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.
I would say that calculating the curvature of the sphere using charts and Christoffel symbols is possibly the most inefficient way of all.
If $x \in S^n$, then the tangent space at the point $x$ is given by
$$ T_x S^n = \{ X \in \mathbb{R}^{n+1} \mid \langle x, X \rangle = 0 \}$$
The curvature tensor of the sphere is then given by
$$R(X, Y)Z = \langle Y, Z \rangle X - \langle X, Z \rangle Y .$$
To see this, use the Gauss formula, for example.
The Ricci curvature is then the trace w.r.t. $X$, i.e.
$$ \mathrm{ric}(Y, Z) = \mathrm{tr} \,R(\cdot, Y)Z = \sum_{i=1}^{n-1}\bigl( \langle Y, Z \rangle \langle e_i, e_i \rangle - \langle e_i Z \rangle \langle Y, e_i \rangle\bigr) = (n-1) \langle Y, Z \rangle$$
where $e_1, \dots e_{n-1} \perp \vec{n}$ is an orthonormal basis of $T_x S^n$.
To get the curvature for spheres of different radius, multiply everything with the inverse radius.
Best Answer
A nice answer for second question is in this post. for first ans third, A Lemma (page 124) of Riemannian geometry: An Introduction to curvature written by John M. Lee states that
Lemma 7.6: The Ricci curvature is a symmetric $2$-tensor field. It can be expressed in any of the following ways: $$R_{ij}=R_{kij}{\ }^k=R_{ik}^{\;\;k}{\ }_j=-R_{ki}^{\;\;\ k}{}_j=-R_{ikj}^{\;\;\;\;k}.$$ I hope this would be helpful.