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What is meant by the word standard in "Euclidean space is special in having a standard set of global coordinates."? Then "A manifold in general does not have standard coordinates" This makes me think standard means something else then 'most common used'. Is R^n special in any sense, as a manifold?

This is from Loring W. Tu - Introduction to manifolds

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Euclidean space has the canonical linear basis $e_1,\ldots,e_n$, where $n$ is the dimension of the space. For manifolds, the existence of a linear basis is not guaranteed - you can only assume that locally, it is homeomorphic/diffeomorphic to $\mathbb R^n$. –  Johannes Kloos Apr 4 '12 at 13:41
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I think the key word is "global," as in: Local coordinates which are determined by one global coordinate system. Essentially, the local coordinates of any two open sets can be "stitched" together to have one coordinate system for the union. This can only be done for arbitrary local coordinates in $\mathbb R^n$. –  Thomas Andrews Apr 4 '12 at 13:42
    
As an aside, this "special" circumstance occurs for the $n$-fold cartesian product $R^n$ for any ring $R$ with unity. For, if $1$ denotes the multiplicative unit in $R$ then any element $r \in R^n$ where $r = (r_1, \dots, r_n)$ can be expressed as $r = r_1(1, 0, \dots, 0) + \dots + r_n(0, \dots, 1)$ –  ItsNotObvious Apr 4 '12 at 14:34
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Usually when we write "$\mathbb{R}^n$" we are thinking of an explicit description of it as $n$-tuples of real numbers. This description "is" the standard set of global coordinates, namely the coordinate functions $x_i$. But this description isn't part of $\mathbb{R}^n$ "as a manifold", in that it contains more information than just the diffeomorphism type of this manifold. What I mean by this is that if I give you an abstract manifold which is diffeomorphic to $\mathbb{R}^n$ -- but I don't tell you an explicit diffeomorphism -- then there isn't a "standard" set of coordinates for it. And if you have a manifold which isn't diffeomorphic to any $\mathbb{R}^n$, then there is no set of global coordinates for it (otherwise you could use those coordinates to produce a diffeomorphism to $\mathbb{R}^n$).

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