A Curvature with different connections on the two-sphere

[pervect]

Suppose we have a 2-sphere, and an associated metric. For specificity
$$d\theta^2 + \sin^2 \theta d\phi^2$$

On this 2-sphere, lets consider two different connections. The Levi-Civita connection, where geodesics are great circles, and a connection with torsion where the curves of constant $\theta$ are geodesics. I don't have a name for this, - for now I'll call it the compass connection.

Let's consider a simple example of a geodesic triangle for each of the two connections. In our example, two of the sides of the triangle will be lines of constant 'longitude' (i.e. curves of constant $\phi$). The third side will be in the case of the Levi-Civita connection a great circle, and in the compass connection a curve of constant latitude (constant $\theta$). Essentially, we are comparing triangles that share the exact same three vertices, but using different connections to draw the sides of the triangle.

The sums of the internal angles on these two different geodesic triangles with different connections on the same geometry will be different. Because geodesics parallel transport themselves, the sum of the internal angles relates to the Riemann tensor, so the Riemann tensor must be different for the two different connections.

The first question is if this argument is correct (I don't see how it can be wrong, but I'm not as good at spotting errors in my thinking nowadays as I once was). Related questions arise as how to go about thinking of this. Does the concept of a surface of uniform curvature require a metric to define? What about the idea of "similar triangles", does that also require a metric? When we talk about "geometry", does changing the connection within the same metric change the geometry? That's a bit of a semantic question, but I'm unclear about standard usage.

 

[Orodruin]

Ah, my favourite example of a metric compatible connection with torsion. (It is discussed in my book, I believe there is even a figure with the geodesics of the connection and at some point I believe the connection coefficients are made explicit.

First of all, it should be noted that what you have is not a connection on the sphere..it is a connection on the sphere minus the poles. Unlike the Levi-Civita connection it cannot be extended to the entire sphere.

Yes, the curvature tensor depends on the connection. For the Leve-Civita connection it is non-zero, but the compass connection is flat and has curvature zero. (The compass returns to the same state if you return to the same point.)

 

[cianfa72]

Let's consider a simple example of a geodesic triangle for each of the two connections. In our example, two of the sides of the triangle will be lines of constant 'longitude' (i.e. curves of constant $\phi$). The third side will be in the case of the Levi-Civita connection a great circle, and in the compass connection a curve of constant latitude (constant $\theta$). Essentially, we are comparing triangles that share the exact same three vertices, but using different connections to draw the sides of the triangle.

Ah ok, so for instance both triangles share the "South-pole" since two sides of them are lines of constant "longitude" $\phi$ (they are geodesics in both Levi-Civita and "compass" connections).

 

[cianfa72]

First of all, it should be noted that what you have is not a connection on the sphere..it is a connection on the sphere minus the poles. Unlike the Levi-Civita connection it cannot be extended to the entire sphere

Why one can't extend the "compass" connection on all sphere?

 

[Ibix]

Why one can't extend the "compass" connection on all sphere?

Which way is east from the north pole?

 

[Orodruin]

Which way is east from the north pole?

Or, put another way, south is in all directions.

 

[cianfa72]

Ah ok so, at all points on the sphere but the poles, fixed the direction parallel to the equator, "compass" connection defines the line of constant "latitude" through it as the line having tangent vectors auto-parallel transported along it. Such a connection basically shouts you "move on in a given direction so that the compass you carry along keeps pointing North".

 

[Orodruin]

Ah ok so, at all points on the sphere but the poles, fixed the direction parallel to the equator, "compass" connection defines the line of constant "latitude" through it as the line having tangent vectors auto-parallel transported along it. Such a connection basically shouts you "move on in a given direction so that the compass you carry along keeps pointing North".

The geodesics of the compass connection are paths with constant compass direction. If you keep a compass and ensure you always walk on the direction SSW, you follow one of those geodesic. Apart from the pure north/south geodesics, these geodesics will not agree with the usual ones.

 

[cianfa72]

The geodesics of the compass connection are paths with constant compass direction. If you keep a compass and ensure you always walk on the direction SSW, you follow one of those geodesic.

Ok, by definition a compass's needle always keeps pointing north. So in my #7 example, starting from a point/location other than the poles, if you always keep walking on either the E or O direction (w.r.t. the compass you carry along with you), you are actually following the geodesic line of constant latitude $\theta$.

 

[Orodruin]

E or O direction

E or W if we speak English.

(w.r.t. the compass you carry along with you), you are actually following the geodesic line of constant latitude.

… but yes

 

[cianfa72]

Using spherical coordinates on a 2-sphere (i.e. latitude $\theta$ and longitude $\phi$ having fixed the radius $r$), which are the coefficients $\Gamma^k_{ij}$ of the compass connection in that chart ?

