# The Unapologetic Mathematician

## 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