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The group $SL(2,\mathbb{C})$, the group of $2 \times 2$ complex matrices with determinent $1$, is a double cover of the Lorentz group. (These transformations can be understood as Mobius transformations on the Riemann sphere, which correspond to the action of Lorentz transformations on a sphere of light rays shooting out from the origin.) The Modular group $SL(2,\mathbb{Z})$ of $2 \times 2$ matrices with integer entries and determinent $1$, is a subgroup of $SL(2,\mathbb{C})$. Does this group therefore correspond to a discrete subgroup of the Lorentz group? What is its significance? How can it be thought of?

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    $\begingroup$ No, it is a discrete subgroup of the double cover. $\endgroup$ – Qmechanic Sep 8 at 14:44
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As emphasized in the other answers, $SL(2,\mathbb{Z})$ is a discrete subgroup of the double cover of the Lorentz group (actually of its connected component), so $SL(2,\mathbb{Z})/\{\pm 1\}$ is a discrete subgroup of the Lorentz group itself. The purpose of this new answer is to say something about the geometric significance of this discrete subgroup.

To get some geometric insight, represent a vector with components $(t,x,y,z)$ as a $2\times 2$ matrix $$ X := \left(\begin{matrix} t+z & x+iy \\ x-iy & t-z \end{matrix}\right) $$ with $X^\dagger=X$ and $\text{det}(X)=t^2-x^2-y^2-z^2$, so that a Lorentz transformation can be expressed as $$ X \mapsto M X M^\dagger $$ with $M\in SL(2,\mathbb{C})$. The fact that $\pm M$ both give the same transformation of $X$ corresponds to the fact that $SL(2,\mathbb{C})$ is the double cover of the Lorentz group.

One thing to notice immediately is that the $y$ component of $X$ is invariant under every $M\in SL(2,\mathbb{Z})$. This follows from the fact that the real and imaginary parts of $X$ are separately self-contained under such a transformation, and from the fact that the reflection $y\to -y$ is excluded from the connected component. Therefore, after quotienting by $\{\pm 1\}$, the set of trasnformations with $M\in SL(2,\mathbb{Z})$ implements a subgroup of the Lorentz group of the three-dimensional spacetime with coordinates $t,x,z$.

Now, recall that the modular group $SL(2,\mathbb{Z})$ is generated by the two transformations $$ \left(\begin{matrix} 0 & 1 \\ -1 & 0 \end{matrix}\right) \hskip1cm \text{and} \hskip1cm \left(\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}\right). $$ The first of these implements the Lorentz transformation $$ \left[\begin{matrix} t \\ x \\ z \end{matrix}\right] \mapsto \left[\begin{matrix} t \\ -x \\ -z \end{matrix}\right], \tag{1} $$ and the second one implements the Lorentz transformation $$ \left[\begin{matrix} t \\ x \\ z \end{matrix}\right] \mapsto \left[\begin{matrix} (3t+2x-z)/2 \\ t+x-z \\ (t+2x+z)/2 \end{matrix}\right]. \tag{2} $$ Notice that the coefficients in this transformation are not all integers, although they are rational. As a check, we can confirm directly that this really is a Lorentz transformation: $$ \left(\frac{3t+2x-z}{2}\right)^2 - (t+x-z)^2 - \left(\frac{t+2x+z}{2}\right)^2 = t^2-x^2-z^2. $$ The geometric significance of (1) is obvious, but (2) is more difficult to visualize.

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  • $\begingroup$ +1. I suspect that the second transformation is a "parabolic" mobius transformation around the $y$-axis. (I.e., it is a member of the little group for a light like particle travelling in the $y$ direction.) I suspect this is because that mobius transformation has a double fixed point. Also, because that generator, when repeatedly composed with itself, never becomes the identity, just like that element of the Lorentz group. $\endgroup$ – user1379857 Sep 8 at 18:29
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I never considered this issue, but I think the answer is positive if, in your view, a discrete subgroup is a subgroup of a topological group which is made of isolated points.

In fact, the canonical projection $\pi : SL(2, \mathbb{C}) \to SO(3,1)_+$ is a surjective Lie-group morphism which is a local Lie-group isomorphism around the identity. As a consequence it is in particular a local diffeomorphism in a neighborhood of every given point in its domain. Hence the subgroup $\pi(SL(2, \mathbb{Z}))$ is made of isolated points and thus it is a discrete subgroup of $SO(3,1)_+$.

Notice that the correspondence between $\pi(SL(2, \mathbb{Z}))$ and $SL(2, \mathbb{Z})$ is still twofold because $-I \in SL(2, \mathbb{Z})$ and $Ker(\pi)= \{\pm1\}$.

The point is that, in general, the $16$ entries of $\pi(A)$ are not in $\mathbb{Z}$ if $A \in SL(2, \mathbb{Z})$.

I cannot see any physical significance of this subgroup, sorry.

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Almost. Note that $SL(2, \mathbb{C})$ is not quite the group of Möbius transformations, but a double cover as well: $\Lambda$ and $-\Lambda\in SL(2, \mathbb{C})$ act identically on $\mathbb P^1(\mathbb C)$, but the action of the quotient $PSL(2,\mathbb C)$ of $S L(2, \mathbb{C})$ by $\langle -I\rangle$ is faithful.

This actually gives an isomorphism between the Lorentz group and the group of Möbius transformations, and the modular group, both acting on the complex projective line (i.e. the Riemann sphere) and on the complex upper half plane (where most of its mathematical interest lies) can be seen as a subgroup of the Lorentz group, as it really should be viewed as $PSL(2,\mathbb Z)$ rather than $SL(2,\mathbb Z)$.

As for the interpretation, see Naber - the Geometry of Minkowski Spacetime for details: the Lorentz group acts on Minkowski space. It actually preserves the light cone $\mathcal C = \left\{x\in\mathbb R^{1,3} \,|\, \eta(x,x) = 0\right\}$, hence it also acts on its quotient consisting of light rays. The latter is isomorphic to the Riemann sphere, in a way that can be made concrete.

It can also be seen as a subset of $\mathbb R^{1,3}$ in a natural way, by intersecting it with the hyperplane with $t = -1$, so that the points are described by

$$x^2 + y^2 + z^2 = t^2 = 1$$

whose elements clearly span $\mathbb R^{1,3}$, so the action or a Lorentz transformation on the cone of light rays determines the full Lorentz transformation.

Making the correspondence concrete (to go in the other direction) is not hard but somewhat cumbersome, and is essentially a stereographic projection.

Since whatever we see is the light that comes to us (all light rays through a point, our pupil, projected onto our retina), a very interesting consequence of this correspondence is that, since both stereographic projection and Möbius transformations preserve circles, anything circular will be seen as a circle in all inertial frames (though centers will not be preserved).

All this has been about the group of Möbius transformations. I don't know if the modular group can by physically singled out.

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