Timeline for references on group representation over local fields / a question on an argument of a Ralph Greenberg's paper
Current License: CC BY-SA 4.0
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| Aug 16, 2018 at 0:26 | comment | added | gualterio | @Venkataramana Thank you again for answering again. But I cannot follow your argument for now. Can you please elaborate your argument? By the way, in the article, the group $\Delta$ is fixed. Do you mean that the idempotents from characters on the 'fixed' group, still form a basis? I'm sorry but I'm not used to the theory of group representations and simple algebras. | |
| Aug 15, 2018 at 3:32 | comment | added | Venkataramana | Well, that is how idempotents arise. If you have an irreducible group representation, the algebra spanned by the group elements in the representation is simple and is hence a matrix algebra over the "commutant' which in the irreducible case, is a division algebra. On the other hand, if you have a matrix algebra $M_r(D)$ as a factor of the group algebra, then $D^r$ is a representation of the group. | |
| Aug 14, 2018 at 17:07 | comment | added | gualterio | Thank you for answering. I guess I need some time to fully understand your answer. I guess you are right in that for any group ring there are orthogonal idempotents forming a basis. But in the article, the idempotents are ones coming from 'representations'. If the field is not algebraically closed there may be representations less than the order of the group. | |
| Aug 14, 2018 at 16:13 | comment | added | Venkataramana | It needs only a small tweaking when the field is not algebraically closed. The group algebra (since char is zero) is a product of matrix algebras $A_i$ over division algebras over the field. Each of these matrix algebras has an identity $e_i$ whose projection to $A_j (j\neq i)$ is zero. Clearly the sum of the $e_i$ is identity on the product algebra. | |
| Aug 14, 2018 at 15:13 | comment | added | gualterio | First of all, thank you very much for the answer. But I guess the problem is not solved. Your general argument works only when the field is algebraically closed(with the assumption on the characteristic) or when the order of the group of roots of unity is a multiple of the order of the group $\delta$. For odd prime $l$, the group of roots of unity of $\mathbb{Q}_l$ has order $l-1$. If $\delta$ is a cyclic group of order prime to $l-1$, then the only character is the trivial one. Then the summation of idempotents is not $1$. | |
| Aug 14, 2018 at 14:01 | history | answered | Filippo Alberto Edoardo | CC BY-SA 4.0 |