Monodromy map

  • Thread starter angy
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  • #1
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Hi!

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

Can anyone help me?
 

Answers and Replies

  • #2
mathwonk
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"a homomorphism ρ:π(X,x)→Sn with transitive image."

huh?
 
  • #3
lavinia
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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.
 
  • #4
mathwonk
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i guess i thought Sn was the n sphere.
 

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