# The Unapologetic Mathematician

## Modules

With the group algebra in hand, we now define a “$G$-module” to be a module for the group algebra of $G$. That is, it’s a (finite-dimensional) vector space $V$ and a bilinear map $A:\mathbb{C}[G]\times V\to V$. This map must satisfy $A(\mathbf{e},v)=v$ and $A(\mathbf{g},A(\mathbf{h},v))=A(\mathbf{gh},v)$.

This is really the same thing as a representation, since we may as well pick a basis $\{e_i\}$ for $V$ and write $V=\mathbb{C}^d$. Then for any $g\in G$ we can write

$\displaystyle A(\mathbf{g},e_i)=\sum\limits_{j=1}^dm_i^je_j$

That is, $A(\mathbf{g},\underbar{\hphantom{X}})$ is a linear map from $V$ to itself, with its matrix entries given by $m_i^j$. We define this matrix to be $\rho(g)$, which must be a representation because of the conditions on $A$ above.

Conversely, if we have a matrix representation $\rho:G\to GL_d$, we can define a module map for $\mathbb{C}^d$ as

$\displaystyle A(\mathbf{g},v)=\rho(g)v$

where we apply the matrix $\rho(g)$ to the column vector $v$. This must satisfy the above conditions, since they reflect the fact that $\rho$ is a representation.

In fact, to define $A$, all we really need to do is to define it for the basis elements $\mathbf{g}\in\mathbb{C}[G]$. Then linearity will take care of the rest of the group algebra. That is, we can just as well say that a $G$-module is a vector space $V$ and a function $A:G\times V\to V$ satisfying the following three conditions:

• $A$ is linear in $V$: $A(g,cv+dw)=cA(g,v)+dA(g,w)$.
• $A$ preserves the identity: $A(e,v)=v$.
• $A$ preserves the group operation: $A(g,A(h,v))=A(gh,v)$.

The difference between the representation viewpoint and the $G$-module viewpoint is that representations emphasize the group elements and their actions, while $G$-modules emphasize the representing space $V$. This viewpoint will be extremely helpful when we want to consider a representation as a thing in and of itself. It’s easier to do this when we think of it as a vector space equipped with the extra structure of a $G$-action.

## 11 Comments	»

1. For whatever reason, I took a course in Module Theory at Caltech, got lost in the forest, and couldn’t remember by the end what I was learning this FOR. Thanks for being so clear. Maybe I’ll get it better this time around…

Comment by | September 15, 2010 | Reply

2. [...] Actions and Representations From the module perspective, we’re led back to the concept of a group action. This is like a -module, but [...]

Pingback by | September 16, 2010 | Reply

3. [...] course, this shouldn’t really surprise us. After all, representations of are equivalent to modules for the group algebra; and the very fact that is an algebra means that it comes with a bilinear [...]

Pingback by | September 17, 2010 | Reply

4. [...] Between Representations Since every representation of is a -module, we have an obvious notion of a morphism between them. But let’s be explicit about [...]

Pingback by | September 21, 2010 | Reply

5. [...] We say that a module is “reducible” if it contains a nontrivial submodule. Thus our examples last time show [...]

Pingback by | September 23, 2010 | Reply

6. [...] I’d like to cover a stronger condition than reducibility: decomposability. We say that a module is “decomposable” if we can write it as the direct sum of two nontrivial submodules [...]

Pingback by | September 24, 2010 | Reply

7. [...] and Kernels A nice quick one today. Let’s take two -modules and . We’ll write for the vector space of intertwinors from to . This is pretty [...]

Pingback by | September 29, 2010 | Reply

8. [...] Now that we know that images and kernels of -morphisms between -modules are -modules as well, we can bring in a very general [...]

Pingback by | September 30, 2010 | Reply

9. [...] and Commutant Algebras We will find it useful in our study of -modules to study not only the morphisms between them, but the structures that they [...]

Pingback by | October 1, 2010 | Reply

10. [...] way of looking at it: remember that a representation of a group on a space can be regarded as a module for the group algebra . If we then add a commuting representation of a group , we can actually [...]

Pingback by | November 1, 2010 | Reply

11. [...] that we’re interested in concrete actions of Lie algebras on vector spaces, like we were for groups. Given a Lie algebra we define an -module to be a vector space equipped with a bilinear function [...]

Pingback by | September 12, 2012 | Reply

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