What is the analogue of $f(x)=e^{-x^2}$ on the torus? What about its Fourier transform? Let $f:\mathbb R \to \mathbb R$ such that $f(x)=e^{-x^2} \ (x\in \mathbb R).$ We know that $f, \hat{f} \in L^{1}(\mathbb R).$

My Question is: What is the natural analogue  function of  $f(x)=e^{-x^2}$ on the torus?
  How to define that the analogue of $f$ on $\mathbb T$?
   We  suppose that function is $g$  defined on $\mathbb T.$ Is it true that $\hat{g}\in \ell^{1}(\mathbb Z)$?

 A: The analogue of the normal distribution on $\mathbb{T}$ is called the wrapped normal distribution (alternative names: wrapped/folded/periodized Gaussian/normal distribution). As Chappers said, the probability density function can be expressed in terms of the Jacobi-$\theta$. But the infinite series representation is the most transparent one: just take the Gaussian (centered for simplicity, s.d. $\sigma$), translate it by multiples of $2\pi$, and add the results. 
$$
f(x)=\frac{1}{\sqrt{2\pi}\sigma}\sum_{k\in\mathbb{Z}} \exp\left(-\frac{(x-2\pi k)^2}{2\sigma^2}\right)  \tag{1}
$$ 
The Fourier series representation is  also a nice one: 
$$
f(x)=\frac{1}{2\pi}+\frac{1}{\pi}\sum_{n=1}^\infty  e^{-n^2\sigma^2/2}\cos nx \tag{2}
$$ 
Both appear, for example, in this book.  The formula (2) is a special case of the Poisson summation formula which says that the Fourier coefficients of a periodized function are obtained by sampling the Fourier transform of the original function. See also the discussion of general wrapped distributions. 
