Is there an analogue of the moduli space of the torus in semi-Riemannian signature? I'm starting to study Riemann surfaces and already met the fact that Riemann surfaces have both a complex structure and a conformal structure that are, in fact, very closely related.
If we consider a Riemann surface one can classify the different conformal structures and if I correctly understand the space whose points label these different conformal structures is the so-called Riemann moduli space.
For the torus the moduli space is $${\cal M}={\cal H}/{\rm PSL(2,\mathbb{Z}})$$
where ${\cal H}$ is the upper half plane in $\mathbb{C}$ and ${\rm PSL}(2,\mathbb{Z})={\rm SL}(2,\mathbb{Z})/\mathbb{Z}_2$. So the distinct conformal structures of the torus are identified by a complex number $\tau$ with ${\rm Re}(\tau)\leq 1/2$, ${\rm Im}(\tau)>0$ and $|\tau|>1$.
Now, all of this is in the context of Riemann surfaces which are equipped with a conformal structure and hence with an equivalence class of Riemannian metrics.
But I would say nothing stops us from picking the manifold $\mathbb{T}^2=S^1\times S^1$ and endowing it with a Lorentzian metric, like $$g=-d\phi\otimes d\phi+d\psi\otimes d\psi,$$
where $\phi$ and $\psi$ are angle coordinate functions on $S^1$ and $(\phi,\psi)$ is the product chart.
Question: Is there an analogue of the moduli space and of the modular parameter for such a Lorentzian torus?
My intuition says that there should be because we can still consider two Lorentzian metrics conformally equivalent if they are a Weyl rescaling of one another, and we can still talk about a conformal structure as an equivalence class of such metrics.
 A: This is a corrected version of my earlier answer.
Many things are different in the pseudo-Riemannian setting. First of all, while every pseudo-Riemannian metric on a surface locally conformally-flat, there are  pseudo-Riemannian metrics on 2-dimensional tori which are not globally conformally-flat (unlike in the Riemannian case):
Sánchez, Miguel, Structure of Lorentzian tori with a Killing vector field, Trans. Am. Math. Soc. 349, No. 3, 1063-1080 (1997). ZBL0873.53046.
I am unaware of any classification of pseudo-conformal structures on tori in full generality. Hence, I will consider only conformal classes of flat pseudo-Riemannian tori: Quotients $T_{\Lambda}$ of ${\mathbb R}^{1,1}$ by lattices, i.e. subgroups $\Lambda$ of ${\mathbb R}^2$ generated by two translations $T_u, T_v$ along two linearly independent vectors $u, v$.
The choice of the generators $T_u, T_v$ of the lattice means that we are actually working with "marked tori;" the space of marked globally conformally flat Lorentzian tori is the Lorentzian analogue ${\mathcal T}_{1,1}$ of the classical Teichmuller space ${\mathcal T}$ of marked conformal structures on 2-dimensional tori. (The space  ${\mathcal T}$ is identified with the upper half-plane $U$, which is a model of the hyperbolic plane; the modular group $SL(2, {\mathbb Z})$ acts on $U$ via linear-fractional transformations and the moduli space ${\mathcal M}$ of conformal structures on the torus is the quotient of $U$ by the action of $SL(2, {\mathbb Z})$. Below, I will explain, to which extent the same holds for
globally conformally flat Lorentzian tori. First of all, I will consider oriented tori and I will assume that conformal maps are orientation-preserving (as it is done in the classical setting).
This orientation will translate into the assumption that the basis $\{u, v\}$ in ${\mathbb R}^2$ has the standard orientation.
One can show that two marked pseudo-Riemannian tori $T_{\Lambda}$,  $T_{\Lambda'}$ (with the given generators $u, v$ for $\Lambda$ and $u', v'$ for $\Lambda'$)
are pseudo-conformally equivalent  if and only if there exists a pseudo-conformal transformation $A$ of ${\mathbb R}^{1,1}$ such that
$$
Au= u', Av=v'$$
where $A$ is an (invertible) linear transformation preserving the given Lorentzian inner product up to a scale factor. It will be convenient for my answer to assume that this inner product is given by the Gram matrix
$$
\left[\begin{array}{cc}
0 & 1\\
1 & 0
\end{array}\right], 
$$
i.e. is $dx dy$ in terms of differentials. Then the matrix $A$ has to be diagonal with the diagonal entries of the same sign. If I were to write $u, v$ as column vectors in a matrix $B$, then the above condition reads:
$$
AB=B'. 
$$
I can also consider the effect of this multiplication on the rows of $B$ and will find that each row gets rescaled. The matrix $B$, of course, contains all the information about the vectors $u, v$. The fact that $u, v$ form a positively oriented basis means simply that $\det(B)>0$. It is convenient (at first) to allow more matrices $B$ than that and assume only that both rows $b_1, b_2$ of $B$ are nonzero. Thus, ${\mathcal T}_{1,1}$ will be contained in a space $X$ obtained from pairs of nonzero vectors $b_1, b_2\in {\mathbb R}^2$
we would be identifying pairs if they differ by rescaling $b_1$ by a positive factor and $b_2$ by a (possibly different) positive factor. This rescaling is done independently for $b_1$ and $b_2$.  The next observation is that the space of nonzero vectors $b\in {\mathbb R}^2$ up to such rescaling is simply the unit circle, since you can just normalize the vector $b$ (divide it by its Euclidean length) to make it a unit vector. Thus, our space $X$ is nothing but the 2-dimensional torus $T^2=S^1\times S^2$ (with points $(z_1, z_2)$, where $z_1, z_2$ are complex numbers of unit absolute value). Now, the torus $T^2$ contains the subset $D$ which corresponds to
pairs of vectors $u, v$ which do not form a basis because they are linearly dependent; in other words, $z_1=\pm z_2$, hence, $D$ consists of two disjoint circles. Removing these two circles, results in two open annuli, one, $A_+$, for the positively oriented bases and the other one for negatively oriented ones. (In fact, we need
to do one more identification, since we can multiply both $u, v$ by  $-1$; this sends $(z_1, z_2)$ to $(-z_1, -z_2)$. This quotient of  $T^2$ is still the 2-dimensional torus.)
Thus, the annulus $A_+$ is our Teichmuller space  ${\mathcal T}_{1,1}$. The trouble comes from an attempt to form the moduli space ${\mathcal M}_{1,1}$ which is the quotient of
${\mathcal T}_{1,1}$ by the action of $SL(2, {\mathbb Z})$. This amounts to "forgetting the basis." Unlike the classical case, this action of $SL(2, {\mathbb Z})$, however, is quite bad: Almost every orbit is dense. Hence, taking the naive quotient we get a highly non-Hausdorff space: almost every point is "infinitesimally close" to any other point.
Algebraic geometers teach us how to work around this problem (the keyword is "stacks"), but I will stop here. The bottom line: Yes, there is a moduli space, just it does not look like a surface (or, for this matter, anything you have seen before in your math or physics classes).
Two more references:

*

*The most comprehensive treatment of pseudo-Riemannian surfaces:

Weinstein, Tilla, An introduction to Lorentz surfaces, De Gruyter Expositions in Mathematics. 22. Berlin: de Gruyter. xiv, 213 p. (1996). ZBL0881.53001.


*This one should be highly relevant, but I do not have access to it:

Smyth, Robert; Weinstein, Tilla, How many Lorentz surfaces are there?, Gindikin, Simon (ed.), Topics in geometry. In memory of Joseph D’Atri. Boston, MA: Birkhäuser. Prog. Nonlinear Differ. Equ. Appl. 20, 315-330 (1996). ZBL0860.53039.
