In the setting as stated in the question, this is a sort of muscle exercise, and the answer will not look nice at all. The things get a little simpler if we put
$$
f=e^{2 \omega}
$$
Using the Koszul formula we obtain
$$
\nabla' _X Y = \nabla _X Y + (X \omega )Y + (Y \omega )X - g(X,Y) \operatorname{grad}\omega \tag{1}
$$
Any two linear connections $\nabla'$ and $\nabla$ are related to each other by
$$
\nabla' _X Y = \nabla _X Y + S(X,Y) \tag{2}
$$
where $S(X,Y)$ is a (1,2)-tensor, the difference tensor of the pair of connections. In our case, the connections are torsion free, and thus the difference tensor is symmetric:
$$
S(Y,X)=S(X,Y)
$$
Moreover, for torsion-free connections $\nabla'$ and $\nabla$ the corresponding curvature endomorphisms are related via their difference tensor $S$ as
$$
R'(X,Y)Z = R(X,Y)Z + \nabla_X S(Y,Z) - \nabla_Y S(X,Z)+S(X,S(Y,Z))-S(Y,S(X,Z)) \tag{3}
$$
where by definition
$$
\nabla_X S(Y,Z):= \nabla_X (S(Y,Z)) - S(\nabla_X Y,Z) - S(Y, \nabla_X Z) \tag{4}
$$
Now it is straightforward to substitute
$$
S(X,Y) = (X \omega )Y + (Y \omega )X - g(X,Y) \operatorname{grad}\omega \tag{5}
$$
into (3) and obtain the expression for the conformal transformation of the curvature endomorphisms. It can be found in W.Kühnel's "Differential Geometry", p.349, and I reproduce the formula with some modifications:
$$
\begin{align}
R'(X,Y)Z &= R(X,Y)Z + g(\nabla_X \operatorname{grad} \omega,Z)Y - g(\nabla_Y \operatorname{grad} \omega,Z)X\\
&+ g(X,Z)\nabla_Y \operatorname{grad} \omega - g(Y,Z)\nabla_X \operatorname{grad} \omega\\
&+ (Y \omega)(Z \omega)X - (X \omega)(Z \omega)Y\\
&- g(\operatorname{grad} \omega, \operatorname{grad} \omega)[g(Y,Z)X - g(X,Z)Y]\\
&+ [(X \omega)g(Y,Z)-(Y \omega)g(X,Z)]\operatorname{grad} \omega
\end{align} \tag{6}
$$
Using this one can now find expressions for the Ricci curvature, the scalar curvature etc.
Formula (6) suggests the outstanding role of $\operatorname{grad} \omega$ in conformal transformations. This actually is an appearance of $d\omega$ which is canonically identified in the Riemannian geometry with $\operatorname{grad} \omega$ by virtue of the musical isomorphisms.
To get more insights one can observe that the difference tensor $S(X,Y)$ has a more symmetric presentation in the "covariant" form:
$$
g(S(X,Y),Z) = d\omega (X) g(Y,Z) + d\omega (Y) g(X,Z) - d\omega (Z) g(X,Y) \tag{7}
$$
From this point it is already not that far from realizing that the (0,4)-curvature tensor will have a nicer expression, and in fact it is so.
An ultimate understanding of the conformal transformation of the curvature is obtained by analyzing the algebraic properties of the curvature tensor, the direction that is better covered in the language of the representation theory.
A canonical list used in the references is given in A.Besse's "Einstein manifolds" on pp. 58-59.
As for the sectional curvature, I think that it would be a very straightforward calculation using formula (6) and the definition
$$
K = \frac{g(R(X,Y)Y,X)}{g(Y,Y)X - g(X,X)Y} \tag{8}
$$
I hope that this answer clarifies the things a little.
This answer attempts to frame a systematic description of the tensorial curvature invariants that arise from algebraic manipulation of $g$ and $Rm$ in terms of representation theory. Among other things, this viewpoint explains fully the exceptional behavior of the decomposition of curvature in lower dimensions.
The symmetries of the curvature tensor $Rm$ of a metric $g$ on a smooth manifold $M$ are generated by the following identities.
- $Rm(W, X, Y, Z) = -Rm(X, W, Y, Z)$ (this follows from the usual definition of $Rm$),
- $Rm(W, X, Y, Z) = -Rm(W, X, Z, Y)$ (this follows from torsion-freeness of the Levi-Civita connection $\nabla$), and
- $\mathfrak{S}_{X, Y, Z}[Rm(W, X, Y, Z)] = 0$, where $\mathfrak{S}_{X, Y, Z}[\cdot]$ denotes the sum over cyclic permutations of $X, Y, Z$ (this is the \textbf{First Bianchi Identity}).
