Eigenvalues of the D'Alembertian operator My question about the spectral theory of the D'Alembertian operator on a Lorentzian manifolds (say the spacetime $M^{3+1}$) given by $$\square = -\partial_{t}^2  + \Delta$$ for the metric $g=(-+++)$. We consider this operator on a $4$-torus (i.e. the quotient of $\mathbb{R}^4$ by a lattice).
Following the analogy with the usual Laplacian, we have a family of eigenfunctions given by $e_m(x^\mu)=e^{2i\pi(x^{\mu},m)_g}$ for $m\in \mathbb{Z}^4$ which are periodic both spacelike and timelike with periods given by the lattice. The torus is compact (for the topology induced on $M^{3+1}$) so necessarily the spectrum is discrete. My questions are:
1) Does the spectrum of $\square$ has a physical interpretation?
2) Can it be used to solve the wave equation $\square u = 0$ on the $4$-torus?
 A: The free particle Schródinger equation is not relativistic:
$$  - \frac{\hbar^2}{2m} \nabla^2 \phi= i \hbar \frac{\partial}{\partial t}\phi $$
Dirac had to figure out a way to make a Lorentz-invariant theory.
\begin{eqnarray*}
E^2 &=& (mc^2)^2 + (pc)^2 \\
\left( - \frac{1}{c^2 }\frac{\partial^2 }{\partial t^2}  + \nabla^2 \right)\phi &=& \frac{m^2 c^2 }{\hbar^2 }\phi^2
\end{eqnarray*}
At this point we could set $c = \hbar = 1$ which are called "natural units" 
$$ ( \square- m) \phi = 0 $$
The probability current density is no longer positive-definite.
$$ ( i\partial - m)\phi = 0 $$
In textbooks these will be solved over a flat space like $\mathbb{R}^4$ with a unique spin-structure.  However, many physical theories are "compactified" to smaller spaces such as $\mathbb{R}^3 \times S^1$ it would not be uncommon to solve a theory on a compact space such as $S^1 \times S^1 \times S^1 \times S^1$ and let the circle radii vary.  
If we set $e(x^\mu) = e^{2\pi i \langle x^\mu , m\rangle}$ then the $m = 0$ solution would need $m_0^2 = m_1^2 + m_2^2 + m_3^2$ and the torus geometry forces $m_0, m_1, m_2, m_3 \in \mathbb{Z}$.  This is a diophantine restriction.
Also discussed on physics stackexchange:


*

*Is there a 2D manifold on which the Dirac equation has a zero mode?

While I'm at it, we should solve the pythagorean quadriple equations $a^2 + b^2 + c^2 = d^2$.  It has solutions parameterized by four integers:
\begin{eqnarray*}
a &=& m^2 + n^2 - p^2 - q^2  \\
b &=& 2(mq + np)\\
c &=& 2(nq - mp)\\
d &=& m^2 + n^2 + p^2 + q^2 
\end{eqnarray*}
Another way to look at it is to organize into three matrices:
$$ \frac{1}{d}\left[ \begin{array}{cccc} 
m^2 + n^2 - p^2 - q^2 & 2(np - mq) & 2(mp + nq) \\
2(mq + np) & m^2 - n^2 + p^2 - q^2 & 2(pq - mn) \\
2(nq - mp) & 2(mn + pq) & m^2 - n^2 - p^2 + q^2\end{array} \right] 
\in SO(3, \mathbb{Q} ) $$
It's known there's relationship between pythagorean triples and spinor groups.  However, different communities might call different things a "spinor".
