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I Drawing a Continuous Symmetrical Grid on a Sphere's surface

  1. Mar 2, 2016 #1
    The motivation for this thread comes from physics, but I'm posting it in the maths section as the question is more of a mathematical one and less concerned with the underlying physics.

    In cosmology, they often talk about closed universes with positive curvature, or non-Euclidean elliptic geometry. The balloon analogy is often used as well. Think of a spherically shaped universe.

    Now, imagine an infinite grid on an infinite plane. An infinite grid consists of tessellating squares. Each square is surrounded by eight other squares.

    Imagine trying to draw a grid on the surface of a sphere. You want the entire surface of the sphere to be covered by the squares of the grid. You want each square to be surrounded by eight other squares.

    You can use convex quadrilaterals instead of squares too. The important issue is to obtain a continuous symmetrical grid.

    Is this mathematically possible?
  2. jcsd
  3. Mar 2, 2016 #2
    I do believe that it is mathematically impossible.
  4. Mar 2, 2016 #3
    Yes indeed, this is impossible. (When you say each square is surrounded by 8 other squares, I am assuming you mean in the same way that a square is surrounded by 8 other squares in a planar grid — 4 squares sharing a side and 4 more sharing just a vertex, right?)

    Any polyhedral structure drawn on the sphere must conform to Euler's formula:

    V - E + F = 2,

    where V is the number of vertices, E the number of edges, and F the number of faces.

    Suppose such a square grid on the sphere were possible. Let's say the number of faces is N, so

    F = N.​

    Each face has 4 edges, giving 4N edges, except that each edge is shared by 2 squares. So the correct number of edges would be

    E = 4N/2 = 2N.​

    As for the vertices, each face has 4 vertices, giving 4N vertices, except that each vertex is shared by 4 squares. So the correct number of vertices is

    V = 4N/4 = N.​

    This means that the left-hand side of Euler's formula is:

    V - E + F = N - 2N + N​

    which is 0, not 2. Hence such a square grid on the sphere is impossible.
    Last edited: Mar 2, 2016
  5. Mar 3, 2016 #4
    thanks for this simple proof.

    that shows that a sphere cannot represent a closed 2-D universe, because for the 2-D space to be continuous, you need a symmetrical grid.
  6. Mar 3, 2016 #5
    By the way, could this be an extension of the Hairy Ball theorem?
  7. Mar 8, 2016 #6


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    are you familiar with the fact that there are exactly 5 regular convex polyhedra? and that the cube is the only one with square faces? this is proved using the euler number method given above. the hairy ball theorem combines this argument with the theorem relating sums of zeroes of vector fiields with euler characteriatics. i.e. each tesellation yields a vector field with zeroes that calculate the euler number, and in the other direction one shows all non degenerate vector fields have the same oriented sum of zeroes.
  8. Mar 8, 2016 #7
    For the hairy ball theorem, just the theorem that states: "For any continuous vector field with isolated zeroes on a closed surface, the sum of the indices at the zeroes equals the Euler characteristic" suffices. (This is due to the mathematician Henri Poincaré.)

    For, the Euler characteristic χ(S2) of the 2-sphere S2 is 2, but if a vector field had no zeroes at all, the sum of the indices at the (nonexistent) zeroes would then be 0 . . . contradiction.

    Maybe you're thinking of calculating the Euler characteristic of S2 using a regular convex polyhedron? Yes, that would work, though there are also many other ways to see that χ(S2) = 2. For example, the Betti numbers of S2 are β0 = 1, β1 = 0, β2 = 1, and so their alternating sum is 1 - 0 + 1 = 2. Alternatively, consider the vector field on S2 tangent to a family of constant-angular-speed counterclockwise rotations about the z-axis, which could be given by

    V(x, y, z) = d/dθ (Rz(θ)) (x, y, z) |θ=0 = d/dθ (cos(θ) x + sin(θ) y, -sin(θ) x + cos(θ) y, z) |θ=0 = (y, -x, 0)

    for (x, y, z) ∈ S2, or in other words for x2 + y2 + z2 = 1.

