I would like to ask the following : Is there any example of a compact conformally flat Riemannian manifold $(M^n,g)$ with $n\geq 4$ which is not flat and has zero scalar curvature?

3$\begingroup$ Do you want an explicit example or a proof that one exists? I think it is likely that there are such $(M^4,g)$, but I don't know how to make an explicit example. The idea for existence is to make a family of compact conformally flat $4$manifolds by taking conformally flat connected sums of compact space forms, some with Yamabe energy $>0$ and some with Yamabe energy $<0$. Varying the parameters in the connected sum should make the Yamabe energy vary, so it should be possible to make one with zero Yamabe energy. Then, by Schoen, it will have a conformal metric with zero scalar curvature. $\endgroup$– Robert BryantJan 13 '16 at 0:29
Take a unit sphere $S^2$ and a hyperbolic surface $X$. Then the product $S^ 2 \times X$ is not flat and has zero scalar curvature. Also it is conformally flat by a paper
Simon Salamon (2009) Complex structures and conformal geometry. In: BOLLETTINO DELLA UNIONE MATEMATICA ITALIANA, vol. 9, pp. 199224
as per editor Holonomia.

2$\begingroup$ @ katz: Just a doubt: your product $S^2 \times X$ is conformally flat as the OP asked ? it is wellknown that $S^2 \times H^2$ is conformally flat where $H^2$ unit disk with the hyperbolic metric i.e. the universal covering of X. So $S^2 \times X$ is locally conformally flat. It is obvious that a multiple of the product metric is flat?. $\endgroup$ Jan 12 '16 at 18:57

1$\begingroup$ @katz: Indeed, your local argument cannot be correct. In dimensions $n>2$, there are many conformally flat metrics with vanishing scalar curvature that are not flat. Essentially, they depend on $2$ arbitrary functions of $n{}1$ variables, and the generic one does not have constant sectional curvature. $\endgroup$ Jan 12 '16 at 20:45

$\begingroup$ In dimensions 4 or higher, conformally flat means the Weyl tensor vanishes. If the scalar curvature also vanishes, there is still the possibility that the tracefree part of the Ricci curvature is nonvanishing. As Robert points out, there are indeed local metrics satisfying this. $\endgroup$ Jan 13 '16 at 0:12

Responding to the OP's comments on the duplicate question, the example that was given in dimension $4$ can be easily generalized to higher dimensions:
If $(M,g_M)$ has constant sectional curvature $+1$ and $(N,g_N)$ has constant sectional curvature $1$, then the product manifold $(M\times N,g_M\oplus g_N)$ is conformally flat. This is proved, e.g., in Besse's book "Einstein manifolds" (see Example 1.167, p. 61). Furthermore, this product manifold has vanishing scalar curvature if and only if $\dim M=\dim N$.


$\begingroup$ @katz: No, I actually could not find a copy online... Can you share one? $\endgroup$ Jan 18 '16 at 15:38

$\begingroup$ Renatto, I didn't read the paper but Bryant seems to think the result is there. What I was wondering about is the much earlier date of Besse. $\endgroup$ Jan 18 '16 at 15:42

$\begingroup$ @RenatoG.Bettiol thanks! I couldnt comment since i was new and i had no reputaion.. $\endgroup$– ChristosFeb 3 '16 at 10:20
It seems to me that the answer is NO. Namely, if $(M,g)$ is compact, conformally flat and has zero scalar curvature then $g$ is a flat metric. Indeed, by the hypothesis there is a smooth function $f:M \to \mathbb R$ such that $e^f g$ is a flat metric on $M$. Then a finite covering of $M$ is a torus. The metric $g$ then lift to such a torus and by the GromovLawson theorem (see Corollary A in http://www.ihes.fr/~gromov/PDF/8%5B26%5D.pdf) the lifted metric is flat hence the original $g$ is flat. QED
As Robert Bryant observed I made a mistake assuming that $e^f g$ is a flat metric on $M$. Indeed, conformally flat means that $e^f g$ has constant sectional curvatures so the sectional curvature can be also 1 or 1. Watching the formula for the scalar curvature of a conformal change
https://en.wikipedia.org/wiki/List_of_formulas_in_Riemannian_geometry
it seems that the problem reduce to the question if the number $(n2)/4(n1)$ is an eigenvalue of the Laplacian of a compact hyperbolic manifold of dimension $n$.
The same formula allows to rule out the flat case without using GromovLawson theorem. Indeed, if $e^f g$ is flat and $g$ has zero scalar curvature then $e^{n2}f$ is harmonic hence constant since $M$ is compact. So $g$ is flat.

3$\begingroup$ @Holonomia: Your argument is not right either, for, along the way, you argue that a compact conformally flat manifold carries a flat metric (i.e., you do not use the hypothesis that the scalar curvature is zero when you derive this). However, this statement is false for dimensions bigger than $2$: $S^1\times S^{n1}$ has a conformally flat metric but has no flat metric when $n>2$. $\endgroup$ Jan 12 '16 at 20:42

1$\begingroup$ @Holonomia: I'm afraid that you are still making a mistake. Having $g$ be conformally flat does not imply that there is an $f$ such that $e^fg$ has constant sectional curvature. Again the natural metric $g$ on $M = S^1\times S^{n1}$ is conformally flat, but, when $n>2$, there is no function $f$ on $M$ such that $e^f g$ has constant sectional curvature. $\endgroup$ Jan 12 '16 at 23:06

2$\begingroup$ @Holonomia: A metric $g$ on $M$ is conformally flat if each point $p\in M$ has an open neighborhood $U$ on which there exists a function $f$ such that $e^f g$ is flat on $U$. This condition really should be called 'locally conformally flat', but the common usage is to simply say 'conformally flat', because, in that way, conformally flat becomes something that one can test locally. Another criterion that is used when $n>3$ is to say that a metric $g$ is conformally flat if its Weyl tensor vanishes. (In dimension $3$, the condition is that its Cotton tensor vanishes.) $\endgroup$ Jan 13 '16 at 0:15

1$\begingroup$ @ThomasRichard: Indeed, you are correct. The product of the unit $2$sphere (with $K\equiv1$) with the hyperbolic metric (with $K\equiv1$) on any compact Riemann surface of genus $g\ge2$ gives a conformally flat metric with zero scalar curvature that is not flat. I forgot about that example when I was proposing a general method to construct families. $\endgroup$ Jan 13 '16 at 9:59

1$\begingroup$ @katz: I learn that $S^2 \times H^2$ is conformally flat time ago in the nice paper by Simon Salamon: Complex structures and conformal geometry. In: BOLLETTINO DELLA UNIONE MATEMATICA ITALIANA, vol. 9, pp. 199224. The issue is related to the twistor of $S^4$ and there are there is also a nice picture. $\endgroup$ Jan 13 '16 at 11:30