## An Example of a Manifold

Let’s be a little more explicit about our example from last time. The two-dimensional sphere consists of all the points in of unit length. If we pick an orthonormal basis for and write the coordinates with respect to this basis as , , and , then we’re considering all triples with . We want to show that this set is a manifold.

We know that we can’t hope to map the whole sphere into a plane, so we have to take some points out. Specifically, let’s remove those points with , just leaving one open hemisphere. We will map this hemisphere homeomorphically to an open region in .

But this is easy: just forget the -component! Sending the point down to the point is clearly a continuous map from the open hemisphere to the open disk with . Further, for any point in the open disk, there is a unique with . Indeed, we can write down

This inverse is also continuous, and so our map is indeed a homeomorphism.

Similarly we can handle all the points in the lower hemisphere . Again we send to , but this time for any in the open unit disk — satisfying we can write

which is also continuous, so this map is again a homeomorphism.

Are we done? no, since we haven’t taken care of the points with . But in these cases we can treat the other coordinates similarly: if we have our inverse pair

while if we have

Similarly if we have

while if we have

Now are we done? Yes, since every point on the sphere must have at least one coordinate different from zero, every point must fall into one of these six cases. Thus every point has some neighborhood which is homeomorphic to an open region in .

This same approach can be generalized to any number of dimensions. The -dimensional sphere consists of those points in with unit length. It can be covered by open hemispheres, each with a projection just like the ones above.

is there a requirement to make sure the maps are compatible? ie if a point’s domain is in more than one map that the composite is continuous (in both directions)?

Comment by scot | February 23, 2011 |

We’re getting to that scot, stay tuned!

Comment by John Armstrong | February 23, 2011 |

[…] Let’s look back at yesterday’s example of a manifold. Not only did we cover the entire sphere by open neighborhoods with homeomorphisms to […]

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