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THESIS

40°

7/

This is to certify that the

dissertation entitled

New Riemannian and Kaehlerian Curvature Invariants
and Strongly Minimal Submanifolds

presented by

Dragos-Bogdan Suceava

has been accepted towards fulfillment
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Ph . D . degree in Mathematics.

Mjor professor

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6/01 c:/ClRC/DateDue.965-p. 15

NEW RIEMANNIAN AND KAHLERIAN CURVATURE
INVARIANTS AND STRONGLY MINIMAL SUBMANIFOLDS

By

Dragos-Bogdan Suceava

A DISSERTATION

Submitted to
Michigan State University
in partial fulfillment of the requirements
for the degree of

DOCTOR OF PHILOSOPHY

Department of Mathematics
2002

ABSTRACT

NEW RIEMANNTAN AND KAI—ILERIAN CURVATURE INVARIANT S AND
STRONGLY MINIMAL SUBMANIFOLDS

By

Dragos-Bogdan Suceava

During the last decade, B.-Y.Chen's fundamental inequalities have been
investigated by many authors from various viewpoints. In Section 2 we provide
an alternate proof for Chen's fundamental inequality associated with classical
invariants. In Subsection 2.5, we obtain an inequality for warped product
manifolds as a consequence of the previous study. Section 3 is devoted to the
study of applications of Chen's fundamental inequality. It is well-known that the
classical obstruction to minimal isometric immersions into Euclidean space
is Ric 2 O. In this section, we present a method to construct examples of
Riemannian manifolds with Ric < O which don't admit any minimal isometric
immersion into Euclidean spaces for any codimension. The study of the relations
between curvature invariants and the topology of the manifold yields in section 4
a Myers type theorem for almost Hermitian manifolds. Chen's fundamental
inequality for Kahler submanifolds in complex space forms is discussed in
Section 5. We provide an extension of the inequality and provide
characterizations of strongly minimal complex surfaces in the complex three
dimensional space. The last section is dedicated to the study of strong minimality

through examples.

iii

To my family

NEW RIEMANNIAN AND KAHLERIAN CURVATURE INVARIANTS AND
STRONGLY MINIMAL SUBMANIFOLDS

Contents

1. Introduction: Chen’s Fundamental Inequalities 1

2. Chen’s Fundamental Inequality with Classical Invariants 8
2.1. The Hypersurface Case 8
2.2. The General Codimension Case 1
2.3. A Remark on Totally Umbilical Points 1
2.4. A Conformal Invariant Related to Chen’s Fundamental Inequality

with Classical Invariants 14
2.4.1. Geometric Inequalities on Compact Submanifolds 18
2.4.2. The Noncompact Case 21
2.4.3 Examples 23
2.5. A Fundamental Inequality for Warped Product Manifolds 25
3. Applications of Chen’s Fundamental Inequality, the General Case 29
3.1. Warped Product of Hyperbolic Planes 29
3.2. Multiwarped Product Spaces 33
4. Curvature and Topology: A Myers Type Theorem for Almost Hermitian
Manifolds 38
5. Chen’s Fundamental Inequality for Complex Submanifolds 45
5.1. New Kahlerian Invariants 45
5.2. Strongly Minimal Submanifolds 46
5.3. An Extension of B.-Y.Chen’s Fundamental Inequality with
K'ahlerian Invariants 48
5.4. Characterizations of Strongly Minimal Surfaces in the Complex Three
Dimensional Space 54
6. A Study of Strong Minimality Through Examples 69
6.1. 62 = 0 on degree two complex surfaces 69
6.2. 8; = 0 on degree three complex surfaces 80

Bibliography 86

1 Introduction to Chen’s Fundamental Inequali-
ties

In the geometry of submanifolds, the following problem is fundamental:

Establish simple relationships between the main intrinsic invariants and the main
extrinsic invariants of the submanifolds.

The first result in this respect was Gauss’ Theorema Egregium, which in 1827
asserted that the Gaussian curvature is an intrinsic invariant. Also concerning this
fundamental problem, Chen’s fundamental inequalities obtained in [14, 20] are the
starting point for many recent papers done by various geometers during the last ten
years or so, as for example one may see in [18], [29], [30], [31], [32], [44], [58], [59],
[60].

We will discuss in this first section the context and the problems we have worked
on in the present dissertation.

Let h denote the second fundamental form of an isometric immersion of a Rie-
mannian n-manifold M ” into an ambient Riemannian space 1171““. Then the mean
curvature vector field is H = (1 / n) trace h. The immersion is called minimal if its
mean curvature vector field H vanishes identically.

The following is a classical basic problem in Riemannian geometry.

Problem: When does a given Riemannian manifold M admit ( or does not admit)

a minimal immersion into a Euclidean space of arbitrary dimension .9

For a minimal submanifold M in a Euclidean space the Gauss equation implies

that the Ricci tensor of the minimal submanifold satisfies
Ri,c(X X) =—Z|h(X ,e,-) )[2 < o (1)

where e1, ..., en is an orthonormal local frame field on I”. This gives rise to the first
solution to the Problem above; namely, the Ricci tensor of a minimal submanifold M
of a Euclidean space is negative semi-definite, and a Ricci—flat minimal submanifold
of a Euclidean space is totally geodesic.

The second solution to the Problem mentioned above was obtained by B.Y. Chen
as an immediate application of his fundamental inequality and his invariants [14, 20].
Based on these facts, it is interesting to construct precise examples of Riemannian
manifolds with Ric < 0, but which do not admit any minimal isometric immersion
into a Euclidean space for any codimension.

Let M n be a Riemannian n-manifold. For any orthonormal basis e1, ..., en of the
tangent space T pM , the scalar curvature is defined to be scal(p) 2 2K}. sec(e,~ A e,-).
For any r-dimensional subspace of T pM denoted L with orthonormal basis e1, ..., er

one may define

scal(L)= 2: sec (6, /\ ej). (2)

1_<_i<j<r
In [20], Chen considered the finite set S (n) of k-tuples (n1, ..., nk) with k _>_ O which
satisfy the conditions: n1 < n, n, 2 2, and n1 + + nk S n. For each (n1, ...,nk) E

S (n) he introduced the following Riemannian invariants:

6(n1, ...,nk)(p) = scal(p) — inf {scal(L1) + + scal(Lk)}(p), (3)

where the infimum is taken for all possible choices of orthogonal subspaces L1, ..., Lk,
satisfying nj =dim Lj, (j = 1, ...,k). Note that the Chen invariant with k = O is
nothing but the scalar curvature.

As in [20], we put,

_ 712(Tl-i-k—l—Z’nj)
_ 2(Tl+k—an) ,

 

b(n1, ...,m,) = %{(n(n — 1) — Eng-(n3- — 1)}

3:1
Chen’s fundamental inequalities obtained in [20] can be stated as follows:
Theorem 1.1 For any n-dimensional submanifold 114 of a Riemannian space form

Rn+m(e) of constant sectional curvature e and for any k-tuple (n1, ...,m,) E S(n), we

have
6(n1, ...,m,) S C(n1,...,nk)|H|2 + b(n1, ...,nk)e. (4)

The equality case of the inequality above holds at a point p E M if and only if there
exists an orthonormal basis e1, ...,en+m at p such that the shape operators of M in
R"+m(e) at p take the following forms: S, = diag(A’1‘, ...,/12,11“ ...,/1..) for r = n +

1, ...,m, where each A; is a symmetric n,- x n,- submatriz such that
trace(A[) = = trace(A’,:.) 2 u,.

The invariants 6(n1, . .. ,nk) became known as the Chen invariants in literature

and inequality (1.4) as Chen’s fundamental inequality. Chen’s fundamental inequality

has many nice applications; for example, one has the following important result as

an immediate consequence.

Theorem 1.2 Let M be a Riemannian n-manifold. If there exists a k-tuple (n1, ..., nk)

in S(n) and a point p E M such that

6(n1,nk)(p)>%.({n n —1)—an(nj —-1)}e, (5)

then M admits no minimal isometric immersion into any Riemannian space form
Rm(e) with arbitrary codimension.

In particular, if 6(n1, ...,nk)(p) > 0 at a point for some k-tuple (n1, ...,m,) E S(n),
then M admits no minimal isometric immersion into any Euclidean space for any

codimension.

We will use the second part of this theorem in our applications. Namely, in the
context of Theorem 1.2, we are interested in the following problem.

Are there examples of manifolds with Ric < 0 , but which have some positive Chen
invariant .9

This is similar to a classical problem 4 mentioned in Peter Petersen’s list of prob-
lems in [2]:

Scalar versus Ricci curvature problem. Are there examples of simply con-
nected manifolds which admit Riemannian metrics of positive scalar curvature, but
do not admit Riemannian metrics of positive Ricci curvature .9

We will solve the Ricci vs. Chen invariant problem in the subsections 3.1 and 3.2.

As far as we know, the scalar vs. Ricci curvature problem is still open.

4

In the section 2 we study Chen’s fundamental inequality associated with classical
invariants, and also consider a few of its algebraic implications. Specifically, we will

provide an alternate proof of the following (see [17]).

Theorem 1.3 Let f : M " —> Rn+m(c) be an isometric immersion of a Riemannian
n-manifold A!" with normalized scalar curvature p into an (n + m)-dimensional Rie-

mannian space form Rn+m(c) of sectional curvature 6. Then
p S IHI2 + 6- (6)
The equality holds at a point p E [W if and only if p is totally umbilical point.

The last two sections of the present work are dedicated to the study of Chen’s
fundamental inequalities for complex submanifolds. The context of our study is the
following.

Let M n be a Kahler manifold of complex dimension n. Let us denote by J its
complex structure. We denote by sec(X A Y) and seal (p) the sectional curvature of
the plane determined by the vectors X and Y and respectively the scalar curvature
at the point p. Consider U a coordinate chart on M and e1,..., eme’f = Je1,...,e‘,", =
J en a local orthonormal frame on U. Then we have at p E U:

scal(p) : Zsec(e,- /\ ej), i,j =1,...,n,1*,...,n*. (7)
i<j

Let 7r C TpM be a plane section. Then r is called totally real if Jr is perpen-
dicular to it. For each real number k, B.-Y.Chen’s Kahlerian invariant of order 2 and

coefficient k at p E M is defined by

62(1)) 2 scal(p) — k inf sec(rr'), (8)

where inf sec(rr") is taken over all totally real plane sections in TPIW.

In [22] the following theorem is proved:

Theorem 1.4 For any Kc’ihler submanifold A!" of complex dimension n 2 2 in a
complex space form Afln+p(4c), the following statements hold:

(I) For each is E (—oo,4] we have

6,: 3 (2n2 + 2n — k)c. (9)

(2) Inequality (9) fails for every k > 4.
(3) (5" = (2n2 + 2n — k)c holds identically for some k E (—00, 4) if and only if M"

is a totally geodesic thler submanifold of Mn+p(4c).

The theorem describes completely, in a forth claim, the pointwise equality situa-
tion in the case k = 4.

In Theorem 5.4 we extend B.-Y.Chen’s fundamental inequality for Kahlerian cur-
vature invariants. In Proposition 5.5 we give a characterization of strongly mini-
mal surfaces in C3. The last section of the present work is dedicated to the study
of strong minimality through examples. Namely, we prove that the Kahler surface
21 + 22 + 2% = n, with It 6 C is strongly minimal in C3, and we prove that on the

Kahler surfaces Azf + Bzé2 + C23 2 O and 2]3 + 2% + 23} = 1 there exist points where

the strong minimality condition is satisfied. This study is inspired by the discussion

on Chen’s Kahlerian curvature invariants from [22], in the context described above.

2 Chen’s Fundamental Inequality with Classical
Invariants

2.1 The Hypersurface Case

We will discuss in this chapter Chen’s fundamental inequality associated with clas-
sical invariants. To clarify the geometrical interpretation in the equality case, we
distinguish two situations: the hypersurface case and the general codimension case.
The present section is dedicated to the codimension one case.

The main goal of this section is to prove the following:

Proposition 2.1 Let Ill" be a hypersurface in a Riemannian (n+1)-manifold M "+1.
Then at every point p E M the following inequality holds:

n(n -— 1)

l <
sca (p) _ 2

H2+ZS—8C(€i/\€j), (10)

i<j

where seal is the scalar curvature of M at p, H is the mean curvature at p, and
g(e, /\ ej) is the sectional curvature on the plane generated by vectors e,- and e]-
tangent to the ambient space M.

The equality holds at p if and only if p is an umbilical point.
We first need the following elementary lemma.
Lemma 2.2 Let g be a real polynomial
g(X) = aoX" + aan‘1 + aan’2 + + an
with a0 # 0 and n > 1. If all the roots ofg are real, then

2 —1
A(g,n) E £1720? — 40002 2 0. (11)

Proof of the lemma: If g has only real roots, then 9’ has also only real roots. After

(n — 2) steps, we know that g("‘2) has only real roots. Hence, we obtain A Z 0.

