-simple eigenvalue of multiplicity2, is a non-simple eigenvalue of multi-
plicity 3 or is a complex
eigenvalue. The
first and
fourth cases will be said to
have simple eigenvalues.
This is a slightly sharper version of the theorem
in [P]. Theorem
[Mai] allows us to use an orthonormal basis in the case of a complex
instead of the more
The
0.5
eigenvalue
complicated basis which Petrov employs.
following proposition can be proved
found in [Mo
2, p. 229
using 0.4.
Another
proof can
be
21
Proposition.
Let M(c) and M(c) denote space forms
of constant curvature
c and c. Suppose
(i)
MS(c) is isometrically immersed
in M"+1(c)
M&c)
is isometrically immersed
in M'1+＼c)
or (iii) M['(c)is isometrically immersed
in M"+1(c).
or (ii)
If c^c
(ii)c<c
and n>2
then the immersion
is umbilical and in case (i) c>c, in case
and in case (iii)c>c.
It is necessary to know
what
form a Lorentzian symmetric
matrix takes if it
36
Martin
commutes
0.6
A. Magid
with one of the four standard Lorentzian symmetric matrices.
Proposition.
Let A and B be square matrices
which
are symmetric
with
respect to a Lorentzian inner product. If, with respect to a pseudo-orthonormal
basis.
'1
1
0
2
?J
lco
*i4
lJks
and
B
commutes
with
A
then
/I
b
0
ft
:
ci
0
dh
0
r>
c*.
dlj
where
0.7
drtii s krxkr.
Proposition.
respect to a
Let A and B be square matrices which
are symmetric
with
Lorentzian inner product. If, with respect to a pseudo-orthonormal
basis,
2
0
1
0
2
0
0
1
X
A=
M*
hhi
^≫-4s
and B commutes
with A, then
Isometric Immersions
b
I 1
-
Ci
0
0
37
of Lorentz Space
o... o
0
0
0
0
0
0
D
dh
.
0
Cka
:
0
6
6
t -1
d＼j
/7s
where
0.8
d＼jis krxkr.
Proposition.
Let A and B be square matrices which
are symmetric
respect to a Lorentzian inner product. If, with respect to an orthonomal
a
with
basis
-P
0
a
≪4>
/3^0
M*
and
B
commutes
with
A
then
r
0
s
r
d＼j
Q>ij /
where
drtji s krxkr.
Propositions 0.6, 0.7 and 0.8 can
a general symmetric
0.9
matrix and setting AB
Partially umbilical immersions.
and there is a globally denned
(i)
6 is everywhere
(ii)
P0=O
(iii) A,=Xld
be obtained by letting B have the form of
and BA
If /: M]-≫R
equal.
is an isometric immersion
normal vector field0 such that
non-zero
J=5fcO
then f(M'j) is contained inside
c≫-./V(M)＼
if
(0,6)>0,
38
Martin
0.10 Corollary.
immersed
A. Mag
id
H￡-A=?tz&￡)
if
LC￡[＼
if (O,0)= O.
(0,0)<O
If 0 in 0.9 Is the mean
minimally in S
and
curvature vector, then f(M") is
* or H sxl.(Such an immersion is called pseudo-
umbilicalby Chen and Yano [Y-C].)
Proofs.
Note firstthat the vector f(x) + 0/X is constant for allx in M.
this vector bv c we
Denoting
have
{6,0)
(f(x)-ctf(x)-c) = W,Oll) =
Because
r-
8 is parallel this is a constant, and so f{x) is contained in the sphere,
hyperbolic space or lightcone with center c.
Moreover, if 0 is the mean
curvature vector of f(M") in R , then 0 is parallel
to the position vector f{x) ―c and so f(M") is minimal
Here
^0
implies (6,6)^-0. Q. E. D.
1. Isometric Immersions
The
with
Parallel Second
Fundamental
classificationof these isometric immersions
facts about submanifolds
The
in either S "1 or H^l[＼
with F*a=0
forms.
Thus,
Rn^R"+k,
will follow from some general
and global, parallel normal
firststep is to classify one-dimensional submanifolds
fundamental
Forms:
fields.
with parallel second
the firstproposition deals with non-null curves which
have parallel mean
curvature vector.
1.1
Let x(s) be a non-null curve in an indefinite euclidean space
RJ
Proposition.
with parallel mean
(i)
curvature vector -q.
If 7]―Q then x(s) is a line segment.
(ii) If (y],yj)^Othen x(s) is part of a circle map.
(iii) If (77,
r/)=Q and ^0
(s,as2+6s + c). The
then a?(s)is a curve in R20ylor /2fa of the form
sl->
firstmap is denoted by U＼ the second by t//.
In case (iii)the metric vanishes on the second
coordinate. R2jA is imbedded
in R3J+1 by sending (x＼ x2)*-*(x2, x＼ x2).
Proof.
