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Study article R. C. Reilly is entitled Applications of Hessian operator in the Riemann manifold had a doubt in the remark, shortly after the theorem 2 of that Article.

The theorem is stated as: Suppose that $M^{n}$ is compact and $N^{n-1}=\partial M$ is empty. If $f:M \to \mathbb{R}$ is a smooth function, then $S_{2}(f)$ cannot be constant unless it vanishes. Moreover if $S_{2}(f)=0$ and $Ric$ is definite on $M$ then $f$ is a constant function., where $M^{n}$, $N^{n-1}$ denote smooth connected, oriented Riemannian manifolds of dimensions $n$ and $n-1$ (respectively), $S_{2}(f)$ is 2-nd invariant of Hessian of a function $f:M \to \mathbb{R}$. And the remark is as follows: If $Ric$ is not definite there may exist nonconstant $f$ such that $S_{2}(f)=0$. The flat tori provide us with numerous simple samples.

In this observation, it refers to the fact that if $Ric$ iis not defined, there are numerous functions that give example that the theorem is not worth.

In my case, I could not get any of these examples, but would like to count on the help of you to give me an idea or an indication of a book which is the subject better.

Already, thank you for attention.

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  • $\begingroup$ The sentence you quoted says "If Ric is not definite," not "if Ric is not defined." There's a big difference. $\endgroup$ – Jack Lee May 17 '16 at 17:42
  • $\begingroup$ In the text, the term is "not definite." I understand that to be in the sense that there is no sign at $ Ric $. In my view, definite means $ Ric \ge $ 0. $\endgroup$ – user98236 May 17 '16 at 17:46
  • $\begingroup$ I was referring to your third paragraph, where you wrote "if Ric is not defined." Usually, a bilinear form is called definite if it is either positive definite or negative definite. A flat torus has $\operatorname{Ric}\equiv 0$, which is neither. $\endgroup$ – Jack Lee May 17 '16 at 17:49
  • $\begingroup$ OK. Indeed, it is if Ric is not definite. $\endgroup$ – user98236 May 18 '16 at 11:59

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