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)0.)

DIFFERENTIAL GEOMETRY
OF
SLAN T SURFACES

By

Yoshihiko Tazawa

A DISSERTATION

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

DOCTOR OF PHILOSOPHY

Department of Mathematics

1989

<2. Ci

'J/A
.JU

Q04

ABSTRACT

DIFFERENTIAL GEOMETRY 0F SLANT SURFACES
By
Yoshihiko Tazawa

We consider immersions of differentiable manifolds into almost Hermitian
manifolds. The Wirtinger angle is a quantity which measures how an
immersed submanifold differs from a holomorphic submanifold. An immersi on
is called a slant immersion if the Wirtinger angle is constant. It is a
generalization of holomorphic submanifolds and totally real submanifolds. A
slant immersion which is neither holomorphic nor totally real is called a
prOper slant immersion.

In this article, we mainly consider slant surfaces of codimension 2. We
first clarify the relation between 2-planes and complex structures of Euclidean
4-space E4 from the view point of multilinear algebra. Combining this with
the Gauss map, we characterize slant surfaces in complex 2-space C2. We
also show that a surface without complex tangent point in a 4—dimensional
almost Hermitian manifold can be a slant surface with any given constant
Wirtinger angle with respect to a suitable almost complex structure. This
shows a big difference between almost Hermitian manifolds and Kahler
manifolds.

Next we show that no compact prOper slant submanifolds exist in any
complex space Cm. This is a similarity shared by prOper slant submanifolds

and holomorphic submanifolds.

Finally, under some additional conditions, we can determine the shapes of
slant surfaces in (2. If a slant surface is contained in a 3-sphere 53, then
it is obtained from a kind of helix in S3 or a great circle by
left-translations along a curve in .53. If a slant surface is contained in a
3—plane, or, more generally, the rank of the Gauss map is less than 2, then
we can do the analogues of the classification of flat surfaces in Euclidean

3—space E3.

ii

ACKNOWLEDGEMENTS

I wish to express my sincere gratitude to my advisor, Professor
Bang-Yen Chen. All of my work has been done with his kind guidance and
many valuable suggestions. I would also like to express my appreciation to
Professor Gerald Ludden and Professor David Blair for their helpful
instruction. Finally, I would especially like to thank Professor Chorng—Shi

Houh who encouraged me to finish my program.

iii

TABLE OF CONTENTS

INTRODUCTION ................................... 1
CHAPTER 1. Preliminaries ............................ 4
§ 1 Notations ................................ 4
§2 Geometry of G(2,4) .......................... 7
§ 3 The Gauss map ............................. 13
§ 4 Slant immersions ............................ 18
CHAPTER 2. A characterization of slant surfaces ............... 24
§ 1 Complex structures on E4 ...................... 25
§ 2 Slant surfaces in c2 ......................... 33
§ 3 Slant surfaces in 4-dimensional almost Hermitian manifolds . . . 40
CHAPTER 3. Compact slant submanifolds in c’" .............. 46
CHAPTER 4. Spherical slant surfaces in C2 ................. 54
§ 1 Geometry of .53 ............................ 54
§ 2 Another Gauss map .......................... 58
§ 3 Classification of spherical slant surfaces ............... 63
CHAPTER 5. Slant surfaces in c2 with rank 1/ < 2 ............ 75
SUMMARY ..................................... 89
BIBLIOGRAPHY .................................. 91

iv

INTRODUCTION

In this article, we consider slant immersions of differentiable manifolds
into almost Hermitian manifolds.

The most natural submanifolds of an almost complex manifold are
holomorphic submanifolds in the sense that they inherit both differentiable
structure and almost complex structure from the ambient space, and the
tangent spaces of the submanifolds are invariant under the almost complex
structure. On the other hand, the notation of totally real immersions (or
anti—invariant immersions, Lagrangean immersions) were introduced in the early
1970's. A submanifold of an almost Hermitian manifold is called totally real
if each tangent space of the submanifold is mapped by the almost complex
structure into the normal space.

Recently B.—Y. Chen defined the term slant immersions as a
generalization of both holomorphic and totally real immersions ([BYCSD. An
immersion of a differentiable manifold into an almost Hermitian manifold is
defined to be a slant immersion if its Wirtinger angle a is constant (Chl
§4). If a a 0, then the immersion is holomorphic and if a a 1r/2, then it
is totally real.

In Chapters 2—5, we will state some preperties of slant immersions which
we have obtained up to now. We consider mainly slant surfaces in complex
2—space (2 expect for Chapter 2 §3 and Chapter 3. In Chapter 2 §3, the
ambient space is an almost Hermitian manifold. Chapter 3 is about slant
submanifolds in complex Spaces of arbitrary dimension and codimension.

Chapter 1 is a preliminary. In §1 we recall the basic formulas of the
differential geometry of submanifolds. §2 is about descriptions of the

Grassmannian G(2,4) as a product of two 2—spheres. The relation between

two descriptions of G(2,4), a quadric Q2 in (P2 and a set Dl(2,4) of
unit decomposable 2—vectors in A2E4, is clarified. §3 is a review of the
generalized Gauss map. In §4 we introduce the definitions and basic
prOperties of slant immersions written in [Bch].

In Chapter 2, we consider slant surfaces with codimension 2. In §1, we
investigate the relation between '2-planes in E4 and complex structures on
E4 (Pr0position 1). This provides a pointwise observation of slant surfaces.
In §2, we consider the Gauss map and characterize slant surfaces in (2
(Pr0position 2). Especially, we show that a non-minimal surface in E4 can
be slant with respect to at most four complex structures on E4. In §3, we
show that any surface without complex tangent points in an almost Hermitian
manifold becomes a proper slant surface with any given constant Writinger
angle with respect to a suitable almost complex structure (Proposition 3). §§2
and 3 show the difference between a Kahler manifold and an almost Hermitian
manifold.

In Chapter 3, we show that a compact pr0per slant submanifold does not
exist in complex space Cm (Pr0position 4). This is a similarity of slant
submanifolds and holomorphic submanifolds.

Most examples of slant surfaces in C2 which we have constructed up to
now are doubly slant and have the rank of the Gauss map less than 2 and
hence flat surfaces (Chapter 2 §2). So, it is natural to consider the problem
of classifying flat slant surfaces in [2. Under some additional conditions, we
can determine the shapes of flat slant surfaces (Chapters 4 and 5).

In Chapter 4, we consider slant surfaces contained in a 3—sphere S 3
in (2. Since 5 3 is a Lie group of unit quaternions, the theory of curves
and surfaces in S 3 is given a Special deveIOpment (§1). In §2, we define

another Gauss map on S 3 using left invariant vector fields and

characterize slant surfaces in S 3 (Proposition 5.). This is a spherical
version of Proposition 2. In § 3, we determine prOper slant surfaces in S 3
(Pr0position 6).

In Chapter 5, we consider a slant surface with the rank of the Gauss
map less than 2. Then, the surface becomes a flat slant ruled surface in' C2
and we can do the analogy of the classical classification of flat surfaces in E3
and determine the shapes of slant surfaces (Proposition 7). In particular, if a
slant surface is contained in a 3—plane in (2, then its shape becomes more
concrete (Pr0position 8).

We are at the starting point of the differential geometry of slant
immersions. We hope it will have a fertile deve10pment similar to the studies

of holomorphic or totally real immersions.

CHAPTER 1.
PRELIMIN ARIES

In this chapter we review and arrange some well—known formulas and
facts which we will use in this article. §1 is a list of formulas of differential
geometry of submanifolds. In §2 we recall the description of the
Grassmannian G(2,4) as a product of two 2-5pheres. § 3 is about the
generalized Gauss map of submanifolds in Euclidean spaces. In §4 we

introduce the definition and basic properties of slant immersions.

§ 1. Notations

We follow basically the definitions and notations of [CBYl] - [CBY3].
Differentiability always means differentiability of class C °°. Listed below are
some formulas which we will use in this article.

Let (M, g) be an n—dimensional Riemannian manifold with the
Rienannian connection V, {ei} 2:1 be a local orthonormal frame field and

{mi} ?=1 be its dual coframe field. The connection form {a};- } is defined

by
i _ i

(1.1) w]. (X) — w(VXej) X 6 TM
i.e.
(1.2) . Vxej = 2 cu;- (X)ei
The curvature form {03 } is defined by

k I
(1.3) - Qj=22RijAw
where
(1.4) R(e ej,ek)el = E lekez'

4

If we put
(1.5) Vs]. ck = 2 1‘37; c2. ,
then
i k

(1.6) wj=E [‘ka I
(1.7) v wk = -2 r", w.

c]. J
The structure equations are given by
(1.8) dw'=-ijl\wj

i _ _ k z‘
(1.9) dwj — 2 wk A w]. + Q]-

Let (M, g) and (M, 3}) be Riemannian manifolds of dimensions 11

and m respectively, and V and V be their Riemannian connections. If
(1.10) 1: : M -1 M

is an isometric immersion, then

(1.11) VXY == VXY+ h(X, Y), X, Ye 5(1)!)

(1.12) VX{= -A€Y+DX{, X65(M), £6.3(M)

where h, A and D are the second fundamental form, the Weingarten map
and the connection in the normal bundle. Let {EA}'2=1 be a local adapted
frame field, i.e., a local orthonormal frame field on M and, if restricted to

M, {tiff}:1 is an orthonormal frame field on M. Let {52A} and {51;}

be its dual coframe field and connection form on M. If we put

(1.14) w' = Ele, i..e, w' = 1’52.
A A

(1.15) “B = DBIM

where

i, j = 1,...,n; A, B = l,...,m

then {82'} and {mi} and dual, {11);} is the connection form with respect

to {ei} and
( ) m s X)
1.16 D = E w e, r: n+1,...,m.
x6. m“ .1 s
If we put
m n . .
h: 2 2: h§.w'caflccr
r=n+l i,j=l J
i.e.,
~ __ r
(1.17) g(h(ez., e]. ), er) - hij ,
then
r _ r

(1.18) hi]. - hji .

r _ r J
(1.19) mi — 2 hij w .
The mean curvature vector H is defined by
(1.20) h = 712- trace )1

_l m n r
— '5 2 ( E hm) 8r.

r=n+l i=1 n
The equations of Gauss, Codazzi and Ricci are given respectively by

(1.21) R(X. Y. z, W) = R(X. Y. z, W) + WM. 2). h(Y. W))
- §(h(X, W): h( Y) 2))

(122) (fax. 1021* = (17th 2) 417,111.13 2)
(1.23) RD(X, Y. e. n) = R(x. Y. c. n) + m5. A,,1x. Y)
where

(124) WWW. 2) = DXh(Y. 2) - 12(va 2) - 12(1ch 2)
(1.25) RD(X, Y)£ = DXDY£ — DYDXg — 0m”);

§ 2. Geometry of G(2,4).

Let Em = (lRm,< , >) be the Euclidean m—space with the canonical

inner product < , >. Denote the canonical basis and orientation by

- (A)
(2.1) {2A}’X___l, %A=(o,...,o,1,o,...,0),
(2.2) w = 9:1 1....11 3m 6 Am 5"".

For each n E {l,...,m}, the space A" Em is a (YD—dimensional real vector
space with the inner product, also denoted by < , >, defined by
(2.3) <X1 A....A Xn’ Y1 A....A Yn> = det [<Xi’ Yj>] and bilinearity.

* n m * . . .
and (A E ) are 1dent1fied 1n a natural way,

Two spaces A"(Em)
namely, for n e A”(E"‘)* and X1,...,Xn e Em,
(2.4) (2(X1 A....A Xn) = 0(X ,...., X11) and linearity.

Let, G(n, m) denote the Grassmannian of oriented n—planes in Em.
Then, G(n,m) is identified with the set Dl(n,m) of unit decomposable
n—vectors in AnEm. The correspondence (0 is given by
(2.5) 1b : G(n, m) -—+ D1(n,m)

w(V) = XI A....A Xn’
where {X 33:1 is a positive orthonormal basis of V.
In particular, if n = 2 and m = 4, G(2,4) is represented as a

product of two 2—spheres as follows. The star Operator

(2.6) * : A2E4 .4 A2E4
is defined by
(2.7) <¢€,17>w=§/11),for£, nEA2E4.

For V e G(2,4), considered as an element of 01(2,4) through w,
(2.8) *V = V‘

where V‘ is the oriented orthogonal complement of V in E4. Since *

is a symmetric involution, AZE4 is decomposed into an orthogonal direct

sum
(2.9) A2E4 = A: E4 e AEE4

where AiE4 are the eigenSpaces of * corresponding to the eigenvalues 11.
Denote

(2.10) xi: A2E4 -+ AEE4

the projections of this decomposition. For a positive orthonormal basis
{8A}j=1 of E4, put

1
1 f2- 1 2 33 4

1
(2.11) < 172 = — (cl/1123 — 62Ae4)

 

_ 1
n3 — —‘/_2__ (cl/164 + e2Ae3)

 

1’4 = (7;: ”1% ‘ 63%)
(2.12) < 715 = E1 (cl/1c3 + c.2Ae4)
. "6 = 31 (cl/\e4 - e2Ae3)

then {171, 172, n3} and {114, 05, 116} are orthonormal bases of AiE‘i
and A3 E4 respectively. In particular, (9)1, (1)2, 973} and {914, 1’75, 916}
obtained from {2A} form canonical bases. For g 6 01(24),
(2.13) M“) = i“ + *5)

at) = as - *t)
and

(2-14) IIW+(€)II = l|7r_(€)ll = 1N7.

