Applied Mathematics E-Notes, 1(2001), 149-154 \begin{matrix}c \\ ◯ \end{matrix} Available free at mirror sites of http://math2.math.nthu.edu.tw/∼amen/

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 vector fields 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 matrix a and a vector b such that for each x ∈ M, Z_{x} = af(x) + b where a^{t} = −εaε.

1 Introduction

Let M and \overline{M} be semi-Riemannian manifolds and g and \overline{g} the metric tensors on M and \overline{M} respectively. Suppose M is isometrically immersed in \overline{M} by the immersion f : M → \overline{M} with the understanding that f(M) is identified with M. 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 map F : I × M → \overline{M}, where I = (−ε, ε), suppose that the submanifolds F_{s}(M) are the semi-Riemannian submanifolds of \overline{M} for all s ∈ I. For the rest of this section we will assume that the induced metrics on F_{s}(M) is non-degenerate for all s ∈ I.

An isometric variation is a smooth map F such that F_{s}(x) = F (s, x) is an isometric immersion for each s ∈ I and F (0, x) = f(x).

Let

g_{s} = (F_{s})^{∗}\overline{g}.  (1)

Then g_{s} is a metric tensor on M. Let T_{2}(T_{p}(M)) denote the tensor algebra of (0, 2)- tensor fields on T_{p}(M). If we consider the following function for all p ∈ M,

\begin{aligned}g_{p} : & I & → & T_{2}(T_{p}(M)) \\ & s & → & (g_{s})_{p} \end{aligned}

∗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. Sabuncuoğlu.

‡Department of Mathematics, Mersin University, Mersin, Turkey

then, (g_{p}, I) is a curve on T_{2}(T_{p}(M)).

DEFINITION 1. The smooth map F is called an infinitesimal isometric variation of M if g′(0) = 0.

EXAMPLE 1. Consider the Minkowski space ℝ_{1}^{3}. The metric tensor \overline{g} on this space is given by the following equality:

\overline{g}_{p}(v, w) = −v_{1}w_{1} + v_{2}w_{2} + v_{3}w_{3}  (2)

for every v, w ∈ T_{p}(ℝ_{1}^{3}). Now suppose

M = {(p_{1}, p_{2}) | |p_{1}| < 1, 0 < −p_{1}^{2} + p_{2}^{2} < 1}  (3)

and

F : I × M → R_{1}^{3}, F = (u, v, t(1 + u^{2} − v^{2})),

where I = (−1/2, 1/2), {u, v} is the Euclidean coordinate system of ℝ_{1}^{2} and T_{p}(M) = S_{p}\{ \frac{∂}{∂u} |_{p} , \frac{∂}{∂v} |_{p}\}. Then F_{0} = f and rank J(F_{s}) = 2. The tensor field g_{s} on M defined by (1) is obtained as

(g_{s})_{p}(X_{p}, Y_{p})
= ⟨X_{p}, Y_{p}⟩ + 4s^{2}[p_{1}^{2}a_{1}(p)b_{1}(p) − p_{1}p_{2}a_{1}(p)b_{2}(p)
−p_{2}p_{1}a_{2}(p)b_{1}(p) + p_{2}^{2}a_{2}(p)b_{2}(p)],  (4)

where X_{p} = a_{1}(p) \frac{∂}{∂u} |_{p} + a_{2}(p) \frac{∂}{∂v} |_{p} and Y_{p} = b_{1}(p) \frac{∂}{∂u} |_{p} + b_{2}(p) \frac{∂}{∂v} |_{p} for all X_{p}, Y_{p} ∈ T_{p}(M). Since,

(g_{s})_{p} = (−1 + 4s^{2}p_{1}^{2}, −4s^{2}p_{1}p_{2}, −4s^{2}p_{1}p_{2}, 1 + 4s^{2}p_{2}^{2}),

we have g′(0) = 0.

The tensor field g_{0} on the manifold M assigns a scalar product to each point of M and the index of g_{0} is the same for all p. Therefore f(M) is a semi-Riemannian submanifold of ℝ_{1}^{3}. The submanifold F_{s}(M) is the semi-Riemannian submanifold of the space ℝ_{1}^{3} for every s ∈ I if the index of g_{s} is the same for all p and the following condition holds,

∀Y_{p} ∈ T_{p}(M), (g_{s})_{p}(X_{p}, Y_{p}) = 0 ⇒ X_{p} = 0.

In (4), if we take Y_{p} = (1, 0) and Y_{p} = (0, 1) then the following system of equations are obtained:

\begin{matrix}(−1 + 4s^{2}p_{1}^{2})a_{1}(p) − 4s^{2}p_{1}p_{2}a_{2}(p) = 0 \\ (−4s^{2}p_{1}p_{2})a_{1}(p) + (1 + 4s^{2}p_{2}^{2})a_{2}(p) = 0 \end{matrix}  (5)

Let △ denote the determinant of system (5), then

△ = 0 ⟺ p_{1}^{2} − p_{2}^{2} = \frac{1}{4s^{2}} .

