The Unapologetic Mathematician

Mathematics for the interested outsider

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 | Algebra, Group theory, Representation Theory | 8 Comments

   

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