Timeline for What are double groups mathematically?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Feb 16 at 14:21 | vote | accept | FusRoDah | ||
| Feb 15 at 23:01 | comment | added | Theo Johnson-Freyd | I have a hard time imagining any experiment that can be sensitive to this overall factor. It is just some overall signs, so it does not affect norm-squared's of wave functions, which is to say it does not affect any probability densities determined by any states (pure or mixed). You can sometimes pick up subtle signs / relative scalar factors by setting up two-slit interference patterns. But I don't think in this case there's any way to arrange the necessary superposition without some magic mirror that spatially-reflects any state that goes into it. | |
| Feb 15 at 22:54 | comment | added | Theo Johnson-Freyd | ... only difference is the fusion (aka tensor product). As I said, if you fuse a fermionic module (one with central character -1) with a bosonic module (one with central character +1), then that doesn't care about whether you think of them as $2_+ G$-modules or $2_- G$-modules. The only change is if you fuse two fermionic modules. If you translate between the Pin groups, the fusion of fermionic modules gets off by a (universal, overall) factor of the determinant $\det : G \to \mathrm{O}(1)$. This is the factor that for example exchanges the vector representation with the pseudovector. | |
| Feb 15 at 22:52 | comment | added | Theo Johnson-Freyd | But that's a dismissive response. Here is what I think the actual answer is. The actual answer is probably: "It doesn't matter which of the Pin groups you use". The reason is that what you care about is the block of Rep(2G) in which the centre acts nontrivially. This block is the same for both Pin groups. Indeed, it is the same as a Rep(G)-module category. Indeed, if you arbitrarily choose a $\sqrt{-1}$, then, writing $2_\pm G$ for the $\mathrm{Pin}_\pm$-cover of $G \subset \mathrm{O}(n)$, what you find is that every $2_+ G$-module is also canonically a $2_- G$-module, and conversely. The ... | |
| Feb 15 at 22:49 | comment | added | Theo Johnson-Freyd | ... canonically to a unitary-and-antiunitary representation of the full Lorentz group. The existence and uniqueness of this extension is what you should think of if you hear people talk about "CRT theorem" (usually called "CPT theorem", but that "theorem" only holds in odd dimensions where P is a product of odd-many R's --- P := diag(-1,-1,...,-1) \in O(n), and R is any reflection in a single direction). So the point is that even if your crystal has a space-reflecting symmetry, the electron cloud has no reason to: it spontaneously breaks spatial reflection. | |
| Feb 15 at 22:46 | comment | added | Theo Johnson-Freyd | ... double group 2G for which the central character is nontrivial. Now you ask: what if the point group includes reflections? You might try to dismiss the question as follows. The electron is a Dirac fermion: it is not invariant under any time-reversing symmetry. In a relativistic unitary quantum field theory, this implies also that the electron is not invariant under and space-reflecting symmetries. I'm telling you something nontrivial: I'm telling you that if you have a unitary representation of the time-preserving but possibly-space-reflecting subgroup of Lorentz group, then it extends... | |
| Feb 15 at 22:42 | comment | added | Theo Johnson-Freyd | @WillSawin Good question. I don't know much chemistry, just a little physics, so I can only speculate. My guess is that what you want to know is the possible eigendistributions of electrons, where you know that you have n electrons per site of your lattice. This is basically a question about the representation theory of the point group: the final answer will be a free module for the translation group, so it will be induced up from a representation of the point group (aka little group). If you have an odd number of electrons, then you need to look at spin representations aka reps of the ... | |
| Feb 14 at 23:36 | comment | added | Peter LeFanu Lumsdaine | Less snarkily put: Chemists are less concerned with such groups in higher dimensions. | |
| Feb 14 at 18:34 | comment | added | Will Sawin | Since some chemical structures are symmetrical under reflections, I would guess sometimes $G$ may be a subgroup of $\mathrm{O}(n)$ instead of $\mathrm{SO}(n)$, in which case the spin group could be replaced with a pin group. But which one? | |
| Feb 14 at 18:33 | comment | added | LSpice | Probably the mathematical content of this answer can do without the funny but unnecessary dig at chemists at the end. | |
| Feb 14 at 18:13 | history | answered | Theo Johnson-Freyd | CC BY-SA 4.0 |