Suppose that we have two functions f and h on a riemannian manifold M. If Hessf =Hessh can we get a relation between both functions.
Without any more information about M,f or g, I'm pretty sure the best you can expect is that (f-g) is in the kernel of Hess. You could get an estimate on the size of (f-g) with a Poincare inequality or something (?). If you have boundary conditions then you could uniquely solve Hess(f-g)=0 for (f-g).
I'm not sure this is what you're looking for though (?)
Thank you Stephen for your time and concern.
What i really want is a relation like h=h(f). For example h=f+c or h=f+linear factors, ....
If the manifold is Rn, and the functions are smooth, then Hess f = Hess g means that Hess(f-g) = 0 so grad(f-g) = constant. That is, f & g differ by a linear function. I'm not quite sure what this translates to for a general smooth Riemannian manifold.
Thank you Stewart, now I have an answer in Rn . The problem is still open in smooth manifolds.
To answer @David Steward's question, on a Riemannian manifold
Hess(f - g) = \nabla (d (f-g) ) = 0
means \theta = d(f-g) is a parallel 1-form, of which there typically are none other than the 0 form. Hence in the generic case d(f-g) = 0, and f and g differ by a locally constant function (i.e. a constant for every connected component).
However since \theta is also exact it is even more restricted than just being parallel.
Since \nabla theta = 0 we have
d^* \theta = div theta = g^{ab}\theta_{a;b} = g^{ab} (f-g)_{;a;b} = 0
i.e. f-g is harmonic. Moreover: by the Weitzenböck formula
0 = \nabla^*\nabla \theta = (d^*d + dd^* +Ric)(\theta) = Ric(\theta)
This shows once more that for generic Riemannian manifold (where Ric is non degenerate) \theta = 0.
Thank you dear Rogier, would you kindly give me a link or a simple reference check this my self.
The Hessian is the covariant derivative of the gradient. Then the gradient, $gr(f-g)$, of $(f-g)$ is a parallel vector field. If it is zero, then f and h differ by a constant. If not, by de Rhan decomposition theorem $M$ splits (locally or globally if $M$ is complete and simply connected) a line, i.e. $M= N\times \mathbb R$ (locally or globally). In this way $grad (f-g)$ is a constant field in the $\mathbb R$-factor of $M$. So $f$ differs from $g$ in a function wich is a multiple of the projection to the line factor $\mathbb R$.
Thus, if $M= \mathbb R ^k\times M'$, where $\mathbb R ^k$ is the Euclidean de Rham factor of $M$, $f-g = T\circ \pi_0$, where $T:\mathbb R ^k \to \mathbb R$ is affine and
$\pi _0$ is the projection from $M$ to the Euclidean de Rham factor $\mathbb R ^k$.
take the coordinates x and y on the plane. Both have a vanishing Hessian !
as explainded above, f-g is constant in the generic case
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