Hello, I was wondering if if has any sense of talking about angles on an arbitrary metric space (where only a distance function with some properties is defined) At first sight it seems to not has any sense, only some metric spaces has angles, namely does that are induced from an inner product space (where angles are defined as usual). But defining an angle as the "distance" (lenght) of a segment of circunference (with an integral) between two elements of the set has no sense? Because we can make (riemann)-integrals on an abstract metric space, can't we? Or that is wrong too? Thanks
No, one can only define angles in inner product spaces. That's exactly what inner product spaces are good for! I don't see how you would start in defining an integral on metric spaces?? To have an integral in a good sense, you must have access to a measure with good properties. I don't see immediately what kind of measure you could take on an arbitrary metric space...
Actually, applying geometric ideas to metric spaces is a very active area of research. Although you don't get a very good analogue of angle per sey, you are able to equip a fairly robust notion of geometry. The notion of (ricci) curvature, for example, is crucial for studying geometries in metric spaces. As far as integration is concerned, probably the nicest and most widely used measure, the Hausdorff measure, actually finds its most natural setting in that of metric spaces. IMO, the best introductory resources for this fascinating field are those of "A course in Metric Geometry" by Burago et al, "Lectures on Analysis on Metric Spaces" by the late Heinonen, and Topics on Analysis in Metric Spaces by Ambrosio and Tilli. Cheers!
You are right, it makes no sense to talk about angles in an arbitrary metric space. I think my confusion came from de Riemannan "metric" on a Riemann manifold. There the word metric is used but really is more similar to a inner product space that to a metric space (angles has sence) Maybe it would be clearer if the riemannian metric tensor would be called the riemannian inner product tensor? (I don't know why they used the word metric on that tensor if it really gives an inner product). Thanks.
You just have to learn the lingo. The Riemannian metric makes each tangent space into an inner product space - but it also makes the manifold into a metric space. If the space is complete, the distance between two points is the length of the shortest path connecting them. since the space is generally curved the idea of angle makes sense only infinitessimaly.
Actually, like shoescreen said, angles can be defined in general metric spaces under some conditions, even without a Riemannian structure, as described in, e.g., "Metric Geometry" by BBI.
This isn't really an answer, oh well. A space of homogeneous type is consists of a set X, a quasi-distance δ i.e. a functions [itex]\delta:X^2 \rightarrow [0,\infty)[/itex] that satisfies [itex]\delta(x,y)=0[/itex] iff [itex]x=y[/itex], [itex]\delta(x,y) = \delta(y,x)[/itex] for all [itex]x,y\in X[/itex], and there is a positive integer [itex]\kappa > 0[/itex] such that [itex]\delta(x,y) \leq \kappa(\delta(x,z)+\delta(y,x))[/itex], and lastly a measure defined on a sigma-algebra of subsets of X which contains the open balls and satisfies the doubling condition: there exists a positive constant A such that for every x in X and every r>0, we have [itex]0 < \mu(B_{2r}(x)) \leq A \mu(B_r(x))[/itex]. This too is a huge area in modern analysis.
Hi I think that you're a little confused: a set [itex]X[/itex] with a function [itex]f:X\times X\rightarrow \mathbb{R}\cup \{0 \}[/itex] holding triangular inequality, simetry and [itex]d(x,y)=0 \leftrightarrow x = y[/itex] is called a metric space, and the function [itex]f[/itex] is called metric. This in the analysis context. In the differential geometry context, if [itex]M[/itex] is a manifold and for [itex]p\in M[/itex], [itex]g_p[/itex] is a "smooth" inner product over the tangent space of [itex]M[/itex], then [itex]g_p[/itex] is called a Riemannian metric on [itex]M[/itex]. I prefer to call "Riemannian inner product" to avoid confusion. The Riemannian metric on a manifold is not a metric as in the analysis context, but we can define the lenght of curves on the manifold as the integral of the norm of the tangent of the curve. The function that measures the infimus of the lengths of curves between two points on the manifold, defines a metric (in the analysis sence) on M. In general, metric spaces are not equipped with Riemannian Metrics.
Not only are angles defined (under specific conditions) in metric spaces that are not inner-product spaces, but there are other bizarre-seeming ideas in Burago's book, like, e.g., a notion for a sequence of metric spaces to converge, and these are not all subspaces of the same space either.
Just wanted to make my response more specific, from some notes (unfortunately, I don't have the book with me, so some details may be wrong, but this is the general layout): Burago defines a comparison angle by using the arccosine function. Given three points x,y,z in a matric space (X,d) , we consider the values of d(x,y), d(x,z), d(y,z), and apply the law of cosines to define the arccos . The idea is that the comparison angle is one that would be formed if we had a triangle in R^2 whose sides had lengths d(x,y), d(x,z) and d(y,z). There is also a notion of an angle between paths a(s),b(t) in X --seen as maps of an interval [o,e) into the space, where the paths have the same initial point. Then the angle between these paths is defined as the angle in a limiting triangle (as in above paragraph), if this limiting angle: lim_>t,s->0 (a(s),p, b(t)) exists, i.e., for different values of s,t, as we approach 0, we will have different triangles (modeled in R^2 ), and an angle for each choice. From this, we will get a sequence of angles, which may or may not converge , as s,t->0 . If the sequence does converge, then the angle exists.
I'm not 100%; it had "Metric Geometry" in the name, and the authors were two Buragos and one Ivanov. Let me see if I can find the actual book. I thought he was kidding when he talked about convergent sequences of metric spaces. Let me check.