I am studying a recent paper in which the author worked on the rectangular, flat 3 tori. It can be realized, the author explained, as $\mathbb{R}^3 \over (L\_1 \mathbb Z \times L\_2 \mathbb Z \times L\_3 \mathbb Z)$ with $L\_j \in (0, \infty),j=1,2,3.$ For notational convenience, we use the coordinates for the standard torus $\mathbb{T}^3 :=
{\mathbb{R}^3 \over \mathbb{Z}^3}$ and incorporate the geometry of the torus into the Riemannian metric, using the corresponding Laplace-Beltrami operator


$$\triangle = \sum\_{j=1}^3 L\_j^{-2} \frac{\partial^2}{\partial x\_2^2}$$


We then define the Schrodinger propagator $e^{i t \triangle} $ by


$$\mathcal{F}(e^{i t \triangle } f)(\xi)=e^{- 2 \pi i t \sum\_{j=1}^3 L\_j^{-2} \xi\_j^2} \hat{f}(\xi), \,\,\, for\,\,\, \xi =(\xi\_1,\xi\_2,\xi\_3)\in \mathbb{Z}^3.$$


I have some difficulties understanding how he scaled the coordinates of the Laplace operator. Also, I can not get how he got the semigroup. Could you please explain for me in details. Thanks in advance.



The semigroup of Laplace-Beltrami operator on 3-flat torus
You are probably just overthinking it since this is basically just a multivariable calculus change of variables.


The transformation $\mathbb{R}^3 \to \mathbb{R}^3$ given by
$$ (x\_1, x\_2, x\_3) \mapsto (y\_1,y\_2,y\_3) = (L\_1 x\_1, L\_2, x\_2, L\_3 x\_3)$$
maps the torus $\mathbb{T}^3 = \mathbb{R}^3 / \mathbb{Z}^3$ to the rectangular torus $\mathbb{R}^3 / L\_1\mathbb{Z} \times L\_2 \mathbb{Z}\times L\_3 \mathbb{Z}$. In other words, this defines a *change of variables* between the standard torus and the rectangular torus.


The change of variables satisfies
$$ \frac{\partial}{\partial x\_i} = L\_i \frac{\partial}{\partial y\_i}$$


So the Laplace operator on the rectangular tori
$$ \sum \left(\frac{\partial}{\partial y\_i}\right)^2 = \sum \frac{1}{(L\_i)^2} \left( \frac{\partial}{\partial x\_i} \right)^2 $$
in the new coordinates.


