# The Unapologetic Mathematician

## Presheaves

Strictly speaking, sheaves are not just about differential topology. And it’s also possible to get away without talking about them explicitly in differential topology. But it’s in differential topology that they really start to make their presence felt.

The best description of a sheaf I ever got doesn’t really translate to text, unfortunately. I forget the originator, as told to me by the professor explaining it — and if he remembers and can fill in the gap this can be recorded for posterity — but it really does capture the essence. A sheaf is a thing where the topology goes this way (moving one’s open hand around in circles at about chest level) and the algebra goes this way (moving one’s hand up and down across chest level). To be a bit more explicit, sheaves are about taking algebraic structures and localizing them to open sets in a topological space. I’ll start with the simplest versions, which are sheaves of sets. Further, I’ll actually start with something simpler: presheaves.

So, a presheaf $\mathcal{F}$ of sets on a topological space $X$ is a choice of a set $\mathcal{F}(U)$ for each open set $U\subseteq X$. We call the members of $\mathcal{F}(U)$ the elements — or the “sections” — of the presheaf “over $U$“. But there’s an important condition on this choice: if $V\subseteq U$ is a smaller open subset of $X$, then we should be able to “restrict” our element $f\in\mathcal{F}(U)$ to an element $f\vert^U_V\in\mathcal{F}(V)$. Thus, if we have an inclusion $V\subseteq U$, then we have a restriction map $\cdot\vert^U_V:\mathcal{F}(U)\to\mathcal{F}(V)$.

These restriction maps are subject to a couple conditions. First of all, if we restrict from $U$ to itself, then we shouldn’t change anything. That is, the restriction map $\cdot\vert^U_U$ is always the identity map. Secondly, if we restrict from $U$ to $V$, and then from $V$ to $W$, we should get the same result as if we just restricted directly from $U$ to $W$. That is, we have the equation $\cdot\vert^V_W\circ\cdot\vert^U_V=\cdot\vert^U_W$. Because of this, we’ll often just write the restriction map as $\cdot\vert_V$, since which subset it came from doesn’t really matter.

We can express this definition more succinctly if we remember that containment of open subsets of a topological space constitutes a partial order, and thus defines a category $\mathrm{Subset}(X)$. The objects are the open sets themselves, and there is a unique arrow from $V$ to $U$ if $V\subseteq U$. If we look closely, we’ll find that what we’ve defined as a presheaf is actually a contravariant functor from this category to the category of sets! For every arrow $V\mapsto U$ we have an arrow $\cdot\vert_U:\mathcal{F}(U)\to\mathcal{F}(V)$ — in the “opposite direction”, since the functor is contravariant. The conditions we impose on the restriction maps just say that they preserve identity arrows and compositions.

Now, there’s nothing inherently special about sets here. We can set up exactly the same construction with any target category to define, say, a presheaf of rings to be a contravariant functor $\mathcal{F}:\mathrm{Subset}(X)\to\mathbf{Ring}$. This assigns a ring $\mathcal{F}(U)$ to every open set $U$, and the restriction maps have to be ring homomorphisms. In the same way we get presheaves of groups, of abelian groups, or of vector spaces over a given field.

The one possibly confusing case is when we talk about a presheaf of modules over a presheaf of rings. In this case, say we have a presheaf $\mathcal{F}$ of rings on a topological space $X$. A presheaf of modules $\mathcal{M}$ over $\mathcal{F}$ assigns an abelian group $\mathcal{M}(U)$ to every open set $U\subseteq X$ — it’s a presheaf of abelian groups — in such a way that $\mathcal{M}(U)$ is a module over $\mathcal{F}(U)$. The restriction map has to work with the restriction map of $\mathcal{F}$, so we have $f(m)\vert_V=f\vert_V\left(m\vert_V\right)$.