Hence, if we denote by 54:2 the 2-spheres of radius 1H? in AEE4

centered at the origin, then

2
(2.15) art: D1(2,4) -o S*
and actually this gives rise to a description
- 2 2
(2.16) 01(2,4) = 5+ x 5_

(cf [S—T] p 360). If we choose an adapted frame {e A} for V in G(2,4);

4

i.e., {(3/1} is a positive orthonormal basis of E such that {(31, 62} is a

positive basis of V, then
(2.17) «m = .W) = twee + em.)
r_(V) = 1r_(lp(V)) = gel/162 - e3Ae4)

A’- E“ a G(2,4)-=2 cites) = 8% x 55

/

 

 

 

 

 

 

 

 

 

FIGURE 1

There is another description of G(2,4) (cf [H-Oll). Let
em) [ y fisCP3

10.

be the canonical projection of the complex 4—space onto the complex projective
Space and define a quadric 02 by
(2.19) o2={(z ,....,z4)|z§+....+ 22 = 0} c c4

a, =112112e 62,) c c P3.
We define a map
(2.21) <I> : G(2,4) -o Q2
as follows. For V e G(2,4), pick a positive orthonormal basis {X, Y} of
V and put Z = X + W. Then, the complex vector Z is contained in
02. Put
(2-22) NV) = [Z 1
Then, (D is well—defined and bijective and hence we may identify G(2,4)
with Q2 through <I> ([H—Ol] p6).

If we define a map

(2.23) (p:CxC-’C4
by
(2.24) ewl. 212,) = (101(w. mg)....e4(w,.w2))

= (1 + w1w2, i(1 - 1111102), w1 - w2, - z'(w1 + w2)),
then (p satisfies
(2.25) 1p? + ~-+cpi = 0
and hence [tp(w , 102)] E 02. On <p(CxC), ([1 is given by
2:34-2:24 ’-z3+z:z4 )
21“”2 ‘1'”2

and the biholomorphic map [tp] from CxC into (22 extends to a

(2.26) 2'10: .---.24) = (

biholomorphic map of (P1 x (Pl onto Q2, when we consider
(wl, w2) 6 (x6 as inhomogeneous coordinates on CP1 x (P1

(2.27) [(0] : (Pl x (P1 -. Q2.

11

We consider the metric on 02 induced from the Fubini-study metric of
CP3 with constant holomorphic curvature 2. We also consider a metric on
(P1 defined, using inhomogeneous coordinate w, by

2 __ 211111212
d3 - 2 2 o
(1+lwl )
Then, [go] becomes an isometry and hence 02 is considered a product of

two 2—spheres of radius 1/42 through [9”]-
(2.29) [tp] : (P1 x (P1 = 5? x 5% -» Q2.
(For details, cf [H—Ol]§2)

 

(2.28)

The following lemma shows the relation between the two descriptions of

the Grassmannian G(2,4) as a product of two 2—spheres.

Lemma 1.
The bijection
(2.30) WOT-10[¢]:S%X5%4SEXSE
is an isomet S2 ' 2 S2 ' 2
ry. 1 is mapped onto 5+ and 2 is mapped onto S_.

(Proof)
Let V e G(2,4) and put
(2.31) W) = (5,. 6-). t. e 53
(2-32) , @(V) = M101. 102)]
Express 5* with the coordinates with reSpect to the canonical basis
{97A}2=1-
l 3 o
= . . . = 2 °
5- (£4 £5 £6) 6- k=4 6k "k

 

l2

 

 

 

 

 

Put
(2.34) A + £8 = (0(a) , w2) i.e. [A + £8] = <I>(V)
Since [AI = |B| it 0 and <A, B> = 0,
(2.35) 1141/) = lit—I12 A A B
Hence I
(2.36) 51 = < 111(V), 971 >
= L2 < A A B, 971 >
IAI1
fl(1+u?1+vl) a b2 — 02b1 + a3b4 — a4b3)
where A = EaAgA, B = EbAeA, and 10]. = uj+ iv]. , j = 1,2.
Therefore,
1 - a? - 11%
£1 = J2 (1+1)? + 617)
By similar calculations,
1 2 2
(2.37) 4 5+ = fl (1+“?+ 11%) (l - ul-vl ,2v1 ,—2a1)
L 5‘ = )2 (1+1); + 1);) (1136346272112)

This shows that the mapping

(2'38) (“1) "2) —’ (51) £21 ‘53)
is the composition of the stereographic projection, the homothety with ratio
IN? and a exchange of the coordinate axes. Note that this is the same
way, except for the change of axes, in which .5‘2 = C P1 was parametrized
and given the metric. Hence, \II o 0-1 o [,9] maps SL1) isometrically onto

2 . . 2
5+. Slmllar for S2.

QED.

13

In this sense, we identify
si x 33 = 01(2,4) = G(2,4) = Q2 == 5% x 5?,
2 _ 2 _

We choose orientations on A384 and Aik" such that {971, 972, 5’73} and

(2.39)

{974, 975, 26} are positive basis respectively and also orientations on S: and

53 corresponding to the exterior normal vectors.

 

 

 

 

 

FIGURE 2

§ 3. The Gauss Map.

Let a: be an immersion of an n—dimensional oriented differentiable

manifold M into Euclidean m—space

14

(3.1) :1: .- M -. Em.
We always identify the tangent spaces of Em with Em itself. Then the

Gauss map, or the generalized Gauss map,

(3.2) V : M -) G(n, m)
is defined by
(3.3) V(p) = TpM c TpEm 5 Em.

Let {8A}’.2=1 be an adapted local frame field on Em, i.e., a local positive
orthonromal frame field so that {411)}?=1 is a positive basis of TpM if
p e M. Then, identifying G(n, m) with Dl(n, m),
(3.4) V(p) = (elA....Aen)(p)

If M is compact, then the Gauss image V(M) is mass—symmetric in
the unit sphere of A" Em, according to Chen—Piccini ([CBY—P] Lemma 3.1),
namely
(3.5) fpeM 1(1))de = o
where u is considered as a A" Em—valued function on M, and (WM is
the volume element of M with respect to the metric induced by the
immersion z.
(3.6) 1/ .- M —-)G'(n,m)=Dl(n,m) c SCI-1(0) c Ann’" = (mm, < , >)

= E”. N = (7:)

We rewrite this as follows for later use.

Lemma 2.
If M is compact, then
(3.7) fM <11, 15> dVM=0

for any n-vector 6 E AnEm .

15

Inthecaseof n=2 and m=4, weput Vt=7ri°ll where ”a

are the projections defined in §1.

 

1%

These maps u+ and u_ are related to the Gaussian curvature G of
M with respect to the induced metric and the normal curvature GD of the
immersion 2: as seen in the following lemma. This lemma is stated by
Hoffman-Osserman ([H—O2] Pr0position 4.5) in terms of the decomposition

G(2,4) = Q2 = 5:12 x S; but we prove it here in our notations.

Lemma 3.

(3.9) Jacobian of 11+ = 5. (G + CD)
Jacobian of u_ = _ % (G __ GD)

(Proof)

Let {2A}A=l be the canonical basis of E4 and {EA}j=1 be a
positive local adapted frame field such that restricted to M, {233:1 is a
positive frame on M. Let {6‘}, {avg}, {eA}, {wi}, {cg} be as in §1.
Put

Bo

16

where (62) is a SO(4)-valued local function of E4. By (3.4) and (3.11)

(3-12) V(p) = 81(2) A 82(2)
AB 0 o
= E t (p) e A e
A<B A B
where
so was wit)
are local functions on M. If we denote by (yAB) 1 5 A < B 5 4 the

coordinate of A2E4 with respect to the canonical basis {‘EAA <3:8}
15 A < B S 4, then, for a function tp on E4 and X 6 TM,

(3-14) (W012 = X02 - u)
6 AB
= 2 43-3—
A<B 6y (X6 )

= (2mm) %) e
01/
Since A2E4 is a vector s ace 6 is identified with ‘3: A ‘3: and

hence

(3.15) WX = 2 (X (AB) %A A °B
A<B

BKX 114)ng. 2401* (25)- (X 2,3)2/2‘ — 5?(X6§‘)J%AA 923
_1(X BAWB eAA eB + AEB-l flA(XflB)° eAA eB

A2
A43
23
=(2 or 21‘s,) A ( 2: 23°e31+ (2 efeA) A ( 2(X 35) t3)
= (VX1) A 32 + 31 MVX ‘32)
Hence by (1.2)

(1.16) V“ X = (E U11(X)8A) A e2 + e1 A (23 Dg(X)eB)
since X 6 TM
(3.17) MX=2w3l3(X)eAe3+w2(X)e144Ae

‘ “1“)32" ‘33 ‘ ”1(X)e2" e4

17

We use here {"a}3=1 defined by (2.11) and (2.12). Then

 

(3.18) w x = yé' [(11% + «13)an + (.111; + «131120173
+ (w? + w§)(X)n5 + («2:13 + w§)(X)n6]
Hence
u+.x = 7:, to»? + ngxng - (-1513 + @0013] -

13-19) u_.x = 5111—4 + wS’XXM, + (w? + w§)(X)n61
By (1.19)

, ”+*(31) = 7; [(h41+h31)’72 + (‘h31+h§1)"3]
(3.20) ”+46? = .1. [(1.4 12+h22)02 + ( 4112+}: 32)?)3]

1461) = ”:2“ h11+h31)'75 + (h31+h§1)”6]
(321) _ u_.(e2) = ; [(-h‘f2+h§2)n5 + (hiz+h~§2)vsl
Since at each 11 e M
(3.22) 11+(p) = 321- (elA e2 + 63A e4)(p) = $17M),

171(1)) is the exterior normal of SE at V+(p) and hence {172, 113} is a
positive orthonormal frame field on $3 with respect to the orientation
defined in §1.

On the other hand, by definitions of G and GD ([CBY2]), and Gauss'

and Ricci's equations,

(3.23) G = R(e1, e2; 32, el)r)
= 23 0' 11"22
(3.24) GD = Del, e2; e4, e3)

3 4 4 4 3 3
= ”120‘22 ‘ hll) ’ ”12(h22 ‘ h
By (3.20)—(3.24), we get (3.9) and (3.10) .
Q.E.D.

18

§ 4. Slant Immersions.

The notion of slant immersions has been defined recently by B.Y.Chen as
a generalization of both holomorphic and totally real immersions ([CBY5]). In
this section we introduce the definitions and some basic pr0perties of slant
immersions written in [CBY5].

Let
(4.1) x : M 4 M
be an isometric immersion of a Riemannian manifold (M, g) with a
Riemannian metric g into an almost Hermitian manifold (M, ‘g, .7) with
an almost complex structure .7 and almost Hermitian metric Z].

For each nonzero vector X tangent to M at p the angle 0(X)
between 3X and the tangent space TpM of M at p is called the
Wirtinger angle.

(4.2) 9(X) = 40.1, TpM), X e TpM

 

 

 

 

 

FIGURE 3

19

If 0(X) is constant, we call immersion a: a general slant immersion.
If a: is a totally real immersion, i.e.,
(4.3) 3(TpM) C T;M for V p E M

P
slant immersion with 0(X) a 1r/2.

where T M denotes the normal Space of M in M, then x is a general

If M is also an almost Hermitian manifold (M,g,J ) and :1: is an

holomorphic immersion, i.e.,

(4.4) n..(JX) = 7(aX) for V X 6 TM

where 1,. is the differential of :c, then :1: is a general slant irmnersion
with 0(X) a 0. Similarly, an anti-holomorphic immersion satisfying

(4.5) 1:*(JX) = —.7(z...X) for V X 6 TM

is a general slant immersion with 0(X) s 0.

In this sense, a general slant immersion is a generalization of totally real
or holomorphic immersions. A general slant immersion with 0(X) ,1 0 is
called a slant immersion and the angle 0(X) is called the slant angle.

A slant immersion with 0(X) )6 1r/2 is called a pr0per slant immersion.

For any vector X tangent to M, we put
(4.5) JX = PX + FY, PX e TM, FX 6 PM
Then, P is a (l,l)—-tensor field on M and F is a T‘M—valued l—form on
M. A prOper slant submanifold which satisfied VP 5 0 is called a Kahlerian
slant submanifold. If (M, g) is a Kahlerian slant submanifold with the slant
angle 0, then, with respect to an almost complex structure 9 defined by
(4.6) 9 = (sec 0) P,

(M, g, 3) becomes a Kahler manifold.
Listed below are some of the lemmas, propositions and theorems in

[CBY2] for later use.

