In studying Riemannian geometry, one learns about the Riemann curvature tensor $$Rm(X,Y,Z,W) = \langle \nabla_X\nabla_YZ -\nabla_Y\nabla_XZ - \nabla_{[X,Y]}Z, W\rangle$$ and its various symmetries. From the Riemann curvature tensor, one can define the Ricci and scalar curvatures, which give us "pieces" of the curvature.

I understand that both the Ricci and scalar curvatures are important ways of measuring curvature. My question is: are these (in any sense) the "only" interesting tensors that come from the Riemann curvature?

That is, I could imagine inventing other curvature tensors by performing various operations on $Rm$. Is there a reason that doing so would be fruitless? Why do we privilege the Ricci and Scalar curvatures? Do they give us all the information that we might want?

A previous question of mine hinted at this, though my thoughts were not quite as clear.

  • $\begingroup$ Physicists might want to add the Einstein tensor to the list. $\endgroup$ Feb 5, 2013 at 10:06
  • $\begingroup$ The Einstein Tensor that I’m familiar with is the one for pseudo-Riemannian manifolds. However, I’m pretty sure that in the Riemannian case, it is still divergence-free, as the Bianchi identities obviously hold. $\endgroup$ Feb 5, 2013 at 10:21
  • $\begingroup$ The tracefree Ricci tensor, $\operatorname{Ric} - \frac{1}{n} R g$ is also important---its vanishing characterizes Einstein manifolds. It is also essentially the projection of curvature onto the subbundle corresponding to the factor $[1, 1] \cong S^2_0 T^*M$ in the decomposition $[2, 2] \oplus [1, 1] \oplus [0]$ of the curvature modules into irreducible $SO(n)$-representations. (The projection for $[2, 2]$ gives the Weyl curvature, the projection for $[0]$ gives the scalar curvature.) $\endgroup$ Oct 23, 2018 at 15:20
  • $\begingroup$ In particular, essentially all curvature invariants (that don't involving derivatives of curvature, like $\nabla R$---which is interesting in its own right) are linear combinations of Weyl, tracefree Ricci, and scalar curvatures. To name one, for $\dim M \geq 3$, the Schouten tensor, which is characterized by $\operatorname{Ric} = (n - 2) P + (\operatorname{tr} P) g$, is important in conformal geometry. $\endgroup$ Oct 23, 2018 at 15:51
  • $\begingroup$ What's missing here are tensors that are nonlinear functions of the Riemann curvature tensor. Such tensors arise naturally when studying the Ricci flow and other geometric PDEs. $\endgroup$
    – Deane
    Mar 31, 2022 at 22:37

4 Answers 4


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 property follows from the usual definition of $Rm$),
  • $Rm(W, X, Y, Z) = -Rm(W, X, Z, Y)$ (this property 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 identity is the 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. The space $\mathsf C$ 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.
  • $\begingroup$ Amazingly clear exposition! Thank you! Do you know a reference for this kind of analysis carried out in the lorrentzian signature case? $\endgroup$ Dec 15, 2020 at 23:17
  • 1
    $\begingroup$ @SaalHardali Not offhand, but the situation is broadly similar. The key difference is that in the Lorentzian setting you're decomposing $\textsf C$ as an $SO(n-1, 1)$-representation rather than as an $SO(n)$-representation, and the representation theories of these groups are different. In particular, if $n=4$, then the analogue of the $10$-dimensional representation $\mathsf W$ is irreducible as an $SO(3,1)$ representation, in contrast to what happens in the Riemmanian case. So (without complexifying, anyway) the notions of self- and anti-self-duality are not available. $\endgroup$ Dec 16, 2020 at 2:55
  • $\begingroup$ Well in that case I guess the question is where does one learn the kind of representation theory that enables one to solve these kinds of problens by oneself? $\endgroup$ Dec 16, 2020 at 3:15
  • 1
    $\begingroup$ A lot of this appears in: Berger, Eric ; Bryant, Robert ; Griffiths, Phillip . The Gauss equations and rigidity of isometric embeddings. Duke Math. J. 50 (1983), no. 3, 803--892. $\endgroup$
    – Deane
    Mar 31, 2022 at 22:34

No, there is (at least) Weyl tensor. Basically Weyl tensor measure how far away the Riemannian metric from being conformally flat. Conformally flat means that the metric is conformal to flat metric. If Weyl tensor vanishes, then it is conformally flat.


No, there's a lot more interesting stuff. One particular subject that is actively studied is identities involving curvature invariants and how formulae equating combinations of curvature invariants constrain the form of the metric.

For example, see this article.

If you are willing to include not strictly tensorial quantities, an obvious thing that you missed is the sectional curvature operator. But a less well-known operator, that is nevertheless very interesting, that relates to the Riemann curvature tensor is the isotropic curvature, which is defined as

$$ (X,Y,Z,W) \mapsto Rm(X,Z,X,Z) + Rm(X,W,X,W) + Rm(Y,Z,Y,Z) + Rm(Y,W,Y,W) - 2 Rm(X,Y,Z,W) $$

This operator featured prominently in the resolution of the differentiable sphere theorem by Simon Brendle and Rick Schoen, see e.g. this paper or this book.


As mentioned by Harald and myself in the comments above, an important tensor that arises from the Riemann curvature tensor is the Einstein tensor. In terms of components, the Einstein tensor $ G_{\mu \nu} $ is defined from the Riemann-curvature tensor as follows: $$ G_{\mu \nu} \stackrel{\text{def}}{=} R_{\mu \nu} - \frac{1}{2} g_{\mu \nu} R, $$ where $ g_{\mu \nu} $ is the metric tensor, $ R_{\mu \nu} $ is the Ricci-curvature tensor and $ R $ is the Ricci scalar.

An important fact about the Einstein tensor is that it is divergence-free. Historically, this is important because when Einstein was attempting to find the correct field equations for gravity, he tried to equate the stress-energy tensor $ T_{\mu \nu} $, which describes density and flux of energy and momentum in spacetime, with a tensor $ G_{\mu \nu} $ that describes the curvature of spacetime. If $ G_{\mu \nu} $ could be found, then one would be able to determine how the distribution of matter and energy in spacetime affects its curvature, as well as how curvature of spacetime affects matter and energy.

As the Riemann-curvature tensor encodes information about spacetime curvature, it was naturally believed that $ G_{\mu \nu} $ should and would be derived from it. Einstein could narrow down his search for $ G_{\mu \nu} $ precisely because of the fact that the stress-energy tensor is divergence-free, which translates into physical terms as conservation of energy. In the paper The Four-Dimensionality of Space and the Einstein Tensor, published in 1972 in Volume 13, Issue 6, of the Journal of Mathematical Physics, David Lovelock showed that the only contravariant $ 2 $-tensors that are divergence-free on one index and that are concomitants of $ g_{\mu \nu} $, together with the first two derivatives of $ g_{\mu \nu} $, are $ g_{\mu \nu} $ itself and $ G_{\mu \nu} $ as defined above.

There are other kinds of tensors that can be derived from the Riemann-curvature tensor. For example, if you Ricci-decompose the Riemann-curvature tensor, then you obtain three kinds of tensors, one of which is the Weyl tensor as mentioned by Paul.


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