The canonical example to keep in mind is continuous, real-valued functions on a topological space. This is a sheaf of real algebras that associates to the open set $U\subseteq X$ the algebra $\mathcal{F}(U)$ of real-valued functions that are defined and continuous there. Clearly we can restrict such a function $f$ to whatever open subset $V$ we want — and, in fact, we have. The nice thing is that this gives us a way of talking about and dealing with functions on our space that may not be defined or continuous everywhere. Just work within a suitable open set where the function does play nice! If you need to work with two functions defined over different open sets, just restrict them both to their common intersection and work there. Many structures we run into in differential geometry will be naturally expressible in terms of presheaves, just like this.

March 16, 2011 Posted by | Topology | 18 Comments

## Bump Functions, part 2

As an immediate application of our partitions of unity, let’s show that we can always get whatever bump functions we need.

Let $U$ be an open subset of $M$, and $V$ be a set whose closure $\bar{V}$ is contained within $U$. I say that there is a nonnegative smooth function $\phi:M\to\mathbb{R}$ which is identically $1$ on $\bar{V}$, and which is supported within $U$.

To find this function, we start with a cover of $M$. Specifically, let $U$ be one set of the cover, and let $M\setminus\bar{V}$ be the other set. Then we know that there is a countable smooth partition of unity subordinate to this cover. That is, for every $k$ we either have $\phi_k$ supported in $U$, or $\phi_k$ supported in $M\setminus\bar{V}$ (or possibly both).

In fact, no matter what countable partition we come up with, we can take all the $\phi_k$ supported within $U$ and add them all up into one function $\phi$, and then take all the remaining functions and add them all up into one function $\psi$. Then $\{\phi,\psi\}$ is a partition of unity subordinate to our cover, and I say that $\phi$ is exactly the function we’re looking for.

Indeed, as a part of a partition of unity, $\phi$ is a nonnegative smooth function, and we know it’s supported in $U$. The only thing we need to determine is if it’s identically $1$ on $\bar{V}$. But for $p\in\bar{V}$ we know that $\phi(p)+\psi(p)=1$, and yet we also know that $\psi(p)=0$, since $\psi$ is supported in $M\setminus\bar{V}$. Thus we must have $\phi(p)=1$, and $\phi$ is indeed our bump function.

March 16, 2011

## Partitions of Unity (proof)

Finally we can prove what we’ve asserted: given any open cover $\{U_\alpha\}$ of a smooth manifold $M$ we can find a countable smooth partition of unity $\{\phi_k\}$ subordinate to it.

So, as we’ve seen we can find a countable atlas $(V_k,x_k)$; we use $x_k$ for the coordinate maps since we’ll want the $\phi_k$ free. We’ve also seen that we have a smooth bump function $\phi$ between the two cubes $C_1(0)$ and $C_2(0)$ in $\mathbb{R}^n$. So let’s define

$\displaystyle\theta_k(p)=\left\{\begin{array}{lc}\phi\left(x_k(p)\right)&p\in V_k\\{0}&p\notin V_k\end{array}\right.$

Now, it’s easily verified that the furthest points from the origin in $C_2(0)$ are the corners, each of which is $2\sqrt{n}$ away. Thus we can tell that $C_2(0)\subseteq B_{3n}(0)$, and the support of $\phi$ is contained entirely within $x_k(V_k)=B_{3n}(0)$. This means that the support of $\theta_k$ is entirely contained within $V_k$ — by the time we get to the edge it’s already smoothly tailed off to zero, and so even though we define it piecewise, $\theta_k$ is a smooth function defined on all of $M$.

Now set up the sum

$\displaystyle\theta(p)=\sum\limits_k\theta_k(p)$

Since $V_k$ is locally finite and $\theta_k$ is supported within $V_k$, this sum is guaranteed to be finite at each point, which makes $\theta$ a smooth function on all of $M$. The ball $B_1(0)$ is contained within the cube $C_1(0)$, so $\phi$ takes the constant value $1$ on this ball. Since the preimages $x_k^{-1}(B_1(0))$ form an open cover of $M$, there is always at least one $k$ for which $\theta_k=1$. In particular, for which it’s not zero, and thus the whole sum is nonzero at $p$.

