Timeline for Exact Definition of Dirac Operator
Current License: CC BY-SA 3.0
12 events
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| Jan 11, 2015 at 16:31 | comment | added | José Figueroa-O'Farrill | (continued) This representation would not be irreducible and would decompose into a number of (non-spinorial) representations. None of these representations would include the one induced from the irreducible Clifford module. | |
| Jan 11, 2015 at 16:29 | comment | added | José Figueroa-O'Farrill | @Jjm Physically, (free) particles are interpreted as irreducible unitary representations of the Poincaré group and equations such as Dirac's,... can be interpreted as projections onto irreducible components of perhaps reducible representations. If you were to take a reducible Clifford module, you would not be describing a single particle. For example, take $X$ to be Minkowski spacetime and induce a representation of the Poincaré group taking the Clifford module to the Clifford algebra itself. (continued below) | |
| Jan 11, 2015 at 8:16 | comment | added | Jjm | @JoséFigueroa-O'Farrill Only one more quetion. Why to work with irreducible modules? Would not be the most natural choice to take $V$ as $Cl(TX)$ itself? A connection in $TX$ leads to a connection in $Cl(TX)$, and we would have a truly "root of $\Delta$", at least when looking at sections of $TX\subset Cl(TX)$. | |
| Jan 10, 2015 at 21:56 | comment | added | José Figueroa-O'Farrill | @Jjm: Clifford algebras were rediscovered in Physics by Dirac, who was looking for (and found!) a square root of the Klein-Gordon equation (basically a wave equation with a mass term). So this is $s=3, t=1$ (or perhaps $s=1, t=3$ since the Physics convention for clifford algebras differs from that of Clifford by a sign). In one case the Clifford module is real (the so-called Majorana spinors) and in the other case it's quaternionic (the so-called Dirac spinors, except that the quaternionic structure was not emphasised). | |
| Jan 10, 2015 at 8:19 | comment | added | Jjm | @JoséFigueroa-O'Farrill Perhaps it is going too far from the initial question, but nevertheless: which is the point of working with modules over the Clifford Algebra? Which is the "first module" (since all this theory comes from Physics) that was used in this setting? | |
| Jan 9, 2015 at 21:53 | comment | added | José Figueroa-O'Farrill | @IgorKhavkine: Ah, sorry -- I think I misunderstood your question. You were asking about the fact that in odd dimension there are inequivalent Clifford modules and whether the Dirac operator depends on which module one takes. The bundles are different as bundles of Clifford modules, but of course equivalent as bundles of spinors. The Dirac operator in principle is different because the bundles are different. | |
| Jan 9, 2015 at 20:50 | comment | added | Igor Khavkine | I could of course also be just missing something elementary and the information you added already answers that somehow... | |
| Jan 9, 2015 at 20:49 | comment | added | Igor Khavkine | José, thanks for adding the info. But I think it might only be tangential to my comment. I essentially summarized what Trautman says on p.15 of these very nice notes. It just seems to me that if I read your definition of the Dirac operator, having fixed the ground field of $Cl(T_xM)$ to either $\mathbb{R}$ or $\mathbb{C}$, the Dirac operator is uniquely fixed, except in the case $\dim M$ odd over $\mathbb{C}$. Then there are two inequivalent choices for the irrep $\Sigma$ (which is otherwise unique). Is there a common way to break this ambiguity? | |
| Jan 9, 2015 at 20:21 | comment | added | José Figueroa-O'Farrill | @IgorKhavkine: I've added some details to the answer. | |
| Jan 9, 2015 at 20:21 | history | edited | José Figueroa-O'Farrill | CC BY-SA 3.0 |
added more details to respond to the comment.
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| Jan 9, 2015 at 19:43 | comment | added | Igor Khavkine | Just wondering how you would make this more precise, when distinguishing complex and real spinor bundles $\Sigma$. In my understanding, if $\dim M$ is even, there is a unique irreducible representation of $Cl(T_xM)$, either over $\mathbb{R}$ or $\mathbb{C}$. But when $\dim M$ is odd, $Cl(T_xM)$ has two possible inequivalent irreducible representations over $\mathbb{C}$. Is there a common terminology that distinguishes between the two corresponding Dirac operators on complex spinors? Over $\mathbb{R}$ both representations are equivalent, so there is no ambiguity. | |
| Jan 9, 2015 at 18:27 | history | answered | José Figueroa-O'Farrill | CC BY-SA 3.0 |