I have seen similar questions, but none of the answers relate to my difficulty, which I will now proceed to convey.

Let $(M,g)$ be a Riemannian manifolds. The Levi-Civita connection is the unique connection that satisfies two conditions: agreeing with the metric, and being torsion-free.

Agreeing with the metric is easy to understand. This is equivalent to the parallel transport associated with the connection to satisfy that the isomorphism between tangent spaces at different points along a path are isometries. Makes sense.

Let's imagine for a second what happens if we stop with this condition, and take the case of $M=\mathbb{R}^2$, with $g$ being the usual metric. Then it's easy to think of non-trivial ways to define parallel transport other than the one induced by the Levi-Civita connection.

For example, imagine the following way to do parallel transport: if $\gamma$ is a path in $\mathbb{R}^2$, then the associated map from $TM_{\gamma(s)}$ to $TM_{\gamma(t)}$ will be a rotation based with angle $p_2(\gamma(s))-p_2(\gamma(t))$, where $p_i$ is the projection of $\mathbb{R}^2$ onto the $i^{th}$ coordinate.

So I guess torsion-free-ness is supposed to rule this kind of example out.

Now I'm somewhat confused. One of the answers to a similar question that any two connections that satisfy that they agree with the metric satisfy that they have the same geodesics, and in that case choosing a torsion-free one is just a way of choosing a canonical one. That seems incorrect, as $\gamma(t)=(0,t)$ is a geodesic of $\mathbb{R}^2$ with the Levi-Civita connection but not the one I just described...

Let's think from a different direction. In the case of $\mathbb{R}^2$, if $\nabla$ is the usual (and therefore Levi-Civita) connection then $\nabla_XY$ is just $XY$, and $\nabla_YX$ is just $YX$. So of course we have torsion-free-ness.

So I guess one way to think of torsion-free-ness is saying that you want the parallel transport induced by the connection to be the one associated with $\mathbb{R}^n$ via the local trivializations.

Except that this seems over-simplistic: torsion-free-ness is weaker than the condition that $\nabla_XY=XY$ and $\nabla_YX=YX$. So why this crazy weaker condition that $\nabla_XY-\nabla_YX=[X, Y]$? What does that even mean geometrically? Why is this sensible? How would say that in words that are similar to "it means that the connection is the connection induced from the trivializations" except more correct than that?

I have seen similar questions, but none of the answers relate to my difficulty, which I will now proceed to convey.

Let $(M,g)$ be a Riemannian manifolds. The Levi-Civita connection is the unique connection that satisfies two conditions: agreeing with the metric, and being torsion-free.

Agreeing with the metric is easy to understand. This is equivalent to the parallel transport associated with the connection to satisfy that the isomorphism between tangent spaces at different points along a path are isometries. Makes sense.

Let's imagine for a second what happens if we stop with this condition, and take the case of $M=\mathbb{R}^2$, with $g$ being the usual metric. Then it's easy to think of non-trivial ways to define parallel transport other than the one induced by the Levi-Civita connection.

For example, imagine the following way to do parallel transport: if $\gamma$ is a path in $\mathbb{R}^2$, then the associated map from $TM_{\gamma(s)}$ to $TM_{\gamma(t)}$ will be a rotation based with angle $p_2(\gamma(s))-p_2(\gamma(t))$, where $p_i$ is the projection of $\mathbb{R}^2$ onto the $i^\text{th}$ coordinate.

So I guess torsion-free-ness is supposed to rule this kind of example out.

Now I'm somewhat confused. One of the answers to a similar question that any two connections that satisfy that they agree with the metric satisfy that they have the same geodesics, and in that case choosing a torsion-free one is just a way of choosing a canonical one. That seems incorrect, as $\gamma(t)=(0,t)$ is a geodesic of $\mathbb{R}^2$ with the Levi-Civita connection but not the one I just described...

Let's think from a different direction. In the case of $\mathbb{R}^2$, if $\nabla$ is the usual (and therefore Levi-Civita) connection then $\nabla_XY$ is just $XY$, and $\nabla_YX$ is just $YX$. So of course we have torsion-free-ness.

So I guess one way to think of torsion-free-ness is saying that you want the parallel transport induced by the connection to be the one associated with $\mathbb{R}^n$ via the local trivializations.

Except that this seems over-simplistic: torsion-free-ness is weaker than the condition that $\nabla_XY=XY$ and $\nabla_YX=YX$. So why this crazy weaker condition that $\nabla_XY-\nabla_YX=[X, Y]$? What does that even mean geometrically? Why is this sensible? How would say that in words that are similar to "it means that the connection is the connection induced from the trivializations" except more correct than that?

