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Constructing a simple orbifold

  1. Aug 12, 2007 #1
    an orbifold is defined as "a space obtained by identifications that have fixed points"

    I am working on a problem that gives a circle C -1 <= x < =1 identified by x ~ x+2 with fundamental domain -1 < x <= 1. This is not an orbifold, right, even though -1 and 1 are identified on the circle?

    The problem then has C/Z_2 where Z_2 is the identification defined as usual x ~ -x. Obviously it has a fixed point at x = 0.

    I am confused about the question: "define the action of this (the Z_2) identification on the circle"

    Does this mean find you take the union of all identified points or the intersection?

    The problem also says that there are two fixed points. How? I can only think of x = 0?
     
  2. jcsd
  3. Aug 13, 2007 #2

    George Jones

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    I'll start with the last question first.

    What about x = -1?

    Remember, you have to consider both equivalence relations.
     
  4. Aug 13, 2007 #3
    How is x = -1 a fixed point in either equivalence relation? It is assigned to -1 +2n for nonzero integer n.

    Our circle contains x=-1 and x = 1 but I did not think either of these were fixed points because -1 ~ 1 but neither is identified with itself, but maybe I do not understand what a fixed point is.

    In my thinking, only the identifications are necessary to determine which points are fixed, not the somewhat arbitrary "shape" of the space. Is that wrong?
     
  5. Aug 14, 2007 #4

    George Jones

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    -1 and +1 are different labels for the same point. Therefore, this point is fixed under the action of x ~ -x.
     
  6. Aug 17, 2007 #5
    I think it is the wording of the problem that is confusing me.
    Does this make any sense to people:

    "Consider a circle C, presented as a real line with the identification x ~ x+2. Choose -1 < x <= 1 as the fundamental domain. The circle is the space -1 <= x <= 1 with the points x = +/-1 identified"

    How can the circle be both the real line in the first sentence and only that interval in the third sentence? This problem is contradictory, right? The circle in the first sentence contain an infinite number of fixed points and the circle in the third sentence contains only two, right?
     
  7. Aug 20, 2007 #6

    George Jones

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    I'm not sure how to explain this.

    As topological spaces, the quotient space [itex]\mathbb{R}/\sim[/itex] is the same as the circle [itex]S^1[/itex]. A *single* element of [itex]\mathbb{R}/\sim[/itex] is an equivalence class of elements of [itex]\mathbb{R}[/itex] that consists of an infinite number of elements of [itex]\mathbb{R}[/itex]. Such equivalence classes often are labeled by one of their of their elements.

    For example, [itex]\left[ 1.5 \right][/itex] (Here, square brackets denote equivalence class, not the closed interval that consists of a single point.) is a single element of the quotient space [itex]\mathbb{R}/\sim[/itex], and also the infinite subset [itex]\left\{ ..., -0.5, 1.5, 3.5, ... \right\}[/tex] of [itex]\mathbb{R}[/itex]. Also, [itex]\left[ 1.5 \right][/itex] and, e.g., [itex]\left[ 23.5 \right][/itex] are the same single element of [itex]\mathbb{R}/\sim[/itex].

    If [itex]x[/itex] ranges over all the elements of the interval (of [itex]\mathbb{R}[/itex]) [itex] \left[-1, 1\right)[/itex], then every element of [itex]\mathbb{R}/\sim[/itex] is hit once and only once.
     
    Last edited: Aug 20, 2007
  8. Aug 28, 2007 #7
    I still the language of the problem is contradictory or at least extremely confusing. I just do not understand what the third sentence adds to the problem. From the first sentence we know that C is the set of equivalence classes of for which [-1,1) provide us a complete set of "representatives". The third sentence just seems meaningless to me. It tells us what the circle "is" even though it was completely described in the first sentence.
     
  9. Aug 29, 2007 #8

    George Jones

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    The third sentence is supposed to give a pictorial, intuitive picture of the equivalence relation; take the string [-1,1], loop it around and join the ends, thus forming a circle.

    This type of presentation is not uncommon - first define an equivalence relation, then give a pictorial description using identification and gluing.
     
  10. Aug 30, 2007 #9
    OK. I will probably need to look at another source on orbifolds anyway. Let's move on.

    a) the action of the second identification is to reduce the circle to a semicircle [0,1] or [0,-1]

    the two fixed points are 0 and -1 =(in some mysterious sense) 1

    A fundamental domain could be either [0, 1] or [0,-1]

    what does he mean "describe the orbifold S^1/Z_2 in simple terms"?

    b)

    the four fixed points are (0,0), (0,1) = (0,-1), (-1,0) = (1,0), and I have no idea what the fourth point is

    okay and isn't a two dimensional sphere a circle? I really cannot visualize this pillowcase right now?
     
  11. Aug 30, 2007 #10

    George Jones

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    Yes.

    What about the corners of the fundamental domain?
    In mathematics, balls and spheres are different things.

    A 3-dimensional ball [itex]B^3[/itex] (like the Earth) is [itex]\{(x,y,z)|x^2 + y^2 + z^2 \le 1 \}[/itex], which has the 2-dimensional sphere [itex]S^2 = \{(x,y,z)|x^2 + y^2 + z^2 = 1 \}[/itex] as its surface.

    A circle (and its inside) is the ball [itex]B^2[/itex], while the 1-dimensional sphere [itex]S^1[/itex] is the ring that is its "surface".
     
  12. Aug 30, 2007 #11
    Okay. Doesn't that make five: (-1,1) ~ (1,-1) as well as (-1,-1) and (1,1) ? The way I thought of the torus, however, neither of those points are touching...

    Okay, but I still do not see the pillowcase.

    And I still do not know how to describe (a) "in simple terms".
     
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