A NNALES DE LA FACULTÉ DES SCIENCES DE T OULOUSE
L UC V RANCKEN
Affine higher order parallel hypersurfaces
Annales de la faculté des sciences de Toulouse 5
esérie, tome 9, n
o3 (1988), p. 341-353
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- 341 -
Affine higher order parallel hypersurfaces
LUC
VRANCKEN(1)
Nous étudions des immersions équiaffines, non-dégénérées
d’une variete afhne
(M,
V) dans satisfaisant = 0. Nous prou-vons que ou bien M est une quadrique ou bien M est plat. Dans le second cas, nous obtenons que M est afhnement equivalente a un graphe d’une
fonction polynomiale de degre 3 ou moins.
Si la connection sur M est celle qui est induite canoniquement par l’im-
mersion, nous generalisons pour des dimensions plus grandes des résultats de K. NOMIZU et de U. PINKALL.
ABSTRACT. - We study non-degenerate equiaffine immersions of an affine manifold
(M, ~!)
into which satisfy V2h = 0 We prove that either M is a quadric or M is flat. In the second case, we obtain that M is afnnely equivalent to a graph immersion of. a polynomial of degree at most 3.If the connection on M is the canonically induced connection, we extend
a result by K. NOMIZU and U. PINKALL to higher dimensions and to higher
order derivatives of h. .
1. Introduction
In
[N - PJ2
K. NOMIZU and U. PINKALLclassify
the affine surfaces inR3 satisfying ~2
h = o. Moreprecisely, they
prove thefollowing
theorem.THEOREM. - Let
MZ
be anondegenerate surface
inR3.
. Let ’~7 be the inducedaffine
connection and let h be the secondfundamental form (affine metric). If ~2h
= 0 but ~h~ 0,
then M2 is congruent to an open subsetof
theCayley surface
z = xy +y3 by
anequiaffine transformation of R3
.Here,
wegive
somegeneralizations
of this theorem tohigher
dimensions.Also we will not restrict ourselves to the induced
(canonical)
affinenormal,
~1~ Faculteit Wetenschappen, Department Wiskunde B-3030 Leuven - Belgique
but we will also consider
general nondegenerate equiaffine
immersions.Definitions and
elementary properties
are described in section 2. For moredetails,
we refer the reader to[N]
and[N - P,1. Especially,
we will prove thefollowing
theorems.THEOREM 1. Let
(~~In, ®, ~~
-~ be anondegenerate equiaffine
immersion. Then(v2h)
= 0if
andonly if
oneof
the twofollowing
statements holds.
(i)
M is an openpart of
anondegenerate quadric
and ~ is theinduced connection.
(ii)
M isflat
and M iscongruent
under anequiaffine transfor-
mation
of
to thegraph
immersion =where F is a
polynomial
in x1, x2, ... , xnof degree
at most 9 andwith nonzero Hessian.
THEOREM 2. Let
M3
be anondegenerate hypersurface
inR4.
Let ’~ bethe induced
affine
connection and let h be the secondfundamental form. If
= 0 but ®h
~ 0,
thenM3
iscongruent
under anaffine transformation
to either one
of
thefollowing graph
immersions.THEOREM 3. Let
f
:®)
~ (Rn+1,D) be a convexequiaffine
immersion with second
fundamental form
h. Then = 0for
somek E
~1, ... , 4~ implies
that M ispart of
anellipsoid
and V coincides with the induced connection or M isaffinely equivalent
to thegraph
immersionof
a convexfunction.
THEOREM 4. Let be a
nondegenerate
convexhypersurface
inwith induced second
fundamental form
h. Then = 0for
k E~l, ... , 4~
irn,plies
that M ispart of
anondegenerate ellipsoid
orof
anondegenerate paraboloid.
2. Preliminaries
Let M be an n-dimensional manifold with an affine connection ~.
Furthermore,
let D denote the standard connection onBy
an affineimmersion
f
:(M, V) 2014~ D)
we mean an immersion for which there existslocally (i.e.
in aneighbourhood
of eachpoint)
a transversal vectorfield ~
such that for alltangent
vector fields X and Y to M thefollowing
formula holds
Clearly h
is asymmetric
bilinear form. If h isnondegenerate,
we say thatthe immersion is
nondegenerate
and if h ispositive
ornegative definite,
wesay that the immersion is convex. It is clear that these definitions do not
depend
on the choice of the transversal vectorfield £.
Let cv be the
parallel
volumeform ongiven by
thedeterminant, f
:(M, ~) ~ (Rn+1, D)
an affine immersionand 03BE
an associated transversalfield,
then we can define a volumeform 03B8 on Mby
However,
it is clearthat,
ingeneral,
this volumeform need not beparallel.
