$L^2(\mathbb{R})$ and $L^2(\mathbb{R}^2)$ are isometrically isomorphic because both are infinite-dimensional separable Hilbert spaces.


If a Hilbert space $H$ is $L^2(\mathbb{R})$ or $L^2(\mathbb{R}^2)$, how do you tell them apart without checking the domain of their elements?


For example, there is a bijection between $\mathbb{R}$ and $\mathbb{R}^2$, so they are set-theoretically isomorphic, but we can tell them by their topology.



What is the difference between $L^2(\mathbb{R})$ and $L^2(\mathbb{R}^2)$?
We can tell $\mathbb{R}$ and $\mathbb{R}^2$ apart by topology, and we can tell $L^2(\mathbb{R})$ and $L^2(\mathbb{R}^2)$ apart by quantum topology. In the sense that a choice of C\*-algebra contained in $B(H)$ amounts to a choice of quantum topology on $H$.


In your case that C\*-algebra would be either $C\_0(\mathbb{R})$ or $C\_0(\mathbb{R}^2)$, realized as multiplication operators in $B(L^2(\mathbb{R}))$ or $B(L^2(\mathbb{R}^2))$. There can't be an isomorphism between the two Hilbert spaces that induces an isomorphism between the two C\*-algebras, because the two C\*-algebras aren't isomorphic --- precisely because $\mathbb{R}$ and $\mathbb{R}^2$ aren't homeomorphic.


My book *Mathematical Quantization* is all about these sorts of analogies.


