Arbitrariness of connection and arrow on sphere

[strauser]

I'm trying to understand connections and their arbitrariness.

Many diff. geom. books or webpages appear to be contradictory. A chapter or page on connections may start off stressing that a connection is an arbitrary method of mapping between tangent spaces, then shortly after, show that nice picture of the "parallel transport" of an arrow on the surface of the sphere.

Now, this confuses me a little. If a connection is genuinely arbitrary, shouldn't we see various pictures of "parallel transport", where the arrow appears to rotate or stretch (smoothly, I guess) as it moves from tangent space to tangent space? Yet I've never seen this - I've only seen the somewhat cliched image where the arrow is transported such that it is parallel in the eyes of an observer sitting outside the manifold (from the pole to the equator, along the equator, and back to the pole)

Am I right in assuming that this picture in fact is showing not some arbitrary connection but specifically a Levi-Civita connection? i.e. that it tacitly assumes some metric on the sphere from which the connection coefficients have been calculated? If so, doesn't this image give a very confusing and limited intuition for what a connection is allowed to do? Or am I myself confused?

 

[Orodruin]

No, you are correct. The typical image shown of parallel transport on a sphere assumes the Levi-Civita connection of the metric induced by the typical embedding into $\mathbb R^3$. You can in general have many other connections on the sphere. My favourite example is the connection on the sphere (with the poles removed) that preserve compass directions. This is a flat (but not torsion free) connection that is also metric compatible. Its geodesics are the curves with constant bearing.

 

[fresh_42]

No, you are correct. The typical image shown of parallel transport on a sphere assumes the Levi-Civita connection of the metric induced by the typical embedding into $\mathbb R^3$. You can in general have many other connections on the sphere. My favourite example is the connection on the sphere (with the poles removed) that preserve compass directions. This is a flat (but not torsion free) connection that is also metric compatible. Its geodesics are the curves with constant bearing.

Not really important, but do you have the defining formula at hand?

 

[Orodruin]

Not really important, but do you have the defining formula at hand?

Just define the two unit vector fields pointing "South" and "East" to be parallel (this is why you need to exclude the poles, these fields are singular at the poles). You can easily compute the connection coefficients from there (only one is non-zero) as it gives you eight equations for the eight coefficients. I don't want to say more since I use it as an example in my book and I am not sure the publisher would appreciate it.

Edit: To clarify that. It might be fine to post it, but I don't want to risk it without clarifying it with the publisher first.

 

[strangerep]

I use it as an example in my book [...]

Is your book available yet?

 

[Orodruin]

Is your book available yet?

Not yet. Depending on the production editing process, it should be out around x-mas.

 

[strauser]

You can in general have many other connections on the sphere. My favourite example is the connection on the sphere (with the poles removed) that preserve compass directions. This is a flat (but not torsion free) connection that is also metric compatible. Its geodesics are the curves with constant bearing.

Thanks. That clarifies things a bit.

Do you have a picture for what this "compass direction preserving" connection does to vectors during parallel transport? I can't visualise it.

I'm also not clear what is meant by "flat but not torsion free". I guess that the flatness means that a vector taken infinitesimally around any closed curve finally maps back to itself (Riemann tensor is identically 0?) but what does the torsion or lack of it do to the vectors? I'm only familiar with it via the formula and have no intuition.

[edit: I forgot to ask this] So what restrictions are there on an arbitrary connection? I guess that it has to be at least continuous and $C^\infty$ (or $C^\omega$?) in the vector components i.e. an observer in $\mathbb{R}^3$ won't see abrupt jumps or changes of direction in a vector being parallel transported in $S^2$?

 

[strauser]

No, you are correct. The typical image shown of parallel transport on a sphere assumes the Levi-Civita connection of the metric induced by the typical embedding into $\mathbb R^3$.

Is this also what is called the pullback metric from $\mathbb R^3$ to $\mathbb S^2$?

 

[Orodruin]

Do you have a picture for what this "compass direction preserving" connection does to vectors during parallel transport? I can't visualise it.

Imagine how a compass needle behaves when you move around on the sphere (assuming, of course, that the magnetic field is an idealised dipole field such that the needle always points north).

I'm also not clear what is meant by "flat but not torsion free". I guess that the flatness means that a vector taken infinitesimally around any closed curve finally maps back to itself (Riemann tensor is identically 0?)

