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Jun 19, 2016 at 13:15 review Suggested edits
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Sep 14, 2015 at 7:05 comment added Duchamp Gérard H. E. @SebastianBurciu The space considered is exactly the space of central functions $G\rightarrow \mathbb{C}$. It is a product $\mathbb{C}^n$. The remaining question is "can it happen that this algebra be generated by a single irreducible character ?". This I do not know.
May 17, 2010 at 19:08 vote accept Sebastian Burciu
May 31 at 12:33
Jan 9, 2010 at 10:15 comment added Sebastian Burciu Next natural question. What is the polynomial space generated by two characters, $\chi$ and $\mu$? Is possible a similar description?
Jan 9, 2010 at 10:10 comment added Sebastian Burciu Yes, the space of all polynomials in $\chi$ is exactly the space you mentioned. Basically given distinct complex numbers $a_1, a_2, ... a_r$ there are polynomials $P_i$ with $P_i(a_j)=\delta_{i,j}$. Then $P_i(\chi)$ give a basis of central idempotents for the space you mentioned. One chooses the $a_i$'s as all distinct group values of $\chi$. And indeed the answer of my first question follows from here.
Jan 8, 2010 at 17:01 comment added Richard Stanley As for your first question, the space of all polynomials in a class function $\chi$ is the space of all functions $f$ that satisfy $f(u)=f(v)$ whenever $\chi(u)=\chi(v)$, a simple consequence of the evaluation of the Vandermonde determinant. Since a faithful character has a different value at the identity element than at any other element of the group, the proof follows.
Jan 8, 2010 at 16:18 comment added Richard Stanley You are right. I meant to say that the character is irreducible.
Jan 7, 2010 at 17:34 comment added Sebastian Burciu I suppose you keep the assumption that $\chi$ is an irreducible character, or at least a "real character"(i.e the character of a representation). Virtual characters with this property of course exist.
Jan 7, 2010 at 1:28 comment added Richard Stanley As a followup to my answer, it's not hard to show that if $\chi$ is a character of a finite group $G$, then any character is a complex polynomial in $\chi$ if and only if $\chi$ takes distinct values on distinct conjugacy classes. I suspect that this property is quite rare. Can anyone make this suspicion more precise? For instance, I feel that a "typical" character takes on small integer values such as $0,1,-1$ quite often.
Jan 4, 2010 at 17:17 comment added Sebastian Burciu What about question 1? Did anyone see this result before?
Jan 2, 2010 at 18:25 comment added Sebastian Burciu Thank you! $\chi$ is constant on the conjugacy classes of (12) and (1234).
Jan 2, 2010 at 18:05 vote accept Sebastian Burciu
May 17, 2010 at 19:07
Jan 2, 2010 at 17:45 history answered Richard Stanley CC BY-SA 2.5