(a)

20

(Lemma 2) Let M be a submanifold of an almost Hermitian
manifold M. Then, VP 5 o if and only if M is locally the
Rienannian product Mlx....ka , where each Mi is either a
Kahler submanifold, a totally real submanifold, or a Kahlerian

slant submanifold.

This lemma comes from the decomposition TpM=D;®- - - @020”,

where 05's are the eigenspaces of the self—adjoint Operator

(4.7)

(b)

((1)

(Lemma 5) If M is a slant surface in (2, then G=GD
identically. 0', GD are as in §3. Hence in this case u_ is

degenerated.

(Proposition 5) Let M be a compact surface. Then we have

1. If the Euler number x(M) at 0, then M admits no slant
embedding in C2.

2. If x(M) = 0, then every slant immersion of M in (2 is
regularly homotOpic to an embedding.

3. If M has positive (or negative) Gauss curvature, then M

admits no slant immersion into (2.

(Lemma 6) If M is a holomorphic surface in (2, then, for any
constant a, 0 < a S 1r/2, M is a slant surface in (E4, J a)
with slant angle a, where J a is the compatible almost complex
structure on E4 defined by

Ja(a,b,c,d) = (cos a)(—c,-d,a,b) + (sin a)(—b,a,d,—c)

21

In [CBY5] and [CBY3], and also in this §4, Cm is regarded as follows.
Cm = (Rzm, < , >, Jo)

(4.7) J(z' "”"3m yl,...,ym) = (—y1,...,—ym, 1:1,....,zm)

' (e) (Theorem 1) Let M be an oriented surface in (2. Then there
is a compatible complex structure 9 on E4 such that M is

holomorphic in (E4, 9) if and only if M is minimal.

(f) (Theorem 3) Let M be a proper slant surface in (2. Then
there is a compatible almost complex structure Jl on E4 so
that M is totally real in (E4, 11) if and only if M is

minimal.

(g) (Theorem 4) Let M be a totally real surface in [2. Then
there is a compatible almost complex structure J1 on E4 so
that M is a pr0per slant surface in (E4, J1) if and only if M

is minimal.

The pr0perties (b), (d) (e), (j) and (g) are all explained by a simple
characterization of slant surfaces as we will see in Ch 2. We will deal with
the compact case (c) in Ch 3.

We also list here the examples of slant surfaces in [CBY5]. These

examples have a common interesting feature as seen in Ch 2.

(Eg 1) For any non-zero constants a and b,
(4.8) :1: (u, v) = (n cos n, b cos v, a sin n, b sin 12)

gives a compact totally real surface in (2 with Vh = 0.

(4.9)

(4.10)

(4.11)

(4.12)

22

Here Vh is defined by (1.24).

(Eg 2) For any a>0

(Es 3)

(Es 4)

(Es 5)

(Es 6)

:1: (u, v) = (n cos n, v, a sin n, 0)

defines a non—compact totally real surface in C2 ’with (M = 0.

For any a, 0<a$ir/2,
z (u, v) = (n cos a, u sin a, v, 0)

defines a slant plane with slant angle a in (2.

For any positive constant 1:,

kn

z (u, v) = (elm cos n cos v, e sin n cos 1),

elm cos u sin 2), aka sin 11, sin 11)

defines a complete, non-minimal pseudo—umbilical pr0per slant
surface in C2 with slant angle cos-10: JIM?) and with

mean curvature e‘ku/JlH—cg.

For any positive number 1:,
2(11, 0) = (n, 1: cos 0, v, k sin v)

defines a complete, flat, non-minimal and non-pseudo-umbilical,

proper slant surface with slant angle cos-10: J1+;2) and
constant mean curvature k/2(1+lc2) and with non-parallel mean

curvature vector.

Let k be any positive number and (9(3), 12(3)) a unit speed

plant curve. then

23

(4.13) :1: (u, v) = (—ks sin 11, 9(3), ks cos n, h(s))

defines a non—minimal, flat, proper slant surface with slant angle

(1: (5+3).

There is another example of a Kahlerian slant submanifold in C4.

CHAPTER 2.
A CHARACTERIZATION OF SLAN‘I‘ SURFACES

In [CBY—M] B.—Y. Chen and J.-M. Morvan characterized holomorphic
surfaces and totally real surfaces in C2 using the description G(2,4)=S:x53
as follows. Let a: : M 4 E4=(|R4, < , >) be an isometric immersion of an
oriented Riemannian surface into E4 and u = (u , u_) be its Gauss map

defined in Ch 1 §3. Then,

(a) z is an holomorphic immersion with respect to some complex
structure J on E4 compatible with < , > if and only if
u+(M) is a singleton.

(b) 1: is a totally real immersion with respect to some complex
structure J on E4 compatible with < , > if and only if

u+(A4) is contained in some great circle in 5:.

The purpose of this chapter is to show that (a) and (b) have a natural
generalization to the case of slant immersions.

In §1 we consider the relation between 2—dimensional linear subspaces of
E4 and complex structures on E4 compatible with < , >. In § 2 we
combine this with the Gauss map and characterize slant surfaces in C2. In
§ 3 we show that most surfaces in an almost Hermitian manifold (M, y, .7)
can be slant surfaces with any given slant angle with respect to some almost
complex structures 31's so that (M, 2'), .71)'s are almost Hermitian. This
shows that the argument about slant surfaces in 4—dimensional almost
Hermitian manifolds does not have much significance and also that there is a

big difference between almost Hermitian manifolds and Kahler manifolds.

24

25

§ 1. Complex structures on E4

Let Cm be the complex m—space with the canonical complex structure

J0.

(1.1) [m = (R2m,< , >,Jo)

where

(1.2) J0(:l:l,yl,...,zm,ym) = (-yl,zl,...,-ym,zm).

If we use the canonical basis {53 (4)3121 of E2m=(82m,< , >) then
(1.3) 10552144:ng and hence Jo%2A="%2A-l for A=1,...,m.

Note that this is different from [CBY3], [KN] and Ch 1 §4, and this is the

only difference between our notations and those of [CBY3]. J 0 is an

orientation-preserving isomorphism of E2m. In this section we consider the
case m = 2.
We denote by f the set of all complex structures on E4 compatible
with < , >, i.e.,
(1.4) 1: (J: E4 .. E4|linear, J 2 = -id,
<JX, JY> = <X, Y> for V X, Y e E4}.
For each J E 1, we can always choose a J-basis {8A}A=1’ i.e., an
orthonormal basis satisfying
(1.5) Jol = e2, Je3 = 124.
Two J-bases of the same J have the same orientation. Hence using the
canonical orientation w = %1A'°°°l\%4. We divide ,1 into two disjoint
subsets:
(1.6) 1+ = {J 6 J lJ-bases are positive}
1' = {J E } lJ-bases are negative}

26

For each J 6 J, we determine a unique 2-vector (J E A2E4 as

follows. Let Q, be the Kahler form of J.
(1.7) QJ(X, Y) = <X, JY> X, Y e E4, 12] e A2(E4*)
Since A2(E4*) is identified with (112134)" by Ch 1(2.4), we can set (J
to be the metric dual of -QJ e (A2E4)* with reSpect to the metric < , >
of A2134 defined by Ch 1(2.3). Hence, for X, Y e E4,
(1.8) <(J, XAY> = -QJ(XAY)

= -QJ(X, Y)

= -< X, JY>

= <JX, Y>

We have the following lemma.

Lemma 4
The mapping
(1.9) C: .1» 424:4; J» c,
determines bijections
(1.10) C ‘ 1+ " 53“?)
c: f .. 53w?)

where SEQ/2') are 2—spheres with radius 42 centered at the origin in AEE“.

(Proof)

Let J e ,t and (51);:l be a J-basis. 11 J 6 fr (or ,a‘),
then {e A} is a positive (or negative respectively) basis and vice versa. By
(1.7)

(1.11) (J = elAe2 + e3Ae4
Hence, by Ch 1(2.3)
(1.12) ”<le = o

27

By (1.10), if J E 1* then *9 = :9 and so (J E AEE4, and hence
(J 6 Sim). The injectivity of the map ( is clear.

Conversely, let f E SEQ/2’). Then, % E 6 Si and hence we can pick
an oriented 2—plane V such that
(1.13) ' V 5 file 5) c G(2,4).
Choose a positive adapted frame {e ”i=1 of V in E4 and define a
complex structure J on E4 by
(1.14) Je1 = e2, Je2 = -el, Je3 = e4, Je4 = —e3.
Then, J e j and (J = 5. Similarly for f.
(1.14’) .I'-e1 = e2, J'e2 = -e1, J'e3 = -e4, fe4 = e3.

Q. E. D.

Through the bijection of Lemma 4, we can identify these sets.
(1.15) z site) 11 site)
2 _ -
all?) = .1“. site) = 1‘.
Next, we consider slant 2—planes in (E4,< , >, J ) for J E 1. Before

that we deform the definition of the slant angle slightly.

Definition 1.
For V E G(2,4) and J E ,4, put
(1.16) aJ(V) = cos-1(—QJ( V)) 6 [0,77]
and call V to be a-slant with respect to J if aJ(V) = c.

The relation between 0(X) of Ch 1(4.2) and aJ(V) is as follows.
Let :1: : M -+ (M,§,J) be an immersion of a 2—dimensional differentiable

manifold M into an 4-dimensional almost Hermitian manifold M. Then,

regarding (Tpfv'l, i1, 3) E (E4. < , >, 7).

28

(1.17) 0(X) = min {a~(T M),7r— a~(T M)}
J 1’ J 1’

for X E TpM.

 

FIGURE 4

If M is oriented, then it has a unique complex structure J
determined by the orientation and the induced metric, with respect to it M
is a Kahler manifold. Hence,

(1.18) :1: is a holomorphic immersion

Z'J z*X = a:*JX v X 15 TM
<7 2*X, 1*JX> = 1, X 5 TM, X]
03(2*X A 2*(JX)) = -1, X 6 TM, ||X1|=1

=1

 

 

 

11 ll

:- a-(T M) = 0
J P
Similarly
(1.19) 1:; anti-holomorphic immersion
: a-(T M) = 7r
J P
(1.20) 1:; totally real immersion

:a~(TM)=n‘/2
J P

29

This argument holds also for dim M > 4, and we note here that the angle
aJ coincides with the angle defined by Chem and Wolfson in [CSS-Wl],
although they look quite different.

The angle (11 can be also described in the following way.

Lemma 5.
If J6 /+, then aJ(V) is the angle between 77+(V) and (J. If
J E )1 then aJ(V) is the angle between 7r_(V) and (J .

(Proof)
Let J e 1+. Then by (1.16), (1.7)
(1.21) cos(aJ(V)) = —QJ(V)
= <(J , V>
= <4). 4+(V1+7_(V)>
= <<,. mm
since (1 E SEQ/2') C AiE4. Note that ”CI“ = J2 and ||7r+(V)|| = 1N2.
Similarly for J 6 f’. Q.E.D.

For each a 6 [0,77] and J 6 f, we define -GV,a to be the set of
all oriented 2—planes in I?)4 which are a—slant with respect to J, i.e.,
(1.22) 0,1,0: {V E G(2,4)] aJ(V) = a}
and also, for each a E [0,7r] and V E G(2,4), we put 4,,“ to be the set
of all complex structures on E4 compatible with the metric with respect to
which V is a—slant, i.e.,
(1.23) Ike: {JG }|01(V) =a}.
Put

Xi. = 4,. n 1*-

30

Then we can "visualize" these sets as follows.

Proposition 1.
(i) If Je y“, then
; 1 2
GJ,a - SJ,o " 5-
where 5.1,“ is the circle on 53 consisting of the 2-vectors which have the
angle a between (J. If J E )1 then
_ 2 1
GJ,a - S+ " 5J,a
where 5.1111 is a circle on SE defined similarly.
(ii) Under the identification of (1.15), J], a is a circle on SEQ/2‘)
consists of the 2—vectors which have the angle a between 7r +(V).

”(V-a is a circle on SEQ/2') defined similarly by 7r_(V).

(Proof)

Direct from Lemma 5.

 

 

 

AiE“ ME"

 

 

 

 

 

 

FIGURE 5

31

;= 1* u ,7- é» stars) usicri)

 

 

4. .-
Jtca. Ace.

 

A115"

 

 

 

 

 

 

FIGURE 6
We state the following lemma which we need in Chapter 4 and 5.

Lemma 6
Let W0 6 G(3,4) and V6 G(2,4) such that VC W0. Then V is
a-slant with respect to J E f if and only if
(1.24) <19, Jn> = - cos a
where i/ and n are positive unit normal vectors of V and W0 in W0

and E4 respectively.

