representations vs modules

Let $G$ be a group and $k$ a field. Recall that a pair $(V,\cdot )$ is a representation of $G$ over $k$, if $V$ is a vector space over $k$ and $\cdot :G\times V\to V$ is a linear group action (compare with parent object). On the other hand we have a group algebra $kG$, which is a vector space over $k$ with $G$ as a basis and the multiplication is induced from the multiplication in $G$. Thus we can consider modules over $kG$. These two concepts are related.

If $𝕍=(V,\cdot )$ is a representation of $G$ over $k$, then define a $kG$-module $\overline{𝕍}$ by putting $\overline{𝕍}=V$ as a vector space over $k$ and the action of $kG$ on $\overline{𝕍}$ is given by

$$(\sum {\lambda }_i{g}_i)\circ v=\sum {\lambda }_i(g_i\cdot v).$$

It can be easily checked that $\overline{𝕍}$ is indeed a $kG$-module.

Analogously if $M$ is a $kG$-module (with action denoted by ,,$\circ$”), then the pair $\underset{¯}{M}=(M,\cdot )$ is a representation of $G$ over $k$, where ,,$\cdot$” is given by

$$g\cdot v=g\circ v.$$

As a simple exercise we leave the following proposition to the reader:

Proposition. Let $𝕍$ be a representation of $G$ over $k$ and let $M$ be a $kG$-module. Then

$$\overline{\underset{¯}{𝕍}}=𝕍;$$

$$\overline{\underset{¯}{M}}=M.$$

This means that modules and representations are the same concept. One can generalize this even further by showing that $\overline{\cdot }$ and $\underset{¯}{\cdot }$ are both functors, which are (mutualy invert) isomorphisms of appropriate categories.

Therefore we can easily define such concepts as ,,direct sum of representations” or ,,tensor product of representations”, etc.

Title	representations vs modules
Canonical name	RepresentationsVsModules
Date of creation	2013-03-22 19:18:59
Last modified on	2013-03-22 19:18:59
Owner	joking (16130)
Last modified by	joking (16130)
Numerical id	4
Author	joking (16130)
Entry type	Definition
Classification	msc 20C99