Let
x(s) be a curve in Rf
with s its arc-length parameter
and X(s) =
dx{s)jds―'.x (s) its unit tangent vector. If D denotes the flat connection in Rf
see that DsX(s) is normal to x(s) and so DsX(s)=a(X(s), X(s)). The
mean
we
curvature
Isometric Immersions
vector
7j{s)
= (X(s),X(s))a(X(s)),X(s)).
stant.
If 57=0
then
If k=￡0 set
and
Y(s) form
DsX(s) = 0 and
Because
orthonormai
r/s) is parallel (r/s), y/s))= &, a
con-
x(s) is a line segment.
Y(s) = Y/(s)lV＼k＼,
so that
an
39
of Lorentz Space
frame
Y(s) is a unit vector
along
x(s) such
normal
to x(s).
X(s)
that
DsX(s)=V＼k＼(X(s),X(s))Y(s)
/?
DtY(s) =
Vl*￨
X(s)
(ii)follows by 0.2 and inspection.
Now
suppose (y],r])
―0 and jy^O. By 0.2 x(s)cR2OA or R＼A, depending
length of X(s). Since x(s) is an immersed
on
the
submanifold
ar(s)
= (s,0(s)) or (0(s),s),
where
the metric vanishes on the second coordinate. Thus
i(s) = (l,<&')
and
a(X(s), X(s)) = (0, <j>") or (0, 0).
The
or (f//,l)
second
possibility
is totally
geodesic.
Since
Fta(X(s),X(s))=O
ft"=0
and 0(s)=as2 + bs + c. Q. E. D.
One
way
to obtain submanifolds
parallel second
fundamental
of an indefinite inner product
space
forms is to consider umbilical immersions.
with
We
will
find, besides the sphere and hyperbolic space, a third type of umbilical immersion,
and from
1.2
these a complete classificationof umbilical immersions.
Lemma.
Suppose/:
Mf->R}+k+*
Rrj+k is projection onto the firstn+k
is an isometric immersion.
If p: Rnj+k+s-
coordinates then pof: Mf -> R]+Ic is an iso-
metric immersion.
Proof.
Choose
x0
in
ordinates
(a?1, ･ ･ ･, xn).
vanishes
in
the
last
Lemma.
fundamental
s
and
let
U
be
a
coordinates.
dxm) -V*
Let/:
form.
coordinate
neighborhood
/(a?1, ･ ･ ･, xn) = (fl, ･ ■･,fn+k, h＼ ･ --,hs),
w'f*dx≫
M?-+R?+,m+s
of
where
x0
bean
5
=((^o/)*^'(^o/)*
isometric immersion
dzhv
53hp
2jM-1
co-
metric
)
3^
Jidxudx*
^v
Q. E. D.
with parallel second
If f(x＼ ■･･,xn) = (f＼･ ･･,fn+m, h＼---,hs) locally, then
dx%dx'dxk
with
the
Clearlv
)
･(
1.3
M
Let
40
Martin
Proof.
A. Magid
If / has parallel second fundamental
form
then
Pka(X, Y) = a(PzX, Y) + a(X, VZY) .
If p is projection onto the last s coordinates then
Vbp°a{X, Y)=poa{rzX,
Y)+poa(X,
FZY).
Replacing X, Y, Z by djdxJ, d＼dxl＼
djdx'1yields the lemma,
since, as is well
I/nmun
d2f
=*W.(lj￨r)
dxjdxk
1.4
Theorem.
mean
Let /: M"k -
curvature
vector
If (v, r})>0
then
/(M)cS￡ci?￡+1
If (v,v)<0
then
f(M)czHZ(zRKl
(iii) If (37,57)^0 then
this last
Both
and
affine spaces
Proof.
case
Rtf1
V
dxk/
+Rf be a complete
(ii)
In
d
＼
3
(
M=Rl
and
＼0XJ
f(M)(zRl;il
as an
have
in
the standard
depending
and by 0.10 the image
the
RnkH
by
^2+S"=*+i
xl*)+ b,z1+
■■･ +
(?/',-･ -,^M)i->(yra+1,y＼---,vn,yn+l).
metric.
Denote
this submanifold
into
which
on the sign of (30,y j). This mapping
by
US.
Ml
is iso-
is umbilical
is contained in a sphere or hyperbolic space.
If (t),7})=Qthe image
so that Ml
Assume
umbilic.
If (r?,
rj)i=Othen by 0.2 there is an Rl+1 or RUl
immersion
Q. E. D.
OX
umbilical immersion.
f(x＼ ･■･xn) = (x1, ■■-,xn,a(-ZU
is imbedded
metrically immersed,
)
d
ry^-O, so that / is not totally geodesic.
(i)
bnxn+c)
/
of Ml is in R^＼
By 1.2 /: Ml -*Rnk is an isometric
Rl.
1.3 allows 11s to determine all /zsuch that
f(x＼
As
in
1.3
we
･ --,xn)
= (x＼
■･ -, xn,h{x＼
■--,xn))
have
･(
i
dx
≫
dx')
= (O,---,O,A,y).