(Proof)
We put W = W n J W. Then, W is a 2—dimensional J-invariant
linear subspace of E4. We choose an orthonormal J-basis {e A} of E4

such that

(1.25)

Then e3 6 W‘L n W and {e1, e2, e3} is a positive orthonormal basis of
W. Let {X1, X2} be a positive orthonormal basis of V. Since V c W,
X1 A X2 is spanned by {elA e2, elAe3, e2Ae3} and hence by (1.21) and
(1.11)
(1.26) cos(aJ(v)) = <(J, V>

= <e1Ae2+e3Ae4, XlAX2>

= <elAe2, XlAX2>
Since W is a Euclidean 3—space, the wedge product A in W is identified

with the usual vector product 1:, more precisely, the map defined by

I: 42W) .. W
(1'27) f(X A Y) = X x Y and linear
is an a isomorphism preserving the inner product. Therefore,
(1.28) cos(aJ(v)) = <e , ix>
= -< J77, 19>

Q.E.D.

We also use the following notation. For V e G(2,4), we denote by

J], and J; the complex structures determined by
(1.29) 1t, = (’17, M)

where C is the bijection in Lemma 4 and hence JT, 6 1* respectively.

§ 2. Slant Surfaces in (2

Let a: be an immersion of a surface M into a 4-dimensional almost

Hermitian manifold (M, g, .7). If we fix a point p in M, then we can

33

use the argument in § 1 about the slant angle of TpM in (TpM , gp)
with respect to Jp. But, in order to compare the situations at different
points, we need some global structure.

In this section we assume M = E4 and choose the parallel
displacement in TM, i.e., the identification of TpE4 and E4, as a "global
structure“. In short we use the Gauss map. We note that the argument in
this section also holds when M is a Riemannian quotient E4/I‘ by some
discontinuous group, since it is parallelizable. In Ch 4 we will consider a
different "global structure" using left invariant vector fields. Another
interesting example of this "global structure" has been given by Micallef and
Wolfson ([M-W]): if M is a Ricci fiat K3 surface then Ai(TM) is a flat
bundle over M and we can use parallel displacement in A3(TM) instead
of TM.

A slant surface in C2 is characterized as follows.

Proposition 2.
Let :1: be an immersion of a surface M into E4.
(2.1) :1: : M -. E4
Then, :1: is a-slant with respect to J E }+ if and only if
2
(2.2) u+(M) c 5.1,", c 5+

where 8‘1,“ is the circle in 5: defined in PrOposition 1. The same holds

replacing + with —.

(Proof)

Direct from Pr0position 1 and the definition of Vt.

34

.a-slant immersion out. 16f"

§J

n. \

 

 

    

Z(M)

 

 

 

4:2 Ai E“

 

 

 

FIGURE 7

The following lemma is shown in [CBY 5] (in the proof of Theorem 1).

We state and prove it again in our notation.

Lemma 9.
Let :1: be an immersion of a surface M into E4. Then, :1: is
minimal and slant with respect to some J 6 }+ if and only if 11+ (M) is

a singleton. The same holds replacing + with -.

(Proof)

Assume :1: is minimal. Then 111 and V2 are both
anti-holomorphic ([CSS 1] and also cf [H—O3]). In particular 111 and 122
are Open maps if they are not constant. Hence by Lemma 1, 11+ and u_

are Open maps if they are not constant. Furthermore, if :1: is slant with

35

respect to J E 1+, then by PrOposition 2 V cannot be an Open map

+
and hence 12+(M) is a singleton.

Conversely, let u+(M) be a singleton {5}. Then 25 e Sig/2') and
we can choose a complex structure J = C-l(2{) 6 1+ determined by
Lemma 4. By Proposition 2, z is 0—Slant, i.e., holomorphic with respect to

J and hence minimal.
Q. E. D.

By PrOposition 2 and Lemma 9, we can say:
(h) The following are equivalent:

(i) :1: is minimal and slant with respect to some J 6 }+ (or

1‘)

(ii) V+(M) (or V_(M)) is a singleton.

(iii) 2: is holomorphic with respect to some J e }+ (or 1.)

(iv) For any aE[0, 77], there exists Jae 1+ (or 1‘) such

that 1: is a—slant with respect to J a'

(ii):(iv) is shown as follows. If 11+(M) is a singleton, we choose
some V e 7r:1(u+(M)) and 1: is a—slant with respect to any J E 1],",
(iv):(ii) follows from (i) below.

If 2: is a non-minimal a—slant immersion with respect to J E }+,
then 11+(M) contains a l-dimensional portion of the circle 5:11,“, hence we
have;

(i) If :1: is not minimal, then :r can be slant with respect to at

most two complex structures tJ E 1+, and at most two

I —
complex structures tJ e f .

36

 

minimal slant

   

tata 11y re at anti-holomorphic

FIGURE 8

These facts give a clear geometric image to (b), (d), (e), (f) and (g) in
Ch 1 §4. And they show that minimal surfaces are completely atypical (cf
[H—O3] p 731) also from the view point of slant immersions.

We define the following term, because this prOperty is common in

examples (Eg 1) - (Eg 6) in Ch 1 § 4.

37
Definition 2.
An immersion :1: : M2 -1 E4 is called doubly slant if it is slant with

respect to one complex structure J E }+ and at the same time slant with

respect to another complex structure J 6 1..

doubly slant, MM) c S},ax8},p

\ 53w-

 

 

5,95

O

    

  

A315" ,1‘.‘ E‘

 

 

 

 

 

 

FIGURE 9

Or equivalently we can say, 2: is doubly slant if and if there exists
V e G(2,4) such that z is slant with respect to I]; and J}, where
ff, is defined by (1.29).

Lemma 10.
If 1: : A42 -1 E4 is a doubly slant immersion, then G 5 GD 2 0.

(Proof)
Direct from Lemma 3 and Proposition 2.
Q.E.D.

38

The next lemma shows how to calculate slant angles and ranks of u, 1/ +

and V.

Lemma 11.

Let :1: : A42 -1 E4 be an immersion of an oriented surface and put

“(P)=0'Jo( TpM)

(2.3)
b(p)=ayl(TpM)
where JO=JO+ o and J1=J; A° defined by (1.29). Let {e1, e2} be
81‘82 ”1 2
a positive orthonormal frame field on M and put
(2.4) e = BE: 3323, 1': 1,2
Then
(25> . = tee-413 + (vie-fie)
1 1
(26) b = (Ali-Bias) - (bit-4231
(2.7) the, 33333333331333” 2A A 23
(Proof)
By (1.21), (1.11) and Ch 1 (3.12), (3.13),
cos a= < (J, v(p)>
o
(2.8) = <%1A%2 + 33A‘1’e4, 23 SABeA AeB>

A<B
= 512 + 534
= mitt-flirt) + vie-4431
Similarly for cos b. (2.7) is Ch 1 (3.5).
Q.E.D.

39

The following is an example of a doubly slant surface with given slant

angles a, b E (0, 7r/2).

(Es 7)
_ Let p, g be a non-zero real numbers and define an immersion 1: by
(2.9) :1: : R x (o, m) -+ E4
:1: (u,v) = (pv sin 71, pv cos u, v sin q u, v cos qu)
Then, using the notations of Lemma 11, :1: is doubly slant with respect to

J0 and J1 and the slant angles are given by

 

cos a = (22+q)//(112+q)(pf+1)

 

(2.10)
cos b = (pZ-q)/1/(22+q)(22+1)
and
(2.11) rank u 5 rank u+ 5 rank V_ E 1

We can also show by an elementary calculation that for any a and b
in (0, 7r/2) there exist non-zero real numbers p and q which satisfy
(2.10).

In fact Eg 1 - Eg 6 of [CBY 5] stated in Ch 1 §4 are all doubly slant

as is seen in the following table.

slant angles rank V rank V + rank u_
Eg 1 a=b=7r/2 2 1 1
Eg 2 a=b=7r/2 1 1 1
Eg 3 a=b€[0,7r/2] o o 0
Eg 4 a€(0,7r/2],b=7r/2 2 I 1
Eg 5 a=b€(0,7r/2) 1 1 1
' Eg 6 a=b€(0,7r/2) not constant 1 1

Eg 7 a, be(0,7r/2) 1 1 1

40

Except for Eg 4, a and b are slant angles with respect to J0 and

’1

b = slant angle with respect to J- .
elAe3

{pEMlg'(p) = h'(p) = 0} and rank V = 2 on M\M’. Note that all

of Lemma 11. In Eg 4, a = slant angle with reSpect to J0 and
InEg6,rankV=1 on M’ =

examples are flat, and proper slant examples have rank V < 2. Note also
that if an immersion x is a-slant with respect to J then x is
(ir-a)—slant with respect to -J.

Eg l — Eg 6 are adjusted here to match our complex structure by

(2°12) (2 1 $2: 33934) 4 (31, I3: $2934l°

§ 3. Slant Surfaces in 4—dimensional Alm03t Hermitian
Manifolds.

Consider an immersion x of a differentiable manifold M into an
almost complex manifold (M, J). Then, a point p E M is called a complex
tangent point if the tangent space TpM of M at p is invariant in TM!
under the action of J. The purpose of this section is to show the following

proposition.

Proposition 3.

Let x be an embedding of an oriented surface into a 4—dimensional
almost Hermitian manifold (M, 9, J). Assume that x has no complex
tangent point. Then, for any angle a E (0, 7r) there exists an almost
complex .7 on M satisfying the following conditions.

(i) (M, g, 7) is almost Hermitian manifold.

41

(ii) x is an a-slant immersion with respect to J.

(Proof)

We follow several steps.

Step 1. An almost complex structure on M is considered as a
cross-section Of a sphere bundle 5'31 M).

Step 2. The tangent bundle TM of M corresponds to a
cross-section of the pull—back 33(1)!) = x*(S_i(M)).

Step 3. We construct a suitable cross-section a of SE( M).

Step 4. We extend a to a cross-section 6 Of 53(M) to obtain

a desired almost complex structure.

Step 1
M has the natural orientation determined by J. We note again that

at each point pEM, (T M, gp) is a Euclidean 4-space and we can apply the

argument of §l. P
The bundle A2(M) Of 2—vectors is a direct sum of two bundles.
(3.1) 42114) = Aim) 0 431M)
where
(3.2) A2(TpM) = Ai(TpM) o 113(TPM), v p e M
in the sense of Ch 1(2.9) and
(3.2) 113017) :2” AE(TPM)
We define two bundles over M by
(32) 531M) = {t e 43.11411 Iél = 1/12}
(34) 331M) = {t e 43.11411 lél = h}

Then a cross—section

42

(3.5) p : M 4 3.210?!)
determines at each point p E M a complex structure (J p) p E }+ on

TpM by Lemma 4 compatible with 9p in the following sense

(3.5) gp((Jp)pX, (Jp)pY) = gp(X, Y), v X, Y e TpM.

And hence p determines an almost complex structure J so that (M, g,

,0
J p) is an almost Hermitian manifold, and vice versa.

Step 2
We consider the pull—backs of these bundles by the immersion x.
(37) 43.144) = 2143(3))
(38) 5.23M) = 27153114»
(3.9) “53014) = 2133.114»

Then the tangent bundle TM of M determines a cross-section r of

2 .
5+(M), i.e.,

(3.10) r : M -1 $31M)

(3.11) r(p) = n+(TpM) for V p E M

where 77+ is a projection of A2(TpM) onto A:(TpM). Note that 2r
is a cross-section of 33(1)!)

(3.12) 27 : M -1 3.24M)

and we have another cross-section x*p, which we also denote by p, of
*2

5+(M).

(3.13) p : M .. Siam

Step 3

Let p E M. By the assumption that p is not a complex tangent

point,

43

(314) 10(2) 14 *2 T(v)

hence p(p) and 27-(p) determine a 2-plane in (Ai(M))p = A3(TpM)

which intersects the circle (17+ 11) p at two points, where ( )1,"-
9 Q

a ) p is the
circle on (83(M)) p determined by Proposition 1 setting V= TpM. We
define 0(p) to be the one of these intersecting points such that p( p), 27( p)
and o( p) lie on an Open hemi—sphere of (3:04)) p as indicated in the
figure. Since p and r are differentiable, the cross-section

(3.15) o : M .. Sim!)

is differentiable.

 

 

 

 

 

 

 

 

 

 

FIGURE 10

Step 4.
For each p E M, we choose an Open neighborhood Up of p in M
such that o|(Up n M) can be extended to a cross-section of 531 M) on

Up‘

. , _, -2 ~
(3.15) op. up S+(M)|Up.

44

We can do this by the local triviality of 531M). We identify here the
manifold M and the embedded image x(M) C M. We put

(3.17) U = u U
26M 1’

and pick a locally finite countable refinement { U1}?=1 Of the Open covering
{Uplp E M} of U. For each i we pick p E M such that U1. C Up
and put
(3.18) 0i = op] Ui .

We choose a differentiable partition of unity {pi} on U subordinate

to the covering {Ui} and define a cross-section 3 of A:(M))| U by

(3.19) 3: U-1 AE(M))| U; 3 = 2 pic]. .

