Timeline for Hausdorff formula for Dilatation operator
Current License: CC BY-SA 4.0
7 events
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| Oct 9 at 14:47 | comment | added | Cosmas Zachos | I have no access to your text, so I can't read it with you. A realization is a set of operators like your (2) (acting on x and its functions, which leaves the origin invariant) that satisfies the conformal algebra, and here the Lie subalgebra (1)-- whose combinatorics were strictly used in this answer. Restricted to your realization (2), the combinatoric Lie identity (1) of mine yields zero. | |
| Oct 9 at 11:35 | comment | added | Amateur Physicist | For example, in Di Francesco's book named Conformal Field Theory. Probably not, because I am missing something. | |
| Oct 9 at 11:31 | comment | added | Cosmas Zachos | “They”? I don’t understand your comment. Do you fully appreciate the difference between a Lie algebra and its realizations/representations? | |
| Oct 9 at 10:22 | comment | added | Amateur Physicist | Thanks so much. My last question is that they usually start with the operator that leave the origin invariant and they call it a different way, for example $\Delta$ for the dilation operator instead of D. In your formula (1) they would conclude that $D=\Delta -x^\rho P_\rho = \Delta -ix^\rho \partial_\rho$. In the logic of translating the operator, why don't we start with $\Delta$ between the exponential? Does the subalgebra that constitute of operator that leave the origin invariant satisfy the same commutation relation as the whole conformal group ? | |
| Oct 8 at 21:21 | history | edited | Cosmas Zachos | CC BY-SA 4.0 |
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| Oct 8 at 18:40 | history | edited | Cosmas Zachos | CC BY-SA 4.0 |
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| Oct 8 at 18:30 | history | answered | Cosmas Zachos | CC BY-SA 4.0 |