# The Unapologetic Mathematician

## Smooth Maps

As usual, we’re going to want our objects of study — smooth (or differentiable) manifolds — to be objects in a category. And a category means we need morphisms. The morphisms between smooth manifolds are smooth maps.

Given two smooth manifolds, $M^m$ and $N^n$, a continuous map $f:M\to N$ is smooth at a point $p\in M$ if we can find a coordinate patch $U\subseteq M$ containing $p$ and a coordinate patch $V\subseteq N$ containing the image $f(U)$ so that the composite function $\phi_V\circ f\circ\phi_U^{-1}$ is smooth as a function from $\phi_U(U)\subseteq\mathbb{R}^m$ into $\phi_V(V)\subseteq\mathbb{R}^n$. We say it’s smooth if it’s smooth at all points.

But wait, maybe we just got lucky when we picked these coordinate patches. Well, it actually doesn’t matter. If $f$ is smooth according to one pair of coordinate patches, it’s smooth according to any other pair. Indeed, if we take another set of coordinates $(U',\phi_{U'})$ around $p$ then the compatibility condition says that the transition function $\phi_U\circ\phi_{U'}^{-1}$ is smooth. And then so is the composite:

$\displaystyle\left(\phi_V\circ f\circ\phi_U^{-1}\right)\circ\left(\phi_U\circ\phi_{U'}^{-1}\right)=\phi_V\circ f\circ\phi_{U'}^{-1}$

But this is just the smoothness condition in terms of $U'$ and $V$. Similarly, if we change coordinates in the target from $V$ to $V'$, the compatibility condition says that $\phi_{V'}\circ\phi_V^{-1}$ is smooth, and so the composite

$\displaystyle\left(\phi_{V'}\circ\phi_V^{-1}\right)\circ\left(\phi_V\circ f\circ\phi_U^{-1}\right)=\phi_{V'}\circ f\circ\phi_U^{-1}$

is smooth as well. The condition for smoothness at a point, therefore, really only depends on the behavior of the function near that point, and not on what particular coordinates we use to attest to its smoothness.

In particular, let’s consider what it means for a function to be smooth from a manifold $M^m$ to the real space $\mathbb{R}^n$. In this case, we can choose the entire space with the identity function as the coordinate patch on the target manifold. Thus a function $f:M\to\mathbb{R}^n$ is smooth at a point $p$ if there is some $U\subseteq M$ containing $p$ with coordinate map $\phi_U:U\to\mathbb{R}^m$ such that the composite $f\circ\phi_U^{-1}:\phi_U(U)\to\mathbb{R}^n$ is smooth.

Finally, just like we have the fancy word “homeomorphism” for isomorphisms of topological spaces, we have the fancy word “diffeomorphism” for isomorphisms of differentiable manifolds.

March 2, 2011

## Atlases and Structures

I just noticed a problem: this was supposed to have gone up Monday…

Now that we know what it means for coordinate patches to be compatible — for various definitions of “compatible” — we can define atlases on a manifold. An atlas is a collection of coordinate patches that cover a manifold.

More specifically, we have various classes of atlases. A piecewise-linear atlas is one whose transition functions are all piecewise-linear; differentiable atlases have $C^1$ transition functions; “smooth” atlases have $C^\infty$ transition functions; analytic atlases have $C^\omega$ transition functions.

In our example of a manifold, we covered the two-dimensional sphere with coordinate patches, and so we have an atlas. Let’s look at it a little more closely and see what kind of atlas we have. Really, all we need to consider is the overlap of two of our patches, since they all look very similar.

So let’s consider the set $U$ where $z>0$ and the set $V$ where $x>0$. The transition from the first to the second will take a point in $\phi_U(U\cap V)$ — the open half-disk of points $(x,y)$ with $x^2+y^2<1$ and $x>0$ — and lift it up to the sphere using the formula

$\displaystyle\phi_U^{-1}(x,y)=\left(x,y,\sqrt{1-x^2-y^2}\right)$

Then the transition function projects this point down by dropping the $x$-coordinate. That is, the transition function can be written down:

$\displaystyle\left[\phi_V\circ\phi_U^{-1}\right](x,y)=\left(y,\sqrt{1-x^2-y^2}\right)$

Now, since $x^2+y^2<1$, both of these component functions are analytic, and so we have an analytic atlas. Of course, since analyticity implies smoothness, we also have a smooth atlas, and a differentiable atlas. We don’t, however, have a piecewise-linear atlas.

A general atlas can be a little constraining, though. In our example, we only have six coordinate patches to work with, and these coordinates may not always be the most efficient ones for our purposes. For one example, our usual latitude and longitude coordinates are perfectly valid, and yet we can’t use them if all we have are the six axial projections!