Since △ ≠ 0 for each p of M, there is no non-vanishing solution of the system (5). Therefore X_{p} = 0. Hence (g_{s})_{p} is nondegenerate.

By setting v_{1} = (1, 0), v_{2} = ((4s^{2}p_{1}p_{2})/(−1 + 4s^{2}p_{1}^{2}), 1) and e_{1} = v_{1}/ |v_{1}| , e_{2} = v_{2}/ |v_{2}| , we have

(g_{s})_{p}(e_{1}, e_{1}) = −1,  (6)
(g_{s})_{p}(e_{2}, e_{2}) = 1.  (7)

By (6) and (7), the index of the metric tensor g_{s} is 1 at each point p. It can easily be seen that the metric tensor g_{s} is symmetric and bilinear. Thus the manifolds F_{s}(M) for s ∈ (−1/2, 1/2) are the semi-Riemannian submanifolds of the space ℝ_{1}^{3}. Therefore F is an infinitesimal isometric variation of M.

2 Preparatory Results

Now, we consider an isometric immersion f : M → \overline{M} as before. Let Z_{x} be the tangent vector of the curve α : I → \overline{M}, α(s) = F (s, x) at α(0), for each x ∈ M, Z_{x} is the initial velocity of the orbit of f(x) under F. The section Z is called the variation vector field of the variation of F.

LEMMA 1. Let M and \overline{M} be semi-Riemannian manifolds and F be an isometric variation of M in \overline{M}. Let f̃: \overline{M} → ℝ_{ν}^{n} be an isometric immersion. Suppose that

F̃ = f̃∘ F.

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

PROOF. The variation F̃ gives the mapping F̃_{s} : M → ℝ_{ν}^{n}, F̃_{s}(x) = F̃(s, x) for all s ∈ I and x ∈ M. Let the metric tensors g, \overline{g}, g̃ be the metric tensors on the manifolds M, \overline{M} and ℝ_{ν}^{3} respectively. Let us suppose

g̃_{s} = (F̃_{s})^{∗}g̃,

then

(g_{p})′(0)(X_{p}, Y_{p}) = 0
⇔ lim_{h→0} \frac{1}{h}\{(g̃_{p})(h) − (g̃_{p})(0)\}(X_{p}, Y_{p}) = 0
⇔ lim_{h→0} \frac{1}{h}\{(g̃_{h})_{p} − (g̃_{0})_{p}\}(X_{p}, Y_{p}) = 0
⇔ lim_{h→0} \frac{1}{h}\{[(F̃_{h})^{∗}g̃]_{p} − [(F̃_{0})^{∗}g̃]_{p}\}(X_{p}, Y_{p}) = 0
⇔ lim_{h→0} \frac{1}{h}\{g̃(f̃_{∗}((F_{h})_{∗}X_{p}), f̃_{∗}((F_{h})_{∗}Y_{p})) − g̃(f̃_{∗}((F_{0})_{∗}X_{p}), f̃_{∗}((F_{0})_{∗}Y_{p}))\} = 0
⇔ lim_{h→0} \frac{1}{h}\{\overline{g}((F_{h})_{∗}X_{p}), (F_{h})_{∗}Y_{p}) − \overline{g}(f_{∗}X_{p}, f_{∗}Y_{p})\} = 0
⇔ lim_{h→0} \frac{1}{h}\{(g_{h})_{p}) − (g_{0})_{p}\}(X_{p}, Y_{p}) = 0
⇔ lim_{h→0} \frac{1}{h}\{g_{p}(h) − g_{p}(0)\}(X_{p}, Y_{p}) = 0
⇔ g′_{p}(0)(X_{p}, Y_{p}) = 0
⇔ [g′(0)]_{p}(X_{p}, Y_{p}) = 0

for all X_{p}, Y_{p} ∈ T_{p}(M). This completes the proof.

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

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

⟨D_{X}Z, X⟩ = 0, for all X ∈ χ(M).  (8)

Here D is the induced connection and ⟨, ⟩ is the fibre metric on the tangent bundle T (\overline{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 space ℝ_{ν}^{n} be I(ℝ_{ν}^{n}) and let φ(I) be a curve in I(ℝ_{ν}^{n}) such that φ(0) = 1. Let us define the mapping μ by the equation

μ(s, x) = φ(s)(f(x)).  (9)

Then

(μ_{s})_{∗}(X) = (φ(s) ∘ f)_{∗}(X) = (φ(s))_{∗}(f_{∗}X)

for X ∈ χ(M). We also have g_{s} = g_{0} since

g_{s}(X, Y) = [(μ_{s})^{∗}\overline{g}](X, Y) = \overline{g}((φ(s))_{∗}(f_{∗}X), (φ(s))_{∗}(f_{∗}Y))
= \overline{g}(f_{∗}X, f_{∗}Y) = [(μ_{0})^{∗}\overline{g}](X, Y) = g_{0}(X, Y)

for all X, Y ∈ χ(M). Thus μ is 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 variation μ defined by (9) then F 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 a and a vector b such that

Z_{x} = af(x) + b, for each x ∈ M

where a^{t} = −εaε.

