The Discrete Group Algebra of a Group G

# The Discrete Group Algebra of a Group G

Let $(G, \cdot)$ be a group over the field $\mathbb{F}$ and let:

(1)
\begin{align} \quad \ell^1(G) = \left \{ f : G \to \mathbb{C} \biggr | \sum_{x \in G} |f(x)| < \infty \right \} \end{align}

For each $f, g \in \ell^1(\mathbb{R}$ and for all $a \in \mathbf{F}$ define the sum $f + g$, the scalar multiple $af$, and the convolution product $f * g$ for each $x \in G$ by:

(2)
\begin{align} \quad (f + g)(x) &= f(x) + g(x) \\ \quad (af)(x) &= af(x) \\ \quad (f * g)(x) &= \sum_{y \in G} f(y)g(y^{-1} \cdot x) \end{align}

Define a function $\| \cdot \| : \ell^1(G) \to [0, \infty)$ for each $f \in \ell^1(G)$ by:

(3)
\begin{align} \quad \| f \| = \sum_{x \in G} |f(x)| \end{align}

The following proposition will tell us that $\ell^1(G)$ with the above operations and norm is a Banach algebra.

 Proposition 1: Let $G$ be a group. Then $(\ell^1(G), \| \cdot \|)$ is an algebra.
• Proof: It is clear that $\ell^1(G)$ with the operation of addition and scalar multiplication, as well with the norm is a normed linear space, and is in fact actually a Banach space. All that remains to show is that the properties of the convolution product makes $\ell^1(G)$ an algebra, and that the properties of the norm make the norm an algebra norm.
• 1. Let $f, g, h \in \ell^1(G)$. Then for every $x \in G$ we have that:
(4)
\begin{align} \quad [(f * g) * h](x) &= \sum_{y \in G} [(f * g)(y) h(y^{-1} \cdot x)] \\ &= \sum_{y \in G} \left [\left [ \sum_{z \in G} f(z) g(z^{-1} \cdot y) \right ] h(y^{-1} \cdot x) \right ] \\ &= \sum_{y \in G} \sum_{z \in G} \left [ f(z)g(z^{-1} \cdot y) h(y^{-1} \cdot x) \right ] \\ \end{align}
• Now let $A = \sum_{x \in G} |f(x)|$, $B = \sum_{x \in G} |g(x)|$, and $C = \sum_{x \in G} |h(x)|$.
• Then $A, B, C < \infty$, and we have that $\sum_{y \in G} \sum_{z \in G} \left | f(z)g(z^{-1} \cdot y) h(y^{-1} \cdot x) \right | = \sum_{y \in G} |h(y^{-1} \cdot x)| \sum_{z \in G} |f(z)g(z^{-1} \cdot y)| \leq \sum_{z \in G} |h(y^{-1} \cdot x)| AB \leq ABC < \infty$. So by Fubini's theorem for double series we can rearrange the order of summation to get:
(5)
