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In page-385 of Roger Penrose's Road to Reality, the following is written:

In our Aristotelian scheme, it is appropriate to think of spacetime as simply the product:

$$ \mathbb{A}= \mathbb{E}^1 \times \mathbb{E}^3$$

and, in page-387,

Galilean spacetime $G$ is not a product space $\mathbb{E}^1 \times \mathbb{E}^3$,it is a fibre bundle with base space $\mathbb{E^1}$ and fibre $\mathbb{E^3}$

Now I am a bit confused, isn't it still a Fiber bundle even in the Aristotelian sense because we have the trivial bundle of $\mathbb{E^1}$ over $\mathbb{E^3}$? The fact product of set are bundle is mentioned in page-329

enter image description here

Source: Road to Reality, page-329

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    $\begingroup$ feels like you misread the text and ended up thinking Penrose said something which wasn't actually said. $\endgroup$
    – peek-a-boo
    May 25 at 8:43

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Yes, $\mathcal{A}:=\Bbb{E}^1\times\Bbb{E}^3$ together with the mapping $\alpha:\mathcal{A}\to\Bbb{E}^1$, $\alpha(t,x):= t$, the obvious projection, does indeed constitute a fiber-bundle $(\mathcal{A},\alpha,\Bbb{E}^1)$ with base space $\Bbb{E}^1$ and typical fiber $\Bbb{E}^3$. Product spaces can very trivially be made into fiber bundles, but so what? Does Penrose ever say it is not a fiber bundle/cannot be considered a fiber bundle/ should not be considered a fiber bundle?

Penrose is saying that the 'Galilean' description of spacetime is as a fiber bundle with $\Bbb{E}^3$-fibers $(\mathcal{G},\gamma,\Bbb{E}^1)$ where $\mathcal{G}$ is a topological space homeomorphic to $\Bbb{E}^4$ (so WLOG you may as well assume it is $\Bbb{E}^4$), but the mapping $\gamma$ is no longer a trivial mapping. That's the difference he's trying to highlight: in the Aristotlean picture, we have a trivial product space, so if you want to consider it as a fiber bundle then the bundle projection is the obvious one, but in the Galilean picture, the projection is not the obvious one. To be really explicit:

  • Total space: homeomorphic to $\Bbb{E}^4$ in both cases
  • Base space: $\Bbb{E}^1$ in both cases
  • Typical fiber: $\Bbb{E}^3$ in both cases
  • The bundle projections $\alpha$ vs $\gamma$ is where the difference lies. The first is trivial (emphasizing that the points in space 'remain the same' for all times, i.e that we can canonically identify the various fibers in the bundle), while the latter is not (we can't identify the different fibers canonically).

Edit:

I noticed a slightly subtle and potentially confusing matter which I didn't emphasize above. It is the difference between trivial and trivializable fiber bundles, where I'm using the first to mean 'equals a product bundle', and the second to mean 'is isomorphic to a product bundle'. Now, it is a theorem in topology that every fiber bundle over a contractible space (eg $\Bbb{E}^1$ or $\Bbb{R}$) is trivializable. The point is that in the Galilean description, we do have a trivializable fiber bundle (because of the topological theorem), but there is no canonical choice of an isomorphism, and hence there is no way we can canonically identify the various fibers.

For example, consider the following lower dimensional situation (so I can draw things). Let $\gamma:\Bbb{R}^2\to\Bbb{R}$, $\gamma(t,x)=t-x$ for example. Then, $(\Bbb{R}^2,\gamma,\Bbb{R})$ is a fiber bundle over the base $\Bbb{R}$, having (in this case) typical fiber $\Bbb{R}^{2-1}=\Bbb{R}$ (and this is not the trivial product bundle, which would have projection $\alpha(t,x)=t$). A sketch is shown below

enter image description here

The blue lines are the various fibers of $\gamma$ (I didn't draw the fibers of the trivial projection $\alpha$, but these are horizontal lines). To emphasize just how different the fibers are (yes, they're all isomorphic to $\Bbb{R}$, but there's no canonical isomorphism between them), consider the black dot in the picture above. Is there really an obvious correspondence with the other fibers? For example, I could say "I like the vertical direction, so the black dot should correspond to the red dots in the other fibers". But then I could ask what's so special about the red vertical line? We might then consider the horizontal purple line and say those dots should be identified when we consider the various fibers. But then someone else might come along and say they like the diagonal (green) line better.

As you can see, this is an extremely arbitrary choice. Later on, this also relates to the fact that in Galilean relativity, we shouldn't single out any one inertial observer (i.e a straight line). All such lines are equally good, and thus no one is preferred, and hence no 'absolute' canonical identification of the various fibers is possible. That's why we're modelling things using a fiber bundle which is not just the product bundle.

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    $\begingroup$ Could you explain in what sense it is different ? Like what are the details of this difference $\endgroup$ May 25 at 10:56
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    $\begingroup$ @Aplateofmomos you should just think of it as being given a bundle $(\mathcal{G},\gamma,\Bbb{E}^1)$. You have not much extra information about the projection $\gamma$ (the absolute time function). With $\gamma$, you can say "the time at the event $p\in\mathcal{G}$ is $\gamma(p)$". And since the fibers are $\Bbb{E}^3$, you can make statements like "if $p,q\in\mathcal{G}$ are events such that $\gamma(p)=\gamma(q)$ (i.e simultaneous) then they are spatially separated by a distance of $|p-q|$ (this makes sense since they live in the same affine-space fiber $\Bbb{E}^3$)". $\endgroup$
    – peek-a-boo
    May 25 at 13:32
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    $\begingroup$ So, it only makes sense to talk about distances provided you're speaking of simultaneous events. By merely imposing a fiber-bundle structure, you can't take two arbitrary events $p,q\in\mathcal{G}$ (with $\gamma(p)\neq \gamma(q)$) and speak of their spatial distance, because $p,q$ live in different copies of $\Bbb{E}^3$, and purely based on the bundle structure, there is no way to canonically identify the fibers. In the product case, we can obviously identify the various fibers: $\{t_1\}\times\Bbb{E}^3\cong \{t_2\}\times\Bbb{E}^3$ simply via the identification $(t_1,x)\mapsto (t_2,x)$. $\endgroup$
    – peek-a-boo
    May 25 at 13:36
  • $\begingroup$ If you say "merely" by imposing a fiber structure then even before in Aristotle case it was a fiber, yet we could compare.. so .. I'm a bit stuck at that point. $\endgroup$ May 25 at 13:48
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    $\begingroup$ @Aplateofmomos In the other case its a trivial fiber bundle. SO its not 'merely' an arbitrary fiber bundle. Trivial fiber bundles, as the name suggests, are trivial in both the math sense and the everyday linguistic sense, so it's a hyper special case which is why we can compare the various $\Bbb{E}^3$ (i.e a canonical isomorphism exists). $\endgroup$
    – peek-a-boo
    May 25 at 13:49

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