# what are Ricci curvatures?

What are Ricci curvatures (in the plural)?

I know the definition of the Ricci tensor ($Ric_{jk} = R_{ijki}$ where $R$ is the Riemann curvature tensor), but I've doing a problem which asks for definitions of the Ricci tensor and then the sectional and Ricci curvatures. It's not in the books by do Carmo, Lee, or Guillemin & Pollack.

More exactly, the problem says:

'Define what is meant by the Ricci tensor on a Riemannian manifold M, proving that it is a rank-2 symmetric tensor. Define the sectional curvatures and the Ricci curvatures. When dim$(M)=3$, show that if the Ricci curvatures are constant at some point, then so too are the sectional curvatures. Show that there is a Riemannian metric on $S^2\times S^2$ for which the Ricci curvatures are constant but the sectional curvatures are not.'

Many thanks for any help with this!

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Where is the problem from? Maybe the definition at the start of the "Properties" section here is relevant: en.wikipedia.org/wiki/Ricci_curvature#Properties – Henry T. Horton Jan 10 '13 at 22:03
I'm pretty sure that "the scalar and Ricci curvatures" means "the scalar curvature and the Ricci curvature". It does not mean there are multiple Ricci curvatures. Pluralisation is a frequent source of ambiguity in mathematical writing. – user53153 Jan 10 '13 at 22:33
@HenryT.Horton: the problem is from a master's level exam paper at Cambridge. Does the Wikipedia article suggest that 'Ricci curvatures' mean all possible values of $Ric(v,v)$ (cf. sectional curvatures)?? – Harry Macpherson Jan 11 '13 at 9:40
@HarryMacpherson Yes. – Willie Wong Jan 11 '13 at 9:41
@PavelM: sorry, I meant sectional curvatures, not scalar! I've corrected it now, and also added the exact words of the question. Unfortunately your explanation doesn't work! – Harry Macpherson Jan 11 '13 at 9:42

To give an answer for the sake of it being here.

The Ricci tensor is defined as usual, as the trace of the Riemann curvature tensor.

Some authors use Ricci curvature to denote the function on the unit tangent bundle: $$UTM \ni v \mapsto \mathrm{Ric}(v,v)\in \mathbb{R}$$

The values of this function is sometimes called the Ricci curvatures.

Constant Ricci curvature at a point $p$ means that for all $v\in UT_pM$ we have $\mathrm{Ric}(v,v) = c$ for some constant $c$. In three dimensions because the Weyl curvature vanishes identically the Riemann curvature tensor is uniquely determined by the Ricci tensor, and the Ricci tensor being symmetric is uniquely determined by the Ricci curvature function. Hence you can in principle write the Riemann curvature tensor in terms of the Ricci curvature (by polarisation) and show that the sectional curvatures are constant.

Or you can use that in $\mathbb{R}^3$ the space of two-dimensional subspaces is itself three dimensional, so one can actually solve a linear system of equations to write sectional curvatures in terms of Ricci curvatures.

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