Is the torus an orbifold?

Question

Reading the book Geometry of Surfaces by Stillwell I notice that he defines an orbifold as the orbit space $S_\Pi = S/\Gamma$ where $S = \mathbb{C}, S^2, \mathbb{H}^2$ and $\Gamma$ is a group generate by the side-pairing transformations of a polygon $\Pi\subset S$.

On Wikipedia and other sites they say that the quotient must be done with a finite group. The thing is that, if you take a square and identify his sides, isn't it done by the group $\mathbb{Z}\times \mathbb{Z}$ which is not a finite group?

Thank you.

Answer 1

You might have missed something, namely that an orbifold is an object that is locally modeled by the quotient of a finite group action. It need not be a global quotient of a finite group action.

Let me quote the following very nice sentence from the Wikipedia:

Like a manifold, an orbifold is specified by local conditions; however, instead of being locally modelled on open subsets of $\mathbb R^n$, an orbifold is locally modelled on quotients of open subsets of $\mathbb R^n$ by finite group actions.

For example, one of these local models of an orbifold might have this form: a point, and a neighborhood of that point, that is modelled on the unit disc in $\mathbb R^2$ modulo the order 2 group generated by a $180^\circ$ rotation.

Here is an example of an orbifold of the form $\mathbb R^2 / \Gamma$ as Stillwell specifies (or, if you prefer, as $\mathbb C / \Gamma$), where the group $\Gamma$ is infinite, and where these order 2 rotation models occur.

Start with the rectangle $[-1,1] \times [0,1] \subset \mathbb R^2$, where the two long sides are subdivided by inserting vertices $(0,0)$ and $(0,1)$. The group $\Gamma$ is generated by the following side pairings:

• $\{-1\} \times [0,1]$ and $\{1\} \times [0,1]$ are identified by translation, identifying $(-1,t)$ to $(1,t)$
• $[-1,0] \times \{0\}$ and $[0,1] \times \{0\}$ are identified by $180^\circ$ rotation, identifying $(x,0)$ and $(-x,0)$
• $[-1,0] \times \{1\}$ and $[0,1] \times \{1\}$ are identified by $180^\circ$ rotation, identifying $(x,1)$ and $(-x,1)$.

The quotient orbifold looks like the cover of a square pillow, and the four corners of that pillow have neighborhoods modelled by the unit disc modulo a $180^\circ$ order 2 rotation group. Intuitively, think of the pillow cover as obtained by appropriately sewing the sides of a piece of cloth corresponding to $[-1,1] \times [0,1]$.