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Given a class function $f: G \to \mathbb Q$, where $G$ is a finite abelian group, is there an easy way to decide whether $f$ is an element of the rational representation ring $R_{\mathbb Q}(G)$, i.e. whether $f$ is a virtual character of some representations of $G$?

If it makes things easier you could also assume that $G$ is a $p$-group and the class function has values in $\mathbb Z$.

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up vote 3 down vote accepted

This essentially boils down to the case of a cyclic group. For a cyclic group of order n, the irreducible representations correspond to the action on $\mathbb Q[\omega_d]$ where $\omega_d$ is a primitive $d^{th}$-root of unity where $d$ divides $n$. So one can easily produce the rational character table and check if your function is a non-negative linear combination of these irreducible characters.

Added: You can now use the orthogonality relations over $\mathbb Q$ to see if something is a character of a rep. One knows the endomorphism algebra of $\mathbb Q[\omega_d]$ as a module over the cyclic group $G$ is $\mathbb Q[\omega_d]$ which has degree $\phi(d)$. So the orthogonality relations (valid over any field of characteristic 0, but in the non-algebraically closed case involves the dimensions of the endomorphism division algebras) can be used to decompose class functions in terms of irreducible characters.

Answer 2 Alternatively, one can build the character table of a finite abelian group $G$ over $\mathbb Q$ in the following way (without reducing to the case of a cyclic group). The Schur index of a degree one character is $1$. So the irreducible characters for $G$ are obtained as follows. Take a complex character $\chi$ and let $\Gamma(\chi)=Gal(\mathbb Q[\chi]:\mathbb Q)$ where $\mathbb Q[\chi]$ is the field generated by the values of $\chi$. Then the sum of the orbit of $\chi$ under $\Gamma$ is a $\mathbb Q$-irreducible character and all such characters are obtained in this fashion. So this describes an orthogonal basis for class functions over $\mathbb Q$ and hence all class functions.

supplement This may be off, but I think a rational-valued class function $f$ is a rational character iff for each complex character $\chi$, one has the inner product $\langle f,\chi\rangle$ is an integer and this integer is constant on the orbit of $\chi$ under $\Gamma(\chi)$.

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Could you explain a little more why "this essentially boils down to the case of cyclic groups"? – Dimitrios Sep 29 '11 at 10:29
    
Every finite abelian group is a direct product of cyclic groups. A representation of a direct product is a tensor product of representations of the factors. So the characters of an abelian group over $\mathbb Q$ will be products of characters of cyclic groups over $\mathbb Q$. So once you compute the character table of a cyclic group, as I explained how to do above, then you can compute the character table of any abelian group. – Benjamin Steinberg Sep 29 '11 at 12:33
    
The formula $R_{\mathbb Q}(G_1\times G_2)\cong R_{\mathbb Q}(G_1)\otimes R_{\mathbb Q}(G_2)$ only holds if the orders of the (finite) groups $G_1$ and $G_2$ are coprime. So especially in the case where $G$ is a $p$-group you can't reduce the problem to cyclic subgroups. I might be wrong though or I didn't get something right. – Dimitrios Sep 29 '11 at 13:11
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One more question concerning your example: $\mathbb Q[\omega_3]\otimes\mathbb Q[\omega_3]\cong \mathbb Q[\omega_3]\times \mathbb Q[\omega_3]$, thus it is not irreducible as an $\mathbb Q(G_1\times G_2)$-module, or is it? – Dimitrios Sep 29 '11 at 16:12
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Your are right of course and what I wrote was dumb. The nontrivial irreducible representations of $(Z/p)^n$ over $\mathbb Q$ all have degree $p-1$. Indeed, this is true for $Z/p$. Now there are (p^n-1)/(p-1) distinct homomorphisms from $(Z/p)^n$ to $Z/p$. So this gives $(p^n-1)/(p-1)$ distinct irreducible reps of degree $p-1$. Thus in the regular representation this accounts for all nontrivial irreducible reps. I think my second answer using Schur index is correct. – Benjamin Steinberg Oct 4 '11 at 16:05

A necessary criterion is that, if $g \in G$ has order $m$, and $k$ is relatively prime to $m$, then $f(g) = f(g^k)$.

Proof: Let $\rho$ be a rational representation. Let the eigenvalues of $\rho(g)$ be $\omega_1$, $\omega_2$, ..., $\omega_N$. Then $f(g) = \sum \omega_i$ and $f(g^k) = \sum \omega_i^k$.

But the $\omega_i$ are all $m$-th roots of unity, and there is a Galois symmetry $\sigma$ of $\mathbb{Q}(e^{2 \pi i/m})$ which acts on $m$-th roots of unity by $\omega \mapsto \omega^k$. So $f(g^k)$ is the image of $f(g)$ under this symmetry. Since $f(g)$ is in $\mathbb{Q}$, this shows that $f(g) = f(g^k)$. QED

One can show that any rational class function as above is a $\mathbb{Q}$-linear combination of rational characters. If you want to know about $\mathbb{Z}$-linear combinations, that sounds harder; I don't know a simple rule.

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