# Surface with prescribed metric

1. Jan 17, 2007

### KarstenKarsten

Consider a Riemannian metric in R^2, e.g. consider g given at a point (x,y) by the matrix
(1 x
x 1+x^2 )

Is there a surface S, embedded in R^3, which has the property that the metric on S which is induced by the Euclidean metric of R^3 coincides with the given metric g? If yes, what representation does S possess?

2. Jan 17, 2007

### Chris Hillman

This isn't HW, is it?

Hi, KarstenKarsten,

Sounds like you are asking for an embedded surface in E^3 which has the stated metric. A good keyword to look up in a good differential geometry textbook should be "Monge patch" or "embedding". In any case, the idea is simple: write down an undetermined basis (two vectors in E^3) for the tangent space to some point of S, form their E^3 inner product, and compare with the desired induced metric on S. See "Coordinate tutorial" at http://www.math.ucr.edu/home/baez/RelWWW/group.html [Broken] for some radially symmetric examples.

Last edited by a moderator: May 2, 2017
3. Jan 17, 2007

### gvk

It looks like it is not embedded surface in E^3 (3D Euclidian space). If the surface is given to us in the form z = f(x, y), then the metric induced on the surface is

$$g_{1,1} = 1 +f_{x}^{2}, g_{1,2}=g_{2,1} = f_{x} f_{y}, g_{2,2} = 1 + f_{y}^{2}$$.

In Karsten case
$$g_{1,1} = 1 , g_{1,2}=g_{2,1} = x, g_{2,2} = 1 + x^{2}$$.

That means $$f_{x} =0$$ and should be $$g_{1,2}=g_{2,1} = 0$$.
Hence, the surface is embedded in some noneucleadian 3D space.

4. Jan 18, 2007

### KarstenKarsten

Thanks!

I realize now that, at least, one can find a conformally equivalent metric
g~_{ij} = c(x) g_{ij}
with c(x) the solution of c(x)^2-c(x)(2+x^2) +1=0. And with g~ be metric of a in E^3 embedded surface. The components of g~ can then also be computed (and are slightly more complicated terms).

5. Jan 19, 2007

### Chris Hillman

Right, you don't need a noneuclidean embedding space.

6. Jan 19, 2007

### KarstenKarsten

For me, it would somehow be counterintuitive that such an embedded surface should not exist.
But can I obtain from the surface with the conformally equivalent metric a surface which actually has the very same metric I started with? How?
(In the very beginning I started with the ansatz, as gvk proposed, of course...)

7. Jan 23, 2007

### gvk

The Riemannian tensor for your metric is $$R_{1,2,1,2}=-1$$. That means your embedded 2D manifold has constant negative curvature. Of course, you always can orthogonize your metric to get conformal presentation in isothermal coords.

Last edited: Jan 24, 2007
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