This is local flatness. This connection is globally flat (i.e., the Riemann tensor is zero everywhere) meaning that any parallel transport around any finite loop homotopic to a point returns the same vector. (In this case, actually all parallel transports around any loop do that.)

but what does the torsion or lack of it do to the vectors? I'm only familiar with it via the formula and have no intuition.

Torsion can be taken as a measure of the failure of consecutive infinitesimal geodesics along two vectors to commute. In other words, given two vectors $X$ and $Y$ following the geodesic in the $X$ direction and then following the geodesic in the (parallel transported) $Y$ direction will not take you to the same point as first following the geodesic in the $Y$ direction and then following the geodesic in the (parallel transported) $X$ direction. For this connection, imagine that you start at the equator. Going a distance $\ell_1$ north and then going a distance $\ell_2$ east does not get you to the same point as going a distance $\ell_2$ to the east and then a distance $\ell_1$ to the north. The torsion $T(X,Y)$ tells you the direction of the mismatch when $\ell_1$ and $\ell_2$ are infinitesimal.

So what restrictions are there on an arbitrary connection? I guess that it has to be at least continuous and C∞C∞C^\infty (or CωCωC^\omega?) in the vector components i.e. an observer in R3R3\mathbb{R}^3 won't see abrupt jumps or changes of direction in a vector being parallel transported in S2S2S^2?

Essentially you need to require that it behaves as you would expect a directional derivative to behave. See https://en.wikipedia.org/wiki/Affine_connection#Formal_definition_as_a_differential_operator for a short introduction.

Is this also what is called the pullback metric from $\mathbb R^3$ to $\mathbb S^2$?

Yes, it is the pullback of the Euclidean metric in $\mathbb R^3$ onto the sphere (given the natural embedding).

 

[strauser]

Imagine how a compass needle behaves when you move around on the sphere (assuming, of course, that the magnetic field is an idealised dipole field such that the needle always points north).

Right. I shall have to draw some pictures to see what that looks like. It's a nice example of a more interesting connection though.

This is local flatness. This connection is globally flat (i.e., the Riemann tensor is zero everywhere) meaning that any parallel transport around any finite loop homotopic to a point returns the same vector. (In this case, actually all parallel transports around any loop do that.)

Yes, I think that I worded my response poorly. I was in fact thinking about transport around a finite loop, but I guess the word "infinitesimal" made my meaning unclear. I see how your connection is globally flat, however.

Torsion can be taken as a measure of the failure of consecutive infinitesimal geodesics along two vectors to commute ... The torsion $T(X,Y)$ tells you the direction of the mismatch when $\ell_1$ and $\ell_2$ are infinitesimal.

I've seen this explanation, though I don't fully understand it yet. However, I was really trying to see if the torsion can be understood by its effects when "integrated up" over a finite curve, rather than by its infinitesimal action. For example, if I transport vectors over a finite curve using a metric compatible connection, then lengths and angles are left unchanged. Is there any comparable effect that, say, an observer in $\mathbb{R}^3$ would see if I were to transport a vector (or a pair of vectors) using a torsion-free or non-torsion-free connection in $S^2$?

 

[Orodruin]

Right. I shall have to draw some pictures to see what that looks like. It's a nice example of a more interesting connection though.

Yes, I think that I worded my response poorly. I was in fact thinking about transport around a finite loop, but I guess the word "infinitesimal" made my meaning unclear. I see how your connection is globally flat, however.

I've seen this explanation, though I don't fully understand it yet. However, I was really trying to see if the torsion can be understood by its effects when "integrated up" over a finite curve, rather than by its infinitesimal action. For example, if I transport vectors over a finite curve using a metric compatible connection, then lengths and angles are left unchanged. Is there any comparable effect that, say, an observer in $\mathbb{R}^3$ would see if I were to transport a vector (or a pair of vectors) using a torsion-free or non-torsion-free connection in $S^2$?

The entire point of manifolds is that you do not have to reference any embedding space. Thus, any argument you make should be based on properties within the manifold only.

The torsion concept may be a bit difficult to visualise in a 2D manifold apart from in the sense I mentioned. Note that it typically does extend to finite displacements, as in the case with the connection on the sphere that we discussed.

 

[strauser]

The entire point of manifolds is that you do not have to reference any embedding space. Thus, any argument you make should be based on properties within the manifold only.

