Timeline for Multiple factors of the character of a representation
Current License: CC BY-SA 2.5
5 events
| when toggle format | what | by | license | comment | |
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| Dec 1, 2014 at 23:28 | comment | added | Geoff Robinson | If you wish to speak in terms of algorithms, then as Jim says, the multiplication on $A = Z(\mathbb{C}G)$ determines the character table. Use the basis of class sums, and for each class sum $C$, write down the matrix (in terms of the standard basis) for the linear transformation of $A$ given by multiplication by $C$, say $M(C)$. Simultaneously diagonalizing the $M(C)$ (which theory tells us can be done) leads to a basis of mutually orthogonal idempotents of $A$. This easily yields the character table, which allows decomposition of all characters. | |
| Nov 18, 2010 at 23:01 | comment | added | Jim Humphreys | But algorithms in character theory depend on the starting point: what do you already know about the group and its characters? What you wrote in the highlighted part of your question is still mysterious to me, since it is usually impossible to know how the sums over classes multiply. | |
| Nov 18, 2010 at 21:02 | comment | added | Denis Serre | I know the book by the other Serre. I even know Jean-Pierre, because we are related (not a secret). But this "canonical" decomposition is not what I look for, for several reasons. First, the elements of the decomposition are multiple of a single irrep. Second, it concerns representations, whereas I should like to work with characters, and we don't know in general to pass from a char to its rep. Third, this canonical decomp is an abstract result, with no algorithm. | |
| Nov 18, 2010 at 19:40 | history | edited | Jim Humphreys | CC BY-SA 2.5 |
added 510 characters in body
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| Nov 18, 2010 at 14:46 | history | answered | Jim Humphreys | CC BY-SA 2.5 |