Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top
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.


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 $V$.


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! :-)


share|cite|improve this question
I'd guess that the definitions are meant such that in your example, $A(V)=\{0,0,0,2,2\}$ and the multiset corresponding to the exterior square is $\{0,0,0,1,2,2,2,2,2,2\}$. In other words, $\phi_i(\sigma)$ occurs $\langle \phi_i, \chi(V) \rangle$-times in $A(V)$. – ladisch Oct 17 '13 at 18:22
@ladisch Thank you very much for that hint! That was exactly what I needed though I really don't think that this definition is a good one. Maybe it is just that I'm not familiar with multisets, but I would definitely say that this definition is not clear. What do you think? – BIS HD Oct 18 '13 at 11:02
Well, the definition of $A(V)$ you cite is, I think, very confusing. – ladisch Oct 18 '13 at 13:00
up vote 0 down vote accepted

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:

$A(V) = (S, \pi)$ with

$S = \{\phi_i(\sigma) | i = 0, \dots, n-1\} \subset \mathbb{Z}/n\mathbb{Z}$ and $\pi: S \to \mathbb{N}_0, \phi_i(\sigma) \mapsto <\phi_i,\chi(V)>$

I will compute the example I've given above with this right definition later.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.