32
$\begingroup$

This has been on the back of my mind for one whole semester now. Its possible that in my stupidity I am missing out on something simple. But, here goes:

Let $M$ be a topological manifold. Now, even though $C^\infty$-compatibility of charts is not transitive, it is true that if two charts glue with all the charts of a given atlas, then they are compatible with each other.

Given this, one may conceivable define an equivalence relation on the atlases of a manifold, and then consider the equivalence classes. But, we do not do this. Instead we define a differentiable structure to be the maximal atlas (which being uniquely built up from a given atlas will be in its equivalence class).

Why do we do this (apart from its arguable simplicity)? Why don't we take the equivalence class of atlases in stead to be the differentiable structure?

| cite | improve this question | |
$\endgroup$
  • 1
    $\begingroup$ It's exactly the same: assume the axiom of choice, so that Zorn's lemma is valid; then every atlas is contained in a maximal atlas, and every pair of subatlases of an atlas are compatible in the sense you describe. $\endgroup$ – Zhen Lin Jun 4 '11 at 15:40
  • 15
    $\begingroup$ Zorn's lemma is a red herring here. Each atlas $A$ is contained in exactly one maximal atlas, and it is easy to desribe it: it is the union of all atlases compatible with $A$, or, if you want, the set of all charts compatible with $A$. I suspect it is this unnecessary appeal to Zorn's lemma that annoys people, and not the consideration of maximal atlases itself... $\endgroup$ – Mariano Suárez-Álvarez Jun 4 '11 at 16:35
  • $\begingroup$ I know it's been a long time since this question was asked, but can somebody help me figure out what exactly is the equivalence relation here? I think one can use something like: two atlases are equivalent if there union is also an atlas. $\endgroup$ – P-addict Jun 28 at 6:48
25
$\begingroup$

If $C$ is one of those equivalence classes, and you make the union of all the elements of $C$, you get an atlas which also belongs to $C$. It is in fact that maximal atlas of $C$. This means that nature was gracious enough to provide a canonical choice of representatives in each equivalence class, and we use it.

In practice, that an atlas be maximal means that anything that could be a coordinate chart is a coordinate chart.

| cite | improve this answer | |
$\endgroup$
22
$\begingroup$

It comes from the same aesthetic as Bertrand Russell's famous definition of $2$ as "the class of all pairs". To be a bit more specific, in set theory and logic some people define numbers as particular sets, e.g. $2 = \{ \varnothing, \{ \varnothing \} \}$. But there are (infinitely many) other possible choices, and from a certain perspective this lack of canonicity is disturbing. (One of my favorite essays in the philosophy of mathematics takes on this issue: Paul Benacerraf's What numbers could not be.) Hence Russell's solution: define $2$ (or the cardinal number associated to any set $S$) as the proper class of all sets which have the same cardinality as $S$.

The problem with this definition is that in order to be canonical we have arranged things so that the formal definition of an arguably simple, concrete mathematical object is something big and complicated. This is exactly what is happening in the definition of an atlas as a maximal collection of coordinate charts. The study of differential topology is not the study of maximal atlases any more than arithmetic is the study of proper classes: it is not fruitful to attempt to describe all the elements of a given maximal atlas, so far as I know. (Gian-Carlo Rota wrote briefly but persuasively on this topic in his Indiscrete Thoughts: he called maximal atlases "polite fictions").

There are other ways to set up the foundations of the subject which avoid making this kind of definition. For instance, a more modern and graceful approach to geometric structures on a space is via a sheaf of functions on that space. It would also be possible to take a more categorical approach.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ @Mariano: I wouldn't go so far as to call the definition annoying, but I can appreciate why people don't like it: it seems to focus attention on the wrong thing. A beginner in the subject might try to think about how to "construct" a maximal atlas -- i.e., in a more explicit way than in your comment above -- and this does not seem to be a good use of their time. $\endgroup$ – Pete L. Clark Jun 4 '11 at 16:48
  • 12
    $\begingroup$ The beginner should be immediately offered a proof of the fact that a compatible atlas can be extended to a maximal one in a unique way (and probably told to forget about the whole issue...)! For someone first encountering the definition of manifolds, an excursion into sheaf theory, or into other more categorical ways of doing this, seem like a worse use of their time :) Presumably, at that point in their life they are about to learn some basic riemannian geometry, or geometric or differential topology, none of which benefit from those approaches. $\endgroup$ – Mariano Suárez-Álvarez Jun 4 '11 at 16:57
  • 1
    $\begingroup$ @Mariano: the approach that you describe in your comment indeed sounds best for a "first course on manifolds", especially if this is an undergraduate course. But I do think that students of geometry -- not just algebraic geometry -- would benefit from being introduced to sheaves earlier on in their career. I find it curious that in subfields of geometry sheaves are de rigueur and in others they don't appear at all in the texts on the subject... $\endgroup$ – Pete L. Clark Jun 4 '11 at 22:42
  • 2
    $\begingroup$ Let me also say that in subjects of geometry which do use sheaves: if you wait until exactly the point that you need to talk about sheaves in order to introduce them -- say you are studying Dolbeault cohomology of complex manifolds -- then you are making things hard for the students, because they need to learn two things at once: (i) sheaves and (ii) the specific, nontrivial application of them to the matter at hand. This makes (i) harder than it should be. $\endgroup$ – Pete L. Clark Jun 4 '11 at 22:45
  • 3
    $\begingroup$ @Pete: that is indeed a problem. But, on the other hand, introducing a technical complication at a point where it is quite unneeded, thereby rendering something else, at an earlier point in mathematical life, harder is surely not the answer :) I think a good place to mention sheaves is in an elementary course of complex functions, for example in the context of analytic continuation, Runge's theorem and friends, &c. $\endgroup$ – Mariano Suárez-Álvarez Jun 5 '11 at 18:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.