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I am confused by some definitions. Forgive the looseness of my language.

A Cartesian space is basically a space of points that can be represented by n-tuples ( and other things, but I won't go into specifics). Open sets are denoted by $|x^i_{0}-x^i|<{\epsilon}$. It is denoted by $R^n$ (according to Dubrovin's book).

Euclidean spaces are Cartesian spaces where there exists co-ordinate systems in which the metric can be represented by the Kronecker delta. Its open sets can be represented as open balls with the usual definition of 'balls'.

Riemannian spaces are Cartesian spaces which have a positive definite quadratic form on its tangent vectors. Euclidean spaces are special cases of Riemannian spaces. So I imagine that open sets are determined by the riemannian metric? or are they determined by euclidean balls?

Now manifolds are spaces, whose subsets can be identified with Euclidean spaces. But such charts can be equipped with additional Riemannian metric.

So I am confused. Does that mean that the charts on manifolds have 2 'metrics', a fundamental one determined via the identification with a region of euclidean space and second additional riemannian metric?

Also open sets on a manifold may be defined by identifying them with the open sets in Euclidean space. So then how does the additional riemannian metric help determine open sets on the manifold?

I am totally confused about the hierarchy of these spaces, if Cartesian is on the top and Riemannian is below and Euclidean even lower than that, than why are manifolds defined first by identification with Euclidean spaces and then given additional structure of a Riemannian metric? Isn't this the inversion of the given hierarchy?

When answering please don't be ambiguous with the words and notations, for it will only compound my confusion.

orange_soda
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    The Kronecker delta is not a metric. – T.J. Gaffney Mar 17 '14 at 07:50
  • Its not just metric in case of the Riemannian manifolds but a "$\textit{smoothly varying}$" metric. – DiffeoR Mar 17 '14 at 08:29
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    Locally (i.e. on a chart) it is always possible to choose a coordinate system such that the matrix repn of metric is nothing but identity matrix, but you cannot say anything globally from that. – DiffeoR Mar 17 '14 at 08:36

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Every smooth $n$-dimensional manifold $M$ comes equipped with an atlas of charts $\phi_\alpha: U_\alpha\to R^n$. Of course, on $R^n$ we have the standard (flat) Riemannian metric, usually denoted $\delta_{ij}$. However, the transition maps $\phi_\beta\circ \phi_\alpha^{-1}$ (typically) do not preserve these metrics. Another way to put is that $\phi_\alpha^*(\delta_{ij})\ne \phi_\beta^*(\delta_{ij})$ on the intersection $U_\alpha\cap U_\beta$. Therefore, one (typically) cannot use the metrics $\delta_{ij}$ to define a Riemannian metric on the entire $M$. For concreteness, you can work out the example of $M=S^2$ and two charts given by stereographic projections. The transition map will be an inversion which does not preserve the flat metric on $R^2$.

In fact, with this definition of a smooth manifold, it is unclear that it admits any Riemannian metric whatsoever; one can, however, give an argument that it does, say, by embedding the entire $M$ in some $R^N$, $f: M\to R^N$, and then taking the pull-back Riemannian metric $f^*(\delta_{ij})$. (There are other constructions, of course.)

Moishe Kohan
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  • @dj_mummy: As I said in my answer, the "first distance" typically does not exist. – Moishe Kohan Mar 19 '14 at 10:37
  • @dj_mummy: I am not sure what you mean by "call the projections Cartesian spaces". A projection (I presume stereographic one) is a function. How does a function become a space? You can look at the graph, of course, but then what? – Moishe Kohan Mar 19 '14 at 10:52
  • I do not know what ruler and scale you are talking about. There are no rulers in Riemannian (or pseudo-Riemannian) geometry. I think, what you should do is to pick up a textbook; my personal favorite is do Carmo's "Riemannian Geometry", I suggest to read carefully the first chapters 0 through 3 to get things straight. Also, please, do not be loose with the language, this is a sure way to get in trouble in mathematics. – Moishe Kohan Mar 19 '14 at 11:03
  • Charts can be defined with values in the Euclidean space (defined axiomatically, without any reference to coorinates) instead of the Cartesian space $R^n$. There is a branch of mathematics dealing with infinite-dimensional Riemannian manifolds where Cartesian coordinates (whatever this means in infinite dimensions) are (frequently) useless and then this more axiomatic approach is followed. But for a physist (and I am guessing, you are a physics student) this might not be a good approach. – Moishe Kohan Mar 19 '14 at 11:20
  • @dj_mummy: One way to say that it is a Hilbert (or, more generally, a Banach) space, i.e., a vector space $V$ equipped with a certain inner product (or norm). (If you do not know what it means, check wikipedia.) For example, take the Banach space $L^p([0,1])$ of $p$-integrable functions on the unit interval, this is a Banach space without any coordinates! The norm here is the $L^p$-norm. – Moishe Kohan Mar 19 '14 at 11:26
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    You need it to define topology, notion of derivative, etc. In finite-dimensional case, you can safely replace one norm by another (say, Euclidean by max-norm), but in infinite dimensions this really makes a difference. – Moishe Kohan Mar 19 '14 at 12:35
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    @dj_mummy: Topology has to come before a Riemannian metric. In order to define a Riemannian metric you need to know what continuous/smooth functions are, that is, you need an atlas with values in some topological vector space. – Moishe Kohan Mar 19 '14 at 19:37
  • @dj_mummy: Very good. – Moishe Kohan Apr 03 '14 at 22:00
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    @dj_mummy: It does not have its own topology defined via a Riemannian metric until you have notion of a smooth curve. To define the latter, you need manifold structure. In the end, one proves that the two topologies agree. – Moishe Kohan Apr 22 '14 at 15:42
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    Yes, of course, this makes sense only for connected manifolds. A proof would be in any textbook on Riemannian geometry. For instance, see M. do Carmo "Riemannian Geometry", page 146, Proposition 2.6. – Moishe Kohan Apr 22 '14 at 16:39
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    That's right: There is no way to convert a Lorentzian manifold to a metric space using its pseudo-Riemannian metric. This does not mean that it does not have interesting metric geometry though: you can consider only space-like paths, define distance functions, etc. However, not every pair of points will be connected by such a path and there will be other issues, like you will get a topology inconsistent with the manifold topology. – Moishe Kohan Apr 22 '14 at 17:51