Is the discretization of the Guassian distribution on torus still a discrete Gaussian distribution? Let $\rho_s(x) = e^{-\pi x^2/s^2}$ be the Gaussian measures, then the discrete Gaussian distribution on $\mathbb{Z}$ could be defined as $D_{\mathbb{Z},s}(x) = \rho_s(x)/\sum_{n\in \mathbb{Z}}\rho_s(n)$.
In [Regev05], the distribution $\Psi_s(r)$ on torus $\mathbb{T}=\mathbb{R}/\mathbb{Z}$ is defined :
$$\Psi_s(r)=\sum_{k=-\infty}^{+\infty}\frac1s\cdot exp\left(-\pi \left(\frac{r-k}{s}\right)^2 \right), r\in \mathbb{T}$$
then, it can be discretized as a distribution on $\mathbb{Z}_p$ :
$$\overline{\Psi}_s(i) = \int_{(i-1/2)/p}^{(i+1/2)/p}\Psi_s(x)dx, \ \ i\in \mathbb{Z}_p$$
What I want to know is: if the $D_{\mathbb{Z},s} \mod p$ is the same as the $\overline{\Psi}_s$ on $\mathbb{Z}_p$ ?
There are two ways to discretize: rounding and conditioning.
The first discretization you use to define $D_{\mathbb Z,s}$ from $\rho_s$ is using conditionning.
The second one defining $\bar \Psi_s$ from $\Psi_s$ is using rounding (the density at integer $x$ of the former distribution is given the total density of the interval $[x-\frac 1 2, x+ \frac 12]$ of the latter distribution).
So, because of this, the two distributions are not the same. However, for large $s$, they would be very close to each other, because they would both be very close to uniform.
To summarize: the moding (ie going from R to the torus) and discretizing do commute, but only if you use the same discretization on both path.