In addition, because / is umbilical,
d
-(
dx1
dxJ
)
＼dxi
)
a￡r ≫=<°.-■･.≫.*,).
It follows easily that h is of the desired form.
The
classificationof umbilical immersions
to determine
trivialnormal
Riemannian
immersions
Q. E. D.
allows us, following Walden
with parallelsecond fundamental
connections in an indefinite euclidean space.
[Wa],
forms and
Later this will allow
Isometric Immersions
us to classifyisometric immersions
41
of Lorentz Soace
Rn->R"rk
with
parallel second
fundamental
forms.
1.5
Theorem.
Riemannian
If /:
manifold
connection,
then
with
and
an
isometric
x C/'1X ■･ ■X ￡/'*X S
UlPCiRl1p'i2 as an umbilic
p―＼, "-,x
as an
umbilic
# = 1, -―,y
Hnr(zR"r*1
as an
umbilic
R^±R＼p+2±Rm<i+1±R?r+＼
r=l,
-―,z
Note
k=2x
immersion
fundamental
of umbilical
Sm≪cRm≪n
Proof.
of
form
and
a
complete
trivial normal
immersions.
ix ･･ ･ X Sm≫ X //*' X ･ ■･ X Hn*
+ y + z and j = x + z.
[Wa].
We
1.6
is
parallel second
/ is a product
f(Mn)=Rl°
Here
Mra->i2"f&
now
prove some lemmas.
Lemma.
Let/:
fundamental
form.
R7} ―>J? be
an isometric immersion
with
If (x＼ -■･,xn)is a global coordinate system
parallel second
with Fa.dj=O,i,j=
1,･ ･･ ,n, then a(dt,dj) is a global parallel normal field.
Proof.
1.7
This is an easy consequence
Lemma.
Let/:
Mj-+R￡
of the definitionof F*a=0.
be an isometric immersion.
If a
global normal
field d is parallel with respect to the normal connection, then Ag commutes
As for all normal
Proof.
Since
vectors B.
R1(X,Y)6
= FjtF^d-
F^F}t0-Fl1x,Y1e=0,this
Codazzi's equation and the non-degeneracy
1.8
Proposition.
fundamental
Proof.
mutes
If/:
Rn-^R^k
By
be shown
1.6 and
is an isometric immersion
with parallelsecond
that, for any two normal vectors f and 6, A^
and
Ao
the other hand, if p￡N°(x)then Ap =
with every shape operator.
sufficientbecause the normal
space is non-degenerate
space. In
this is
space equals N°(x)0 Nl(x).
If the metric restricted to N＼x) is degenerate
CiNHx).
from
connection, i.e.,i?i=0.
If the metric restricted to the first normal
entire normal
follows
1.7, if j is in the firstnormal space N＼x) then Ar com-
with every other shape operator. On
0 and Ap commutes
result
of the metric on M.
form, then it has trivialnormal
It must
commute.
with
then N°(x)+N1(x) is not the
this case there is a unique lightlike direction d in N＼x)
Choosing a lightlike d with (d,S)= l we
see that N°(x),N＼x) and 8 span
42
Martin A. Magid
the normal space.
tors commute.
Since At commutes
with itselfit is clear that all shape opera-
Q. E. D.
1.8 allows us, in conjunction with 1.5, to classifyisometric immersions
of Rn
infn J?'Hfcwith nnrallpi sprnnH ftmrlampnfsi fnrmc
1.9
Theorem.
fundamental
(i)
If/:
Rn->R?+k
is an Isometric immersion
parallel second
form, then / has one of the following forms:
/ is a product of circle maps
/=s,x
or
with
･ ■■xsjxld: R'x---
f=hxsix---xsjxld:
and a totallygeodesic map.
xRixRn-J-+S1x
RlxR'X
■■■xS'x/T-'c/Jr*
■･ ･ xRlxRn
j
l
-+ FP xSl x ･ ･ ■x S1x Rn-J-lr. R?+k .
(ii) / is a product of a flat umbilical map, circle maps, and a totallygeodesic
f=uxslx---xsjxld:
RmxRlx---xRxxRn-m-j
-> Um x S1x ･･ ･ xSl x /2B-m >c/2f+*
≪ is one of the flat umbilical immersions
1.10
Example.
x2―yz/2) with
where
Rl,2
Let /: R2^R'i2
of Theorem
1.4.
be the isometric immersion
metrics ( + , +) and ( + , +,0,0).
The
associated mapping
>Rtby {y＼y2,y*,yi)＼->(y＼yi,y＼y'i,y＼yi)
neither has
connection nor splitsinto a product of umbilics. This shows
the rprpivincrsnnrp must
1.11
Lemma.
isometric
If
d is a
immersion
of
with
global
R"
respect
with
parallel
parallel
normal
second
entries
Proof.
