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$\text{Suppose we have} \nabla R = 0 $, where R represents curvature tensor, Prove that Ricci curvature and scalar curvature are constant.

Chen Jie
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  • What did you try? – Moishe Kohan May 03 '14 at 14:50
  • I write down the condition in component， I have on ideal how to tansform the covariant derivative？ the scalar curvature is easy since I could calculate the derivation and it is zero, and I got the conclusion, but for Ricci ,it doesn't work. – Chen Jie May 03 '14 at 14:57
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    The big picture is, $R$ is constant, hence so are its contractions. In detail: the covariant derivative fulfills the product rule of differentiation, and $g_{ij|k}=0$. This should get you further if you insist on looking at components. – ccorn May 03 '14 at 15:02
  • I will have a try at first, thank you. – Chen Jie May 03 '14 at 15:04
  • Does $Ric_{ij|h} = 0$ means $Ric = 0$？ can you prove it? – Chen Jie May 03 '14 at 15:18
  • $\operatorname{Ric}_{ij|h}=0$ colloquially means that the Ricci tensor is constant (which is to be shown, as the question asks). – ccorn May 03 '14 at 15:22
  • Wait, it seems I have led you from the right track here. The covariant derivative being zero is generally called parallel, not constant. You can (supposedly) figure out that the covariant derivatives vanish, but you cannot conclude constant from parallel? – ccorn May 03 '14 at 19:46
  • @ccorn: For scalar functions parallel and constant are equivalent, while for other tensors there is no real notion of "constant" other than parallel - I'm guessing the question meant to say "Ricci curvature is parallel". – Anthony Carapetis May 03 '14 at 22:24
  • yes， my problem is I cannot conclude constant from parallel, but the problem says clearly "constant"? – Chen Jie May 07 '14 at 11:32
  • What does $\nabla R$ mean? – rmdmc89 Jun 06 '19 at 17:54

1 Answers1

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I am assuming a Riemannian manifold, and I am going to use its Levi-Civita connection.

Contrary to my initial assumption, the constant in the problem statement is not just supposed to mean that the covariant derivative vanishes. And you already know that covariant differentiation and contraction commute. I apologize for having used such a simple-minded interpretation in the comments.

To fix the wording, a tensor field with vanishing covariant derivative is called parallel here, not constant. In fact, as Ted Shifrin has pointed out, any locally Riemannian symmetric space features a parallel Riemann curvature tensor field, but this does not make it a space of constant curvature.

We only require a parallel Riemann curvature tensor field, but seek to establish a constant Ricci tensor field and a constant scalar curvature. The latter is no problem, as for scalar fields, parallel is still the same as constant.

For the Ricci tensor field, interpreted as a family of bilinear forms, being constant in the strong sense means $$\exists\lambda\in\mathbb{R}: \forall p\in M: \forall X\in \operatorname{T}_p M: \operatorname{Ric}(X,X) = \lambda\langle X,X\rangle$$ That property characterizes an Einstein manifold.

So you are essentially asking whether every locally Riemannian symmetric space is an Einstein manifold. The answer is: No, but you seem to be close.

An easy counterexample to the proposition in question is $$M = S\times T\quad\text{where}\quad S=\mathbb{S}^2,\ T = \mathbb{R}$$ the Riemannian product of the unit sphere $S$ and a line $T$.

For $p\in M$, decompose $p=(s,t)$ where $s\in S$, $t\in T$. Likewise in the tangent space, the composition of $M$ allows you to assemble $X\in \operatorname{T}_p M$ as $X=(X_s,X_t)$ where $X_s\in \operatorname{T}_s S$ and $X_t\in \operatorname{T}_t T$. Furthermore, both $(X_s,0)$ and $(0,X_t)$ are in $\operatorname{T}_p M$, and we have $$\left\langle(X_s,X_t),(X_s,X_t)\right\rangle_p = \langle X_s,X_s\rangle_s + \langle X_t,X_t\rangle_t$$ where the inner products are those of $\operatorname{T}_p M$, $\operatorname{T}_s S$ and $\operatorname{T}_t T$ as indicated.

If you go through the calculations, you will find that $$\begin{align} \nabla R &= 0 \\ \operatorname{Ric}\left((X_s,0),(X_s,0)\right) &= \left\langle(X_s,0),(X_s,0)\right\rangle_p \neq 0 &\text{for }& X_s\neq 0 \\ \operatorname{Ric}\left((0,X_t),(0,X_t)\right) &= 0 \neq \left\langle(0,X_t),(0,X_t)\right\rangle_p &\text{for }& X_t\neq 0 \end{align}$$ which shows that $M$ is locally symmetric, but not an Einstein manifold.

Now you might suspect that the statement of your question was only a near miss, and you might want to additionally require some sort of irreducibility of the Riemannian manifold $M$. You may also want to require that $M$ be complete and simply connected to get a global symmetric space from the local one. Then look up e.g. the following:

  1. Chapter XI (symmetric spaces), theorem 8.6 + corollary 8.7 from: Shoshichi Kobayashi and Katsumi Nomizu: Foundations of differential geometry, Vol.2. Wiley 1969
  2. Section 2, corollary 3 from these Lecture notes on symmetric spaces by J.-H. Eschenburg

These concretize what irreducibility of $M$ shall mean and ultimately establish an Einstein manifold.

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