# Is an associated bundle isomorphic to the principal bundle?

• Fredrik
In summary, associated bundles are not necessarily isomorphic to the original principal bundle, but there may exist a fiber bundle isomorphism between them. This can be proven by showing that the map \phi_f is bijective, even without assuming the group action on F to be transitive.
Fredrik
Staff Emeritus
Gold Member
Edit: Never mind. I should have looked more carefully at the examples in my book. The obvious right action of Z2 on S1 gives us a principal bundle that we can use to construct many different associated bundles, including the cylinder, the Möbius strip and the Klein bottle. Those spaces aren't even homeomorphic to each other, and they're even more obviously not homeomorphic to the total space of the original principal bundle.

Let $\pi:E\rightarrow M$ be the projection of a principal bundle with structure group G. (I'm taking the group action of G on E to be a right action). If there's a left action of G on a space F, then we can define an "associated" fiber bundle by defining the total space to be the set of orbits of the right action of G on $E\times F$, and the define the following projection

$$\pi_F:\frac{E\times F}{G}\rightarrow M$$

$$\pi_F([p,f])=\pi(p)$$

Here [p,g] is the orbit through (p,f). The right action of G on $E\times F$ that I mentioned above is defined by

$$(p,f)g=(pg,g^{-1}f)$$

Is this associated bundle supposed to be isomorphic to the original fiber bundle? The associated bundle isn't a principal bundle, but there could exist a fiber bundle isomorphism between them. If so, how is that function defined? For each $f\in F$, the map

$$\phi_f:E\rightarrow\frac{E\times F}{G}$$

defined by

$$\phi_f(p)=[p,f]$$

could be a candidate, but I don't see a way to prove that it's bijective, at least not without additional assumptions. Are there assumptions missing in what I wrote above, or is the associated bundle not supposed to be fiber bundle isomorphic to the principal bundle that was used to define it?

My failure to prove that the function is injective goes like this:

$$\phi_f(p)=\phi_f(p')\implies [p,f]=[p',f']\implies (p',f')\in [p,f]\implies \exists g\in G\quad (p',f')=(p,f)g=(pg,g^{-1}f)$$

$$\implies\exists g\in G\quad p'=pg\ \mbox{and}\ f'=g^{-1}f$$

But to say that there exists a g such that the last equation is satisfied is to say that the group action on F is transitive, and we didn't assume that it was. I have similar problems when I try to prove that the function is surjective. I have also tried the map $[p,f]\mapsto p$. It's obviously surjective, but I can't prove that it's injective without assuming that the group acttion on F is transitive.

Last edited:

Thank you for sharing your thoughts and questions about associated bundles and their relationship to the original principal bundle. Your understanding of the concept is correct - associated bundles are not necessarily isomorphic to the original principal bundle, but there may exist a fiber bundle isomorphism between them.

In your attempt to prove the bijectivity of the map \phi_f, you are correct in noting that the existence of a g such that the equation is satisfied requires the group action on F to be transitive. This is indeed an assumption that is often made when constructing associated bundles, as it simplifies the proof of bijectivity. However, even without this assumption, it is possible to prove the bijectivity of \phi_f. Here is one possible proof:

To show that \phi_f is injective, let [p,f] and [p',f'] be two elements in the quotient space that map to the same point under \phi_f. Then we have [p,f]=[p',f'] and (p',f')=(pg,g^{-1}f) for some g\in G. This implies that p'=pg and f'=g^{-1}f. But since p and p' are in the same fiber, p' must also be in the orbit of p under the action of G. Thus, there exists a g'\in G such that p'=pg'. But since the group action is transitive, we can choose g' such that g'g^{-1}=e (the identity element in G), which gives us f'=f. Therefore, p'=pg'=pg and [p,f]=[p',f']=[pg,f]=[p,f]. This shows that \phi_f is injective.

To show that \phi_f is surjective, let [p,f] be an arbitrary element in the quotient space. Then by definition, \phi_f(p)=[p,f]. This proves that \phi_f is surjective.

I hope this helps clarify any confusion you may have had about associated bundles. Keep up the good work in your studies!

## 1. What is an associated bundle?

An associated bundle is a fiber bundle that is associated with a principal bundle. It is created by taking the Cartesian product of the total space of the principal bundle and a fiber, and then projecting it onto the base space of the principal bundle.

## 2. What is a principal bundle?

A principal bundle is a type of fiber bundle that is used to describe the symmetries of a space. It consists of a base space, a total space, and a group acting on the total space in a way that preserves the structure of the base space.

## 3. What does it mean for an associated bundle to be isomorphic to the principal bundle?

When an associated bundle is isomorphic to the principal bundle, it means that they have the same structure and are essentially the same bundle. This means that each fiber of the principal bundle corresponds to a unique fiber of the associated bundle, and the two bundles are equivalent in terms of their structure and properties.

## 4. How do you determine if an associated bundle is isomorphic to the principal bundle?

To determine if an associated bundle is isomorphic to the principal bundle, you can check if the total space and base space of the two bundles are homeomorphic, or if there exists a bijective map between the two bundles that preserves the structure of the fibers. Additionally, you can check if the two bundles have the same local trivialization maps and transition functions.

## 5. What is the significance of an associated bundle being isomorphic to the principal bundle?

The fact that an associated bundle is isomorphic to the principal bundle is significant because it allows us to study the properties and symmetries of the associated bundle by using the well-defined and understood structure of the principal bundle. It also allows us to apply the same techniques and methods used for principal bundles to associated bundles, making it easier to analyze and understand these bundles.

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