Note that by constructions Of ”i and ‘0'

(3.20) EIM = 0.

Since AMP). 12(2)) < I and a,(p)=o(p) for V p 6 M.
(3.21) 405(1)), p(p)) < 71' for VP 6 U, n M v,.

Since h'(p) is a finite linear combination of oi(p)'s with positive
coefficients, (3.21) means

(322) 3(2) t 0, 45(2), 10(2) )< r for V p 6 M.

By the continuity of 3, we can pick an Open neighborhood W of M
contained in U such that

(323) 3(a) t 0, 45(4). 10(0)) < r for V q 6 W.
and we can define a cross—section 6 of Sid/1)] W by

(3.24) a : we Siam) w; a = E/ma
satisfying

(3.25) 40‘7“), p(a)) < r for V qE M.

Finally we consider an Open covering (W, M—M} of M and local

cross-sections

45

a: w 4 Siam w
(3°26) p : M-M —. 33(M)|M-M
and repeat the same argument using a partition of unity subordinate to {W,

M-M} and get a cross-section 6

(3.27) a : M -» SEW)
satisfying
(3.28) 77' M = 6

Note that this is possible by (3.25).
Now, it is clear that the almost complex structure .7 corresponding to

6 in the sense of Step 1 is the desired one.
Q. E.D.

CHAPTER 3
COMPACT SLANT SUBMANIFOLDS IN c'"

In this chapter we prove the following.

PrOposition 4.
Let x be a general slant immersion of l-dimensional differentiable
manifold M into the complex m-space 0'". If M is compact, then x

is a totally real immersion.

In other words, there exists no compact proper slant submanifold in (m

just as in the case of holomorphic submanifolds:
(j) There exists no compact holomorphic submanifold in 0’".

Since a slant submanifold is a generalization of both holomorphic and
totally real submanifolds, this similarity is no surprise. (j) is shown in several
ways. From the view point of complex manifolds, (j) is a consequence of the
maximal modulus principle. From the view point of differential geometry, (j)

comes from

(k) Holomorphic submanifolds are minimal.

and

(1) There exists no compact minimal submanifold in E",

and (l) is shown by the maximal principle of harmonic functions or by the
existence of a point where the mean curvature vector does not vanish.
We note that our proof of PrOposition 4 contains another proof of (j)

basically based on Stokes' Theorem.

46

47

We note also that there is a compact prOper slant surface in a complex
torus and, as is well-known, a compact totally real submanifold in complex

spaces as seen in the following examples.

(Eg 8) Let 'x : R2 -1 C2 = (E4,J ) be the proper slant plane of Ch 1
§4 Eg 3. Let {e14}:=1 be a basis of E4 such that e1 =
f(a/Bu) and e2 = 1*(6/6‘v). Let I‘ be the lattice generated
by {eA}. Then, x induces a prOper slant immersion of

31.51 = (11/1)x(11/z) into 1%.

(Eg 9) Let 1 5 lg m. For each j= 1,...,l., choose a closed curve
' 1 — ..... ° x o o o o x 1 _)
0].. s .. (xj, yj) plane. Then 01:: x01 . 31 S
(m is a totally real immersion.
To prove Proposition 4, we consider two cases:
(Case 1) l = m = 2
(Case 2) general dimensions.
In case 1, Proposition 4 comes directly from Proposition 2 and Lemma 2.

In case 2, the idea is the same but we need some lemmas.

(Proof of case 1)

Let c2 = (E4, J) and (J bethe 2—vector in 11:13“ defined by

Ch 2 (1.7) and (1.8). By Lemma 2 and Ch 2 (1.21)
(1) o = fng < 17(1). c) > W

= fpeM 608(oJ(V(p))) «WM
vol (M) COS(OJ(V))

48

because x is slant and hence aJ(V) is constant. Hence aJ(V) = 7r/2,
which means x is a totally real immersion.

(case 1 Q.E.D.)

To consider the second case, we set some notations similar to the ones in

case 1. We put

(2) c’" = (112“, < , >, J)

(3) (2(X, Y) = < X, JY >, n e A2(E2m)*
Then by the identification of Ch 1(2.4)

(4) Q" E A2n(E2m)* = (A2nE2m)*,

and we define (J to be the metric dual of (42)" in A2nE2m with
respect to the inner product < , > defined by Ch 1(2.3)

(5) <4, 11> -_- (-1)“o"(q) for v n e 112%:2’".
For VEG(l, 2m) and a E[0, 7r/2], we call V a—slant if
(6) AUX, V) = a for XEV, X at 0.

Lemma l2

(7) fln(X1A....AX2n) =

(9%]! deg sgn(6)Q(X0(l), Xam) ..... 0(Xa(2n_1), Xa(2n))
2n

2m
for X1"""X2n E E
where S2“ is the symmetric group of order 2n and sgn denotes the

signature of permutations.

(Proof)
Let “(4)321 be an orthonormal frame of E2," and {41A} be its

dual coframe. Put

49

A
(8) 9 — A 33: ”AB” A“ 1 8713 - ‘PBA
Then
2m
(9) 0(X, Y)— oABw HOW (Y)
A, 3:1
and
n
(10) f2 (X1,...,X2n)A A
111A
1 2 n-lA 2n
— [(21p to At.) )A ----- A(2<p 1.11 X,
AIA A2n—1A2n )l( 3X2")
= gm ..... [wAlA ..... A A2n(X X )]
Al A2 _1¢A1A2 WA221—1’4211 w 1’ 2"
,... n
gm
_ A1,. .A2n—1 A1A2 A2n—1A2n
A A
1
[1251! M253 mow 1(XAI))~~9.2"(X(2,,))1
1 2 sgn(o)[( 21p wA1(X )WA2(X )) "
=(2_)'! 17 A1142 6(1) 6(2)
A A
2n-1 2n
" (38 1PA A “’ (Xo(2n—1) “’ (Xo(2n)))]

2n—l 2n
By (9), (10) and Ch l(2.4), we get (7).

Q.E.D.
The following Lemma 13 is shown in [CBY5] but we prove here again to

make our argument clear.

Lemma 13
Let VE G(l, 2m) and TV:
If. V is a—slant in Cm = ( E27", J) with a ,e r/2, then the linear

E27" -1 V be the orthogonal projection.

endomorphism JV of V defined by
(11) JV = (sec a)(7rV o J] V)

is a complex structure compatible with <, >| V. In particular, I is even.

50

(Proof)
Put
(12) P=7rVo(JlV): V-rV
(13) N=J|V-P:V-1VJ‘
(14) s = P2 : V 4 v*
Then
(15) Jl V = P + N.
By a simple computation using (15), we get
(16) <SX, Y> = <X,SY>,
(17) <PX,Y> = -<X,PY> for V X, Ye V.
Since V is a-slant
(18) a = AUX, V) = A(JX,PX), X e V, X a o,
and hence
(19) ||PXI|=||XII cos a for V XE V.

By (16) S has real eigenvalues {Ai}l=1' Let {3i}:=1 be corresponding

orthonormal eigenvectors, i.e.,

(20) 3(ei) = Aiei Vi

Put

(21) P(ez-) = E Pij e]-

then by (17)

(22) Pi]. = —Pfi

Hence

(23) Ai = <S(ei),ei>
= <P2(ei),ei>
= - Evil)? 5 o.

On the other hand, by (19), (20) and (14)

51

(24) IAil = ||S(ei)|| cosza for Vi,
i.e.,

(25) A‘- = —c032a Vi,

(26) SX = —c032aX for V XEV

and hence JV defined by (11) is a complex structure on V. Since
(27) IIJVXu2 = secza ||PXl|2 = "X"2 for v XEV
namely, J V is compatible with < , >| V.
(2.13.1).

Lemma 14
Let V EG(2n,2m). If V is a—slant with a at 7r/2, then

_ n
(29) <(J,V>—pncosa

where '“n is a non-zero constant determined by n.

(Proof)
Let J V be the complex structure on V defined by (11). Let X be
a unit vector in V and put Y = JVX E V. Then, using P defined by
(12);
Q(X,JVX) = (l(X,Y)

= <X, JY>

= <—JVY, JY>

= IIJVYII llJYll 404.4(va m

= cos 4(PY,JY)

= - cos 4W. JY)

=-0080

i.e.

(30) Q(X,JVX) = - cos a for V XEV with [le] = 1

If ZEV,Z1JVX,

52

Q(X,Z) = <X, JZ >
= <X, PZ >
=cosa<X, JVZ>=0
i.e., ‘
(31) O(X,Z) = 0 for X, Z E V with 21 JVX.
We choose an orthonormal J V—basis {ea}i__':1 on V, i.e.,
(32) e2k = JV e2h—l’ k = 1,...,n
(33) V = elA....Ae2n.
We fix a notation for indices by
(34) 275 = 213—1, 2531' = 21: for k = 1,...,n

Then, by (30) and (31),
(35) (l(ea, ch) = 45b cos a for a < b

Using (7), (33) and (35), we compute fln(V) as follows
(35) (Zn)! o"(V) = (2n!) n (elA...Ae2n)

2 sgn(o)fl(eo,(1),ea(2)) ..... 9(e0(2n_1),e0(2n))

065%
gn 61. ..... 2n ( ) (
0 e e ...... f) e
_ a ..... a a ’ a
a1,...,a2n—l 1 2n 1 “2 2
§n 612......2n 9( ) (
_ _ e ,e_ ...... it e
01,...,a =1 alal'"anan a1 a1
n
n 2 2 61 ..2n 0( )
2 ..... - - e e- .
- a a .. a a a ’ a
al<a1 an<an 1 1 n n 1 1
1 . .2n
2n(-cos a)" E_ ..... 2_ 60 6. a a
al<a1 a <an l 1 n n
2n(—cos a)" n! 2 b ..

2"(-cos a)" n!

_ - a a
al<al<...<an<an 11

53

Hence

(37) Q"( V) = (—1)npncosna

where

(38) p" = 2nnl/(2n)!

Note that, if n = 1, then ”n = 1 and (37) is nothing but Ch 2 (1.16).

Q.E.D.
Now we can prove the second case of PrOposition 4.

(Proof of case 2)
Assume that the immersion x is not totally real. Then, by Lemma 13,

l is even. Put 1 = 2n. By Lemma 2.

(39) o = fp€M< V(p), (J > de

Let a be the slant angle of x. By assumption a 1!: r/2. So by Lemma
14

(40) 0 =fMpn cosna dVM

= pnvol(M) cosna,
but this contradicts to cos a at 0. Hence x is a totally real immersion.

Q.E.D.

CHAPTER 4
SPHERICAL SLANT SURFACES IN 1:2

As we have seen in Ch 2 §2, examples Egl-Eg7 are all flat slant
surfaces. In Chapters 4 and 5 we consider flat slant surfaces under slightly
stronger assumptions, Spherical slant surfaces and slant surfaces with the rank
of the Gauss map less than 2. Both surfaces are flat slant surfaces. Under
these additional assumptions, the shapes Of slant surfaces become clearer.

In this chapter we consider Spherical slant surfaces in (2, namely a
slant surface contained in a 3-sphere in (2. Slant angles are invariant under
parallel translations and homotheties, so, without loss of generality, we can
assume that the 3-Sphere is the unit sphere centered at the origin. Our

3-—the Lie group of unit

argument depends on a special structure of S
quaternions. We review this in §1. In §2, we define a map analogous to the
Gauss map to characterize spherical slant surfaces. In §3, we will see
Spherical prOper slant surfaces are two families of surfaces which we will

temporarily call helical cylinders and circular cylinders in S 3.

§ 1. Geometry of S3.

This section is a short review of the geometry of the 3—sphere written in

[S1] vol. 4 Ch 7.
R4 is considered the non-commutative division algebra of quanternions
generated by {1, i, j, k}
(1.1) l = (1,0,0,0), i= (0,1,0,0),
j = (0,0,1,0), I: = (o,o,o,1)

54

55

satisfying

I 1=unit

(1.2) . i°j= k= —j-i, j-k= i= —k-j, k-i= j= —i-lc
1'2 =j2 = k2 =-1

 

L

Then, S 3 is the lie group consisting of quaternions of norm 1. We can

also regard S 3 as a subgroup of 0(4) through the identification

a —b -c —d l
(1.3) a+bi+cj+dk —-1 b a -d c
c d a -b
d —c b a
L

 

 

For each p E .53, the left translation Lp
(1.4) LpzS3-oS3;q-1p-q
is an isometry satisfying
(1.5) d(qu, q) = const for V q E S 3
where d is the distance on S 3 induced from E4, and hence Lp is the
analogous of a translation in E3.
We identify TpE4 with E4 as usual. We put
(1.6) X1 = (0,1,0,0), X2 = (0,0,1,0), X3 = (0,0,0,1)
then {Xi}:l=l is an orthonormal basis of TlS 3. For Y=(0, yl, y2, 373) E

T15 3 and p = (a, b, c, d) E .S' 3, Lp4Y is calculated by

r a —b -c —d l 0
(1.7) t(Lp*Y) = b a -d c y1
c d a -b y2

. d -6 b a 1 L 3’3 .

