How much of differential geometry can be developed entirely without atlases? - MathOverflow [closed]

How much of differential geometry can be developed entirely without atlases?

2010-02-10T08:41:12Z

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.

Answer by Orbicular for How much of differential geometry can be developed entirely without atlases?

2010-02-10T09:02:55Z

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).

Answer by Pete L. Clark for How much of differential geometry can be developed entirely without atlases?

2010-02-10T09:39:17Z

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.

Answer by babubba for How much of differential geometry can be developed entirely without atlases?

2010-02-10T09:59:23Z

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).

Answer by Dmitri Pavlov for How much of differential geometry can be developed entirely without atlases?

2010-02-10T10:03:20Z

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.

Answer by Kevin Lin for How much of differential geometry can be developed entirely without atlases?

2010-02-10T10:10:18Z

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.

Answer by Dmitri Pavlov for How much of differential geometry can be developed entirely without atlases?

2010-02-10T10:22:35Z

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.).

Answer by S1 for How much of differential geometry can be developed entirely without atlases?

2010-02-10T18:12:15Z

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...).

Answer by Emerton for How much of differential geometry can be developed entirely without atlases?

2010-02-10T18:55:44Z

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.]

Answer by BMann for How much of differential geometry can be developed entirely without atlases?

2010-03-04T00:56:56Z

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.

Answer by Martin Gisser for How much of differential geometry can be developed entirely without atlases?

2010-09-11T12:43:05Z

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.

Answer by unknown (google) for How much of differential geometry can be developed entirely without atlases?

2010-10-15T06:43:15Z

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.

Answer by Patrick I-Z for How much of differential geometry can be developed entirely without atlases?

2011-01-04T10:32:12Z

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 :-)