Since $\theta$ is never zero, we can divide by it. We define $\phi_k(p)=\theta_k(p)/\theta(p)$. These are smooth functions on all of $M$, and their sum is everywhere exactly $1$. Thus the $\phi_k$ form a partition of unity subordinate to $\{V_k\}$. And since $\{V_k\}$ refines $\{U_\alpha\}$, the partition is subordinate to this cover as well.

March 14, 2011

## Bump Functions, part 1

Dealing with an unexpected breach of my GMail address book kept me busy yesterday. But at least I get this one up before today’s activities.

Now we come to the heart of our partitions of unity: the bump functions. These are like smooth analogues of characteristic functions. A characteristic function $\chi_S$ is defined as $1$ on a set and $0$ off of it. We can use them (and have!) to “mask” off a function $f$; multiply $f$ by $\chi_S$ and suddenly $f$ is supported on $S$. But doing this introduces some nasty discontinuities.

A bump function $\phi$ fixes the problem by smoothly tailing off to zero between an inner set $V$ and an outer open set $U$ that contains the closure of $V$. Then the product $f\phi$ will be at least as smooth as the original function $f$ was, except in the case of analytic functions. On $V$, $\phi$ is identically $1$, and so $f(x)\phi(x)=f(x)$ for points $x\in V$. Outside of $U$, $f(x)\phi(x)=0$.

So let $C_\epsilon(0)$ be the open cube in $\mathbb{R}^n$ consisting of those vectors with each of their components in the interval $(-\epsilon,\epsilon)$. We will start by constructing a bump function between $C_1(0)$ and $C_2(0)$.

The real core here is the function

$\displaystyle h(x)=\left\{\begin{array}{lc}e^{-\frac{1}{x}}&x>0\\{0}&x\leq0\end{array}\right.$

I leave it to you to verify that this function is, in fact, smooth at $x=0$; show that each derivative of the function on the right is zero at this point. It’s clearly not analytic, though, since its Taylor series at this point sums to the zero function.

Now, consider the function $h(2+x)h(2-x)$. If $x\geq2$ or $x\leq-2$, one or the other factor is zero, and so the product is supported inside $[-2,2]$. We can also write down $h(2+x)h(2-x)+h(x-1)+h(-x-1)$, which is everywhere strictly greater than zero, meaning we can divide by it:

$\displaystyle f(x)=\frac{h(2+x)h(2-x)}{h(2+x)h(2-x)+h(x-1)+h(-x-1)}$

If $x\leq1$ then $h(x-1)=0$, while if $x\geq-1$ then $h(-x-1)=0$. So on the whole interval $[-1,1]$, this quotient is exactly $1$.

Therefore $f$ is a bump function between the intervals $(-1,1)$ and $(-2,2)$ in $\mathbb{R}$. For $\mathbb{R}^n$, just define

$\displaystyle\phi(a_1,\dots,a_n)=f(a_1)\dots f(a_n)$

and we have a suitable bump function between the cubes $C_1(0)$ and $C_2(0)$ for any dimension.

March 12, 2011

## Atlases Refining Covers, part 2

Again with the late posts…

Now, armed with our two new technical assumptions, we can prove the existence of the refining covers we asserted yesterday.

Since $M$ is (now) known to be locally compact, Hausdorff, and second countable, there must exist a countable basis $\{Z_k\}$ for the topology of $M$ with each closure $\bar{Z}_k$ compact. Basically we can start with a neighborhood of each point that has compact closure and whittle it down to a countable basis, using the Hausdorff property to make sure we keep compact closure.

We will construct a sequence of compact sets inductively. Let $A_1=\bar{Z}_1$, which is compact by assumption. Given $A_i$ already defined, let $j$ be the first index for which $A_i\subseteq Z_1\cup\dots\cup Z_j$, and define $A_{i+1}=\bar{Z}_1\cup\dots\cup\bar{Z}_j\cup\bar{Z}_{j+1}$. Then $\{A_k\}$ is a sequence of compact sets with $A_k\subseteq\mathrm{int}(A_{k+1})$, and whose union is all of $M$. Define $A_0$ to be the empty set.

