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$L^2(\mathbb{R})$ 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.

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    $\begingroup$ The question isn't really well-defined. When you say "how do you tell them apart" you must be using some structure aside from the Hilbert space structure, as you observe. Note that the same problem persists if your replace $L^2$ with $L^p$, so this is not so much a phenomenon about Hilbert spaces as it is a phenomenon about measure spaces $\endgroup$
    – Yemon Choi
    Commented Nov 30, 2019 at 5:47
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    $\begingroup$ The isometric translation actions are natural structure that sets them apart. In this sense they are not non-isomorphic as dynamical systems, they are not even comparable, since the acting groups are different. $\endgroup$
    – Ville Salo
    Commented Nov 30, 2019 at 6:51
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    $\begingroup$ You can compare the orbit relations though, dunno if they are orbit equivalent. $\endgroup$
    – Ville Salo
    Commented Nov 30, 2019 at 6:54

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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.

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    $\begingroup$ Is there an easy way to do this without going to quantum topology? $\endgroup$
    – Praphulla Koushik
    Commented Nov 30, 2019 at 7:07
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    $\begingroup$ Is there an easier way to distinguish $\mathbb{R}$ from $\mathbb{R}^2$ than using topology? I'm not being facetious: the questions aren't well defined but however you interpret them they should have the same answer. $\endgroup$
    – Nik Weaver
    Commented Nov 30, 2019 at 7:11
  • $\begingroup$ Seeing them as vector spaces over $\mathbb{R}$, we can distinguish $\mathbb{R}$ and $\mathbb{R}^2$.. Do we have something similar when dealing with $L^2(\mathbb{R})$ and $L^2(\mathbb{R}^2)$?? I agree that the question is not well defined... Can we distinguish by structures other than quantum topology? $\endgroup$
    – Praphulla Koushik
    Commented Nov 30, 2019 at 8:17
  • $\begingroup$ I'm afraid distinguishing them as quantum vector spaces is going to be harder ... you'd need to bring Hopf algebras in to characterize the group structure, to start with. $\endgroup$
    – Nik Weaver
    Commented Nov 30, 2019 at 13:10

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