The Unapologetic Mathematician

Mathematics for the interested outsider

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 M^m and N^n are m- and n-dimensional smooth manifolds, respectively, then we can come up with an atlas that makes the product space M\times N into an m+n-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 (U,\phi_U) on M and another (V,\phi_V) on N, then we surely have U\times V\subseteq M\times N as an open subset of the product space. We just define

\displaystyle\phi_{U\times V}=\phi_U\times\phi_V:U\times V\to\mathbb{R}^m\times\mathbb{R}^n=\mathbb{R}^{m+n}

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

\displaystyle\phi_{U'\times V'}\circ\phi_{U\times V}^{-1}=(\phi_{U'}\times\phi_{V'})\circ(\phi_U\times\phi_V)=(\phi_{U'}\circ\phi_U^{-1})\times(\phi_{V'}\circ\phi_V^{-1})

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

So we have an atlas making M\times N a smooth manifold. It should also be clear that its dimension is m+n, 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 \pi_M:M\times N\to M, projecting on the first factor by \pi_M(p,q)=p. We pick a coordinate patch U\times V on M\times N and a coordinate patch U' on M. We set up the composite:

\displaystyle\begin{aligned}\phi_{U'}\circ\pi_M\circ\phi_{U\times V}^{-1}&=\phi_{U'}\circ\pi_M\circ(\phi_U^{-1}\times\phi_V^{-1})\\&=\phi_{U'\circ\phi_U^{-1}}\end{aligned}

which is one of the transition functions from the atlas on M. Clearly this is always smooth, and so the projection \pi_M:M\times N\to M 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 P with smooth maps f:P\to M and g:P\to N, do we get a unique smooth map (f,g):P\to M\times N? 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 W in P and a patch U\times V in M\times N. We need to know if the composite

\displaystyle\phi_{U\times V}\circ(f,g)\circ\phi_W^{-1}:\phi_W(W)\to\mathbb{R}^{m+n}

is smooth. But the target of this composite is \mathbb{R}^{m+n}, and a function to this real space will be smooth if and only if each component is. In particular, the first m components and the last n components must all be smooth, which means that our function is smooth if and only if both projections

\displaystyle\begin{aligned}\pi_{1,m}\circ\phi_{U\times V}\circ(f,g)\circ\phi_W^{-1}&=\phi_U\circ f\circ\phi_W^{-1}\\\pi_{m+1,n}\circ\phi_{U\times V}\circ(f,g)\circ\phi_W^{-1}&=\phi_V\circ g\circ\phi_W^{-1}&\end{aligned}

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

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

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March 7, 2011 - Posted by | Differential Topology, Topology

4 Comments »

  1. [...] 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 | Reply

  2. [...] 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 | Reply

  3. 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 | Reply

  4. thanks; fixed.

    Comment by John Armstrong | August 2, 2012 | Reply


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