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Chapter 18: Problem 17

Show that a space is locally flat if and only if there exists a local basis of vector ficlds \(\left\\{e_{t} \mid\right.\) that are absolutely parallel, \(D e_{1}=0\).

Short Answer

Expert verified
The statement is true. In a locally flat space, around each point, there is a region behaving as a flat space. In this region, we can find a set of absolutely parallel vector fields. Conversely, around a point where a set of absolutely parallel vector fields exists, the space behaves as a flat one, signifying locally flatness of the space at that point.

Step by step solution

01

Understand the definition of 'Locally Flat'

To start with, it is crucial that 'locally flat' is understood. A locally flat space is one where, around any given point, there is a region which behaves just like a flat space. This is a considerable generalization of a flat space.
02

Understand the definition of 'Absolutely Parallel Vector Fields'

Next, understand what is meant by 'absolutely parallel vector fields'. A vector field is called absolutely parallel if its covariant derivative vanishes. This means, for a set of vector fields {e_{t}}, if D e_{t} = 0 for each t, then this set of vector fields is absolutely parallel.
03

Proving the if part of the statement

Consider a locally flat space, and pick a point in it. Around this point, there is a region behaving like a flat space. In this region, one can always find a set of vector fields {e_{t}}, which are constant, and hence, their covariant derivative is zero. This means the set of vector fields is absolutely parallel.
04

Proving the only if part of the statement

Now let's say that there exists a set of absolutely parallel vectors fields {e_{t}} around a point. The behavior of objects around this point will be the same as in a flat space since there's no change in the direction of indices. So this region behaves as if it is a flat space, meaning the space is locally flat at this point.

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