It is traditional to say that a representation of a group $G$ over a field $F$ is "exactly like" a module over the group ring $F[G]$.

I think it is inaccurate. I think a module over $F[G]$ encodes more than a representation of $G$. I will give an example of two modules over $F[G]$ giving rise to the same group representation of $G$.

I will give the simplest example:

Let $G$ be the trivial group of order $1$, and let $F=\mathbb{C}$. I can give the abelian group $M=\mathbb{C}$ a structure of a $\mathbb{C}[G]$-module in different ways. I can let each $z\cdot 1_G\in\mathbb{C}$ act on $m\in M$ either by $(z\cdot 1_G)\cdot m=zm$, or by $(z\cdot 1_G)\cdot m=\overline{z}m$.

Is it true to say that there is a bijective correspondence between the following collections?

  1. The collection of pairs $(\rho,\theta)$, where $\rho$ is a group representation of $G$ over $F$, and $\theta$ is a field endomorphism of $F$, and
  2. The collection of modules over $F[G]$
  • $\begingroup$ Letting a complex number act by scalar multiplication by its conjugate is not "legal", that is, it is not $\mathbb C$-linear. That is a fatal objection if the representations you wish to consider are to be on complex vector spaces. $\endgroup$ Nov 15, 2013 at 0:01
  • $\begingroup$ @paulgarrett: But does it define a $\mathbb{C}[G]$-module? $\endgroup$
    – Gils
    Nov 15, 2013 at 0:02
  • 1
    $\begingroup$ In a "technical" sense, it does not define a $\mathbb C[G]$-module, because the usual convention (!) for the latter is that $\mathbb C$ acts $\mathbb C$-linearly. That is, one is not "at liberty" to fool around with the action of complex numbers. I do agree, this might be under-emphasized in some presentations, all the worse that one might reasonably think of the complex numbers as an "optional" extension of the "more natural" real numbers. $\endgroup$ Nov 15, 2013 at 0:06
  • $\begingroup$ @paulgarrett: I understand. Thanks. $\endgroup$
    – Gils
    Nov 15, 2013 at 0:10

1 Answer 1


It is possible to define a $\mathbb{C}[G]$-module structure like this, but the structure is not compatible with the $\mathbb{C}$-vector space structure on $M$. When you define a representation, you start with a vector space, so the action of the base field is prescribed.

So these two module structures are two different representation:

  • the usual one;
  • the one that starts with a different vector space over $\mathbb{C}$, that you might call $\overline M$, and the action you described.

As for your more general question: you have two categories, the category of $\mathbb{C}[G]$-modules and the category of representations of $G$ (where objects are pairs $(V, \rho)$, $V$ a vector space, $\rho : G \to End(V)$). Then these two categories are equivalent.

  • $\begingroup$ Thanks. I upvoted, but did not accept, because this does not address my question in the gray box. $\endgroup$
    – Gils
    Nov 15, 2013 at 0:12
  • $\begingroup$ FWIW your question doesn't really makes sense. Your are asking for a bijection between two things that are not sets. And if they were sets the only invariant is cardinality. THe correct notion is categorical equivalence; I will edit my question accordingly. $\endgroup$ Nov 15, 2013 at 0:14

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