## Product Manifolds

*More* drafts that didn’t go up on time!

Next we want to show that we have (finite) products in the category of manifolds. Specifically, if and are - and -dimensional smooth manifolds, respectively, then we can come up with an atlas that makes the product space into an -dimensional smooth manifold, and that it satisfies the conditions to be a product object in our category.

So, we have our topological space already. What atlas should we put on it? Well, if we have a coordinate patch on and another on , then we surely have as an open subset of the product space. We just define

If and are another pair of coordinate patches we can set up the transition function

Each of these factors is smooth since each is a transition function from one of the two smooth atlases we already know on and . Since smoothness is determined component-by-component, it follows that the product mapping is smooth as well.

So we have an atlas making a smooth manifold. It should also be clear that its dimension is , as asserted. But is it a product object? To see this, we need to consider the projections, which are the same as the ones we get from the underlying topological spaces. The first question is: are these projections smooth maps?

Well, let’s consider , projecting on the first factor by . We pick a coordinate patch on and a coordinate patch on . We set up the composite:

which is one of the transition functions from the atlas on . Clearly this is always smooth, and so the projection is a smooth map of manifolds. The same is true of the other projection as well.

Now, is this universal? That is, if we have some other manifold with smooth maps and , do we get a unique smooth map ? Obviously we have a unique *continuous* map, by just considering everything in sight as a topological space and forgetting the manifold structure. The question is whether this is smooth.

So, pick a coordinate patch in and a patch in . We need to know if the composite

is smooth. But the target of this composite is , and a function to this real space will be smooth if and only if each component is. In particular, the first components and the last components must all be smooth, which means that our function is smooth if and only if both projections

are. But these are both smooth since we assumed that and were smooth maps.

Thus the product manifold really is the product in the category of smooth manifolds, as we asserted.

[…] and be smooth manifolds, with the -dimensional product manifold. Given points and we want to investigate the tangent space of this product at the point […]

Pingback by The Tangent Space of a Product « The Unapologetic Mathematician | April 27, 2011 |

[…] To be a little more explicit, a Lie group is a smooth -dimensional manifold equipped with a multiplication and an inversion which satisfy all the usual group axioms (wow, it’s been a while since I wrote that stuff down) and are also smooth maps between manifolds. Of course, when we write we mean the product manifold. […]

Pingback by Lie Groups « The Unapologetic Mathematician | June 6, 2011 |

I think you meant \pi_M (p,q)=p. Small typo, you have \phi_M (p,q)=p.

Comment by Dustin Bryant | August 2, 2012 |

thanks; fixed.

Comment by John Armstrong | August 2, 2012 |