Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

We may define a topological manifold to be a second-countable Hausdorff space such that every point has an open neighborhood homeomorphic to an open subset of $\mathbb{R}^n$. We can further define a smooth manifold to be a topological manifold equipped with a structure sheaf of rings of smooth functions by transport of structure from $\mathbb{R}^n$, since $\mathbb{R}^n$ has a canonical sheaf of differentiable functions $\mathbb{R}^n\to \mathbb{R}$, with a canonical restriction sheaf to any open subset. This gives a manifold as a locally ringed space. (Of course this definition generalizes to all sorts of other kinds of manifolds with minor adjustments).

Then the questions: If we totally ignore the definition using atlases, will we at some point hit a wall? Can we fully develop differential geometry without ever resorting to atlases?

Regardless of the above answer, are there any books that develop differential geometry primarily from a "locally ringed space" viewpoint, dropping into the language of atlases only when necessary? I looked at Kashiwara & Schapira's "Sheaves on Manifolds", but that's much more focused on sheaves of abelian groups and (co)homology.

Edit:

To clarify (Since Pete and Kevin misunderstood): It's easy to show that the approaches are equivalent, but proofs using charts don't always translate easily to proofs using sheaves.

share|improve this question
14  
I am a Riemannian geometer, and I am curious what you are trying to accomplish. The basic reason why some differential geometry is done in coordinates is that using the correct coordinates adapted for the problem at hand makes computations much easier and gives invaluable insight like the Jordan normal form gives insight into matrices. Try computing curvature tensor of complex projective space without coordinates. I know only a few cases where sheaf theoretic language helps but the the most part I do not see why bother. It is more efficient to use other methods. –  Igor Belegradek Feb 10 '10 at 16:47
5  
@Igor: I prefer to leave computations as the "last step", so to speak. That is, proofs should be coordinate free when possible. –  Harry Gindi Feb 10 '10 at 17:23
6  
It's like in linear algebra, where choosing coordinates can actually obscure the important points. –  Harry Gindi Feb 10 '10 at 17:25
6  
@Igor: I would agree that trying to study a Riemannian manifold via its sheaf of smooth functions has limited potential, but on the other hand, I do think a lot of Riemannian geometry is best done without co-ordinates. In co-ordinates, you have a lot of extra baggage that is a pain in the neck, including Christoffel symbols. As for your example of complex projective space, there are definitely clean and easy ways to compute its curvature without using co-ordinates. –  Deane Yang Feb 10 '10 at 18:37
5  
@Harry: there are many cases when working in coordinates makes things much easier; in other situations coordinate-free arguments work better. I think sticking to either approach will inevitably "obscure important points". –  Igor Belegradek Feb 10 '10 at 19:11
show 7 more comments

closed as no longer relevant by Harry Gindi, Akhil Mathew, Alex Bartel, Andrey Rekalo, Qiaochu Yuan Jan 15 '11 at 14:27

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

12 Answers

up vote 23 down vote accepted

There is the book by Ramanan "Global Calculus" which develops differential geometry relying heavily on sheaf theory (you should see his definition of connection algebra...). He avoids the magic words "locally ringed space" by requiring the structure sheaf to be a subsheaf of the sheaf of continuous functions (hence maximal ideal of stalks = vanishing functions).

share|improve this answer
    
This book looks really good! –  Grétar Amazeen Feb 11 '10 at 15:51
    
i currently have this book checked out from my library. havent spent enough time with it though. –  Sean Tilson Jun 17 '10 at 14:45
    
A nitpick: It is not true that every subsheaf of rings of the sheaf of continuous functions gives a locally ringed space. For example, take the real line (with its standard topology) and let $\mathcal{O}(U)$ be the functions on $U$ which, on each connected component of $U$, are restrictions of polynomials. Then the stalk at any point is $\mathbb{R}[x]$, which is not a local ring. However, it is true that each stalk contains a canonical maximal ideal, of the functions which vanish at that point. This may be enough for many purposes. –  David Speyer Jan 4 '11 at 13:48
add comment

This is a comment, not an answer, but is too long to fit in the comment box: having read the question, answers, and comments, I don't quite follow the intent of this question:

We can define a manifold to be a locally ringed space in which each point has a neighbourhood isomorphic to an open subset of ${\mathbb R}^n$ (or even just ${\mathbb R}^n$ itself) with its sheaf of smooth functions (plus second countability and Hausdorffness, if you like). As was remarked by Dmitri, the collection of all such will then form an atlas, but one doesn't need to say this.

