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Gaussian curvature:
$R = g^{ab} R_{ab} = g^{ab} R_{(ab)}$

Gaussian curvature from Riemann curvature:
$R = {R^{ab}}_{ab}$

Gaussian curvature from the Ricci curvature one-form:
$R = R_a \cdot e^a$
$= R_{ab} e^b \cdot e^a$
$= R_{ab} g^{ab}$
Now I could either here assume that $e^a \cdot e^b = g^{ab}$ based on the inverse of $e_a \cdot e_b = g_{ab}$, though it seems some sources don't like that, so lets just use the differential form definition of an inner product:
$R = \sigma \star (R_a \wedge \star e^a)$

Gaussian cuvature from Riemann curvature 2-form:
$R = {\Omega^a}_{b} (e_a) \cdot e^b$
$= \sigma \star ({\Omega^a}_{b} (e_a) \wedge \star e^b)$

An easier way to define this would be using the contravariant-indexed Ricci one-form:
$R = R^a(e_a)$ $= {R^a}_b e^b(e_a)$ $= {R^a}_b \delta^b_a$ $= {R^a}_a$
$= g^{ab} R_{bc} e^c (e_a)$ $= g^{ab} R_{bc} \delta^c_a$ $= g^{ab} R_{ba}$

As interior product:
$R = e_a \lrcorner R^a$
$= e_b \lrcorner e_a \lrcorner \Omega^{ab}$
$= g^{bc} e_b \lrcorner e_a \lrcorner {\Omega^a}_c$

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