Timeline for Action of the Casimir on highest weight modules for Kac-Moody algebra
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 24, 2021 at 4:50 | vote | accept | tudong | ||
| Feb 24, 2021 at 0:46 | comment | added | SamJeralds | @tudong You are correct, the summation in the final term of $\Omega(v \otimes v)$ should have been over $\Delta$ and not $\Delta^+$; this has been edited. And yes, these are the dual basis, with the convention that $e_{-\alpha}=f_{\alpha}$. To see why the Casimir distributes the way that it does, you can work out for any $e,f$ in $\mathfrak{g}$ that $ef(v \otimes v) = e(fv \otimes v + v \otimes fv)=efv \otimes v +fv \otimes ev +ev \otimes fv +v \otimes efv$; doing this for each term and appropriately handling the scalar action coming from the basis of $\mathfrak{h}$ will give it to you. | |
| Feb 24, 2021 at 0:45 | history | edited | SamJeralds | CC BY-SA 4.0 |
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| Feb 24, 2021 at 0:05 | comment | added | tudong | Thanks for your explanation. Could you say something more about the relation $\Omega(v\otimes v)=(\Omega(v))\otimes v+v\otimes (\Omega(v))+2\sum_{\alpha\in\Delta^+\cup\{0\}}\sum_{i=1}^{n_\alpha}e_\alpha^i(v)\otimes f_\alpha^i(v)$. I am confused in this relation. And why it is $\Delta^+$? It seems to me it will be $\Delta$. In this case, $e_\alpha^i$ and $f_\alpha^i$ are the dual basis of $g$, just as $\{u_j\}$ and $\{u^j\}$. | |
| Feb 22, 2021 at 18:08 | history | answered | SamJeralds | CC BY-SA 4.0 |