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

Definition of surface

  1. Oct 24, 2006 #1

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Definition of surface from Pressley's "Elementary differential geometry":

    A subset S of [itex]\mathbb{R}^3[/itex] is a surface if, for every point P in S, there is an open set U in [itex]\mathbb{R}^2[/itex] and an open set W in [itex]\mathbb{R}^3[/itex] containnign P such that [itex]S\cap W[/itex] is homeomorphic to U.

    I do not find it evident that this definition reproduces the intuitive notion of a surface. For instance, it is not obvious that for S a "solid" (such as a full sphere: {[itex](x,y,z)\in \mathbb{R}^3:x^2+y^2+z^2\leq R^2[/itex]}), we can't find a collection of homeomorphisms that cover S.

    Is there a result somewhere in mathematics that says something like that?
     
    Last edited: Oct 24, 2006
  2. jcsd
  3. Oct 24, 2006 #2

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    Any open set that intersects that ball will contain some subset homeomorphic to R^3, or the upper half of R^3. And neither of those is homeomorphic to an open set in R^2: they have different dehram cohomology if you want to be fancy, or removing one point from a disc creates something not simply connected, whereas a ball minus one point is still contractible.
     
  4. Oct 25, 2006 #3
    Well, a solid (3-dimensional object) in [itex]\mathbb{R}^3[/itex] is not a surface in [itex]\mathbb{R}^3[/itex]! If you'd take the sphere [itex]S^2=\{(x,y,z)\in \mathbb{R}^3:x^2+y^2+z^2=R^2\}[/itex] it's a different story.
     
  5. Oct 26, 2006 #4
    A solid sphere is not a surface. A surface is a 2D object. So it does not make sense to talk about things like the volume of a surface. When we talk about the 2D sphere, S^2, we mean the outer shell. Think of thing like hollow balls, hollow donuts, and deformed peices of paper. You can also have weird surfaces like the mobius strip.
     
  6. Oct 26, 2006 #5

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    so I can see the latex:

     
  7. Oct 27, 2006 #6

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    cliowa and mtiano,

    If a set of points in R^3 is a surface or not is determined by the definition we make of "surface", regardless of what our intuitive notion of a surface is. Of course, we will try to rigorously define what a surface is so that it matches what we intuitively regard as a surface, i.e. some kind of continuous 2D sheet in R^3.

    In particular, we want our definition to exclude certain things (sets) such as a ball, a single point and a curve. What I'm saying in the OP is that it is not evident that the above cited definition do in fact exclude such things. For it to exclude such things would mean that no collection of homeomorphism from opens of R² to R^3 can cover these objects.

    I was asking if there exists a thm that says just that. Matt grime answered by saying that there is such a thing as a dehram cohomology and since R^3 and opens of R² have different deram cohomology, there cannot exist a homeomorphism btw them.

    At least, this is what I got from it.
     
    Last edited: Oct 27, 2006
  8. Oct 27, 2006 #7

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    Invoking cohomology really is unnecessary. The dimension of something, locally, can be characterized by what you need to remove to disconnect it.
     
  9. Oct 27, 2006 #8

    mathwonk

    User Avatar
    Science Advisor
    Homework Helper
    2015 Award

    it is not trivial that R^n is not homeomorphic to R^m for n < m. a nice proof for n = 2, is the one matt gave. i.e. remove a point, and use loops to study the result.

    but if n = 3, you need to use spheres to stuy the result, not as well studied in elementary math.

    as matt says, the question is one of what disconneects soemthing.

    but there is a higher notion of connectivity, involving spheres instread of paths, that lets you always restrict to removing a point, instead of removing a higher dimensional subset.

    this can be studied by integrating higher dimensional angle forms, an elementary name for the generators of de rham cohomology.
     
  10. Oct 27, 2006 #9

    mathwonk

    User Avatar
    Science Advisor
    Homework Helper
    2015 Award

    a solid ball is not a surface becuase it has a border or boundary, the points on the surface of the ball.

    the nbhd of a\such a point remains contracftible after removal of the point, unlike a nbhd of a point of a surface.
     
  11. Oct 27, 2006 #10

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    As a heuristic as to why it can't be a trivial result I would like to put forth the observation that there are at least two topologies in which they are homeomorphic - the discrete and indiscrete. So there has to be something important in what the open sets are. Actually, I think I made a little white lie when I said that you don't need to invoke (co)homology and need only look at subsets that disconnect. I suspect there is some decent homology theory using the codimension r subspaces as cycles, so I was probably using a homology argument in disguise.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Definition of surface
  1. Definition of a square (Replies: 5)

  2. Definition of manifold (Replies: 4)

  3. Manifold definition (Replies: 42)

Loading...