We know that, if we have a surface $z=f(x,y)$ with Euclidean space being ambient manifold, the induced metric is as follows (in matrix form): $$g=\begin{bmatrix} 1+\left ( \frac{\partial f(x,y)}{\partial x} \right )^2 & \frac{\partial f(x,y)}{\partial x}\frac{\partial f(x,y)}{\partial y} & \\\ \frac{\partial f(x,y)}{\partial x}\frac{\partial f(x,y)}{\partial y} & 1+\left ( \frac{\partial f(x,y)}{\partial y} \right )^2 & \end{bmatrix}$$

Now, consider a problem of prescribing an induced metric, i.e. assume we have a metric

$$h=\begin{bmatrix} a(x,y) & b(x,y)\\\ b(x,y) & c(x,y) \end{bmatrix}$$

The question is, when metric $h$ is an induced metric, i.e. when there exist a surface $z=f(x,y)$ for which an induced metric is $h$?

An example would be a metric $$h=\begin{bmatrix} 2 & 1\\\ 1 & 2 \end{bmatrix}$$

for which there exist a surface $z=-(x+y)$ with an induced metric being $$g=\begin{bmatrix} 2 & 1\\\ 1 & 2 \end{bmatrix}$$

An obvious obstruction is that the following identity has to be satisfied by the givien metric: $$a(x,y)c(x,y)-a(x,y)-c(x,y)+1=b^2(x,y).$$ This identity follows from requirement that $$\frac{\partial f(x,y)}{\partial x}\frac{\partial f(x,y)}{\partial y}=b(x,y).$$

Ths problem occured when toying with several examples of positive definite Hessians (with minus sign), i.e. when (minus) Hessian for some other surface $z=\phi(x,y)$ is positive definite. For example, if $z=-(x^2+xy+y^2)$, then the minus Hessian is $$h=\begin{bmatrix} 2 & 1\\\ 1 & 2 \end{bmatrix}$$ This matrix can be taken as a first fundamental form of the surface $z=-(x+y)$, which has a zero Gaussian curvature. It seems that often when negative Hessian is taken as a first fundamental form, Gaussian curvature (for a surface with induced metric being equal to this negative Hessian) is equal to zero. I'm not claiming that this is a general phenomena, it's just an interesting observation (it is important from the point of view of optimization as Hessian often play a significant role there).

I looked for litereture on "prescribed metric" problem, but found none. Probably I wasn't looking in the right place. A link to a paper, of a monograph considering this problem would be appreciated.

I'm familiar with the problem of prescribed curvature. Probably one possible way would be to look for prescribed Ricci curvature tensor problem. But at this stage I think it would be an overkill. And maybe the class of possible solutions (taking into consideration the obstructions) is way too small and simple to be of value: it looks like negative Hessians for quadratic surfaces might a class of solutions.


Assume $b\ne0$.

Then you know $\alpha=\tfrac{\partial f}{\partial x}$ and $\beta=\tfrac{\partial f}{\partial y}$ up to sign. You should check $$\frac{\partial \alpha}{\partial y}=\frac{\partial \beta}{\partial x}.$$ If this is true you can restore $f$ by integrating.

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  • $\begingroup$ Yes, thanks, but I know this. This is a very easy part. But there is additional condition on the metric that we want to prescribe. Namely, $\alpha\beta = b$. $\endgroup$ – Tomas Nov 23 '15 at 9:41
  • $\begingroup$ @Tomas: Once you know a, b, and c you know α and β up to sign. In particular αβ=b. $\endgroup$ – Anton Petrunin Nov 23 '15 at 10:12
  • $\begingroup$ So, $\alpha = \pm \sqrt{(a-1)}$, and $\beta = \pm \sqrt{(c-1)}$. Then there is a condition that $\alpha \beta = \sqrt{(a-1)(c-1)}=b$. In other words, if $ac-a-c+1=b^2$, then I can restore the surface $z=f(x,y)$. If this condition is not satisfied, then a given matrix $h$ there is no surface (at least not with Euclidean space as ambiant manifold) with $h$ as an induced metric $\endgroup$ – Tomas Nov 23 '15 at 12:45
  • $\begingroup$ @Tomas, almost right --- in addition you need to check that $\tfrac{\partial \alpha}{\partial y}=\tfrac{\partial \beta}{\partial x}$ and if yes then yes. $\endgroup$ – Anton Petrunin Nov 23 '15 at 13:14

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