# Why is the Brillouin zone a 2-torus?

I have been reading through several references on topological band theory, for example:

1. https://arxiv.org/abs/1510.07698
2. http://www.damtp.cam.ac.uk/user/tong/qhe/qhe.pdf
3. Topological Insulators and Topological Superconductors by A. Bernevig.

Each of these claim the the Brillouin zone in two-dimensions is a 2-torus, with the reasoning being that a shift by a reciprocal lattice vector does not matter due to the discrete lattice symmetry. For example, in the case of a square lattice with spacing $$a$$, the crystal momenta can be restricted to the compact set $$-\frac{\pi}{a}\leq k_x\leq\frac{\pi}{a}, \qquad -\frac{\pi}{a}\leq k_y\leq\frac{\pi}{a}.$$ I do not understand why this implies that the Brillouin zone can't be mapped to some other kind of closed surface, such as a sphere, which will also satisfy the periodic boundary conditions. I understand that a similar question has been asked here:

However, this does not address my question as to why the Brillouin zone cannot be taken to be some other closed surface.

To be slightly less abstract, if you plot the BZ as a rectangle in the $$(k_x,k_y)$$ plane, the statement that it is a torus means that we identify each point on the left side of the rectangle with the corresponding point on the right side, and each point on the bottom with the corresponding point on the top. This reflects the fact that the points on the left and right edges differ by $$\Delta k = \frac{2\pi}{a}$$, and are therefore the same point from the perspective of Bloch's theorem.
The statement that the BZ is a sphere, on the other hand, would mean that we identify the entire boundary as a single point. That is to say, every point in the BZ of the form $$(k_x, \pm \frac{\pi}{a})$$ or $$(\pm \frac{\pi}{a},k_y)$$ would be considered to be physically identical. But this is not the kind of identification we want, because of course e.g. the points $$(\frac{\pi}{a},0)$$ and $$(0,\frac{\pi}{a})$$ should represent different wavevectors.