Is there any way to uniquely characterise manifolds ( their geometrical and topological properties) without refering to charts or to a particular hyperspace containing the manifold. For example could the 2 sphere be defined as a set of points on which all geodesics are its circles of equal lenght ,which in turn are just sets which are the codomain of the exponential map. Could a metric tensor be defined on such a manifold in a coordinate free way. This is just an example of an attempt , but the idea is to use intrinsic curvatures and topological properties to define manifolds. The point is , how does one characterise the geometry of a manifold ( all the intrinsic curvatures of differential geometry , gaussian curvature most importantly) without refering to a basis or an embedding space and how that would work on some concrete exaples, like spheres or ellipsoids.
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$\begingroup$ math.stackexchange.com/q/2010035/127263 $\endgroup$– WojowuCommented Jan 27, 2019 at 16:29
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4$\begingroup$ Anything that involves a preferred class of geodesics is going to have a lot more structure than a differentiable manifold. $\endgroup$– Steven LandsburgCommented Jan 27, 2019 at 20:15
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1$\begingroup$ It's not clear, at least to me, whether you're asking about a way to define, without using coordinates or an embedding into Euclidean space, the topological structure of a smooth manifold or the geometric properties of a Riemannian manifold without using coordinates. A smooth manifold has no geometric properties, only topological ones. In particular, there is no concept of curvature on a manifold per se. However, if the manifold has a Riemannian metric or is emedded into Euclidean space, then the metric or embedding has geometric properties such as curvature. $\endgroup$– Deane YangCommented Jan 27, 2019 at 22:51
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$\begingroup$ How is that not clear to you. I specifically asked for things like gaussian curvature, metric tensor etc..which surely isn't purely topological. PS geometrical and topological aspects of geometric objects are never completely independent.. $\endgroup$– SheldonCommented Jan 27, 2019 at 23:37
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2$\begingroup$ Normally, when we say simply “a manifold”, we don’t assume it to have any geometric structure on it. The sphere, purely as a manifold, has no specific geometric properties. However, the unit sphere in $\mathbb{R}^n$ does. $\endgroup$– Deane YangCommented Jan 28, 2019 at 3:57
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May be Alexandrov spaces is what you are looking for. See, for example, http://www.math.psu.edu/petrunin/papers/akp-papers/shiohama_1.pdf as an introduction.
