# Ricci-flat vs Riemann-flat

Why for the n-dimensional Riemannian manifold (with $$n=2$$ or $$3$$) the Ricci curvature tensor and the Riemann tensor are the same, while for $$n>3$$ not?

For example if I have 2-manifold (or 3-manifold) $$M$$ that is Ricci-flat, then it is also Riemann-flat, but if I have 4-manifold $$M$$ that is Ricci-flat, may be not Riemann-flat.

• @DietrichBurde Stronger condition? If the curvature tensor is $0$, then then in all dimensions the Ricci tensor vanishes; but the converse fails in dimensions $\ge 4$. To the OP, I wouldn't say the Ricci tensor is the same as the Riemann curvature tensor in dimension $3$, but it does completely determine it. Commented Jan 3, 2020 at 22:12
• It is unclear what you mean by "why": Are you asking for specific examples? Then it would be a duplicate of math.stackexchange.com/questions/953328/… Commented Jan 4, 2020 at 14:07

Here are the precise claims.

• When $$n=2$$, one has $$\operatorname{Ric}=\frac{1}{2}Rg$$ and $$R_{ijkl}=\frac{1}{2}R(g_{il}g_{jk}-g_{ik}g_{jl})$$
• When $$n=3$$, one has $$R_{ijkl}=g_{il}R_{jk}-g_{ik}R_{jl}-g_{jl}R_{ik}+g_{jk}R_{il}-\frac{1}{2}R(g_{il}g_{jk}-g_{ik}g_{jl})$$

So you can directly see that when $$n=2$$, zero scalar curvature implies zero Ricci curvature and zero Riemann curvature, and that zero Ricci curvature implies zero scalar curvature and hence zero Riemann curvature.

And when $$n=3$$, zero Ricci curvature implies zero Riemann curvature but zero scalar curvature does not necessarily imply zero Riemann curvature or zero Ricci curvature. A standard counterexample, written in a single coordinate chart, is the "Riemannian Schwarzschild manifold" $$\frac{dr^2}{1-\frac{2M}{r}}+r^2\,d\theta^2+r^2\sin^2\theta\,d\phi^2$$ where $$M$$ is any real number; it has zero scalar curvature but nonzero Ricci curvature and Riemann curvature.

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To prove the above formulas:

• when $$n=2$$, let $$e_1,e_2$$ be a $$g$$-orthonormal basis of $$T_pM$$. Then the definition of $$R$$ and then of $$\operatorname{Ric}$$ says $$R=\operatorname{Ric}(e_1,e_1)+\operatorname{Ric}(e_2,e_2)=\operatorname{Rm}(e_2,e_1,e_1,e_2)+\operatorname{Rm}(e_1,e_2,e_2,e_1)$$ so that $$R=2\operatorname{Rm}(e_1,e_2,e_2,e_1).$$ This directly shows that $$R_{ijkl}=\frac{1}{2}R(g_{il}g_{jk}-g_{ik}g_{jl})$$ holds when evaluated on $$(e_1,e_2,e_2,e_1)$$. By the simple symmetries of the Riemann tensor, it also holds when evaluated on $$(e_1,e_2,e_1,e_2)$$, on $$(e_2,e_1,e_1,e_2)$$, and on $$(e_2,e_1,e_2,e_1).$$ And both sides are trivially zero when evaluated on $$(e_i,e_j,e_k,e_k)$$ when $$i=j$$ or $$k=l$$. This covers all possibilities, so the given formula holds for any input. One trace of it gives $$\operatorname{Ric}=\frac{1}{2}Rg.$$

• The same sort of proof works when $$n=3$$ but is a bit more complicated. Let $$W_{ijkl}$$ denote the difference of the LHS and the proposed RHS; it is easy to check $$g^{il}W_{ijkl}=0.$$ Let $$e_1,e_2,e_3$$ be a $$g$$-orthonormal basis of $$T_pM$$ and evaluate $$g^{il}W_{ijkl}=0$$ on $$(e_1,e_1)$$; it says that $$W(e_1,e_1,e_1,e_1)+W(e_2,e_1,e_1,e_2)+W(e_3,e_1,e_1,e_3)=0.$$ The first term vanishes since the original LHS and RHS both vanish when evaluated on $$(e_1,e_1,e_1,e_1).$$ So $$W(e_2,e_1,e_1,e_2)=-W(e_3,e_1,e_1,e_3).$$ Repeating the same proof but starting from $$(e_2,e_2)$$ and $$(e_3,e_3)$$, we have $$W(e_1,e_2,e_2,e_1)=-W(e_3,e_2,e_2,e_3)$$ and $$W(e_1,e_3,e_3,e_1)=-W(e_2,e_3,e_3,e_2).$$ And the definition of $$W$$ shows easily that $$W(e_a,e_b,e_b,e_a)=W(e_b,e_a,e_a,e_b).$$ So (writing $$W_{abcd}$$ to abbreviate $$W(e_a,e_b,e_c,e_d)$$) there is $$W_{2112}=-W_{3113}=-W_{1331}=W_{2332}=W_{3223}=-W_{1221}=-W_{2112}.$$ So $$W_{2112}=0$$, and likewise $$W_{abba}=0$$ for any $$a$$ and $$b$$. It is easy to see from the definition of $$W$$ that $$W_{aabc}=0$$ and $$W_{bcaa}=0$$ for any $$a,b,c.$$ With a little thinking, since $$a,b,c$$ are only between $$1$$ and 3, the only possibly nonzero components are $$W_{abca}$$ where $$a,b,c$$ are all distinct. To see that these vanish, evaluate $$g^{il}W_{ijkl}=0$$ on $$(e_b,e_c)$$ to get $$W_{1bc1}+W_{2bc2}+W_{3bc3}=0.$$ Supposing, for instance, that $$(b,c)=(2,3)$$, this shows that $$W_{1231}=0.$$ If $$(b,c)=(1,3)$$, it shows that $$W_{2132}=0$$. And so on.

I learned the argument from page 276-277 of Hamilton's paper "Three-manifolds with positive Ricci curvature" but the proof probably goes back to the early 1900s.

• thank you very much for yours fantastic answer! While for $n>3$ what happens?
– user333046
Commented Jan 13, 2020 at 15:28

@youler's answer addresses the OP's question for $$n < 4$$, but I felt that something was missing that is crucial for understanding the answer to this question:

For example if I have 2-manifold (or 3-manifold) $$M$$ that is Ricci-flat, then it is also Riemann-flat, but if I have 4-manifold $$M$$ that is Ricci-flat, may be not Riemann-flat.

The answer is that the Riemann tensor can be decomposed, where the traceless component is the Weyl tensor. Generally, the Weyl curvature is the only component of curvature for Ricci-flat manifolds. The Ricci tensor measures how much an object's volume changes as it moves along a geodesic, and the Weyl tensor measures how much an object's shape changes as it moves along a geodesic. This is why (essentially) a necessary condition for a Riemannian manifold to be conformally flat is that the Weyl tensor vanishes.

When you have a Ricci-flat manifold that is also not Riemann-flat, it is because the Weyl tensor is nonzero. A classic example is the Schwarszchild metric in vacuum, which is Ricci-flat but is not Riemann-flat because the Weyl tensor is non-vanishing.