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The Wikipedia's article for geometry is somehow overwhelming. To make things clear, allow me to ask some questions:

  1. I wonder if "geometry" can be defined as the study of a metric space (possibly with or without other structures)?

  2. Any thing more general than metric space (such as uniform spaces and topological spaces) is not in the scope of "geometry"?

  3. Does "geometry" assume the set under study to have some algebraic structure?

    Also there is algebraic geometries.

  4. Is the underlying set a topological vector space, normed space, inner product space, or even Euclidean space?

  5. Since projective and affine spaces are pure algebraic concepts without metrics, why are there "projective geometry" and "affine geometry"?

Thanks and regards!

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Differential geometry doesn't generally assume a metric. Even if we insist that a differential structure alone is not sufficient to merit the label geometry, once we include a connection, we have geodesics, and I expect you will want to call it geometry. But there need not be a metric. –  yasmar Mar 17 '12 at 12:14
    
@yasmar: Thanks! (1) So are you saying it is connection that merit the label geometry? (2) Can this view be unified with geometries on inner product space, on (pure algebraic) projective and affine spaces? –  Tim Mar 17 '12 at 12:17
    
Hi @Tim, I would certainly not go so far as to say that a connection is what is needed for geometry. In fact, I think the question in the title of your post is a good one, but I doubt there is any universally accepted answer. I would like to see more opinions expressed. I think 'geometry' can happen even in the absence of a metric, and a connection is an example, but there must be other examples of structures that don't imply a connection, or even a manifold. I can't competently answer your second question in your comment. I hope somebody else will. –  yasmar Mar 17 '12 at 18:18

2 Answers 2

A partial (i.e. highly incomplete) answer to your questions on algebraic geometry:

Is the underlying set a topological vector space, normed space, inner product space, or even Euclidean space?

It depends - in a lot of algebraic geometry, you think about particular subsets of affine or projective spaces, which are not equipped with standard metrics, norms or inner products. In more abstract settings you might even be dealing with schemes, which are a big generalisation of these spaces. Very loosely, they can be thought of as being constructed by turning rings into geometric objects and then gluing them together. For example, $n$-dimensional affine space over a field $K$ is the geometric object associated to the polynomial ring $K[x_1,\ldots,x_n]$, and projective space can be obtained by gluing together some affine spaces in a particular way.

It is definitely too strong to insist that the underlying space is Euclidean, because that would be ignoring all of the interesting geometry on spheres and hyperbolic spaces, among others. (Here I mean geometry in the more traditional sense of theorems about distances, angles, intersections of lines and so on, but there is a lot more to spherical and hyperbolic geometry as well).

Since projective and affine spaces are pure algebraic concepts without metrics, why are there "projective geometry" and "affine geometry"?

I may have got the wrong idea from your question here, but I've taken it to be asking what kinds of geometric theorems you can have without metrics. A number of the big theorems in algebraic geometry are primarily algebraic in nature, but are motivated by geometric questions, while others are more recognizable as being geometric theorems.

One example is as follows. The main object of study in affine and projective algebraic geometry is an algebraic variety, which is a subset of affine or projective space defined by an ideal of polynomial equations. In particular, any curve in the plane ($2$-dimensional affine space) defined by a single polynomial is an algebraic variety. Up to worrying about degenerate cases, we can say that the degree of the curve is the degree of the polynomial defining it. Then Bézout's theorem states that two "general" curves of degrees $d_1$ and $d_2$ intersect in $d_1d_2$ points (there is a lot of interesting mathematics involved in saying exactly what "general" means). So this is an example of a fairly strong theorem, which says a lot about how curves behave, without any reference to a metric.

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+1 Thanks! I wonder what "turning rings into geometric objects and then gluing them together" means? Especially what are "geometric objects"? Do they assume metric or further structures on their sets? –  Tim Mar 16 '12 at 0:10
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Here what I mean is to take the spectrum of a ring, which yields a set with a topology and a sheaf of functions, but no metric. See en.wikipedia.org/wiki/Spectrum_of_a_ring, although this article is fairly technical. Again the topology is the one which is supposed to make sets defined as the zero set of an ideal closed, so it is not induced by any metric. –  Matt Pressland Mar 16 '12 at 0:18
    
Thanks! (1) So the kind of "geometry" in your reply is purely topological and algebraic. Why isn't it called "algebraic topology" instead of "algebraic geometry"? (2) Can it be unified with the other kind of "geometry" such as for inner product space? –  Tim Mar 16 '12 at 2:08
    
One answer to (1) is that the notion of isomorphism for topological spaces is much weaker than that for the geometric objects I'm describing here. One way to think of this is to imagine the unit disk in the plane, and think of raising the origin up into a point, so you get a cone. Topologically, that space is homeomorphic to the disk, so the algebro-topological invariants are the same as for the disk, but the geometry somehow "knows" that the origin is now special, so the algebro-geometric invariants reflect this. As for (2), I don't know. It's certainly not something I've seen. –  Matt Pressland Mar 16 '12 at 8:48
    
About (1), isomorphism for topological spaces means bijective continuous mapping with continuous inverse. But for the geometric objects you described earlier, what is the isomorphism? –  Tim Mar 16 '12 at 9:09

I would say that "geometry" in a strict sense needs a notion of length AND some kind of inner product (I'd say a quadratic form). Thus the study of most manifolds is geometry since it's usually equipped with a standard Riemanian or Lorentz form (it's the case in particular for projective or affine spaces on $\mathbb{C}$ or $\mathbb{R})$.

I wouldn't say that the study of general, abstract metric spaces is "geometry" in that sense (and therefore much less so for general topological spaces), but I think it's not something set in stone.

Also there is a very broad acceptation of the term "geometry" as in "algebra, analysis, probabilities and geometry", and since general topology fits better in geometry than in analysis or algebra, you could say that it belongs there, but for me it's a slight abuse of language.

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+1 Thanks! "Metric spaces" at Wikipedia says "the geometric properties of the space depend on the metric chosen, and by using a different metric we can construct interesting non-Euclidean geometries such as those used in the theory of general relativity." It leads me to think maybe "geometry" only or at least assumes metric spaces. –  Tim Mar 15 '12 at 23:15
    
the metrics they speak of are derived from a quadratic form of signature (1,3) (a Lorentz form) which gives more structure than just a metric. also the term "non-euclidian geometries" refer to the notions of curvature and geodesics, for which you need more structure than just a distance. that's why I said that in my answer –  Glougloubarbaki Mar 15 '12 at 23:18
    
So "projective geometry" and "affine geometry" don't just assume pure algebraic projective and affine spaces, do they? How about algebraic geometry? –  Tim Mar 15 '12 at 23:22
    
I would say that algebraic geometry does have a "pure" notion of affine and projective spaces, without metrics or preferred co-ordinates; in fact the first definition of affine space I was given as an undergraduate was that it is a Euclidean space with no metric and no distinguished origin. I've tried to give a flavour of this in my answer, but there's an enormous amount that could be said on this subject so it's certainly not a complete answer. –  Matt Pressland Mar 15 '12 at 23:30
    
I don't know too much about algebraic geometry, and Matt Presland gave an extensive answer below for that. as for projective geometry and affine geometry, as I said, it's not so much that you "assume", but when the fields are $\mathbb{C}$ or $\mathbb{R}$ (ie with a nice topology), you have some natural metric. –  Glougloubarbaki Mar 15 '12 at 23:33

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