Principle Fibre: Definition & Examples

[RoboticMezon]

Hello everyone!

I'm fully aware of the complete and utter stupidity if this question, but I am also aware about how unable I am to find any text in which this is explained so that even yours truly can understand. (In fact, I have found NOTHING so far...another proof of my lack of mental processes?...Who knows! :) )

What is a fibre bundle's 'typical fibre'?

Am I right in inferring from the name that it is one fibre of the bundle which 'represents', as it were, all the rest? Something like a representation element of an orbit in a group? Or am I more stupid than I had originally thought? (I would find that amazing...I doubt you can get stupider... :) :) )

(Seriously, at our uni they toss up new terms without thinking that they should, perhaps, maybe, be explained.)

Thanks in advance!

4R

 

[lavinia]

Hello everyone!

I'm fully aware of the complete and utter stupidity if this question, but I am also aware about how unable I am to find any text in which this is explained so that even yours truly can understand. (In fact, I have found NOTHING so far...another proof of my lack of mental processes?...Who knows! :) )

What is a fibre bundle's 'typical fibre'?

Am I right in inferring from the name that it is one fibre of the bundle which 'represents', as it were, all the rest? Something like a representation element of an orbit in a group? Or am I more stupid than I had originally thought? (I would find that amazing...I doubt you can get stupider... :) :) )

(Seriously, at our uni they toss up new terms without thinking that they should, perhaps, maybe, be explained.)

Thanks in advance!

4R

not sure what a typical fiber means - but all of the fibers are homeomorphic -

Where did you read this?

 

[element4]

Well, do you know the definition of a fiber bundle? The typical fiber is related to the property going under the name "local triviality". Let $$E\stackrel{\pi}{\rightarrow}M$$ be a bundle over the manifold $$M$$, and let $$U\subset M$$ be an open set. Furthermore let $$V$$ be a (fixed) topological space (vectorspace for vectorbundles).

The bundle is locally trivial if there is a homeomorphism $$\phi:\pi^{-1}(U)\rightarrow U\times V$$ such that

commutes. So each fiber is in some sense "identical" to the topological space $$V$$, and therefore it is the "typical fiber".

 

[lavinia]

Well, do you know the definition of a fiber bundle? The typical fiber is related to the property going under the name "local triviality". Let $$E\stackrel{\pi}{\rightarrow}M$$ be a bundle over the manifold $$M$$, and let $$U\subset M$$ be an open set. Furthermore let $$V$$ be a (fixed) topological space (vectorspace for vectorbundles).

The bundle is locally trivial if there is a homeomorphism $$\phi:\pi^{-1}(U)\rightarrow U\times V$$ such that

commutes. So each fiber is in some sense "identical" to the topological space $$V$$, and therefore it is the "typical fiber".

that's what I said.

 

[element4]

Well, do you know the definition of a fiber bundle? The typical fiber is related to the property going under the name "local triviality". Let $$E\stackrel{\pi}{\rightarrow}M$$ be a bundle over the manifold $$M$$, and let $$U\subset M$$ be an open set. Furthermore let $$V$$ be a (fixed) topological space (vectorspace for vectorbundles).

The bundle is locally trivial if there is a homeomorphism $$\phi:\pi^{-1}(U)\rightarrow U\times V$$ such that

commutes. So each fiber is in some sense "identical" to the topological space $$V$$, and therefore it is the "typical fiber".

There is a typo here, replace $$V$$ with $$F$$, and then $$F$$ is the typical fiber.

that's what I said.

Sorry, I wasn't contradicting you. Just giving a few more details, in order to show precisely what people call the typical fiber.

 

[RoboticMezon]

Well, do you know the definition of a fiber bundle? The typical fiber is related to the property going under the name "local triviality". Let $$E\stackrel{\pi}{\rightarrow}M$$ be a bundle over the manifold $$M$$, and let $$U\subset M$$ be an open set. Furthermore let $$V$$ be a (fixed) topological space (vectorspace for vectorbundles).

The bundle is locally trivial if there is a homeomorphism $$\phi:\pi^{-1}(U)\rightarrow U\times V$$ such that

commutes. So each fiber is in some sense "identical" to the topological space $$V$$, and therefore it is the "typical fiber".

Yes, I'm aware of the definition, thank you. :)

Actually, its used in an article I've read in the very beginning under the 'Theoretical Background' section; right in the second sentence of this section. I think I can give you the link:

http://arxiv.org/abs/1012.4662"

Thanks a bunch for your help!

4R

 

[Fredrik]

Note that the fibers are sets of the form $\pi^{-1}(b)$, where b is a member of the base space. The definition implies that every fiber is homeomorphic or diffeomorphic to F. (Homeomorphic if we're dealing with topological spaces, diffeomorphic if we're dealing with manifolds). So if anything deserves to be called "the typical fiber", it's F.

 

[RoboticMezon]

Note that the fibers are sets of the form $\pi^{-1}(b)$, where b is a member of the base space. The definition implies that every fiber is homeomorphic or diffeomorphic to F. (Homeomorphic if we're dealing with topological spaces, diffeomorphic if we're dealing with manifolds). So if anything deserves to be called "the typical fiber", it's F.

*headdesk*


I just knew it was going to be 'trivial'.

Thank you so much for your help!
It's a clear as day now, or would be if it wasn't raining...