Timeline for Quantum group Uq(sl(2))
Current License: CC BY-SA 3.0
2 events
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| Mar 24, 2012 at 1:17 | comment | added | Ryan | Thanks for the reply. That makes sense, and it seems easy if one starts with the result and then recovers the appropriate relation, but what what would be the process of arriving at $KX = q^{2}XK$ just starting from $[H,X] = 2X$? | |
| Mar 24, 2012 at 0:31 | history | answered | Mariano Suárez-Álvarez | CC BY-SA 3.0 |