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Any answer to my questions may assume arbitrary knowledge of differential geometry, which I will be happy to learn in order to understand the most appropriate type of formulation for the theory.

I would like to consider Special Relativity as the theory of a general Lorentz manifold $(M,g)$ where $M$ is diffeomorphic to (the standard diff. structure of) $\mathbb{R}^4$ and the metric (of constant Lorentzian signature $(-+++)$) is constant over all of $M$ in the coordinate-free sense. I do not, however, want to assume that $M=\mathbb{R}^4$ and that $g=(\text{usual matrix})$ because I find it unphysical to apply "linear transformations" to positions in spacetime.

In particular, I demand that (for this discussion) the theory be formulated with arbitrary smooth coordinates (in the sense of differential geometry) where we can of course choose global coordinates, but no single coordinate system is in any sense "more fundamental" than another or "a priori innertial". This is as opposed to the case of $\mathbb{R}^4$, where one chooses as innertial the coordinate system $id$, which returns positions in spacetime as the tuples of numbers that they are, together with any other coordinate system which is connected to $id$ by a Poincarré transformation.

I interpret physically the manifold as the set of all events in spacetime, and the metric as designating causality; in particular the metric dictates what (smooth) curves in $M$ are space-, time-, and lightlike. In order to specify past and future we assume the existence of a global vector field $X\in TM$, which is defined to be future-directed at each point. Together $X$ and $g$ specify the past and future light cones for each point $p\in M$ as subsets of $M$ (which is not considered as a vectorspace) as well as all(?!) other concepts relevant to causality, such as Cauchy surfaces and the like.

My questions:

Question 1. In the next step of this theoretical setup, which is not completely clear to me, one defines a linear action of the Lorentz group$^1$ (I don't think Poincarré makes sense here?!) on each tangent space $T_pM$. How does one set up this representation and justify physically why both the group and the chosen representation are meaningfull and the correct choice?

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Question 2. Is it possible with the above prerequisites to meaningfully define when a diffeomorphism $x:M\to\mathbb{R}^4$ is an "innertial coordinate system"? Would this be equivalent to requiring that all "$g$-Levi-Civita-connection-straight" curves be mapped by $x$ to straight curves in $\mathbb{R}^4$?

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Question 3. Assuming the above makes sense, what is the connection of the Poincarré group with coordinate changes between innertial coordinate systems?


Footnotes:

$^1$ I consider all these groups just as mathematical objects - Lie groups with a priori no physical significance. I want to include the physical significance by arguing why these groups with their appropriate actions on the appropriate objects are physically meaningfull.

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  • $\begingroup$ I can answer half of question 1: the Lorentz group is the isometry group of the tangent spaces. $\endgroup$ – Javier Nov 16 '16 at 21:44
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    $\begingroup$ Also, what do you mean by constant metric in a coordinate free sense? $\endgroup$ – Javier Nov 16 '16 at 21:56
  • $\begingroup$ @Javier Now that you're saying it, I'm not actually sure what I mean by constant metric. There is no way to compare the values of a section at different points... I guess I would have to settle for the statement "there exists a coordinate system such that the representation of $g$ is the constant Minkowski-matrix"? Unless anyone sees a way of making this precise without such a hack... $\endgroup$ – Adomas Baliuka Nov 16 '16 at 22:08
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    $\begingroup$ You could ask for the metric to be flat; according to Wikipedia $\mathbb{R}^n$ is the universal cover of any complete flat manifold, so if you ask that spacetime be diffeomorphic to $\mathbb{R}^4$ you're done. $\endgroup$ – Javier Nov 16 '16 at 22:11
  • $\begingroup$ The problem starts with your assumption: SR is a Lorentz invariant theory- it is not enough to provide a general Lorentzian manifold, but it also requires the Poincare group to be the isometry group. In your formulation, $g=$constant does not guarantee such a condition. So the setting you proposed is not sufficient for one to formulate SR. $\endgroup$ – user110373 Nov 16 '16 at 22:14
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First off, your abstract setting is slightly off - it doesn't make sense to demand that $g$ is "constant in the coordinate-free sense" because $g$ takes values in $T^\ast_p M \otimes T^\ast_p M$ at each point, so you cannot compare the values directly to find out whether they are "constant". A more natural requirement for SR seems to be that $(M,g)$ be flat, non-compact and complete, which all are abstract, coordinate-free requirements making no reference to the groups you want to examine. Also an abstract, but somewhat "cheaty" way for your purpose appears to be to just require $(M,g)$ to be isometric to standard Minkowski space directly. Let me just remark that some might consider the *symmetries to be the more fundamental objects, and claim that requiring the Poincaré group to be the global isometry group is the "correct" abstract assumption. Now, for the appearance of the Lorentz and Poincaré groups in your setting:

