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How many coordinate charts does it take to cover a surface?

  1. Jun 13, 2007 #1
    I was wondering about this, I've never seen any general theorem. Obviously it takes more than one, but I would think that in general it can take quite a few, for I can't see how to cover a torus with only 2.

    Is there any sort of general result on this?
  2. jcsd
  3. Jun 13, 2007 #2


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    well if you think about the charts as faces of a triangulation, and their overlaps as edges, you begin to see that the number of such faces and edges must be related to the euler characteristic.

    i guess an open cover can be thought of a vertices of a triangulation, and pairwise overlaps as edges, and triple overlaps as faces. and in some sense then you need eulers formula to hold among these vertices edges and faces.

    i.e. for a torus V-E+F = 0. so if you only had two vertices and one edge, you would get 1 instead of 0. this is called cech homology theory.

    of course something is fishy here since this is making a sphere llok like a 1 manifoild instead of a 2 manifold, but the cover of a sphere by two sets has a circular overlap, and to get all the invariants right you want the overlaps themselves to be contractible I guess.

    but this is an idea. see some books on topology.
    Last edited: Jun 13, 2007
  4. Jun 13, 2007 #3
    If we ask ourselves how many simply connected charts we need to cover a surface.

    If you consider non-compact surface, we may need infinitely many coordinate charts. So we shall consider only the case of compact surfaces.

    Compact surfaces are classified by genus up to homeomorphism. Thus we ask ourselves how many charts we need to cover a surface of genus g. Such a surface can be constructed from a polygon with 4g sides by identifying pairs of edges in an appropriate way. From this it is apparent that we need at most 2g+1 charts, and we can take these charts to be simply connected.

    As for a torus. It is obvious that we can't cover it with two simply connected open subspaces since if we can, we can not find generators of the fundamental group of the torus as elements of the fundamental groiups of these subspaces.
  5. Jun 15, 2007 #4


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    why is it apparent? it seems false already for g = 0. and a torus seems to need four from your description since the two circles omitted by the rectangle need three more.

    I am not following. thanks for this post. can you say a bit more?
  6. Jun 15, 2007 #5

    The construction described of surfaces of genus g is valid only when g is positive. Thus the case g=0 should be omitted.

    Ah, you are right. Even in the case g>0, we may need 2(g+1) charts in general. That is one cover as a nbd of a point, 2g covers as nbds of edges, and one more to cover the remaining part.

    But as for a torus, isn't three enough? There will be three overlapped parts. For curves g>2, we may still need only 2g+1 covers....I haven't check this, though.
  7. Jun 15, 2007 #6
    AHHH! I did it again. In the last paragraph, I mean surfaces g > 2, not curves. I tend to say a curve when I speak of a genus....
  8. Jun 15, 2007 #7
    one more thing. I restricted my attention to compact orientable manifold of dim 1.

    What I wanted to say equivalent to that they are CW-complexes.
  9. Jun 15, 2007 #8
    again agina... cpt orientable mfds without boundaries.
  10. Jun 18, 2007 #9


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    for a torus i got a rectangle, a small disc, and two thin rectangular strips, for four.
  11. Jun 18, 2007 #10


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    i guess your approach is to look at a 4g sided polygon with edges indentified pairwise, and take the open polygon as one chart, then one thin rectangular strip for each of the distinct 2g edges, then one disc over the common point they all share, for 2g+2.

    but a sphere cannot be represented this way, so for a sphere you get, hey 2g+2 again! ok i believe 2g+2, but not three for a torus, (unless you convince me).

    at least you have proved 2g+2 is an upper bound it seems.
  12. Jun 19, 2007 #11
    2g+2 is no more than an upper bound. You are right.
  13. Jun 19, 2007 #12


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    you know it dawns on me we need to define a coordinate patch. because they do not need to be simply connected. for example we could cover a torus by three of them if we allow coordinate patches homeomorphic to an annulus, and indeed an annulus can be embedded in the plane.

    so it seems we could use one rectangular patch and two annular poatches. can we do better? yes!! i claim we can use two annular patches.
  14. Jun 19, 2007 #13
    That is what I thought when I first read the question posted in this thread. When DeadWolfe said he couldn't see how a torus can be covered by two coordinate charts, I said to myself, "he must be considering only the simply connected charts."

    So I limited ourselves to consider only simply connected nbds, or his question doesn't make sense. (I said so in my first post, in its first sentence, which is an unfinished sentence, though....)
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