    Then it's very easy to see that the index at each of the two zeroes of V (which occur at (0, 0, ±1)) is equal to 1, so the sum is 2.
    Last edited: Mar 8, 2016
  9. Mar 9, 2016 #8
    I don't understand the connection here. What do you mean by "continuous", and why does one need a symmetrical grid for a space to be such?
  10. Mar 16, 2016 #9
    Imagine being a 2-D creature living on the 2-D surface. You draw a box around yourself and surround it with eight other boxes. In a closed universe, you can map out all of space, and draw boxes to cover everywhere. But you can't do that for a sphere, hence a sphere can't represent a closed 2-D universe. We have to use a torus instead.
  11. Mar 16, 2016 #10
    What you are describing is essentially a coordinate chart. Who's to say that the chart has to be global? The observable universe only represents a local neighborhood in relation to the entire universe.
    Also, what does any of this have to do with the universe being closed?

    Based on what I can gather from the Wikipedia article, there's nothing wrong with a 3-sphere universe. The 3-sphere doesn't have a global coordinate chart, so in this respect it is not different from the 2-sphere. Does your reasoning imply that the 3-sphere model isn't a plausible model for the universe?
    Last edited: Mar 16, 2016
  12. May 11, 2016 #11
    a closed 2D surface can be in the form of a torus or a sphere for example.

    the torus meets my grid requirements, a sphere does not.

    I'm still trying to visualize how a 2D creature living on the surface of the sphere would see the Cartesian space around it.

    For a torus it is really easy. A sphere is really tricky.
  13. May 11, 2016 #12
    "I'm still trying to visualize how a 2D creature living on the surface of the sphere would see the Cartesian space around it."

    Of course, that depends on what dimension of Euclidean space you want to imagine your surface in. And each manifold has a lowest-dimensional Euclidean space that it can exist in. For example, a closed non-orientable surface cannot be constructed in 3-space (without self-intersections): such require a minimum of 4 dimensions to exist in.

    And, I suspect most of the readers of this forum live on the Earth, so we already have some idea what it's like to live on the surface of a sphere!

    And even those manifold, like the torus, that can exist in one Euclidean space can often exist with a wider range of metrics in a higher-dimensional Euclidean space: For example, the torus cannot exist with constant Gaussian curvature in 3-space, but in 4 dimensions (and any higher number), it can exist with constant curvature and as a result with more symmetry.

    By the way, although the 3-sphere S3 cannot have a global coordinate chart (and technically, no closed manifold can have that, exactly) . . . it does have something very nice nevertheless — it has a global continuous field of bases for its tangent spaces at each point. This can even be chosen so as to be orthonormal at each point. For example, if we let

    S3 = {(x, y, z, w) ∈ R4 | x2 + y2 + z2 + w2 = 1},​

    then one choice of a global orthonormal field of bases for its tangent spaces (called a "frame field") is given by

    V1(x, y, z, w) = (-y, x, -w, z)

    V2(x, y, z, w) = (-z, w, x, -y)

    V3(x, y, z, w) = (-w, -z, y, x),​

    which is often very useful to have. (An interesting fact is that every closed, orientable 3-manifold has such a continuous orthonormal frame field. The only other dimension where this is true is the trivial case of 1-manifolds.)
  14. Oct 4, 2016 #13
    Imagine you live on a planet which has an entirely flat, homogeneous surface. You have a sharpie with you. You use that sharpie to draw a square on the ground.

    Next, you surround that square with 8 squares, and you keep expanding the grid. You'll be dividing the surface of that planet into squares.

    Of course, as you have proven in post #3, there is a limit to your grid.

    We can also divide the volume around us into a grid, with each cube being surrounded by 26 other cubes.

    But, if our 3D spatial universe is imprinted on the surface of a glome, will it be impossible to surround each cube with 26 other cubes and still cover the entire surface of the glome? That would be rather astounding.

    And, if you traveled far enough in one direction you would end up where you started. That would be extremely astounding too.
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