\begin{align} \quad [(f * g) * h](x) &= \sum_{z \in G} \sum_{y \in G} \left [ f(z)g(z^{-1} \cdot y) h(y^{-1} \cdot x) \right ] \\ &= \sum_{z \in G} \left [ f(z) \left [ \sum_{y \in G} g(z^{-1} \cdot y)h(y^{-1} \cdot x) \right ] \right ] \\ & \overset{(\dagger)} = \sum_{z \in G} \left [ f(z) \left [ \sum_{y \in G} g(z^{-1} \cdot zy)h((zy)^{-1} \cdot x) \right ] \right ] \\ &= \sum_{z \in G} \left [ f(z) \left [ \sum_{y \in G} g(y)h(y^{-1} \cdot z^{-1} \cdot x) \right ] \right ] \\ &= \sum_{z \in G} \left [ f(z) (g * h)(z^{-1} \cdot x) \right ] \\ &= [f * (g * h)](x) \end{align}
• (Note that if $v : G \to G$ is the function defined by $v(y) = zy$ then $v$ is bijective and so the equality at $(\dagger)$ comes from the permutation of the terms in the sum, which yields the same sum since the series is absolutely convergent.)
• Thus, $f * (g * h) = (f * g) * h$ for every $x \in G$.
• 2. Showing that $f * (g + h) = f * g + f * h$: Let $f, g, h \in \ell^1(G)$. Then for every $x \in G$ we have that:
(6)
\begin{align} \quad [f * (g + h)](x) = \sum_{y \in G} f(y)(g + h)(y^{-1} \cdot x) = \sum_{y \in G} [f(y)g(y^{-1} \cdot x) + f(y)h(y^{-1} \cdot x)] = \sum_{y \in G} f(y)g(y^{-1} \cdot x) + \sum_{y \in G} f(y)h(y^{-1} \cdot x) = (f * g)(x) + (f * h)(x) \end{align}
• Thus, $f * (g + h) = f * g + f * h$.
• 3. Showing that $(\alpha f) * g = \alpha (f * g) = f * (\alpha g)$: Let $f, g \in \ell^1(G)$ and let $\alpha \in \mathbf{F}$. Then for every $x \in G$ we have that:
(7)
\begin{align} \quad [(\alpha f) * g](x) = \sum_{y \in G} \alpha f(y) g(y^{-1} \cdot x) = \alpha \sum_{y \in G} f(y) g(y^{-1} \cdot x) = [\alpha (f * g)](x) \end{align}
• And also:
(8)
\begin{align} \quad [\alpha (f * g)](x) = \alpha \sum_{y \in G} f(y) g(y^{-1} \cdot x) = \sum_{y \in G} f(y) \alpha g(y^{-1} \cdot x) = [f * (\alpha g)](x) \end{align}
• Therefore $(\alpha f) * g = \alpha (f * g) = f * (\alpha g)$. All of this shows that $\ell^1(G)$ with convolution multiplication is an algebra over $\mathbf{F}$.
• We will now show that norm is further an algebra norm. Let $f, g \in \ell^1(G)$. Then:
(9)
\begin{align} \quad \| f * g \| = \sum_{x \in G} |(f * g)(x)| = \sum_{x \in G} \left | \sum_{y \in G} f(y)g(y^{-1} \cdot x) \right | \leq \sum_{x \in G} \sum_{y \in G} |f(y)||g(y^{-1} \cdot x)| &= \sum_{y \in G} \sum_{x \in G} |f(y)||g(y^{-1} \cdot x)| \\ &= \sum_{y \in G} \left [ |f(y)| \sum_{x \in G} |g(y^{-1} \cdot x)| \right ] \\ &= \sum_{y \in G} |f(y)| \sum_{x \in G} |g(x)| \\ &= \| f \| \| g \| \end{align}
• So $\| \cdot \|$ is an algebra norm on $\ell^1(G)$. Thus we conclude that $(\ell^1(G), \| \cdot \|)$ is a Banach algebra. $\blacksquare$

