Covering action

  • Thread starter mich0144
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  • #1
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Hatcher defines actions as:

Given a group G and a space Y , then an action of G on Y is a homomorphism ρ from G to the group Homeo(Y ) of all homeomorphisms from Y to itself. Thus to each g ∈ G is associated a homeomorphism ρ(g) : Y →Y.

now the following problem:
Letφ:R2 →R2 be the linear transformation φ(x, y) = (2x, y/2). This generates an action of Z on X = R2−{0}. Show this action is a covering space action and compute π1(X/Z).

I don't understand the question, going by the definition Z is G and each integer is associated with a homeomorphism from R2−{0} to R2−{0}. So how exactly does the linear transformation generate the action. Is φ the homeomorphism (kx,y/k) associated with k in Z (in this case 2) ?
 

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  • #2
quasar987
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Consider the following less abstract but equivalent definition of group action:

Given a topological group G and a topological space X, an action of G on Y is an action of the group G on the set Y which is continuous in the obvious sense that the map G x Y --> Y is continuous from the product space G x Y to Y.

The axioms of a group action on a set means that the obvious associated map G-->{Maps Y-->Y} is a homomorphism, and in fact it takes its values in Homeo(Y) since g^-1 is sent to a continuous map Y-->Y which is the inverse of the map that is the image of g.

With this in mind, observe that a linear map L:R²-->R² induces a Z-action on R²-0 simply by k*(x,y) := L(x,y)+(k,k).

Then I believe all you have to do is check condition (*) of page 72 and apply Prop 1.40.
 
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  • #3
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hmm thanks, so you're saying that the associated homeomorphism is L(x,y)+(k,k). so it maps a point (1,1) to (2,1/2) + (1,1) = (3,3/2) under the integer 1. How do you know this is the generated action aren't there others?
 
  • #4
quasar987
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Well, if the author is not more precise, then it means he's talking about the obvious action. This is the most obvious one to me.
 
  • #5
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I agree quasar I'v never see a problem where they don't specify the generated action no matter how obvious it is, for ex another action where integers are associated to powers of φ(x, y) also fits the definition of a homomorphism this seems pretty natural as well, so I have no idea.
 
  • #6
quasar987
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What do you mean "powers of φ(x, y)"?
 
  • #7
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as in each integer a is associated to (φ(x,y))^a. This works as well it seems and it's easy to see this is properly discontinuous, you can just pick your neighborhood U to be an interval less than 2x on the right endpoint and all the power translates will take it out of the neighborhood.
 
  • #8
quasar987
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What is x^a for x in R²?
 
  • #9
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x as in the x coordinate from (x,y) in R² so k in Z corresponds to taking x,y to x*2^k, y/2^k
 
  • #10
quasar987
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See this is not a group action because Z is an additive group, so its neutral element is 0. And with your proposed action, 0 acting on (x,y) gives (1,1) and not (x,y) as it should.
 
  • #11
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I don't understand how does 0 give (1,1), (φ(x,y))^0 gives (x*(2^0), y/(2^0)) = (x,y) In other words this is when phi is applied 0 times so nothing happens to (x,y) of course.
I think you thought it was (x*2)^k. I should have added parenthesis to begin with.
 
  • #12
quasar987
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Ok, I see. Well he is probably talking about your action then.

I just realized mine isn't good because 1 acting on (½,2) goes to (0,0) which he deliberately removed from the domain!
 
  • #13
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yea I think that's the implied one I just got to calculate the actual fundamental group when i I get around to it thanks for the help.
 

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