Example of a Manifold which has One Non-zero Component of Ric which equals the Scalar Curvature

Question

I am wondering if there is a simple example of a manifold such that, given a value for the scalar curvature $R$, I can find a manifold such that the Ricci tensor has all zero components except for one component which takes the value $R$.

I know this could be achieved using a warped product of two metrics to separate out one coordinate and then solve the differential equations so that the first coefficient vanishes, but obviously the coefficient of the non-zero component needs to be $R$.

Answer 1

Here's a reasonably explicit construction, similar to the one in my answer to your related question, that works in some cases, namely when $R$ is strictly positive and we work in a neighborhood of any point not critical for $R$. The latter condition guarantees that there are coordinates $(x_1, \ldots, x_{n - 1}, y)$ for which $R$ is a function of $y$ alone.

If we take $\bar g := \sum_{i = 1}^{n - 1} dx^i \otimes dx^i$ (the standard Euclidean metric on $\Bbb R^{n - 1}$) and $dy^2$ to be the standard metric on $\Bbb R$, then the Ricci tensor of $$g := F(y) \bar g + G(y) dy^2$$ is $$\operatorname{Ric} = \frac{2 F'' F G - F' G' F + (n - 3) (F')^2 G}{4 F G^2} \bar g - \frac{3 (2 F'' - (F')^2 G - F' G' F)}{F^2 G} dy^2 .$$

We want this to coincide with $R \,dy^2$ for some scalar function $R$, which amounts to asking (1) that the coefficient of $\bar g$ vanish and (2) the coefficient of $dy^2$ be our prescribed function $R$.

This is a nonlinear system of two o.d.e.s in unknown functions $F, G$, but we can still solve it as explicitly as one might hope. The solutions are given by $$F(y) = C \exp \left[\pm \frac{2}{\sqrt{(n - 1) (n - 2)}} \int_{y_0}^y \sqrt{R(t)} \,dt\right], \quad G(y) = C F'(y)^2 F(y)^{n - 3}, \quad C \in \Bbb R,$$ and this defines a metric if $C > 0$.

Answer 2

Since you asked for a simple example, I was curious about what happens for a 2D manifold. Let's try to fix the Ricci scalar $R$ to be some value $S$. (Note that it is twice the Gaussian curvature in this case $K$, i.e. $R = 2K$).

In 2D, we have (see: [1], [2], [3]): \begin{align} R_{abcd} &= \frac{R}{2}(g_{ac}g_{bd} - g_{ad}g_{bc}) \\ R_{\alpha\gamma} &= g^{mn} R_{n\alpha m\gamma} \\ &= g^{mn}\frac{R}{2}(g_{ac}g_{bd} - g_{ad}g_{bc}) \\ &= \frac{R}{2}(2g_{\alpha\gamma} - \delta^m_\gamma g_{\alpha m}) \\ &= Kg_{\alpha\gamma} \\ R &= \text{tr}_g (\text{Ric})=g^{ij}R_{ij} = 2K \end{align} You are requiring two constraints:

1. $S = R$

2. Every component of $R_{ij}$ is zero except one.

Notice that in this case you are requiring every entry of the metric tensor to be zero except one entry (which must be on the diagonal, since $g$ is symmetric) that must equal $1$. Such a metric is not invertible (nor positive definite)!

I guess in 2D, asking for this is too harsh of a constraint on the metric tensor.