# Proving that an action is transitive in the orbits

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• Andres316
In summary, we need to prove that if p(e) = b then p*(π1(E,e)) is a normal subgroup of π1(B,b). This can be done by showing that the group action of Δ(p) on the fiber above b is transitive, which implies that p*(π1(E,e)) is a principal G-bundle with G as a discrete group. By the homotopy lifting property, we can construct a unique lift of any path from b to another point b' in the fiber. The deck transformations must then permute these lifts, and if a path representing an element of p*(π1(E,e)) is conjugated, we can use the homotopy lifting property to construct a lift thatf

#### Andres316

<Moderator's note: Moved from General Math to Differential Geometry.>

Let p:E→ B be a covering space with a group of Deck transformations Δ(p). Let b2 ∈ B be a basic point.
Suppose that the action of Δ(p) on p-1(b0) is transitive. Show that for all b ∈ B the action of Δ(p)on p-1(b) is also transitive.

I have to prove that if e ∈ E is such that p(e)=b then p*(π1(E,e)) is a normal subgroup of π1(B,b) but I do not know how to do this, could someone help me please? Thank you very much.

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I have to prove that if e ∈ E is such that p(e)=b then p*(π1(E,e)) is a normal subgroup of π1(B,b) but I do not know how to do this, could someone help me please?
Can you show the steps from the original statement to normality? I assume you left out some information: Do we have a principal fiber bundle here? And try to use LaTeX, that makes reading a lot easier. A quick guide is here:
https://www.physicsforums.com/help/latexhelp/

I think this works.

Surround the base point ##b_0## by an open neighborhood ##U## whose inverse image is a collection of disjoint open sets ##O_{i}## that are each mapped homomorphically onto ##U##. For each point in the fiber above ##b_0## there is a unique ##O_{i}## containing it. This means that the deck transformations act transitively on these open sets as well. So they act transitively on any fiber above a point within ##U##.

Draw a path from ##b_0## to any other point ##b_1##. Cover the path with open sets like ##U##.

Andres316
I think this works.

Surround the base point ##b_0## by an open neighborhood ##U## whose inverse image is a collection of disjoint open sets ##O_{i}## that are each mapped homomorphically onto ##U##. For each point in the fiber above ##b_0## there is a unique ##O_{i}## containing it. This means that the deck transformations act transitively on these open sets as well. So they act transitively on any fiber above a point within ##U##.

Draw a path from ##b_0## to any other point ##b_1##. Cover the path with open sets like ##U##.

What do you mean by "Cover the path with open sets like ##U##", explain a little more please, thank you.

What do you mean by "Cover the path with open sets like ##U##", explain a little more please, thank you.

Around each point on the path there is an open set whose inverse image is a disjoint collection of open sets homeomorphic to it. Cover the path with sets like this. Since the path is compact there is a finite sub cover. You need to convince yourself that there is a set that overlaps ##U## and then another that overlaps the second set until the entire path is covered. If the deck transformations are transitive on one of these open sets they are transitive on any that overlap with it.

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Can you show the steps from the original statement to normality? I assume you left out some information: Do we have a principal fiber bundle here? And try to use LaTeX, that makes reading a lot easier. A quick guide is here:
https://www.physicsforums.com/help/latexhelp/

Showing that the group action is transitive on the fibers is proving that this is a principal G-bundle with G a discrete group.

Showing that the group action is transitive on the fibers is proving that this is a principal G-bundle with G a discrete group.
Yes, I misread the question, because I thought that a horizontal transitivity on sections had to be shown and was distracted by the normal subgroup. How does it fit in here?

Yes, I misread the question, because I thought that a horizontal transitivity on sections had to be shown and was distracted by the normal subgroup. How does it fit in here?

I think transitivity implies that ##p_{*}(π_{1}(E,e))## is normal.

Here is an attempt at a proof:

The key is the homotopy lifting property. This guarantees a unique lift of any path(a path can be thought of as a homotopy) with start point ##p(e)## to a path in ##E## with a given start point in the fiber above ##p(e)##. If the deck transformations are transitive then they must permute these lifts.

In general the lift of a closed path may not be closed but will connect two different points in the fiber. But any path representing an element of ##p_{*}(π_{1}(E,e))## will always lift to a closed path.

If ##γp(λ)γ^{-1}## is a conjugate of a path in ##p_{*}(π_{1}(E,e))## then the homotopy lifting property allows one to construct a lift at ##e## in stages:

- First lift ##γ^{-1}## to a path that starts at ##e## and whose end point ##e^{'}## is in the fiber above ##p(e)##.

-Next lift ##p(λ)## to start at ##e^{'}##. By transitivity of the deck transformations and uniqueness of lifts, this lift is the image of ##λ## under some deck transformation. Its end point is ##e^{'}## and it projects to ##λ##.

- Finally, lift ##γ## to start at ##e^{'}##. By uniqueness, this lift is the inverse of the lift of ##γ^{-1}## and so returns to ##e##.

The composition of the three lifts is a closed path at ##e## that projects to ##γp(λ)γ^{-1}##.

Note:
- In general the fundamental group starting at different points in the fiber projects to conjugate subgroups in the base.

- If ##B## is locally arcwise connected then the reverse is also true. Normal implies that the deck transformations are transitive on fibers.

.

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Andres316
I know the formal definition of ##pi_*(\pi_1(E,e)) ## , but, is there a nice way of representing it/ understanding it, i.e., a practical way?