 

 

 

 

where t( ) denotes the transposed matrix. Let {Xi} be the left invariant

extension of {X i}’ i.e.,

56

(1.8) 3’1“?) = LP...X, p e s 3
then, every tangent vector of S 3 is spanned by {X 1.}.

Since S 3 is a space of constant curvature, the analogy of the curve
theory in Euclidean Space holds. Suppose c(s) is a curve in S 3

parametrized by arclength. We denote by V’ the Riemannian connection of

s 3 and put

(19) 1(3) = 5(3) = “(a/63)
(1-10) “(3) = Mimi"

and if n 15 0, put

(1-11) "(3) = (1/“(3))V2(3)t
(1-12) 5(3) = “3) " "(3)

(1.13) 7(3) = -< V’t(s)b, «(3) >

where x denotes the vector product in T ms 3 determined by the metric

and orientation. Then the Frenet—Serret formulas hold. If we set

(1.14) Ks) = its) was»,
then we have the following equations.
(1.15) 2 fig! = o
(1.16) 1 = mob”?
(1.17) n = n-IEfi’ X,
(1.18) b = 16-12932.
where
’ 91 = fzfs' " 13f2'
(1.19) « 92 = «bfl' ‘ Us,
93 = f1f2' ‘ J5211'

 

(1.20) V’tb = -n + 2 (gt/10’1"

57

and also the following hold.

(m) If 1'51,

then b is left invariant along c. If re -1, then b is

right invariant along 0.

(n) A flat surface M in S 3 is "in general" a translation surface

(1.21)

{C(8)°7(t)}

where c and 7 are curves in S 3 parametrized by arclength

satisfying one of the following conditions:

0)

(iii)

Torsions of c and 7 are +1 and -1 respectively,
c(0)=7(0) and the osculating planes at s=h0 coincide.

(In this case, at each point c(so)-7(t0) the binormals of the
s—curve c(s)-7(to) at 3:30 and the t-curve c(sO)-7(t) at
t=tO are normal to M in S3.)

c has torsion r=l and 7 is geodesic. (In this case, at
each point c(so)-7(to) the binormal of the s—curve
c(s)-7(to) at 3:30 is normal to M in S 3.)

c is a geodesic and 7 has torsion r=—1. (In this case, at
each point c(so)-7(to) the binormal of the t-curve
c(sO)-7(t) at 1:10 is normal to M in s 3.)

c and 7 are both (distinct) geodesics. (In this case,
s—curves and t—curves are geodesics intersecting at a constant

angle.)

Here, "in general" means that we avoid the case in which the curvature

nofcor

7

has isolated zeros.

58

§ 2 Another Gauss Map

Let :1: be a spherical immersion of an oriented surface M into C2 =

4
(E . 1,)
(2.1) a: : M .. s 3 = 503(1) c c2
We define 1;, p, r) and u as follows. Let I; be the complex structure

If, E }- defined by Ch 1 (1.29) with V = 31A932, where {SEA} is the
canonical basis of E4, i.e.,

(2.2) J; : ‘31-’32, 2’24 -%1, 334 -%4, ‘34-» 933.

Let p be an isometry of E4 defined by

(2.3) ¢(a, b, c, d) = (a, b, d, e)

Let 1), i/ be the positive unit normal vector field of S 3 in E4 and of
11(M) in S 3 respectively.

(2.4) r] = unit normal of S 3 C E4

(2.5) i/ = unit normal of :1:(M) C S 3

Using the notation of § 1, we state the following Lemmas.

Lemmza 15
(2'6) JOWQ) = Rqakxls
(27) 130(4) = 11¢qu = 31(4)

for VqES3

Hence, Jon and Jan are right-and left-invariant respectively.

(Proof)

Let q = (a, b, c, d) E S 3. Then
(2.8) Jar](q) = (-b, a, —d, c)
(2.9) Ignaz) = H». a, d, -c)

59

Define a curve 7 in S 3 by

(2.10) 7(3) = cos 3 + sin 3 i

Then

(2.11) 5(0) = X1

and hence

(2.12) 12,..x, = [$502, 0 c<s))1,=o
= £(cos s + sin s :)(a + bi + cj + dlc)]3___0
= (-b, a, -d, c)

(2,13) quxl = %[(a + bi + cj + dk)(cos s + sin 3 013:0
= (-b, a, -d, c)

Hence we get (2.6) and (2.7)

Q.E.D.

Lemma 16
The following are equivalent.

(i) z is a—slant with respect to J 0

(ii) < 19(p), J01)(1:(p)) > = -cos a for V p E M.

(iii) «is 9(1)). 5w o 1112)) > = -cos a for V p E M.

(iv) w o z is (ir—a)-slant with respect to J3.
(Proof)

If we apply Lemma 6 pointwise, we get the equivalence of (i) and (ii).
Since ¢ is an orientation-reversing isometry, —w,..i/ is the positive unit
normal of w o 1:(M) in S 3. Hence, using (2.7), we apply again Lemma 6
pointwise and get the equivalence of (iii) and (iv). By (2.3), (2.8), (1.7) and
(1.8)

60

(2.14) Mona» = am» for v «1 e s 3,

which shows the equivalence of (ii) and (iii).

Q.E.D.

Now we define two maps analogous to the Gauss map in E3 which
translate each unit normal of a surface in S 3 to the unit tangent sphere at
the unit 1 = (1,0,0,0) using left—translations instead of parallel translations.
Recall that left-translations in S 3 are the analogues of parallel translations
in E3 (§1).

Definition 3
Let a: be a spherical immersion a: : M a S 3 C E4 of an oriented

surface M. We define two maps gt from M to the unit 2—sphere S 2

in T15 3

(2.15) 9*: Maszc T153
by

(2.16) 9+0») = (LMP,).)‘1<w.z>(p))
(2.17) up) = (Lmrrloon
for V p E M.

We show here two examples.

(Es 10)
Let M = Sle1 be a flat torus in S 3 parametrized by

(2.18) 1(n,v) = —1- (cos a, sin n, cos v, sin 1))

61

Then

(2.19) i/(u, v) = -1— (cos a, sin n, —cos v, -sin 1))
42—

(2.20) w*19(u, v) = L (cos a, sin a, —sin v, -cos v)

Hence, the images of g* are great circles perpendicular to X1.

(Es 11)
Let M = .52 be a totally geodesic 2-sphere in S 3 parametrized by
(2.23) 1:(a,v) = (cos n cos 1:, sin n cos v, sin 11, 0)
Then
(2.24) i/(n,v) = (0,0,0,1)
(2.25) wad/(u,v) = (0,0,1,0)
Therefore
(2.26) g+(u,v) = (0, sin v, cos n cos 1), -sin n cos 2))
(2.27) g_(u,v) = (0, -sin 11, sin n cos v, cos n, cos 2))

Hence, 9+ and g_ are isometries.

Now, we can state the spherical version of Proposition 2. As before, we
define circles in 32 C TlS 3 perpendicular to X1:
(2.28) S: = {X E T15 3| "X" = l, < X, X1> = —cos a}, a E [0, 7r]

Then, the following proposition characterizes spherical slant surfaces.

Pr0position 5.
Let a: : M -i S 3 C E4 be a spherical immersion of an oriented surface
M. Then
(i) a: is a—-slant with respect to J 0 if and only if
1 3
(2.29) g+(M) C So C TlS .

62

(ii) 2: is a—slant with respect to J; if and only if
(2.30) g_(M) c six”) c T15 3.

(Proof)
Direct from Lemma 16, (2.7) and Definition 3.

Spherical slant Suffice Q.E.D.

 

 

 

c2 n53; E3

 

 

 

 

 

 

FIGURE 11
Corollary 1.
If g+(M) is contained in a circle S1 in S 3, then 3: is slant with
respect to a complex structure It, determined by Ch 2(1.29), where
(2.31) V = (l, 0, 0, 0) A Z
(2.32) z e T15 3, z 1 51, "z" = 1
Similar for g_(M).

(Proof)
If suffices to consider an orthogonal transformation of E4 which leaves
(1, O, 0, 0) fixed and maps (0, 1, 0, O) to Z.
Q.E.D.

63

Corollary 2
9+(M) or g_(M) cannot be a singleton.

(Proof)
If g+(M) is a singleton, then by Corollary 2 '2 is a holomorphic
immersion with respect to some complex structure on E4 and hence - :r(M)

is minimal in E4. But spherical submanifolds cannot be minimal in E4.

Q.E.D.

§ 3 Classification of Spherical Slant Surfaces

As will be seen in Pr0position 6, spherical prOper slant surfaces are
something like "helical cylinders". A generalized helix in Euclidean 3-space
E3 is a curve such that the angle between its tangent vector 23(3) and a
fixed vector v E E3 is constant. In other words, if we extend v E E3 s
T 0E3 to a global vector field it on E3 by parallel translations, then a
general helix is a curve c(s) satisfying
(3.1) < c(s), ir(c(s)) > = const.

We will define a "helix" in S 3 replacing parallel translations with left
translations. A cylinder in R3 is a surface obtained by parallel translations
of a curve along a straight line. We will define a "cylinder" in S 3 as a
surface obtained by left translations along another curve in S 3.

Let
(3.2) c : I -' S 3

be a curve parametrized by arc length 3 and put

64

3 -
(3-3) 5(8) = (S) = ,3 IXS) X,(0(S))
:1
as (1.14).
Definition 4
(1) We call c(s) a helix in S 3 with the axis vector X1 if
r f1(s)= b
(3.4) ‘ 5(3) = it cos (a’ s + so)

 

k f3(s) = a sin (a’ s + so)

where a, b, a’, s are constants satisfying

0

mm fi+fi=1

(ii) We call an immersion x : D -1 S 3 ofa domain D of R2 a
helical cylinder if x(t,s) is a flat translation surface 7(t) - c(s)
described in §1 (11) and c(s) is a helix in S 3 with the axis
vector X1 defined in (i).

(iii) We call an immersion x : D -1 S 3 ofa domain D of R2 a
circular cylinder in S 3 if x(t,s) is a flat translation surface
c(s)-7(t) of type (ii’) in (n) which satisfies for some t0

(3.6) <Lc(s)*b(to)’ Jon(c(s)-7(to))> = -cos a
for Vs
where a is a constant with cos 0360, :1 and b(t) is the
binormal of 7(t).

Of course we can define a helix with an arbitrary left invariant vector
filed as its axis but we don't need it in this article. The following lemma
shows the existence of helixes in S 3. Then, the existence of a helical

cylinder reduced to the existence of the curve 7( t) satisfying the conditions

65

of §1(n), but this is guaranteed by the Existence and Uniqueness Theorem
of the curve theory in S 3 ([81] vol. 4 p 35).

The existence of a circular cylinder in S 3 is a pending problem.

Lemma 17
Let I be an open interval containing 0 and fz(s), i = 1,2,3, be
differentiable junctions on I satisfying
(3.5) [12+ [22+ [32: 1
Then, for any point p06 S 3, there exists a curve c(s) defined on a

neighborhood I’ of 0 in I satisfying

3 -
(3-6) C’(S) = ,2 f,(8) X,(C(S)).
i=1

(Proof)

Considering the curve L: o c if necessary, we can aassume without
0

loss of generality p 0 = 1 = (l, 0, 0, 0).

First we assume such c(s) exists and put

on c(s) = (an), as). 2(8). as» e s 3
Then
(3.8) as) = (ms), y’(s). ms), w’(s))

By (1.3) and (1.6)
(3.9) >3 c(s) Saws» = 2 f,(s)L,(,,.x,

= 1.4st Its) X,»

= 1343,40. 11(3), to), 5(a)

66

Hence, by Ch 1 (5.7)

‘ F
Fx’ x -y -2 —w l 0
(3.10) y’ = y x -w 2 fl
z’ z w x —y [2
w’ w -z x
L 4 l. y J L. f3

 

 

 

 

 

 

Conversely, consider a system of lst order ordinary linear differential

equation (2.10) with the initial condition

(001) (0(0). 1(0). 2(0). 0(0)) = (1.0.0.0) = 1 e s 3.
Then, it has a unique solution in a neighborhood I' of 0 in I. Put
(3.12) c(s) = (0(s). as). 2(0). 0(3)). s e I-

then c(s) is a curve in E4 with 0(0) 6 S 3. Hence, in neighborhood I’
of 0 in 1",

(3.13) (z(s))2+ ------- + (u(s))2 # 0, s e 1'.
Put
(3.14) ,\ = M3) = ”(3)2... ....... +(w(s))2)1/2
Then, (1/A)(x,y,z,w) 6 S 3 and hence

F x -y -z -w
(3.15) 71V y x -w z 6 0(4)

2 w x —y
L w -z y 1: J

 

 

Put this matrix A and multiply A_1 = tA from the left on both sides of
(3.10), then

F x -y -z -w x’ f 0

(3.16) y x —w z y’ = A2 f1
2 w x -y 2’ [2

w -z y x J w Lf3

 

 

 

 

 

 

67

Picking up the first component,

(3.17) xx’ + yy’ + zz’+ ww’ = 0
hence
(3.18) g; (:2 + y2 + 22+ w?) = 0

which. means, together with (3.11),
(3.19) :2 + y2 + 22+ w2 = 1
Therefore cl 1’ is a curve on S 3. Tracing back (3.10) and (3.9), c(s)
satisfies (3.6)
Q.E.D.

