General Setting:
In a paper I'm working on, the author uses multisets to describe the representation theory of the cyclic group $G = C_n = <\sigma>$ of order $n$. Let $\varphi_i$ denote the absolutely irreducible characters of $G$, defined by the action of $\sigma$ on the $n$-th primitive root of unity $\zeta_n$ by $\sigma \cdot \zeta_n = \zeta_n ^{\varphi_i(\sigma)}$. So $\varphi_i(\sigma)$ are just elements of $\mathbb{Z}/n\mathbb{Z}$.
For a representation $V$ of a group $G$ he denotes by $$A(V) = \{<\varphi_i, \chi(V)> \varphi_i(\sigma), i = 0, \dots, n-1\}$$ the multiset corresponding to the decomposition of the $G$-module $V$. Here, $\chi(V)$ denotes the character of $V$ and the scalar product is just the usual inner product.
For an arbitrary multiset $A = \{a_0,\dots,a_n\}$ he then defines the exterior square as $$\bigwedge\nolimits^2(A) = \{a_i + a_j, i<j\}$$ which corresponds to the definition found for example in this paper - though I haven't really found a reference for this notation.
Claim:
The main point now is the following: He claims that the multiset associated to the exterior square representation of $V$ is equal to the exterior square of the multiset associated to $V$.
Problem:
The problem I'm having with this is now, that the multiset associated to $V$ has at most $n$ elements, corresponding to the $n$ absolutely irreducible characters of $C_n$ over a suitable field extension whereas the exterior square of this multiset does have exactly $\frac{1}{2}n(n-1)$ elements. So why should this be a multiset associated to the exterior square of a representation? I think the exterior square of any representation (character) just has at most $n$ irreducible characters as constituents with maybe different multiplicities. Where am I wrong with this?
Example (Character of a cyclic group of order 3):
To make my question clear, let's work out an example: Let $n=3$ and $\chi(V) = 3 \varphi_0 + 2 \varphi_2$ be a the character decomposition of $V$ in the absolutely irreducible characters $\varphi_0$ (trivial) and $\varphi_2$ (the character sending $\sigma \mapsto \zeta_3^2$.) The multiset associated to $V$ is then $$A(V) = \{0,0,1\}$$ because $$<\varphi_0, \chi(V)> \varphi_0(\sigma) = 2 * 0 = 0$$ $$<\varphi_1, \chi(V)> \varphi_1(\sigma) = 0 * 1 = 0$$ $$<\varphi_2, \chi(V)> \varphi_2(\sigma) = 2 * 2 = 4 \equiv 1 \mod 3$$
The exterior square associated to this multiset is then $$\bigwedge\nolimits^2 A(V) = \{0,1,1\}$$ for obvious reasons. If we calculate the exterior square of our representation $V$, we get the following character decomposition: $$\chi(\bigwedge\nolimits^2 V) = 3\varphi_0 + \varphi_1 + 6\varphi_2$$ so the associated multiset is $$A(\bigwedge\nolimits^2 V) = \{0,0,1\}$$ because $$<\varphi_0, \chi(V)> \varphi_0(\sigma) = 3 * 0 = 0$$ $$<\varphi_1, \chi(V)> \varphi_1(\sigma) = 1 * 1 = 1$$ $$<\varphi_2, \chi(V)> \varphi_2(\sigma) = 6 * 2 = 12 \equiv 0 \mod 3$$ So there obviously the multiset which are supposed to be the same, do not agree, since
$$A(\bigwedge\nolimits^2 V) = \{0,0,1\} \neq \{0,1,1\} = \bigwedge\nolimits^2 A(V)$$
Is my example alright? Do you see the problem or am I misscalculating or -understanding something?
Thank you very much! :-)
Tom
As it turns out, the definition of the multiset $A(V)$ associated to the representation $V$ was formally incorrect. Let me give a formal correct definition:
I will compute the example I've given above with this right definition later.