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Is there any sort of general result on this?

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- #1

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Is there any sort of general result on this?

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mathwonk

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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.

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.

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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.

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mathwonk

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I am not following. thanks for this post. can you say a bit more?

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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.

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- #7

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What I wanted to say equivalent to that they are CW-complexes.

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again agina... cpt orientable mfds without boundaries.

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mathwonk

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for a torus i got a rectangle, a small disc, and two thin rectangular strips, for four.

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mathwonk

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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.

- #11

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2g+2 is no more than an upper bound. You are right.

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mathwonk

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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.

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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....)