Let {x＼ ･ ･ -,xn} be a global coordinate
the immersion
has
into R＼,
trivial normal
that in 1.8 and 1.9
Kp nf Ip^cf T nrpnt?
constant
Since
(x, y)＼->(x,y, xy,
to a flat coordinate
vector
field associated
fundamental
system
system
on
such
form
to
then
an
Ae has
R?.
that Fd.dj=0, l<f,i<w.
F*a = 0
Q = A9(Ftid,)= Pai(A9dJ)= F3t(alidl+' ■■+anjdn) = {dialj)dl+ ･ ･ ■+(dianJ)dn .
Therefore,
diakj = Q for all i, j, k.
In order
to classify isometric
it is necessary
to determine
those
Q. E. D.
immersions
from
with
F*≪=0
from
R＼ to R＼ or R＼. For
Rf
to R?+2 or R"+'
this we
need
some
ovomnioc
1.12
Example.
Complex
C and let a+ib = K
circle of radius k. Fix a
＼ C2 can be identified with
non-zero number
R＼ by
c + id=n
sending (x+iy,u
in
+ iv) to
Isometric Immersions
(x,u,y,v).
The
ping {x,y)
with
of Lorentz Space
metric on R＼is ({x,u,y,v),(x,u,y,v)) = x2-＼-n2
―yz―v2. The
+x iy = z＼
parallel second
which comes from
> k(cosz,sin z)gC2 is an isometric immersion
fundamental
form.
Let /
be the complex
multiplication by i on C1. Then
basis ＼Tc.￡＼
for the normal
1.13
and
AjS =
L
"1
a
curvature vector y=al;-＼-bJt;
and Act-nue― ―h.
tained in SH^c^cF)
Example,
of R＼into R＼
structure on R＼
there is a natural orthonormal
＼
mean
map-
st>ace such that
a]
The
43
or HK-V'^TcF)
f : R＼-+ R＼ where
By
0.9, f'(Rf) is con-
or LC;＼.
R＼ has the standard metric.
V 2f'(x, y)―((l+c)siny, (l+c)cosy, (l-c)s'my, (I―c) cosy)
+ (x + cy)(―cos y,sin y,cos y, ―sin y),
This mapping
has, with
space {37,77},
where
respect to a
v is the mean
pseudo-orthonormal
cgR .
basis for the normal
curvature vector, shape operators
m; a - 4; a
The basis for T{Rf) is {d/dx.d/dy}. Since 4*_c,= /2,f'(R＼)is contained in S23(V-2c)
or Hi( ― V2c) or LCV
The
shape operators associated to the immersions
sphere or hyperbolic space have non-simple real eigenvalues and
Thus
by
the fundamental
theorem
into the
non-zero trace.
of surfaces they are the unique immersions
with these shaoe ooerators.
1.14
curve
Example.
x(v) with
5-scroll over the null cubic C in R＼([G 1]). In R＼ take a
null
a null frame, that is, a set of vector fieldsA(v), B{v), C(v) such
that t(v)= A(v), (A,A) = (B,B) = (A,Q
satisfy the following system
= (B,Q
= O
and
(A,B) = (C,Q
= 1. If these
of equations
A＼v)= ki{v)A(v)+ k2(v)C(v)
B(v) = k1(v)B(v)
C(v)=-ki(v)B(v)
then
fr(u,v)―x(v)-ruB(v) is a Lorentz
If ^!=0and^2^l
then y4= 0 and the curve is called the null cubic C. ([B],p. 240).
In this case we have
1.14'
Example.
surface in Rl called a Z?-scrollover x(v).
the 5-scroll over the null cubic C.
^'-scrollover the null cubic C
1.14. with obvious modifications.
in R＼. This
is the same
as
44
Martin A. Mag id
1.15
Theorem.
fundamental
(i)
If /: R＼-> R＼ is an isometric immersion
form
/is
with
parallel second
then, up to a rigid motion, / has one of the following forms.
totallygeodesic
(ii) / is a product of one-dimensional maps
(iii) /: R＼--+K＼clR＼
is a 5-scroll over the null cubic C.
Note:
In case (iii)the mean
curvature
vector is zero, even
though the im
mersion is not totally geodesic.
1.16
Lemma.
If/:
R)->R)
allel second fundamental
or R)+l j = 0,1 Is an isometric immersion
form
and the mean
normal
space has dimension less than two.
Proof
of 1.16:
fi, f2 be an
Assume
that the firstnormal
orthonormal
canonical form.
basis of N{x).
Since Ah commutes with
Lo
Ah
01
-a A
it
-b＼
IS
or
Case
of 1.15:
space is two-dimensional
the first
and
let
Using 0.4 put A$1 into the appropriate
There
o
ro 11
Lo oj
/n
/3 oj
of the form
or
i-o
6-1
lo
ol
calculation of a on the basis of TX{M)
at most one-dimensional.
Proof
then
Thus
Lo
A
curvature vector ^=0
with par-
r
°
or
L-r
shows
]
.
that the firstnormal
spaces are
Q. E. D.
are two cases to consider 7?―0 and wi=Q.
