I think that the Riemann tensor is zero only in the presence of a flat metric. However, the Ricci Tensor and the Ricci Scalar, are unknown to me, whether they are zero only in the presence of a flat metric. Of the three tensors, Riemann Tensor, Ricci Tensor, and Ricci Scalar, which ones are only zero in a flat metric?
[Math] Of the three tensors, Riemann Tensor, Ricci Tensor, and Ricci Scalar, which ones are only zero in a flat metric
riemannian-geometry
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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.
Let me give a simplified overview of the background related to the question. The entire story can be found in the celebrated paper M. Atiyah, R. Bott, V.K. Patodi, "On the heat equation and the index theorem", which is the canonical reference for the subject.
The Riemannian curvature $R_{abcd}$ and the Ricci tensor $R_{ab}$ mentioned in the question are constructed out of the metric in the sense that in a coordinate patch $(U, x^i)$ they are given by a universal formula involving the partial derivatives of the components of the metric. Explicit expressions can be found, e.g. here. Moreover, the formulas turn out to be polynomial in the partial derivatives of $g_{a b}$ of all orders $\ge 0$ and the inverse metric $g^{a b}$. A formula like that gives components of a tensor if they transform under change of coordinates in the tensorial way. These observations give rise to the following notion.
Let $(M,g)$ be a Riemannian manifold of dimension $n = \dim M$.
Definition 1. A metric invariant (also known as a Riemannian invariant, or an invariant of the Riemannian structure) is a section $P(g)$ of a tensor bundle over $M$, such that for any diffeomorphism $\phi \colon M \to M$ the naturality property holds: $$ P(\phi^* g) = g^* P(g) $$ and in any coordinate patch $(U,x^i)$ (that also gives the coordinate trivialization of the tensor bundle where $P(g)$ has values), the components of $P(g)$ are given by a universal polynomial expression in the list of formal variables $\{ g^{i j}, \partial_{k_1} \dots \partial_{k_s} g_{i j} \}$, $s \ge 0$.
Remark 1. If P has values in $\mathbb{R}$, we have a scalar valued invariant.
Remark 2. In a similar fashion we can give a definition of a metric invariant differential operator.
Examples. The metric tensor $g_{a b}$ and its inverse $g^{a b}$, the Riemann curvature tensor $R_{a b}{}^{c}{}_{d}$, the Ricci tensor $R_{a b}$ are tensor valued metric invariants. The scalar curvature $R = g^{a b} R_{a b}$ is a scalar valued tensor invariant. More scalar valued examples are mentioned in another question of the OP. The Levi-Civita connection $\nabla^g$ of the metric $g$ is a metric invariant differential operator. See e.g. Jack Lee, Riemannian Manifolds. An Introduction to Curvature.
As @Jack Lee has pointed in the comments, using the Levi-Civita connection and the Riemannian curvature one can construct many tensor-valued metric invariants, and taking the complete contractions we obtain lots of scalar valued metric invariants. This can be formalized as follows.
Definition 2. A curvature invariant is a linear combination of partial contractions of the iterated covariant derivatives (with respect to the Levi-Civita connection $\nabla$ associated to the metric $g$) of the Riemannian curvature.
More examples can be found here.
The question in the consideration can now be reformulated as follows.
Can all the metric invariants be obtained as curvature invariants?
The answer is known to be positive. This is a consequence of the First Fundamental Theorem of the classical invariant theory. The key geometric tool that is used to reduce this problem to a problem of the representation theory of the orthogonal group is the normal (or geodesic) coordinates. See the details in the aforementioned paper.
This also holds for the metric invariant differential operators.
Best Answer
If the metric is flat, of course, they're all zero, because $\text{Riemann}=0$ $\implies$ $\text{Ricci}=0$ $\implies$ $\text{scalar}=0$. But the converse depends on the dimension. In dimension $1$, every metric is flat, and the Riemann, Ricci, and scalar curvatures are always zero. In dimension $2$, if the scalar curvature is zero, the metric is flat. In dimension $3$, if the Ricci curvature is zero, the metric is flat, but there are non-flat metrics with zero scalar curvature. In dimensions $4$ and up, there are plenty of examples of non-flat metrics with both Ricci and scalar curvatures equal to zero (K3 surfaces in algebraic geometry, for example).