Luckily, the usual latitude and longitude coordinates are compatible (where they overlap) with each of the existing six patches, and so there’s no problem in throwing them into our atlas. And then we have, say, all the European Petroleum Survey Group maps that people working with GIS systems are familiar with; these are compatible too, and so we can throw them in as well. Any coordinate patch that’s compatible with all the ones we’ve seen before can be thrown in, growing our atlas and making it more and more useful.

But wait! maybe the order matters. Let’s say we have an atlas $\mathcal{A}$ and two patches — $(U,\phi_U)$ and $(V,\phi_V)$ — that we want to add. I say that if $U$ and $V$ are both compatible with $\mathcal{A}$, then they are compatible with each other.

Indeed, first we restrict them both down to $(U\cap V,\phi_U)$ and $(U\cap V,\phi_V)$. Now, any patch $(A,\phi_A)\in\mathcal{A}$ that intersects $U\cap V$ intersects both $U$ and $V$. And on the intersection, we know that the transition functions $\phi_A\circ\phi_U^{-1}$ and $\phi_V\circ\phi_A^{-1}$ are in our required class — differentiable, smooth, et cetera — and so their composition is as well. But this composition is

$\displaystyle\left(\phi_V\circ\phi_A^{-1}\right)\circ\left(\phi_A\circ\phi_U^{-1}\right)=\phi_V\circ\phi_U^{-1}$

which is the transition function from $U$ to $V$. So we don’t even need to worry about the order in which we add new patches into our atlas, so long as we start with an atlas that already covers the whole manifold.

What happens when we throw all the possible patches in? An atlas which already contains every coordinate patch that is compatible with it is what we’re really after here. In the $C^1$ case, we call this a “differentiable structure”; in the $C^\infty$ case we call it a “smooth structure”; and so on. Actually, some authors say “differentiable structure” and mean $C^\infty$ transition functions, but the meaning is usually clear from context.

A topological manifold equipped with such a structure — a maximal atlas of some class — is our real object of study. In the smooth case, which we are most concerned with, we call it a “differentiable manifold” or sometimes a “smooth manifold”, when we want to distinguish the $C^1$ and $C^\infty$ cases.

March 2, 2011

## Standard Differentiable Structures

It’s high time we introduced the “standard” smooth structures on real vector spaces, which are (of course) our models for all other smooth manifolds.

The easiest one to discuss is $\mathbb{R}$. The standard smooth structure is given by starting with an extremely simple atlas: the single coordinate patch $U$ contains all of $\mathbb{R}$, and the coordinate map $\phi_U:U\to\mathbb{R}$ is just the identity! But we don’t just have this patch, of course. We also have all coordinate patches which are compatible with it.

Since the inverse $\phi_U^{-1}$ is again the identity, it’s easy to pick these out. A coordinate patch is a subset $V\subseteq\mathbb{R}$ and a real-valued function $\phi_V$ on $V$. The two transition functions between $U$ and $V$ are $\phi_V\circ\phi_U^{-1}=\phi_V$ and $\phi_U\circ\phi_V^{-1}=\phi_V^{-1}$. Both of these functions must be differentiable for the patch to fit into the standard differentiable structure. They must both be smooth to fit into the standard smooth structure.

Now, consider a finite-dimensional real vector space $V$. Again, the standard smooth structure starts with using all of $V$ as a coordinate patch. For the coordinate map, we can choose any linear isomorphism $T:V\to\mathbb{R}^n$. Of course, we know that finding such a $T$ is equivalent to picking a basis — given a basis of $V$ we can just send it to the standard basis of $\mathbb{R}^n$, and given a linear isomorphism we can use the preimages of the standard basis to get a basis of $V$.

So there’s a choice to be made: which linear isomorphism to start with. Does it matter? no! If $T_1$ and $T_2$ are two such linear isomorphisms, then $T_2\circ T_1^{-1}$ is a linear automorphism on $\mathbb{R}^n$. And clearly this transition function is smooth. Thus all the possible choices are compatible with each other and generate the same smooth — and thus the same differentiable — structure.

We say “standard” structures here. This is because — and I know this might sound sort of hard to believe — there actually do exist “nonstandard” or “wild” differentiable structures as well. The proofs establishing these examples are tremendously complicated and I’m not about to go into them now. But the fact remains: there do exist manifolds which are homeomorphic to $\mathbb{R}^4$ — they are equivalent to $\mathbb{R}^4$ as topological spaces — and yet they are not equivalent as differentiable manifolds. Any homeomorphism from one topological space to the other will not be smooth.

March 2, 2011