Trivial fiber bundle vs product space

[cianfa72]

TL;DR

Is the trivial fiber bundle of topological spaces the same as the product space ?

I've a doubt about the following: take the trivial fiber bundle of the base space $B$ and the fiber $F$ vs the product $B \times F$ of topological spaces $B$ and $F$.

Are they really the same ?

As far as I can tell, in the trivial fiber bundle each fiber in the bundle in a distinct/separated copy of the topological space called "the fiber space" $F$, whereas in the product $B \times F$ it is the same copy of $F$ that, let me say, enters multiple times in the product (i.e. all the copies of $F$ are actually identified).

Then, it's true that they are canonically identified, but that's another story....

 

[martinbn]

What is the definition of a trivial bundle that you have seen?

 

[cianfa72]

What is the definition of a trivial bundle that you have seen?

https://en.wikipedia.org/wiki/Fiber_bundle#Trivial_bundle

I'd add that my doubt comes from this picture of $\mathbb E^1 \times \mathbb E^3$ in Penrose book:

 

[martinbn]

https://en.wikipedia.org/wiki/Fiber_bundle#Trivial_bundle

Seems quite clear and unambiguous.

I'd add that my doubt comes from this picture of $\mathbb E^1 \times \mathbb E^3$ in Penrose book:

What is your doubt?

 

[cianfa72]

I believe I got the point. A trivial bundle is different from a global trivializable bundle. The former is a product while the latter is just homeomorphic to a product, however there isn't a canonical/natural homeomorphism.

So $\mathcal A = \mathbb E^1 \times \mathbb E^3$ is a product (i.e. a trivial fiber bundle). It has the two canonical projections on the factors.

Penrose then defines Galilean spacetime $\mathcal G$ as a (trivializable) affine fiber bundle over the base $\mathbb E^1$ and fibers modeled as $\mathbb E^3$. Here there is only the projection on the base space defined by the structure. Since there isn't the other canonical projection of the product, there is no natural way to identify points on the fibers, i.e. absolute time but not absolute space.

See also Aristotelian-vs-galilean-relativity-in-terms-of-bundles

 

[martinbn]

I believe I got the point. A trivial bundle is different from a global trivializable bundle. The former is a product while the latter is just homeomorphic to a product, however there isn't a canonical/natural homeomorphism.

So $\mathcal A = \mathbb E^1 \times \mathbb E^3$ is a product (i.e. a trivial fiber bundle). It has the two canonical projections on the factors.

Penrose then defines Galilean spacetime $\mathcal G$ as a (trivializable) affine fiber bundle over the base $\mathbb E^1$ and fibers modeled as $\mathbb E^3$. Here there is only the projection on the base space defined by the structure. Since there isn't the other canonical projection of the product, there is no natural way to identify points on the fibers, i.e. absolute time but not absolute space.

See also Aristotelian-vs-galilean-relativity-in-terms-of-bundles

I don't think this has anything to do with mathematics. A tribual bundle is a trivial bundle whether it is given as a product or it is diffeomerphic to a product. A spesific realization as a product may be important to physics, but that is a different question.