Talk:Aufgaben:Problem 7
Did we prove the orthonormality of characters? meaning are you sure we can use them, it would make proof 13 much shorter...
Okay that doesn't make sense 13 (a) is the proof of that fact :). I see that in the ML from series 9, this was used so it's probably aright. Carl (talk) 14:30, 17 June 2015 (CEST)
I don't understand what is going on in step 0, when you calculate the number of conjugacy classes. Everything else looks good I think.
Carl (talk) 16:55, 3 July 2015 (CEST)
With alternative Step 5 you can do this exercise without using the orthogonality of the characters, if in Step 4 you can show that the tenor product of an irreducible rep. and a one dimesional rep. is again irreducible (shouldn't be very difficult). Though it seems unlike that this would be the way to go, the exercise gets to long for the exam, making this irrelevant :)
Carl (talk) 23:27, 6 July 2015 (CEST)
In step 5: Is there a reason why we can't just propose that
- \(\rho^{(5)} = \rho^{std \otimes std} - \rho^{triv} - \rho^{std} - \rho^{ sgn \otimes std }\)
defines another irrep and then just show that it indeed is irreducable? This would save a few lines.
--Nik (talk) 15:57, 30 July 2015 (CEST)
No, but you could claim that \( \rho^{std \otimes std} \simeq \rho^{triv} \oplus \rho^{std} \oplus \rho^{ sgn \otimes std } \oplus \rho^{(5)} \), which is probably what you mean. This was the previous version, I just thought that this was to much to propose, but for no real reason.
Carl (talk) 11:33, 31 July 2015 (CEST)
Ok, will use that then. Thanks!
--Nik (talk) 10:38, 1 August 2015 (CEST)
In Step 4 it's enough to show that \(\left(\chi^{sgn\otimes std }, \chi^{sgn\otimes std }\right)_G = 1 \), isn't it?
I think she used that if it's 1 then it's irreducible as well in her solution for S3. If not one would have to show that it's also zero with the missing representation, not?
Beni (talk) 11:13, 1 August 2015 (CEST)
On second thought I can't see why it is not enough that the character is different, so yes it should be enough.