Proof: Now we can prove the proposition. Let p E A! and 77 E TgLM. Let
{61, ..., en} be an orthonormal basis of TPM in which so = S is in diagonal form, i.e.
S(e,) = A,e,~, i :- 1, ..., n, where A1, ..., A" are the shape operator’s eigenvalues. Then,

taking 9 in the lemma to be the characteristic polynomial of S, we have

71H 2 A1+ + An = —a1, (12)

ZAiAj = (12, (13)

i<j
a0 = 1. (14)

On the other hand ( see for example [28], pg. 131 ) we have the following well-known

fact
886(61' /\ 8]“) — Ede,- /\ Cj) 2' AiAj, (15)

for any i < j. Therefore the inequality in the lemma becomes

2(—nT—:—1—)(nH)2 —— 4 Z{sec(e, /\ ej) — §e_c(e,- /\ ej)} 2 O (16)
n(n—1)H2 —2§:sec(e,/\e,) +ZZ's—ei(ei/\ej) 2 o (17)

and since the sum in the second term is nothing but scal (p) we obtain, after a division
by 2, the claimed inequality. The equality case holds if and only if /\1 :- = An =

H / n, i.e. when the point p is an umbilical point.

The following fact comes from Lemma 1 of [17] for the hypersurface case:

Corollary 2.3 Let p = 23cal(p)/n(n — 1) denote the normalized scalar curvature
of a hypersurface A!" isometrically immersed in a Riemannian space form R"+I(c).

Then we have the inequality:
R S H 2 + c, (18)

at every point p E M.

The equality holds if and only if p is an umbilical point.

Proof: For any i # j, i,j E {1, ...,n}, we use the fact that Ede,- /\ e,-) = e in the

inequality in the previous proposition.
2.2 The General Codimension Case

We have discussed in the previous section the hypersurface case. We present in this
section an alternate proof of Lemma 1 of [17] in the general codimension case, i.e. as
it was obtained in [17]. One of the main points of this result is that the codimension
is arbitrary. We emphasize that the term in the left hand side of (19) is an intrinsic

quantity and the terms in the right hand side term are extrinsic quantities.

Proposition 2.4 Let M " be isometrically immersed in a Riemannian manifold Mn+m.

Let sec, s_eE, and scal(p) be the sectional curvature of M, the sectional curvature of

10

M, and the scalar curvature of M at p, respectively. Then the following inequality

holds.

n(n — 1)
2

scal(p) g |H|2 + L‘s—62m A e,) (19)

i<j
Proof: The argument in the proof also uses Lemma 1. Let {{1, ...,{m} be an
orthonormal frame of T1,,i M at p. Let us denote by X1”, ..., A; the eigenvalues of the
shape operator 8, = 55. Then X," + + A; 2: trace(h') and if S, is in diagonal form
then the characteristic polynomial of S, has the coefficient corresponding to /\"‘2
equal to

—Z AIA; (20)

1<i<j<n

If S, is not in diagonal form, then the quantity 2w” MA; is the sum of the 2 x 2 minors

in the matrix (hfjhsiqgn, since two equivalent matrices have the same characteristic
polynomial. In fact, if g = det(S,. — AI") = O, is the real polynomial in lemma 2.2,

then

=2 (h:.h;,-— <h:->2), (21>

1<i<j<n

—a1—— 2: h,,, (22)

Clo = 1, (23)

23mg: Zea-hi4 p?) (24)

11

and the inequality A( g, n) 2 0 is, in fact, for our choice of g :

2(7), —' 1) n r) r r 2
——n__(; hii)2 _ 4 Z( (hiihjj— hij) ) Z 0 (25)

1<i<j<n

or, summing for r : n +1,...,n + m:

n+m
n,(n -1)|H|2— 2:: Z (h;,h;,— ))> 0 (26)
i<jr=n+1

To express the last sum we need the Gauss’ equation (and this is the major
difference with respect to the hypersurface case). Let us denote by R the curvature
tensor of AI and by R the curvature tensor of M. Then, for any X ,Y 6 TPM, we

have

< R(X, Y)X,Y >=< R(X,Y)X,Y > — < h(X,X),h(Y,Y) > +|h(X,Y)|2 (27)

or, if {e1, ..., en} is an orthonormal frame at p E M :

n+nt
360(6i A 62') — 370M A 63‘) = Z (hiihij— (hijlzl (28)
r=n+1

and with this substitution the inequality become

2 Zsec(e, /\ e,) g n(n —1)|H|2 + 223%(e, /\ ej) (29)

i<j i<j
which is the inequality we had to prove, after a division by 2.

Lemma 1 of [17] is the following result, to which we refer as Chen’s fundamental

inequality with classical invariants.

12

Corollary 2.5 Let f : M n —+ R"+m(e) be an isometric immersion of a Riemannian
n-manifold 1U" with normalized scalar curvature p into an (n + m)-dimensi0nal Rie-

mannian space form R"+"’(e) of sectional curvature 6. Then
p S IHI2 + e. (30)
The equality holds at a point p E A! if. and only if p is totally umbilical point.
Proof: For any i 7.5 j, i,j E {1, ..., n} we use that Wk, Aej) = c in the inequality
proved in the proposition.

One can state the following immediate consequence which is more or less in the

same spirit as the obstruction results obtained in [20].

Corollary 2.6 Let M n be a Riemannian n-manifold and Mn+m be a Riemannian
(n + m)-manifold. If the scalar curvature of M is greater than the scalar curvature

of every n-plane section L of MM“, then M admits no minimal immersion into M.

2.3 A Remark on Totally Umbilical Points

We recall the fact that the inequality A(g, n) 2 O has also been used in [63]. Let us

define
2 2 2
o,- = 2||S,-]| — g(trace(Si)) , (31)

where S,- is the shape operator in the normal direction g,- and HS,- | l2 = (A[)2+...+(Ai,)2.
Let us denote by L,- the length of shape operator’s spectrum in the direction of the

normal vector 5,, i.e. the distance on the real axis between the greatest and the

13

smallest of the shape operator’s eigenvalues in the normal direction 6,. Then the

following result was proved in [63]:

Theorem 2.7 Let M " be a submanifold in a Riemannian manifold 151"“? For any

p 6 AI and for any normal basis {1,...,§m we have:

 

2
—— 1' < Li < i . 2
\ln(n_1)"(P)— (p)- o<p> (3)
The equality holds if and only ifp is a totally umbilical point of M in M.

One may get from the previous inequalities the following.

Corollary 2.8 Let 1W", n > 2 be a submanifold of a Riemannian manifold Mn+m. If
for some p E [W there exists a normal directioné such that 05(p) > 0, then the point

p cannot be a totally umbilical point.

Since the double inequality (32) was obtained by the same procedure as Chen’s
basic inequality involving the classic invariants, they have in common the proof basisd
on the idea A(g,n) 2 O, as it is presented in the previous two sections. In fact, the
main idea used in both cases is that the shape operator’s characteristic polynomial
has only real roots. The algebraic background of the next section is also related to

the study of Chen’s fundamental inequality with classical invariants.

2.4 A Conformal Invariant Related to Chen’s Fundamental
Inequality with Classical Invariants

In the classic matrix theory, the spread of a matrix has been defined by Mirsky in [47]

and then mentioned in various references, as for example in [46].

14

Let A E Mn(C), n 2 3, and let A1, ,An be the characteristic roots of A. The

spread of A is defined to be 3(A) = max,”- |/\,- — A,]. We denote by ”All the Euclidean

norm of the matrix A, i.e.: ”A”2 = 23:, |a,—,~|2. We also use the classical notation
E2 for the sum of all 2-square principal subdeterminants of A. If A E Mn(C), then

we have the following inequalities (see for example [46])

.(A) 3 (211/1112 - file/1W”, (33>
3(A) s x/EIIAII. (34)
If A 6 Much), then
1/2
S(A) g [2 (1— 3;) (mi)2 — 4E2(A) , (35)

with equality holding if and only if n —— 2 of the characteristic roots of A are equal to
the arithmetic mean of the remaining two.

Consider now an isometrically immersed submanifold M n of dimension n 2 2 in a
Riemannian manifold (M"+’, g). Then the Gauss and Weingarten formulae are given
by

VXY = VXY + h(X, Y),
fo = —A§X + DXE,
for every X, Y E F(T M ) and 6 E I‘(VM). Take a vector 7) in the normal space to

M at the point p and consider the linear mapping An : TpM -—> TpM. Consider the

eigenvalues A3,, .., A: of Ar We put

1372(1)) = , Slip (Xi) -,=i1{1f (/\‘ ). (36)

Then Ln is the spread of the shape operator’s in the direction 17. We define the spread

of the shape operator at the point p by

L6?) = sup Ln(p)- (37)

nEuph!

Let us remark that when M 2 is a surface we have

Lie) = (so) - Aiw)’ = 401100))? — Km),

where V is the normal vector at p, H is the mean curvature, and K is the Gaussian
curvature. In [7] it is proved that, for a compact surface M 2 in IE2“, the geometric
quantity (IH |2 — K )dV is a conformal invariant. As a consequence, one obtains that
LEdV is a conformal invariant for every compact orientable surface in IE2“.

Let 5”“, ..., (n+3 be a local orthonormal frame in the normal fibre bundle uM. Let

us recall the definition of the extrinsic scalar curvature from [9]:

6:13 :2n—(—_n _ 1) Z Z An+rAii+r

r=1 i<j

In [9] it is proved that, for every submanifold M " of a Riemannian manifold (M, g), the
geometric quantity (IH |2 — ext)g is invariant under any conformal change of metric.
When M is compact (see also [9]), this result implies that, for an n-dimensional
compact submanifold M of a Riemannian manifold (M, g), the geometric quantity
f(|H[2 — ext)3'dV is a conformal invariant.

Let us prove the following fact.

Proposition 2.9 Let M n be a submanifold of the Riemannian manifold (M, g). Then

spread of shape operator is a conformal invariant.

16

Proof: The context and the idea of the proof are similar to the one given in [3,
pp.204-205]. Consider a nowhere vanishing positive function p on M. Suppose that

we have a conformal change of metric in the ambient space M given by

Let. us denote by h and h" the second fundamental forms of M in (1171, g) and (M, g“),

respectively. Then we have (see [13])
g(AEX. Y) : 9(A5X, Y) + g(X. YEW/,5),

where U is the vector field defined by U = (dp)#. Let e1, ..., en be a local basis of the
principal normal directions of Ag with respect to g. Then p"1e1, ..., p‘len, form a local

orthonormal frame of M with respect to g*, and they are the principal directions of

Ag. Therefore

L"(p) = sup LE. = sup sup 0%)” — '_inf (Air
J_l,...,n

5.6VPMiHE'H-=1 {‘EVPM;||£’II.=1 i=1,...,n

= sup [ sup (A; +g(U,§)) — . inf (A; +g(U,g))[ =

{EUPM;||£||=1 i=1,...,n J: ,...,n

: sup [sup (A9331 op] ——-L(p).

£EupM;H£||=1 i=1,... ,n _ ,...,n

This proves the proposition.

When M is a compact surface, both L and L2dV are conformal invariants.

17

The shape discriminant of the submanifold M in NI with respect to a normal
direction 7) was discussed in [63]. Let An be the shape operator associated with an

arbitrary normal vector 77 at p. The shape discriminant of r) is defined by
2 2 2
D" = 2||A,,|| — g(trace A") , (38)

where “A,”2 = (A3,)2 + - - - + (A32, at every point p E M C 117.

The following pointwise double inequality was proved in [63]:

 

S D,,, (39)

We will use this inequality later on. The proof of this fact is algebraically related
to the proof of Chen’s fundamental inequality with classical curvature invariants (see

[17]). The alternate proof of this result is presented in [64].

2.4.1 Geometric inequalities on compact submanifolds

In this section, we study the relationship between the spread of the shape operator’s
spectrum and the conformal invariant from [9]. The main result of the present section

is Proposition 2.10. For its proof we need a few preliminary steps.

Proposition 2.10 Let M” be a compact submanifold of a Riemannian manifold

Mn“. Then the following inequality holds:

(f1. LdV)2 (vol(M))2Tn—3 5 2n(n — 1) (Law? _ extfide. (40)

Equality holds if and only if either n = 2 or M is a totally umbilical submanifold
of dimension n 2 3.

18

Before presenting the proof, let us describe what this inequality means. For any

conformal diffeomorphism <15 of the ambient space M, the quantity

(A ,..,, MW) 2 (v01 (96(111) ) 2.112

is bounded above by the conformal invariant geometric quantity expressed in (40).

 

First, let us prove the following.

Lemma 2.11 Let M" C NP” be a compact submanifold and p an arbitrary point
in AI. Consider an orthonormal normal frame {1,...,£, at p and let DO, be the shape

discriminant corresponding to {0, where a = 1, ..., 5. Then we have

1 ’ 2
m EDQ —— [H] - ext. (41)

Proof: Since

1 3 n i
11:2:(3...) 5.,

—1

ext: _n-_n(2—- 1) i Z ALA?”

