Infinitesimal Isometric Variation of Semi-Riemannian Submanifolds ∗ †

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Infinitesimal Isometric Variation of Semi-Riemannian Submanifolds ^∗ ^†

Leyla Onat

^‡

Received 11 October 2000

Abstract

Infinitesimal isometric variations of submanifolds of a semi-Riemannian manifold are considered. We define variation vectorfields of submanifolds and study their properties. It is shown that an isometric variation of a submanifold of the semi-Euclidean space is trivial if, and only if, there exist a matrixaand a vector bsuch that for eachx∈M, Zx=af(x) +bwherea^t=−εaε.

1 Introduction

Let M and M be semi-Riemannian manifolds and g and g the metric tensors on M and M respectively. Suppose M is isometrically immersed in M by the immersion f :M →M with the understanding thatf(M) is identified withM. In this paper all manifolds are assumed to be C^∞.

Many authors have studied the infinitesimal rigidity of Riemannian submanifolds.

Variation vector field is the main tool of this subject. Our objective is to investigate the variation vector field of semi-Riemannian submanifolds.

For the smooth mapF :I×M→M ,whereI= (−ε,ε),suppose that the subman- ifoldsFs(M) are the semi-Riemannian submanifolds ofMfor alls∈I.For the rest of this section we will assume that the induced metrics on Fs(M) is non-degenerate for alls∈I.

Anisometric variation is a smooth mapF such thatFs(x) =F(s, x) is an isometric immersion for eachs∈I andF(0, x) =f(x).

Let

gs= (Fs)^∗g. (1)

Then gs is a metric tensor on M. Let T2(Tp(M)) denote the tensor algebra of (0,2)- tensorfields onT_p(M).If we consider the following function for allp∈M,

gp: I → T2(Tp(M)) s → (gs)p

∗Mathematics Subject Classifications: 53B25, 53C40.

†The material in this paper forms part of the author’s Ph. D. Thesis under the supervision of Dr.

A. Sabuncuo˘glu.

‡Department of Mathematics, Mersin University, Mersin, Turkey

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then, (gp, I) is a curve onT2(Tp(M)).

DEFINITION 1. The smooth mapF is called an infinitesimal isometric variation ofM ifg(0) = 0.

EXAMPLE 1. Consider the Minkowski spaceR³1.The metric tensorgon this space is given by the following equality:

g_p(v, w) =−v1w1+v2w2+v3w3 (2) for every v, w∈Tp(R³1).Now suppose

M ={(p1, p2)| |p1|<1, 0<−p²₁+p²₂<1} (3) and

F :I×M→R³₁, F = (u, v, t(1 +u²−v²)),

where I= (−1/2,1/2),{u, v}is the Euclidean coordinate system of R²1 and Tp(M) = Sp{∂^∂u p, _{∂v p}^∂ }.Then F0 =f and rankJ(Fs) = 2. The tensor fieldgs on M defined by (1) is obtained as

(gs)p(Xp, Yp)

= X_p, Y_p + 4s²[p²₁a₁(p)b₁(p)−p₁p₂a₁(p)b₂(p)

−p2p1a2(p)b1(p) +p²₂a2(p)b2(p)], (4) where Xp =a1(p) _{∂u p}^∂ +a2(p)_{∂v p}^∂ andYp=b1(p) _{∂u p}^∂ +b2(p) _{∂v p}^∂ for allXp, Yp∈ Tp(M).Since,

(gs)p= (−1 + 4s²p²₁,−4s²p1p2,−4s²p1p2,1 + 4s²p²₂), we have g(0) = 0.

The tensor field g₀ on the manifold M assigns a scalar product to each point of M and the index of g₀ is the same for all p. Therefore f(M) is a semi-Riemannian submanifold of R³1. The submanifold Fs(M) is the semi-Riemannian submanifold of the space R³1 for everys∈I if the index of gs is the same for allpand the following condition holds,

∀Y_p∈T_p(M), (g_s)_p(X_p, Y_p) = 0⇒X_p= 0.

In (4), if we takeYp= (1,0) andYp= (0,1) then the following system of equations are obtained:

(−1 + 4s²p²₁)a₁(p)−4s²p₁p₂a₂(p) = 0

(−4s²p₁p₂)a₁(p) + (1 + 4s²p²₂)a₂(p) = 0 (5) Let denote the determinant of system (5),then

= 0⇐⇒p²₁−p²₂= 1 4s².