Now, we can write

$\displaystyle M=\bigcup\limits_{i\geq0}\left(A_{i+1}\setminus\mathrm{int}(A_i)\right)$

so for every point $p$ we can find a chart $(V_p,\phi_p)$ sending $p$ to $0\in\mathbb{R}^n$ and with $\phi_p(V_p)=B_{3n}(0)$, $V_p\subseteq U_\alpha$ for some $\alpha$, and $V_p\subseteq\mathrm{int}(A_{i+2})\setminus A_{i-1}=W_i$ for some $i$.

Indeed, we can surely find some chart around $p$, and intersecting it with some open $U_\alpha$ — which should contain $p$ — and with the open $W_i$ — likewise — still gives us a chart. We can subtract off whatever offset we need to make sure that this chart sends $p$ to $0$. Then we can take a ball of some radius around $0$ and let $V_p$ be its preimage. Scaling up the coordinate map lets us expand this ball until its radius is ${3n}$. Messy, no?

So now the collection of all the preimages $\phi_p^{-1}(B_1(0))$ as $p$ runs over $A_{i+1}\setminus\mathrm{int}(A_i)$ is an open cover of this compact set, and thus it contains a finite subcover, which we write as $P_i$. Taking the union of all of the $P_i$ gives a countable cover $V_k$ of $M$ refining $U_\alpha$. Each $V_k$ is the domain of a chart with $\phi_k(V_k)=B_{3n}(0)$, and the collection of preimages $\phi_k^{-1}(B_1(0))$ covers $M$, as asserted.

The only thing we haven’t shown here is that $V_k$ is locally finite. But since each point $p\in M$ must lie in one of the $A_{i+1}\setminus\mathrm{int}(A_i)$, so $W_i$ is an open neighborhood of $p$ that intersects at most finitely many $A_i$, and each $A_i$ can intersect at most finitely many $V_k$, so $W_i$ touches at most finitely many of them itself.

Got all that? We’re not out of the woods yet…

March 11, 2011

## Atlases Refining Covers, part 1

A bit late, but at least I got it up today!

Our first step in finding partitions of unity subordinate to a given cover $\{U_\alpha\}$ is actually to set up a nice atlas.

We want a countable differentiable atlas $(V_k,\phi_k)$ where the collection $\{V_k\}$ is a locally finite refinement of the cover $\{U_\alpha\}$. Locally finite we just covered yesterday; recall that being a refinement means that each $V_k$ is contained in some $U_\alpha$.

Getting more technical, we will also require that the image $\phi_k(V_k)\subseteq\mathbb{R}^n$ is $B_{3n}(0)$ — the open ball of radius $3n$ centered at the origin. That is, $\phi_k(V_k)=B_{3n}(0)$ consists of exactly those vectors with length strictly less than $3n$. Further, we will define $W_k$ to be the inverse image $\phi_k^{-1}(B_1(0))$ of the ball of radius $1$, and we will require that the collection $\{W_k\}$ is also a cover of the manifold $M$.

Now, in order to pull this off, we actually need to add some technical restrictions to our definition of a manifold. Since $M$ is locally homeomorphic to $\mathbb{R}^n$, and $\mathbb{R}^n$ is locally compact — each point has some open neighborhood with compact closure — the same is true of $M$. We don’t however, know that $M$ is Hausdorff or second-countable, both of which we’ll need.

I want to give some counterexamples, showing how these conditions can fail, and why the pathologies they prevent don’t adhere to our intuitive notion of “manifold”. Luckily, both of them are one-dimensional, so they aren’t impossible to visualize.

First, we have the line with a doubled origin. Take two real lines and glue them together by identifying each nonzero number on each line with the corresponding nonzero number on the other line. But do not identify the two zeroes. What’s left is one line, but it has two zero points “on top of each other”. Every nonzero number clearly has a nice open neighborhood that looks just like a regular interval — just stay away from the zero points — and we can also set up patches that look like the interval $(-1,1)$, one containing each of the two zeroes. These last patches are still open, and so every point has an open neighborhood homeomorphic to an interval.