I. 7/=0. By 1.16 the firstnormal
space is zero or one-dimensional.
If it
is zero-dimensional, then / is totally geodesic. If it is one-dimensional, let f span
the first normal
space.
Then
by
0.2 f{R＼)c.R＼and As has rank one and trace
zero, so
A
[G 1] shows
Case
that this is the 5-scroll over the null cubic.
II. rj-t-0.Put An into one of the three possible canonical forms.
We
will
see that Ar) can only be diagonal. Let r< be the unit vector in the direction of rt
and choose another unit normal
vector ? such that fir;.
Isometric Immersions
H
A, =
45
since An and At commute.
But fx= 1/2 trace At
＼i
= (￡,?)=<>.
A(
of Lorentz Space
then ^4?
If
{/,/} is the pseudo-orthonormal
are represented,
then
o=(A,,
the Gauss
equation
basis with respect to which Av and
yields
/,i)(Av, I,i)-(Av, i,iy+iAi,
ixaJ,
i)-(A(i, iy
= ―(const. 2.Y.
This
implies
that
Similarly,
/I―0,
if Av. =
contradicting
＼
then
L―b
b ―c
37^0.
U
＼
0
fl
I
c
0
At =
.
a＼
The
Gauss
equation
gives
a―
O contradicting ry^O.
If Av is diagonalizable with distinct eigenvalues, then all other A( are diagonalizable and 0.3 shows
that /: R＼->R＼ can be decomposed
/,x/2: R[xR'-*R＼xR＼
as
or
R＼xR2
or
R＼xR＼
Each ft has parallel mean
curvature vector, so by 1.1
fl(R＼)=S[;URi)=Ri
or
/i(J*{)=SI;/I(J21)= S1
or
A(Ri)=R＼;A(Ri)=Si.
If An―ll% then there is a
operator A
Once
of this minimal
again, the Gauss
final case the map
1.17
Theorem.
fundamental
then, up
/is
(ii)
/ is a product
and
(v)
/ is as in 1.13
/: R＼-+R＼AcR＼
+ ey + g.
k^O,
and
must
be diagonalizable. In
this
Q. E. D.
is an
isometric
immersion
to a rigid motion, / has one
with
parallel second
of the following
forms.
maps
circle, as in 1.12
/ is a Z?-scrQll over
The
shape
can, a priori,be one of the three types.
that A
of one-dimensional
(iv)
Proof.
of R＼ into S^(c). The
totally geodesic
(iii) / is a complex
(vi)
immersion
If /: R＼-*R＼
form,
immersion
equation shows
splitsas above.
(i)
cy2+dx
minimal
the null cubic, as in 1.14, 1.14'
where
(x, y)＼-+{q(x,y), x, y, q(x, y)) and q(x, y) ―axi+bxy
+
rjis lightlike.
proof is divided
case III n=0.
into three
cases:
case
I (■//,
9)--￡(),
c ase
II (rp rj)= O
46
Martin
Case I. {rj,> ?)^0. Suppose
A. Magid
firstthat ATlis diagonal. If the eigenvalues
distinct then, as before, the map
splitsinto a product of one-dimensional maps,
giving (ii). If the eigenvalues are equal, say to /, then
into S＼or H＼. By examining
the Gauss
A of this minimal immersion
can only be of the form
respect
curvature
to an orthonormal
c>0
then
basis {e,f}
0 = f + (-/32/-l)
= c'+ /32. Both
occur.
2>0
and
c = 0 if i<0.
We
1.12.
We
^
= ^
of
/
see that
= 17^=^
to
the
normal
metrics, bundle
immersions,
bundle
imply
have
have
and
motion
a non-simple
Av =
This isometric immersion
r
≪
―
/:
curvature
let c = l/VT,
the given
Nf->Nfl,
from
from
an isometry,
operators.
fii
a]
Assume
C _3
a=￡Q and /3^=0. Trace
space so that {^,^} forms
/ and
By
one image
the
which
preserves
onto
if
the /' in
and
normal
the uniqueness
e<0
d=0
0=
bundle
bundle
theorem
for
the other.
If it did,
the
Gauss
equation
then
o
-r
A^＼
a complex
circle
that
[°o
or
to
ri
oj
(2)
1
0
]or (3)[_
0
/31
Oj
/3
A, = 0 because (yj,r ?)= 0. Choosing
a pseudo-orthonormal
basis, and
3y in the normal
using the fact that
we see that
+-"
If (1) holds the map
G,-J≪(2>
E
a≫*> l:
0
splitsand the possibilitiesh ave been classified.If (2) holds,
the isometric immersion
is equivalent to 1.13. If case (3) holds y = 0 and the map
is equivalent to a complex
must
maps:
is also rigidly equivalent
Case II. (^,yj)=O and ^0.
We
two
real eigenvalue.
L-(3
trace A*=2,
has constant
constant
that (rj,-n)=0.
If
where
If H?
If SI has
{r]f,Vf)= {rlfl,rJf,),
O^/M5?/-.3?/-)
a map
shape
(/,/)=―1.