Since = 0 for each p of M, there is no non-vanishing solution of the system (5).

ThereforeXp= 0.Hence (gs)pis nondegenerate.

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By setting v1 = (1,0), v2 = ((4s²p1p2)/(−1 + 4s²p²₁),1) and e1 = v1/|v1|, e2 = v2/|v2|,we have

(gs)p(e1, e1) =−1, (6)

(gs)p(e2, e2) = 1. (7)

By (6) and (7),the index of the metric tensorg_s is 1 at each pointp.It can easily be seen that the metric tensorg_sis symmetric and bilinear. Thus the manifoldsF_s(M) fors∈(−1/2,1/2) are the semi-Riemannian submanifolds of the spaceR³1.Therefore F is an infinitesimal isometric variation ofM.

2 Preparatory Results

Now, we consider an isometric immersionf :M→Mas before. LetZxbe the tangent vector of the curveα:I→M ,α(s) =F(s, x) atα(0),for eachx∈M, Zxis the initial velocity of the orbit of f(x) underF.The sectionZ is called the variation vectorfield of the variation ofF.

LEMMA 1. Let M and M be semi-Riemannian manifolds andF be an isometric variation ofM in M .Letf :M→Rⁿν be an isometric immersion. Suppose that

F =f ◦F.

Then F is an infinitesimal isometric variation if, and only if, F is an infinitesimal isometric variation.

PROOF. The variationF gives the mappingF_s:M→Rⁿν, F_s(x) =F(s, x) for all s∈Iandx∈M. Let the metric tensorsg, g, g be the metric tensors on the manifolds M, M andR³ν respectively. Let us suppose

g_s= (F_s)^∗g, then

(gp) (0)(Xp, Yp) = 0

⇔ limh→01

h{(gp)(h)−(gp)(0)}(Xp, Yp) = 0

⇔ limh→01

h{(gh)p−(g0)p}(Xp, Yp) = 0

⇔ limh→01

h{[(Fh)^∗g]p−[(F0)^∗g]p}(Xp, Yp) = 0

⇔ limh→01

h{g(f_∗((Fh)_∗Xp), f_∗((Fh)_∗Yp))−g(f_∗((F0)_∗Xp), f_∗((F0)_∗Yp))}= 0

⇔ limh→01

h{g((Fh)_∗Xp),(Fh)_∗Yp)−g(f_∗Xp, f_∗Yp)}= 0

⇔ limh→01

h{(gh)p)−(g0)p}(Xp, Yp) = 0

⇔ limh→01

h{gp(h)−gp(0)}(Xp, Yp) = 0

⇔ g_p(0)(Xp, Yp) = 0

⇔ [g(0)]p(Xp, Yp) = 0

for allX_p, Y_p∈T_p(M).This completes the proof.

Now we state the following lemma for the immersed semi-Riemannian submanifold.

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LEMMA 2. The variationF is an infinitesimal isometric variation if, and only if,

DXZ, X = 0, for allX ∈χ(M). (8)

Here D is the induced connection and , is the fibre metric on the tangent bundle T(M).

Indeed, it can easily be shown by using Lemma 1. Moreover, one can be find the proof for the Riemannian submanifolds in [1] and [2].

3 Trivial Isometric Variations

Let the group of isometries of the spaceRⁿν beI(Rⁿν) and letϕ(I) be a curve inI(Rⁿν) such thatϕ(0) = 1.Let us define the mappingµby the equation

µ(s, x) =ϕ(s)(f(x)). (9)

Then

(µs)_∗(X) = (ϕ(s)◦f)_∗(X) = (ϕ(s))_∗(f_∗X) forX ∈χ(M).We also haveg_s=g₀ since

g_s(X, Y) = [(µ_s)^∗g](X, Y) =g((ϕ(s))_∗(f_∗X),(ϕ(s))_∗(f_∗Y))

= g(f_∗X, f_∗Y) = [(µ₀)^∗g](X, Y) =g₀(X, Y) for allX, Y ∈χ(M).Thusµis an isometric variation.

DEFINITION 2. Let M be a semi-Riemannian manifold and F be an isometric variation of M. If the variation vector field of F coincides with the variation vector

field of the isometric variationµdefined by (9) thenF is said to be trivial.

THEOREM 1. A variation of a submanifold of the semi-Euclidean space is trivial if, and only if, there exits a matrix aand a vectorbsuch that

Zx=af(x) +b, for eachx∈M where a^t=−εaε.

