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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.

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  • $\begingroup$ If the scaling in the paper is as given, it is wrong. The factors should be $L_j^{-2}$ instead of $L_j^{-1}$. (This is a fairly innocent typo that probably doesn't effect the main result at all.) $\endgroup$
    – Willie Wong
    Commented Mar 1, 2022 at 17:48
  • $\begingroup$ I also don't know what you mean by "how he got the semigroup": the formula for $e^{it\triangle}$ is almost a definition. You may wish to edit your question to clarify what it is that you want to know. $\endgroup$
    – Willie Wong
    Commented Mar 1, 2022 at 17:50
  • $\begingroup$ @WillieWong It was my typo, and I edited it. My question is, what is the motivation that made him consider such a definition? $\endgroup$
    – Mr. Proof
    Commented Mar 2, 2022 at 3:15
  • $\begingroup$ Which definition? The Schroedinger propagator's definition is standard. Given arbitrary $f\in L^2(\mathbb{T}^3)$, you have that $e^{it\triangle}f$ solves the Schroedinger equation $i \partial_t \phi + \triangle \phi = 0$ with initial data $f$. (Its formula comes from taking the spatial Fourier transform of the equation, which due to the separability becomes a system of decoupled ODEs.) $\endgroup$
    – Willie Wong
    Commented Mar 2, 2022 at 13:34
  • $\begingroup$ @WillieWong I am confused with the Laplacian with these weights! is it compensating the weights in the tori? How can I show this relation if I am right? $\endgroup$
    – Mr. Proof
    Commented Mar 3, 2022 at 0:54

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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.

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  • $\begingroup$ Indeed. I were overthinking. $\endgroup$
    – Mr. Proof
    Commented Mar 4, 2022 at 0:24

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