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In the paper Master Equation, I do not understand the basic notion introduced in Line 6 of Page 18:

"We work in the d-dimensional torus (i.e., periodic boundary conditions) that we denote $\mathbb T^d := \mathbb R^d/ \mathbb Z^d$."

I wonder what the exact meaning of $\mathbb R^d/ \mathbb Z^d$ is, and why it is torus?

user79963
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3 Answers3

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The quotient set ${\mathbb T}:={\mathbb R}/{\mathbb Z}$ is the set of real numbers modulo $1$. The real numbers $x\in[0,1[\>$ form a complete set of representatives. Since $1\sim0$ under this equivalence relation the point $x$ moving along the interval $[0,1[\>$ actually arrives at the starting point when it reaches the right end of the interval. It follows that ${\mathbb T}$ is actually a circle, or $1$-sphere $S^1$.

The set ${\mathbb T}^2$ then is the cartesian product of two circles, as is the surface of a donut. The latter is called a ($2$-dimensional) torus. You can find many videos on the internet where it is shown how a square becomes the surface of a donut under glueing of its parallel sides.

By analogy ${\mathbb T}^d\approx (S^1)^d$is called a $d$-dimensional torus.

Christian Blatter
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If you are Googling here, then this superset answer I've might be of interest: Intuition behind normal subgroups to get an introduction to the subject of quotient groups.

As a summary of that, I like to think about things like this (focusing on dimension 1, but higher dimensions are analogous):

  • $\mathbb{Z}$ is a normal subgroup of $\mathbb{R}$. This is implied by the notation $\mathbb{R}/\mathbb{Z}$, which is not defined if the second group is not a normal subgroup of the first
  • when you specify a normal subgroup of a group, that uniquely identifies an homomorphism. In this case, the homomorphism is what Christian mentioned at https://math.stackexchange.com/a/2234054/53203 , i.e. a map that maps each real number to its non-integer part in [0, 1[, e.g. 2.34 to 0.34, 5.27 to 0.27 and so on. I'm not going to justify that in detail, see the superset answer for an example, but it is not hard to convince yourself of it
  • once you understand what the homomorphism implied by the given normal subgroup, things become easy: the resulting quotient Q = G/N is just the image of G by the homomorphism specified by N (+ the induced group structure). You can then verify that that image behaves exactly like the torus, i.e. they are isomorphic.
Ciro Santilli OurBigBook.com
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I try to add some visual intuition on why is torus defined the way it is.

In the simplest case, $n = 1$, you have $T = \mathbb{R}/\mathbb{Z}$.

The $\mathbb{R}$ is like the "material" of which the torus is made. The stuff behind the quotient sign (the $\mathbb{Z}$) is the "glue" that you use for creating the shape of the torus.

Imagine you have a string and wind it around. You start at the end of the string and make a circle which has perimeter equal to 1. Then you make another circle and another and another, as you roll the string. All circles have length 1.

Imagine now the string represents the $\mathbb{R}$ and $\mathbb{Z}$ represents the "frequency" of gluing the parts of the $\mathbb{R}$. Here you make a circle of length 1, so the real line meets itself at some point after each new integer.

This makes the round shape of a donut. In higher dimensions, it is more tough to visualize of course, but I hope you get the idea.

Tereza Tizkova
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