)a=1 i<j

we have

(H)2 — ext: n2 :Z(,\;)— 2(n_ ——1—)Z§:,\w. (42)

01:1 i=1 02:1 i<j

A direct computation yields

D, = 2(n_n—_1_) :(X;)— :Z A; A3,. (43)

ni<j

Summing from a = 1 to a = s in (43) and comparing the result with (42) one may
get (41).
From the cited result in [9] and the previous lemma, we have

19

Corollary 2.12 If M is a compact submanifold in the ambient space M, then
s i
j E D, dV
M (121
is a conformal invariant.

Let us remark that for n = 2 this is a well-known fact.

Lemma 2.13 Let M be a submanifold in the arbitrary ambient space M. With the

previous notations we have
4(1le - ext) 3 21.3.02) 3 2n<n -1)(|H|" — ext)
at each point p E M. The equalities holds if and only if p is an umbilical point.

Proof: This is a direct consequence of Lemma 2.11 and (39).
Proof of proposition 2.10 : Let p be an arbitrary point of M and let 770 be
a normal direction such that L(p) = Ln0 (p). Consider the completion of 770 up to a

orthonormal normal basis 170 2 n1, - - - , 7],. Then we have

L200) = L300?) 5 2123(1)) S 2n(n -1)(|Hl2 - ext)- (44)

0:1

By applying Holder’s inequality, one has

(fM [All/>23 ([14 L2dV) (vol(M)).

Applying Hblder’s inequality one more time yields

[M (|H|2 -— ext) dV _<_ ([14 (|H|2 — ext)% dv)% (v0l(M))nT-2

20

Therefore, by using the inequality established in lemma 2.13, we have

(fir [Ail/)2 S (A LQdV) (vol(M)) S 2n(n —1)vol(]l/1)/M(|H|2 —— ext) dV 3

 

g
g 2n(n — 1) (201(M))2"n‘2 UM (|H|2 —— ext)? dV) " .

Let us discuss when the equality case may occur. We have seen that we get identity

if n = 2.

Now, let us assume n 2 3. The first inequality in (44) is an equality at p if there
exist 3 — 1 umbilical directions (i.e. La(p) = 0 for s = 2, ..., n). The second inequality
in (44) is an equality if and only if p is umbilical point (see [63]). Finally, the two
Héilder inequalities are indeed equalities if and only if there exist real numbers 0 and
u satisfying L(p) = 0 and [H |2 —ext = u at every p E M. The first equality conditions
impose pointwise L(p) = 0, which yields 0 = u = 0. This means that M is totally

umbilical.

2.4.2 The noncompact case

Let M be an n-dimensional noncompact submanifold of an (n + d)-dimensiona1 Rie-

mannian manifold (M, g).

Proposition 2.14 Let M" C Mn” be a complete noncompact submanifold and

m, ...,n; a local orthonormal basis of the normal bundle. Suppose that ZALAfi Z 0

and La 6 L2(M). Then

/(|H|2 — ext)dV < 00.
M

21

Proof: We use the inequality (39). It is sufficient to prove locally the inequality:

d
[le—extg 2D,-

i=1

This is true since the following elementary inequality holds:

 

d
2n 2 . 2
d t j )2 d 2 _ __ :
(A322) +. .+ (A) _IZAQ A2 _ +)...+(A2)] ”[202”.
1i<j i=1
This is equivalent to
d d
n(n—1)Z()\’)2 —2n2:/\'/\j<2n—1)2Z(x\')2)—4(n—1)Z/\‘/\j
i=1 i<j i=1 i<j

01'

(n2 —3n+2) ){Z(A;,) 2}+2(n2 —2n+2)Z,\;‘,A{, 20,
i<j
which holds by using the hypothesis and that n 2 2.
The inequality is the a-component of the invariant inequality we are going to

prove. By adding up d such inequalities and by considering the improper integral on

M of the appropriate functions, the conclusion follows. This is due to

d a
/ (IHI2 — ext)dV s / Zadv s (i): / Lgdv
M M i=1 i=1 M
by the first inequality in (39).

In the next proposition we establish a relation between fM[L(p)]2dV and the

Willmore—Chen integral, fM(|H]2 — ext)dV, studied in [9].

Proposition 2.15 Let M " C Mn” be a complete noncompact orientable submani-
fold. If L(p) E L2(M), then fM(|H|2 — ext)dV < oo.

22

Proof: By a direct computation, we have

/M(|H|2 — ext)dV (n _1) ———-—/M :29; —,\J' )2dV < (45)

(1:1 i<j

712(n—_———1)/IWZZL2WV:%LL2@MV

a=1i<j

2.4.3 Examples

Let us look now at two examples. First, let us consider the antenatal defined by

’U . ’U
fc(u, v) = (ccosu cosh —,c smu cosh —, v).
c c

Using the classical formulas for example from [62] one finds

1
A1 = ——)\2 = - cosh"2 1)-
c c

Therefore, we have

°° °°2 _2v °° e‘dt
[mL(p)dv=/_m;cosh Edv=4/-oo-e—2t—:—1-=47r<oo.

Let us consider the pseudosphere whose profile functions are given by (see, for

example [37])

c1(v) 2 ae_”/“

= [v V1 — e‘2t/“dt
0

23

for 0 S v < 00. For simplicity, let us consider just the ”upper” part of the pseudosphere.

We have

ev/a

‘ /1 _ e—2v/a,

A1:

 

a

—1
A2 = — (aev/“VI — e—Qv/a) .

Remark that

1°°dy

—___:(x)

 

oo et/a
LdV = dt =
[M [0 a\/1 — 6‘2”“

A natural question is to find a characterization for surfaces of rotation that have
finite integral of the spread of their shape operator.
Consider surfaces of revolution whose profile curves are described as c(s) = (g(s), s)

(see, for example, [62]). Then we have the following.
Proposition 2.16 Let M be a surface of rotation in Euclidean 3-space defined by
1/)(8, t) = (g(s) cos t, g(s) sint, 3).

Then the integral of the spread of the shape operator on M is finite if and only if there
exist an integrable C°°(R) function f > O which satisfies the following second order

differential equation:

NICO

-m/’ = 1 + (y')2 i f(8)y(1 + (302)

24

For the proof, we use the classical formulas from [37, p.228]. We have for A1 =
lawn-dim, and respectively for A2 = kmmue, :

A1 ___ _yll
[1 + (y’)2]3/2’
1
A = , .
2 yll + (y’)2ll/‘2

 

 

Then, the condition that the integral is finite means that there exists an integrable

function f > 0 such that

[R |/\1-/\2lds= Lf(s)ds.

If we assume that f E C°°, then the equality between the function under integral
holds everywhere and a straightforward computation yields the claimed equality.

For example, for the catenoid f (s) = 0.

2.5 A Fundamental Inequality for Warped Product Mani-
folds

For a warped product N1 x f N2, we denote by D1 and D2 the distributions given
by the vectors tangent to leaves and fibers, respectively. Thus, D1 is obtained from
tangent vectors of N1 via the horizontal lift and D2 obtained by tangent vectors of N2
via the vertical lift. Let (b : N1 x f N2 —> R'flc) be an isometric immersion of a warped
product N1 x f N2 into a Riemannian manifold with constant sectional curvature c.
Denote by h the second fundamental form of 45. The immersion (b is called mixed
totally geodesic if h(X, Z) = 0 for any X in ’D1 and Z in D2.

The problem of suitable conditions on isometric immersions in space forms is

analyzed and explained in [23]. Let us consider N1 x f N2 be the warped product of

25

two Riemannian manifolds and let n1 and n2 respectively their dimensions. We will
use the notation n : n1 + n2.

The following inequality for warped product spaces is proved in [23].

Theorem 2.17 Let L : N1 x, N2 -—> Rm(c) be an isometric immersion of a warped
product into a Riemannian m-manifold of constant sectional curvature c. Then we

have

Af (”1 I ”2)2 2
— < ———H +71, 46

where n,- = dim Ni, i = 1,2, H2 is the squared mean curvature of 45, and A is the
Laplacian operator of N1.

The equality sign of (46) holds identically if and only ifi : N1 fo2 —) I?"(c) is a
mixed totally geodesic immersion with trace h1 = trace hg, where trace hl and trace ’12

denote the trace of h restricted to N1 and N2, respectively.

Several applications of this theorem are given in [23].

The classification of immersions from warped products into real space forms sat-
isfying the equality case of (46) is obtained in [24].

Here, we prove the following inequality, in the same spirit, but whose proof will
use a different argument, namely the idea from our proof to Chen’s fundamental

inequality with classical invariants.

Proposition 2.18 Let i : N1 x N2 —+ M "+m be an isometric immersion of a warped
product manifold into a Riemannian manifold M. Then at every point p E M the
following inequality holds :

26

A —- 1
n27f + scal(Tle) + scal(TpNg) g n_(n_2__)H2 + Z sec(e, /\ e,-), (47)

i<j
where seal is the scalar curvature corresponding to the indicated tangent space with

respect to the warped product metric.

Proof: The following relation was proved in [64], it was also proved in section 2.2.

2 sec(e, /\ ej) g Zigg—BH2 + Z séc(e,- /\ ej). (48)

i<j i<j
The left hand side term of the above inequality can be written in detail as
Zsed (eiAej)— ziised eiAes) )(+Zsec eiAej) )+Zsec(es/\et) (49)
i<j 3:1 1': 1 i<j s<t
where i, j = 1, ...,n1 and s,t 2 n1 + 1, ...n are the subscripts corresponding to the
tangent spaces to N1, respectively N2, at every point T p(N1 x f N2).
We have (see, for example, [5] or [23]):

:nZsec( eiAes) )=n2A—f,

s=1i=1f

Z sec(e, /\ ej) = scal(TpNil,

i<j

Z sec(e3 /\ et) 2 scal(TpNg).

s<t

Replacing these quantities in (48) we get the claimed inequality.

27

The equality holds if and only if the following relation holds at every point:

n(n — 1)H2 = 2 Z[sec(ei /\ ej) — séc(e,— /\ ej)]. (50)

i<j
In the case when in relation (47) the ambient space is a space form, we get the

following.

Corollary 2.19 Let L : N1 x N2 —+ Rn+m(c) be an isometric immersion of a warped

product into a simply-connected space form. Then at every point p E M :

n2? + scal(TpN1)+ scal(TpNg) g in—2-_1)_(H2 + c), (51)

28

3 Applications of Chen’s Eindamental Inequality:
the General Case

3.1 Warped Product of Hyperbolic Planes

In the Introduction we have explained that we are interested to construct explicit
examples of Riemannian manifolds with Ric < 0, but which have some positive Chen
invariants.

For such construction, we use the notion of warped product metrics introduced
by Kruckovic in 1957 and by Bishop and O’Neill in [5] in Sections 3.1 and 3.2. (A
reference on warped product metrics is in [1], which is, in particular, useful in the
calculation on Ricci curvature of a warped product metric. Another reference is [56].
A discussion in the context of manifolds with nonpositive curvature, based mainly on

[5], can be found in [57].)

Let us consider two copies of the hyperbolic plane (H 2, go). The first has coordi-
nates (:17, y) with y > 0 and has metric go 2 (1 / y2)(d:z:2 + dyg). Let u and v denote the
coordinates of the second copy of the hyperbolic plane with v > 0. We consider the
open subset U = {(23, y) E H 2[y > e/ 2}, for sufficiently small 5 > 0. On the product

manifold (U x f H 2, 9) we consider the warped product metric g 2 go + f 290, i.e.,
1 f2 $.31)
g : ?(day2 + dyz) + —(1-)2—-(du2 + dv2), (52)

where f is a positive differentiable function. We use the subscripts 1, 2, 3, 4 corre-

sponding to the coordinates 2:, y, u, v, respectively. At every point p E M, we use the

29

following notation for the tangent vectors

Laban,

3
(9m 0y ’ Bu

we claim the following: There erist diflerentiable functions f on (U x; H 2,g)

such that Ric < O and 6(2, 2) > O everywhere.

A straightforward computation gives

 

 

sec(a.c /\ 8y) 2 —1, (53)

sec(o, A on) = sec(o, /\ a) = M”, y) (2% — gig) (54)
sec(a, /\ a.) = sec(By A a.) = — f (:21, y) (52,—: + 9%) (55)
sec(au A o.) = —?,—(:Ty—) — 7:47:27) [Ci—i)? + (gag-)2] (56)

Therefore, the half of scalar curvature at p = (:13, y, u, v) is given by

scal(p) = —l —

 

 

1 2y2 [02f 02f] (57)

f’(x.y) ‘ rm) 5.2 + 6y?

___y’_ g 2 + :31 2
f’($.y) 59: 0y
Using eventually Proposition 9.106 from [1] and the fact that the components of the

Hessian of a function (15 are given in general by :

02¢ 305 r
(2222 = W " 5;: r

30

the values of the Ricci tensor are :

 

 

 

 

,1. _ _i 2 fl _ 2 (if
R (81,81) — y’ + yf(I,y) 0y f(IIzy) 013” (58)
f , _ 1 2 of 62f
Ric(()y,8y) _ 73 — m (5y- + y—a—yg) . (59)
. . . 1 2f ,, 52f 52f
ch(ou,ou) = ch(o.,,o,,) = —fi — y if; y) [ (911:2 + 5?] (60)
_fi (if. 2 5f 2
v2 (0x) + (5;)
Ric(01.,6y)= -f 2 52f 2 6f (61)

(23.31) 03/516 — yf ($.31)???
Ric(81,5u) = Ric(8x,8v) = Ric(6y, Bu) 2 Ric(5y,0v) = Ric((9u,0v) = 0 (62)

To complete our example, we choose a function “close” to 1 which has the desired
properties: Ric < 0 at every point p = (2:, y, u, v), but at least one of Chen invariants
is strictly positive.