PROOF. If φ(I) is a curve in the space I(ℝ_{ν}^{n}), then we have,

φ(s)(f(x)) = α(s)f(x) + β(s),  (10)

where f(x) ∈ ℝ_{ν}^{n}, α(s) is a semi-orthogonal matrix and β(s) ∈ ℝ_{ν}^{n}. Substituting s = 0 in (10), we obtain

f(x) = α(0)f(x) + β(0).  (11)

The points p_{0}, p_{1}, …, p_{n} can be chosen so that the vectors \overrightarrow{p_{0}p_{i}}, (1 ≤ i ≤ n) form a base of the vector space ℝ_{ν}^{n} in the manifold f(M). Then by using (11), we get the following equalities

[I − α(0)](\overrightarrow{0p_{0}}) = β(0),
[I − α(0)](\overrightarrow{0p_{i}}) = β(0), 1 ≤ i ≤ n.

Therefore,

[I − α(0)](\overrightarrow{p_{0}p_{i}}) = 0.  (12)

By (12), we have I = α(0) and β(0) = 0.

Let λ be the variation which is defined by the curve φ. Let us denote the variation vector field of λ by Z_{1}. The orbit of the point f(x) in the variation λ is the curve

s → α(s)f(x) + β(s).

Thus we have,

(Z_{1})_{x} = α′(0)f(x) + β′(0).

Since α(s) is a semi-orthogonal matrix,

(α(s))^{t} = ε(α(s))^{−1}ε,  (13)

or simply, α^{t} = εα^{−1}ε. We have

(α^{t})′ = ε(α^{−1})′ε.  (14)

Since α is a semi-orthogonal matrix, |det α| = 1. Thus if we let α^{−1} = α̃, then αα̃ = I. Now

αα̃ = I ⇒ α′α̃ + αα′ = 0 ⇒ α′(0)α̃(0) + α(0)α′(0) = 0.

Since α(0) = I and αα̃ = I, then α̃(0) = I. From this, it follows that

α′(0) = −α′(0).  (15)

By (14), we find

(α′(0))^{t} = ε(α^{−1}(0))′ε.

Using α^{−1}(0) = α̃(0) and considering the equality (15) one can obtain

(α′(0))^{t} = −ε(α′(0))ε.

Let us denote a = α′(0) and b = β(0) then

(Z_{1})_{x} = af(x) + b.

Consider the variation F of M in ℝ_{ν}^{n}. If F is trivial then the variation vector field of F coincides with the variation vector field of λ. Therefore the variation field of F is in the form

Z_{x} = af(x) + b.

Conversely, suppose that the variation vector field of F is in the form

Z_{x} = af(x) + b

such that a is a matrix satisfying the equality a^{t} = −εaε. i.e., a is an element of the Lie algebra on O_{ν}(n). The matrix a is in O_{ν}(n) if, and only if, exp(sa) is in O_{ν}(n) such that |s| is sufficiently small [3]. In this case the deformation F takes the form

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

Let us take φ(s) = exp(sa). Then φ(0) = I, and φ(s) is an isometry for all s. The proof is complete.

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

EXAMPLE 2. Let M be a unit disc in ℝ_{1}^{2} and f : M → ℝ_{1}^{3} be an immersion. Consider the following isometric variation of M.

μ : I × M → ℝ_{1}^{3}, 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) ∈ O_{1}(3) to be

α(s) = [ \begin{matrix}cosh s & 0 & sinh s \\ 0 & 1 & 0 \\ sinh s & 0 & cosh s \end{matrix} ] .

Then,

[φ(s)]u = (u_{1} cosh s + u_{3} sinh s, u_{2}, u_{1} sinh s + u_{3} cosh s).

Let us denote the variation vector field of μ by Z_{1}. It is easily seen that Z_{1} = (0, 0, u). Now we consider the variation F of M given by

F : I × M → ℝ_{1}^{3}, F = (u, v, tu), I = (−\frac{1}{2}, \frac{1}{2}).

Then F is an infinitesimal isometric variation of M. Let Z be a variation vector field of F. Then Z = (0, 0, u). Since Z_{1} and Z coincides, F is a trivial deformation.

References

[1] Goldstein, R. A. and Ryan, P. J., Infinitesimal rigidity of submanifolds, J. Differential 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