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.


1 Answer 1


A mapping from the Brillouin zone to a sphere cannot be continuous and invertible (i.e. a homeomorphism) because continuous, invertible maps preserve topological properties, and the torus and the sphere are topologically distinct.

More specifically, the 2-sphere is simply connected, which means that any curve drawn on the surface can be shrunk to a point without leaving the surface. This is sometimes casually stated as, "you can't lasso a basketball." On the other hand, the 2-torus is not simply connected. If you tie a string around the small axis (by looping it through the center of the doughnut), then you can't shrink it down; the same is true of the long axis.

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.

To summarize, we are led by Bloch's theorem to identify the points on each edge of the BZ with the points on the opposite edge (but to make no additional identifications). This already specifies the BZ as being homeomorphic to a torus. A different closed surface would have different points identified with one another, but this would conflict with the physical motivation for identifying different wavevectors in the first place.

  • $\begingroup$ if the lasso happens on a definitive side of the surface and the lasso is allowed to depart it but not cross it (as real lassos do), the small axis of the torus can only be lassoed from the outside, while the long axis can only be lassoed from the inside. Possibly not relevant to your answer, but I thought it was an interesting thing about the lasso intuition $\endgroup$
    – lurscher
    Nov 4, 2019 at 16:44
  • $\begingroup$ @lurscher Yes - the requirement that the lasso not leave the surface is central to the intuition-building. $\endgroup$
    – J. Murray
    Nov 4, 2019 at 17:12

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