Consider f (:8, y) = e5 ”“3”. For this specific function one gets by direct compu-

tation that

Ric(6u, a.) = Ric(6v, o.) = (63)

_v2(1+ y2)2 [(1+ y2)2 + 2€y2(5 _ y)e2earctany] < O

This last conclusion shows us that the only minor we need to study is the one

corresponding to subscripts 1 and 2.

31

The canonical basis we’ve considered is not an orthonormal one. To complete

the computation on an orthonormal basis let us take e1 2 yBx, 62 = yay, e3 =

(v/f(:z:,y))0.., 64: = (v/f(:r.y))05. Then
Ric(e1,e1) : y2Ric(0I,8x),
Ric(e1, e2) 2 yQRic(6a‘, 0y),
Ric(e2, e2) = y2Ric(8y, 6y),
Ric(e3,e3) = (v2/f2)Ric(8u,8u) < 0,
Ric(e4,e4) = (v2/f2)Ric(6v,8v) < 0.

To see that Ric < O, we have to study the 2 x 2 minor:

2y 0f 23/2 5931

Ric(e1,e1) = —1 + —— — —

 

f 0y f 011:”
Ric(e1,e2) = Ric(e2, e1) 2 —g%iai2éfm — a???
R’iC(€2,€2) : —1 — 271’25 — 273/2327];
or, for the considered function:
2ey

R' , =—1 —,
ZC(€1 61) + 1+y2

Ric(e1, e2) 2 Ric(e2, el) = 0,

_ 25y _ 2ey2(e - 2y)
1 + y2 (1 + y?)2

 

Ric(eg, e2) 2 -—1

On the other hand, since on U we get sec(az /\ By) < sec(am /\ 0“), sec(az A By)
< sec(r’?y /\ Bu), sec(Bu /\ 81,) < sec(ax /\ (9.)), sec(au A 0,) < sec(ay /\ By), the smallest

32

values of sec(e, /\ e,-) on the considered basis are sec(c’?z /\ By) and sec(au /\ 0”), we

have on U:

6(2, 2) 2 23ec(6I /\ on) + 2sec(8y /\ 0,) (64)

 

 

:_ 312 (02f a2r):_2ey2(e—2y)>0

f(r.y) 017’ 0—315 (1+y’)2

The last inequality allows us to apply Theorem 1.2 to obtain the following :

Proposition 3.1 For sufiiciently small 6 > O, the Riemannian manifold
NI ___ (U X [12,90 + (e2earctan y)90)

cannot be isometrically immersed in any Euclidean ambient space Em as a minimal

submanifold for any codimension, even though Ric < 0.

One may obtain similar result by applying the same construction with some other
warping functions on an appropriate open set U C H 2.

Let us notice that one doesn’t need a specific computation for 6(2,2) to apply
Theorem 1.2. An estimate as in the relation (64) is sufficient to obtain the obstruction

to minimal immersions into a Euclidean space of any codimension.

3.2 Multiwarped Product Spaces

Let us now consider a multiwarped product of hyperbolic spaces defined as follows.
Let us use a similar notation U = {(11:,y) E H 2[y > 1228} to the previous section.
Consider the product manifold of U with n warped copies of the hyperbolic plane

H2, endowed with coordinates (z,y,u1,v1, . .. ,umvn) with y, v1, . .. ,vn > 0. At an

33

arbitrary point of the product manifold let 171, 02, .. . , 772n+2 denote respectively the

tangent vectors:

8 8

flwflii __
817’y’8 8111,1118 ’8un’8vn.

The multiwarped product metric on (U x11 H 2 x f, .. x In H 2, g) is defined by

g=—(dir2 +dy2) +Zf )(ud2 ,,-2+dv) (65)
where f1(r, y), . .. , fn(.7:, y) are positive differentiable functions.
We claim the following: There are some choices of f1, . . . , fn which satisfy Ric < O

everywhere and at least one of Chen invariants is positive.

 

 

 

 

By direct computation we have, for i = 1,. . . ,n :

sec(m A 772) = —1, (66)

8 ,- 82 ,-
S€C(771 /\ 722m) = 560(771 A 772i+2l = Ki?“ 8);: - y as; )1 (67)

y 6fi 62fi
sec(nz /\ 7721+1) = 360(772 A 772.42) = —m (83; + 31 33/2 , (68)

1 y2 6r.- 2 6f.- 2
. '1 /\ i = — — — — ,

886(7)) +1 772 +2) f3(:v. y) fi2($: y) [( 8:12) + (at! (69)
Sec(7)2i+2 /\ 772j+2l = 360(fl2i+2 /\ 7723'“) = 360(772i+1 /\ 772)“) = (70)

 

 

_ y2 [af__.-_a_f.~ + alt-go]
fi(x.y)fg($, y) 5’96 (9m 02151; ’

scal(p) = -—1 —— 2E2“) 7.2 [(%) + (22)“ (71)

34

 

 

 

 

" 02f. o2f- " 1 of-of- of-of-
__ .2 . r _ .2 _ _t_1 .__,
2y 2(0$’+0’y) 4y 2 fifjlax (993+3y fijl’

i.j=1;i¢j

 

 

 

 

8 , 82 ,-
Ricmml): — +-;-— f.- 10y( f -y 631;) (72)
RZC(7]2,7]2)— _ —— —' " :2— fi 1(ay'l' f+ gay-Z) (73)

RiC(772i+2,7)2i+2) = RiC(fl2z‘+1, 772i+1) = ‘—

 

y of. 6f.- 2
2‘é[(rxl 40.)] <72
fi<fi+_a2_f.)_2y2f.2 " 1 [af__.a_f_j+6f.§§]

2 3332 8y2 of i .,f,f—_j 811283: 831811 '

v- =1;1¢J
A long computation yields the other terms of the matrix of Ric tensor. Let us

 

 

1

explain how to compute Ric(771,172). We need to compute terms of the type Rklkz. We

distinguish three cases: k = 1, k = 2 and k # 1, 2. Then

 

1 (9sz 1 8f]:
R1112”: 2212:”: affirm-n5;-

A similar discussion is taking place for every element of the matrix of Ric tensor,

to yield that all non-diagonal terms vanish everywhere, except

, n 1 82f)c 1 8f)c
R , z _2 E __ + __ . 75
26(771 772) k=1 [fk 8y8a: yfk 8a: ( )

For a specific example let us consider f,(:c, y) = f (1:, y) = eE ”“3“” fori = 1, . . . ,n.
To simplify the computations one may choose 0 < e S 1 / n. For the orthonormal basis
we work with, let us denote as above el = $1771, e2 2 gm, and e2k+1 = (vk/fk)r72k+1,

e2k+2 = (pk/farm”, for k = 1,. .. ,n, respectively.

35

For the subscript 3 to 2n, the Ric matrix is in diagonal form at every point.

Through a direct computation, we obtain, for i = 1, . . . ,n, that

. , 1
RlC(€21+1,€2i+1) = Ric(e2i+2, e2i+2) = __

f2
(2771+1)y2 8f 2 y262f
7%) ‘75?”-

In order to estimate Chen invariant, we compute the sectional curvatures as fol-

(76)

 

lows, fori,j 7- 1,... ,n,i7€j:
360(721 A 772) = ‘1, (77)
e
380(771 A 772i+1) I 360(771 A 7)2:’+2l = J3 > 0: (78)
1 + y
air/(312 - 8y - 1)
88C /\ i = 86C A i = > 0, 79
(772 772 +1) (272 772 +2) (1 + y2)2 ( )
1 €2y2
sec ,- A ,- =———————<0, 80
(772 +1 772 +2) f2 (1+ y2)2 ( )
E23/2
3800721“ A 772j+1) = 3660721” A 772j+2) = 860(U2i+1 A 772j+2) = -W < 0. (81)

In fact, one can easily obtain that

384772;“ A 772i+2) < 36C(’72i+2 A 722j+2)- (82)

This allows us to obtain the estimate of the (2,2, . .. ,2)-order Chen invariant (2

repeats n + 1 times) such that

5(2, - -- ,2) Z 7(1)) — 330(771 A 772) + E: 366072141 A 772i+2) = (83)
i=1
2eny2
: __ 2 _
(1+y2)2( y en) >0

Thus, by applying Theorem 1.2, we have proved the following.

36

Proposition 3.2 The Riemannian manifold (U x H 2 x x H 2, g), endowed with the
metric given by (65) with f,(:1:,y) = eemtany, i = 1,... ,n, cannot be isometrically
immersed as a minimal submanifold into a Euclidean space for arbitrary dimension,

even though Ric < 0.

The same procedure with some other functions f,- may also give rise to other
specific examples of Riemannian manifolds whose Chen’s invariants obstruct mini-
mal immersions via Theorem 1.2, although the classical invariants do not provide

obstruction to minimal immersions.

37

4 Curvature and Topology: A Myers Type Theo-
rem for Almost Hermitian Manifolds

The classic S.B.Myers’ theorem (see [48]) asserts that a complete Riemannian mani-
fold AI that satisfies the condition Ricp(v,v) Z r‘2 > 0, for every point p E M and
for any unit vector v E TpM, is compact and its diameter is less than or at most equal
to 7rr. The condition Ricp(v, v) 2 0 everywhere and a Ricci curvature condition along
geodesic rays from a point pg 6 M has been studied by Calabi in [6]. For some other
references 011 the topic one may see for example [28].

Let us consider (M 2", J, ( , )) an almost Hermitian manifold with curvature tensor
R. To establish the notations, let us consider just for this section the following sign

convention for the curvature tensor
R(X,Y)Z = -VxVyZ + VnyZ + V[x,y]Z,

for any tangent vector fields X, Y, Z 6 TM. The Ricci tensor will be denoted by Ric

and the sectional curvature by sec. The holomorphic sectional curvature is given by

 

(R(JX, X)JX, X)
(X.X>2 '

H(X) = sec(XA JX) =
The main result of this section is Theorem 4.1. Chronologically, the first result of
Myers type in Kahlerian context was established by Tsukamoto in [68]; his result
states that a complete 2n—dimensional Kahlerian manifold M whose holomorphic
sectional curvature is greater than or equal to a > O is compact and has the diameter

less than or equal to n/fi. Furthermore, under the mentioned hypothesis, M is

simply connected.

38

A result of Myers type for nearly Kahler manifolds, with the holomorphic cur-
vature condition, has been proved by A.Gray in [35]. Gray also proved in [36] a

corresponding result for almost Hermitian manifolds as follows.

Theorem A Let A12" be a complete almost Hermitian manifold. Assume that the

holomorphic sectional curvature of M satisfies:
H(X)-||(VxJ)X||2||X||'2Za>0. (84)

for all X 6 TM! and all p E 1%. Then M is compact and the diameter of M is not

greater than n/fi. Furthermore, M is simply connected.

A theorem of Myers type for locally conformal Kahler manifolds has been proven
by Vaisman in [69]. A generalization of Myers’ theorem for contact manifolds has
been proven by Blair and Sharma in [3].

Recall that in [33] Gallaway established the following fact, mentioned also in [28].

Theorem B Let M n be a Riemannian manifold. Suppose there exist constants
a > 0 and c > 0 such that for every pair of points in M 2" and minimal geodesic 'y

joining these points having unit tangent '7’ (t), the Ricci curvature satisfies:

Rico's), r6» 2 a + 3’; (85)

along 7, where f is some function of the arc length t satisfying If (ill S 6 along 7.

Then M 2" is compact and:

 

diam(M2") g g [c + \/(c2 + (1(17. — 1))] (86)

39

Furthermore, the universal covering of M 2" is compact, with diameter bound as in

(86) and the fundamental group of A12" is finite.

For c = 0 one may get the classic Myers theorem. It’s natural to think about a
result similar to Theorem B in the almost Hermitian context, i.e. the context from
Theorem A. The curvature condition we study is inspired from A.Gray’s Theorem A.

We establish the following (see [66]).

Theorem 4.1 Let M 2" be a complete almost Hermitian manifold. Suppose there exist
constants a > 0 and c > 0 such that for every pair of points in M 2" and minimal
geodesic ’y joining these points having unit tangent 'y’ (t), the holomorphic Sectional

curvature satisfies:

Hort» 2 a + 2%- (87)

along 7, where f is some function of the arc length t, satisfying If (15)] S C along 7.

Then M 2" is compact and
diam(M2") _<_ Z— [c + (c2 + a)] . (88)

Furthermore, the universal covering of M 2" is compact, with diameter bound as in

(88), and M 2" is simply connected.

