Timeline for Infinitesimal deformations of a discrete group inside Lie groups vs. algebraic groups
Current License: CC BY-SA 3.0
5 events
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| Jul 21, 2012 at 1:51 | comment | added | paul garrett | Oh, sorry to be opaque: just the obvious, to my mind, namely, the upper-triangular unipotent real matrices, versus those modulo the matrices with 1's on the diagonal, and integers (!) in the upper right corner. That is, the commutator is converted from $\mathbb R$ to $\mathbb R/\mathbb Z$. From a physical/practical viewpoint (all the better), the distinction between the two is unsurprising, obvious, ... but/and, abstractly, it does require a little something to "abstractly" [sic] distinguish these. I should emphasize that I've not thought about this much... Said almost more than I know. :) | |
| Jul 21, 2012 at 1:35 | comment | added | Earthliŋ | Would you mind expanding on what other incarnations of the Heisenberg group you are thinking of? | |
| Jul 21, 2012 at 1:31 | comment | added | paul garrett | Aha! Ok, but/and unfortunately I don't know anything about these! Barely aware... :) | |
| Jul 21, 2012 at 0:51 | comment | added | Earthliŋ | Thanks. The point of the above reasoning with tangent spaces is that $Z^1(\Gamma,\mathfrak g)$ is general non-trivial, but for Weil rigidity et al., $B^1$, which coincides with the space of inifinitesimal deformations induced by the conjugation action, equals $Z^1$, so that we don't obtain any interesting deformation. I am obviously only interested in the case, where a lattice can be deformed in an interesting way. For example, the (standard incarnation) of the Heisenberg group does admit deformations, and I presume other, less algebraic incarnations of it will admit deformations, too. | |
| Jul 21, 2012 at 0:41 | history | answered | paul garrett | CC BY-SA 3.0 |