Connection coefficients

[kent davidge]

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 realize 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?

 

[Orodruin]

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 realize 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.

 

[kent davidge]

Does Relativity theory allow for manifolds with torsion?

 

[pervect]

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 realize 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 ...

 

[kent davidge]

Oh, I see
I'm actually thinking of the $\Gamma$ as $\Gamma^\kappa{}_{\mu \nu} = \partial_\mu e_\nu (x)$. Is it ok?

 

[Orodruin]

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.

 

[kent davidge]

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 $x$.

 

[Orodruin]

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.

 

[kent davidge]

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 $e$?

 

[haushofer]

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).

 

[haushofer]

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.

 

[Orodruin]

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 $X \in T_pM$.

 

[Jianbing_Shao]

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 ?

 

[Orodruin]

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.

 

[kent davidge]

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.

 

[Jianbing_Shao]

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?

 

[Orodruin]

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.

 

[Jianbing_Shao]

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,

 

[Ibix]

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.

 

[Pencilvester]

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.

 

[Orodruin]

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.

 

[haushofer]

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.

 

[stevendaryl]

''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 $B$.

 

[TeethWhitener]

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?

 

[Orodruin]

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.

 

[TeethWhitener]

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.

Yes, thank you. I realize now that it doesn’t make sense. But if you define your coordinates and define a tangent space basis at every point, does that determine a connection, since the connection tells you how the basis changes from point to point?

 

[stevendaryl]

Yes, thank you. I realize now that it doesn’t make sense. But if you define your coordinates and define a tangent space basis at every point, does that determine a connection, since the connection tells you how the basis changes from point to point?

That's not quite right. Think about the surface of the Earth. On every point except the North and South pole, you can define a basis by the following:

• Let $e_y$ be a unit vector pointing North.
• Let $e_x$ be a unit vector tangent to the surface of the Earth which is perpendicular to $e_y$ (so that $e_x$ is 90 degrees clockwise away from $e_y$, looking down on the Earth)

This gives you a basis for each point on the Earth (well, except two---nobody lives there, anyway). But it doesn't define the connection. What you need to define the connection is a way to relate the tangent spaces of nearby points. If you have two points $A$ and $B$ that are close by, you need a way to rewrite a vector $V$ defined at point $A$ to a corresponding vector $V'$ defined at point $B$.

I think your idea for defining the connection by just giving a basis at each point might have been to say: Define $V'$ at $B$ by:

• $V'^x = V^x$
• $V'^y = V^y$

That's simple enough, but it can't possibly work on a curved surface.

 

[Orodruin]

This gives you a basis for each point on the Earth (well, except two---nobody lives there, anyway). But it doesn't define the connection.

This is not entirely true. You can use these two vector fields to define a connection (in fact, it is one of my favourite examples of a metric compatible connection with non-zero torsion and Example 9.15 and some follow-up examples in my book discuss it and its geodesics). What it does not do is to let you define the Levi-Civita connection (which, apart from being metric compatible, is torsion free). If you have set of fields that are linearly independent at each point, you can define a connection by defining them to be parallel. This is the reason that you need to remove points from the sphere - it is not parallelisable.

That's simple enough, but it can't possibly work on a curved surface.

This is a bit of chicken-or-egg. It is the connection that determines whether the surface (or, more generally, manifold) is curved or not. Of course, the connection discussed above has zero curvature. A vector parallel transported around a loop always has the same length and bearing, since the connection is metric compatible. (And it is metric compatible because the basis defined is orthonormal everywhere ...)

 

[Orodruin]

But if you define your coordinates and define a tangent space basis at every point, does that determine a connection, since the connection tells you how the basis changes from point to point?

You do not need the coordinate system, but if you have a set of basis vector fields, then you can define a connection by defining how the connection acts on those fields. In particular, you could define those fields to be parallel. However, there is no guarantee that this connection will have particular properties, such as being the Levi-Civita connection.

