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# 4 Manifolds all of whose geodesics are lines

In order to tie up the first two sections we need to have a notion analogous to lines in a general Riemannian manifold--this is provided by geodesics or energy minimising paths.

To a stationary observer placed on the manifold it would appear that a body travelling along an energy minimising path is subject to no acceleration. The translation of this into differential geometric terms is DX(t)(X(t)) = 0 where X(t) is the tangent vector at time t. By the theory of second order ordinary differential equations there is a unique geodesic starting at a point p with initial velocity for any choice of p and .

A result of Whitehead shows that for any point p there is a small region M surrounding it so that there is a unique geodesic in M joining any pair of points in M. The notion of between-ness is defined by saying that B lies between A and C in M if B lies on the (unique) geodesic joining A and C. Extending the geodesic within M beyond B and before A gives us the line'' joining A and B. We then easily check that Veblen's axioms of local geometry other than those involving planes are satisfied. In particular we have trouble verifying the Pasch axiom--numbered 7 in the list of Veblen's axioms given in section 1.

Let us therefore make the additional assumption that these axioms dealing with planes are satisfied; we will show that this imposes a restriction on the curvature of the Riemannian manifold M, which is satisfied if and only if M is a convex region in one of the classical'' geometries--Euclidean, Hyperbolic or Projective.

According to the results of section 1 we can choose coordinates on M in such a way that the geodesics are mapped into lines. In terms of these coordinates we see that a geodesic must be an accelerated path:

(X(t)) = DX(t)(X(t)) - (X(t))(X(t)) = - (X(t))(X(t))

being the acceleration at time t. Since the geodesics are lines it follows that acceleration is proportional to X(t); since this is to happen for geodesics in all directions

()() = .

for some scalar depending on . In addition we have the torsion-free condition and the linearity of in each variable. It follows that = < , X > is a linear functional associated with some vector field

()() = ( < (p), > . + < (p), > . )

Thus we have

Let X and Y be orthogonal then

K(X, Y, X, Y) = < R(X, Y)X, Y > = < DX(), X > < Y, Y > - < , X > 2 < Y, Y >

Thus the sectional curvature becomes

(X, Y) =

where the right hand side does not depend on Y! Thus by symmetry it does not depend on X either.

Now we have (X, Y)(p) = (p) depends only on the point p. Thus we have

K(X, Y, Z, W) = . ( < X, W > < Y, Z > - < X, Z > < Y, W > )

and hence

R(X, Y)(Z) = . ( < Y, Z > X - < X, Z > Y)

Now applying the second identity proved by Bianchi

So that if U = X is orthogonal to V and W we get

0 = DW . < U, U > V - DV . < U, U > W

Finally, since have assumed that we are in (at least) three dimensions we can assume that V and W are linearly independent and < U, U > 0, so that we obtain DW = 0. Since W is arbitrary this just means that is the constant function. Now, by applying the result of Cartan and Hadamard mentioned in the previous section we see that our local Riemannian space M has to be contained in one of the classical'' geometries--Euclidean, Hyperbolic or Projective.

Next: 5 Conclusion Up: Axiomatic and Coordinate Geometry Previous: 3 Riemannian Geometry
Kapil Hari Paranjape 2002-11-21