Non-torsion-free connection on a manifold given a metric tensor?

Question

The torsion tensor is given by $$T(X,Y)=\nabla_X Y -\nabla_Y X -\partial_X Y+\partial_Y X$$ For vector fields $X,Y$ on a manifold with connection $\nabla$. My question, which may seem silly, is: what connection? Naturally, it can't be the Levi-Civita connection since the torsion would be, by definition 0. So, assuming the manifold is equipped with a pseudo-Riemannian metric tensor: what connection is used in the computation of the torsion tensor? Another definition based on a coordinate system: $$T^k_{ij}=\Gamma^k_{ij}-\Gamma^k_{ji}-\gamma^k_{ij}$$ $$\gamma^k_{ij}e_k=[e_i,e_j]$$ Where $[,]$ is the Lie bracket and $e_i$ are the basis vectors. This definition doesn't make sense to me either since I was taught that the christoffel symbols are symmetric in their lower indices.

Answer 1

A smooth manifold $M$ may be equipped with many different geometric structures. For example:

• A Riemannian manifold is a pair $(M,g)$, where $g$ is a Riemannian metric.
• A manifold-with-connection is a pair $(M, \nabla)$, where $\nabla$ is an affine connection on $M$ (not necessarily torsion-free).

It turns out that on a Riemannian manifold $(M,g)$, there is a "best" connection to choose, the Levi-Civita connection. The Levi-Civita connection is characterized by being both (1) compatible with the metric, and (2) torsion-free.

To reiterate: once a metric $g$ is chosen, there are many connections $\nabla$ we can work with, but the Levi-Civita is the best one (in a certain precise sense).

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Given an affine connection $\nabla$, any whatseover, we can define its associated torsion tensor by $$T^\nabla(X,Y) = \nabla_XY - \nabla_YX - [X,Y].$$ We say that a connection is torsion-free iff $T^\nabla = 0$. Some connections (like the Levi-Civita connection) have this property, but others don't.

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Finally, given an affine connection $\nabla$, any whatsoever, and a coordinate chart, we can talk about the associated Christoffel symbols, denoted $\Gamma^k_{ij}$. Again, they depend on both the choice of connection and the coordinate chart.

It is a fact (exercise) that $\Gamma^k_{ij} = \Gamma^k_{ji}$ in every coordinate chart if and only if $\nabla$ is torsion-free.