As Pete Clark says, what I've said so far is evident.

But it seems that another aspect of the question is whether one can always avoid working in coordinates. This seems to have nothing to do with atlases.

E.g. in arguments in the locally ringed space set-up, one will certainly in many instances verify that a property can be checked locally, and then verify it on Euclidean space with its natural smooth structure. (Just as in the theory of schemes, one often shows that a property is local, and then checks it in the affine case.)

Now one can ask: can one avoid the latter kinds of arguments? This seems unlikely: manifolds are defined to be locally Euclidean (no matter which of the possible formalisms one is using), and so if one is proving theorems about manifolds, one will have to use this somewhere. For example, one can surely define the tangent sheaf in a coordiante free way, but to prove that it is locally free of rank equal to the dimension of the manifold, one is going to reduce to a local computation and then appeal to calculus on Euclidean spaces; there is no other way!

[EDIT: The last sentence may be too categorical of a declaration; see Dmitri Pavlov's answer for a suggestion of a more substantially algebraic reformulation of the notion of manifold.]

share|improve this answer
    
We can't reduce it to commutative algebra by using the module of kahler differentials? –  Harry Gindi Feb 10 '10 at 19:06
5  
"Invoking smoothness" means "doing calculus locally on Euclidean space". Somewhere, there is a property of smooth functions on Euclidean space that you are going to have to use, that will be a matter of calculus. It is certainly possible to develop the foundations in a more coordinate-free way than most texts; the first of Spivak's five volumes is good for this (since he goes carefully over all the possible approaches to the various definitions). But the local Euclidean nature of manifolds is intrinsic to the definition, and will always have to be used somewhere in some way. –  Emerton Feb 10 '10 at 20:04
1  
@Harry: What does "purely algebraic" mean? The local topological structure of Euclidean space is quite crucial in differential geometry. –  Deane Yang Feb 11 '10 at 2:46
1  
Dear Deane, I just mean that at some point you have to refer back to Euclidean space and use its analytic properties in some way. I agree that it is a fairly tautological point. I was just trying to get my mind around what was behind the objection to atlases (which for me just evokes a collection of Euclidean opens covering the manifold), and pointing out that somewhere you have to exploit the locally Euclidean nature of the situation. –  Emerton Feb 11 '10 at 3:16
4  
How do you know that there is no other way? I mentioned in my answer that “One might hope that these additional conditions can be formulated in terms of dimensions of some vector bundles constructed from this algebra (e.g., tangent bundle, jet bundle, connections etc.).”. In other words, the fact that the tangent bundle is locally free of rank equal to the dimension of the manifold might very well become a part of a new definition of a manifold. –  Dmitri Pavlov Feb 11 '10 at 8:12
show 4 more comments

I would say that it is a priori clear that ALL of differential geometry (and differential topology, complex geometry, etc.) can be developed using the language of locally ringed spaces rather than atlases.

"Atlas" was never an important technique in the subject; it is just a definition, a sort of reification of the idea that there is some specific "structure" which corresponds to the ability to tell when a function on an abstract manifold is smooth or not. But the main idea has never been anything other than the following simple one: for a function to be smooth, it has to be compatible with each of the coordinate charts (i.e., a differentiability condition on the composite maps) which also have to be compatible with each other and cover the manifold in question. [And one easily generalizes from smooth functions on a smooth manifold to smooth maps between smooth manifolds.]

The awkward part of the definition of atlas comes when we depart from the simple (and obviously necessary) conditions above and say "Now an atlas is a maximal set of such charts [or possibly an equivalence class of sets of charts]". This part of the definition has been explicitly made fun of by Gian-Carlo Rota in his [I believe; I haven't gone back to check this point] Indiscrete Thoughts, in which he refers to the concept as a "polite fiction".

