Timeline for What does the set of all fundamental coweights look like?
Current License: CC BY-SA 4.0
8 events
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| Nov 4, 2023 at 13:41 | comment | added | Sam Hopkins | Again in Type $A_n$, the hyperplane arrangement you mention at the end is (basically) the so-called "resonance arrangement." This is a classic example of a hyperplane arrangement about which it is very hard to say anything. See e.g. the prior MO questions mathoverflow.net/questions/62764 and mathoverflow.net/questions/372427. | |
| Nov 4, 2023 at 7:28 | history | edited | Dr. Evil | CC BY-SA 4.0 |
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| Nov 4, 2023 at 7:22 | comment | added | Dr. Evil | Thanks a lot Sam. | |
| Nov 4, 2023 at 7:21 | history | edited | Dr. Evil | CC BY-SA 4.0 |
added 107 characters in body
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| Nov 3, 2023 at 13:01 | comment | added | Sam Hopkins | In Type A, something special happens: every fundamental weight is minuscule. So you can characterize $X$ as the set of those nonzero coweights $\omega^{\vee}$ which have $(\omega^{\vee}, \alpha) \in \{0,\pm1\}$ for all roots $\alpha \in \Phi$. But this characterization does not extend to other types. | |
| Nov 3, 2023 at 2:32 | comment | added | Sam Hopkins | In general, the size of the $W$-orbit of $\omega_i^\vee$ is $\#W/\#W_i$ where $W_i$ is the maximal parabolic subgroup corresponding to node $i$ (i.e., the Weyl group of the Dynkin diagram we get by deleting node $i$). It’s not clear you can get a better formula for $\#X$ than just summing this over all $i$. | |
| Nov 3, 2023 at 2:21 | comment | added | Sam Hopkins | If $\Phi$ is of Type $A_n$ then $\# X = 2^{n+1}-2$. | |
| Nov 3, 2023 at 2:04 | history | asked | Dr. Evil | CC BY-SA 4.0 |