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In, it says that 6 charts can be used to make an atlas for a sphere. But the text shows that you have a chart for the northern hemisphere, and you can make a similar chart for the southern hemisphere. Hence, these two charts cover the entire sphere.

What am I doing wrong?

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charts should be open so the two hemispheres does not cover all sphere – old Jul 30 '11 at 19:44
Those two charts don't cover the great circle on the equator. So you need four more charts to fully cover that circle. You can cover the sphere with just two charts using stereographic projection. – JSchlather Jul 30 '11 at 19:45
up vote 18 down vote accepted

The northern hemisphere and southern hemisphere don't cover the entire sphere. The sphere is $$\mathbb{S}^2=\{(x,y,z)\in\mathbb{R}^3\mid x^2+y^2+z^2=1\}.$$

The northern and southern hemispheres are, respectively, $$\mathbb{S}_N^2=\{(x,y,z)\in\mathbb{S}^2\mid z>0\},\qquad\mathbb{S}_S^2=\{(x,y,z)\in\mathbb{S}^2\mid z<0\}.$$ These miss the equator $\{(x,y,z)\in\mathbb{S}^2\mid z=0\}$. Adding "east" and "west" hemispheres $$\mathbb{S}_W^2=\{(x,y,z)\in\mathbb{S}^2\mid x>0\},\qquad\mathbb{S}_E^2=\{(x,y,z)\in\mathbb{S}^2\mid x<0\}$$ still doesn't get everything: we are missing the points on the equator $(0,1,0)$ and $(0,-1,0)$. Finally, adding the last two hemispheres (east and west, only rotated 90 degrees) covers the entire sphere.

This raises the question, why are we defining our hemispheres with $>$ and $<$? Perhaps we could instead use $\leq $ and $\geq$, and this would let us cover the sphere with two hemispheres?

The answer is that a chart of a manifold needs to be a homeomorphism between an open subset of the manifold with an open subset of $\mathbb{R}^n$. The sets $$\{(x,y,z)\in\mathbb{S}^2\mid z\geq 0\},\qquad\{(x,y,z)\in\mathbb{S}^2\mid z\leq 0\}$$ are not open in the topology of $\mathbb{S}^2$ (which is the subspace topology inherited from $\mathbb{R}^3$). So we can't use them as coordinate neighborhoods in the manifold structure of $\mathbb{S}^2$.

It does warrant mentioning, however, that we can cover the sphere using only two charts, via stereographic projection. The two open subsets of $\mathbb{S}^2$ acting as our coordinate domains are $$\mathbb{S}^2-\{(0,0,1)\},\qquad\mathbb{S}^2-\{(0,0,-1)\}$$ and for each, we project a line from the removed point to the plane, which one can check gives a continuous map. It is a tedious (but important) exercise to demonstrate that the smooth structure determined by stereographic projection is the same as that of the hemispheres (i.e., they are compatible atlases).

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Thanks for the answer. – quentin Jul 30 '11 at 20:34
Why can't we just have the hemispheres plus an extra epsilon (still open) to cover the equator? Or is it just that the stereographic projection is nicer? – Harry Stern Jul 31 '11 at 4:54
@Harry: Dear Harry, You can certainly do what you suggest (have the hemispheres plus epsilon). Perhaps stereographic projection is just more traditional? (Personally I prefer your suggestion, as being more intuitive.) Regards, – Matt E Jul 31 '11 at 6:17
While it is true that you can cover ${\mathbb S}^2$ with two charts I feel more at ease if individual charts as well as intersections of charts are homeomorphic to disks. For this you need 4 charts. – Christian Blatter Jul 31 '11 at 9:52
@Harry: When I think about the sphere for qualitative arguments, I tend to imagine the sort of charts you suggest. But if you ever actually want to do computations, stereographic projection is almost certainly easier to deal with than "hemispheres enlarged by $\epsilon$." Also (and this is the real reason I used it in my comment), when you want to give someone an example of a covering of the sphere by two charts, "stereographic projection" is faster to say and less likely to be disputed than "hemispheres enlarged by $\epsilon$." – Charles Staats Aug 2 '11 at 0:55

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