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is of one of two types:

(I)

ds2 = dx2 + adx22 + 2ẞdxdx3 + ydxз2,

a, B, y being functions of x1 alone, the operators of the group being X1 =d/dx2, X2 = d/dx3;

X2

(II) ds2 = dx12 + adx22 + 2(ẞ — αx2) dx2dx3 + (αx22 - 2ẞx2 + y)dx32, a, B, y being functions of x1 alone, the operators of the group being X1 = d/dxз, X2 = X2 e*3d/dx2.

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where the Christoffel quantities are calculated with respect to the linear element of the space-time continuum. When this linear element is taken in the form Vidxo2-ds' as in § 1, the equations Bik = 0 reduce to the six equations2

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ik

where a is the cofactor of a

Σaik Bik = 0,

i, k

in the determinant of the quantities a divided by this determinant; the Christoffel symbols are calculated with respect to (1), and

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By assuming that S is referred to a triply-orthogonal system, so that

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any 3-space3), and making use of the relations

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we find that when Bik = 0, then all the functions (il, hk) are equal to This being the condition that the space S is euclidean, we have the theorem:

zero.

A necessary and sufficient condition that a 3-space be euclidean is that the functions Bik = 0.

3. If we put

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δ = αγ – β', Α = β' γ = βγ', Β = γ' α - γα', C = αβ = αβ', (where primes indicate differentiation with respect to x1) we find that all of the Christoffel quantities for the form (I) vanish except the following:

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The problem reduces to the integration of (3), when a, ß, y are subject to the condition (6). We separate the discussion into the four cases, when V is independent of x and x3; of either x2 or x3; of x1; involves all three variables. Expressing the conditions of integrability of (3) we are led to linear partial differential equations of the first order and these are ultimately solved. The first and second cases are the only ones giving solutions; in the course of the investigation it is shown that in both cases a change of variables can be made so that ß €0. The solutions

of the first type are:

=

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where k is any constant.

Solutions of this kind have been found by

Kasner. For the second case the linear element of S may be written

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where a and b are constants. This is the solution found by Levi-Civita3

and called the quadrantal solution.

4. For the linear element (II) the expressions for the Christoffel symbols which are different from zero are:

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where

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B22 L, B23 = M - x2 L, B33 = L x22 − 2 M x2 + N,

=

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2

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28

28

B"

B

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Y

+

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αγ

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4 δ

28

δ

2

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We consider the system of equations (3) as in § 3, taking up separately the five cases where V is independent of x2 and x3; Of X2; Of X1; Of x3; involves all three variables. Only the first two cases lead to solutions.

For the first case we have the linear element of S

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and

V = √a a -1/2 — 1'

where a is a constant. This is the longitudinal solution obtained by Levi-Civita."

For V independent of x two cases arise, according as ẞ can be made equal to zero by a transformation of coördinates, or not.

When ẞo, the functions a and y must satisfy the three equations

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When k

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d y2 + a dx2 − 2 a x2 dx2 dx3 + ( a x2 + v)dxż. (14)

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1, we solve the second of (12) for a'/a and substitute in the third of (12). The first integral of the resulting equation is

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where a is an arbitrary constant and b = ± (1 − k) √1 +k + k2 . Then we have

(15)

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812+28 (α'y' - 8/2) + 24 ad = 0,

where k denotes an arbitrary constant.

δ = αγ- β,

(17)

14 = 0,

δ

(18)

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which is consistent with (18). Eliminating a from the first of (18) and (19), we get

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If we introduce dependent and independent variables t and @ by

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To the evident solutions of this equation, t = 5/2, t = - 5/3, correspond the solutions:

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excluded since we assume that 8 > 0 in accordance with the theory. If we put t dt/de = y, and take y for dependent variable, the above equation may be replaced by

y

dy

dt

+7 (7+13) + (2+5) (3+5) (1+3)=0.

When a solution of this equation is known, the corresponding functions a, B, y can be found by quadratures.

5. Making use of the formulas of Bianchi we find the following expressions for the principal curvatures of the spaces (7), (8), (11), (14)

− =

- K1 = (1 + k) R2 = (1 + 1) R2 = k(+1)* 1

K1

K2

1 =

K

=

=

(1 + k + k 2) 2 x 12 i

(7*)

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2a

K1 = − / (1 + √3), K2 = 2, K3 = −3 (1−√3). (14*)

γ

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The principal curvatures for k± 1 in (12) can be obtained explicitly, but their forms are quite involved.

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