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Identification question

  1. Aug 10, 2007 #1
    What is the space resulting from the identification (x,y) ~ (x+2piR, y+2piR)? How is it different from the space resulting from
    (x,y) ~ (x+2piR, y)
    (x,y) ~ (x, y+2piR), which is a two-dimensional torus (a donut)
     
  2. jcsd
  3. Aug 10, 2007 #2

    f-h

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    It's a cylinder, rolled up along the diagonal.
     
  4. Aug 10, 2007 #3
    What diagonal?
     
  5. Aug 10, 2007 #4

    CompuChip

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    I'd say the diagonal indicated on the attachment (up to symmetry :tongue:)
     

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    Last edited: Aug 10, 2007
  6. Aug 10, 2007 #5
    Sorry, I don't see any attachment. Was that just a joke?
     
  7. Aug 10, 2007 #6

    CompuChip

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    No, I forgot to click upload. Apparently you posted while I added it.

    (BTW: Post [itex]13^2[/itex] for me)
     
  8. Aug 10, 2007 #7
    [itex]13^2[/itex]

    Why would you ask me to post that?
     
  9. Aug 10, 2007 #8

    jim mcnamara

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    He meant he hit post #169, he did not want you to post anything.
     
  10. Aug 10, 2007 #9

    f-h

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    Take (x,y) to (x+a,y+a) and you move up parallel to the diagonal between the x and the y axis. if you identify after a = 2Pi then you get a cylinder rolled along that diagonal.
     
  11. Aug 10, 2007 #10

    George Jones

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    ehrenfest, what happens if you switch to coordinates x' = (x + y)/sqrt(2) and y' = (x - y)/sqrt(2)?
     
  12. Aug 10, 2007 #11
    I see why it is different than the other identification! The R has to be the same for both x and y.

    If you switch to light-cone coordinates, then it is a cylinder rolled around the y' axis, right?
     
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