# The Unapologetic Mathematician

## Partitions of Unity

And, finally, one to go up today!

A partition of unity is a useful, though technical, tool that helps us work in local coordinates. This can be a tricky matter when we’re doing things all over our manifold, since it’s almost never the case that the entire manifold fits into a single coordinate patch. A (smooth) partition of unity is a way of breaking the function with the constant value $1$ up into a bunch of (smooth) pieces that will be easier to work with.

More specifically, a partition of unity is a collection of nonnegative smooth functions $\phi_\alpha:M\to\mathbb{R}$ indexed by some set $\alpha\in A$, subject to two conditions. First: the collection of supports $\{\mathrm{supp}(\phi_\alpha)\}_{\alpha\in A}$ is a locally finite cover of $M$, which takes a bit to unpack.

The support $\mathrm{supp}(f)$ of a real-valued (or vector-valued) function is the closure of the set on which it takes nonzero values. In other words, the complement of the support is the largest open set on which $f(p)=0$.

To say that a collection of sets is a locally finite cover means that every point $p\in M$ is contained in at least one of them, and that $p$ has some neighborhood which intersects only finitely many of them. For instance, the collection of all intervals $[n-1,n+1]$ centered at integers $n$ is a locally finite cover of $\mathbb{R}$. Every real number is within $1$ of some integer, and around each real number we can draw a small neighborhood that meets at most three of these intervals (why three?).

The other condition is that the sum

$\displaystyle\sum\limits_{\alpha\in A}\phi_\alpha=1$

That is, if we add up all these functions we get the function with constant value $1$. But we made no restriction on the index set, so how do we know that this sum remotely makes sense? Because we evaluate it at each point

$\displaystyle\sum\limits_{\alpha\in A}\phi_\alpha(p)$

and we know that the supports of $\phi_\alpha$ form a locally finite cover! That is, there is some neighborhood $N$ of $p$ which intersects at most finitely many of the $\mathrm{supp}(\phi_\alpha)$. For all of them $N$ doesn’t intersect, we are absolutely certain that $\phi_\alpha(p)=0$, and so our big sum really only involves at most finitely many terms at each point!

As an example, consider the function $\phi_0$ defined by

$\displaystyle\phi_0(x)=\left\{\begin{array}{cc}0& x\leq-1\\\cos(\frac{\pi}{2}x)^2&-1

This is a differentiable — though not smooth — function supported on the interval $[-1,1]$. We can slide this over to define $\phi_n=\phi_0(x-n)$, getting a differentiable function supported on $[n-1,n+1]$. From here, it’s an exercise to verify that this is a partition of unity. We must check that on the interval $[n,n+1]$ we have $\phi_n(x)+\phi_{n+1}(x)=1$.

March 7, 2011

## 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.

March 7, 2011

## Open Submanifolds

Eek! None of these drafts went up on time!

In principle, we know what a submanifold should be: a subobject in the category of smooth manifolds. That is, a submanifold $S$ of a manifold $M$ should be another manifold, along with an “inclusion” map which is smooth and left-cancellable.

On the underlying topological space, we understand subspaces; first and foremost, a submanifold needs to be a subspace. And one easy way to come up with a submanifold is just to take an open subspace. I say that any open subspace $S\subseteq M$ is automatically a submanifold. Indeed, if $(U,\phi_U)$ is a coordinate patch on $M$, then $(U\cap S,\phi_U\vert_{U\cap S})$ is a coordinate patch on $S$. The intersection $U\cap S$ is an open subset, and the restriction of $\phi_U$ to this intersection is still a local homeomorphism. Since the collection of all coordinate patches in our atlas cover all of $M$, they surely cover $S$ as well.

As a quick example, an open interval in the real line is automatically an open manifold of $\mathbb{R}$, and so it’s a manifold. Any open set $U$ in any $n$-dimensional real vector space is also automatically an $n$-manifold.

More generally, it turns out that what we want to consider as a “submanifold” is actually somewhat more complicated, and we will have to come back to this point later.

March 7, 2011