I have seen similar questions, but none of the answers relate to my difficulty, which I will now proceed to convey.

Let $(M,g)$ be a Riemannian manifolds. The Levi-Civita connection is the unique connection that satisfies two conditions: agreeing with the metric, and being torsion-free.

Agreeing with the metric is easy to understand. This is equivalent to the parallel transport associated with the connection to satisfy that the isomorphism between tangent spaces at different points along a path are isometries. Makes sense.

Let's imagine for a second what happens if we stop with this condition, and take the case of $M=\mathbb{R}^2$, with $g$ being the usual metric. Then it's easy to think of non-trivial ways to define parallel transport other than the one induced by the Levi-Civita connection.

For example, imagine the following way to do parallel transport: if $\gamma$ is a path in $\mathbb{R}^2$, then the associated map from $TM_{\gamma(s)}$ to $TM_{\gamma(t)}$ will be a rotation based with angle $p_2(\gamma(s))-p_1(\gamma(t))$.

So I guess torsion-free-ness is supposed to rule this kind of example out.

Now I'm somewhat confused. One of the answers to a similar question that any two connections that satisfy that they agree with the metric satisfy that they have the same geodesics, and in that case choosing a torsion-free one is just a way of choosing a canonical one. That seems incorrect, as $\gamma(t)=(0,t)$ is a geodesic of $\mathbb{R}^2$ with the Levi-Civita connection but not the one I just described...

Let's think from a different direction. In the case of $\mathbb{R}^2$, if $\nabla$ is the usual (and therefore Levi-Civita) connection then $\nabla_XY$ is just $XY$, and $\nabla_YX$ is just $YX$. So of course we have torsion-free-ness.

So I guess one way to think of torsion-free-ness is saying that you want the parallel transport induced by the connection to be the one associated with $\mathbb{R}^n$ via the local trivializations.

Except that this seems over-simplistic: torsion-free-ness is weaker than the condition that $\nabla_XY=XY$ and $\nabla_YX=YX$. So why this crazy weaker condition that $\nabla_XY-\nabla_YX=[X, Y]$? What does that even mean geometrically? Why is this sensible? How would say that in words that are similar to "it means that the connection is the connection induced from the trivializations" except more correct than that?

I have seen similar questions, but none of the answers relate to my difficulty, which I will now proceed to convey.

Let $(M,g)$ be a Riemannian manifolds. The Levi-Civita connection is the unique connection that satisfies two conditions: agreeing with the metric, and being torsion-free.

Agreeing with the metric is easy to understand. This is equivalent to the parallel transport associated with the connection to satisfy that the isomorphism between tangent spaces at different points along a path are isometries. Makes sense.

Let's imagine for a second what happens if we stop with this condition, and take the case of $M=\mathbb{R}^2$, with $g$ being the usual metric. Then it's easy to think of non-trivial ways to define parallel transport other than the one induced by the Levi-Civita connection.

For example, imagine the following way to do parallel transport: if $\gamma$ is a path in $\mathbb{R}^2$, then the associated map from $TM_{\gamma(s)}$ to $TM_{\gamma(t)}$ will be a rotation based with angle $p_2(\gamma(s))-p_2(\gamma(t))$, where $p_i$ is the projection of $\mathbb{R}^2$ onto the $i^{th}$ coordinate.

So I guess torsion-free-ness is supposed to rule this kind of example out.

Now I'm somewhat confused. One of the answers to a similar question that any two connections that satisfy that they agree with the metric satisfy that they have the same geodesics, and in that case choosing a torsion-free one is just a way of choosing a canonical one. That seems incorrect, as $\gamma(t)=(0,t)$ is a geodesic of $\mathbb{R}^2$ with the Levi-Civita connection but not the one I just described...

Let's think from a different direction. In the case of $\mathbb{R}^2$, if $\nabla$ is the usual (and therefore Levi-Civita) connection then $\nabla_XY$ is just $XY$, and $\nabla_YX$ is just $YX$. So of course we have torsion-free-ness.

So I guess one way to think of torsion-free-ness is saying that you want the parallel transport induced by the connection to be the one associated with $\mathbb{R}^n$ via the local trivializations.

Except that this seems over-simplistic: torsion-free-ness is weaker than the condition that $\nabla_XY=XY$ and $\nabla_YX=YX$. So why this crazy weaker condition that $\nabla_XY-\nabla_YX=[X, Y]$? What does that even mean geometrically? Why is this sensible? How would say that in words that are similar to "it means that the connection is the connection induced from the trivializations" except more correct than that?