Levi-Civita connection on a Riemannian surface

We are going to compute the connection form for a Riemannian surface with an orthonormal frame . And also the curvature.

Given a coframe which is orthonormal with respect to the metric $g$ , we can compute the connection 1-form of the Levi-Civita connection directly from the structure coefficients of the coframe.
Suppose you have an orthonormal basis $X_{1}, X_{2}$ of tangent vectors (frame), and its dual coframe $ω^{1}, ω^{2}$ , i.e., $ω^{i} (X_{j}) = δ_{j}^{i}$ . The structure coefficients of the frame/coframe are defined by the commutation relations of the base vectors: $[X_{i}, X_{j}] = - T_{i j}^{k} X_{k}$ . Also we have that
$d ω^{1} = T_{12}^{1} ω^{1} \land ω^{2}$ and $d ω^{2} = T_{12}^{2} ω^{1} \land ω^{2}$ .
On the other hand, the connection 1-forms of the Levi-Civita connection, denoted by $Θ_{j}^{k}$ , are defined by the relation: $\nabla_{X_{i}} X_{j} = Θ_{j}^{k} (X_{i}) X_{k} = Θ_{j i}^{k} X_{k}$ , where $\nabla$ is the covariant derivative associated with the Levi-Civita connection.
The Levi-Civita connection is compatible with the metric, which means that the covariant derivative of the metric is zero. In terms of the connection 1-forms, this translates into the condition: $Θ_{j i}^{k} + Θ_{k i}^{j} = 0$ (see here).

Now, since the connection is torsion-free

Θ_{j}^{k} (X_{i}) - Θ_{i}^{k} (X_{j}) = - T_{i j}^{k}

(see also here).

We can write the 8 equations:

$Θ_{1}^{1} (X_{1}) - Θ_{1}^{1} (X_{1}) = - T_{11}^{1} \to 0 = 0$
$Θ_{1}^{2} (X_{1}) - Θ_{1}^{2} (X_{1}) = - T_{11}^{2} \to 0 = 0$
$Θ_{2}^{1} (X_{1}) - Θ_{1}^{1} (X_{2}) = - T_{12}^{1}$
$Θ_{2}^{2} (X_{1}) - Θ_{1}^{2} (X_{2}) = - T_{12}^{2}$
$Θ_{1}^{1} (X_{2}) - Θ_{2}^{1} (X_{1}) = - T_{21}^{1}$
$Θ_{1}^{2} (X_{2}) - Θ_{2}^{2} (X_{1}) = - T_{21}^{2}$
$Θ_{2}^{1} (X_{2}) - Θ_{2}^{1} (X_{2}) = - T_{22}^{1} \to 0 = 0$
$Θ_{2}^{2} (X_{2}) - Θ_{2}^{2} (X_{2}) = - T_{22}^{2} \to 0 = 0$
and we conclude:
$Θ_{2}^{1} (X_{1}) = - T_{12}^{1}$ from 3,
$Θ_{1}^{2} (X_{2}) = T_{12}^{2}$ from 4,
and from this two last equations:
$Θ_{2}^{1} (X_{2}) = - Θ_{1}^{2} (X_{2}) = - T_{12}^{2}$
$Θ_{1}^{2} (X_{1}) = - Θ_{2}^{1} (X_{1}) = T_{12}^{1}$
Therefore, the connection matrix is

Θ = (\begin{matrix} 0 & - T_{12}^{1} ω^{1} - T_{12}^{2} ω^{2} \\ T_{12}^{1} ω^{1} + T_{12}^{2} ω^{2} & 0 \end{matrix})

This construction is related to Cartan's first structural equation.

Since $X_{1}, X_{2}$ is an orthonormal frame then the connection is a $o (2, R)$ -valued 1-form $Θ$ . This is formalized by considering the associated connection in the principal bundle called orthonormal frame bundle. Or something like that...

Also this could be done with the approach of Christoffel symbols...

Curvature

Since it is a particular vector bundle connection and therefore a particular connection on a fiber bundle, it has a curvature of a connection which is a 2-form of a particular kind. On the other hand, the Riemannian metric gives rise to the Riemann curvature tensor. Both ideas are related.

The curvature form, which is an $o (2, R)$ -valued 2-form, is given by the exterior derivative $Ω = d Θ + Θ \land Θ = d Θ$ . To compute $d Θ$ observe that

d (T_{12}^{1} ω^{1} + T_{12}^{2} ω^{2}) = d T_{12}^{1} \land ω^{1} + d T_{12}^{2} \land ω^{2} + T_{12}^{1} d ω^{1} + T_{12}^{2} d ω^{2} =

= (X_{1} (T_{12}^{1}) ω^{1} + X_{2} (T_{12}^{1}) ω^{2}) \land ω^{1} + (X_{1} (T_{12}^{2}) ω^{1} + X_{2} (T_{12}^{2}) ω^{2}) \land ω^{2} +

+ T_{12}^{1} T_{12}^{1} ω^{1} \land ω^{2} + T_{12}^{2} T_{12}^{2} ω^{1} \land ω^{2} =

= (X_{1} (T_{12}^{2}) - X_{2} (T_{12}^{1}) + (T_{12}^{1})^{2} + (T_{12}^{2})^{2}) ω^{1} \land ω^{2} .

Ω = (\begin{matrix} 0 & - (X_{1} (T_{12}^{2}) - X_{2} (T_{12}^{1}) + (T_{12}^{1})^{2} + (T_{12}^{2})^{2}) ω^{1} \land ω^{2} \\ (X_{1} (T_{12}^{2}) - X_{2} (T_{12}^{1}) + (T_{12}^{1})^{2} + (T_{12}^{2})^{2}) ω^{1} \land ω^{2} & 0 \end{matrix})

This can be renamed to

Ω = (\begin{matrix} 0 & K ω^{1} \land ω^{2} \\ - K ω^{1} \land ω^{2} & 0 \end{matrix})

where $K$ is the Gauss curvature (see here).