Understanding Monodromy Map of a Covering Space: Can Anyone Help?

[angy]

Hi!

Suppose we have a topological space $X$, a point $x\in X$ and a homomorphism $\rho:\pi(X,x) \rightarrow S_n$ with transitive image. Consider the subgroup $H$ of $\pi(X,x)$ consisting of those homotopy classes $[\gamma]$ such that $\rho([\gamma])$ fixes the index $1\in \{1,\ldots,n\}$. I know that $H$ induces a covering space $p:Y\rightarrow X$. However, I can't understand why the monodromy map of $p$ is exactly $\rho$.

Can anyone help me?

 

[mathwonk]

"a homomorphism ρ:π(X,x)→Sn with transitive image."

huh?

 

[lavinia]

Suppose there is a transitive group action on a set of points. And let H be the stabilizer of a point. Then the action of G on the coset space, G/H, is isomorphic to the action of G on the set of points.

G acts transitively - via the monodromy action -on the fiber of the covering corresponding to the subgroup,H. H is the stabilizer of the fiber under this action.

 

[mathwonk]

i guess i thought Sn was the n sphere.

 

[blue_raver22]

I am familiar with the concept of covering spaces and their monodromy maps. The monodromy map is a fundamental tool for understanding the behavior of covering spaces and their relation to the base space. In this case, it seems that you are struggling to understand why the monodromy map of the covering space induced by the subgroup H is exactly equal to \rho.

To help you understand this, let's break down the definition of the monodromy map. The monodromy map is a homomorphism from the fundamental group of the base space to the automorphism group of the fiber. In this case, the base space is X and the fiber is the set of all points in Y that map to x\in X. The monodromy map is defined by taking a loop in X based at x and lifting it to a loop in Y based at some point in the fiber. This lifted loop then gives us an automorphism of the fiber, which is exactly what \rho represents in this case.

Now, let's consider the subgroup H of \pi(X,x) that you mentioned. This subgroup consists of homotopy classes of loops in X that fix the basepoint x and have a transitive image under \rho. This means that any loop in X based at x will have the same image under \rho, which is exactly what we need for the monodromy map to be equal to \rho. So, the reason why the monodromy map of the covering space induced by H is exactly \rho is because the subgroup H satisfies the necessary conditions for the monodromy map to be equal to \rho.

I hope this explanation helps you understand the connection between H, the induced covering space, and its monodromy map. If you have any further questions, please don't hesitate to ask for clarification. As scientists, it is important for us to have a thorough understanding of these concepts in order to further our research and contribute to our field.