Connection coefficients

  • #1
931
56
I was thinking about connection coefficients. Can we choose a coordinate system so that we have a Levi-Civita connection? Or is it a thing imposed on us by the manifold?

EDIT: I realise that, from a geometric perspective, since we'd be dealing with a manifold we would have no torsion around any point as long as the region is small. So my question is more like can we choose a coordinate system so that it vanishes everywhere?
 

Answers and Replies

  • #2
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,829
6,652
The connection coefficients belong to the connection you impose on your manifold and the Levi-Civita is a particular connection that can be applied (i.e., the unique metric compatible and torsion free connection). If your connection is not the Levi-Civita connection, you will not have the Levi-Civita connection's connection coefficients.

EDIT: I realise that, from a geometric perspective, since we'd be dealing with a manifold we would have no torsion around any point as long as the region is small. So my question is more like can we choose a coordinate system so that it vanishes everywhere?
If you do not have a torsion free connection you cannot make torsion vanish. Torsion is a type (1,2) tensor and its vanishing or not does not depend on the coordinate system, only on the imposed connection. "Vanishing locally" is not a thing that non-zero torsion would do. It is a tensor and only depends on the point of interest, not on a region around a point.
 
  • #3
931
56
Does Relativity theory allow for manifolds with torsion?
 
  • #4
pervect
Staff Emeritus
Science Advisor
Insights Author
9,953
1,134
I was thinking about connection coefficients. Can we choose a coordinate system so that we have a Levi-Civita connection? Or is it a thing imposed on us by the manifold?

EDIT: I realise that, from a geometric perspective, since we'd be dealing with a manifold we would have no torsion around any point as long as the region is small. So my question is more like can we choose a coordinate system so that it vanishes everywhere?
I'm not sure I understand the question. Suppose we have a flat Euclidean plane. One can choose any of an infinite number of coordinate systems (for instance Cartesian and polar, to name two of the infinite set) with the Levi-Civita connection.

The way your question is worded, it sounds like you are regarding the connection as a consequence of the coordinate system, whereas I think of the connection as a tool that allows one to define a derivative operator in a manner that's independent of one's particular choice of coordinates, which are therefore free to be chosen at one's convenience.

See for instance wiki:

In the branch of mathematics called differential geometry, an affine connection is a geometric object on a smooth manifold which connects nearby tangent spaces, and so permits tangent vector fields to be differentiated ....
 
  • Like
Likes kent davidge
  • #5
931
56
Oh, I see
I'm actually thinking of the ##\Gamma## as ##\Gamma^\kappa{}_{\mu \nu} = \partial_\mu e_\nu (x)##. Is it ok?
 
  • #6
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,829
6,652
Oh, I see
I'm actually thinking of the ##\Gamma## as ##\Gamma^\kappa{}_{\mu \nu} = \partial_\mu e_\nu (x)##. Is it ok?
No. First of all, the partial derivative of a vector is meaningless (unless you have a unique way of relating vectors in different tangent spaces, such as in Euclidean space, but in a general manifold you don’t). Resolving this is the entire point of introducing a connection.

Second, your indices don’t match, you are missing a basis vector. By definition of the connection coefficients are
$$
\nabla_\mu e_\nu = \Gamma_{\mu\nu}^\rho e_\rho.
$$
Note that what appears here is the connection, not the partial derivative.
 
  • #7
931
56
partial derivative of a vector is meaningless
They are functions of ##x##. So I thought that's fine. I would like to know why we can't study their variation with ##x##, i.e. their derivatives wrt ##x##.
 
  • #8
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,829
6,652
They are functions of ##x##. So I thought that's fine. I would like to know why we can't study their variation with ##x##, i.e. their derivatives wrt ##x##.
Because in order to create the difference you should take the limit of you would need to take the difference of vectors in different tangent spaces. As you were told in an earlier thread the tangent spaces of different points are different vector spaces. This is what the connection is for. It essentially tells you how nearby tangent spaces fit together.
 
  • #9
931
56
Ok I understand that. What I think is confusing me is that the basis vectors are typically written like ##e(x)##, that seems to imply they are functions of ##x##. So is it more correct to write just ##e##?
 
  • #10
haushofer
Science Advisor
Insights Author
2,472
860
Does Relativity theory allow for manifolds with torsion?
Yes. This happes for e.g. fermionic fields. In supergravity one usually has torsion due to the gravitino field(s).
 
  • Like
Likes kent davidge
  • #11
haushofer
Science Advisor
Insights Author
2,472
860
Ok I understand that. What I think is confusing me is that the basis vectors are typically written like ##e(x)##, that seems to imply they are functions of ##x##. So is it more correct to write just ##e##?
Why would that be more correct? Basis vectors can, as you say, be functions of the coordinates you use to cover your manifold.
 
  • Like
Likes kent davidge
  • #12
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,829
6,652
The basis is a function of the coordinates (or more correctly, of the point in the manifold), but a vector field (or in more mathematical language, a section of the tangent bundle) assigns to each point ##p## in the manifold an element of ##T_p M##, i.e., a vector in the tangent space of ##p## itself.