Proof: Let us consider two points p,q E M. Let '7 be a minimizing geodesic
parametrized by arc length t that joins p and q, 7 : [0,1] —> M, the length of 7 being
I. Let us consider the vector field (as for example in [36]):

at

va) = (sin—f) H(t), (89)

40

for any t E [0, I]. Then let us consider the proper variation of 'y in direction of V. We

denote by E the energy functional given by:

l
= / ”MW

Synge’s second variation formula (see for example [28]) yields:

$9; 0) = — f (140%.! + R<v’<t),V(t>>r(t>) dt (90)

or, replacing the expression of V(t) from relation (89):

—-/<<><><>>
— f ((smj ) m ), Bert). (will) J7’(t))i’(t)> =

2 l 1
= 77;— sin2 (3%) dt —/ sin2 (I?) (7’,R(J7',7’)J7') dt.
0 o

The curvature term in the last equation is the holomorphic sectional curvature sec( J 7’/\

7’) and we may use the condition (87) to get:

IdQE 7r2 al I 7rt df
__ <____ '2 — —. 2
2 dt2 (0) - 21 2 Am (l)dtdt (9)

Thus, if I > 7r(c + x/cz—+d) / a then the variation would minimize the length of 7,
contradicting the fact that 7 is minimizing. Hence, the length of 7 is bounded above
by this quantity, therefore (88) holds.

To see the last claim of the theorem, let’s assume the contrary (the argument is
the same as in [36]). Then there exists a non-trivial free homotopy class of 100ps
which contains a non-trivial minimal geodesic 70, defined on [0,l]. Assume that 70
has unit speed. The deformation of 70 in the direction of V0(t) = sin(7rt/l).]7(’)(t)

yields, by the second variation formula, since the length of 70 is bounded above by

7r(c+ V62 + a)/a:
1d2E

5W0» < 0,

therefore 70 cannot be a minimal geodesic. Therefore M is simply connected.

Corollary 4.2 Let M 2" be a complete Kahler submanifold in a complex space form
M2("+k)(e). Suppose there exist constants a > O and c > 0 such that for every pair of
points in M 2” and minimal geodesic 7 joining these points having unit tangent 7’ (t),

the second fundamental form h satisfies along 7:

2|lh(’7’(t),7’(t))ll2 + a +% s a. (94)

where f is some function of the arc length t, satisfying [f(t)] S 0 along 7. Then M 2”

is compact and:
diam(M2n) g g [c+ (c2 + a)] (95)

Furthermore, the universal covering of M 2" is compact, with diameter bound as in

(95), and M 2" is simply connected.

42

Proof of Corollary 4.2 : It is known (see, for example, [51]) that

H(X) = s — 2 Zgaax X)? = e - 2llh(7’(t),7’(t))ll2-

Then we apply Theorem 4.1.
Let us remark the Corollary’s hypothesis cannot be relaxed to a = c = 0. For

example, in the case 5 = 0 there exist complex totally geodesic noncompact subman-

ifolds.

Let us remark that Myers’ Theorem can be stated in terms of Chen’s invariants.
In [18] B.—Y.Chen introduced also the following string of Riemannian curvature in-

variants.

A

(5(n1,n2, ...,m,) = scal(p) — sup{scal L1 + + scal Lnk}, (96)

where L1, L2, Ln, are mutually orthogonal linear spaces of dimension n1, n2,

nk. With this notation, we can state Myers’ Theorem as follows.

Theorem 4.3 Let (M, g) be a Riemannian manifold such that, at every point p E M,

the condition: (f(n — 1) Z a2 > 0 holds. Then M is compact.

Proof: One may write, for any unit vector,
Ric(v,v) 2 3(n — 1) Z a2 > 0.

Thus, the hypothesis from Myers’ theorem is verified.
In general, the positivity of a certain Chen invariant doesn’t imply compactness.

For example, 6(2, 2, ..., 2) Z a"2 doesn’t imply compactness, as one may see from the

43

following example. Consider [W 2 82(1) x R2 x X R2, where R2 is taken n times.

In this case 6(2, 2, ..., 2) = 1 > O at every point, but M is not compact.

44

5 Chen’s Fundamental Inequality for Complex Sub-
manifolds

5.1 New Kahlerian Invariants

Let M n be a Kahler manifold of complex dimension n. Let us denote by J its complex
structure. We denote by sec(X /\ Y) and seal (p) the sectional curvature of the plane
determined by the vectors X and Y and respectively the scalar curvature at the point
p. Consider U a coordinate chart on M and 61,..., eme’f = Je1,...,e;'l = Jen a local

orthonormal frame on U. Then we have at p E U:

scal(p) = 2880(81' /\ 6]”), i,j = 1, ...,n, 1*, ...,n”. (97)

i<j

Let 7r C TpM be a plane section. Then it is called totally real if Jr is perpen-
dicular to 7r. For each real number k, B.-Y.Chen’s Kahlerian invariant of order 2 and

coefficient k at p E M is defined by

62(p) = scal(p) — kinf sec(rr’"), (98)

where inf sec(rr’) is taken over all totally real plane sections in TpM.

In [22] the following theorem is proved:

Theorem 5.1 For any Ka'hler submanifold M" of complex dimension n 2 2 in a
complex space form Mn+p(4c), the following statements hold:

(1) For each k E (—oo,4] we have

6,: 3 (2n2 + 2n — k)c. (99)

45

(2) Inequality (99) fails for every k > 4.
(3) 6,: = (2n2 + 2n — k)c holds identically for some k E (—00, 4) if and only if .M"

is a totally geodesic Kahler submanifold of Adn+p(4c).

The theorem describes completely, in a forth claim, the pointwise equality situa-

tion in the case k = 4.

5.2 Strongly Minimal Submanifolds

It is known ( see for example [51]) that the shape operator of a Kahler submanifold

M n in NW” satisfies:
A“, 2 JA,, JA, 2 —A,J, (100)

for r = 1, ...,p, 1", ...,p", and where we use the well-known convention A, = A6.

Therefore the shape operator of M " takes the form

At All _All Al
A0: ( All _XI )) AG" :( Ala A: )) 021,-",1) (101)

where A; and AZ, are n x n matrices. The condition (101)implies that every Kahler

submanifold M n is minimal, i.e. trace Ac, 2 trace Aa— = O, a = 1, ..., p.
Definition: A Kahler submanifold M " of a Kahler manifold Mn” is called

strongly minimal if at each point there exists an orthonormal frame e1,..., eme; =

J e1,...,e:, = J en such that the shape operator satisfies the conditions
trace A; = trace A2 = 0, a = 1, ...,p.

This class of submanifolds was introduced and studied by B.-Y.Chen in [22]. From

[22] we have the following two results.

46

Theorem 5.2 [22] A complete Kahler submanifold 1W” (n 2 2) in CPn+p(4c) satisfies

the equality
5:; = 2(n2 + n — 2)c (102)

identically if and only if
(1) AI" is a totally geodesic thler submanifold, or

(2) n 2 2 and A12 is a strongly minimal thler surface in CP2+p(4c).

Theorem 5.3 [22] A complete Kahler submanifold M n (n 2 2) of (CM? satisfies
6},” = O identically if and only if

(1) M" is a complex n-plane of CT”, or

(2) IV!" is a complex cylinder over a strongly minimal K ahler surface M 2 in C"+P
(i.e. 1V!" is the product submanifold of a strongly minimal K dihler surface M 2 in CW2

and the identity map of the complex Euclidean (n — 2)-space CW2).

Among the examples studied in [22] let us mention a nontrivial example: The
complex surface N 2 in (C3 given by the equation 2% + 2.3 + z}; = 1 is a strongly minimal
Kahler surface.

The above mentioned results and examples motivate our present study of the
strongly minimal submanifolds. One of the problems we discuss in the present dis-

sertation is the characterization of strongly minimal surfaces in C3.

47

5.3 An extension of B.-Y. Chen’s Fundamental Inequality
with Kahlerian Invariants

In the present section we extend the inequality (99) to orders higher than 2. Let
us motivate first this generalization. As we have mentioned before, the first form of
B.-Y.Chen’s fundamental inequality in Riemannian context has been given in [14],
in 1993, and the string of B.-Y.Chen’s fundamental inequalities has been obtained
in [20], in 2000. It is natural to ask what could be the most general statement one
may get from the geometric idea of B.-Y.Chen’s fundamental inequality for Kahler
submanifolds in space forms, presented in [22].

An l-dimensional linear subspace L, C TpM is called totally real if J L; is orthogonal
to L. For each real number k one may extend the above invariants to Kc'ihlerian

invariant of orderl and coefi‘lcient k by

me) = scal(p) — ,—f—, ,,icr,§M[scaz<Lz(p))1. (103)

where L; runs over all totally real linear subspaces in TpM, and the scalar curvature

of a linear subspace is

scal[L1(p)]= Z seem-Am),

15i<jgz
for the orthonormal basis 771, ..., 171 in L).
Let us now assume that M n is a Kéihler submanifold of complex dimension n in a
complex space form M "+19 (4c). We are closely following the notations from [22] unless
stated otherwise.

Let us consider the orthonormal basis e1,..., eme’f = Je1,...,e‘,’, = Jen such that

48

L, : span { e1,..., e,}, where L; achieves the infimum for scal[L], with dimL = I. If

R is the curvature tensor of .M, then the Gauss equation is

(R(X,Y)Z, W) : (h(X,W),h(Y, Z)) — (h(X,Z),h(Y,W)) + (104)

+c{(X, W) (Y, Z) — (X, Z) (Y, W) + (JY, Z) (JX, W) —
— (JX, Z) (JY, W) + 2 (X, JY) (JZ, W) ,

where h is the second fundamental form.

The result we prove is the following.

Theorem 5.4 Let M ” be a Kahler submanifold of complex dimension n 2 2 in a
complex space form Mn+P(4c). Let (5,2,, be the Chen’s Kc'ihlerian invariant of orderl
and coefficient k. Then we have

(1) For any 2 S l g n, the following inequality holds
g, 3 (2n2 + 2n — (12))c. (105)

The equality case forl = 2 has been described in [22]. Equality holds at every point
for a fixed l _>_ 3 if and only if M n is a totally geodesic submanifold.

(2) For any h E [0,4] the following inequality holds
7' 2 k l
61.; S 2n + 2n — 4 (2) c. (106)

Equality holds at every point for a fixed I Z 3 if and only if M n is a totally geodesic

submanifold.

49

Proof: Let us discuss first claim (1) of the theorem. Taking X = W = ek and

Y : Z = e, for 1 S k, s S t in the Gauss’ equation one gets
p
sec (ek /\ es) =[Z hfkh; W) +h"; 0" —( 2;)2] + c. (107)
0:1
Consider l E {2, 3, ..., n}, the dimension of the totally real space Lf. Then

scal(Lf) = Z (R(e,,, e,)e,, e.) = (108)

lgk<sgl

=2 2 {Mm—0:3)1+[h:*h::—<h3>‘>]}+(3c)

a=1 l<k<s:_<_l

For the following computations, we use (hjc’k)2+ (h? 82) > 4230,12?” the similar

relations for a’ and that l 2 2 implies
2( :.>’2—_—( 3:.)’. (109)

As in relation (3.5) from [22] one may compute the quantity

4n (n+1)c—2scal=4Z{||A; ||2+[|A”| ”2} > (110)
a=1
p I
_423{23(hf§) )2+2 2 (h )2+23(h°* )+2 2: (h;;)"~’=*}
(:21 1'21 1<i<j<l 1<i<j<l

=4i[lsqs[]h_’:1:12]+2Z>]+

i<j

4:{ : (L—r‘: [+223 }

i<j

50

P —2h3hTJ a 2

2 4X Z 77‘ + 2W) +
p -2h3’lh3? at 2

+4 2 2 Ti— + 2031': l 3

0:1 ISi<jSl

4: Z_.{7:—21’h’h” - (hi->21] +

P
—2 at at 0* 2
+2 {—l_1lhiihjj —(hii) l]:
0:1 lSi<jSl
8 P
— a a a 2 an: at at 2 _
_ Vii: _ [h,.,.h,.,. — (12,3) + he he — (hi1) ] —

Therefore we have proved

 

2n(n +1)c — (§)c Z scal — scal(Lf) (111)

l—l

for every totally real l-dimensional space LI'. Therefore
2,: S [2n(n + 1) - (3)16- (112)

From the sequence of inequalities above, it is clear that equality holds everywhere

if and only if (Mi-)2 + (hj’g)2 = —2h§h;¥j, for any disjoint pair i,j from 1 to l and

51

2 : fi. The equality case for l = 2 has been completely described by B.-Y.Chen in
[22]. For I 2 3 the fact that h,,- = 0 is immediate.

