# Homework Help: Differential geometry hypersurface problem - need help starting

1. Oct 12, 2007

### quasar_4

differential geometry hypersurface problem - need help starting!!

1. The problem statement, all variables and given/known data[/b

Let $$f\in C^{\infty}$$$$(R\R^{2})$$ and let S be the set of points in $$R\R^{3}$$ given the graph of f. Thus, $$s={{(x,y,z=f(x,y))|(x,y)\in$$$$R\R^{2}}$$.

a) Show that this set of points can be viewed as a regular level surface.

b) Let X=(x,y,z) be a point on this surface. Find a basis for the tangent space TxS.

c) Give a cover for this surface.

2. Relevant equations

Ehm. Not really "equations", per se. We will need the coordinate basis for R3 which is $$\partial$$x, $$\partial$$y, $$\partial$$z.

3. The attempt at a solution

I am soo, so confused. I guess our F=f(x,y). To show that something is a regular level surface, I believe that we have to show that not all partial derivatives vanish at the point x on the surface. If we had an actual f(x,y) I would compute F*, the differential, and show that it is not simultaneously 0 at some point x on the surface. But, we don't have an explicit f(x,y). So I have no idea how to show this, or to show equivalently, the the mapping from TxU (for our open set U) to T$$_{F(x)}$$S is surjective.

I am TOTALLY lost as to how to come up with a basis. It seems that we should compute the kernel of F* and its basis, but I'm not clear on what we do with this basis.

If I could figure out parts a and b, I MIGHT be able to figure out part c. I know that a cover is the union of surface patches. But then I'd have to start by making coordinate surface patches first... would I be defining some map or something??

I'm terribly confused. :yuck: If anyone can help that would be marvelous!

2. Oct 12, 2007

### HallsofIvy

No. A "level curve" for a function f(x,y) would be a curve in the xy-plane. In order that a surface in 3 dimensions be a level surface, it must be of the form F(x,y,z)= constant. In this case, given z= f(x,y), you know that z-f(x,y)= 0.

What can you say about the partial derivatives of z-f(x,y)?

Given that the surface is defined by z= f(x,y), any tangent vector is of the form $\vec{i}+ f_x\vec{k}$ or [tex]\vec{i}+ f_y\vec{k}[/itex].