Conformal transformation of curvature tensor I'm asked to calculate the curvature Riemann and Ricci tensors and the curvature scalar of a metric that comes from a conformal transformation from a flat metric, that is, of a metric $g'=\Omega^2g$ where $g$ is flat.
Now, I've been able to calculate the Riemann tensor of the $g'$ metric in terms of the other one, getting to this result:
$$R'{_{abc}}^d=R{_{abc}}^d+2C{_{c[a}}^d\nabla_{b]}+2\nabla_{[a}C{_{b]c}}^d+2C{_{c[a}}^eC{_{b]e}}^d$$
Where $R{_{abc}}^d$ are the components of the Riemann tensor in the $g$ metric.
Now for the Ricci, I just contract $b$ and $d$ indices in both sides and the same for the curvature scalar. I don't know though, if the $g$ metric is flat, what to get besides those expressions.
Thank you in advance.
EDIT:
I calculated that by using $R{_{abc}}^d\omega_d=(\nabla_a\nabla_b-\nabla_b\nabla_a)\omega_c$ for  $1$-forms $\omega$.  And $C{_{abc}}^d$ is the components of the difference tensor of both cov. derivatives. Now, obviously, the terms that go like $C\nabla$ must vanish because it wouldn't be a linear operator. That means that it must be tru for $1$-forms that 
$$C{_{c[a}}^d\nabla_{b]}\omega_d$$
must be $0$, but I can't see why.
 A: As far as I can see, the OP uses the difference tensor
$$
C_{a b}{}^c := - \delta_a{}^c \Upsilon_b - \delta_b{}^c \Upsilon_a + \Upsilon^c g_{a b}
$$
that appears in the expression
$$
\widehat{\nabla}_a \omega_b = \nabla_a \omega_b + C_{a b}{}^c \omega_c
$$
where $\nabla_a$ and $\widehat{\nabla}_a$ are the Levi-Civita connections of the metrics $g_{a b}$ and ${\widehat{g}}_{a b} = \Omega^2 g_{a b}$ respectively. Here $\Upsilon_a := \nabla_a \log \Omega$, that is the same thing as $\mathrm{d} \omega$ in the settings of this answer. The difference formula in the above display is just a restatement of the equation (1) in that answer in the abstract index notation.
The formula for the curvature $\widehat{R}$ of $\widehat{g}_{a b}$ should then look as
$$
{\widehat{R}}_{a b c}{}^d = R_{a b c}{}^d + 2 \, \nabla_{[a} C_{b] c}{}^d + 2 \, C_{[a| c}{}^e C_{e| b]}{}^d 
$$
which is a direct application of the general difference formula: 

Theorem (The difference formula for curvatures). 
  Let $\nabla$ and $\nabla'$ be two arbitrary connections in the tangent bundle of manifold $M$.
  Let $K$ and $K'$ be the curvatures  of the connections $\nabla$ and $\nabla'$ respectively, and $A$ be the difference tensor of the pair $(\nabla',\nabla)$, that is $ \nabla' = \nabla + A$. Then
  $$
K' = K + \nabla \wedge A + A \wedge A
$$

Proof. For the sake of simplicity I will do the calculation for the case of torsion-free connections $\nabla$ and $\nabla'$ that is sufficient for our purposes (we deal with the Levi-Civita connections in the main question). In this situation the curvature $K$ of the connection $\nabla$ satisfies the identity $K_{a b c}{}^d \omega_d = 2\,\nabla_{[a} \nabla_{b]} \omega_c$, as we discussed here.
Writing everything out using the definitions, we get
$$
\begin{align}
K'_{a b c}{}^d \omega_d & =  \nabla'_a \nabla'_b \omega_c - (a \leftrightarrow b) \\
& =  \nabla'_a (\nabla_b \omega_c + A_{b c}{}^d \omega_d ) - (a \leftrightarrow b) \\
& =  \nabla'_a \nabla_b \omega_c + (\nabla'_a A_{b c}{}^d) \omega_d + A_{b c}{}^d \nabla'_a \omega_d - (a \leftrightarrow b) \\
& = \nabla_a \nabla_b \omega_c + A_{a b}{}^d \nabla_d \omega_c + A_{a c}{}^d \nabla_b \omega_d \\
& \quad + (\nabla_a A_{b c}{}^d + A_{a b}{}^e A_{e c}{}^d + A_{a c}{}^e A_{b e}{}^d - A_{a e}{}^d A_{b c}{}^e) \omega_d \\
& \quad + A_{b c}{}^d ( \nabla_a \omega_d + A_{a d}{}^e \omega_e ) - (a \leftrightarrow b)
\end{align}
$$
Collecting the terms, we rewrite the above equation as
$$
\begin{align}
K'_{a b c}{}^d \omega_d & = \nabla_a \nabla_b \omega_c + \underbrace{A_{a b}{}^d \nabla_d \omega_c}_{\text{sym. in a, b}} + \underbrace{A_{a c}{}^d \nabla_b \omega_d + A_{b c}{}^d \nabla_a \omega_d}_{\text{sym. in a, b}} +  \underbrace{A_{b c}{}^d A_{a d}{}^e \omega_e}_{\text{cancel}}  \\
& \quad + (\nabla_a A_{b c}{}^d) \omega_d  + \underbrace{A_{a b}{}^e A_{e c}{}^d \omega_d}_{\text{sym. in a, b}}  + A_{a c}{}^e A_{b e}{}^d \omega_d  - \underbrace{A_{a e}{}^d A_{b c}{}^e \omega_d}_{\text{cancel}}  - (a \leftrightarrow b)
\end{align}
$$
The underbraced terms vanish for the reasons indicated in the subscripts, so we obtain the equality
$$
K'_{a b c}{}^d \omega_d = 2 \, \nabla_{[a} \nabla_{b]} \omega_c + 2 \, (\nabla_{[a} A_{b] c}{}^d) \omega_d + A_{[a| c}{}^e A_{|b] e}{}^d \omega_d
$$
which is the explicit form of the claim. QED.
A detailed account of this topic can be found in J.Slovak's dissertation here.