If this volumeform is
parallel,
we say that the immersionf
is anequiaffine
immersion. It is proven in
[N - P]i
that if M admits aparallel volumeform,
the associated transversal vector
field ~
can be chosen in such a way that thisparallel
volumeform coincides with the induced volumeform(i.e. ~
can be chosen such that the immersion becomes an
equiaffine immersion).
Furthermore,
it isimmediately
clear that if~1
and~2
are both associated transversal vector fields such that the immersion isequiaffine
then~1
=where c is a constant on M. Such vector fields are called affine normal vector fields. From now on, we will
always
work with affine normal vector fields.The associated bilinear form is then called the affine second fundamental form. From
[N - P]i
we also recall thefollowing
formulas forequiaffine
immersions.
where S is a
(1.1)
tensor field on M. We call S the affineShape
operator.The
equations
ofGAUSS,
CODAZZI and RICCI are thengiven by
and
where
(Vh)
is definedby
Finally,
weproof
thefollowing proposition
from[N - P]i,
which we will uselater on.
PROPOSITION 1. -
Suppose
thatf
:~03B8)
~ Rn+1 is anequiaffine
immersion with S _ 0. Then
f
isaffinely equivalent
to thegraph
immersionof
a certainfunction
F : l~’~ ----~ R.Proof .
-By assuming
a transversal vectorfield 03BE
to beequiaffine,
S _ 0implies
that =0,
is a constantparallel
vector field. Let H =be a
hyperplane
in which is transversal to~.
Let 7r : Rnbe the
projection along
the directionof ~
so that 7r of :
: ~~ --~ Rn is anaffine immersion with
image W,
an open subset ofThen,
we can finda differentiable function F : : ~l ~ --~ R such that = + Thus
f
isaffinely equivalent
to agraph
immersion.Now,
we will define the induced connection and the canonical affine normal. For moredetails,
see[N] .
Let Mn be ahypersurface
in Forany choice of transversal vector field
~,
we can define an affine connection V and a bilinear form hby (2.1).
Whether h isnondegenerate
or not does notdepend
on the choice of the transversal vector field. Therefore we call the immersionnondegenerate
if there exists a transversal vector field for which thecorresponding
h isnondegenerate.
In that case, h induces avolumeform, given by
itsdeterminant
which we will denoteby
wh.Then,
in[N],
it isproved
that there exists aunique
choiceof ~
such that(i)
=0,
where 8 is the volumeform definedby (2.2),
and
(ii)
the volumeform 03C9h = 03B8 isparallel
withrespect
to V.It is
immediately
clear that the immersion defined this way is indeed anequiaffine
immersion and V is called the induced connectionand ~
is calledthe canonical affine normal vector field.
3. Main results
Let T be a
(0, k)
tensor field on M and let R denote the curvature tensor on M.Then,
we define a(0,
k +2)
tensor field R T on Mby J!’ T(Xi,X2,...,X~+2)
="’T(R(,X1 ~ X2 ).~3, X4, ... , Xk~.2 ~ - I’(Xg, .~(X1, X2 )X4, X5, ... , Xk+2 ) ...
- T (X 3 ~ -x4 ~ ... ~ X 2 ~Xk-E-2 ~
.Then,
we can also defineT,
m > 2by
(7!- . . I’~(~’1, ...
= RZ’)~(~’1 ~ ... ,
LEMMA 1. - Let M be an
equiaffine
immersion with secondfundamental form
h. ThenProof.
. Let x1, x2, ... , , x~+~ betangent
vectors at apoint
p of M. Wecan extend these vectors
locally
to the vector fieldsX2 ,
... ,Xk+2
whichare
parallel
at thepoint
p. PutThen,
wefind, using
several times that is zero at thepoint
p thatLEMMA 2. Let M be an
equiaffine submanifold of Rn+1
with associated secondfundamental form
h. Then(~2kh)(x1, x2, ... , x2~+2)
= 0for
alltangent
vectors x~ , ... , x2k+2 to Mimplies
thatand that
Proof
.First, by applying
Lemma1,
we obtain thatand
Hence, by subtracting
these twoequations
andby applying
Lemma 1 oncemore, we find that
Now let us assume that
(R~ .
= ~.Then,
we know that for alltangent
vectors x2, ... , z2 k thefollowing equations
hold :and
Hence, by subtracting
these twoequations
andby applying
Lemma 1 oncemore, we find that
This proves the first part of this lemma. The
proof
of the second part issimilar, starting
from the fact that(~2kh) -
0implies
that alsoh) - ®.
.proof
of thefollowing
lemma iscompletely
similar to theproof
ofLemma 2.
LEMMA 3. Let M be an
equiaffine submanifold
with associated secondfundamental form
h. Thenx2k+3)
= 0for
alltangent
vectors x1, ... , x2k+3 to Mimplies
thatand that
Now,
we can start theproof
of the main theorems.Proof of
theorem 1 - Let p EM. Since h isnondegenerate,
there exists abasis e2,...,
en~
of such that =E=b=j,
where Ei E~-1,1~.