The more formal way of writing this would be that if you have a manifold ##M##, then a vector field is a map ##\sigma: M \to TM##, where ##TM## is the tangent bundle of ##M## (i.e., the union of all tangent spaces), such that ##\pi(\sigma(p)) = p## if ##\pi## is the projection function ##\pi: TM \to M## with the property that ##\pi(X) = p## if ##X \in T_pM##.
 
  • Like
Likes vanhees71 and kent davidge
  • #13
I was thinking about connection coefficients. Can we choose a coordinate system so that we have a Levi-Civita connection?
It is an interesting question, I think the answer to this question is 'yes', the reason is: if we start from a flat metric ##\eta_{ij}=I##, and the corresponding Levi-Civita connection ##\Gamma=0##. Then under a coordinate transformation ##x\rightarrow x'##, the flat metric ##\eta_{ij}## is a tensor field, so it changes to a new metric ##g_{ij}(x)##, and the connection ##\Gamma## corresponding to ##\eta_{ij}## also changes to a new connection ##\Gamma'##, and we can get:
$$\partial e'(x)=\Gamma'(x) e'(x)$$
The key problem of this question is that what kind of ##\Gamma'## can we get through coordinate transformation, can we get
an arbitrary connection field ?
 
  • Like
Likes kent davidge
  • #14
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,829
6,652
It is an interesting question, I think the answer to this question is 'yes', the reason is: if we start from a flat metric ##\eta_{ij}=I##, and the corresponding Levi-Civita connection ##\Gamma=0##. Then under a coordinate transformation ##x\rightarrow x'##, the flat metric ##\eta_{ij}## is a tensor field, so it changes to a new metric ##g_{ij}(x)##, and the connection ##\Gamma## corresponding to ##\eta_{ij}## also changes to a new connection ##\Gamma'##, and we can get:
$$\partial e'(x)=\Gamma'(x) e'(x)$$
The key problem of this question is that what kind of ##\Gamma'## can we get through coordinate transformation, can we get
an arbitrary connection field ?
No. The correct answer is that the connection has nothing to do with the coordinate system. If your connection is the Levi-Civita connection it will be the Levi-Civita connection in all coordinates. If it is not it will not be the Levi-Civita connection in any coordinates.
 
  • #15
931
56
The correct answer is that the connection has nothing to do with the coordinate system
After learning about the concept of bundle, it became more clear to me that coordinates don't play a role in this case.
The key problem of this question is that what kind of ##\Gamma'## can we get through coordinate transformation, can we get an arbitrary connection field ?
As of my current understanding, locally the manifold is to be thought of as a product space, one is called the base space and the other is the fiber.

The connection appears as an element of the structure group, which maps basis vectors into basis vectors. AFAIK they are elements of ##GL_n(\mathbb R)##. Now the coordinates have to do with the parametrization one implements on the manifold. It gives a way for one to know how two connections when expressed in coordinates, are related. I would like someone to correct me if I'm wrong.
 
  • #16
Yes. This happes for e.g. fermionic fields. In supergravity one usually has torsion due to the gravitino field(s).
No. The correct answer is that the connection has nothing to do with the coordinate system. If your connection is the Levi-Civita connection it will be the Levi-Civita connection in all coordinates. If it is not it will not be the Levi-Civita connection in any coordinates.
Ok, if you think ''the connection has nothing to do with the coordinate system.'' then how can you explain the change rule of the connection field under the coordinate transformation?
For example if we can change a connection field to another arbitrary connection field through coordinate transformation, then what relationship between the connection and the coordinate transformation? Do you still think ''connection has nothing to do with the coordinate system.''?

Another question, what is the relation between Levi-Civita connection and connection? how can you find out if a connection is a Levi-Civita connection or a non-Levi-Civita connection?
 
  • #17
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,829
6,652
Ok, if you think ''the connection has nothing to do with the coordinate system.'' then how can you explain the change rule of the connection field under the coordinate transformation?
I said the connection has nothing to do with the coordinates (and it doesn’t), not that the transformation of the connection coefficients do not follow the appropriate transformation rules. Since connection coefficients depend on the coordinates, obviously the transformation rules do. However this tells you nothing about the connection as such. Do not confuse the connection with the connection coefficient transformations.

You can have different connections using the same coordinates and you can have the same connection expressed in different coordinates. The transformation rules describe how the latter works but has nothing to do with the former.

For example if we can change a connection field to another arbitrary connection field through coordinate transformation, then what relationship between the connection and the coordinate transformation? Do you still think ''connection has nothing to do with the coordinate system.''?
Again, you do not change the connection by changing coordinates. You are just changing the coordinate system it is expressed in.

It is a fact that the connection is independent of the coordinates, it is not a matter of thinking, it is a property of the connection on the fibre bundle in question.


Another question, what is the relation between Levi-Civita connection and connection? how can you find out if a connection is a Levi-Civita connection or a non-Levi-Civita connection?
The Levi-Civita connection is the unique torsion-free and metric compatible connection, ie, no other connection satisfies these requirements.
 