But this space is not Hausdorff! Any two open sets, each containing one of the zeroes, must intersect, and yet these two points are not the same. If the doubled portion were an interval, we’d see the line fork in half on either side of the doubled section, and the forking point would clearly not have any neighborhood that looked like an interval, and so it would clearly not be a manifold. But when we shrink down and only double a point, there is no “forking point”, and we can’t use that to rule this case out. So instead we say that manifolds must be Hausdorff.

The other pathological example is the “closed long ray”. This is less obviously pathological, and second-countability is mainly around so that we can get countable sequences and series and such to make our lives easier down the road. Anyhow, to get our hands on it is sort of technical. We start with the half-open interval $[0,1)$ and the “first uncountable ordinal$\omega_1$. Actually, any uncountable well-ordered set will do [commenter Stevie Hair below isn’t so sure, and he has a good point] but the first one that arises is the most convenient.

Now, we take the product $\omega_1\times[0,1)$ and give it the “lexicographic order”. That is, we compare pairs $(o,x)$ and $(p,y)$ by first comparing $o$ and $p$ in the order from $\omega_1$. If they’re different, we use that order. If they’re the same, though, we move on to compare $x$ and $y$ in the usual way. We then give it the “order topology”, similar to the way we constructed the topology on the rational numbers.

The upshot is that it’s like we’ve strung together an uncountable number of copies of the interval $[0,1)$. Within each copy, obviously, it looks like $\mathbb{R}^1$, and where we glue two copies together it does too, just like we have no trouble going from $[0,1)$ to $[1,2)$. It’s even Hausdorff already, so the previous condition doesn’t rule this case out. The problem is that it’s not second-countable. There’s no countable collection of open sets that generate the whole topology. It’s simply too big to be described without bringing uncountable numbers of sets into the picture.

So, we add the Hausdorff and second-countable conditions to our definition of a manifold, and move forward.

March 10, 2011

## Partitions of Unity Subordinate to a Cover

We know what a partition of unity is, but not all partitions of unity are very useful. For instance, the single function defined by $\phi(p)=1$ for all points $p\in M$ is a partition of unity all on its own — its support is $M$ itself, which is clearly a locally finite cover of $M$, and it adds up to the constant unit function. But we can’t really do anything with it.

What we need is a partition of unity subordinate to an open cover. That is, given a collection $\{U_\alpha\}$ of open sets that cover $M$, we want a partition of unity $\{\phi_\beta\}$ such that for every $\beta$ there is some $\alpha$ so that $\phi_\beta$ is supported in $U_\alpha$. In particular, we can let $\{U_\alpha\}$ be the collection of coordinate patches in a smooth atlas, so each of the functions $\phi_\beta$ “lives in” a single local coordinate system.

But do any such things exist? Remember, except for the trivial example above I haven’t actually given any examples of a smooth partition of unity at all. The example last time was differentiable, and even twice-differentiable, but not smooth. So this is a nice concept, but it might well be vacuous.

Still, all is not lost: I say that given any open cover of a smooth manifold, there is a countable smooth partition of unity subordinate to that cover. In particular, given any smooth structure on a manifold we can always find a partition of unity with each function supported completely within a single coordinate patch. The proof of this fact, however, is one of the few really annoying, fiddly, technical bits in differential geometry. It will take a few days of doing, and I fully understand if you’d rather just skip it. All you really need to know is: whenever we need a partition of unity to break global things defined over our entire manifold up into nice chunks that fit into coordinate patches, we can do it.

However, I should point this out: analytic manifolds are not nearly so forgiving. The basic (but sketchy) idea is that in order to construct our partitions of unity we’ll need to create “bump” functions sort of like the one we did last time, but ones that are smooth instead of just twice-differentiable. This means using a piecewise definition, just like last time, and at the edge of a piece we’ll have points such that in any neighborhood of that point we need two different definitions of the function. But if the function is supposed to be analytic, then the definition that works on one side should keep working on the other side, and so we can’t make the bump functions we need.

This is a big reason why people stop at smooth manifolds rather than working with analytic ones, despite the fact that analytic functions are arguably “nicer”. Unfortunately, this also means that not everything we do carries over quite so easily to complex manifolds — based on complex vector spaces — which must always be analytic.

March 8, 2011

## 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.