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First, i am the one who asked about the existence of compact manifolds of dimension $n\geq 4$ which are conformally flat, non-flat, with zero scalar. Due to the fact that i couldn't comment because i was unregistered i had to write down a "new" question. First, i have to thank you all and especially @Robert Bryant for providing a method of existence of such manifolds of dimension $4$ and also the explicit examples $\tilde{M}^4=\mathbb{S}^2\times M^2$, where $Μ^2$ is a compact Riemann surface of genus $g\geq 2$ with a constant hyperbolic metric $(K\equiv-1)$. Maybe its trivial but i want to ask the following: Can i have such examples in dimensions greater that $4$? I just want to know, in the general case, how large the class of such manifolds really is!

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    $\begingroup$ As noted in one of the comments there, the same method (product of sphere and hyperbolic space) works to give an example in each even dimension. $\endgroup$
    – Mikhail Katz
    Commented Jan 13, 2016 at 15:14
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    $\begingroup$ A link to the previous question is here mathoverflow.net/questions/228254/… $\endgroup$
    – j.c.
    Commented Jan 13, 2016 at 15:19
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    $\begingroup$ @Christos, in all dimensions you have compact quotients of the hyperbolic space. Conformal flatness is a local property. $\endgroup$
    – Mikhail Katz
    Commented Jan 13, 2016 at 15:23
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    $\begingroup$ @Christos: Well, the idea I outlined in the comment to the first question, i.e., taking connected sums of locally homogeneous examples with positive and negative Yamabe energy to approximately balance out the Yamabe energy and then using the parameters in the conformal connect sum operation to perturb the Yambe energy to zero would probably work in any dimension if it works in dimension $4$, so I suspect that the answer is that there is a positive dimensional moduli space in each dimension greater than or equal to $3$. However, a rigorous proof should be written down and checked to be sure. $\endgroup$
    – Robert Bryant
    Commented Jan 13, 2016 at 16:37
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    $\begingroup$ @Christos: I included an answer to your question regarding higher dimensional examples in mathoverflow.net/questions/228254/… $\endgroup$
    – Renato G. Bettiol
    Commented Jan 18, 2016 at 5:27

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