(Before this read relationship parallel transport, covariant derivatives and metrics)
Given a Riemannian metric $g$ on a manifold $M$, the Levi-Civita connection is a vector bundle connection $\nabla$ on the tangent bundle (linear connection) which is compatible with $g$, in the sense that:
1. it preserves the metric, i.e., $\nabla g=0$. This condition means that the inner product of two vectors remains constant when they are parallel transported. On the other hand, remember that a vector bundle connection induces (and indeed is uniquely determined by) a principal connection on the frame bundle (see this). Well, the condition $\nabla g=0$ means that this vector bundle connection induces a principal connection also on the orthonormal frame bundle. In addition, this condition implies antisymmetry for the connection matrix (see below).
2. it is torsion-free in the sense of covariant derivatives. I think that this condition can be interpreted in the sense that the Cartan geometry that appears from "pasting together" $\nabla$ and the solder form $\theta$ is a torsion-free Cartan geometry.
Theorem. Every pseudo-Riemannian manifold $(M,g)$ has a unique Levi-Civita connection.
$\blacksquare$
Proof Wikipedia
$\blacksquare$
In a local chart $\{x_1,\ldots,x_n\}$ we have a local chart frame, and we can express the Levi-Civita connection as a matrix of 1-forms (see here). This gives rise to the Christoffel symbols.
Here appears an algorithmic way to compute the connection 1-form, and the curvature, from a coframe. Basically, it uses Cartan's first structural equation and these two properties:
In an orthonormal frame $e_i$, the metric is simple: $g(e_i, e_j) = \delta_{ij}$, where $\delta_{ij}$ is the Kronecker delta.
The connection 1-forms are defined by the covariant derivative:
$$ \nabla e_j = \Theta_{j}^k e_k. \tag{2} $$We want to show that $\Theta_{j}^i = - \Theta_{i}^j$ for all $i$ and $j$.
To do this, consider the formula (1) in the orthonormal basis, setting $X = e_k$, $Y = e_i$, $Z = e_j$, we have:
$$e_k (\delta_{ij}) = g(\nabla_{e_k} e_i, e_j) + g(e_i, \nabla_{e_k} e_j).$$
$$0 = g(e_k \lrcorner\Theta_{i}^l e_l, e_j) + g(e_i, e_k \lrcorner\Theta_{j}^m e_m) = e_k\lrcorner \Theta_{i}^j + e_k\lrcorner \Theta_{j}^i,$$
for every $k$. Hence, we find $\Theta_{i}^j = - \Theta_{j}^i$.
2. We have $e_i\lrcorner\Theta_{j}^k-e_j\lrcorner\Theta_{i}^k=-T_{ij}^k$
Since the connection is torsion-free we have
$$ \nabla_{e_i} e_j-\nabla_{e_j} e_i =[e_i,e_j]= -T_{ij}^k e_k, $$which leads to $$ \left(e_i\lrcorner\Theta_{j}^k-e_j\lrcorner\Theta_{i}^k\right) e_k=-T_{ij}^k e_k $$and we have the result. $\blacksquare$
We show two examples:
Example 1: connection form for LV in a surface.
Example 2: connection form for LV in R3
Two metrics can have the same Levi-Civita connection and hence the same geodesics, but differ in other properties like volumes, angles, etc. In more formal terms, this question involves projective equivalence of Riemannian metrics, a topic studied in differential geometry. Two metrics are projectively equivalent if they have the same unparametrized geodesics. There are nontrivial examples of projectively equivalent but distinct Riemannian metrics, even in dimension 2. See also the discusion here.
(Here we are going to use Penrose abstract index notation)
Derivative operators (in the sense of particular vector bundle connection on $TM$) and Riemannian metrics give two different ways to check the constancy of a field along a curve. Given $g_{ab}$ and $\nabla$, in principle not related, and a field $\lambda^a$ we could establish that it is constant, as we know, with the expression:
$$ \xi ^ { n } \nabla _ { n } \lambda ^ { a } = 0 $$where $\xi^n$ is the tangent field to a given curve $\gamma$.
On the other hand, would be convenient to say that $\lambda^a$ is constant if $g_{ab} \lambda ^ { a } \lambda ^ { b }$ is constant along $\gamma$, i. e.,
$$ \xi ^ { n } \nabla _ { n } \left( g _ { a b } \lambda ^ { a } \lambda ^ { b } \right) = 0 $$We would say that a connection $\nabla_n$ is compatible with a metric $g_{ab}$ if for any $\gamma$ and $\lambda^a$ the first condition of constancy imply the second one. This can be say more simply thanks to the following proposition:
Proposition
The derivative operator $\nabla$ and the metric $g_{ab}$ are compatible if and only if
$$ \nabla_a g_{bc}=0. $$$\blacksquare$
At the same time, this is equivalent to saying that
$$ \nabla_n g^{bc}=0 $$Given a manifold $M$ it can be shown that there exists a lot of derivative operators. But given a manifold with a metric, it can be shown that there is only one of them that is compatible with the metric and torsion free: the Levi-Civita connection.
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Author of the notes: Antonio J. Pan-Collantes
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