Now, fix a point $p \in M$ and denote $\Bbb V := T_p M$. The above symmetries together imply that $Rm$ takes values in the kernel $$\mathsf C := \ker B$$ of the map $$\textstyle B : \bigodot^2 \bigwedge^2 \Bbb V^* \to \bigwedge^4 \Bbb V^*$$ that applies the symmetrization $\mathfrak{S}_{X, Y, Z}$ appearing above to the last three indices. This is an irreducible representation of $GL(\Bbb V)$ of (using the Weyl dimension formula) dimension $\frac{1}{12}(n - 1) n^2 (n + 1)$, $n := \dim \Bbb V = \dim M$.
The stabilizer in $GL(\Bbb V)$ of the metric $g_p$ on $\Bbb V$ is a subgroup $SO(\Bbb V) \cong SO(n)$, and we can decompose $\mathsf C$ as an $SO(n)$-module. This is a typical branching problem, and this working out this particular decomposition amounts to working out the various ways $g_p$ can be combined invariantly with an element of $\mathsf C$. Forming the essentially unique trace of $\mathsf{C} \subseteq \bigodot^2 \bigwedge^2 \Bbb V^*$ is the $SO(n)$-invariant map $\operatorname{tr}_1 : \bigodot^2 \bigwedge^2 \Bbb V^* \to \bigodot^2 \Bbb V^*$. The kernel of this map is an $SO(n)$-module $\color{#0000df}{\mathsf{W}}$. Likewise, we have an $SO(n)$-invariant trace $\operatorname{tr}_2 : \bigodot^2 \Bbb V^* \to \color{#df0000}{\Bbb R}$, and the kernel of this map is an $O(n)$-module $\color{#009f00}{\bigodot^2_{\circ} \Bbb V^*}$.
In all dimensions $n \geq 5$, we have $$\textstyle{\mathsf{C} \cong \color{#0000df}{\mathsf{W}} \oplus \color{#009f00}{\bigodot^2_{\circ} \Bbb V^*} \oplus \color{#df0000}{\Bbb R}},$$ and all of these representations are irreducible. In terms of highest weights as $SO(n)$-representations, $$\textstyle{\color{#0000df}{\mathsf{W}} = \color{#0000df}{[0,2,0,\ldots,0]}, \qquad \color{#009f00}{\mathsf{\bigodot^2_{\circ} \Bbb V^*}} = \color{#009f00}{[2,0,0,\ldots,0]}, \qquad \color{#df0000}{\Bbb R} = \color{#df0000}{[0, 0, 0, \ldots, 0]}} ,$$ and these modules have the dimensions indicated in
$$\frac{1}{12}(n - 1) n^2 (n + 1) =
\color{#0000df}{\underbrace{\left[\tfrac{1}{12}(n - 3) n (n + 1) (n + 2)\right]}_{\dim \mathsf W}}
+ \color{#009f00}{\underbrace{\left[\tfrac{1}{2} (n - 1) (n + 2)\right]}_{\bigodot^2_{\circ} \Bbb V^*}}
+ \color{#df0000}{\underbrace{1}_{\dim \Bbb R}} .$$
- The projection of $Rm_p \in \textsf{C}$ to $\color{#0000df}{\mathsf{W}}$ is the Weyl curvature $\color{#0000df}{W}$ at $p$, the totally tracefree part of $Rm_p$. Replacing a Riemannian metric $g$ with the conformal metric $\hat{g} := e^{2 \Omega} g$ gives a metric with Weyl curvature $\color{#0000df}{\hat{W}} = e^{2 \Omega} \color{#0000df}{W}$, so we say that $\color{#0000df}{W}$ is a covariant of the conformal class of $g$. The condition $\color{#0000df}{W} = 0$ is conformal flatness of $g$.
- The projection of $Rm_p$ to $\color{#009f00}{\bigodot^2_{\circ} \Bbb V^*}$ is the tracefree Ricci tensor $\color{#009f00}{Ric_{\circ}}$ of $g$ at $p$. The condition $\color{#009f00}{Ric_{\circ}} = 0$ is just the condition that $g$ is Einstein.
- The projection of $Rm_p$ to $\color{#df0000}{\Bbb R}$ is the Ricci scalar $\color{#df0000}{R}$ of $g$ at $p$. The condition $\color{#df0000}{R} = 0$ is scalar-flatness of $g$.
