I am looking for the following tensor field $T^i_j$ on closed riemann manifold $M$ which is dependent only from metric tensor $g_{ij}$ and its Riemann cirvature tensor $R^i_{jkl}$ and moreover quadratic by $R^i_{jkl}$ which have zero divergence:
$$
\nabla_i T^i_j = 0,
$$
For example, the Einstein tensor have zero divergence but it is linear by Ricci tensor and hence linear by Riemann curvature tensor. On the other hand the following tensor:
$$
T^i_j = R_j^i R,
$$
is quadratic by $R^i_{jkl}$ but in general have non zero divergence in the sense given above.
Quadratic from Riemann tensor with zero divergence
curvaturedifferential-geometrygeneral-relativityriemannian-geometrytensors
Related Solutions
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
There might be two ways to construct the expected tensor.
First, perhaps you may want to have a look at $f(S)$ gravity, where $S$ means the scalar curvature. This method considers the following functional of the metric $g$ $$ I[g]=\int_{M}f(S)\,{\rm d}V, $$ where $M$ stands for the closed Riemannian manifold, ${\rm d}V$ is its invariant volume form, and $f$ denotes some smooth function. In the special case $f(S)=S$, this functional reduces to the Einstein-Hilbert action. Now, apply calculus of variation to $I[g]$ with respect to $g$, and Noether's theorem guarantees its functional derivative $$ \frac{\delta I[g]}{\delta g^{\mu\nu}}=f'(S)\,{\rm Ric}_{\mu\nu}-\frac{1}{2}\,g_{\mu\nu}\,f(S)+\left(g_{\mu\nu}\,g^{\sigma\tau}\nabla_{\sigma}\nabla_{\tau}-\nabla_{\mu}\nabla_{\nu}\right)f'(S) $$ to be divergence free, where ${\rm Ric}$ is the Ricci tensor, and $\nabla$ means the Levi-Civita connection. Finally, by choosing, e.g., $f(S)=S^2$, you may obtain a divergence-free tensor with its main part (first two terms in the last formulation) quadratic in the Riemann curvature tensor. Unfortunately, as you will see, this result includes additional derivative terms, hence is not a fully quadratic expression.
Second, perhaps you may want to explore a most general form of the expected quadratic tensor $$ T=\left<C,R\otimes R\right>, $$ where $C$ is a $g$-isotropic tensor, $R$ means the Riemann curvature tensor, and $\left<\cdot,\cdot\right>$ denotes the tensor contraction operator. Its entrywise form reads $$ T_{\xi\eta}=C_{\tau\rho\xi\eta}^{\mu\nu\sigma\alpha\beta\gamma}\,R_{\mu\nu\sigma}^{\tau}\,R_{\alpha\beta\gamma}^{\rho}. $$ Here $g$-isotropic means that $C$ remains identical under all isometries that preserves $g$. Therefore, if $C$ is rank-two, it must be $$ C_{\mu\nu}=K\,g_{\mu\nu} $$ with $K$ being a constant; if $C$ is rank-four, it must be $$ C_{\mu\nu\alpha\beta}=K_1\,g_{\mu\nu}\,g_{\alpha\beta}+K_2\,g_{\mu\alpha}\,g_{\nu\beta}+K_3\,g_{\mu\beta}\,g_{\nu\alpha} $$ with $K_1$, $K_2$, $K_3$ being constants. Now that $C$ is rank-ten from above, and you will have 945 $K_j$'s (of course, provided that $R$ is partly anti-symmetric, many of these constants vanish). Nevertheless, since $C$ is made up of $g$, it appears to be a "constant" tensor as you apply $\nabla$ to it, i.e., $\nabla_{\mu}C_{\cdots}=0$. Hence your goal reduces to explore the symmetric property of this rank-ten $C$ upon the requirement $$ \nabla^{\xi}T_{\xi\eta}=0. $$ Unfortunately, I am afraid this is so huge a task that even a special trial takes much more than expected.