Undergrad Why does a parallelizable manifold imply zero Riemann tensor?

[Silviu]

Hello! Can someone explain to me why does a parallelizable manifold implies zero Riemann tensor? In the book I read this is mentioned but not proved. This would imply that parallel-transporting a vector would be path independent. But I am not sure how to show it. Thank you!

 

[George Jones]

Parallelizable does not imply zero Riemann tensor; Parallelizable implies that it is possible to define a connection such that the resulting Riemann tensor is zero.

 

[lavinia]

Hello! Can someone explain to me why does a parallelizable manifold implies zero Riemann tensor? In the book I read this is mentioned but not proved. This would imply that parallel-transporting a vector would be path independent. But I am not sure how to show it. Thank you!

Look at what your book says again.

A closed flat Riemannian manifold can not be simply connected. A fundamental theorem says that its fundamental group must contain a subgroup isomorphic to $Z^{n}$. So in fact, any parallizable closed manifold with a finite fundamental group e.g. $S^3$ and $RP^3$ can never be made flat.

But this is far from the whole story. For instance every orientable closed 3 manifold is parallelizable.

The fundamental group of a closed flat Riemannian manifold has a special structure.

It is an extension of an n-dimensional free abelian group by a finite group

$0→Z^{n}→π_{1}(M)→G→1$

and has no elements of finite order. For instance, for a torus $G$ is the trivial group.

- There are parallelizable flat Riemannian manifolds that are not tori. For them parallel translation of vectors around closed loops depends upon the homotopy class of the loop..

 

[Silviu]

Look at what you book says again.

A closed flat Riemannian manifold can not be simply connected. A fundamental theorem says that its fundamental group must contain a subgroup isomorphic to $Z^{n}$. So in fact, any parallizable closed manifold with a finite fundamental group e.g. $S^3$ and $RP^3$ can never be made flat.

But this is far from the whole story. For instance every orientable closed 3 manifold is parallelizable.

The fundamental group of a closed flat Riemannian manifold has a special structure.

It is an extension of an n-dimensional free abelian group by a finite group

$0→Z^{n}→π_{1}(M)→G→1$

and has no elements of finite order. For instance, for a torus $G$ is the trivial group.

- There are parallelizable flat Riemannian manifolds that are not tori. For them parallel translation of vectors around closed loops depends upon the homotopy class of the loop..

Thank you for your reply. However I am a bit confused. I attached the part of my book that I took this from (it is from Nakahara, Geometry Topology and Physics). As far as I understand, parallelizable manifold, implies zero Riemann tensor. Could you please explain this to me a bit further (i.e. what am I reading wrong)?

 

[lavinia]

I will have to think about what the book is saying.

What I said is correct. I am assuming a Levi-Civita connection and your book is not. That has to be what is going on.

 

[Orodruin]

The key passage is:

A vector $V_p \in T_pM$ is defined to be parallel to $V_q \in T_qM$ if ...

This defines the connection and therefore the curvature tensor. There is no a priori reason why this connection should be the Levi-Civita connection or even a metric compatible connection.

 

[Silviu]

The key passage is:

This defines the connection and therefore the curvature tensor. There is no a priori reason why this connection should be the Levi-Civita connection or even a metric compatible connection.

Oh, I misunderstood this. So by this they mean that by parallel transporting a vector between 2 points, you keep it having the same coordinates in the respective tangent planes. I thought that this is a general property of a parallelizable manifold (i.e. independent of connection). Thank you!