A First Course in String Theory - Quick Calculation 2.5

1. Apr 17, 2009

1. The problem statement, all variables and given/known data
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?

2. Relevant 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.

3. 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!

2. Apr 18, 2009

George Jones

Staff Emeritus
3. Apr 18, 2009

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

4. Apr 19, 2009

dx

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

5. Apr 19, 2009

George Jones

Staff Emeritus
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 $y' = \left(x - y \right)/\sqrt{2}$.

6. Apr 20, 2009

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.