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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 $Ric$ is not definite, 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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On the torus $T^2$ with the coordinates $x,y$ and the flat metric $g= dx^2 + dy^2$ take any function $f(x)$. Its hessian is given, after raising the index, by the (1,1)-tensor $f''(x) dx\otimes \frac{\partial } {\partial x}$ so the operator $S_2(f)$ (which in dimension two is simply the determinant of this tensor) is zero though the function is not always zero.

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  • $\begingroup$ Vladimir, thank you for the clarification of the doubt. $\endgroup$ Commented May 18, 2016 at 11:58

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