In dimension $4$, all of the general case still applies, except for the fact that $\color{#0000df}{\mathsf{W}}$ is no longer irreducible: The ($SO(4)$-invariant) Hodge star operator induces a map $\ast : \color{#0000df}{\mathsf{W}} \to \color{#0000df}{\mathsf{W}}$ whose square is the identity, so $\color{#0000df}{\mathsf{W}}$ decomposes as a direct sum $\color{#007f7f}{\mathsf{W}}_+ \oplus \color{#007f7f}{\mathsf{W}}_-$ of the $(\pm 1)$-eigenspaces of $\ast$. The vanishing of the projections $\color{#007f7f}{W_{\pm}}$ are respectively the conditions of anti-self-duality and self-duality of the metric (since these depend only on the Weyl curvature, they are actually features of the underlying conformal structure). The decomposition into irreducible $SO(4)$-modules is
$$\textstyle{\mathsf{C} \cong \color{#007f7f}{\mathsf{W}_+} \oplus \color{#007f7f}{\mathsf{W}_-} \oplus \color{#009f00}{\bigodot^2_{\circ} \Bbb V^*} \oplus \color{#df0000}{\Bbb R}} .$$
In highest-weight notation,
$$
\color{#007f7f}{\mathsf{W}_+} = \color{#007f7f}{[4] \otimes [0]}, \qquad \color{#007f7f}{\mathsf{W}_-} = \color{#007f7f}{[0] \otimes [4]}, \qquad
\textstyle{\color{#009f00}{\bigodot^2_{\circ} \Bbb V^*} = \color{#009f00}{[2] \otimes [2]}} , \qquad
\color{#df0000}{\Bbb R} = \color{#df0000}{[0] \otimes [0]} .$$
In particular, $\color{#007f7f}{\mathsf{W}_+}$ and $\color{#007f7f}{\mathsf{W}_-}$ can be viewed as binary quartic forms respectively on the $2$-dimensional spin representations $\mathsf{S}_{+} = [1] \otimes [0]$ and $\mathsf{S}_- = [0] \otimes [1]$ of $SO(4)$, which gives rise to the Petrov classification of spacetimes in relativity. The respective dimensions are
$20 = \color{#007f7f}{5} + \color{#007f7f}{5} + \color{#009f00}{9} + \color{#df0000}{1}$.
In dimension $3$, the curvature symmetries force $\color{#0000df}{\mathsf{W}}$ to be trivial, but the other two modules remain intact. (So, $\color{#0000df}{W} = 0$, but in this dimension conformal flatness is governed by another tensor.) The decomposition into irreducible $SO(3)$-modules is thus
$$\textstyle{\mathsf{C} \cong \color{#009f00}{\bigodot^2_{\circ} \Bbb V^*} \oplus \color{#df0000}{\Bbb R}},$$ and in particular, if $g$ is Einstein, it also has constant sectional curvature.
In highest-weight notation,
$$
\textstyle{\color{#009f00}{\bigodot^2_{\circ} \Bbb V^*} = \color{#009f00}{[4]}}, \qquad
\color{#df0000}{\Bbb R} = \color{#df0000}{[0]},
$$
and the respective dimensions are $6 = \color{#009f00}{5} + \color{#df0000}{1}$.
Finally, in dimension $2$, $\color{#009f00}{\bigodot^2_{\circ} \Bbb V^*}$ is also trivial, so $\mathsf{C} \cong \color{#df0000}{\Bbb R}$, that is, the curvature is completely by the Ricci scalar $\color{#df0000}{R}$, which in this case is twice the Gaussian curvature $K$.
These invariants account for all of the invariants one can produce by pulling apart $Rm$, but of course one can produce new tensors by taking particular combinations of them. Some important ones, including some mentioned in other answers, are combinations of $\color{#009f00}{Ric_{\circ}}$ and $\color{#df0000}{R}$, giving rise to distinguished tensors in $\bigodot^2 \Bbb V^*$:
- The Ricci tensor, which is of fundamental importance to Riemannian geometry, is $$Ric = \operatorname{tr}_1(Rm) = \color{#009f00}{Ric_{\circ}} + \frac{1}{n} \color{#df0000}{R} g.$$
- The Einstein tensor, which arises in relativity, is $$G = \color{#009f00}{Ric_{\circ}} - \frac{n - 2}{2 n} \color{#df0000}{R} g .$$
- The (conformal) Schouten tensor, which appears in conformal geometry, is (for $n > 2$) $$P = \frac{1}{n - 2} \color{#009f00}{Ric_{\circ}} + \frac{1}{2 (n - 1) n} \color{#df0000}{R} g .$$
The vanishing of any of these three tensors is equivalent to vanishing of the other two and is equivalent to $g$ being Ricci-flat.
Remark One can construct many more interesting, new curvature invariants by allowing for derivatives of curvature and its subsidiary invariants. To name two:
- The vanishing of the derivative $\nabla Rm$ of curvature is the condition that $g$ be locally symmetric.
- Skew-symmetrizing $\nabla P$ on the derivative index and one of the other two indices gives the Cotton tensor $\color{#9f009f}{C}$, which satisfies $(3 - n) \color{#9f009f}{C} = \operatorname{div} \color{#0000df}{W}$. In dimension $3$, vanishing of $\color{#9f009f}{C}$ is equivalent to conformal flatness. In dimension $n \geq 4$, vanishing of $\color{#0000df}{W}$ implies vanishing of $\color{#9f009f}{C}$ but not conversely, giving rise to a weaker variation of conformal flatness called Cotton-flatness.
Best Answer
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:
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.