By Lemma
2~~2 h)
= 0implies
that l~ . h = 0.Hence,
we obtain for i~ j
that
Therefore ei , e2 , ... , en are
eigenvectors
of S. ThusS ei
=a&e=, ~i
E~1,2,...,n~.
But thenimplies
that S = AI.Now, (2.6)
becomesFrom
this, using
the connectedness of M and the fact that the dimension of M is at leasttwo,
we obtain that A is a constant on M.Let us now assume that
a ~
0.Then,
we have to prove that M is apart
of aquadric. Therefore,
due to ageneralisation
of the classical Berwald theorem([N - PJ3, [D - V]) ,
it is sufficient to prove that(Vh)
= 0 on M.In order to do so, we know from Lemma 2 that
(.R ~ ( V l~))
= 0. Thisimplies
for
mutually
different indicesi, j
and k thatAlso,
for different indices iand ,,
we obtain thatand that
Using linearity
and thesymmetry
of(~h~,
we then obtain that(Vh)
= 0.Hence M is an open
part
of anondegenerate quadric
and the connectioncoincides with the induced connection.
Let us now assume that A = 0.
By using
theproposition,
we may assume that the immersionf
isaffinely equivalent
to an openpart
of agraph
immersion.
Furthermore,
from theproof
of theproposition,
we see that wemay assume that
where the third condition can be realized
by
a translation and the fourth conditionimmediately
follows fromtaking
as the transversalhyperplane
inthe
proposition
thetangent plane
at theorigin.
Thenby
astraightforward computation
we find thatand
where ~i
= = and~ {1, 2, n} Using
theseequations,
itimmediately
follows from(Q~h~ _
0 that the fourth derivative of F must be zero. Hence F is apolynomial
ofdegree
less then orequal
to 3. Sincethe immersion is
nondegenerate,
F must also benondegenerate.
From theimposed conditions,
we also see that F has no constant and no linear term.This
completes the proof
of this theorem.Proof
ot theorem 2 . - We will use the same notations as in theproof
ofTheorem 1. Since
(Vh)
is notidentically
zero, we deduce from Theorem 1 thatf
isaffinely equivalent
with thefollowing graph.
where
Fi,
i E{2,3}
is ahomogeneous polynomial
in ofdegree
i,
o E dom~F2
+F3) and ç = ~0, 0, 0,1).
Since we assume the immersionto be
nondegenerate,
we know that det 0.Therefore, by
awell known
algebraic property,
we may assume thatF ~ (0) -
wherefi
E ~ -1,1 ~ .
Henceby,
if necessary,taking -~
as affinenormal,
we mayassume that either
or
It is then
clear,
that we canalways apply
a rotation likewhere
pi
+p2
=1,
withoutchanging
the form ofF2. By applying
such arotation in the
(a;i,
x2)-plane
and another one in the(x2
, x3)-plane,
we mayassume that F = F2 + F3 is
given by F(xi,
x2,x3)
=xi
+03B1x22
+03B2x2x3
+where since M is not a
quadric
not all the ai are zero.Notice
also,
that the last rotation didchange
the form ofF2.
Then the conditionthat ç
is the canonical affine normalimplies
that is anon-zero constant. Let A = det
Looking
at the term inxi
of A weimmediately
obtain that a4 = 0.Let us first assume that ai = 0.
Then,
we canapply
another rotation in the(x2,x3)-plane
such that also a2 = 0. But thenby looking
at the termsin
x2 and x33
we obtain that a4 . a6 =0 and a3 . a7 = 0.By,
if necessary,interchanging
x2 and x3 thisgives
threepossible
subcases.subcase 1 a3 = a4 = 0. In this case, the
problem
becomesessentially
2-dimensional and in the same way as in
[N - P]2
we obtain the desired result.subcase 2 a3 = a6 = 0. Then it follows from the term in xl and the term in x1x2 that either a4 = 0 or a = a5 = 0. If a4 =
0,
we canapply
subcase1 to obtain the solution.
Hence,
we may assume that a = a5 = 0. From the term inx3
we then deduce that a7 = 0.Therefore,
we know that F can be written aswhere since F is
nondegenerate (3 ~
0. If ~4 == 0 or a8 =0,
then we canapply
the affine transformationgiven by
to obtain the desired form.
However,
if both a4 and a8 are different from zero, we firstapply
the affine transformationgiven by
such that
Applying
the affine transformationgiven by
then
completes
theproof
in this case.subcase 9 a6 = a7 = 0.
Looking
at the term inx1 gives
a4 = 0. Thisreduces this case to the
previous
case.Now,
we may assume that 0. Then we know that a4 = 0. From the term inx3
we find that a2 . a8 = 0.Again,
we have to consider several subcases.subcas e 1 a2 = 0.
