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It is very well known that any Galilean transformation on $\mathbb{R}^3\times\mathbb{R}$ can be uniquely written as composition of a maps of the following type:

  1. Uniform motion: $(x,t)\to(x+tv,t)\quad ,\quad v\in\mathbb{R}^3$
  2. Translation of origin: $(x,t)\to(x+x_0,t+t_0)\quad,\quad x_0\in\mathbb{R}^3,t_0\in\mathbb{R}$
  3. Orthogonal maps: $(x,t)\to(Qx,t)\quad,\quad Q\in \mathbb{O}(n)$

(all definitions are that of Arnold) I am trying to prove this. I got to this Galilean transform: $(x,t)\to(x+g(t)\hat i,t)$ where $g:\mathbb R\to\mathbb R$ is any additive nonlinear map that $g(0)=0$. As far as I can observe, this function satisfies all criteria for a Galilean transformation:

  1. It is a well-defined affine map
  2. The map is injective and surjective
  3. It preserves time intervals
  4. It preserves distance in simultaneous sections

But it's easy to see that it cannot be written as a composition of uniform motions, translations, and rotation of coordinate axes. Have I missed something in the definition of Galilean transformation or should one include some notion of continuity in the definition?

Any help or reference would be appreciated.

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  • $\begingroup$ It probably says somewhere that it is a linear transformation? $\endgroup$
    – Roger V.
    Commented Oct 31 at 10:58
  • $\begingroup$ The general settings involves an affine space and Galilean transformations are defined as affine-maps preserving time intervals and the Euclidean structure of simultaneous sections. There is no linear structure to impose linearity conditions on the maps $\endgroup$
    – Aryan
    Commented Oct 31 at 11:21
  • $\begingroup$ Section The galilean group and Newton's equations seems to be rather explicit about parallel displacement and linear time... I am not a mathematician, but my guess is that some physics intuition/background is injected somewhere, even if the book aspires to mathematical rigor. $\endgroup$
    – Roger V.
    Commented Oct 31 at 11:30
  • $\begingroup$ Yes it is precise and I understand the reasons he imposes the conditions. Yet it seems insufficient to prove the claim he makes for the decomposition. $\endgroup$
    – Aryan
    Commented Oct 31 at 13:38

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As far as I can observe, this function satisfies all criteria for a Galilean transformation:

It is a well-defined affine map; The map is injective and surjective; It preserves time intervals; It preserves distance in simultaneous sections.

Considering all that, it seems that your map $g(t)$ matches all criteria to be represented as $$ g(t) = \alpha t + \beta $$ For real scalars $\alpha$ and $\beta$. Check it here that this is more general representation of an affine map. Considering $g(0) = 0$ as you said, we should have $\beta =0$. So we have $g(t) = \alpha t$, which can be matched with uniform motion by $v = \alpha \vec i$.

Please tell me if I understood your question right, because I believed that you already found it.

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  • $\begingroup$ I can prove, given the definition I mentioned (that is the same as that of Arnold's) that any Galilean map can be reduced, by composition with a translation of origin and an orthogonal map) to a map $(x,t)\to(g(x),t)$ such that $g$ is an additive map. Now the theorem would imply that one should be able to reduce this to identity map via a uniform motion, which as you mentioned, is equivalent to $g$ being linear. All things said, we both agree that $\beta=0$ holds. But there is no reason that an additive map is also linear. Elementary examples can be constructed via a Hamel basis on $\mathbb R$ $\endgroup$
    – Aryan
    Commented Oct 31 at 13:25
  • $\begingroup$ Maybe try this math.stackexchange.com/questions/152632/… $\endgroup$
    – Ruffolo
    Commented Oct 31 at 13:30
  • $\begingroup$ My problem is that such a map is a counterexample to what you claim, that these conditions provide a criteria for the map to be represented as $Gt+\beta_0$ where $G\in \mathbb O(n)$ and $v_0\in\mathbb R^3\times\mathbb R$, or equivalently in our case, for $g$ to be represented as $\alpha t+\beta$ where $\alpha,\beta$ are scalars. So I suspect there shall be other conditions on the map to make this work. If you have a proof for your criteria, I'll be glad if you could mention the reference. $\endgroup$
    – Aryan
    Commented Oct 31 at 13:32
  • $\begingroup$ that's the problem. As I mentioned in my question, I suspect that there should be some notion of continuity on the map to make this work since as you mentioned, linearity of an additive map is deduced only if it satisfies continuity or some other condition such as local boundedness, local monotoneness, etc. But this would not be easy since in the settings described in Arnold's text, no clear order or topology can be equipped on the affine space we're working on. $\endgroup$
    – Aryan
    Commented Oct 31 at 13:35
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    $\begingroup$ Oh! On the Wikipedia article you referenced, there is an extra condition that not only the lifted map is well-defined, but it is also linear. This is not a part of original definition of an affine transform but it would be sufficient to take care of the linearity issue, specifically since $g(tx)-g(0)=t(g(x)-g(0)),\quad\forall t\in\mathbb R$ where the scalar multiplication is meaningful in the ambient vector space structure. In the $\mathbb R^3\times\mathbb R$ case, this implies that $g:\mathbb R\to\mathbb R^3$ is linear. This is probably the extra condition Arnold has imposed. Thanks! $\endgroup$
    – Aryan
    Commented Oct 31 at 13:47

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