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How many fixed points on a circle S/Z2?

  1. Mar 5, 2008 #1
    I am working on Zweibach's First Course in String Theory and question 2.4 asks: Show that there are two points on the circle that are left fixed by the Z2 action. (For those without the text, the circle is the space -1 < x <= +1, identified by x ~ x + 2. And the Z2 mod imposes the x ~ -x identification on the circle.)

    I know that x = 0 is one of the fixed points, but the other alludes me. Just a guess, is it the center??? but that is not in the fundamental domain!


    ...eek!


    Thanks in advance,
    -LD
    _______________________________________
    my bread: http://www.joesbread.com/ [Broken]
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    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Mar 5, 2008 #2

    tiny-tim

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    Hi Living_Dog!

    x = 1 is fixed.

    It goes to x = -1, and x = x + 2, so -1 = 1. :smile:
     
  4. Mar 5, 2008 #3
    After thinking about it I think you are saying that:

    x ~ -x so 1 goes to -1 for the Z2 mod.

    THEN

    x ~ x + 2 takes the -1 and it goes to -1 + 2 = +1 = itself.

    Great. I see that now, but is this the general approach? First the mod and then the identification on the fundamental domain (f.d.)?

    Let's take the f.d. identification 1st, namely:

    x ~ x + 2 takes +1 to 3, which on the space of the circle, -1 < x <= +1 is 1 ... hmmm. This goes to itself already. Why include the Z2 mod?

    I'm sorry I am asking such basic questions but I only had 1 topology course a million years ago, no group theory, and ... am not that bright to begin with! :blushing: (Tomorrow I am getting Introduction to Compact Transformation Groups by Berdon. Hopefully that will help.)


    Thanks,
    -LD
    _______________________________________
    my bread: http://www.joesbread.com/ [Broken]
    my faith: http://www.angelfire.com/ny5/jbc33/
     
    Last edited by a moderator: May 3, 2017
  5. Mar 6, 2008 #4

    tiny-tim

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    :smile: … sorry to make you think! … :smile:

    The two mods just happen to have the same invariant sets.

    But if you defined, for example, x ~ -x + .1, then they wouldn't!

    I think you can do them in either order …

    btw, I don't know what the examiners' view on this is, but the reason I've been writing "=" instead of "~" is that, once you've used ~ to create the space, the "two points" are one point! :smile:
     
  6. Mar 6, 2008 #5
    You mean since -1 < x <= 1 has the same range as 0 < x <= 2?

    But why include the Z2 mod if the ... unless one is out to fix two points (requiring both id's) instead of only one.

    Now that I think of it, there is no fixed point for the id x ~ x + 2, yes?


    Thanks for your help. Don't apologize for making me think. I teach a course on critical thinking. :smile:

    -LD
    _______________________________________
    my bread: http://www.joesbread.com/ [Broken]
    my faith: http://www.angelfire.com/ny5/jbc33/
     
    Last edited by a moderator: May 3, 2017
  7. Mar 6, 2008 #6

    tiny-tim

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    Sorry - maybe I'm using no-standard terminology - by "invariant", I just meant any fixed point (any point that ~ itself).

    Hurrah! :smile:
     
  8. Mar 6, 2008 #7
    I see:

    fixed = invariant

    points = set

    I always was a fan of mathematics... it seems so useful. :smile:


    Thanks!

    -LD
    ________________________________________
    my bread: http://www.joesbread.com/ [Broken]
    my faith: http://www.angelfire.com/ny5/jbc33/
     
    Last edited by a moderator: May 3, 2017
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