The Unapologetic Mathematician

Mathematics for the interested outsider

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<x<1\\{0}&1\leq x\end{array}\right.

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 Posted by | Differential Topology, Topology | 8 Comments

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 Posted by | Differential Topology, Topology | 4 Comments

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 Posted by | Differential Topology, Topology | 2 Comments

   

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