On each tangent space, there is a natural action of $\mathrm{GL}(4)$ simply because it's a four-dimensional vector space. The Lorentz group $\mathrm{SO}(1,3)$ is simply the subgroup of $\mathrm{GL}(4)$ that leaves the pseudo-inner product on that tangent space invariant. There is no need to "define" its action, it comes naturally in this fashion. No representation is chosen, this is naturally the fundamental representation of $\mathrm{SO}(3,1)$.

An inertial coordinate system is simply an isometry of $(M,g)$ and $(\mathbb{R}^4,\eta)$ where $\eta$ is the standard Minkowski metric. Note that this is different from a general chart because charts are not required to be isometries, and indeed maps geodesics in $M$ to straight lines in $\mathbb{R}^4$

The notion of the Poincaré group now takes the idea of the Lorentz group as the group of "isometries" of the pseudo-inner product and makes it global: The Poincaré group is the group of isometries of $(M,g)$, or, equivalently, of $(\mathbb{R}^4,\eta)$. Here, an isometry of $M$ is a diffeomorphism $f : M\to M$ such that $f_\ast g = g$.

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  • $\begingroup$ +1 for "charts are not required to be isometries": this is often overlooked and it is indeed the point of requiring the Poincaré group to do that job. $\endgroup$ – gented Nov 16 '16 at 22:28
  • $\begingroup$ Thank you very much for your answer. I have two questions. First: If $x$ is some arbitrary coordinate system of $M$ (not necessarily innertial), what is the action of a general Lorentz matrix $\Lambda\in S(1,3)$ on the vector $\frac{\partial}{\partial x^1}+5\frac{\partial}{\partial x^2}$ at some point? Basically, I'm worried about there not being a canonical choice of basis for the tangential spaces to fix a representation. Question 2: Is there really a isomorphism of groups between $$\{f:M\to M: \text{ is a diffeomorphism and }f_* g=g\}$$ and the Poincaré group? $\endgroup$ – Adomas Baliuka Nov 16 '16 at 22:31
  • $\begingroup$ @AdomasBaliuka 1. To answer that for a point $p\in M$ one would need to look at the form $g_p$ takes in its matrix representation for the basis $\frac{\partial}{\partial x^i}\rvert_p$ and determine which matrices leave it invariant (or which basis transformation makes $g_p$ into $\eta$, which is equivalent information). Note that by choosing the coordinates $x$ you already make my answer uncanonical. There is no canonical choice of basis, but the notion of "subgroup of GL(n) which preserves the pseudo-inner product" doesn't need a basis, so it fixes the Lorentz group & action uniquely. 2. Yes. $\endgroup$ – ACuriousMind Nov 16 '16 at 22:40
  • $\begingroup$ Thank you very much for your answer! I still can't really imagine how the isomorphism is constructed. Do you know any sources for this? $\endgroup$ – Adomas Baliuka Nov 16 '16 at 22:54
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    $\begingroup$ @AdomasBaliuka Use the isometry to $\mathbb{R}^4$ and then pick your poison e.g. from this MO thread. $\endgroup$ – ACuriousMind Nov 16 '16 at 23:06

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