# The Unapologetic Mathematician

## The Group Algebra

A useful construction for our purposes is the group algebra $\mathbb{C}[G]$. We’ve said a lot about this before, and showed a number of things about it, but most of those we can ignore for now. All that’s really important is that $\mathbb{C}[G]$ is an algebra whose representations are intimately connected with those of $G$.

When we say it’s an algebra, we just mean that it’s a vector space with a distributive multiplication defined on it. And in our case of a finite group $G$ it’s easy to define both. For every group element $g\in G$ we have a basis vector $\mathbf{g}\in\mathbb{C}[G]$. That is, we get every vector in the algebra by picking one complex coefficient $c_g$ for each element $g\in G$, and adding them all up:

$\displaystyle\sum\limits_{g\in G}c_g\mathbf{g}$

Multiplication is exactly what we might expect: the product of two basis vectors $\mathbf{g}$ and $\mathbf{h}$ is the basis vector $\mathbf{gh}$, and we extend everything else by linearity!

$\displaystyle\left(\sum\limits_{g\in G}c_g\mathbf{g}\right)\left(\sum\limits_{h\in G}d_h\mathbf{h}\right)=\sum\limits_{g,h\in G}c_gd_h\mathbf{gh}$

We often rearrange this sum to collect all the terms with a given basis vector together:

$\displaystyle\sum\limits_{g\in G}\sum\limits_{g_1g_2=g}c_{g_1}d_{g_2}\mathbf{g}=\sum\limits_{g\in G}\left(\sum\limits_{h\in G}c_{gh^{-1}}d_h\right)\mathbf{g}$

Neat, huh?

And we can go from representations of a group to representations of its group algebra. Indeed, if $\rho:G\to GL_d$ is a representation, then we can define

$\displaystyle\rho\left(\sum\limits_{g\in G}c_g\mathbf{g}\right)=\sum\limits_{g\in G}c_g\rho(g)$

Indeed, each $\rho(g)$ is a matrix, and we can multiply matrices by complex numbers and add them together, so the right hand side is perfectly well-defined as a $d\times d$ matrix in the matrix algebra $M_d$. It’s a simple matter to verify that $\rho:\mathbb{C}[G]\to M_d$ preserves addition of vectors, scalar multiples of vectors, and products of vectors.

Conversely, if $\rho:\mathbb{C}[G]\to M_d$ is a representation, then we can restrict it to our basis vectors and get a map $\rho:G\to GL_d$. The image of each basis vector must be invertible, for we have

$\displaystyle\rho(\mathbf{g})\rho(\mathbf{g^{-1}})=\rho(\mathbf{gg^{-1}})=\rho(\mathbf{e})=1$

September 14, 2010 -

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