A Riemannian Einstein 3-manifold has constant sectional curvature. I know a proof of a stronger theorem I read somewhere (it was iff) that makes use of the Weyl tensor but for just this implication it should be possible to use far less machinery, namely the $r$ endomorphism of the tangent space, defined as $<r(x),y>=Ric(x,y)$ and the isomorphism $\phi$ from the tangent space to the alternating 2-forms on it defined as $\phi e_i=e_j\wedge e_k$ for every $(ijk)$ cyclic permutation of $(123)$, where $\{e_i\}_{i=1,2,3}$ is a local orthonormal frame of $TM$. In particular these ingredients alone and the scalar curvature should determine algebraically the curvature operator $\mathcal{R}$ from the 2-forms on the tangent space to itself. I struggled to find any book that approaches this matter in such way, but I know from an exercise from an old course I took years ago. it should be possible to prove it this way. Anybody knows how? Any reference helpful towards this approach would also be helpful.

  • $\begingroup$ There are only 6 components in each tensor, so this can be proved by just writing out the equations that give the Ricci tensor in terms of the curvature tensor and solving for the components of the curvature tensor. Or you can define the symmetric tensor $Q_{ij} = R_{\phi e_i, \phi e_j}$ and observe that the trace-free part is equal to the trace-free Ricci tensor and the trace is a multiple of the scalar curvature. $\endgroup$
    – Deane
    Commented Apr 11, 2021 at 15:20

2 Answers 2


Assume $(\mathcal{M}^3, g)$ is Einstein, i.e $\operatorname{Ric}_{\mathcal{M}} = \lambda g$. By the Bianchi identity $\mathrm{d} \operatorname{Scal} = 2 \operatorname{div}(\operatorname{Ric})$ where $\operatorname{Scal}: \mathcal{M} \to \mathbb{R}$ is the scaIar curvature. If $\mathcal{M}$ is Einstein we then have $\mathrm{d}(\lambda \operatorname{tr}(g)) = n \cdot \mathrm{d} \lambda = 2 \operatorname{div}( \lambda g)$, which in coordinates can be written as

$$ 2 g^{i j} \nabla_{i}\left(\lambda g_{j k}\right) \mathrm{d} x^{k}=2 g^{i j} g_{j k} \nabla_{i} \lambda \mathrm{d} x^{k}=2 \mathrm{~d} \lambda = n \cdot \mathrm{d}\lambda $$

where I'm using Einstein notation. Hence if $n \geq 3$ and $\mathcal{M}$ is connected $\mathrm{d} \lambda \equiv 0$, and therefore $\lambda$ is constant. Then for an arbitrary $p \in \mathcal{M}$ and a geodesic frame $\{e_1, e_2, e_3\}$ around $p$, we get $$\sum_{i = 1}^{3} R(X, e_i, Y, e_i) = 2 \lambda g(X, Y)$$ Making $X = Y = e_j$ with $j \in \{1, 2, 3\}$ and evaluating at $p$ we then get $$\lambda = R_{1212}(p) = R_{1313}(p) = R_{2323}(p)$$ It follows from Schur's lemma that $\mathcal{M}$ has constant sectional curvature (and in particular all it's curvatures are constant as well).

  • $\begingroup$ Thank you very much for your help, I have only one question: what is the Schur lemma mentioned in the end? We didn t study it back then. What role does it play in the conclusion? Also, connectedness wasn t stated in the exercise, but it may be carelessness. Is connectedness also required to trigger Schur s lemma? $\endgroup$
    – xyz
    Commented Apr 13, 2021 at 0:19
  • $\begingroup$ Schur's lemma states that if the curvature is pointwise constant (which is what I showed in the end) then it must be globally constant (assuming connectedness, otherwise it would be constant in each connected component) as well. If you don't require connectedness, the result may not be true but of course it will be true in each connected component. If you think about it connectedness is actually a very weak restriction to impose. $\endgroup$ Commented Apr 13, 2021 at 23:55
  • $\begingroup$ I see, thanks again for the help. I need to work a bit longer on this, as I think connectedness may not be required, maybe the lecturer wasn't just careless about it. In fact in Lee Riemannian manifolds, the very similar Problem 8.14 the connectedness hypothesis is not mentioned. $\endgroup$
    – xyz
    Commented Apr 14, 2021 at 3:10
  • $\begingroup$ Btw I think your second last displayed equation may have a $\lambda$ without factor of 2, hence all a half $\lambda$ in the sectional curvatures in the last displayed equation. I was thinking probably it is just a matter of definitions whether connectedness is necessary or not. The lecturer from this old course defined Einstein with $Ric=\lambda g$ where $\lambda$ is a constant. It seems you define it with $\lambda$ a real function instead, and then show its constant by connectedness. I just looked at the proof and in fact Schur’s lemma alone does not require connectedness. $\endgroup$
    – xyz
    Commented Apr 14, 2021 at 13:49
  • $\begingroup$ However, if I am not mistaken you showed actually more, because the sectional curves are all pointwise constant, but also at all points they are equal to the same constant $\lambda$ because the manifold is Einstein. Doesn t this prove that the curvature is globally constant even without using Schur’s lemma? $\endgroup$
    – xyz
    Commented Apr 15, 2021 at 11:43

EDIT: Based your comments, here is how to map your notation to mine: Given an orthonormal frame $(e_1,e_2,e_3)$, denote the components of the curvature tensor by $$ R_{ijkl} = R(e_i,e_j,e_k,e_l) = \langle R(e_i,e_j)e_l,e_k\rangle $$ Define the symmetric tensor $Q$ by $$ Q_{ab} = R(e_i,e_j,e_k,e_l), $$ where $(a,i,j)$ and $(b,k,l)$ are cyclic permutations of $(1,2,3)$. This is essentially the same as what you denote by $\phi^{-1}\circ\mathcal{R}\circ\phi$.

