Newbie: Group Representation $\Leftrightarrow$ Left Module over the Group Ring

I am trying to understand the equivalence between group representations, $(V, \rho)$, and left modules over the group ring $F[G]$. Can you explain explicitly why it is the same?

My progress: Consider a group $G$. If $(V, \rho)$ is a representation of $G$, we can take $V$ to be a left module over $F[G]$ by defining: $gv = \rho(g)v$. So given a representation we can get a left module over $F[G]$.

The other direction is more confusing. Given a left module $M$ over $F[G]$, what is the vector space? Is $M$ necessarily a vector space? (The concept of a module is rather new to me).

• A vector space over $F$ is the same thing as an $F$-module (exercise!). An $F[G]$-module must, in particular, be an $F$-module, so it's an $F$-vector space. – Qiaochu Yuan Jun 16 '11 at 14:09
• Check also that an $F[G]$-module map is the same thing as a $G$-equivariant $F$-linear map between the representation spaces. – t.b. Jun 16 '11 at 14:24
• Ok. I got it. Thank you both. – Alba Jun 16 '11 at 14:46
• It would be great if you posted what you have as an answer. Then people could look at it and leave comments. – t.b. Jun 16 '11 at 14:52

Suppose we have a $F[G]$-module, $M$. Note that $F[G]$ contains an isomorph of $F$ within it (as the elements $\alpha 1_G$). So for our underlying abelian group we can still use $M$, and for an element $m \in M$, and $\alpha \in F$, we can define:

$\alpha m = (\alpha 1_G)m$, where the right-hand side is the $F[G]$-action: $F[G] \times M \to M$.

It should be clear from this definition that this does indeed define an $F$-module structure on $M$ (the module axioms are readily verified, as a direct consequence of $M$ being an $F[G]$-module). Now an $F$-module is a vector space over $F$ (modules and vector spaces share the same axioms, you see, so the preceding is true by definition).

So our vector space is just $M$, as you conjectured. So how to get a representation out of this? We define $\rho(g)$ to be the mapping $m \mapsto (1_Fg)m$. This mapping is invertible, with inverse $\rho(g)^{-1}$ being the map $m \mapsto (1_Fg^{-1})m$ (we are appealing to the fact that $(1_F1_G)$ is the unit of the ring $F[G]$, so that $(1_F1_G)m = m$).

There are a couple of details left to verify: we need to show that $\rho$ is a homomorphism, and that $\rho(g)$ is $F$-linear.

Let $m,m' \in M, \alpha,\beta \in F$. Then $\rho(g)(\alpha m + \beta m') = \rho(g)((\alpha 1_G)m + (\beta 1_G)m')$

$= (1_Fg)((\alpha 1_G)m + (\beta 1_G)m) = (1_Fg)((\alpha 1_G)m) + (1_Fg)((\beta 1_G)m')$

$= ((1_Fg)(\alpha 1_G))m + ((1_Fg)(\beta 1_G))m' = ((\alpha 1_G)(1_F g))m + ((\beta 1_G)(1_F g))m'$

$= (\alpha 1_G)((1_Fg)m) + (\beta 1_G)((1_Fg)m') = (\alpha 1_G)(\rho(g)(m)) + (\beta 1_G)(\rho(g)(m'))$

$= \alpha(\rho(g)(m)) + \beta(\rho(g)(m'))$, as desired, $\rho(g)$ is $F$-linear.

For arbitrary $g,h \in G, m \in M,\ \rho(gh)(m) = (1_Fgh)m = ((1_Fg)(1_Fh))m = (1_Fg)((1_Fh)m)$

$= \rho(g)((1_Fh)m) = \rho(g)(\rho(h)(m)) = (\rho(g) \circ \rho(h))(m)$, so $\rho(gh) = \rho(g) \circ \rho(h)$, thus $\rho$ is a homomorphism from $G$ to $GL_F(M)$.

I always think this stuff is done incredibly formally, as the other answers suggest. Basically the idea is that we are lazy and want to ignore the rho!

Ok so we have a representation $\rho: G \longrightarrow GL(V)$. So each $\rho(g)$ is a linear map on $V$.

So instead of writing $\rho(g)v$ as the linear map applied to $v\in V$, why not just write $gv$, and declare this as a "linear action" of $G$ on $V$ (of course there are certain axioms etc that follow but the upshot is that we get something like a "module over a group", a $G$-module. You can add in scalars to get a ring $F[G]$ and in this way you get an actual module over a ring, a $F[G]$-module).

This is ALL that is going on here!

(So to go the opposite way we just define $\rho(g)$ to be the linear map $v\mapsto gv$, and then we get a representation of $G$.)

• I know exactly what you mean: the formalism obscures the underlying truth we're trying to get at: (linear) group representations and F[G]-modules are two different lenses for looking at the same beast. Using the module axioms, for example, we can regard $1_Fg$ as simply "$g$" and $1_F1_G$ as simply $1$ (or $e$, if you prefer). A great deal of the rigamarole above is more succinctly put as "elements of F and G commute", and the notation $g.v$ is a lot less unwieldy than the "rho notation". – David Wheeler Jun 13 '12 at 3:03