Timeline for Can the product of a simple and a non-simple indecomposable representation be semisimple?
Current License: CC BY-SA 3.0
13 events
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| Jun 30, 2015 at 21:32 | comment | added | Giuseppe Sellaroli | @DavidHill: I'm considering the infinite dimensional representations given by Harish-Chandra in his paper "Infinite Irreducible Representations of the Lorentz Group". | |
| Jun 29, 2015 at 23:46 | comment | added | David Hill | As Jim Humphreys suggests, it would be helpful to know what kind of "generic irreducible representations" $\sigma$ you are considering? Do they have any special properties? The fact that you have information about the semi-simplicity of $\rho\otimes\sigma$ leads me to think there is more relevant information available. | |
| Jun 25, 2015 at 12:48 | history | edited | Giuseppe Sellaroli | CC BY-SA 3.0 |
added update
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| Jun 24, 2015 at 17:36 | comment | added | Will Sawin | Yes, that's the dual. Because $\rho$ is finite dimensional, $1$ is a summand of $\rho^\vee \otimes \rho$, so $\pi$ is a summand of $\rho^\vee \otimes \rho \otimes \pi$. | |
| Jun 24, 2015 at 15:12 | history | edited | Giuseppe Sellaroli | CC BY-SA 3.0 |
fixed typo
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| Jun 24, 2015 at 15:06 | comment | added | Giuseppe Sellaroli | @WillSawin: Is $\rho^\vee$ the dual representation in your notation? Can you elaborate on why this has to happen? | |
| Jun 24, 2015 at 13:15 | comment | added | Will Sawin | If $\rho$ is finite-dimensional, the reverse has to happen also - $(\rho \otimes \pi)$ is semsimple but $\rho^\vee \otimes (\rho \otimes \pi)$ is not. | |
| Jun 24, 2015 at 12:49 | history | edited | Giuseppe Sellaroli | CC BY-SA 3.0 |
added alternative wording for the question
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| Jun 24, 2015 at 3:38 | comment | added | Giuseppe Sellaroli | @JimHumphreys: I added some motivation for my question, to explain the consequences. I'm expecting not to find any example as well, but I need to be sure. | |
| Jun 24, 2015 at 3:35 | history | edited | Giuseppe Sellaroli | CC BY-SA 3.0 |
added a motivation section
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| Jun 23, 2015 at 23:25 | comment | added | Jim Humphreys | I'm not sure what the answer is, but does it have consequences (either way)? There tend to be lots of exotic infinite dimensional representations of such Lie algebras, so it's hard to predict what is possible. There don't seem to be examples in the BGG category of such semisimple tensor products with $\rho$ finite dimensional and $\pi$ of course not; but this category of modules is very restrictive. If I had to guess, I'd expect to find no examples of the type you describe. | |
| Jun 23, 2015 at 20:58 | review | First posts | |||
| Jun 23, 2015 at 21:09 | |||||
| Jun 23, 2015 at 20:58 | history | asked | Giuseppe Sellaroli | CC BY-SA 3.0 |