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I was learning about covariant and contravariant vectors due to special relativity, and it occured to me that we don't live in $\mathbb{R}^4$. I'll explain myself better. Consider the space of polynomials of degree $\leq n$. This is a vector space, and choosing ${1,x,x^2,...,x^n}$ as a basis makes perfect sense: those objects exist regardless of their coordinates. However, when we set up a coordinate system in classical mechanics (for example) we have a preesisting space (the "physical world"), in which we choose a point (which is going to be the $0$ of our identification with $\mathbb{R}^3$) and then a basis with which we are able to give coordinates to that point. So far so good, except that I have no idea what the "physical world" is without resorting to an $\mathbb{R}^3$ description in the first place.

Now, I thought of thinking about the "physical world" in a manifold sense: you have a topological space $(M, \tau)$ and you construct an atlas with one chart. In that sense, we can talk about coordinates and when we choose a reference point we are equipping the topological space of a manifold structure to talk unambiguously of points.

My question is: what's that topological space? What is the space I am equipping with a manifold structure?

Gennaro Marco Devincenzis
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    Can't you just say that classical mechanics by definition uses a three-dimensional affine space (plus a time axis) as its model of reality? (In other words, it is taken as an axiom.) Then the question of what physical space "really is" belongs to a different theoretical level, "above" classical mechanics. Other models of reality might be used in other theories. Within general relativity, for example, reality (spacetime) is a certain kind of manifold, etc. – Hans Lundmark Apr 28 '15 at 13:01
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    I've never had those kind of delusions of grandeur while phrasing the question; I realize that "what the universe really is" is a question that is way out of my depth. My only concern was how to talk of the physical world (in particular in classical and special relativistical mechanics) without resorting to an $\mathbb{R}^4$ description, and you've kind of answered to that by mentioning affine spaces, which I haven't studied and I don't really know. – Gennaro Marco Devincenzis Apr 28 '15 at 13:08
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    Affince spaces are not very difficult to learn if you know vector spaces. An affine space is a almost like a vector space, it's just that you haven't decided yet where to put the origin. The "points" in mechanics form an affine space, while the "vectors" form a vector space. Once you (arbitrarily) pick an origin $O$, you can identify each point $P$ with the vector $\vec{OP}$. This question might be helpful: http://math.stackexchange.com/questions/185768/what-are-affine-spaces-for – Hans Lundmark Apr 28 '15 at 15:02
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    @Gennaro: You might enjoy perusing The Road to Reality by Roger Penrose. If memory serves, he addresses this type of question (particularly, Gallilean versus Lorentzian spacetime) along the lines Hans Lundmark sketches. :) – Andrew D. Hwang Apr 28 '15 at 17:10
  • Thank you for the reference. I'll certainly do that. – Gennaro Marco Devincenzis Apr 28 '15 at 17:12

1 Answers1

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So, I answer the question since I've found the answer I was looking for and it might be useful to somebody else sooner or later. I'll treat the case of special relativity. Axiomatically, we assume that $M^4$ is an affine space, that is a triad $(A^n, V, \vec \cdot)$ where $A^n$ is a set whose elements are called points, $V$ is an $n$-dimensional vector space and $\vec \cdot$ is an operation with the following properties:

  • $\forall P\in A^n, v \in V \exists! Q\in A^n: \vec{PQ}=v$
  • $\vec{PQ}+\vec{QR}=\vec{PR} \space \forall P,Q,R \in A^n$

Now, once the origin $O$ is fixed there clearly is a bijection $f:V \rightarrow \mathbb{R}^n$. The euclidean topology naturally induces a topology on $A^n$, namely the open sets are the counterimages of the open sets of $\mathbb{R}^n$. We'll call the function $f$ "system of cartesian coordinates" of origin $O$ and axes $e_1,...,e_n$. Moreover we can equip, using this function, $A^n$ of a structure of topological and differentiable $C^\infty$ manifold. We want more structure though, and since we have a simmetrical bilinear form on $V$ ($\operatorname{diag}(1,-1,-1,-1)$) we'd like to transport it to $A^n$. Since it is a differentiable manifold we have no problem in taking the tangent spaces $T_pA^n$(which are essentially $\mathbb{R}^n$) and constructing an isomorphism $L_p: T_pA^n \rightarrow \mathbb{R}^n$ which sends the canonical basis of the tangent space in the canonical basis of $\mathbb{R}^n$. Then we define, $\forall u, v \in T_pA^n$:

$$g_p(u,v)=g(L_pu,L_pv)$$

And that's as far as I got. Please correct me if something's off.

Gennaro Marco Devincenzis
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