Isomorphism between $\mathbb{T}$ and $\mathbb{R}^n/\Gamma$?

Question

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 $\mathbb{T}$?

Answer 1

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.

Answer 2

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:

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