A First Course in String Theory - Quick Calculation 2.5

[adartsesirhc]

Homework Statement

Zwiebach's A First Course in String Theory
Quick Calculation 2.5: Consider the plane (x,y) with the identification

(x,y)\rightarrow (x+2\pi R,y+2\pi R).

What is the resulting space?

Homework Equations

A one dimensional line with identification x\rightarrow x+2\pi R is a circle.

A plane with identifications
(x,y)\rightarrow (x+2\pi R,y)
(x,y)\rightarrow (x,y+2\pi R)
is a torus.

The Attempt at a Solution

The identification is just the combination of both of the torus identifications... but how would I join two boundaries with a single identification? And what would be the fundamental domain? Thanks!

 

[George Jones]

Here's a hint:

https://www.physicsforums.com/showthread.php?p=1271030#post1271030.

 

[adartsesirhc]

Hmm... I clicked on the link, but it appears to be invalid.

 

[dx]

It may be easier to think of it this way: (x,y) \rightarrow (x,y) + 2\pi R (1,1).

 

[George Jones]

Hmm... I clicked on the link, but it appears to be invalid.

That's strange. I made the link on my computer at work, and it works here at home.

Another method: try coordinates x' = \left(x + y \right)/\sqrt{2} and y' = \left(x - y \right)/\sqrt{2}.

 

[Tian WJ]

Dear adartsesirhc, I wish I could do some help. I'm very angry with the administrator's slow permission, and I have to repost via quick reply.

As you know, the Caretessian plane of $xoy$ is compact for the identification:
$$
x\sim x+2\pi R
y\sim y+2\pi R
$$
where x and y vary independently and individually, and the fundamental domain is
$$
x\in (0, 2\pi R), y\in (0, 2\pi R).
$$
However, for the identification of
$$
(x,y)\sim (x+2\pi R, y+2\pi R),
$$
the plane will no longer be compact. And because $x$ and $y$ alter together by the vector $\lambda(2\pi R, 2\pi R)$, $\lambda \in R$, there's no limited fundamental domain which could generate the 2-dimensional space via the identification.

In another sense, nevertheless, we could also attain a generalized fundamental domain. Firstly set the 2D Cartessian coordinates on the plane, with pionts $A(-\pi R, 0)$, $A'(0, -\pi R)$, $C(0, \pi R)$, $C'(\pi R, 0)$. see figure 1:

(figure 1)

and the so-called domain is the space between the lines $x+y=-\pi R$ and $x+y=\pi R$ and goes to infinity along the line. Notice that although the domain is confined between the two lines, there's no constraints along the lines, so the it is an infinite band.

Now, we could glue the boundary lines of $A-A'$ with $C-C'$, see figure2:

(figure 2)

All in all, we should notice that, for the identification of $(x,y)\sim (x+2\pi R, y+2\pi R)$, the 2D space is no longer compact due to co-movement of x and y.

As a natural extension, for n-dimensional space located $(x_1, x_2, x_3, ..., x_n)$, if the space is to be compact, the identifications for each variable should be independent to one another. Only in this way, can a limited domain exist to cover the whole space via the transformation of identifications.

Sorry that I couldnot get the two figures uploaded, please contact tianwj1@gmail.com for them.