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Differential geometry hypersurface problem - need help starting

  1. Oct 12, 2007 #1
    differential geometry hypersurface problem - need help starting!!

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

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

    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 [tex]\partial[/tex]x, [tex]\partial[/tex]y, [tex]\partial[/tex]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[tex]_{F(x)}[/tex]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. jcsd
  3. Oct 12, 2007 #2


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    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 [itex]\vec{i}+ f_x\vec{k}[/itex] or [tex]\vec{i}+ f_y\vec{k}[/itex].

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