Lemma. 18
Let c(s) be a curve parametrized by arclength. Then, the following are
equivalent.

(A) c(s) is a helix with the axis vector X1 of the form

 

11(s) = 0
(3.20) ‘ f2(s) = a cos ((-2/b) s + so)
133(3) = a sin ((-2/b) s + so)
where
(3.21) a2 + b2 = 1, ob ¢ 0
(B) c(s) satisfies
(3.22) 1(8) 5 —1
(3.23) <b(s), Xl(c(s))> a a, a at £1, 0
(Proof)
((3) =>(A))

Assume (B) holds. By Frenet—Serret formulas and (3.22)
(3.24) V’tb = n

68

By (3.24), (1.20) and (1.17),

(3.25) (g/x)’ = 241/14, 2': 1,2,3
By (3.23) and (1.18)

(3.26) a = gl/n

By (3.25) and (3.26)

(3.27) 2fi’/n = (a)’ = 0
Hence f1 = const. Put

(3.28) b = f1 (const).

By (1.18), (3.26), (1.19) and (3.28)

(3.29) b = 0X1 - (bf3’/)s)Z\’2 + (312’ /n)}\’3.
By (1.16) and (3.28)

(3.30) 0:2 = ((2')2 + (13')2
By (1.17) and (3.28)

(3.31) n = (f2’/n)}~(2 + (f3’/n))~(3
By “b“ = 1, (3.29) and (3.30),

(3.32) a2 + I)2 = 1

By ||t|| = 1, (1.14), (3.28) and (3.21)

(3.33) 322+ f32 = a2

By (1.15) and (3.28)

(3.34) 12-5, + f3]; = 0

By (1.18) and (3.29)

(3.35) g1 = an, g2 = -bf3’, g:3 = bfz’

By (3.33) we can put
f2 = [12(3): a cos 0, 0 = 0(3)
(336) f3 = f3(s) = a sin 0
By (3.30) and (3.36)
(3.37) n = |a0’| # 0

69

because we are considering the curve c with binormal b. By (3.35) and
(3.36)
a |a0’|

.5?
II

(3.38) g2 = a b cos 0-0’

 

g3 = -a b sin 0-0’
By (3.25), (3.37) and (3.38)

(3.39) (-a b cos 000’/|a0’|)’ = -2 a sin 0-0’/|a0’|

Hence

(3.40) sin 00 (b0’ + 2) = 0.

By (3.37) sin 0(3) has isolated zeros so by continuity,

(3.41) b0’ + 2 = 0

By assumption a at :1 and hence by (3.32) b at 0, so

(3.42) 0 = - 1233 + 30, so = const.

Hence, by (3.28), (3.36) and (3.42) we get (3.20) together with (3.21) =
(3.32).

((A) :5 (B)) straightforward.
Q.E.D.

Now, we can state and prove the following proposition which determine

spherical proper slant surfaces.

Proposition 6
Let x be a spherical proper slant immersion of an oriented surface M
into complex 2—3pace.
x: M4 s3 = 303(1) c c2 = (13“, J0)
Then, x(M) is locally a helical cylinder in S 3 or a circular cylinder in

S 3. Conversely, a helical cylinder and a circular cylinder (if exits) in S 3

70

. 2
are proper slant surfaces in C .

(Proof)
First we note that the isometry w has the following properties:
(3-44) 10(0-0)=¢(0)-¢(p) for 1). <1 6 S 3
(3.45) X e .3 (S 3) is left (or right) invariant.
<—__->- ax is right (or left respectively) invariant.
(3.46) 7(woc)=-Tc for a curve 3 in S 3
(3.47) b is the binormal of c in s 3.

:1?- - (bxb is the binormal of woe.
Assume x is prOper slant immersion with the slant angle 0:. Since x
is spherical, its normal curvature GD vanishes and hence by Lemma 3 and
Pr0position 2, x(M) is a flat surface in S 3. Therefore x(M) is locally a
flat translation surface '
(3-48) 1(1)!) = {CM-7(t)}

described in §1 (n). We follow the four cases in (n).

Cases (i) and (ii)

With a suitable choice of orientations we can assume that the binormal
of s-curves are positive unit normals of 1(M)’ in s 3. Let b(s) be the
binormal of c(s). Then,

(3.49) i/(c(s)) = b(s).
By Lemma 16 (iii) and (3.49),
(3.50) <w,.K b(s), X1(iboc(s))> = —cos a

Let 'E = 1110c and b be the binormal of 3. Then, by (3.47) and (3.50),
(3.51) <b(s), X1(E(s))> = cos a

71

By 70:1 and (3.46)
(3.52) r=-1

Since x is prOper slant,

(3.53) cos a a! 0, t 1

By (3.51), (3.52), (3.53) and Lemma 18, E .is a helix in S 3 with a, b,
a’ in (3.4) determined by

(3.54) a = cos a, b = sin a, a’ = -2/sin a.
By (3.49)
(3-55) (i/JozXM) = {11’ (0(8)-7(t))}

= {1147(0) - ¢(0(8))}
= {(10090) ° 3(3)}
Since (wox)(M) is also flat, (3.55) shows that (wox)(M) is a helical
cylinder. Note that if 7 has torsion -1, then $07 has torsion 1 by (3.46)
and if 7 is a geodesic, then 11.07 is also a geodesic.
Conversely, if (wax)(M) is a helical cylinder
(3-56) (100$)(M) = {770) - 3(8)}
such that E is a helix satisfying (3.54) and '7 is a geodesic or has torsion
r = +1.
Put c = 1005, 7 = 1120?. Then by (3.44),
1W) = {0(8) ° 7(t)}

and rc=1. Let b and b be the binormals of c and Z. By Lemma 18,

(3.57) <b(s), XI(E(3))> = cos a
By (3.47)
(3.58) <b(s), w*(X1('c'(s)))> = -cos 0

Since M, = J v.

72

(3.59) <b(s), Jon(c(s)> = —cos 0
Since the binormals of s—curves are the normals ii of x(M) in S 3 and

x(M) is a translation surface,

(360) 9(48)-7(t))=R,(,)*b(s)
Since Jon is right invariant, (3.55) and (3.56) means
(3-61) «c(s)-7(t)), Jon (c(s)-7(t))> = -cos a

which shows, by Lemma 16, that x(M) is a—slant.

Case (ii’)
Let x(M) = {c(s)-7(t)} where c is a geodesic and 7 has torsion
r=—1. With suitable choice of orientations, the binormal of any t—curve is the

positive unit normal vector of x(M), i.e.,

(3-62) l"(C(19) ‘ 7(t))=Lc(s)*b(to)
By Lemma 16 and (3.62),
(3.63) <Lc(s)*b(to), J07) (c(s)-7(t0))> = —cos a for V s

for any fixed to, which shows that x(M) is a circular cylinder in S 3
defined in Definition 4.
Conversely, let x(M) be a circular cylinder in S 3 satisfying

(3°64) <i’(pl)i Jo’lU’l)> = <Lc(31)*b(t1)i JO" (131))
Since 7 has torsion -l and hence its binormal is right invariant,
(3.65) b(t ) = R _ ‘ ,..b(t)

1 [70,) 107(5)]

Jon is also right invariant, so

(3.60) 10701) = R[ ].Jon(c(sl)-~)(to))

--1
700) "(01)
Hence, usmg Lpo Rq = HQ 0 Lp for V p, q E 53,

73

(3-67) <1A’(Pl)r JOWP1)> = <Lc(sl)*b(to)l J07? (C(31)‘7(t0))>
= " C03 0'

by (3.63). Hence x is a—slant.

Case (iii)
Let x(M)={c(s)-7(t)} where c and 7 are geodesics in S 3 and
c(0)=7(0). Since s—curves and t—curves are geodesies intersecting at a constant

angle, the immersion (s, t) -0 c(s)o7(t) can be extended to a global

immersion
(3.08) v R2 . s 3. as. t) = c(s)-7(t)
Since c and 7 are periodic, y induces an immersion of torus
(3.69) y = 12 = (IR/2r?!) .. (ll/2d) a s 3
c and 7 are great circles and hence we can write
f ”cos 3
sin 3
(3.70) c(s) = A 0
0
‘ L
”cos t
sin t
7(t) = B 0
0

 

 

with suitable A, B 6 0(4). Let X and i’ be the vector field along
y(lR2) determined by s—curves and t—curves reSpectively. Then, using ( 1.7)
and its right version, we can see that the components of X and i" at
c(s)~7(t) with respect to the global frame field {Xi}?=1 are polynomials of
sin 3, cos 3, sin t and cos t with coefficients in ll. The same holds for
ix = X x i" and Jon = l/Agxl. Hence if we put

(3.71) Fist) = «(c(s)-7(t)). JOU(C(S)-7(t))>

then F(s,t) is a polynomial of sin s,....,cos t.

74

2

Since x: M -+ S 3 C C is a—slant, we can choose an open domain U

in R2 such that

(3.72) y=le:U-)S3CC2
is proper slant. Then, by Lemma 16,

(3.73) Hat) = —cos or = const on U
and hence,

(3.74) F(s,t) = -cos a on R2

which shows y is prOper slant globally on R2 and hence y: T2 —+ S 3 C
C2 is a proper slant immersion, contradicting Pr0position 4.

Therefore, 2 can be prOper slant only in cases (i), (ii) and (ii’) and
hence, Proposition 6 is proved.

Q.E.D.

CHAPTER 5
SLANT SURFACES m (2 WITH RANK u<2

Let x : M -0 C2=(E4, J 0) be a slant immersion of an oriented surface.
In this chapter, we consider slant surfaces with the rank of the Gauss map
less than 2.

(1) rank V < 2.

We note that proper slant surfaces among Eg l—Eg 7 have this prOperty (Ch2
§2). Eg 4 has rank u = 2 and prOper slant with respect to J a but totally
real with respect to J1. We also note that rank 11 < 2 means rank 12+
< 2 and rank u_ < 2, hence, by Lemma 3,

(2) G 5 GD; 0

and x(M) is a flat slant surface in (2.

What we are going to do here is a version of the classification of flat
surfaces in E3. As was pointed out by Spivak ([Sl] vol.4 Ch 4, [82]), the
classical "classification" of flat surfaces was not complete. Likewise, if we try
here to classify flat slant surfaces with rank u < 2 completely, we cannot
avoid some messy argument. But it is not our main concern, so we will just
consider typical surfaces and will not go into the problem of gluing pieces of
these surfaces together. The result is stated in Proposition 7.

If we assume some additional conditions, the shapes of these surfaces
become more rigid. For example if x(M) is slant and contained in some

3-plane in E4, then rank u<2 and we get Proposition 8.

We first prove the following lemma.

75

76

Lemma 19
If x is slant and rank u < 2, then x(M) is a flat ruled surface in

E4.

(Proof)
Let 7, V and R, R be the connections and curvature tensors of E4,
M with respect to the induced metric of M. By GD 2 0, (bars are

simultaneously diagonalized, so we can choose an adapted from {e A} such

that

(3) (h3,)= [" 0]. (h4,)= [g 2.]
By Gauss' equation Chl (1.21),

(4) be + de = 0

Put

(5) M1= {p 6 Mlfllp) t 0} Open C M
(6) Mo = Interior of(M—M1)

If M0#0, thenby(4)and b+c= d+e=0, b=c=d=e=0
an M0, i.e., x(M 0) is totally geodesic in E4 and hence

(7) 2(M0) = a portion of 2-plane in E4.

In the following assume M = M1 (i.e., we don't think about 6M1).
Let e3 = H/lIHH. Then, since rank u < 2,
(8) bc=0,d=e=0
by a direct computation using Chl (3.7) and Ch 1 (1.19). Without loss
generality c = 0, b it 0, i.e.,

(9) 0.3.. ) =

b 0 4
U [ J (h..)=0

00 ‘J

77

Put

(10) a); = Awl + ya?

(11) (4)2 = lwl + mw2

Then,

(12) Mel, e.2)e1 = R(ei, e2)e1 - (ezb + Ab)e3 + bme4
(13) Mel, e2)e2 = R(e1, e2)e1 +' b pe3

Since 13:0, 12:0 and b$0

(14) e2b + Ab = 0

(15) m = p = 0

(16) (762.22 = ,ue1 = 0

which means the integral curves of e2 in x(M) is geodesics in E4, i.e.
straight lines.
Therefore, x(M) is a flat ruled surface.
Q.E.D.