Now let’s prove the claim (2) of the theorem 5.4. It has been proved in [22] that
seal 5 (2n2 + 2n)c. (113)

We multiply this inequality by p > 0 and we add it to (105) term by term. We get

 

(p +1)scal— inf scal(Lf) g [(p +1)(2n2 + 2n) — (12)]c (114)

l — 1
Dividing both sides by p + 1 > 0 we get

1
p+1

 

 

4
seal — inf scal(L?) 3 [(2n2 + 2n) —

(l — 1)(p+ 1) (2)]c. (115)

By denoting :45 = k, we get p = i — l = 5%. Using this in (115) we can write the

result as

 

k inf scal(L?) 3 [(2n2 + 2n) —- E(’2)]c, (116)

scal—l_1 4

which is the claimed inequality. From the equality case in relation (3.3) in [22], claim
3 of Theorem 2 from [22] and the claim ( 1) of present theorem, if the equality holds
at every point for some k 6 [0,4] and some I between 2 and n, then M 2 is a totally
geodesic submanifold. In detail, the argument can be written as follows, for k 6 (0,4).

(The argument is practically the same as in [22].)
k k k
[271.2 + 2n — 1(2):] C = (1 ‘ Z) 631+ 16;) S (117)

S (1 — 2) (2n2 + 2n)c + 2 [2n2 + 2n — (12)] c = [2n2 + 2n — 2(3)] c.

52

From this equality, in particular one gets 65,, = (2n2 + 2n)c. It is shown in [22] that

this equality at every point implies that the submanifold AF is totally geodesic.

Let us remark that the implication of being totally geodesic from the equality
doesn’t have a statement which is similar to (4) of Theorem 2 in [22]. In fact, for the
case I = 2, the equality situation has already been completely discussed in [22]. In
this sense, the equality case for n 2 3, 3 S l S n, is different from the situation for
l = 2 where strongly minimal complex surfaces appear naturally. We have seen in the
generalization that a natural cut off in the expression of 4n(n + 1)c — 2scal matches
an expression obtained from Gauss equation. This match points out the sharpness of

inequality in the case l = 2 studied in [22].

53

5.4 Characterizations of Strongly Minimal Surfaces in the
Complex Three Dimensional Space

In this section we consider a complex surface M 2 C (C3. The coordinates of the ambient
space are 21, z2, 23; for j = 1, 2, 3. Put 2,- : x,- + iyj. We suppose M is embedded so

that there exists d E hol((C3) such that M = {z 6 C3] gb(z) = 0} = V(qfi) and

£92_(3¢ 03” 92
02—

821 ’ 022 ’ 0.23
never vanishes on .M.

Let us assume that p : (21(1), 23, zg) is a nonsingular point on .M. Two unit normal

vectors at p are E and J{, where

10?):

i: ||0¢/02||5'

By definition, M 2 is called strongly minimal in C3 if the second fundamental form can
be written pointwise as follows. There exist an open neighborhood U C M of p such
that there exists two orthogonal unit length vector fields X and Y on U, such that
the second fundamental form with respect to the orthonormal basis {X , Y, J X , JY }

can be written in the form:

(2) we a(z) d<z>
A z (z) —a(z) d<z> —c(z)
3 (z) d<z> —a(z) —b<z>

z) —c(z) —b(z) a(z)
and, respectively,

—c(z) —d(z) a(z) b(z)

A“: —d<z> a(z) 3(2) —a(z)

a(z) b(z) C(z) d(z)
b(z) -a(z) d(z) -c(z)

where a, b, c, d are real analytic functions on U.

54

Suppose that the strongly minimal submanifold is realized on U as a graph man-
ifold, i.e., 23 : f(21, 22). Let us consider an open set V C C2 and w : V —) C3 such

that 02(21, 2;) = (21, 22, f(21, 22)). we also have (15(21, 22, 23) = f(21, 22) — 23. Then

e. - a“ = (1.2%) e -— a” = (0.1.25).

_ 'o—zl oz1 2 — 0—22 022

and Jej 2 ie,, wherej = 1, 2. To express f as a function of a(z),b(2),c(2),d(2), one

may use relations of type:
a(Z) = (A,X,X) = (h(X,X).€) = (VXX - VxX.€),

b(z) = (AsY,X> = (h(Y.X),E> = (Vs/X -— VYX.€).
C(z) = (AEJX,X) = (h(JX,X),g) = (VXJX — VXJX,§),
d(2) = (AgJY, X) = (h(JY,X),§) = (vXJY — Vme).

We use below these equations in the proof of the parametric equations of a strongly
minimal surface.

We consider the real and complex parts of the function f as follows.

Z3 : f(zl) Z2) : U($1,$2,y1,y2) + iU($1,$2, 311,92).

We use the notation:

a».
6171

:: uitl)

and the other similar notations. In fact, we have

um = vy vx. = —u (118)

J” J y):

55

since f is holomorphic with respect to both variables. Let us compute el and e2 in

terms of function u and its derivatives:

(90)
=—= 10 I,0,0,— ,,
61 821 (a 7u’l “’31)
(90)
62 Z 8—22 2 (03131141223070) —uy2)7

where the first three components correspond to the real part, and the last three
components correspond to the imaginary part of el and eg, regarded in (C3. Let us

compute

_ 62f
V3181 = (0,0, 5;?) (119)

To compute the projection Velel of Velel to TPM, one needs to compute every term

of the expression:

 

 

_ e1 e1 — 62 62
V8 6 = <Ve e , > + <Vele, > + 120
. 1 *1 HelH Hedi 1 H62“ Heal ( ’

- J61 > Jel <— J62 > J82
V816) + V816 3—
< 1 HJedl ”Jedi 1 HJeal HJeal

In the following considerations, one may use the Cauchy-Riemann equations and

 

 

the fact that u = Re f is harmonic, one may express everything in terms of u.

Let us remark that the harmonicity of u can be written as

“Ii-1'1 + ”mm = 0: ”$2112 + ”312312 Z 0'

We get the following expressions for the covariant derivatives on the complex
surface.

56

__ ureul‘lxl + “yiuriyi

Vale] — , (1,0,uat ,0,0, -—uy ) + (121)
1+ 1131+ 2112/1 I 1

 

“127113111 + uyg ”3313/1

u um — uI u
1+ U2 + U2 (1) 0’ Ill/1:2) 0) 0) _uyz) + y] 1x1 1 x1y1(0, O,Uy1, 1, 0, ux1)+
$2 312

2 2
1+ u$1 + uyl

 

 

“1121111931 _ uI2UI1y1(O 0 1 O )
1+ U2 + 212 3 3 uyz: 3 auxz 3
$2 "312

 

“’1‘ “'1' 1‘ +uy1uy1$2
V. = 1 12 1,0, ,,0,0,— , 122
182 2(1-+113:1 +1312“) ( u 1 My ) + ( )

 

U12u$132 + uy2uy1$2 “111 ”171352 _ ”$1 uylxz
2(1+ ug, + 11.5,, ( 2 ”2) 2(1+ 2131+ ugl) ( 3" 3‘)

 

uyzuzi-‘Bz — “32%132
0, 0, , 1’ 0) ux 1
2(1+ ug, + «1,2,, ( u?” 2)

 

ux um + ’11,, U3;
2 1 + ugl + u?“ 1

”327113232 + “212 “$2y2

lioaux 30101—71 +
1 +71%: +1352 ( 2 312)

 

 

U at x — “a: um
yl 2 22 12 292(0,0,uy1,1,0,u$1)+
1 + 11,,1 + uy1

57

 

31.1 Jr: 2'2 + 2 -y-(030’uy23130>ux2)-
1112 um

The other expressions for covariant derivative may be deduced as follows.

Ve281 : Vel 82,

VJele1= JVele1+[Je1,el] = JVelel,

Vjezeg = JV8262 + [J€2,€2] : JVQBQ,

VJe1€2 = JVe2€1 + [J€1,€2],

VJegel = Jve1€2 + [J62381],

VJ81J61= JVelJel + J[J€1,€1] = -—Ve,el,

VJelJeg = JV62Je1+ J[J81,€2],

VJe2J€1 = VJelJeg + [J62, J61],

VJ82J62 = JVJ8262 + J[J82,€2] = JVJ3282.

58

Let us use the notation suggested by the example in the work [22] and let us
consider two vector fields X = ((11,ag,ag,fi1,fig,fig) and Y : (71,72,73,61, 62,63) on

the open neighborhood U. We put

a : (01,02,013), fl 2 (filafi23fl3)7 ’7 : (Pl/1772,73), 6 : (61:62:63)-

The fact that X and Y are tangent vector fields can be expressed as

X = (a1 + ifil)(1,0, 52:?) + (a2 + 2‘32)(0, 1, 352-) (124)
Y =(71+i61)(1,0,g—Zf1)+(72 +152)(0,1,-g;fZ-) (125)

and, for the third component, one may get by direct computations the following.

03 = alu$1 + filuyl + ozgux2 + figuyz, (126)

(B3 : fllull — aluyl + 1821112 - a2uy2- (127)

The conditions g(X, Y) = g(JX, Y) = 0 can be written

0171 + 0272 + 01373 + 3151 + 5252 + 3353 = 0, (128)

‘7151 — 7232 — 7353 + 0151 + 01252 + 01353 = 0, (129)

where aj, Bj, 73-, (Sj, are real analytic functions on the open set U C M, for j = 1, 2, 3.
(The conditions g(X, JY) = g(JX, JY) = 0 are insured by the relations above and

the conditions g(X , J X ) = g(Y, J Y) = 0 yield trivially.)

59

The fact that X and Y are unit vector fields can be written as

a§+a§+a§+fif+fi§+fi§=L (130)

7f+722+7§+6f+6§+6§=L (131)

We have the following parametric equations of strongly minimal complex surfaces

in C3.

Proposition 5.5 Let u be the real part of a holomorphic function f(zl, 22, 23). Then
the complex surface V( f ) = M is strongly minimal if, for every point p E M, there
exist an open coordinate neighborhood U C M of p such that on U there exist four
real analytic functions a, b, c, d and two orthonormal tangent vector fields X and Y

such that g(X,X) = g(Y, Y) = 1, g(X,Y) = g(X, JY) = 0

X 2(01102)a3)fil)fl21183)1 Y :(71a’721’73161a62763)

a :(01202aa3)a16 : (51:1827fi3):7 : (71172373)76 : (51,62,615)?

such that on U we have
"11(2)“ + “12:1 + ”(132 + “:1 + 11:2)1/2 : (a? _ I8?)u33131 + 2alflluxlyl + (132)

(0102 _ ,8152)ux1x2 + (alfl2 + 021601111122 + (a; "‘ 5%)Ux222 + 2a2182ux2yg
: —(712 — 6f)ux1x1 — 27161114131311 - (7172 _ 6162)u$1$2

_(7162 + 7261)uy1$2 _ (73 _ 6%)“.‘5212 — 272621112312,

60

—b(Z)(1 + U31 + 11-32 + U32“ "l‘ ”(1:2)1/2 ‘2 ((1171 — ,8161)UII$1 'l‘ (133)

1
(5171 + 0150111513,, + 5(0271 + 0172 " £3251 — 5152lux1z2+

1
5(5271 + 5172 + 0261 + 0162)uy1x2 + (0272 - 3252)uz2z2 + (5272 + a252lux2y2,

—c(z)(1 + “:1 + U32 + u; + ail)”2 = —201fi1u11x1 + (of — 512M313“ — (134)

(alfli’ + 02/81)u13112 + (0102 — fi1fi2)uy1x2 _ 20218271112132 + (0% — ’83)”:ng :
271612113111 + (6? _ 712)ux1y1 + (7162 + 7261)u$1x2+

(6162 — 7172)“!1122 + 27262u$2152 + (6; — 7%)1‘3323123

d<z><1 + u; + at + u; + at)” -= (c1161 + mount - (135)

1
”(01171 — 31501121311 + 5(0152 + 0251 + 3172 + ,3271)ux12:2+

1
5(5251 + 5152 — 01271 — 0172)uy1x2 + (0252 + [3272)U12x2 + (5252 — 01272)’Ux2y2-

Proof: Let us remark that the normal unit vector field g has the form (see for

example [70])

g = (1+ ”3:, + “:2 + u; + u§2)_1/2(u1,,u32, —1,uy,,uy2,0) (136)

61

Applying this fact, let us consider the expression which yields the first entry
in the second fundamental form operator: g(AéX, X) = a(z). Now, Let us use a
computational idea presented as relation (3.2) in [70] respectively relation (5.12) in

[22] to get
0f — " _ an
g(—H;,;H xxx-kit ,X) =a, (137)
where

6%

'k 2 .
3 823-82,,

 

This can be written in detail as

 

 

. 6% 65¢ ta" .
89¢ O1 — 251 5;? 6:19;; 0 011 + 1,31
"H‘a—H-lg 012 — 1'32 6626?, ‘92 O , 012 + W2 = 0(2)-
Z a .— .B 21 4:2 632 + .fl
3 Z 3 0 0 0 O3 2 3

Let us remark that in general one can use as basis of the tangent fibre bundle on

U the orthonormal frame {X , Y, J X , JY } This means, for our computations, that
um" = (u,X)X + ('0, Y) Y + (v,JX) JX + (v,JY) JY.

In fact we need just g(vm", X) = (i), X) . (We denote consistently by ( , ) the scalar
product in R6.)