Anyway, the point is that you want to be able to say what a smooth map between manifolds is. If we agree on what such maps are, then we are talking about the same concrete category of manifolds and smooth maps. You can certainly check that the notion of LRS gives rise to the same category -- but via a formalism which you, I and much of the contemporary mathematical world probably views as more graceful than that of atlases -- and that's really all that matters.

share|improve this answer
    
That's what I'd hoped to hear. The question is: things like the inverse function theorem seem like they're more easily stated using atlases (at least I don't know of a good statement using locally ringed spaces). I'd like to see all of the common theorems stated in terms of LRS's, whence came the book request. –  Harry Gindi Feb 10 '10 at 9:45
2  
I actually don't see how atlases come up in the statement of the inverse function theorem [or even what this result has to do with manifolds at all, really; you can apply it to maps between manifolds, but by nature it's a local result]. Do you have a specific formulation or reference in mind? –  Pete L. Clark Feb 10 '10 at 9:52
2  
This reminded me of the message that I try to get across in my "comparative smootheology" talks (that's talks, not article): that the point (ha ha) of smooth manifolds is to define what "smooth map" means and for that we only need to agree on what smooth curves are - the whole "atlas" and "chart" stuff is just to ensure that we agree on this. Interestingly, if you read Kriegl and Michor's book then you get the impression that they regard IFT as being what separates "hard" and "soft" geometry (my words). –  Andrew Stacey Feb 10 '10 at 11:04
3  
Andrew, I have to agree with you. What's the point of doing differential geometry without a being able to do analysis on them? I enjoy synthetic reasoning, but when looking at how far they got, I'd rather rely on a theory with an inverse function theorem. (Anyone ever thought about sheaves with values in Hilbert/Banach manifolds?) –  Orbicular Feb 10 '10 at 11:41
3  
I don't really follow the debate going on here. So what if a few key tools such as the IFT need to be proved using local co-ordinates. Once you have it, it can be stated and used in a co-ordinate-free environment. Virtually all of PDE theory is proved using local or global co-ordinates, but once you have developed the necessary tools and restated the theorems in a co-ordinate-free form, you can often successfully avoid co-ordinates when applying the PDE theorems to differential geometry. –  Deane Yang Feb 11 '10 at 2:42
show 4 more comments

This matter is discussed in Tennison's "sheaf theory". He defines manifolds your way and writes (p. 90):
"...the above definition is in accordance with the more usual definitions in terms of atlases of charts with transition maps of the appropriate kind[...], with two possible exceptions. Some authors may require that $X$ have a countable basis of open sets. Other authors may insist that $(X,\mathcal{O}_X )$ satisfy a separation (hausdorff) condition..."
You might also be interested in the treatment "Smooth Manifolds and Observables" by Jet Nestruev, where manifolds are characterised by their ring of smooth functions (it has to be "geometric").
Last but not least you might take a look at "Algebraic Geometry over C-infinity rings" by Dominic Joyce,
http://arxiv.org/abs/1001.0023
where "classical" differential geometry is widely generalised (in the sense of Spivak's derived manifolds).

share|improve this answer
2  
Let me just point out that the "exceptions" referred to above have nothing to do with charts (which are local in nature), but only give restrictions on the underlying topological space. In particular, the question of whether one wants manifolds to be second countable and/or Hausdorff comes up in just the same way in the classical formulation as well. (I guess the point of the remark is that it is less "classical" not to require these conditions as a part of the definition.) –  Pete L. Clark Feb 10 '10 at 9:47
add comment

Your definition of a smooth manifold still uses atlases in a slightly disguised way because it amounts to saying that a smooth manifold is a topological manifold with an open cover whose elements are equipped with an isomorphism of the restriction of the structure sheaf and the standard sheaf on R^n. This open cover is nothing else but an atlas.