Therefore, \begin{align*} Q_{11} &= R_{2323}\\ Q_{12} &= R_{2331}\\ Q_{13} &= R_{2312}\\ Q_{22} &= R_{3131}\\ Q_{23} &= R_{3112}\\ Q_{33} &= R_{1212} \end{align*} On the other hand, the 6 components of the Ricci tensor are \begin{align*} \newcommand{\Ric}{\mathrm{Ric}} \newcommand{\Ric}{\mathrm{Ric}} \Ric_{11} &= R_{1212} + R_{1313} = Q_{33} + Q_{22}\\ \Ric_{12} &= R_{1323} = -Q_{21}\\ \Ric_{13} &= R_{1232} = -Q_{31}\\ \Ric_{22} &= R_{2121} + R_{2323} = Q_{33} + Q_{11}\\ \Ric_{23} &= R_{2131} = -Q_{32}\\ \Ric_{33} &= R_{3131} + R_{3232} = Q_{22} + Q_{11} \end{align*} From this, you can see that \begin{align*} \newcommand{\tr}{\operatorname{tr}} \Ric_{ij} &= -Q_{ij} + (\tr Q)\delta_{ij}\\ \end{align*} Taking the trace of this, we get $$ \tr \Ric = 2\tr Q $$ Therefore, \begin{align*} Q_{ij} &= - \Ric_{ij}+ (\tr Q)\delta_{ij}\\ &= -\Ric_{ij} + \frac{1}{2}(\tr \Ric)\delta_{ij}. \end{align*} Therefore, the Riemann curvature tensor is uniquely determined by the Ricci tensor. Moreover, if the metric is Einstein, then $$ \Ric = \frac{1}{3}(\tr \Ric)\delta = \lambda \delta $$ and $$ Q = \frac{1}{6}(\tr \Ric)\delta $$ and, by the contracted first Bianchi identity, $\tr\Ric$ is constant. From this, the sectional curvatures satisfy, if $(i,j,k)$ is a cyclic permutation of $(123)$, $$ K(e_i,e_j) = R_{ijij} = Q_{kk} = \frac{1}{6}\tr \Ric = \frac{1}{2}\lambda. $$

$Q$ is an interesting tensor. It looks like minus the Einstein tensor. Unlike the Ricci tensor, it is not transversally elliptic, and therefore, the analog of the Ricci flow but using $Q$ or $-Q$ instead does not give a geometric heat flow.

  • $\begingroup$ Thank you for the reply. Unfortunately I may not have the same background, my knowledge of the topic is very limited, so I cannot fully understand the reply. It would help to know what is the tensor Q you mention. $\endgroup$
    – xyz
    Commented Apr 13, 2021 at 0:16
  • $\begingroup$ I define the tensor $Q$ in my answer. Using your notation, the components of the Riemann tensor can be written as $\mathrm{Rm}(e_i\wedge e_j, e_k\wedge e_l)$. The tensor $Q$ is defined to be, using your notation, $$Q_{ij} = \mathrm{Rm}(\phi e_i, \phi e_j)$$ The first set of displayed equations in my answer lists this for all possible values of $i$ and $j$. $\endgroup$
    – Deane
    Commented Apr 13, 2021 at 1:02
  • $\begingroup$ Sorry but I still do not quite understand: isn't $Rm$ (I denote it as simply $R$) defined as a (0,4)-tensor? Shouldn t it have 4 tangent vectors as arguments? I kept working on the problem, and managed to prove, using cyclic permutations, maybe equivalently to what you did, that $\phi^{-1}\circ\mathcal{R}\circ\phi=-r+\frac{1}{2}SI_3$, where $S$ is the scalar curvature, $I_3$ the identity matrix. How does one move on from there, assuming the metric is Einstein? $\mathcal{R}(x\wedge y)=R(x,y,e_2,e_3)e_2 \wedge e_3+R(x,y,e_3,e_1)e_3 \wedge e_1+R(x,y,e_1,e_2)e_1 \wedge e_2$. $\endgroup$
    – xyz
    Commented Apr 13, 2021 at 11:24
  • $\begingroup$ Yes. My $Q$ is your $-r + \frac{1}{2}I_3$, where $r$ is the Ricci tensor. So you already know what my answer states: The Riemann tensor is completely determined by the Ricci tensor. Now suppose the metric is Einstein. Then the scalar curvature is constant and therefore $r = cI_3$ and $S = 3c$, where $c$ is a constant. Therefore $-r+\frac{1}{2}SI_3 = \frac{1}{2}cI_3$. This implies that the Riemann curvature tensor has constant sectional curvature. In particular, given $i \ne j$, $R(e_i\wedge e_j, e_i\wedge e_j) = c$, and, if $k \ne i, j$, then $R(e_i\wedge e_j,e_i\wedge e_k) = 0$. $\endgroup$
    – Deane
    Commented Apr 13, 2021 at 14:32
  • 1
    $\begingroup$ Yes that is exactly what I meant, the whole $\phi$ and $\mathcal{R}$ business was to prove the fact that the Riemann tensor was determined by the Ricci tensor; the last step, of proving the curvature constant is much more trivial per se, but the exercise was asking to go through these constructions, which was not so trivial, notationally speaking. Thank you so much you really did help me understand the different notations in the subject as well. $\endgroup$
    – xyz
    Commented Apr 14, 2021 at 17:25

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