 

[stevendaryl]

This is not entirely true. You can use these two vector fields to define a connection (in fact, it is one of my favourite examples of a metric compatible connection with non-zero torsion and Example 9.15 and some follow-up examples in my book discuss it and its geodesics).

What's the connection? Do you just assume that parallel transport of a vector $V$ leaves the components $V^x$ and $V^y$ unchanged?

 

[Orodruin]

What's the connection? Do you just assume that parallel transport of a vector $V$ leaves the components $V^x$ and $V^y$ unchanged?

You assume that your defined basis is parallel, i.e., if your basis is $\{X,Y\}$, then $\nabla_V X = \nabla_V Y = 0$ for all $V$ (which is what it means for $X$ and $Y$ to be parallel fields). In effect, if you write a vector $V = V^x X + V^y Y$, then yes, $V^x$ and $V^y$ constant along a line would lead to the vector being parallel transported along it. This completely defines a flat connection on the manifold (of course, some manifolds do not allow flat connections because they do not allow you to define vector fields that are linearly independent everywhere). It also gives you sufficient information to uniquely compute the connection coefficients. In particular, if $X = X^a \partial_a$, then
$$
0 = \nabla_a X = [(\partial_a X^b) + \Gamma_{ac}^b X^c]\partial_b,
$$
implying that $\Gamma_{ac}^b X^c = - \partial_a X^b$ with the corresponding relation for each of your basis vectors. You therefore have $N^2$ equations per basis vector (one for each combination of $a$ and $b$), leading to a total of $N^3$ relations to determine the general $N^3$ connection coefficients in whatever coordinate system you have chosen. Note that, hardly surprising, if you choose the coordinate basis of some coordinate patch, the connection coefficients all vanish as the components are constants.

I want to be very clear with the fact that this is just a connection on the manifold. In general there will be other possible connections. Some will be metric compatible, some will be torsion free, and only one will be both (the Levi-Civita connection). The connection defined by the procedure above will generally not be either metric compatible or torsion free (but it will be flat and it might be metric compatible or torsion free, depending on how you chose your fields).

 

[Orodruin]

Just to give a concrete example (other than the compass connection on a sphere, which I worked out in my book), consider the orthonormal basis
$$
X = \partial_r, \quad Y = \frac{1}{r}\partial_\phi
$$
on the plane in polar coordinates with the origin excluded. Requiring that $\nabla_a X = \nabla_a Y = 0$ leads to the relations
\begin{align*}
0 &= \nabla_a X = \nabla_a \partial_r = \Gamma_{ar}^b\partial_b, \\
\quad 0 &= \nabla_a Y = \nabla_a[(1/r)\partial_\phi] = [\partial_a (1/r)] \partial_\phi + \frac{1}{r}\Gamma_{a\phi}^b \partial_b.
\end{align*}
The first relation yields $\Gamma_{ar}^b = 0$ and the second with $a = \phi$ yields $\Gamma_{\phi\phi}^b = 0$. For $a = r$, the second relation becomes
$$
\Gamma_{r\phi}^b \partial_b = - r \frac{\partial(1/r)}{\partial r} \partial_\phi = \frac{1}{r}\partial_\phi,
$$
from which we can read off $\Gamma_{r\phi}^r = 0$ and $\Gamma_{r\phi}^\phi = 1/r$.

Edit: We can also explicitly check that this connection is indeed metric compatible. We generally have
$$
\nabla_a g_{bc} = \partial_a g_{bc} - \Gamma_{ab}^d g_{dc} - \Gamma_{ac}^d g_{bd}.
$$
Since the only non-zero connection coefficient is $\Gamma_{r\phi}^\phi$, we can write this for $a = \phi$ as
$$
\nabla_\phi g_{bc} = \partial_\phi g_{bc} = 0,
$$
since none of the metric components depend on $\phi$. For $a = r$ we instead obtain
$$
\nabla_r g_{bc} = \partial_r g_{bc} - \frac{1}{r}(\delta_b^\phi g_{\phi c} + \delta_c^\phi g_{b\phi}).
$$
For the off-diagonals, e.g., $b = r$, $c = \phi$, we obtain
$$
\nabla_r g_{r\phi} = \partial_r 0 - \frac{1}{r} (0 r^2 + 1\cdot 0) = 0.
$$
For the diagonals, we have
$$
\nabla_r g_{rr} = \partial_r 1 - \frac{1}{r}(0+0) = 0 \quad \mbox{and} \quad
\nabla_r g_{\phi\phi} = \partial_r r^2 - \frac{1}{r}(r^2 + r^2) = 2r - 2r = 0.
$$
Consequently, the connection is compatible with the standard metric in the plane.

