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 - Posted by John Armstrong | Algebra, Group theory, Representation Theory

8 Comments »

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This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).

I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.

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