Thus one still needs an atlas-free definition of a smooth manifold. One possible way to do this is to define the category of smooth manifolds as the opposite category of the full subcategory of the category of real algebras consisting of real algebras satisfying certain properties, e.g., the intersection of kernels of all homomorphisms to R must be 0. One might hope that these additional conditions can be formulated in terms of dimensions of some vector bundles constructed from this algebra (e.g., tangent bundle, jet bundle, connections etc.).

share|improve this answer
1  
I'd be very interested in seeing something like this. –  Harry Gindi Feb 10 '10 at 10:23
1  
If you have a look at Toen's notes (a masters course on stacks) he recovers the definition of manifold just as a particular sheaf on the site consisting of open subsets of euclidean spaces (with smooth maps between them) and coverings being jointly surjective families of open immersions. It's all part of a general theory of "Geometric Contexts" (although theory is probably a too big a name). In the same vein he recovers schemes from affine schemes (of course!). –  babubba Feb 10 '10 at 13:39
2  
"Thus one still needs an atlas-free definition of a smooth manifold." Why? Not "Why are locally ringed spaces not truly atlas-free?" I understand that and that was essentially my response, but why would be it be desirable to speak of manifolds in a way that avoids mentioning that they are locally isomorphic to Euclidean space? What does one hope to gain by doing so? Anything in manifold theory itself? –  Pete L. Clark Feb 11 '10 at 6:45
1  
@Dmitri I think a lot of the most important analytic properties of smooth manifolds-particularly those that arise in geometric analysis-would suffer from such a treatment. And we won't even bring up the crucial applications of manifold theory to modern physics,which are legion. I think it can be done,I just think it's going to cause more headaches then it would solve. –  Andrew L Sep 11 '10 at 20:41
3  
@Andrew L: Do you have any concrete examples of properties of smooth manifolds that would suffer from such a treatment? By the way, when applications of smooth manifolds to physics are expressed in coordinate-free form, they become more easy to understand. –  Dmitri Pavlov Sep 12 '10 at 13:53
show 7 more comments

It is an easy exercise to show that the standard definition of a manifold is equivalent to the ringed spaces definition of a manifold. (Try it!)

The latter definition is nice because the sheaf condition gets rid of the need for all that "maximal atlas" business.

share|improve this answer
1  
@Kevin: Yes, that's exactly what I was getting at in my answer. +1 for your exhortation "Try it!" which would be my advice as well. –  Pete L. Clark Feb 10 '10 at 10:20
    
I've put an explanation in the OP. Proving equivalence is pretty straightforward. The problem is, sometimes proving something about a mathematical object is easier in one formalism than another one. For example, the functor of points approach to algebraic geometry allows us to get lots of proofs for free, but some proofs are still easier using locally ringed spaces (consider the algebraic and analytic structures of a complex variety). –  Harry Gindi Feb 10 '10 at 10:33
    
And of course I've "tried it!" it was on homework 2 or 3. –  Harry Gindi Feb 10 '10 at 10:34
    
Rather, the analytic structure associated with the variety. –  Harry Gindi Feb 10 '10 at 10:58
8  
Well, you don't really remove the maximal atlas, you just hide it in another box. It's fairly impossible to check whether an atlas is maximal, that's obviously true. But on the other hand - do you know all open sets of R^n, together with all smooth functions on them? In practice you are dealing with an open cover only (Cech-style). The same is true for atlases! –  Orbicular Feb 10 '10 at 11:36
add comment

Apart from Nestruev's book which is good but unfortunately very elementary, I recommend you to take a look at Ramanan's Global Calculus. Ramanan almost manages to avoid coordinates except for a few places and yet is able to prove nontrivial theorems.

share|improve this answer
add comment

There is another way to develop differential geometry without atlases, and even without charts, that is Diffeology. I'm not sure this is the right answer to your question but it worths looking at.

Comment: There are many ways to develop differential geometry without atlases. You may change the category of differentiable manifolds for a larger one. This is the case for Diffeology, or Differential Spaces à la Sikorski. These two approaches correspond to the two ways you may interpret the smooth structure: in the first case the smooth structure is characterized by the smooth parametrizations in the space (called plots), this is the "in-way", in the second case the smooth structure is characterized by the smooth functions from the space into the field of real numbers $\bf R$, the "out-way". In some sense they are "dual" but not equivalent approaches. The "intersection" of these two approches gives the so-called Frölicher spaces (reflexive diffeological spaces). The diffeological way gives a richer category than the Sikorski's one. For example it gives a non trivial structure for quotients (like spaces of leaves of a dense foliation, for example) where Sikorski structure is trivial. The two of them gives access to infinite dimensional spaces, without necessarily modeling these spaces on topological vector spaces. You can develop a whole theory of homotopy, cohomology, differential calculus and De-Rham cohomology, groups, fiber bundles etc. in diffeology without loosing much of what you have learned in manifold differential geometry.

