Definition of a symmetric connection

[Angelos K]

Hi, all,

According to my script, a connection \nabla_v is symmetric if the following holds (I assume for every pair of vectors):

\nabla_v w - \nabla_w v =[v,w]

What is the idea behind that? Why are we interested in that kind of symmetry (not for instance 0 instead of the commutator)?

Thanks for any advice!
Angelos

 

[Ben Niehoff]

First of all, you can't set the RHS to zero, because it is impossible for that to hold for all vector fields v, w.

This equation simply states that the torsion vanishes.

If we travel around a small parallelogram whose sides are geodesics, our intuition says that we ought to end up where we started. If a parallelogram does fail to close, we attribute that to the failure of the vector fields defining its sides to commute; and the remaining gap is given by the commutator.

In the presence of torsion, this does not hold. Small parallelograms may fail to close even if the vector fields commute! The torsion gives this additional gap.

Your equation expresses the idea that the failure of any parallelograms to close is due precisely to the failure of the vector fields to commute, and that there is no additional gap we need to account for.

 

[Altabeh]

Hi, all,

According to my script, a connection \nabla_v is symmetric if the following holds (I assume for every pair of vectors):

\nabla_v w - \nabla_w v =[v,w]

What is the idea behind that? Why are we interested in that kind of symmetry (not for instance 0 instead of the commutator)?

Thanks for any advice!
Angelos

Suppose that L is a smooth scalar field then from basic calculus you remember that clearly \partial_a\partial_b L=\partial_b\partial_a L. But it is necessary to note that this doesn't follow when the ordinary derivatives are replaced by the covariant derivatives. To wit, \nabla_a\nabla_b L and \nabla_b\nabla_a L are not equivalent generally. The reason is that you can simply show there is a tensor T_{ab}^c known as the torsion tensor such that for any scalar field of class C^\infty we have

(\nabla_a\nabla_b -\nabla_b\nabla_a) L=T^c_{ab}\nabla_c L.

If T^c_{ab}=0, then the connection is said to be torsion-free (torsionless) and obviously it follows that the connection is symmetric because

\nabla_a\nabla_b L=\nabla_b\nabla_a L.

But how does this imply a symmetry of connection in two lower indices? Let us calculate the torsion T^c_{ab} in terms of the connection \Gamma^c_{ab}. Recalling that \nabla_a U_b=\partial_a U_b -\Gamma^c_{ab} U_c for any covariant vector field U_b. Hence if one sets U_b=\nabla_b L=\partial_b L, we get

\nabla_a\nabla_b L=\partial_a \partial_b L -\Gamma^c_{ba} \partial_c L ,

and

\nabla_b\nabla_a L=\partial_b \partial_a L -\Gamma^c_{ab} \partial_c L .

By subtracting the first from the second we obtain

(\nabla_b\nabla_a -\nabla_b \nabla_a )L =T^c_{ab}\nabla_c L ,

where

T^c_{ab}=-2\Gamma^c_{[ab]} .

You must know that the difference of two connections is always a tensor, so is the torsion. Therefore a torsion-free spacetime has this property that its connection is symmetric.

AB

 

[Angelos K]

thanks!

Thank you so much both, that really helped, I'm starting to get the idea.

I'll sit down right now and calculate a couple of things about it.

@AB Thanks for mentioning the Christoffel-property. It seems that they use it a lot!