Timeline for Non-commutative harmonic analysis on the discrete Heisenberg group
Current License: CC BY-SA 4.0
4 events
| when toggle format | what | by | license | comment | |
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| Jan 4, 2022 at 6:16 | comment | added | Yemon Choi | Another thought: could you encode your problem into a "reduced" version of the integer Heisenber group, by which I mean: identify the centre with ${\bf Z}$, and then quotient out the whole group by a central copy of $n{\bf Z}$ for some large integer $n$? This would yield a group that is virtually abelian, where once again there is an explicit description of the Plancherel formula (worked out in detail in Folland's A Course in Harmonic Analysis, 2nd ed.) | |
| Jan 4, 2022 at 6:07 | comment | added | Yemon Choi | As a substitute, would you be able to do anything for your intended applications if you worked with "step functions" defined on the real Heisenberg group? (There one has a very explicit Plancherel formula) | |
| Jan 4, 2022 at 5:32 | comment | added | Terry Tao | I think the best you can do in this direction is take the Fourier transform with respect to the action of the centre of the Heisenberg group (which is isomorphic to ${\mathbf Z}$), and any partition of the Pontryagin dual of that centre (which is isomorphic to ${\mathbf R}/{\mathbf Z}$) into countably many measurable pieces induces a decomposition of the form you ask for (though it would be rather artificial to identify the resulting Hilbert spaces as L^2 spaces). I suspect that these are essentially the only such decompositions, though I don't know a particularly slick way to establish this. | |
| Jan 4, 2022 at 3:56 | history | asked | LeechLattice | CC BY-SA 4.0 |