In fact, in 1.12
of /', covering
connections,
there is a rigid
Av cannot
would
with
have
A,f = Ari/, Aj;f = A^r,
so that we
minimally
Oj
0 = cr-[(-/32)/-l] = c-/S2.
then
d=l/V'-/(,
R＼ immerses
equations, we see that the shape operator
'"[.-/!
with
are
circle.
also consider what
happens
if /!,= (). This implies that the first
normal space is one-dimensional and lightlike,so f{RX)C.K＼A. 1.3 shows
that
Isometric
Immersions
of Lorentz
Space
47
f(x, y) = (q(x, y), x, y, q(x, y))
and
q{x, y) = axz + bxy + cx2+dx
Case
III. rj―O.
By 1.16
is not totally geodesic.
is contained
ro
Let
the first normal
0 span
in Rl or R＼ and
A9
+ ey + g.
space is one-dimensional
the first normal
must
have
rank
space.
one
and
If (0,0)^0
trace
zero.
if the map
then
f(Rf)
Thus
As =
in
. In
Ae = 0 and
For
this case
the map
the immersion
falls into (vi).
completeness
to R*
with
1.18
Theorem.
we
If/:
to 1.14 or 1.14'.
fundamental
R^^-R*
If (0,0)―Q
then
Q. E. D.
state the classification of isometric
parallel second
fundamental
is equivalent
immersions
from
R2
form.
is an isometric immersion
with
parallel second
form, then / is either totally geodesic or a product of one-dimensional
maps.
2. The K-Dimensional
Here
we classify isometric immersions
second fundamental
in
Classification.
forms.
All such
maps
from
R% into Rln'2or R"+2 with parallel
break into two
the first class are quadratic in nature, like the flat umbilical immersion
lightlike mean
curvature vector in 1.4. Immersions
ducts of the identity map
or 1.18 or the B-D
2.1
Example.
B-D
and one of the two-dimensional immersions
scrollimmersion
A, B
Consider a null curve x(v) in
x(v) such
and E=―bB―D.
x(v) is a quintic Q.
that x(v)= A(v),
unit timelike, D, E unit spacelike, C,D,E±A,B,
and C, D, E are mutually orthogonal. In addition, suppose
C+E,
in 1.15, 1.17
of R＼into R＼ given below.
scroll over the null quintic Q.
are lightlike, C
with
in the second class are pro-
R＼ with a frame A(v), B(y), C(v), D(v) and E{v) along
t)
classes. Immersions
(A,B) = 1
A = bE, B=0,
C――D,
Note that the fifthderivative of A is zero, so that
Let f(u,v,w) ―x(v)+ uB(v) + wD(v).
mersion of Rl into R＼ with {3/3w,d/dv,ajdw) forming
This is an isometric im-
a pseudo-orthonormal
R＼. If f=―w>jB(y)+ C(v) and fi = ―wB(v) + E(v) then {￡,
f 1} form
an
basis of
orthonormal
basis of the normal space, Pf = 0 = F1^1
A(=
"0
o
r
b
1"
0
0
o
0
0
.0
1
0.
1
0.
with respect to {djdu,d/dv,djdw}.
2.2
Lemivia. Let /: M? ->R]Tk
be an isometric immersion
with
P*a=0
and
a
48
Martin
definite first normal
space. If x^M
A. Magid
and ￡ is a parallel normal fieldin a neigh-
borhood of xo such that A=xo has a complex
neighborhood
Proof.
eigenvalue, then / decomposes,
in a
of x0, as /,x/t: M^xM^-^R^xR'lf.
Let a(x)±ifi(x),Xx(x),*■･, Am(x) be the distinct eigenvalues of A^ at x.
First we
show
that h{x) is constant and that
T* = lXeTx(M):
A(xX=hX}
is a parallel distributionon M".
Let XeT＼
ZzT{M).
Then
O = FziAsX)-At(FzX)
= Vz(XX)-Ae(VzX)
= {Z＼)X+W2X)-A*{VZX)
^(zx)x+x(Pzxy
where (
)･*■
denotes the orthogonal complement
= k(FzX).
let Cx-(Tp)Ln
YqC
(X,FZY)
mute
=0
of T1. Thus
and
･ ･･ n(T>
)1. C is also a parallel distribution. In fact,let
Xe T* c･ ･ ■c TV
so FzYgC.
Therefore
Then
M? = M?xMn-t
0 = Z(X, F) = (FZX, Y) + (X, FZY) =
locally. Because
for all normal vectors 0, each A≪ has the same
whenever
by 0.3.
Let
(ZX) = § and AS(FZX)
It follows that each T^ is parallel and each lk is constant.
Now
ZgTM,
M(Fzxy)
form
A< and Ag com-
by 0.8 and so a(X, Y)
X is tangent to M? and Y is tangent to M
2.
The
theorem
follows
Q.E. D.