 

[TeethWhitener]

That's not quite right. Think about the surface of the Earth. On every point except the North and South pole, you can define a basis by the following:

• Let $e_y$ be a unit vector pointing North.
• Let $e_x$ be a unit vector tangent to the surface of the Earth which is perpendicular to $e_y$ (so that $e_x$ is 90 degrees clockwise away from $e_y$, looking down on the Earth)

This gives you a basis for each point on the Earth (well, except two---nobody lives there, anyway). But it doesn't define the connection. What you need to define the connection is a way to relate the tangent spaces of nearby points. If you have two points $A$ and $B$ that are close by, you need a way to rewrite a vector $V$ defined at point $A$ to a corresponding vector $V'$ defined at point $B$.

I think your idea for defining the connection by just giving a basis at each point might have been to say: Define $V'$ at $B$ by:

• $V'^x = V^x$
• $V'^y = V^y$

That's simple enough, but it can't possibly work on a curved surface.

You do not need the coordinate system, but if you have a set of basis vector fields, then you can define a connection by defining how the connection acts on those fields. In particular, you could define those fields to be parallel. However, there is no guarantee that this connection will have particular properties, such as being the Levi-Civita connection.

Thank you both for your answers. I’m just trying to feel out an intuition for the connection, per the definition in post 6:
$$\nabla_{\mu}e_{\nu} = \Gamma^{\rho}_{\mu\nu}e_{\rho}$$
Since the connection tells you how the basis vectors change from point to point on the manifold, I was thinking that specifying a basis at every point would give the connection a fortiori. Of course it’s not necessarily the Levi-Civita connection, but as long as it meets some minimal criterion of smoothness, would it still be a connection?

 

[Orodruin]

Thank you both for your answers. I’m just trying to feel out an intuition for the connection, per the definition in post 6:
$$\nabla_{\mu}e_{\nu} = \Gamma^{\rho}_{\mu\nu}e_{\rho}$$

Note that what I wrote down there was the connection coefficients in the basis $e_\mu$ and what I computed in the previous post was the connection coefficients in the holonomic basis for the polar system.

Since the connection tells you how the basis vectors change from point to point on the manifold, I was thinking that specifying a basis at every point would give the connection a fortiori. Of course it’s not necessarily the Levi-Civita connection, but as long as it meets some minimal criterion of smoothness, would it still be a connection?

As I discussed above, it is one way of specifying the connection to define how it acts on a set of vector fields that are independent everywhere. If you know this you can write any vector field as a linear combination of those (with coefficients that are functions) and then apply the product rule. Of course, you need not specify that your basis is parallel, it is also fine to use a basis that is not parallel and specify how the connection acts on it.

 

[martinbn]

Another example:

Let $M=\mathbb R^3$ be the Euclidean space with the usual metric and usual coordinates $(x,y,z)$, for convenience number them $(x_1, x_2, x_3)$. Then consider the connection with Christoffel symbols

$\Gamma_{12}^3=\Gamma_{23}^1=\Gamma_{31}^2=1, \quad\Gamma_{21}^3=\Gamma_{32}^1=\Gamma_{13}^2=-1$

All other symbols are zero. Then it is obviously not symmetric, but it is compatible with the metric (and has minimizing geodesics). This is an exercise in Lee's book.