Ps. in order to be a valid chart map (onto, one-to-one), spherical coordinates maps the sphere minus an entire "meridian" line (including north and south poles) onto the $\mathbb R^2$ open rectangle $-\pi/2 \lt \theta \lt \pi/2, -\pi \lt \phi \lt \pi$.

 

[Orodruin]

Using spherical coordinates on a 2-sphere (i.e. latitude $\theta$ and longitude $\phi$ having fixed the radius $r$), which are the coefficients $\Gamma^k_{ij}$ of the compass connection in that chart ?

$\Gamma_{\theta\varphi}^\varphi = \cot(\theta)$. The rest are zero (including $\Gamma_{\varphi\theta}^\varphi$). This is shown in Example 9.14 of my book.

Ps. in order to be a valid chart map (onto, one-to-one), spherical coordinates maps the sphere minus an entire "meridian" line (including north and south poles) onto the $\mathbb R^2$ open rectangle $-\pi/2 \lt \theta \lt \pi/2, -\pi \lt \phi \lt \pi$.

This does not preclude defining the connection on the entire sphere minus the poles though. Having a chart that covers the sphere is not the issue. The connection can be smoothly extended to the meridian, but not to the poles.

 

[cianfa72]

This does not preclude defining the connection on the entire sphere minus the poles though. Having a chart that covers the sphere is not the issue. The connection can be smoothly extended to the meridian, but not to the poles.

Ah ok, you mean that the $\Gamma_{\theta\varphi}^\varphi = \cot(\theta)$ function is well-defined everywhere in the chart map's open image $-\pi/2 \lt \theta \lt \pi/2, -\pi \lt \phi \lt \pi$. Technically $\Gamma_{\theta\varphi}^\varphi$ in this chart isn't defined on the meridian left out in the mapping (simply because the chart does not include it), however it can be smootly extended to it taking, for instance, another spherical coordiantes chart map that excludes this time a different meridian in the mapping.

 

[Ibix]

$\Gamma_{\theta\varphi}^\varphi = \cot(\theta)$. The rest are zero (including $\Gamma_{\varphi\theta}^\varphi$). This is shown in Example 9.14 of my book.

At risk of asking you to reproduce the book, I think there are eight $\Gamma^i_{jk}$ here, and the usual symmetry in the lower indices is absent because that's a consequence of torsion-freeness. I think we insist that $\nabla_ig_{jk}=0$, which gives us six constraints ($g_{10}=g_{01}$, so we don't get eight). Then we insist that $U^i\nabla_iU^j=0$ when the $U^j$ are parallel to the coordinate basis vectors, thereby specifying what we mean by geodesics. That gives us two more constraints. That makes eight, unless there's some degeneracy I've missed, and a little algebra gives us your result. Right?

 

[Orodruin]

At risk of asking you to reproduce the book, I think there are eight $\Gamma^i_{jk}$ here, and the usual symmetry in the lower indices is absent because that's a consequence of torsion-freeness.

That’s just the thing, the connection is indeed not torsion free so you a priori have 8 connection coefficients to be determined.

I think we insist that $\nabla_ig_{jk}=0$, which gives us six constraints ($g_{10}=g_{01}$, so we don't get eight).

You do not need to assume that the connection is metric compatible, but if you have a metric it is reasonable to do so.

In this case, the connection is indeed metric compatible (parallel transported vectors preserve their inner product as they remain at the same angle to north and maintain their length.

If you require both metric compatibility and that the connection is torsion free, then you have the Levi-Civita connection as the unique connection satisfying both.

Then we insist that $U^i\nabla_iU^j=0$ when the $U^j$ are parallel to the coordinate basis vectors, thereby specifying what we mean by geodesics. That gives us two more constraints.

You need to require that $\nabla_{X_i} X_j = 0$ where the $X_i$ are normalized fields proportional to the usual coordinate directions at every point. That the $X_i$ are orthonormal everywhere ensures metric compatibility. Each of these equations have two components - so in total 8 equations (2 components x 2 choices each for i and j) to fix all of the connection coefficients.

 

[Orodruin]

To describe things briefly but somewhat more hands on:

The orthonormal fields are $X_1 = \partial_\theta$ and $X_2 = \tfrac{1}{\sin\theta}\partial_\varphi$. This gives, for example,$$\nabla_{X_2}X_1 = \tfrac{1}{\sin\theta}\nabla_\varphi \partial_\theta
= \tfrac{1}{\sin\theta}\Gamma^a_{\varphi\theta}\partial_a = 0$$Identification immediately leads to $\Gamma^a_{\varphi\theta}=0$. Similarly for the other four.