We give the above Banach algebra a special name.

 Definition: Let $G$ be a group. Then the linear space $\ell^1(G)$ with the algebra norm $\| \cdot \| : \ell^1(G) \to [0, \infty)$ defined for all $f \in \ell^1(G)$ by $\displaystyle{\| f \| = \sum_{x \in G} |f(x)|}$ is called the Discrete Group Algebra of $G$.

Given a group $G$ we have proven in proposition $1$ that the discrete group algebra of $G$, $\ell^1(G)$ is an algebra. The following proposition tells us more - that is, the discrete group algebra is actually a Banach algebra.

 Proposition 2: Let $G$ be a group. Then the discrete group algebra $\ell^1(G)$ is a Banach algebra.
• Proof: Let $\epsilon > 0$ be given and let $(f_n) = (f_1, f_2, ...)$ be a Cauchy sequence in $\ell^1(G)$. We will first find a candidate limit $F$ for this Cauchy sequence. We will show that $F$ is a member of $\ell^1(G)$ and then that $(f_n)$ norm converges to $F$.
• For this given $\epsilon$ there exists an $N \in \mathbb{N}$ such that if $m, n \geq N$ then:
(10)
\begin{align} \quad \| f_m - f_n \| = \sum_{x \in G} |f_m(x) - f_n(x)| < \epsilon \end{align}
• Note that for each particular $g \in G$ we have that:
(11)
\begin{align} \quad |f_m(g) - f_n(g)| \leq \sum_{x \in G} |f_m(x) - f_n(x)| < \epsilon \end{align}
• So for every $g \in G$, $(f_n(g))$ is a Cauchy sequence of complex numbers and hence converges to some number $z_g \in \mathbb{C}$ (by the completeness of $\mathbb{C}$). Define $F : G \to \mathbb{C}$ for each $g \in G$ by:
(12)
\begin{align} \quad F(g) = \lim_{n \to \infty} f_n(g) = z_g \end{align}
• Showing that $F \in \ell^1(G)$: Since $(f_n)$ is a Cauchy sequence it is a norm bounded sequence, and so there exists an $M > 0$ such that $\| f_n \| \leq M$ for every $n \in \mathbb{N}$.
• There is now an important observation to make. For each $n \in \mathbb{N}$, $\| f_n \| = \sum_{x \in G} |f_n(x)|$ is a sum of nonnegative real numbers that converges. Therefore $|f_n(x)| = 0$ for all by a countable subset $G_n$ of $G$. Let $G^* = \bigcup_{n=1}^{\infty} G_n$. Then $G^*$ is countable, and for each $n \in \mathbb{N}$ we have that:
(13)
\begin{align} \quad \| f_n \| = \sum_{x \in G} |f_n(x)| = \sum_{x \in G^*} |f_n(x)| \end{align}
• Let $(x_k)$ be any enumeration of $G^*$. Then for any $n \in \mathbb{N}$:
(14)
\begin{align} \quad \| f_n \| = \sum_{x \in G} |f_n(x)| = \sum_{k=1}^{\infty} |f_n(x_k)| \end{align}
• Note that any enumeration $(x_k)$ of $G^*$ is fine since the series above is absolutely convergent and rearrangement of terms in the series does not affect the sum.
• Now, for each $K \in \mathbb{N}$ we have that:
(15)
\begin{align} \quad \sum_{k=1}^{K} |F(x_k)| = \sum_{k=1}^{K} \lim_{n \to \infty} |f_n(x_k)| = \lim_{n \to \infty} \sum_{k=1}^{K} |f_n(x_k)| \leq \lim_{n \to \infty} \sum_{k=1}^{\infty} |f_n(x_k)| = \lim_{n \to \infty} \| f_n \| \leq M \end{align}
• Taking the limit as [[$K \to \infty$}] shows us that:
(16)
\begin{align} \quad \| F \| = \sum_{k=1}^{K} |F(x_k)| \leq M < \infty \end{align}
• So $F \in \ell^1(G)$.
• Showing that $(f_n)$ norm converges to $F$: Let $\epsilon > 0$ be given and let $(f_n)$ be a Cauchy sequence in $\ell^1(G)$.
• We make a slight modification to the notation given above. Since $F \in ell^1(G)$ and $f_n \in \ell^1(G)$ for all $n \in \mathbb{N}$ we have that $f_n - F \in \ell^1(G)$ for all $n \in \mathbb{N}$ since $\ell^1(G)$ is a linear space. So $\| f_n - F \| < \infty$ for all $n \in \mathbb{N}$. As mentioned before, this means that in the series $\| f_n - F \| = \sum_{x \in G} |f_n(x) - F(x)|$, all but countably many $x \in G$ are such that $|f_n(x) - F(x)| > 0$. For each $n \in \mathbb{N}$ we now let $G_n$ be the set of $x \in G$ such that $|f_n(x) - F(x)| > 0$ and we now let $G^* = \bigcup_{n=1}^{\infty} G_n$. Then again, $G^*$ is countable and we take any enumeration $(x_k)$ of $G^*$ so that for every $n \in \mathbb{N}$:
(17)
\begin{align} \quad \| f_n - F \| = \sum_{x \in G} |f_n(x) - F(x)| = \sum_{k=1}^{\infty} | f_n(x_k) - F(x_k) | \end{align}
• Now since $(f_n)$ is Cauchy, for the given $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $m, n \geq N$ then:
(18)
\begin{align} \quad \| f_m - f_n \| = \sum_{x \in G} |f_m(x) - f_n(x)| = \sum_{k=1}^{\infty} |f_m(x_k) - f_n(x_k)| < \epsilon \end{align}
• Then for any fixed $K \in \mathbb{N}$ and $m, n \geq N$ we have that:
(19)
\begin{align} \quad \sum_{k=1}^{K} |f_m(x_k) - f_n(x_k)| < \epsilon \end{align}
• Fixing $K \in \mathbb{N}$ and $n \geq N$, we see that as $m \to \infty$:
(20)
\begin{align} \quad \sum_{k=1}^{K} |F(x_k) - f_n(x_k)| = \lim_{m \to \infty} \sum_{k=1}^{K} |f_m(x_k) - f_n(x_k)| \leq \epsilon \end{align}
• Since this is true for all $K \in \mathbb{N}$ we have that for all $n \geq N$:
(21)
\begin{align} \quad \| F - f_n \| = \sum_{k=1}^{\infty} |F(x_k) - f_n(x_k)| \leq \epsilon \end{align}
• So $(f_n)$ norm converges to $F$. Thus every Cauchy sequence in $\ell^1(G)$ norm converges in $\ell^1(G)$ and so $\ell^1(G)$ is a Banach algebra. $\blacksquare$