Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Surfaces without an explicit representation.

  1. May 2, 2013 #1
    Ok this question may be kinda stupid, but here goes.

    Do any surfaces exist for which a parametric form is possible, but cannot be described explicitly due to their highly irregular shape? (Or vice-versa)
     
  2. jcsd
  3. May 2, 2013 #2

    WannabeNewton

    User Avatar
    Science Advisor

    Let ##S\subseteq \mathbb{R}^{3}## be a regular surface. Then, by definition, for any ##p\in S## there exists a neighborhood ##V\cap S,V\subseteq \mathbb{R}^{3}## of ##p## in the subspace topology and a homeomorphism ##\varphi:U\subseteq \mathbb{R}^{2}\rightarrow V\cap S## such that ##D\varphi## has full rank; the pair ##(V\cap S,\varphi)## is called a local parametrization of ##S##. So yes, by definition there always exists a local parametrization of a regular surface. Note however that there need not always exist a single, global parametrization; the standard example is the 2-sphere ##S^{2}## which needs at least two local parametrizations to cover it fully.
     
  4. May 2, 2013 #3
    Suppose you have the surface given by the equation
    (x^5y^2+3x^3y^4z^2 - 5)^2 = 1.

    At most points, a parametrization of this surface exists in principle, but that does not mean you can explicitly find it. And if someone comes on here and solves this equation to prove me wrong, then simply add a few more terms, preferably involving non elementary functions so that you can't in fact solve this explicitly.

    This is kind of like the problem of finding antiderivatives. Every continuous function has an antiderivative, but you cannot necessary write it down in terms any more explicit than "the antiderivative of the original function".
     
  5. May 2, 2013 #4

    WannabeNewton

    User Avatar
    Science Advisor

    The surface you described is a regular value of ##f:\mathbb{R}^{3}\rightarrow \mathbb{R},(x,y,z) \mapsto (x^{5}y^{2} + 3x^{3}y^{4}z^{2}-5)^{2}##. Let ##p = (x_{0},y_{0},z_{0})\in f^{-1}(1)##. Consider the map ##F:U\rightarrow \mathbb{R}^{3}## given by ##F(x,y,z) = (x,y,f(x,y,z))##. Then, ##DF(p) = \begin{pmatrix}
    1 & 0 & 0\\
    0& 1 &0 \\
    \partial_{x}f|_{p}& \partial_{y}f|_{p} & \partial_{z}f|_{p}
    \end{pmatrix}##. If ##\nabla f## does not vanish identically on ##f^{-1}(1)## then this obviously has full rank and the inverse function theorem then guarantees that there exists a neighborhood ##V## of ##p## such that ##F:V\rightarrow F(V)## is a diffeomorphism. So as long as the gradient is nowhere vanishing, you have an explicit form for the parametrization by projecting onto the plane. All you have to do is check that the gradient doesn't vanish identically on the surface. This is a basic result from the classical theory of surfaces - see proposition 2 on page 59 of Do Carmo "Differential Geometry of Curves and Surfaces".
     
  6. May 3, 2013 #5
    Hi WannabeNewton. Your explanation is very clear, but I don't think you got the point of my comment. Most students don't consider existence theorems to qualify as "explicit". So while the implicit function theorem shows that x and y are local coordinates, it does not give an explicit representation of the parametrization in terms of x and y. To do that you must solve the original equation for z=z(x,y).

    Looking at my formula again, I see that this can be done, but as I said before, we can easily modify it to make it so the solution is not expressible in terms of elementary functions. So then the parametrization exists but cannot be written down explicitly. I realize that this answer depends on the subjective definition of "explicit", but I think one could find an example (or demonstrate the existence of an example), no matter how broadly you stretch the term. (Unless you go all the way and consider something guaranteed by an existence theorem to be explicit, which sort of breaks the word in my opinion)
     
  7. May 3, 2013 #6

    WannabeNewton

    User Avatar
    Science Advisor

    Oh I see what you mean by explicit! So like an actual formula for the function given by the implicit function theorem; yes that would be troublesome in general. I was interpreting explicit as simply the existence of, as guaranteed by the things mentioned above.
     
  8. May 4, 2013 #7
    NOW we're talking ;)

    I'm sorry I'm still a student, so most of what you said was quite hard to take in, WannabeNewton (awesome name btw)

    Thanks though, and thank you Vargo :)
     
  9. May 4, 2013 #8

    WannabeNewton

    User Avatar
    Science Advisor

    Hi james (and thanks hehe)! Yes in general we do have to rely on existence theorems as explicit constructions might not always be apparent, as Vargo noted above very clearly. It's a very nice question :)
     
  10. May 4, 2013 #9

    lavinia

    User Avatar
    Science Advisor
    Gold Member

    One can ask a related question which is whether one can find a parameterization of a neighborhood of a surface in 3 space that has certain special properties.

    A classical theorem - see Struik's book on classical differential geometry - says that any linear PDE of the form

    Af[itex]_{xx}[/itex] + Bf[itex]_{xy}[/itex] +Cf[itex]_{yy}[/itex] + Df[itex]_{x}[/itex] + Ef[itex]_{y}[/itex] = 0

    where A,B,C,D,and E are arbitrary functions ( continuously differentiable?) of the parameters

    can be transformed into the equations for either conjugate or asymptotic coordinates on a surface. Three independent solutions to these transformed equations give the surface's coordinates in three space.

    So finding these coordinate systems on all surfaces is equivalent to solving all possible linear PDE's of the above form.
     
    Last edited: May 4, 2013
  11. May 8, 2013 #10
    If we care only up to diffeomorphism then yes, since every surface has a "pants decomposition" and presumably there is an explicit parametrization of a pair of pants (though I don't know how to write one).

    If we care about isometry (i.e. for every embedding into Euclidean space) then certainly not, since the set of such embeddings is uncountable, and the set of things we can write down is not.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Surfaces without an explicit representation.
  1. Adjoint representation (Replies: 1)

Loading...