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Related to a previous question I am asking furthermore a proof for the following:

Question 1: If $\chi$ is a faithful irreducible character of a finite group $G$ then the regular character of $G$ is a polynomial with integer coefficients in $\chi$?

I know this fact is true since there is a generalization of it for Hopf algebras in Corollary 19 of the paper FSU96-08 from here.

The proof from that paper is a little complicated using some (although elementary) results on norms and inner products.

I was wondering if anyone knows a different proof of this.

Using the Stone - Weierstrass method mentioned in the previous question, I am asking further if the following is true:

Question 2: If $\chi$ is a faithful irreducible character of a finite group $G$ does any character of $G$ is a complex polynomial in $\chi$?

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  • $\begingroup$ Sorry for the question, but what exactly is a "faithful irreducible character"? The trace form of a faithful irreducible representation? $\endgroup$
    – darij grinberg
    Commented Jan 2, 2010 at 16:14
  • $\begingroup$ Yes, the character of an irreducible representation which has trivial kernel. $\endgroup$
    – Sebastian Burciu
    Commented Jan 2, 2010 at 18:07

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I put an answer (due to Blichfeldt, not me) to essentially this question at your earlier question. To address the problem raised by Richard Stanley, one result I know in this direction is by John Thompson: if $\chi$ is an irreducible character of a finite group $G$, then there are more than $|G|/3$ elements at which the value taken by $\chi$ is either zero or a root of unity.

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I assume that by "faithful irreducible character" you mean the character of a faithful (i.e., trivial kernel) irreducible representation. In this case, the answer to Question 2 is negative. For instance, the irreducible character $\chi$ of the symmetric group $S_4$ indexed by the partition (3,1) is faithful and has the same value on two different conjugacy classes of $S_4$, so the same will be true of any complex polynomial in $\chi$.

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  • $\begingroup$ Thank you! $\chi$ is constant on the conjugacy classes of (12) and (1234). $\endgroup$
    – Sebastian Burciu
    Commented Jan 2, 2010 at 18:25
  • $\begingroup$ What about question 1? Did anyone see this result before? $\endgroup$
    – Sebastian Burciu
    Commented Jan 4, 2010 at 17:17
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    $\begingroup$ 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. $\endgroup$
    – Richard Stanley
    Commented Jan 7, 2010 at 1:28
  • $\begingroup$ 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. $\endgroup$
    – Sebastian Burciu
    Commented Jan 7, 2010 at 17:34
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    $\begingroup$ 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. $\endgroup$
    – Richard Stanley
    Commented Jan 8, 2010 at 17:01
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Here is a short proof of the weaker version of the statement from Question 1 (giving a polynomial with rational coefficients). Let's think of characters as functions on conjugacy classes. Then $\chi(1)=n={\rm dim}(V)$, and $\chi(g)$ for $g\ne 1$ has smaller absolute value than $n$ (since the representation is faithful and eigenvalues of $g$ in $\chi$ are roots of 1). In particular, $\chi(g)\ne n$. Now let P be the interpolation polynomial such that $P(n)=|G|$ and $P(x)=0$ for any other value $x$ of $\chi$. Then $P(\chi)$ is the regular character, and it's easy to see that $P$ has rational coefficients.

However, there seems to be a counterexample to the statement that $P$ can be chosen to have integer coefficients. Namely, take $G=A_5$, and $\chi$ the 5-dimensional character. Its values are well known to be $5,0,1,-1$, so we can take $P_0=(x^3-x)/2$, and any other polynomial which works will be of the form $P=P_0Q$, where $Q$ is another polynomial (as $P$ must vanish at $0,1,-1$). If $P$ has integer coefficients, then $Q/2=P/(x^3-x)$ must have integer coefficients, so values of $Q$ at integers are even. On the other hand, we must have $Q(5)=1$, contradiction.

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  • $\begingroup$ Thank you for the answer. Corollary 19 from the paper FSU96-08, mentioned above, states indeed that it should be a polynomial in \chi with rational coefficients, not integers. Sorry about that! $\endgroup$
    – Sebastian Burciu
    Commented Feb 2, 2010 at 14:30

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