2

In the Analysis on Manifolds via the Laplacian page $51$, they define the torus as $\mathbb{T} = \mathbb{R}^n/\Gamma$. This quotient is unintuitive that it defines the torus. Is there exists a natural isomorphism between $\mathbb{T}$ and $\mathbb{R}^n/\Gamma$? Could anyone be able to explain to me rigourously why the quotient defines the torus $\mathbb{T}$?

  • If they use it as definition then there is nothing to show – Hagen von Eitzen Jun 29 '16 at 12:32
  • What definition of the torus are you starting from? – Lee Mosher Jun 29 '16 at 12:32
  • @LeeMosher The definition I use is simply $\mathbb{T} := \mathbb{R}^n/\Gamma$. The thing here is that it is not clear that we could glue the edge of the "parallelepiped" just with the definition. It must exists an isomorphism between the torus itself and the quotient. –  Jun 29 '16 at 12:37

2 Answers2

1

Yes. It's actually $\mathbb{R}^n / \Gamma$, not just $\mathbb{R}/\Gamma$.

$\Gamma$ is an $n$-dimensional lattice. Quotienting amounts to "gluing" two parts of the space together so they become the same. When you quotient $\mathbb{R}^n$ by an $n$-dimensional lattice $\Gamma$, you are in effect rolling $\mathbb{R}^n$ into a torus.

For example, take $n=2$. Then you can visualize $\mathbb{R}^2$ and an $n$-dimensional lattice like $\mathbb{Z}^2$. The creation of the torus involves rolling and gluing $\mathbb{R}^2$ in two directions— say vertically along the lattice and horizontally along the lattice. The end result is that $\mathbb{R}^2$ wraps repeatedly around into a torus shape.

Or, visually, imagine if every square of the lattice $\Gamma \subset \mathbb{R}^2$ were labeled like this:

torus fundamental polygon

And imagine rolling it to align A with A and B with B. You end up with a torus, rolled many times over.

user326210
  • 17,287
  • Do you mean $\mathbb{Z}^2$ rather than $\mathbb{N}^2$ ? – Max Jun 29 '16 at 03:32
  • Yes, thanks! Fixed. – user326210 Jun 29 '16 at 03:58
  • Thanks for your answer, but I have already read that argument. The thing here is that it is not clear that the quotient allows us to glue the two parts. Clearly, there exists an isomorphism allow us to glue those part. What is this isomorphism? –  Jun 29 '16 at 12:34
0

The point is to understand an isomorphism between the circle $S^1$ and $\mathbb{R/Z}$ first. If you think of $S^1$ as the unit circle in $\mathbb{C}$ then the isomorphism can be defined using the map $e^{2\pi i x}$ and applying the isomorphism theorem for groups.

Mikhail Katz
  • 42,112
  • 3
  • 66
  • 131
  • Thanks, your comments make me understand better the issue. I will try to understand myself using the isomorphism theorems (probably the third). –  Jun 29 '16 at 12:40