PROOF. Ifϕ(I) is a curve in the spaceI(Rⁿν), then we have,

ϕ(s)(f(x)) =α(s)f(x) +β(s), (10) where f(x)∈Rⁿν,α(s) is a semi-orthogonal matrix andβ(s)∈Rⁿν. Substitutings= 0 in (10),we obtain

f(x) =α(0)f(x) +β(0). (11)

The points p0, p1, . . . , pn can be chosen so that the vectors −−→p0pi, (1 ≤ i≤ n) form a base of the vector space Rⁿν in the manifold f(M). Then by using (11), we get the following equalities

[I−α(0)](−→0p0) =β(0), [I−α(0)](−→0pi) =β(0), 1≤i≤n.

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Therefore,

[I−α(0)](−−→p0pi) = 0. (12) By (12),we have I=α(0) andβ(0) = 0.

Letλbe the variation which is defined by the curveϕ.Let us denote the variation vectorfield ofλbyZ1.The orbit of the pointf(x) in the variationλis the curve

s→α(s)f(x) +β(s).

Thus we have,

(Z₁)_x=α(0)f(x) +β(0).

Sinceα(s) is a semi-orthogonal matrix,

(α(s))^t=ε(α(s))⁻¹ε, (13)

or simply,α^t=εα⁻¹ε.We have

(α^t) =ε(α⁻¹)ε. (14)

Sinceαis a semi-orthogonal matrix, |detα|= 1.Thus if we letα⁻¹=α,thenαα=I.

Now

αα=I⇒α α+αα = 0⇒α(0)α(0) +α(0)α(0) = 0.

Sinceα(0) =I andαα=I,thenα(0) =I.From this, it follows that

α(0) =−α(0). (15)

By (14),we find

(α(0))^t=ε(α⁻¹(0))ε.

Usingα⁻¹(0) =α(0) and considering the equality (15) one can obtain (α(0))^t=−ε(α(0))ε.

Let us denote a=α(0) andb=β(0) then

(Z₁)_x=af(x) +b.

Consider the variation F ofM in Rⁿν.IfF is trivial then the variation vectorfield of F coincides with the variation vectorfield ofλ.Therefore the variationfield ofF is in the form

Zx=af(x) +b.

Conversely, suppose that the variation vector field ofF is in the form Z_x=af(x) +b

such that ais a matrix satisfying the equality a^t=−εaε.i.e., a is an element of the Lie algebra onO_ν(n).The matrixais inO_ν(n) if, and only if, exp(sa) is inO_ν(n) such that |s|is suﬃciently small [3]. In this case the deformationF takes the form

F(s, x) = exp(sa)f(x) +sb.

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Let us takeϕ(s) = exp(sa).Thenϕ(0) =I,andϕ(s) is an isometry for alls.The proof is complete.

Now we present the following example for illustrating the previous theorem.

EXAMPLE 2. Let M be a unit disc in R²1 and f : M → R³1 be an immersion.

Consider the following isometric variation ofM.

µ:I×M→R³1, µ(s, x) =ϕ(s)f(x),

where ϕ(s) is an isometry which is determined by the equality [ϕ(s)]u=α(s)u.If we take α(s)∈O1(3) to be

α(s) =



 coshs 0 sinhs

0 1 0

sinhs 0 coshs



.

Then,

[ϕ(s)]u= (u1coshs+u3sinhs, u2, u1sinhs+u3coshs).

Let us denote the variation vectorfield ofµbyZ1.It is easily seen thatZ1= (0,0, u).

Now we consider the variationF ofM given by

F:I×M→R³1, F = (u, v, tu), I = (−1 2,1

2).

Then F is an infinitesimal isometric variation of M. LetZ be a variation vectorfield ofF. ThenZ= (0,0, u).SinceZ1andZ coincides,F is a trivial deformation.

References

[1] Goldstein, R. A. and Ryan, P. J., Infinitesimal rigidity of submanifolds, J. Diﬀer- ential Geometry, 10 (1975), 49-60.

[2] Dajczer, M. and Rodriguez, L., Infinitesimal rigidity of Euclidean submanifolds, Ann. Inst. Fourier. Grenoble, 40(4), (1990), 936-949.

[3] O’Neil, B., Semi-Riemannian geometry with applications to relativity, Academic Press, New York, 1983