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I'm looking for an example of $n$-dimensional affine space that is isomorphic with $\mathbb{R}^n$ as affine space but not with respect to other properties (for example it has different ordering etc.) Thank you.

EDIT: in practice, I am looking for an affine space that has more or less structure than $\mathbb{R}^n$. Intuitively $\mathbb{R}^n$ has "more structure" than a canonical affine space because, by its field properties, it has a special point (that is the zero with respect to addition). I need an example of affine space different from $\mathbb{R}^n$ but having the same dimension.

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It would help if you explained what exactly you mean by "$n$'dimensional affine space". I, for one, do not know what you are asking about! –  Mariano Suárez-Alvarez Jun 22 '12 at 19:33
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I don't seem to recall there being a natural ordering (ie topologically significant) ordering on $\mathbb{R}^n.$ –  Tim Duff Jun 22 '12 at 19:54
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Are you referring to isomorphism of $\mathbb{R}$-modules? –  Tim Duff Jun 22 '12 at 19:59
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It's not too difficult to add some structure to $\Bbb R^n$ (the vector space). You can make it a topological space, a differentiable manifold, a Clifford algebra, you can add some order on it if you want, etc. –  anon Jun 23 '12 at 1:03
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You can make $\Bbb R^n$ as an affine space with two different types of topologies (eg Euclidean versus Zariski), or make them $\Bbb R$-algebras in two inequivalent ways (eg Clifford algebras with quadratic forms of different signatures) ... –  anon Jun 24 '12 at 3:31
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up vote 4 down vote accepted

If you look on the Wikipedia page http://en.wikipedia.org/wiki/Affine_space you will find an example mentioned in the very first paragraph. Did you look there? Just take any $n$-dimensional subspace of a (larger) vector space and add a fixed vector to it (a shifted subspace). That's an $n$-dimensional affine space. (For example, any line in the plane -- not necessarily containing the origin -- is a 1-dimensional affine space.) If this doesn't contain 0 then it is not isomorphic to ${\mathbf R}^n$ as a vector space since it doesn't have a natural structure as a vector space.

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Trivial but good. –  Flast9 Jun 24 '12 at 5:20
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Take the space of possible energy values of $n$ independent particles. This is an affine space of dimension $n$ because you can add a vector in $\mathbb{R}^n$ to any $n$-tuple of energy values, but there is no distinguished origin because there is no distinguished value of energy that we can call zero.

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