A flat ruled surface in E4 is, "in general", a cylinder, a cone or a

tangent developable ([81] vol. 4 p 127). Let us consider slant surfaces of each

type.

(fie—A) M = a—slant cylinder
In this case
(17) x(M) = c x l
where c(s) and l( t) are integral curves of e1 and e2 through
po = c(o) = 1(0) 6 x(M) and l is a straight line. If x(M) is a—slant,
(18) cos a = -Q(e1, 32)
= <el, -.l (32>

= <c’(3), —J e2>

78

If we denote by W the orthogonal complement of l in E4, then
c(s) is a curve in W and -J32 is a fixed vector in W, hence c(s) is
a generalized helix in W whose tangents have a constant angle a between

-Je2.

5! ant cylinder"

 

 

 

 

 

FIGURE 12

(Case B) M = a—slant cone.
Let c(s), (t) be the integral curves of 31 and 82 through a point
p0= c(o) = l(o). Without loss of generality, ||p0||=1 and the vertex of the

cone is the origin of E4. Since

(10) c(s)/(c(s)) = e2(c(s)) for Vs
l.e.

(20) <c(s), c’(s) > = 0 for V3
hence

(21) MS)“ = const = 1

which shows c(s) is a curve on S 3:3 §(0).
If x(M) is a—slant, then as before,
(22) cos a = <c’(s), -J e2(c(s))> for V3

79

But

(23) J e2(c(s)) = J n(c(s))

where r) is the exterior normal of S 3 and hence by the argument of Ch 4
§2, (11) o c)(s) is a generalized helix in S 3 with the axis vector field X1.

slant cone

 

 

 

 

 

FIGURE 13

(Case C) M = a—slant tangent deveIOpable. Put
(24) 1(3) = 0(3) + (t-S)C’(8)

where c(s) is a curve in E4 parameterized by arc length.

 

Put
' c(s) = c'(s)

(25) , 101(0) = "01(8)"

02(8) = (1/~1(8))vi(8)
then
(26) <vlrvg> = 0. Ilvlll = ””2“ = 1-
Put -

61 (at) = 02(0)
(27)

82 (at) = 01(3)

80

then, {e1, 32} is a positive orthonormal frame on M.

Assume x is a—slant, then
(28) cos a = - Ofelllez)
= <u2(s), —J 01(3)>
= <0'(s)/uvi(s)u. -J v1(s)>
We consider 01(3) a curve in S 3. Then (28) means
(29) cos a = <t(3), -J i)(vl(3))>
where t is the tangent of the curve a (3). Hence, by Ch 4§1, (w 0 0(3))

is a generalized helix in s 3 with the axis vector field 3‘1-

slant tangent developable

//’.

61(le (t-AIC (A)
r ref—:—
C(A) \C

FIGURE 14

 

 
  

  

 

S

 

 

 

 

In cases B and C, a generalized helix in s 3 with axis vector field

XI is defined, analogous to Euclidean case, by Definition 4 with (3.4)
replaced by

f1(3) = b
(3") (t,.(s))2 + (f3(s))2 = 1—02.

81

In each case A-C, the converse is also true. Summing, up, we have the

following proposition.

Proposition 7
If x(M) is a slant surface with rank V<2, then x(M) is a flat ruled
surface in (2. Furthermore,
(i) A cylinder in C2 is a slant surface if and only if it is of the
form c(s) x l(t) where, l is a straight line generated by a unit
vector, say e, and c is a generalized helix with axis Je in a
3-plane perpendicular to 1.
(ii) A cone in C2 is a slant surface if and only if it is of the form
(modulo translations)
tc(s)
where (w o c)(s) is a generalized helix in S 3 with axis vector
field X1.
(iii) A tangent developable
148,0 = 0(3) + (t-8)6’(8)
is a slant surface if and only if w o c’ is a generalized helix in

S 3 with the axis vector field X1.

Next, we consider slant surfaces in (2 contained in a hyperplane in

E4. We note first the following lemma.

Lemma 20
Let x : M -+ C2 = (E4, Jo) be a slant immersion of an oriented

surface. If x(M) is contained in some W 6 G(3,4), then rank u < 2

 

82

and x is doubly slant with the same slant angle.

__._ slant surface in a hyperplane

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

FIGURE 15

(Proof)
As in the proof of Lemma 6, we choose a positive orthonormal basis

{eA} of E4, such that

(31) e1, c2 = Joel E Wn JW,
where n is the positive unit normal of W in E4. Put
(33) AZW = {2-vectors of W},
2
(34) GW = G(2,4) n A W

A2W is a 3—dimensional linear subspace of A2E4 spanned by {cl/\e2,

83

e2Ae3, e3Ae1}. Any 5 E A2W satisfies {A5 = 0, i.e., 5 is decomposable,
and hence GW is the unit sphere in A2W. For a E [0, 1r], put

(35) GW,a = {V E G’Wl V is a-slant with respect to Jo}
Then, by Ch 2(1.26), GW a is a circle on GW = .S' 2 expressed by
(36) GW,a = {V E GWI <V, e1Ae2> = cos 0}
Put as before
(37) J = f
1 81A€2

by Ch 2(1.29).
Then, using notations of Ch2 §1,

_ l 2
(38) ”+(GW,a ) — 8J0“; C 3+,
_ 1
(39) wow,“ ) _ 3J0,“ c sf
If x is a—slant with respect to J0 and x(M) C W, then
(40) ”(M C Gw,a
which implies
(41) rank V < 2,

and by (39) and Preposition 2, x is also a-slant with respect to J1.
Q.E.D.

Remark

If we identify A2W with Euclidean 3-space E3 Spanned by {31, e2,
e3} through the isometry X A Y -0 X x Y, where x is the usual vector
product, then V : M -) G W C A2W is nothing but the classical Gauss map
g : M -o S 2 C E3. So, x(M) is a slant surface if and only if g(M) is
contained in a circle S: on S 2 U W = E3 perpendicular to elxe2 =

(42) S; = {Z 6 WI < Z, -Jn > = cos a}

84

By Lemma 20, we consider slant surfaces contained in a hyperplane
according to the three cases of Pr0position 7. We choose a local frame field
{e A} used in the proof of [BYC5] Theorem 2 as follows. Let P, F be as
in Ch 1 (4.5) and t, f be as in [BYCS], namely,

(43) JOY = tY + f Y for Y E TLM

where tY 6 TM and f Y 6 TM. Pick a local unit vector field e on M
which takes value in T;M n W at p 6 M. Let a be the slant angle of
x(M) and assume x(M) is proper slant. Then, we can put

l e1 = te/ltel

e.2 = (see a) Pel
(44) ‘ c3 = (cosec a) Fel
c4 = (cosec a) Fez

 

b

then {eA} is an adapted frame field on M and satisfies

(45) e3(p) E T;M n W
(46) te3 = —(sin a) el , te4 = -(sin a) e2

fe3 = —(cos a) e4 , fc4 = (cos a) e3.
Since )7 e4 = o,

4 _.

(47) (hij) - 0.
By [BYCS] Proposition 2 and Lemma 3, we can use A FXY = A FYX’ and
hence

4 * 0
(48) (h,,) =[0 O]

(47) and (48) means our frame {eA} coincide with the one we chose in the
proof of Lemma 19 and all equations there are also valid here. As before, we
consider the case a it 0. By 7X(Je4) = 0 and (46),

(49) -(sin a) 7X432 - (cos a) A3X = 0

85

hence

(50) algal) = -b cat a
and by (10) and (14)

(51) e2b = b2 cat a.

Case (A’): a slant cylinder in W

(52) x(M) = c x l C W

where l is a straight line generated by some unit vector v and c is a
curve in a 2—plane W’ in W perpendicular to it Then, x(M) is
totally real with respect to J W" So 2(M) is either a portion of a
2—plane (the case of Ch 2 §2(h)) or a non—minimal totally real cylinder (the
case of Ch 2 §2 (i)) and hence J0 = Jive, or JW"

Case (B’): a slant cone in W.

(53) 2(M) = (t c(s)) c W
where c(s) is a curve in
(54) s 2 = s 3 n W.

Assume x(M) is prOper slant and define {eA} by (44). Let t, n, b, n
and r be the tangent, normal, binormal, curvature and torsion of c(s) in

W = 3". Then,

(55) e1(s, t) = c’(s) = (1/t)6/8s
(56) e2(s, t) = c(s) = 8/0t
(57) e3(s, t) = e1(s, t) x c.2(s, t)

where x is the vector product in W = E3. Hence

86

(58) b = <Vele1, e3 >

= (1/t) (Va/63 e1, e3 >
=(1/t) < t’,tx c>
=—(n/t)<n,txc>
=-(n/t)<nxt,c>
= -(n/t) < b, c >.
Differentiating < c, c > = 1 by s twice,
(59) < n, c > = -1/K.

note that n it 0 since c is Spherical. Differentiating again,

(60) < b, c > = -(1/r)(1/x)’.
By (58) and (60)

(61) b = (-Ic’/nr)(1/t).
Hence by (51), (61) and (56)

(62) x’ = nr tan 0.

By (59), (60) and < t, c> = 0

(63) e = -(1/Is)n - (1/r)(1/n)’b
Since [cl = 1,

(64) T2534 = 1:212 + (Ic’)2
By (62) and is it 0

(65) , r2(x2-1—tan2rr) = o

(65) and (62) imply
(66) r = 0
which means c(s) is a circle on S 2 and hence x(M) is a circular cone.

By the remark after Lemma 20, we can see that the axis of this cone is

-JO 7).

87

Case (0’): a slant tangent developable in W.
(67) o(srt) = 6(8) + (t-8)6’(S)
Assume x be proper slant and {eA}, t = 771, n = 772, b, n = Kl, r be as

in case (B’) and case (C). Then,

(68) e1(s, t) = n(s) = (l/t—s) 6/63
(69) (32(3, t) = t(s) = d/dt
(70) e3(s, t) = el(s, t) x e2(s, t) = —b(s).
Hence
(71) Val c1 = (1/(t-s)x)n'(s)
= -1/(t-s)t + r/(t—s)n b
= —1/(t-s)e2 — r/(t—s)ne3
so by (9)
(72) b = - r/(t—s)n
By (72) and (69)
(73) e2b = r/n(t—s)2
By (51) and (72)
(74) r/n = tan a = const.

Hence, c(s) is a generalized helix in W and x(s, t) is a helicoid. The

axis of this helix is —Jon.

In each case A-C, the converse holds. For any circular cone in W,
we can choose a complex structure J on 12‘.4 such that -Jr) is the axis of
this cone, and the same for a helicoid.

Thus we have proved the following Proposition.

88

Proposition 8

Let x : M -+ C2=(E4, J0) be a proper slant immersion of an oriented
surface M. If x(M) is contained in a 3—plane W, then x(M) is a flat
ruled surface in W. And,

(A) A cylinder in W is a proper slant surface with respect to a
complex structure J on E4 if and only if it is a portion of
2-plane.

(B) A cone in W is a proper slant surface with respect to a complex
structure J on E4 if and only if it is a circular cone.

(C) A tangent developable in W is a proper slant surface with respect

to a complex structure J on E4 if and only if it is a helicoid.

(1)

(2)

(3)

(4)
(5)

(7)

SUMMARY

We have proved the following:
The set of 2—planes in C2=(E4, Jo) with constant Wirtinger angle a

is described as Sb a x 33 where G(2,4):S}r x 33 is the
O,

decomposition of the Grassmannian into the product of two 2-spheres of
radius 1N7 in the eigenspaces of the star Operator of AZE4 and

1 . . . 2 .
S Jo’ a is a Circle m S + determined by J0 and a.
An immersion x of an oriented surface M into C2 is a slant

immersion with the Wirtinger angle a if and only if V +(M) C 5.1, a
o,

where V = r 0 V, V is the Gauss map and r is the projection

+ + +
of G(2,4) onto Si.
Any surface without complex tangent points in a 4—dimensional almost
Hermitian manifold is a proper slant surface with given Wirtinger angle
with respect to a suitable almost complex structure.
No compact prOper slant submanifolds exist in complex spaces Cm.
For a surface contained in 5'3 of E4, we can define another Gauss
map by means of left-invariant vector fields. A surface in S3 is a
slant surface with respect to a complex structure if and only if the image
of this Gauss map is contained in a circle.
A surface in 53 is a slant surface if and only if it is a "helical
cylinder" where "helixes" and "cylinders" in .53 are defined by
analogies of those of E3 replacing parallel translations with left
translations.

If a slant surface in C2 has the rank of the Gauss map less than 2,

then it is a flat slant ruled surface in C2 and we can apply the

89

90

classification of flat ruled surfaces in E4. In particular a proper slant
surface contained in a 3-plane in C2 is, in general, a portion of a

2-plane, a circular cone or a helicoid.

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