From the POV of making a rigorous argument, yes. From the POV of developing intuition, I can't agree. After all, this thread is based on the existence of a diagram representing parallel transport in a 2D manifold embedded in a 3D manifold. And it's a very common diagram, of course, presumably because it gives enough intuition to be useful for developing a rigorous understanding of the topic.

The torsion concept may be a bit difficult to visualise in a 2D manifold apart from in the sense I mentioned. Note that it typically does extend to finite displacements, as in the case with the connection on the sphere that we discussed.

I've since found some useful threads on math.stackexchange and mathoverflow which I need to study properly. However, it seems that torsion in a 3D manifold can be related to the rotation of a basis of vectors as they move along a geodesic (or something along those lines - I don't fully understand it). I have no idea what torsion means in 2D though, when applied over a finite displacement.

 

[lavinia]

A nice connection on the sphere minus its north and south pole is the connection induced by the cylindrical Mercator projection. The geodesics are also the Rhumb lines and the connection is also flat. Another flat connection is induced from stereographic projection. Here you only need to remove one pole. This connection is also flat.

- A connection on the entire sphere (put the poles back) cannot be globally flat. So a connection on the sphere can not be arbitrary.

If one deforms the sphere, one gets another Levi-Civita connection on the underlying topological sphere. Parallel translation will look different on a deformed sphere than on the standard sphere. Try drawing parallel translation for an ellipsoid.

The number of possible Levi-Civita connections on a topological sphere is uncountable since there are uncountably many deformations of the standard sphere. Each has a different parallel translation. I wonder if there are metrics that are so wild that no region - no matter how small - can be visualized in 3 dimensional space. Four dimensions would be required for every tiny neighborhood. I wonder what parallel translation would loo like for such a metric.

 

[lavinia]

Here is an example of a connection on the plane that may help demonstrate the effect of torsion on parallel translation.

This connection will be compatible with the standard Euclidean metric $<∂x,∂x>=<∂y,∂y> = 1$, $<∂x,∂y> = 0$.
Metric compatibility constrains covariant differentiations to satisfy:

$∇_{∂x}∂x = h∂y$ for some function $h$
$∇_{∂y}∂y = g∂y$ for some other function $g$
$∇_{∂x}∂y = -h∂x$
and $∇_{∂y}∂x = -g∂x$

(For instance, $0 =∂x<∂x,∂x> = 2<∇_{∂x}∂x,∂x>$ so $∇_{∂x}∂x$ is perpendicular to $∂x$.)

The torsion term $∇_{∂x}∂y-∇_{∂y}∂x -[∂x,∂y]$ equals $g∂x-h∂y$ because the bracket $[∂x,∂y] = 0$.

The only way that this torsion term can be zero (to make the connection torsion free) is if $h$ and $g$ are both zero. In that case, one gets the standard connection on the Euclidean plane.

Choose for example $h=-1$ and $g=0$. The covariant derivatives are

$∇_{∂x}∂x = -∂y$
$∇_{∂y}∂y = 0$
$∇_{∂x}∂y = ∂x$
and $∇_{∂y}∂x = 0$

Here are a couple of examples of parallel translation for this connection.

Parallel translation along a horizontal straight line:

Let $V$ be a unit length vector field along a horizontal straight line. $V$ can be written as $V(x) = cos(f(x))∂x + sin((f(x))∂y$
Its covariant derivative with respect to $∂x$ is $sin(f(x))(1-f_{x})∂x + cos(f(x))(f_{x}-1)∂y$. $V$ is parallel when $f_{x} = 1$ so $f(x) = x+c$ where $c$ is a constant determined by the vector that one wants to parallel translate.

So $V(x) = cos(x+c)∂x + sin(x+c)∂y$ rotates counterclockwise with respect to the $xy$ coordinates and also with respect to an observer traveling from left to right along the line. By comparison, in the Levi-Civita connection, $V$ would be constant with respect to the $xy$ coordinate system.

Parallel translation along a circle:

In this case one wants to solve $0 =∇_{-sin(t)∂x+cos(t)∂y}cos(f(t))∂x+sin(f(t))∂y$. A similar calculation as for the horizontal straight line gives

$f(t)= cos(t) + c$

so $V(t) = cos(cos(t) + c)∂x + sin(cos(t)+c)∂y$.

$V$ swings back and forth along the circle but does not make a full rotation with respect to the $xy$ coordinate system. For instance, for $c=0$, $cos(cos(t))$ oscillates between $1$ and about $.54$ and $sin(cos(t)$ oscillates between about $-.841$and $+.841$. Again by contrast, in the Levi_Civita connection, $V$ is constant with respect to the $xy$ coordinates .

