How many fixed points on a circle S/Z2?

[Living_Dog]

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
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[tiny-tim]

Hi Living_Dog!

x = 1 is fixed.

It goes to x = -1, and x = x + 2, so -1 = 1.

 

[Living_Dog]

Hi Living_Dog!

x = 1 is fixed.

It goes to x = -1, and x = x + 2, so -1 = 1.

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! (Tomorrow I am getting Introduction to Compact Transformation Groups by Berdon. Hopefully that will help.)Thanks,
-LD
_______________________________________
my bread: http://www.joesbread.com/
my faith: http://www.angelfire.com/ny5/jbc33/

 

[tiny-tim]

After thinking about it I think you are saying …

… sorry to make you think! … ​

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?)

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

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

is this the general approach? First the mod and then the identification on the fundamental domain (f.d.)

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!

 

[Living_Dog]

… sorry to make you think! … ​

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

You mean since -1 < x <= 1 has the same range as 0 < x <= 2?

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!

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.

-LD
_______________________________________
my bread: http://www.joesbread.com/
my faith: http://www.angelfire.com/ny5/jbc33/

 

[tiny-tim]

You mean since -1 < x <= 1 has the same range as 0 < x <= 2?

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

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

Hurrah!

 

[Living_Dog]

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

I see:

fixed = invariant

points = set

Hurrah!

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


Thanks!

-LD
________________________________________
my bread: http://www.joesbread.com/
my faith: http://www.angelfire.com/ny5/jbc33/