V and
W
be vector spaces. For a bilinear map
a: Vx
V-^-W
define the
relative nullity space of a, N(a), by
N(a) = {XeV:
a(X, Y) = 0 for all Y in V].
a is said to be flat with respect to an inner product (, ) on W
if
(a(X, Z), a(Y, W)) = {a{X, W), a(Y, Z))
for all X
2.3
Proposition.
(i)
W
Y, Z, W in F.
If a:Vx
(ii) If a: Vx
then have
([Mo 2])
V
then dim iV(a)>dim
and dimW=2
We
>W
is flatwith respect to a positive definiteinner product on
V-dimW.
V'-*W
is flat with respect to a Lorentzian inner product on W
then either dim N(a)>6＼mV-dimW
or a(X, Y) is null for all X, Y
in V.
Note:
This
will be referred to as Moore's nullity result
Isometric Immersions
2.4
Theorem.
fundamental
If/:
form
R" ->Rfi2 is an isometric immersion
We
Case
f=AxId:
R＼xRn-z-+R*xir-2
f=IdxA:
R?-*xR2-+R'r*xR*
maps
I. j?=0.
We
will see below
the first normal
Now
Suppose
2
or
consider the two cases ??=0 and r/^0.
zero canonical form.
-+RtxRn
parallel second
of 1.15 or 1.18.
that in this case the firstnormal
0- or 1-dimensional. If it is 1-dimensional, then
spans
with
then
where /i is one of the two-dimensional
Proof.
49
of Lorentz Space
space then
Ae
/QR?)cjRr+1
space is
non-trivially. If 0
can be put into the only rank one, trace
It is clear then that / can be decomposed
as fiXld:
R{xRn
z
where fx is the B-scroll over the null cubic.
it is shown
that the first normal
space
that it is two-dimensional and let {d1}
cannot
be
two-dimensional.
be an orthonormal basis of the
firstnormal space.
We
will assume
As can be put into one of the canonical forms and then show
that the first normal space is not two-dimensional.
If At were
diagonal with respect to an orthonormal basis, then by minimality
and 2.3 it has at most two non-zero eigenvalues ±/tand the eigenspaces are onedimensional.
The
same
reasoning shows that this is also true for A^,
the eigenspaces are the same
and in fact
as those of As. If this is true, then the firstnormal
space is one-dimensional.
If A? had a non-simple eigenvalue of multiplicity2 then combining
imality and the fact that Ae and Ax
0
1
0
0
commute,
and
A< =
2.3, min-
we have
0
b
0
0
0
0
0
0
･･･CjQ
"-0
0
0
A^ ―
0≫_2
On-Z
0
6
with
respect
to a pseudo-orthonormal
basis
{/,/,eu ･ ･･, en
6 + 6?1, a(ej, ej)―0 and
a(ej, /)= cJ-cJ-. Flatness
the first normal
is one-dimensional.
If At
space
has a complex
eigenvalue
then
A^
6
of a gives
does
z}. This
gives
≪(/,/)=
0 = (c/f-＼c/s1) ―c} so that
also and
so
50
Martin
0
/?
B
0
A. Magid
o
r
-r
A,=
and
o
A;±
On-2
The
firstnormal
On-2
space has dimension one.
If As had a non-simple eigenvalue of multiplicity3 we see that
0
0
1
0
b
c
0
0
0
0
0
0
0
1
0
0
c
0
JL =
and
.4fi =
On-,
0K_3
with respect to a pseudo-orthonormal
basis {l,!,e,eu ･･■,en_3}.Here
a(i,e)=￡+ d;1
and a(e,e)= 0 so that l+c2 = 0, which is impossible.
Case
II. r;=￡0.Here
we
show
that At/ is diagonalizable. Choose rf to be a
unit vector in the direction of t]and y/1 to be a unit vector perpendicular to rj.
Suppose
firstthat An. is of the form
X
t
0
X
X
Ar
*=￡0
･^ra-2
with
respect
to a pseudo-orthonormal
we
see that either Ai^O
We
have
spanned
gives
each
Therefore
and
≪(/,/)= 0 and
by
/ and
and
//= 0 = l
possibly
one
a(ejo,ejo)= /ljo7}',
a(eja,l)=0
so AT. cannot
have
basis
or A=0
≪(/,l ) = ty'+nr)1-.
/ shows
c/ = 0 and
all Aj=O
a non-simple
The
Thus
and
exactly
Gauss
However,
one
equation
^0=￡0 for some
djdo^0.
and
{/, /,eu ■■-,en-%}. By
applied
;0.
result,
to the
Moore's
trace A^=0
a(!J) = trf'+ by]L. From
eigenvalue
Moore's
ZJoi=O.
forces
this we
of multiplicity two.
plane
result then
djojo
O.
get ^0 = 0
Isometric Immersions
X
Suppose
Aj,
0
c
0
a
0
0
c
X
of Lorentz Space
Then
2.3
implies
51
X ― Q ―Xj, contradicting
^1
AH
If An had a complex
dimensional argument
_S
eigenvalue, then by
shows
2.2 the
map
splits, but the two-
this splittingcan't occur.