  • Like
Likes Pencilvester
  • #18
I said the connection has nothing to do with the coordinates (and it doesn’t), not that the transformation of the connection coefficients do not follow the appropriate transformation rules. Since connection coefficients depend on the coordinates, obviously the transformation rules do. However this tells you nothing about the connection as such. Do not confuse the connection with the connection coefficient transformations.
''connection has nothing to do with the coordinates'' but the connection coefficients do, connection coefficients changes under the coordinate transformation and connection do not,
So I can draw a conclusion if connection has nothing to do with the coordinates ,obviously it also has nothing to do with the connection coefficients,
 
  • #19
Ibix
Science Advisor
Insights Author
2020 Award
7,393
6,475
If you have a point with coordinates ##x^{\mu}## and another with coordinates ##x^{\mu}+\delta x^\mu##, then the connection tells you which vector in the tangent space of ##x^\mu+\delta x^\mu## is defined to point in the same direction as a vector in the tangent space of ##x^\mu##. Of course, if we change coordinates then the numbers we use identify the two points change, from ##x^\mu## to ##x^{\mu'}##. But they're still the same points. The numbers we use to describe the vectors change too. But the vectors are still the same. So the numbers we use to describe the connection must change. But the connection is still the same. How could it be otherwise? The concept of "pointing in the same direction" is the same whatever coordinates we use - at least in physics where we take out cues from how stuff works in the real world.

At least, that's my understanding.
 
Last edited:
  • #20
184
42
So I can draw a conclusion if connection has nothing to do with the coordinates ,obviously it also has nothing to do with the connection coefficients,
Your choice of connection does not depend on coordinates. You choose a connection and a set of coordinates independently, and your connection coefficients are determined by those two choices.
 
  • #21
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,829
6,652
So I can draw a conclusion if connection has nothing to do with the coordinates ,obviously it also has nothing to do with the connection coefficients,
No. I don’t know where you got that idea from as it does not logically follow. That is like saying that the components of a vector have nothing to do with the vector itself. The vector itself is independent of whatever basis you choose, but the vector components depend both on the vector and on the basis. Same thing for the connection coefficients, they depend on the connection and on your choice of coordinates.
 
  • #22
haushofer
Science Advisor
Insights Author
2,472
860
Ok, if you think ''the connection has nothing to do with the coordinate system.'' then how can you explain the change rule of the connection field under the coordinate transformation?
Maybe too much, but another 2 cents: look at post #6, where the connection coefficients are defined. This definition is coordinate independent. That already should answer your question. Of course, after that you can derive the transformation law under a coordinate transformation when you change coordinate charts.

Look at it this way: a car is defined by certain properties. Let's take these properties to be the number of wheels (4), engine power, and weight. After that you can describe such a car in different colour patterns, like black, blue or red. But the colour just changes the description of a specific type of car. The colour cannot change this type. Here, the analogy is car <--> connection and colour <--> coordinates.
 
  • #23
stevendaryl
Staff Emeritus
Science Advisor
Insights Author
8,460
2,625
''connection has nothing to do with the coordinates'' but the connection coefficients do, connection coefficients changes under the coordinate transformation and connection do not,
So I can draw a conclusion if connection has nothing to do with the coordinates ,obviously it also has nothing to do with the connection coefficients,
I think that maybe it's clearer that the connection is coordinate-independent if you think in terms of parallel transport. You have a starting point, ##A##. You have a starting vector, ##V_A##. You take the vector on a journey along a path ##\mathcal{P}(s)## to get to a point ##B##. All along the way, you try to keep the vector pointing in the "same direction" and having the "same magnitude". You end up with a vector at point ##B##: ##V_B##. The connection is what tells you that ##V_A## and ##V_B## are the "same" vector (or as close to being the same as you can get in following the path ##\mathcal{P}(s)##.

There is nothing in this description that in any way depends on coordinates. You have defined a vector at ##B## as a function of a vector at ##A## and the path to get from ##A## to ##B##.
 
  • #24
TeethWhitener
Science Advisor
Gold Member
1,932
1,360
Following this discussion with some interest and I wanted to make sure I was thinking about this correctly. If you choose a set of coordinates and a connection, then choosing a tangent space basis at one point on the manifold determines it everywhere else, right?

Follow up: if the above is true, does choosing a set of coordinates along with a basis at every point on the manifold determine the connection?
 
  • #25
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,829
6,652
then choosing a tangent space basis at one point on the manifold determines it everywhere else, right?
No, this is not correct. The parallel transport of the basis from the point you chose it at to another point will generally depend on the path unless your manifold is flat and simply connected.
 
  • Like
Likes TeethWhitener

Related Threads on Connection coefficients

  • Last Post
Replies
6
Views
2K
Replies
13
Views
1K
Replies
14
Views
3K
  • Last Post
Replies
2
Views
827
  • Last Post
Replies
3
Views
746
  • Last Post
Replies
3
Views
2K
Replies
14
Views
2K
  • Last Post
Replies
3
Views
4K
  • Last Post
Replies
3
Views
2K
Top