Constructing a simple orbifold

  • Thread starter ehrenfest
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In summary, the problem asks if the circle C -1 <= x < =1 identified by x ~ x+2 with fundamental domain -1 < x <= 1 is an orbifold. The problem states that x = +/-1 is a fixed point, but x = -1 is not. The problem also says that there are two fixed points. How? I can only think of x = 0?
  • #1
ehrenfest
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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?
 
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  • #2
I'll start with the last question first.

ehrenfest said:
I can only think of x = 0?

What about x = -1?

Remember, you have to consider both equivalence relations.
 
  • #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?
 
  • #4
-1 and +1 are different labels for the same point. Therefore, this point is fixed under the action of x ~ -x.
 
  • #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?
 
  • #6
ehrenfest said:
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?

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.
 
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  • #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.
 
  • #8
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.
 
  • #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?
 
  • #10
ehrenfest said:
fixed points are (0,0), (0,1) = (0,-1), (-1,0) = (1,0)

Yes.

I have no idea what the fourth point is

What about the corners of the fundamental domain?
isn't a two dimensional sphere a circle?

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".
 
  • #11
George Jones said:
Yes.



What about the corners of the fundamental domain?
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...

George Jones said:
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".
Okay, but I still do not see the pillowcase.

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

1. What is an orbifold?

An orbifold is a mathematical object that is constructed by taking a space and dividing it into smaller pieces in a specific way. This division creates symmetries within the space, giving it a unique structure.

2. How is an orbifold different from a manifold?

An orbifold differs from a manifold in that it allows for singularities, or points where the space is not smooth. These singularities are essential for creating the symmetries within an orbifold.

3. What are the basic steps for constructing a simple orbifold?

The basic steps for constructing a simple orbifold involve choosing a space, identifying the symmetries that should exist within that space, and then defining the orbifold by specifying how the space is divided and how the symmetries are reflected in that division.

4. Can an orbifold have more than one type of singularity?

Yes, an orbifold can have multiple types of singularities. This can result in a complex and unique structure with a variety of symmetries.

5. How are orbifolds used in physics?

Orbifolds are used in physics to study and understand the behavior of certain physical theories, such as string theory and quantum gravity. They also have applications in other fields, such as topology and geometry.

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