Finally, if An is diagonalizable,it has at most two non-zero entries, and the
map
splits into a product
dimensional maps.
of the identity map
and
one
of the possible two-
Q. E. D.
If the normal
space is Lorentzian, then there are more
possibilities.
2.5 Theorem. If/: R?-+Rlt+2 is an isometric immersion
fundamental form, then / is one of the following.
(i) f=flXId:
RlxRn-z->mxRn
(ii) f=f1xld:
Rf
with parallelsecond
2
2xR2->RT2xRt
where /, is one of the two-dimensional
RlxRn-3->RlxRn-s
where f, is the B-D scrollover the null
maps.
(Hi) f^f.xld:
quintic.
(iv) / is a quadratic map:
(x＼･■-,xn)i->(q(x),x1,---,xn,q{x)) with
q(x＼■
･･,xn)= j＼aijxixj
+ J＼biXl+c
.
Proof.
We
consider
three
cases.
Case
I:
0?,j?)=￡0,Case
II: (9,37)=0, rj^O,
Case
III: ry= 0.
Case
I. We
show
that An cannot have any non-simple real roots, and that if
it has simple roots it splits as in (i) and (ii).
Let rf be a unit vector in the direction of rjand let 3?xbe a vector orthogonal
to v satisfying (≫',
v')+(vL,vL)=0.
(X
0
Define o'―(v',rf)a nd r=(≫■■■,
w1). If
t
' ft
b
X
0
ft
A,,=
h
＼
with respect to a pseudo-orthonormal
An-2
Ci---cTC_2>
"
＼
Cn-2
basis {/, /,eu ･ ･･, en_2} with
>
^^0,
then
Moore's
52
Martin
result
and
says
one
duct
of
Since
has
or
the
trace
rank
^=^0
identity
≪(/,/)= 0
that
so
Ar
x^O
and
<2
and
map
and
and
all dij = O.
cannot
^4,/
rank
map
same
a
trace.
Then
then
non-simple
that
/?= 0 = t-1= ■･ ･ = cn-z
would
two-dimensional
r//2 = 0 implying
Then
either
/
split into
map,
/j = 0.
By
a(eio, /) = 0, a(eh, eJo)= /.Jo<x7]'
and
would
real
imply
^0=0,
eigenvalue
of
so
2.3
which
assume
and
prox = 0.
the
≪(/,i) = ah/
cannot
multiplicity
the
fact
+ br7]L
happen.
3, non-zero
trace
<2.
If Av>
the
have
id
If ,2^0
a non-existent
Q = (2JoG7)/,atrf + bT7]L) = AJoat which
and
Mag
non-zero
t/jy= O = /!i= Ci.
≪(/,/) = ryr/,
AiJ-=0
A.
is diagonalizable
splits
is
true
into
a
when
then,
product
A*,
has
as
of
a
above,
it has
the identity
complex
map
one
and
or
two
non-zero
entries
a two-dimensional
map.
and
The
eigenvalue.
Case II. >y^0 and (rhjy)=O.
If
^4,^0 and has simple eigenvalues the map splits into the product
identity map and an appropriate two-dimensional map.
of the
When
X
1
0
X
A,=
where
fx^O
This
we
A*=
9 is a lightlike normal
then
known
then
h
Cj ―Q―dij and
two-dimensional
for
some
would
have
a(l,i)
Av
On
(y,r))= l we
to a product
the other
possibly
djojo,
contradicting
has a non-simple
hand,
cjo=￡Q, while
r)+ br],a(eh,l) = cjoy]and
O = (^+br], dJojl/j)
When
and
satisfying
reduces
immersion.
j0, djgj0^0
give
vector
the map
eigenvalue
that
/L=O = /Lj. If
/.≪
cannot
map
be zero.
all the other
and
a
If /j= 0
entries are 0.
<x(eh,eJo)= dJohrj. Because
trace
of
see
of the identity
a is flat
A^O.
multiplicity
three
then
A* is of the
form
ft
b
0
d
･ ' -Cn-2
0
0
I1
Ct
dij
Cn-%
Using the arguments
trace A^O.
above, we
have
pt―Q―dij,contradicting once
again that
Isometric Immersions
of Lorentz Space
53
If Ay ―Q it is easy to see using 1.3 that / is quadratic in nature and
we
get
(iv).
Case
III. rj^O.
If the first normal space is one-dimensional
and
definite then the immersion
is the product of the identity and the ^-scroll. If the firstnormal
dimensional and lightlike,the immersion
If the normal
space is two-dimensional, arguments
that if {-,I1} is an
orthonormal
space is one-
is quadratic.
similar to those above show
basis of the first normal space then the only
canonical form possible for A- and A;1 is
0
0
1
0
b
1
0
0
0
0
0
0
0
10
0
10
^v
At =
and / splitsas the product of the identity map
and the B-D
scroll over Q.
Q. E. D.
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