Confusion on the intuition of a connection

Question

At 6:27 of this lecture by Frederic Schuller, it is remarked that only when we fix the connection (by providing Christoffel's) that we get the specific shape of the sphere that we usually imagine, and, perhaps a choice of a different connection would have ended up in a sphere:

This is all good and fine. But then I saw this video by Eigen Chris, and 17:55 of it, he says that one can set all the Christoffels while on the sphere to zero to get another connection named the boring connection:

Now this is contradictory with the Schuller lecture's to me because of two points:

1. In Schuller's lectures, it is found that the connection fixes the shape, but here Eigen Chris has concluded that a single shape can have more than one connection.

2. At 1hr 4mins of the 7th lecture of the GR course by Frederic Schuller, it is mentioned that it is impossible to have a connection with Gammas vanishing on each coordinate chart on a hyperbolic surface. I am supposing this would extend to the sphere case as well.

Picture:

How do I resolve the apparent contradiction here?

Answer 1

Here's what we can say

• A given smooth manifold $M$ can have many possible connections.
• If a manifold $M$ can be covered by a single coordinate chart, then you can transport the trivial connection of $\Bbb{R}^n$ onto the manifold $M$. In other words, using that single chart, we can set all the $\Gamma=0$, and thus we have a trivial/flat connection on $M$. However, the sphere cannot be covered by a single chart ($S^2$ is compact), and also $S^2$ does not admit a flat connection.

I briefly watched the lectrue, and it seems the point he's trying to make is that the unit sphere $S=\{(x,y,z)\in\Bbb{R}^3\,:\,x^2+y^2+z^2=1\}$ and the the ellipse $E=\{(x,y,z)\in\Bbb{R}^3\,:\, (x/a)^2+(y/b)^2+(z/c)^2=1\}$ (with $a,b,c>0$ and not all equal to $1$) are diffeomorphic manifolds and in that sense we can't 'tell them apart'. However, from an extrinsic point of view, once you consider their embedding in $\Bbb{R}^3$, they are clearly different sets. We might then ask what intrinsically makes them 'different'. One approach is to consider their induced metric $g_S$ and $g_E$; then they are not isometric as Riemannian manifolds.

Also, you seem to be using the word 'shape' in more than one sense hence your confusion.

Answer 2

Just to add to what others have said - you can set the Christoffel symbols to zero in one coordinate patch and get a local connection in that patch. But for $S^2$, you wouldn't be able to set them to zero in all coordinate patches covering $S^2$ simultaneously - one can show that in going from a coordinates $x^i$ to $\bar{x}^i$, the Christoffel symbols must transform like $$ \bar{\Gamma}^i_{\ jk} = \frac{\partial\bar{x}^i}{\partial x_m}\frac{\partial x^n}{\partial \bar{x}^j}\frac{\partial x^p}{\partial \bar{x}^k}\Gamma^m_{\ np} + \frac{\partial^2x^m}{\partial\bar{x}^j\partial\bar{x}^k}\frac{\partial\bar{x}^i}{\partial x^m} $$ in order to define a consistent global connection. For the sphere, $\bar\Gamma^i_{\ jk}$ and $\Gamma^m_{\ np}$ can't both be zero, since the last term doesn't vanish no matter what two coordinate patches you choose to cover $S^2$ (assuming a minimal cover of two patches here). So Eigen Chris' video is somewhat misleading in that respect.