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
$\newcommand{\R}{\mathbb{R}}$ Even though you're asking about manifolds in general, it suffices to understand this when the manifold is just an open set $M \subset \mathbb{R}^n$. The fact that it is a subset of $\R^n$ means there is already a set of coordinate functions $x^k: M \rightarrow \R$. However, there can be other coordinate functions $y^k: M \rightarrow \R$. For example, if $M = (0,1) \times (0,2\pi)$, the standard coordinates would be $x^1(a,b) = a$ and $x^2(a,b) = b$. Another set of coordinates would be $y^1(a,b) = a\cos b$ and $y^2 = a\sin b$.
When studying geometry or physics, coordinates are a necessary evil because we need to be able to measure and compare things. However, geometric properties and physical laws should not depend on the coordinates used. For example, physical laws should not depend on whether we are measuring lengths in meters or feet.
The first step towards defining a Riemannian metric is understanding what a tangent space is. The key idea here is that every point $p \in M$ has its own tangent space $T_pM$. It consists of all possible velocity vectors of curves. If we fix one set of coordinates, then $T_pM$ is obviously isomorphic to $\R^n$. If we switch to a different set of coordinates, then $T_pM$ still isomorphic to $\R^n$, but the isomorphism is different from the first one. The crucial observation is that the map $\R^n \rightarrow T_pM \rightarrow \R^n$ defined using the two different sets of coordinates is linear. What this allows us to do is to view $T_pM$ as an abstract vector space, where a set of coordinates defines an isomorphism $T_pM \rightarrow \R^n$. Put differently, a set of coordinates implies a basis of $T_pM$, commonly denoted as $(\partial_1, \dots, \partial_n)$.
Now recall that a dot or inner product on an abstract vector space $V$ is a bilinear function $g: V \times V \rightarrow \R$ such that $g(v,v) \ge 0$, with equality holding only if $v = 0$. A Riemannian metric is simply an inner product defined on each $T_pM$. It is important to note that we are not assuming any relationship between the inner products on two different tangent spaces $T_pM$ and $T_qM$.
Next, recall that, given a basis $(v_1, \dots, v_n)$ of $V$, an inner product $g$ is uniquely determined by the positive definite symmetric matrix $[g_{ij}]$, where $$ g_{ij} = g(v_i,v_u).$$
So if we choose a set of coordinates, $(x^1, \dots, x^n)$ on $M$, then, at each point, we have a basis $(\partial_1, \dots, \partial_n)$ of $T_pM$, which then allows us to write the Riemannian metric at $p$ as the matrix $[g_{ij}(p)]$, where $$ g_{ij}(p) = g(p)(\partial_i, \partial_j) $$ Here's the next important point: This matrix is a matrix of functions that depend on the point on the manifold. On top of that, the matrix depends on which coordinates you used. What's important is that the matrix changes in just the right way, so that, no matter what coordinates you use, the matrix, used with the basis of $T_pM$ defined by the coordinates, defines the same abstract inner product on $T_pM$.
To study this further, I encourage you to look at fundamental examples such as the standard sphere, hyperbolic space, surfaces of revolution in $\R^3$, and graphs of functions in $\R^n$.