Is there a practical formula for quantization on arbitrary subsets of $\mathbb{R}^n$

Question

On $\mathbb{R}^n$, there is a convenient formula for quantization in terms of the Fourier transform. More precisely, given a (sufficiently nice) function $a :\mathbb{R}^{2n}\to\mathbb{C}$, $$ \operatorname{Op}_h(a)u(x) = \frac{1}{(2\pi h)^n}\int_{\mathbb{R}^n}\int_{\mathbb{R}^n}e^{\frac{i}{h}\langle x-y, \xi\rangle}a(x,\xi)u(y)\,\mathrm{d}y\mathrm{d}\xi $$ Quantization can also be defined on subsets of $\mathbb{R}^n$. On such subsets, is there still a convenient formula that one can work with?

Answer 1

In fact, there is. If it's okay, I'll do away with the semi-classical formulation and talk microlocally.

If $\Omega\subset\mathbb{R}^n$ is open, then we can define a local version of the Kohn-Nirenberg symbol class $S^m_{loc}(\Omega\times\mathbb{R}^n)$ as the set of all $a\in C^\infty(\Omega\times\mathbb{R}^n)$ so that for any compact $K\subset \Omega$ and every multi-indices $\alpha,\beta,$ there exists $C_{K,\alpha,\beta}$ so that $$|\partial_\xi^\beta\partial_x^\alpha a(x,\xi)|\leq C_{K,\alpha,\beta}\langle\xi\rangle^{m-|\alpha|}$$ for all $(x,\xi)\in K\times\mathbb{R}^n.$ One can prove that if we take such a symbol, then the corresponding pseudodifferential operator sends

1. $C_c^\infty(\Omega)\rightarrow C^\infty(\Omega)$
2. $\mathcal{E}'(\Omega)\rightarrow\mathcal{D}'(\Omega)$
3. $H^{s+m}_{comp}(\Omega)\rightarrow H^s_{loc}(\Omega)$ for all real $s$

There's a bit of an issue, in that it doesn't generate an operator algebra. To get around this, one can utilize a locally-finite partition of unity on $\Omega$ (conjugate by cutoffs and sum over all cutoffs with overlapping support). This gives us a properly-supported pseudo-differential operator, and the symbol will differ from the original by a symbol which corresponds to a smoothing operator. This class of operators actually sends all of the relevant spaces to themselves, and we can generate a nice algebra here (e.g. we can compose in the same way).