If $ ||Ric||^{2}=\frac{S^{2}}{n} $can be concluded $ Ric=\frac{S}{n} $? where Ric is Ricci tensor and S is scalar curvature.
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1I think it is wrong. $Ric$ is tensor, how tensor equal to scalar ? – Enhao Lan Aug 05 '17 at 12:43
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1By "$Ric = \frac{S}{n}$", do you mean $Ric = \frac{S}{n} g$? In any case (assuming the norm is the pointwise norm, not the $L^{2}$ norm), I suspect "no": The type of counterexample I'd seek is a product of Einstein metrics of differing Einstein constants. – Andrew D. Hwang Aug 05 '17 at 13:10
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I'm sorry for the typo mistake. It should be Ric=S/n g – mirahmad mirshafeazadeh Aug 06 '17 at 05:10