📂Representation Theory

Group Algebra

Definition¹

The vector space generated by the following formal sums, with a finite group $\braket{G, \cdot}$ taken as a basis, is called the group algebra of $G$.

$$ \mathbb{C}[G] = \left\{ \sum_{i} a_{i}g_{i} : a_{i} \in \mathbb{C}, \quad g_{i} \in G \right\} $$

Explanation

Note that the ‘sum’ referred to in the definition is a formal sum that has nothing to do with the group $G$’s operation or its algebraic structure. More generally, instead of the complex numbers $\mathbb{C}$ one can define the same construction over any field $\mathbb{F}$.

Equality

Two elements $\sum a_{i}g_{i}$ and $\sum b_{i}g_{i}$ are equal precisely when each corresponding coefficient $a_{i}$ and $b_{i}$ are equal.

$$ \sum_{i} a_{i} g_{i} = \sum_{i} b_{i} g_{i} \iff a_{i} = b_{i} \quad \forall i $$

Addition

On $\mathbb{C}[G]$ we can define the natural addition below, and $\braket{\mathbb{C}[G], +}$ itself becomes an abelian group.

$$ \sum_{i} a_{i} g_{i} + \sum_{i} b_{i} g_{i} = \sum_{i} (a_{i} + b_{i}) g_{i} $$

This is called natural because, viewing $G$ as a basis, the sum of two vectors is expressed by the sum of their coordinates (coefficients).

Inner product

Since $\mathbb{C}[G]$ is a vector space, in particular a finite-dimensional vector space, we may naturally consider the following inner product. Thus $\mathbb{C}[G]$ is an inner product space.

$$ \Braket{\sum_{i} a_{i} g_{i}, \sum_{i} b_{i} g_{i}} = \sum_{i} a_{i} \overline{b_{i}} $$

Multiplication

In general, multiplication between two vectors is not naturally defined. However, in the group algebra $\mathbb{C}[G]$ we can define multiplication using the product of $\braket{G, \cdot}$ as follows.

$$ \left( \sum_{i} a_{i} g_{i} \right) \cdot \left( \sum_{j} b_{j} g_{j} \right) = \sum_{k} \left( \sum_{g_{i} \cdot g_{j} = g_{k}} a_{i}b_{j} \right) g_{k} $$

Although the formula looks complicated, it means the product is defined similarly to the product of two polynomials. The multiplication in $\mathbb{C}[G]$ is not generally commutative; if the multiplication in $G$ is commutative then the multiplication in $\mathbb{C}[G]$ is also commutative. Since the associative law holds, it is an associative algebra.

Examples

Consider the group algebra $\mathbb{Z}_{2}[\braket{a}]$ for the modular group $\mathbb{Z}_{2}$ and the cyclic group $\braket{a} = \left\{ e, a \right\}$. Its elements are all formal sums of the following form.

$$ 0e + 0a, \quad 1e + 0a, \quad 0e + 1a, \quad 1e + 1a $$

For brevity (and by intuition) this can be written simply as follows.

$$ 0,\quad e, \quad a, \quad e + a $$

The results of addition are given in the table below.

$$ \begin{array}{c|cccc} + & 0 & e & a & e + a \\ \hline 0 & 0 & e & a & e + a \\ e & e & 0 & e + a & a \\ a & a & e + a & 0 & e \\ e + a & e + a & a & e & 0 \end{array} $$

Multiplication is computed as follows.

$$ \begin{align*} e \cdot a &= (1e + 0a) \cdot (0e + 1a) \\ &= (1\times 0)e\cdot e + (1\times 1)e\cdot a + (0\times 0)a\cdot e + (0\times 1)a\cdot a \\ &= 0e + 1a + 0a + 0e = 1a \end{align*} $$

Summarizing all cases in a table gives the following.

$$ \begin{array}{c|cccc} \cdot & 0 & e & a & e + a \\ \hline 0 & 0 & 0 & 0 & 0 \\ e & 0 & e & a & e + a \\ a & 0 & a & e & e + a \\ e + a & 0 & e + a & e + a & 0 \end{array} $$