Timeline for Counting adjoints in the symmetric or antisymmetric square of a Lie group representation
Current License: CC BY-SA 4.0
4 events
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| Nov 3, 2022 at 13:36 | comment | added | Callum | @BrunoLeFloch Ah yes I overlooked the spin representations and the like. Not surprising that $A$, $D$, $E$ are the exceptions to be honest. | |
| Nov 1, 2022 at 21:19 | comment | added | Bruno Le Floch | E.g., for $D_n$, consider the representation with $\mu=\varpi_{n-1}+\varpi_n$ namely the representation $R=\Lambda^{n-1}V$ where $V$ is the defining representation. Then $S^2 R$ and $\Lambda^2 R$ both contain $\text{adj}(G)$, contrarily to your conjecture, in contrast to the case of all $R=\Lambda^kV$ for $k\leq n-2$, for which your conjecture is correct. I still need to work out a bunch of details and see if there are natural generalizations, and I will then write an answer to my own question in a month or so. | |
| Nov 1, 2022 at 21:15 | comment | added | Bruno Le Floch | Thank you very much! I found the answer and will write it up ASAP. I explicitly determined the $\text{adj}(G)$ inside $R\otimes\overline R$, labeled by non-zero entries $\mu_i$ of the highest weight $\mu=\sum_i\mu_i\varpi_i$ in the basis of fundamental weights $\varpi_i$. If $\varpi_i\leftrightarrow\varpi_j$ under conjugation then any (real) representation with $\mu_i=\mu_j>0$ will have a pair of $\text{adj}(G)$ in $S^2R$ and $\Lambda^2R$. If $\varpi_i$ is conjugation-invariant then the corresponding $\text{adj}(G)$ sits in $\Lambda^2R$ or $S^2R$ if $R$ is real/quaternionic. | |
| Oct 30, 2022 at 22:23 | history | answered | Callum | CC BY-SA 4.0 |