Timeline for Real-valued character in Block with cyclic defect has at most two constituents modulo $p$
Current License: CC BY-SA 3.0
9 events
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| Jun 15, 2016 at 7:45 | comment | added | Matthias Klupsch | @JayTaylor : You are right, this looks better. I edited it. | |
| Jun 15, 2016 at 7:44 | history | edited | Matthias Klupsch | CC BY-SA 3.0 |
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| Jun 13, 2016 at 15:00 | comment | added | Jay Taylor | @MatthiasKlupsch I think from an English perspective I'd slightly modify your formulation as follows: "If $\chi \in \mathrm{Irr}(B)$ is real valued then there are at most two real-valued irreducible Brauer characters which are constituents of $\widehat{\chi} = \chi|_{G_{p'}}$." This, for me, is then quite clear. | |
| Jun 13, 2016 at 14:42 | comment | added | Frieder Ladisch | Maybe better, but I'm not completely sure... I still find my formulation, or Feit's original formulation, less ambiguous. Of course it is a question of english language, and it's not my native language, either. | |
| Jun 13, 2016 at 14:12 | comment | added | Matthias Klupsch | @FriederLadisch You are right, I tried to fix it. Do you think it is Ok now? | |
| Jun 13, 2016 at 14:07 | history | edited | Matthias Klupsch | CC BY-SA 3.0 |
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| Jun 13, 2016 at 13:22 | comment | added | Frieder Ladisch | I think your formulation is still ambiguous and can be mistaken to say that $\widehat{\chi}$ is the sum of at most two irreducible Brauer characters, and these are real valued. But the theorem says: At most two of the irreducible Brauer constituents of $\widehat{\chi}$ are real valued (but there may be others). | |
| Jun 13, 2016 at 10:04 | vote | accept | Matthias Klupsch | ||
| Jun 13, 2016 at 8:51 | history | answered | Matthias Klupsch | CC BY-SA 3.0 |