Timeline for Category of typical representations for Lie superalgebras
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Jun 14, 2019 at 15:16 | comment | added | GA316 | Thanks for your kind reply and it is very helpful. Have a nice day :) | |
| Jun 14, 2019 at 13:25 | comment | added | Thorsten Heidersdorf | Yes, this will only happen for $\mathfrak{osp}(1|2n)$. The superdimension is the dimension of the even part minus the dimension of the odd part. Projective representations always have superdimension 0 if your category is not semisimple (so in particular the typical ones). No, $\mathfrak{osp}(1|2n)$ does not have nontrivial one-dimensional irreducible representations. This Lie superalgebra is simple and hence the kernel of a representations is either zero or the entire Lie superalgebra as for ordinary simple Lie algebras. | |
| Jun 14, 2019 at 9:21 | history | bounty ended | GA316 | ||
| Jun 14, 2019 at 5:36 | vote | accept | GA316 | ||
| Jun 13, 2019 at 4:37 | comment | added | GA316 | Nice answer with excellent references. Thank you. So for $B(0,n)$ only this category will be closed under tensor product? Can you please explain your answer to the question (3) little more. I am not familiar with the term super dimension. please explain. Also, are you saying in the case of $\mathfrak g = B(0,n)$ there non-trivial one-dimensional modules? Is it mean to say, $\frac{\mathfrak g}{[g,g]}$ is non-trivial? Kindly explain to me. Thank you, | |
| Jun 12, 2019 at 16:30 | review | First posts | |||
| Jun 12, 2019 at 17:06 | |||||
| Jun 12, 2019 at 16:26 | history | answered | Thorsten Heidersdorf | CC BY-SA 4.0 |