In this context, we have

_ _ a [B
Xij : (aluxixi + IBlu-lel + 3211931132 + gait/1132, (138)

011 ,31
3- + 7mm + azuxzxz + flzuzm, 0,

62

(32 012 011 51 O
aluriyi —' filuri-‘n _ 3711112 + gut/1332) gum-’52 — —2_u11$2 + 02113.2” — B'Zuxzxzv )

Computing the 6—dimensional scalar product we get the claimed equation. Similar
computations prove the other analogous equalities.

Now, let us study the Gauss and Codazzi equations of a strongly minimal complex
surface into (C3. Following [61] and [50], the computational idea is to write explicitly
relevant relations of the complex surface in (C3. In [61] there are defined and studied
the symmetric covariant tensors h and k and the tensor field 3, of type (0,1), such

that the Gauss and Weingarten formulae are

my = VXY + h(X, Y)£ + k(X, Y)Jg, (139)

We 2 —AX + s(X)J{. (140)
With these notations, the Gauss and Codazzi equations (see for example [61]) are
R(X, Y, Z, W) = g(AX , Z )9(AY, W) - g(AX, W)9(AY, Z) + (141)

+g(JAX, W)g(JAY, Z) — g(JAX, Z)g(JAY, W),

(VXA)Y — (VyA)X = s(X)JAY — s(Y)JAX (142)

The Gauss equation has been used to prove Proposition 6 in [22]. It is also the

main idea in the following.

Proposition 5.6 Let {X ,Y} a pair of orthonormal tangent vectors X ,Y E TPM,

such that g(X , Y) = g(X,.IY) = O, with respect to which the shape operator of the

63

manifold .M = V((fi) on the open set U satisfies the strong minimality condition. Then

we have

Ric(X, X) = may, Y) = —2(a2 + b2 + c2 + d2).
Proof: We compute, by the Gauss equation, that
sec(X /\ Y) = g(AgX, X)g(A£Y, Y) — g(AEX, Y)g(A£Y, X)+
+g(JA5X, Y)g(JA5Y, X) — g(JAgX, X)g(JA£Y, Y).

Either using this relation or using relation (5) from [61], we get the claimed fact.
As a remark, the condition [A’ ,Ag] -_-. 0 proved also in Proposition 6 from the
cited work is satisfied identically once we prescribe the shape operator in the form

Ag, as we did.

Proposition 5.7 Let U be an open neighborhood of a regular point M such that on U
there exists a pair of orthonormal tangent vector fields X and Y, with the property that
at every point g(X , Y) = g(X, JY) = 0, satisfying the strong minimality condition.

If s is the tensor field of type (0,1) defined by the Weingarten formula
VXE =3 —A£X + S(X)J€,

then the following relations hold:

X(c(z)) — JX(a(z)) = s(X)a(z) + s(JX)c(z); (143)

X(a(z)) + JX(c(z)) = —s(X)c(z) + s(JX)a(z); (144)

64

Y(c(z)) — JY(a(z)) = s(Y)a(z) + s(JY)c(z); (145)

Y(a(z)) + JY(c(z)) = —s(Y)c(z) + s(JY)a(z). (146)
Proof: It is convenient to work with the following form of the Codazzi equation:
Vx(A£Y) —' Vy(A£X) + A£([Y,X]) = S(IL‘)JA£Y — S(Y)JA£X. (147)

Let us prove for example the third equation from the ones stated above. We write
the Codazzi equation in Y and J Y and multiply on the right by Y (we understand

by multiply the product given by the metric ( , ) in (C3). We get
(Vy(A£JY), Y) — (VJy(A£Y), Y) = s(Y) (JAgJY, Y) — s(JY) (JAgY, Y). (148)

Using JY = iY and the metric property of the Riemannian connection on the

submanifold U, we have

Y (AgY, JY) — (AgY, VyJY) - JY (AEY, Y) + (AgY, VJYY)
= —s(Y)a(z) — s(JY)c(z).

Now, one can use the fact that: JVyY = VyJY = Vin = iVyY to simply
the expression. Furthermore, the shape operator A5 has a prescribed form on the

considered basis. Therefore, we find

Y(—c(z)) — (AgY, JVyY) — JY(—a(z)) + (AgYJVyY)

65

: —,3(Y)a(z) — s(JY)c(z).

This proves the relation (145).

Similarly one can prove by the same steps the other equations.

We have used so far four cases of the Codazzi equation: in X and J X multiplied
by X and JX, and in Y and J Y multiplied by Y and JY. Let us now use Codazzi
equation in X, J X multiplied by Y, respectively JY, then Codazzi equation in Y,

J Y, multiplied by X, respectively J X .

Proposition 5.8 Let U be an open neighborhood of a regular point M such that on U
there exists a pair of orthonormal tangent vector fields X and Y, with the property that
at every point g(X , Y) = g(X, JY) = O, satisfying the strong minimality condition.

If s is the tensor field of type (0,1) defined by the Weingarten formula
Vxfi = —A£X + s(X)J§,

then the following relations hold:

X(d(z)) — JX(b(z)) = s(X)b(z) + s(JX)d(z); (149)
X(b(z)) + JX(d(z)) = —s(x)d(z) + s(JX)b(z); (150)
Y(d(z)) — JY(b(z)) = s(Y)b(z) + s(JY)d(z); (151)

66

Y(b(z)) + JY(d(z)) : —s(Y)d(z) + s(JY)b(z). (152)
The proof is similar to the one given in the previous proposition.

Corollary 5.9 Let U be an open neighborhood of a regular point M such that on U
there exists a pair of orthonormal tangent vector fields X and Y, with the property that
at every point g(X,Y) : g(X, JY) = O, satisfying the strong minimality condition.

Ifs is the tensor field of type (0,1) defined by the Weingarten formula
Vxé = -A£X + S(XlJé,
then we have the following relations:
(X + Y)(d(z) + b(z)) + (JX + JY)(d(z) — b(z)) = (153)

(800 + S(Y))l(b(3) - (1(2)) + z(WI) + d(2))1
(X + Y)(c(z) + a(z)) + (JX + JY)(c(z) — a(z)) = (154)
(800 + S(Y))l(a(2) - C(2)) + 2(0(2) + a(2))1
(X + Y)(a(z) + b(z) + c(z) + d(z)) + (JX + JY)(c(z) + d(z) — a(z) — b(z)) =

(155)
(300 + s(Y))[(a(2) + b(2) - c(z) - d(z) + z'(a(2) + b(2) + C(Z) + d(2))l-

Proof: Straightforward linear computations from the previous two propositions.

67

Proposition 5.10 Let U be an open neighborhood of a regular point M such that on U

there exists a pair of orthonormal tangent vector fields X and Y, with the property that

at every point g(X, Y) : g(X, JY) = 0, satisfying the strong minimality condition.

Suppose that at least one of the analytic functions a, b, c, d is nonvanishing everywhere

on U, say a 7£ 0 on U. Then we have

 

X ( (1(2) _ ib (z)) _ b(z) + id(z)

"’ (1(2) + iC(Z)X(C(Z) — ia(z))?

Y<d<z>-z~b<z»=:2:1:::::;

 

Y(c(z) — ia(z)).

Proof: We have the relations:

X(d(Z)) - JX(b(Z)) = S(X)(b(Z) + 24(3)),

X(C(Z)) - JX(a(Z)) = S(X)(a(Z) + z'C(Z))-

(156)

(157)

Solving the second equation for s(X) and replacing in the first, then using J X = iX

and keeping in mind that a 75 0 everywhere on U, we get the claimed result. Similar

relations hold true in Y.

68

6 A Study of Strong Minimality Through Exam-
ples

6.1 63 = 0 on degree two complex surfaces

In the previous section we have seen a few characterizations of strongly minimal
complex surfaces in (C3, as for example the parametric equations, in Proposition 5.5.
We keep the same notations in the present section, which is consistent with the
notation of [22].

In [22] it was proved that zf + 23 + 2% 2 1 is a strongly minimal surface. We
compute here locally the functions a,b,c,d and see how the parametric equations

look like in this case. First of all, we have

f(Z1,Z2)=(1- 2? - Z§)1/2.
where we work locally on the principal branch of the complex radical. Since 21- 2
xj + iyj, j = 1,2, 3, we will denote C by

C = 1 - 2i - 3% = (1‘ 33:12 _ 33% + Eli + 313) + i(—2$1y1 — 232%)-

If we denote by 6 the polar angle of the complex number C then, on the principal

branch of the complex radical, we get

6 0
(1/2) = _ ' ' _
C |C|(cosz+ism2),

or, by a direct computation, we have

“21.22) = %(\/|C| + Rec +M<I — Rec),

69

where Re( 2 l — x? — x3 + y? + yg. Therefore, we find

VIC] +ReC, (158)

"($1, $27y11 y'Z) :

Sl"

”(371,152,3/113/2): :15 ICI — Rec. (159)

It was established by Chen in [22] that, for a position vector x = (a1 + ibl, a2 +

tbg, a3 + ib3) E V( f), the basis realizing the strong minimality condition is given by

 

1 ()2 20.1
62— , ,1 160
1 n(mml ( )

 

e,- —1— “’2 a (161)

Therefore, by a direct computation, we get

a(z )= 41?,— fir‘ (L—:“ + 1) (162)

 

a,+b§
6(2) -—- 41%11-1 (163)
C(z)=o (164)
6(2): —g—ll 11-1 (01“,: 1) (166)

at every point x = (a1 + ib1,a2 + ibg, a3 + ib3) E V(qfi).

Let us study here the following generalization of B.-Y.Chen’s example presented

in [22]: what are the sufficient conditions for a complex surface given by Az1 + Bz2+

7O

02% = l to be strongly minimal .9 The general answer is stated below in Proposition
6.1. The existence of points where the strong minimality is observed is proved in
Proposition 6.2.

Let 1le? be the complex surface in (C3 defined by:
M2 = {z e (CW/1212+ 323 + 02; = 1, A ,1 0,13 71 0}, (166)

We have seen in the previous section that A12 is strongly minimal if at every point
p E M we can prove the existence of two vectors X, Y E TpM such that the following

system holds

g(X, Y) = g(X, JY) = 0, (167)
g(X,X)=9(Y,Y)=1. (168)

g(X.€) =9(X.J€) =9(Y,£) =9(Y.J€) =0. (169)
g(A5X, X) + g(AEY, Y) = 0, (170)
g(AJgX, X) + g(Ang, Y) = o. (171)

From equations (167, 168), using the notation X = a + ifi and Y = ’y + i6, we get

(0.)) + (6.5) = 0. (0.5) = ([317), (172)

“all2 + IIBII2 =1, llvll2 + Hill2 = 1- (173)

71

Let us remark that 3% : 2(Az1, 822,023) never vanishes on M 2. The unit normal

vector field is:

11315)) 3%: : mgr—”WA“ ‘ “013W — z(2)6001; - 163)). (174)

 

5:

where Ilgfll :- 2(A2(a1+b1)2+B2(a2+b2)2+C(a3+b3)2)1/2- The conditions 90(16):

g(Y, {) = 0 yield the equations

A0101 + BCLQQQ + 00303 — Ablfil - Bbgflg — 0113,83 = 0, (175)

A0171 + 30272 + 00373 — Ab161 — Bb262 — Cb3d3 = O. (176)
The conditions g(X, J5) : g(Y, J6) = 0, yield the equations:

Ablal + Bb202 + Cb303 + Aalfil + 302,82 + ca3fi3 2’ 0, (177)

Ab1’71 + Bb2’7’2 + Cb3’73 + [10,161 '1' 30.262 + 00363 = 0. (178)

Let us compute now the matrix:

85 f
623-62,,

 

= 2diag{A, B, C}. (179)
By direct. computation we see that g(AgX, X) + g(AgY, Y) = 0 is equivalent to

4(04‘ —- £3? + 7i - 51H 3(03 - fl? + ”)3 - 53) + C(Og - 53 + 732 - 53) = 0- (180)
Using the properties of the complex structure J, from

g(Ajgx, X) + g(Ang, Y) = 0,

72

we get.
g(AgJX,X) + g(AgJY, Y) = O. (181)
This equation can be written as
2401/31 + B02132 + C0353 + A7161 + B7262 + C7363 = 0. (182)

Besides the above presented equations, we have also the constraints that describes

that p = (a1 + ib1,a2 + ibg, a3 + ib3). These two relations are

A(af — 5%) + 3(63 — 53) + C(a§ — 53) .—. 1, (183)

Aalbl + 302122 + C(13b3 Z 0. (184)
Therefore we have proved the following.

Proposition 6.1 The manifold M 2 = {z 6 (33/242? + B222, + (1'25;2 = 1} is strongly

minimal in (C3 if and only if the following system of equations admits a solution in

a, fi, '7, 6:
(0.7) + (3,5) = 0. (185)
(0,5) = (6,7). (186)
||a||2+|lflll2= ||7l|2+||5||2 =1, (187)
Aalal + Bagag + Cagag — Ablfll — Bbgfig — Cb3,83 = 0, (188)

73

A0171 + 80272 + 00373 — Ablél -- Bbgdg — Cb3d3 = 0. (189)

Ablal + 80202 + Cb303 + A0151 + 802,132 ‘1' 003,83 = 0, (190)

Ab171 + Bb272 + Cb373 + A0151 + B0262 + 00363 = 0. (191)

A(Oi-flf+)12-5i)+ 3(03 -fl§ +73 ~53) +C(a§ 433 +732 -5§) =0- (192)

A(0131 + 7151) + B(C1252 + 7252) + C(O3fi3 + 7353) = 0 (193)
This result is needed to prove the following.

