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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!2012-06-22
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    I don't seem to recall there being a natural ordering (ie topologically significant) ordering on $\mathbb{R}^n.$2012-06-22
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    Are you referring to isomorphism of $\mathbb{R}$-modules?2012-06-22
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    @MarianoSuárez-Alvarez: I mean that its underlying vector space has finite dimension equal to $n$2012-06-22
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    Isomorphism in what sense?2012-06-23
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    @QiaochuYuan: isomorphic as affine spaces.2012-06-23
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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.2012-06-23
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    @Flast9: what do you mean by "other properties"? Are you really just looking for a construction of an affine space isomorphic to $\mathbb{R}^n$ which genuinely does not come with an origin?2012-06-23
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    @QiaochuYuan: if I had known one (particular property) I would have not asked for it!! So, I need an affine space different from the real one, where "different" means that there is at least a good property that holds for one and not for the other. The question was originated from a discussion in which an affine space was introduced at a certain time in the discourse, and one of my colleague said "don't read affine, read $\mathbb{R}^n$ directly"... I believe that mathematicians would have not defined affine spaces if they were exactly the same as $\mathbb{R}^n$ spaces.2012-06-23
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    I have a little knowledge of field theory or algebraic geometry, but algebraic geometers sometimes define $n$-dimensional affine spaces as $n$-products of a field with some property. In virtue of this, we can restate the question asking for a field that is affine with the reals but different for (at least) some non affine property.2012-06-23
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    @Flast9: what do you mean by "affine with the reals"? I really cannot understand what you are asking.2012-06-23
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    @QiaochuYuan: for example consider two structures $A$ and $B$. There is the possibility that $A$ is isomorphic with $B$ with respect to a certain operation/relation (say that they are isomorphic as groups) but not with respect to another one (for example they may not be isomorphic as rings). So we have found that $A$ and $B$ are _different_ even if they behave the same way as groups. Now, let $A$ be $\mathbb{R}^n$ and $B$ a set s.t. $A$ and $B$ behave the same way as affine spaces. Show me a $B$ that is _different_ from $A$.2012-06-24
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    @Flast9: different with respect to what?!2012-06-24
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    @QiaochuYuan: anything that makes them different should be good. In fact I can't find an example of $n$-dimensional affine space that has a feature not shared by $\mathbb{R}^n$2012-06-24
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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) ...2012-06-24
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    @anon: your answer is the best fitting with respect to what I wanted to know, but I need an example in which one of the two affine spaces is not $\mathbb{R}^n$.2012-06-24

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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.2012-06-24
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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.