Well, there is a lot say, may be too much for this discussion :-)

share|improve this answer
    
By the way, this is related to the definition mentioned above, as appearing for instance in Toen's master course notes on algebraic stacks: a diffeological space is a concrete sheaf on the category of smooth open balls, and a smooth manifold is a locally representable such sheaf. –  Urs Schreiber Jan 5 '11 at 0:20
    
Hey Urs, yes sure! However, I must admit that my intuition in diffeology never came from the formal categorical properties, even if I never ignored this point of view, but it came from my familiarity with ordinary differential geometry (for example, when I had to define what is a diffeological fiber bundle to include any quotient of kind $G/H$). I tried stay the closest as possible to the classical vocabulary, just to spare my energy and focus on the geometrical intuition. But I agree, times changing :) –  Patrick I-Z Jan 5 '11 at 7:41
    
Yesm sure. It just occurred to me that all these commenters here are happily chatting about smooth analogs of algebraic stacks, but not so much about your diffeology. Which is kind of curious, because its two ways to talk about the same thing, essentially. –  Urs Schreiber Jan 5 '11 at 11:55
add comment

Just as everyone else is saying, I don't think we really gain anything (at least if we stick to just studying differential and Riemannian manifolds). In essence, both approaches tell us exactly what the smooth functions on our manifold look like, and differential geometry is mainly concerned with the (analytic) behavior of these functions on our manifold and what they tell us about the manifold.

For example, if one wants to know that geodesics exist, or that parallel vector fields exist, the problem reduces to one of finding unique solutions of differential equations locally (and then using the uniqueness to patch everything together). I don't see an advantage of one side over the other when developing the theory.

On the other hand, there may be more advanced topics which would be better looked at from a sheaf theoretic viewpoint, but I can't say much about this.

share|improve this answer
add comment

Take also a look at "Theory of Lie Groups I" by C. Chevalley (Princeton UP 1946), chapter III. That approach (differential structure defined by choosing a ring of functions) is further developed in R. Penrose & W. Rindler, "Spinors and Space-Time 1" (Cambridge UP 1984), section 4.1 ff.

share|improve this answer
add comment

There is something called abstract differential geometry "developed" by Mallios et al (wikipedia). Perhaps you should skim through the first chapter of his book "Modern Differential Geometry in Gauge Theories: Maxwell fields". It looks pretty nice (despite the physical title). But I don't know how readable or how awesome all this is, I did not read it (there are too many nice explicit things to figure out...).

share|improve this answer
    
I'm sceptic about Mallios' generalization, in particular since he claims to do "physics"... –  Orbicular Feb 10 '10 at 18:22
    
Nah, I did not have the impression that the first(!) chapter has been written by a "physicist". I think most of the stuff has also been discussed in "Geometry of vector sheaves: an axiomatic approach to differential geometry" by Mallios and this is entirely mathematics... –  user717 Feb 10 '10 at 19:32
add comment

I'm on a similar quest and have a tentative answer (I'm currently working it out). My definition of a generalized differential manifold would be: A locally ringed space with nontrivial cotangent sheaf. Covariant derivatives can be defined using only the cotangent sheaf, and there is a dual version of the fundamental lemma of Riemannian geometry for this. (My creed: forget tangent space.)

The abstract differential geometry of Mallios et al looks quite attractive to me, but I haven't yet read it in detail (lacking time, money, and a useable math library). One immediate problem I see with their definition of vector bundle, being the usual suspect, a locally finite and free sheaf module. But a major motivation for using sheaves is infinite dimensional applications, like path space (I suspect path space should be introduced before geodesics, Jacobi fields, General Relativity, perhaps Brownian motion... Dunno how sheaves connect to uniform spaces...).

I have an incomplete copy of yummy lectures by Jürgen Bingener, Regensburg 1983/4 on basics of calculus and Riemannian geometry in the language of ringed spaces. Need to contact him and ask for more.

share|improve this answer
add comment

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