curvaturedifferential-geometryriemannian-geometry
I see the scalar curvature $R$ as an indicator of how a manifold curves locally (the easiest example is for a $2$-dimensional manifold $M$, where the $R=0$ in a point means that it is flat there, $R>0$ that it makes like a hill and $R<0$ that it is a saddle point).
Are there analogous interpretations for the Riemann tensor $Rm$ and for the Ricci curvature $Rc$? I tried to think about it but I really can't get anything making sense.
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.
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 $$\frac{1}{12}(n - 1) n^2 (n + 1) = \color{#0000df}{\underbrace{\left[\tfrac{1}{12}(n - 3) n (n + 1) (n + 2)\right]}_{\dim \mathsf W}} + \color{#009f00}{\underbrace{\left[\tfrac{1}{2} (n - 1) (n + 2)\right]}_{\bigodot^2_{\circ} \Bbb V^*}} + \color{#df0000}{\underbrace{1}_{\dim \Bbb R}} .$$
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 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:
Using the definition of the Riemann curvature tensor $$R_{XY}Z=\nabla_{[X,Y]}Z-[\nabla_X,\nabla_Y]Z,$$ we can compute that $$R^a_{\;bcd}=\partial_d\Gamma^a_{cb}-\partial_c\Gamma^a_{db}+\Gamma^a_{de}\Gamma^e_{bc}-\Gamma^a_{ce}\Gamma^e_{bd},$$ (note that my convention is $R_{\partial_c\partial_d}\partial_b=R^a_{\;bcd}\partial_a$). Now using our definitions of the Ricci curvature $$R_{ab}=R^c_{\;acb}$$ and of the scalar curvature $$R=R^a_{\;a},$$ you would just substitute and calculate.
I'm not sure I understand your second question, since anything expressed in terms of Christoffel symbols is also an expression in coordinates (recall that $\nabla_{\partial_a}\partial_b=\Gamma^c_{ab}\partial_c$).
Best Answer
We consider the Riemann tensor first. A crucial observation is that if we parallel transport a vector $u$ at $p$ to $q$ along two different pathes $vw$ and $wv$, the resulting vectors at $q$ are different in general (following figure). If, however, we parallel transport a vector in a Euclidean space, where the parallel transport is defined in our usual sense, the resulting vector does not depend on the path along which it has been parallel transported. We expect that this non-integrability of parallel transport characterizes the intrinsic notion of curvature, which does not depend on the special coordinates chosen.
It is useful to say that in this sense visualization of the first Bianchi identity is very easy:
We can give a quantitative geometric interpretation to the sectional curvature tensor in any dimension. Let M be a Riemannian n-manifold and p ∈ M. If $\Pi$ is any $2$-dimensional subspace of $T_pM$, and $V \subset T_pM$ is any neighborhood of zero on which $\exp_p$ is a diffeomorphism, then $S_\Pi := \exp_p(\Pi \cap V)$ is a $2$-dimensional submanifold of $M$ containing $p$ (following figure), called the plane section determined by $\Pi$. Note that $S_\Pi$ is just the set swept out by geodesics whose initial tangent vectors lie in $\Pi$. We define the sectional curvature of $M$ associated with $\Pi$, denoted $K(\Pi)$, to be the Gaussian curvature of the surface $S_\Pi$ at $p$ with the induced metric. If $(X, Y)$ is any basis for $\Pi$, we also use the notation $K(X, Y)$ for $K(\Pi)$.
We can also give a geometric interpretation for the Ricci and scalar curvatures. Given any unit vector $V \in T_pM$, choose an orthonormal basis $\{E_i\}$ for $T_pM$ such that $E_1 = V$ . Then $Rc(V, V )$ is given by
$$Rc(V,V)=R_{11}=R_{k11}^k=\sum_{k=1}^{n} Rm(E_k,E_1,E_1,E_k)=\sum_{k=2}^{n}K(E_1,E_k)$$
Therefore the Ricci tensor has the following interpretation: For any unit vector $V \in T_pM$, $Rc(V, V )$ is the sum of the sectional curvatures of planes spanned by $V$ and other elements of an orthonormal basis. Since $Rc$ is symmetric and bilinear, it is completely determined by its values of the form $Rc(V, V )$ for unit vectors $V$ .
Similarly, the scalar curvature is
$$S=R_j^j=\sum_{j=1}^n Rc(E_j,E_j)=\sum_{j,k=1}^{n}Rm(E_k,E_j,E_j,E_k)=\sum_{j\ne k}K(E_j,E_k)$$
Therefore the scalar curvature is the sum of all sectional curvatures of planes spanned by pairs of orthonormal basis elements.