By looking
at the terms inx1, x21x2
andx21x3,
we obtainBut
then,
since A is a constant itimmediately
followsthat ai = 0.
Hence,
in this case, we obtain a contradiction.subcase 2 ag = 0 = a4 and
0 ~
a2.Then,
we findby explicitly calculating
the determinant A that F must have thefollowing
form.=
(1/2~(xi - x3
+(1/2~a~x2txmf- x3 - ~x2~2.
By applying
one after another thefollowing
affinetransformations,
weget
also in this case the desired form.First,
weapply
thefollowing
afhnetransformation. _
Up
to ascaling factor,
F has then thefollowing
form.Then
applying
the affine transformationgiven by
and a
rescaling completes
theproof.
Proof of
theorem 9.. - Let p E M. Since M is convex, we know thatthe second fundamental form is
positive
ornegative
definite. Since(2.7) holds,
we then know that hand S aresimultaneously diagonalisable,
i.e.there exists a basis e2, ... , of
TpM
such thath(ei, ej)
=~03B4ij
andSei
=aieg,
where E E{-1,1}, 03BBi
E R andi, j
E{1,2,...,n}.
Let us nowassume that
(Vkh)
=0,
k 4. In all these cases, we know that~~4h)
= 0.But
then,
from Lemma 2 it follows thatR2 . h
= 0.Therefore,
we obtainthat
From
this,
it follows that S =AI, A
E R. In the same way as in theproof
of Theorem
1,
it follows from(2.6)
that A is a constant on M. If A = 0 onM,
the theorem isproved.
Therefore,
we may assumethat ~ ~
0. In this case, we must prove that M is a part of anondegenerate quadric.
In order to do so,by
thegeneralized
BERWALD theorem
([N - P]3, [D - V]),
it is sufficient toproof
that(Vh)
= 0.If
(V3h)
=0,
we know from Lemma 3 that R . = 0. Fromthis,
wededuce
just
as in theproof
of Theorem 1 that(Vh)
= 0.If
( ~4 h )
=0,
we know from Lemma 2 thatR2
= 0.Now,
we takev, u E
TpM
such thath(v, v)
=h(u, u)
= E andh(u, v)
= 0.Then,
we havethat
and
Hence for all v,
(Vh)(v, v, v)
= 0. Since(Vh)
issymmetric,
linearization thenimplies
that(Vh)
= 0. Thiscompletes
theproof
of Theorem 3.Proof
ottheorem 4 .
.By
Theorem 3 we know that either M ispart
ofa
quadric,
in which case, theproof
of the theorem is obvious or S = 0 andM is
amnely equivalent
to thegraph
immersionwhere P is a convex
polynomial
in z2 , ... , X n ofdegree
at most 5. SinceS =
0,
P satisfies in aneighbourhood
of o thefollowing
differentialequation
where
Pij
is defined as in Theorem 1. Since P is apolynomial function,
itis then clear that we may assume that P is convex on the whole of Rn and satisfies this differential
equation
on the whole of Rn.Therefore,
we obtainby
a theorem of POGORELOV([P], [C - Y]) ,
which is ageneralisation
of thetheorem of CALABI
[C]
that M is part of anondegenerate paraboloid.
References
[C]
CALABI (E.).2014 Hypersurfaces with maximal affinely invariant area, American Journal of Mathematics, t. 104, 1982, p. 91-126.[C -
Y]
CHENG (S.Y.), YAU (S.T.). 2014 Complete Affine Hypersurfaces. Part I. The Com- pleteness of Affine Metrics, Communications on Pure and Applied Mathematics,t. Vol. XXXIX, 1986, p. 839-866.
[D - V]
DILLEN (F.), VRANCKEN (L.). 2014 Parallel and Locally Symmetric Hypersurfaces ofAffine Spaces, preprint.
- 353 -
[N]
NOMIZU (K.). - On completeness in affine differential geometry, Geometriae Dedi- cata, t. 20, 1986, p. 43-49.[N - P1]
NOMIZU (K.), PINKALL (U.). 2014 On the geometry of affine immersions, Math. Z.,t. 195, 1987, p. 165-178.
[N - P2]
NOMIZU (K.), PINKALL (U.).- Cayley Surfaces in Affine Differential Geometry, preprint.[N - P3]
NOMIEU (K.), PINKALL (U.). 2014 Cubic form theorem for affine immersions, Procee- dings Oberwolfach Conference on Affine Differential Geometry, November1986,Resultate der Mathematik, t. 13, 1988, p. 338-362.
[P]
POGORELOV (A.V.). 2014 On the improper affine hyperspheres, Geometriae Dedicata,t. 1, 1972, p. 33-46.
(Manuscrit reçu le 8 septembre