# The Unapologetic Mathematician

## 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 - Posted by | Differential Topology, Topology

1. […] 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 […]

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2. Paragraph 5:

the open half-disk of points (x,y) with x^2+y^2=0 instead be x>0?

Paragraph last-1: “…contains all every coordinate…”

Comment by cclingen | March 3, 2011 | Reply

3. Hmmm… should say should it be x=0 instead?

Comment by cclingen | March 3, 2011 | Reply

4. Thanks for catching those. Yes, it should have been $x>0$ as I’d said earlier in that graf.

Comment by John Armstrong | March 3, 2011 | Reply

5. […] if and are – and -dimensional smooth manifolds, respectively, then we can come up with an atlas that makes the product space into an -dimensional smooth manifold, and that it satisfies the […]

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6. […] so that is supported in . In particular, we can let be the collection of coordinate patches in a smooth atlas, so each of the functions “lives in” a single local coordinate […]

Pingback by Partitions of Unity Subordinate to a Cover « The Unapologetic Mathematician | March 8, 2011 | Reply

7. […] our extension of the implicit function theorem in hand, we have another way of getting at the sphere, this time as a […]

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8. […] shows that we can find an atlas on all of whose patches have compatible orientations. Given any atlas at all for , either use a […]

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