 

[Joker93]

Here is an example of a connection on the plane that may help demonstrate the effect of torsion on parallel translation.

This connection will be compatible with the standard Euclidean metric $<∂x,∂x>=<∂y,∂y> = 1$, $<∂x,∂y> = 0$.
Metric compatibility constrains covariant differentiations to satisfy:

$∇_{∂x}∂x = h∂y$ for some function $h$
$∇_{∂y}∂y = g∂y$ for some other function $g$
$∇_{∂x}∂y = -h∂x$
and $∇_{∂y}∂x = -g∂x$

(For instance, $0 =∂x<∂x,∂x> = 2<∇_{∂x}∂x,∂x>$ so $∇_{∂x}∂x$ is perpendicular to $∂x$.)

The torsion term $∇_{∂x}∂y-∇_{∂y}∂x -[∂x,∂y]$ equals $g∂x-h∂y$ because the bracket $[∂x,∂y] = 0$.

The only way that this torsion term can be zero (to make the connection torsion free) is if $h$ and $g$ are both zero. In that case, one gets the standard connection on the Euclidean plane.

Choose for example $h=-1$ and $g=0$. The covariant derivatives are

$∇_{∂x}∂x = -∂y$
$∇_{∂y}∂y = 0$
$∇_{∂x}∂y = ∂x$
and $∇_{∂y}∂x = 0$

Here are a couple of examples of parallel translation for this connection.

Parallel translation along a horizontal straight line:

Let $V$ be a unit length vector field along a horizontal straight line. $V$ can be written as $V(x) = cos(f(x))∂x + sin((f(x))∂y$
Its covariant derivative with respect to $∂x$ is $sin(f(x))(1-f_{x})∂x + cos(f(x))(f_{x}-1)∂y$. $V$ is parallel when $f_{x} = 1$ so $f(x) = x+c$ where $c$ is a constant determined by the vector that one wants to parallel translate.

So $V(x) = cos(x+c)∂x + sin(x+c)∂y$ rotates counterclockwise with respect to the $xy$ coordinates and also with respect to an observer traveling from left to right along the line. By comparison, in the Levi-Civita connection, $V$ would be constant with respect to the $xy$ coordinate system.

Parallel translation along a circle:

In this case one wants to solve $0 =∇_{-sin(t)∂x+cos(t)∂y}cos(f(t))∂x+sin(f(t))∂y$. A similar calculation as for the horizontal straight line gives

$f(t)= cos(t) + c$

so $V(t) = cos(cos(t) + c)∂x + sin(cos(t)+c)∂y$.

$V$ swings back and forth along the circle but does not make a full rotation with respect to the $xy$ coordinate system. For instance, for $c=0$, $cos(cos(t))$ oscillates between $1$ and about $.54$ and $sin(cos(t)$ oscillates between about $-.841$and $+.841$. Again by contrast, in the Levi_Civita connection, $V$ is constant with respect to the $xy$ coordinates .

Hello. I tried to do the analogous thing on a sphere by choosing a connection such that
$∇_{∂x}∂x = 0$
$∇_{∂y}∂y = 0$
$∇_{∂x}∂y = ∂x-∂y$
and $∇_{∂y}∂x =-(∂x-∂y)$.
This gives the geodesic equation $γ_i''=0$ which are "straight lines", meaning that all small and big circles on the sphere are geodesics with respect to this connection.
Not, this torsion-full connection seems unnatural to me. It's like not even having a sphere.
And, in reality, what distinguishes the manifold that I am parallel transporting on from the Euclidean plane? I just labelled the coordinates by φ and θ, but other than that, due to not using a metric and because I took this arbitrary connection, I did not use anywhere the fact that it is a sphere.
So, what is the intuition behind this?

Lastly, in these video lectures by Schuller: ,
if you go to 51:11 he says that choosing another connection for the sphere(other than the Levi-Civita connection) would give the straightest lines(geodesics) for other manifolds, such as an ellipsoid. I am confused by this. How can I know, without having a metric(just by having a smooth-not a Riemannian-manifold), what my manifold looks like?
And how does torsion comes into play? I mean, if I want to find a torsion-full connection on a sphere, how can I do so while making sure that my manifold is still the sphere(since Schuller implies that the connection determines the manifold-something which I do not understand)?

Thank you.