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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?

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    $\begingroup$ 2) Yes. Any $L^2$ solution can be written (by Fourier series) as something with frequency support on those $m\in \mathbb{Z}^4$ satisfying $m_0^2 = m_1^2 + m_2^2 + m_3^2$. $\endgroup$
    – Willie Wong
    Commented Feb 19, 2016 at 17:47
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    $\begingroup$ 1) Physical interpretation requires connecting to physics. So for example, you can in fact solve the Schrodinger equation on $\mathbb{T}^4$ with the Minkowski metric (e.g. arxiv.org/abs/math/0503205). In this case you have a very strange momentum operator, but the physical interpretation is "the same". $\endgroup$
    – Willie Wong
    Commented Feb 19, 2016 at 17:50
  • $\begingroup$ Thanks Willie, when you say that the frequency is supported by those $m$ such that $m_0^2= m_1^2+ m_2^2 + m_3^2$, I am quite perplex, because this would implies that the eigenvalues are all null ? Maybe you mean that$m_0^2 \neq m_1^2+ m_2^2 + m_3^2$? $\endgroup$
    – usermathphy
    Commented Feb 20, 2016 at 17:34
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    $\begingroup$ If $\Box u = 0$ then by definition it is a zero-eigenvalue eigenvector of the D'Alembertian. $\endgroup$
    – Willie Wong
    Commented Feb 22, 2016 at 13:55

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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:

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".

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  • $\begingroup$ I think that we need to distinguish here between physics and math. In mathematics, the D'Alembertian can be written for spaces of arbitrary dimension, but the corresponding second-degree inhomogeneous differential equation can only be factored in 2-, 4-, or 8-space, because only these have a multiplicative norm (A. Hurwitz, 1895).For physics the multiplicative norm is required, because of the necessary conservation of being and movement (energy). (Continued next comment) $\endgroup$
    – Edgar Mueller
    Commented Oct 21, 2017 at 20:39
  • $\begingroup$ My articles e-periodica.ch/cntmng?pid=fng-001:2016:105::185 and e-periodica.ch/cntmng?pid=fng-001:2014:103::267 only focus on physics, where the spectrum of the D'Alembertian can be obtained using Dirac's coup. For spaces without a multiplicative norm, the "solutions" may not have a physical meaning. $\endgroup$
    – Edgar Mueller
    Commented Oct 21, 2017 at 20:39

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