Proposition 6.2 On the complex manifold L12 = {z E C3/Azf+Bz%+Cz§ = 1, A #

0, B # 0} there exists points of strong minimality. At these points (if; = 0.

Proof: Inspired by the solution given by B.-Y.Chen in [22], we will look for solu-
tions satisfying the conditions: 02 = O, 61 = [33 = 0, 71 = '73 = 0, 62 = 0. Two of the

equations of the system vanish identically. The other eight equations are:

01151 + 01353 = 5272, (194)

a? + 03 + 622 = 1, (195)

)3 + 6% + 63 = 1, (196)
Aalal + Ca3c13 — Bbgflg = O, (197)

74

80272 — Ab161 — Cb363 Z 0, (198)

Ablal + Cb303 + 802,32 2 0, (199)
Bb272 + A0161 + 00363 = 0, (200)
4(04 flit) +B()§ -fi§) +C(a§ -5§) = 0- (201)

Let us consider points with a = (a1, 0,0), and b = (0, b2, 0). The constraint is Aaf —
Bbg = 1. Let us assume further that al 76 0, b2 75 0.
With this assumption, two of the equations vanish identically. The remaining

system has six equations:

01151 + 0353 = 3272, (202)

af + a3 + 03 = 1, (203)

)3 + 6,”- + 6; = 1, (204)

A1115, — 35262 = 0, (205)

86272 + 46,61 = 0, (206)

AM? - 617+ 86% — 6%) + C(a§ - 6%) = 0- (207)

75

Let us solve for (11 in (205) and for 61 in (206):

 

_ Bb2fi2
Bl>272
5 = _ , 2
1 A01 ( 09)

We obtain a system with four equations and with four unknowns, 03, 62, 72 and

(53, as follows.

 

_B_27l:_1__§f:72 + 0363 = (3272, (210)
3:22.33 + at + (3% = 1+ (2“)
7% + $72523 + 62_ _ 1 (212)
8—323032- 5+) 66% — 63) + C(aé - 63> = 0- (213)

This system can be written equivalently as:

 

13263

(1363 = (I. + A202 ———§)fi272, (214)

13253
(1142+ ) a, (215)

13252
@+A,a§)fi+fi=r (mm

01

2 2

((3,2 )3) —(Ba,2 b2 —B) + C(a§ — 63) = 0. (217)

76

In this setting the constraint is Bbg + 1 = Aaf. This system admits the solution
(in particular for A = 1 and B = 1 the result is consistent to the one obtained by

B.-Y.Chen in [22] for all the points of the surface with A = B = C = 1)

 

 

Bbg 1
a = 1 ,0, — , 218
(¢2(A2a3 + B265) V2) ( )

 

 

 

 

A01
2 0, _ .2 ,0 , 219
B ( \/2(A2a3+B2b3) ) ( )
Aal
= O, , ,0 , 220
7 ( \/2(A2a3+B2b3) ) ( )

 

 

13b2 1
6: — ,0,— . 221
( \/2(A2a3+82b3) «2) ( )

This concludes the proof of Proposition 6.2.

If the previous result proved the existence of points satisfying the strong mini-
mality condition in C3 for a specific class of surfaces, the next Theorem describes an

example where the strong minimality holds at every point.

Theorem 6.3 The K c'ihler submanifold given by
M2={zEC3/zl+zg+z§=n,n€C}
is strongly minimal in (C3.

Proof: We use consistently the notations from [22], as well as everywhere in this
current section. Let f(z1,22,23) = 2:1 + :52 + :53, then 35 = (1,1,2z3). For z =
(21,22,292 (a1+z°b1,a2 + z'b2,a3 + ib3), we get Ilgfll = (2 + 4(a3 + b3)?” > 0-

77

We compute also 29%;sz = diag(0, 0,2).
The submanifold is strongly minimal if and only if at every point z there exists

two vectors X and Y such that

g(X, X) = g(Y, Y) = 1, (222)

g(X, Y) = g(X, JY) = 0, (223)

g(X,€) =9(X,J€) =9(Y,€) =g(Y,J€) =0, (224)
g(AgX, X) + g(AgY, Y) = 0, (225)
g(Ajéx, X) + g(AJEY, Y) = o. (226)

From the first equation of the system we keep just g(X, X) = g(Y, Y), and in the
last step we will normalize the basis obtained. With this adjustment, the system be-

comes, using the same notation convention as before, i.e. X = (01,02, 03,,81,,82,,63),

Y 2 (71272173161? 62’ 63)

a§+a§+a§+fif+fii+fl§=vf+7§+7§+5i+5§+5§, (227)
(an) + (fl, 6) = 0, (228)

(0,5) = (Kim), (229)

011 + a2 + 2a3a3 — 2b3fl3 = 0, (230)

78

21 + (32 + 2a3fl3 + 2b3a3 = 0, (231)

71 + ’72 + 2a373 — 2b363 = 0, (232)
61 + 62 + 2a363 + 2b373 = 0, (233)
03 — (33 + 73 — 63 = 0, (234)
01353 + 7353 = 0- (235)

A solution for this system is obtained if one is taking ,63 = 73 = 0. With this

choice, one may elliminate the unknown

(11 = —C¥2 — 261303 ,61 = —,82 — 2b303, (236)
’71 Z —’72 + 2b363 (51 = —62 — 20.363. (237)
Setting 012 = 0, 03 = 1 and 63 = —1 one may get, through a direct computation:
C1 = (—2a3, 0, 1), (238)
V2 4b2 V2 4b

(3 = (—b3 - ———‘2* 3,—b3 + ——3 3,0), (239)

2 4b2 V2 41)2
’7 = (—b3 + ——_2*:—_3'a _b3 — +20): (240)
6 = (0,2a3, —1). (241)

79

One may verify directly that the system (227)—(235) is satisfied by the above
solution. Therefore, after normalization, the basis satisfying the strong minimality

condition is

. ,/2 4123 ,/2 4123
X=(2+4a§+4b§)'1/2(—2a3—i(b3+2 + ————), i(— —b3+ + ———),1), (242)

. 2 4b2 2 4b2
2 z (2 2 4.2 + 422-22 (.2. + _Vgx, -2. _ _Vgs . 2..., _.) . (2.2)

6.2 6’" = 0 on degree three complex surfaces

The surfaces 21 + 212 + 23 = 1 and 23 + 23 + 2% = 1, as we have seen, are strongly
minimal. We mentioned above sufficient conditions for A23 + 823 + 023 = 1 to be
strongly minimal. Let us extend our discussion to complex hypersurfaces of higher

algebraic degree; we prove the following.

Theorem 6.4 On the complex surface M given by the algebraic equation zf+z§+z§ =

1 there exists points where the strongly minimality condition in C3 is satisfied.

Proof: Using the notations from [22], we get f (z) = 23 + 23 + z; — 1; therefore

6f
52 = (323,323,323).

Let us prove first that Hgfll 7g 0 at every point p. We have
II— i-II —- 3M y?)2 + ($3 - 21%)2 + (23 — .213)2 + 4:23:23 + 4w2y2 + 42:32:31:

80

31(17? + 31W + ($3 + 313.)2 +017}? + 213)2 > 0-

The unit normal vector 5 (written as a real vector field) is given by

3 *1/2
5: (Z(£?+yz2)2) (Ii 912332— 312,373_93,—21313/12—21'292,—2$3y3)- (244)

£21
Let us consider a point p in C3 whose position vector is given by (a1 +ib1, a2 +ib2, a3 +
ib3). The tangent space to our complex surface M at p is the set of all vectors of the

form

Z =(u1 + i221, U2 + iv2,U3 + iv3)

which satisfy the following conditions:

of .67 _
4 .—;->- «mm—a)

and these conditions yield the equations:

121(503 — y?) + 222(253 — y?) + u3(a:§ — 313) — 2$1y1v1 — 2xgy2v2 — 22:3y3v3 = 0, (245)

2u1x1y1 + 2u2$2y2 + 2u3$3y3 + 121(933 — y?) + 122(233 — yg) + v3(:c§ — y3) = 0. (246)

The condition that the point p = (21, 22,23) lies on the complex surface M is

expressed by the following two equations:

2&3 —3ajb§ 1:, (247)

3
2(3a3b, — b3) = 0. (248)
j=1

81

Now let us study the shape Operator, using the formula (discussed in [70] and applied

in a similar setting in [22]):

 

85f tan
A W— — - 1 —— 24
2 HZ— :‘H {W (021.6221)} ( 9)
First note that
2 621 0 0
06; = 0 62:2 0 (250)
22’ 2k 0 0 6.23

The product in braces yields

2
xaf

5— : (6(01 — 2,81)(1131 —iy1),6(a2 - 2,62)($2 — iy2)76(a3 _ ifl3)($3 - 2313)) Z
zJ-sz

(251)

( 6(0111131 — 513/1) — (Wall/1+ 511131),6(02$2 '- 523/2) — 6i(012312 + 32232),
6(0131153 — 53113) “ 62(033/3 + 53153) )-
The first condition for strong minimality is g(AfX, X) + g(AgY, Y) = 0:
_ 05f tan _ 65f tan
X Y Y = 2 2
g ({XBzJ-sz} , ) +9 ({ dzJ-sz} ’ 0 ( 5 )

This equation can be written also as

 

 

3
2[ [3(0' - 182M —20jfijbj +( _12— 6.?)(13’ _ 27j6jbj] : 0 (253)

j=1
The second condition for strong minimality is g(AJ£X, X) + g(A “Y, Y) = 0. Since
A J5 = J A5 = —JA€ (see for example [61]), this is the same as discussing: g(AEJ X , X )+

g(AEJY, Y) = 0. This can be written as

- 65f tan 2f _—
gmm} ,X)+g({:,yaz2mk} ,Y)_o (254)

82

 

 

or, breaking down the computation, as

3
Elan—0152' — Ojbj) + 5103115 — 0102') + 71(—aj5j — ijj) + 51(51191 — 71%)] = 0
1:1
(255)

and, grouping terms as in the similar relation above, we get

3
Z[—2ajajflj + bj(fi]2 - 012) — 2aj’7j6j + bj((512- — 732)] = O. (256)

i=1
In fact, to prove the strongly minimality of [VI in C3 is equivalent to find an ortho-

normal basis {81,62}, (31 = (a + w), 82 = (7 + 7L6), satisfying the following system:

(Ida? —- bf) + 02(ag — bi) + 03(ag — b3) — 2a1b151 — Zagbgfig — 2a3b333 = 0, (257)
ma? — 6%) + 72(ag — 63) + 73(63 — 63) — 2a1b161 — 2a2b252 — 2636363 = 0 (258)
2516161 + 2626262 + 2636363 + ma? — bf) + 62(63 — 63) + 63013 — 63) = 0, (259)

27101171 + 27261.2()2 + 273a3b3 + 61(03 - b?) + (52(03 — b3) + 63(03 — b3) = 0, (260)

31 [(6]? — 6]?)aj — 2aj6jbj + (7,? — 6})aj — 27,.6jbj] = 0, (261)
j:
3

;[-2ajaj,3j + 51(5)? - 032-) - 203'71'53' + M5,? — 732)] = 0- (262)

(a, 7) + ((3, 6) = 0, (263)

(0,5) = (3,7), (264)

83

af+a§+a§+fif+fl§+fi§=7f+7§+7§+6f+6§+6§=1 (265)

a? — 3616f + a; — 36262 + 63 — 3a3b§ = 1, (266)

36$!)1 — 63‘ + 36352 —- 63 + 36353 — 63 = 0, (267)

Given two vectors a, b 6 1R3 with constraints (266), (267), the equations above (257)-
(265) are an undetermined system which admits some nontrivial solutions 01,5, 7, 6 E
R3.

At the point (1,0,0) the system implies from relation (257) that a1 = 0. From
this and the next three relations we deduce that: a = (0,012,03), fl = (0,62,53),
7 = (0,72,73), 5 =(0162163)'

The system we need to solve now is

llall2 + llfill2 =1, (268)
”7“2 + “(5”2 =1, (269)
(01,7) + (£5) = 0, (270)

(065) = (fin). (271)

This system admits the solution:

a = (0,—}—§,0), 6 = (0,0,%) (272)

84

7 = (6%,0), 6 = (0,0,——%). (273)

This means the strongly minimality condition is satisfied at (1, 0, O).
For the points of type (a1,0,0), (O,a2,0), (0, 0, a3) a similar system admits the
same solutions. Therefore along these curves the stronlgr minimality condition is

satisfied on the orthonormal basis el = a + 2'6, 62 = 'y + i5. This concludes the proof

of theorem 6.4.

85

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