$\ell^1(\mathbb{Z})$ with discrete convolution and the Gelfand transform

Question

I want to show the following:

$\ell^1(\mathbb{Z})$ with the discrete convolution $*$ is isomorphic to a subalgebra of $C(T)$ where $T = \{z \in \mathbb{C} : |z| = 1\}$.

The following theorem seems to be handy,

Let $\Lambda$ be the maximal ideal space of a commutative Banach algebra $A$. The Gelfand transform is a homomorphism of $A$ onto a subalgebra $\tilde{A}$ of $C(\Lambda)$, whose kernel is $\text{rad } A$. The Gelfand transform is therefore an isomorphism if and only if $A$ is semisimple.

In view of this it seems that we would like to have $T$ to be the maximal ideal space and that $\ell^1(\mathbb{Z})$ is semisimple. Using that there is a one-to-one correspondence between the complex homomorphisms of $\ell^1(\mathbb{Z})$ and the maximal ideals of the space, I am guessing that a good start would be to determine all complex homomorphisms. How could we determine this?

Answer 1

Let $\chi$ be a continuous character of $\ell_1(\mathbb{Z})$, then $$ \chi(c) =\chi\left(\sum\limits_{n=-\infty}^{+\infty} c_n\delta_n\right) =\sum\limits_{n=-\infty}^{+\infty} c_n\chi(\delta_n) $$ for all $c\in\ell_1(\mathbb{Z})$. Denote $a_n=\chi(\delta_n)$. Since $\delta_n*\delta_m=\delta_{n+m}$ then $a_na_m=a_{n+m}$. Now it is clear that $a_n=a_1^n$. Note that $|a_n|\leq\Vert\chi\Vert\Vert\delta_n\Vert=\Vert\chi\Vert$ for all $n\in\mathbb{Z}$, so $|a_1|=1$. Denote $z=a_1\in\mathbb{T}$, then $$ \chi(c) =\sum\limits_{n=-\infty}^{+\infty} c_n\chi(\delta_n) =\sum\limits_{n=-\infty}^{+\infty} c_nz^n $$

Now after necessary identifications Gelfands transform is nothing more that discrete Fourier transform: $$ F_\mathbb{Z}:\ell_1(\mathbb{Z})\to C(\mathbb{T}): c_n\mapsto \left(z\mapsto\sum_{n=-\infty}^{+\infty} c_n z^n\right) $$

Answer 2

Note that the group ring $\mathbb C[\mathbb Z]$ embeds into $\ell^1(\mathbb Z)$ as the subring of finitely supported sequences. Furthermore, its image is dense.

Thus a continuous homomorphism $\ell^1(\mathbb Z) \to \mathbb C$ is determined by its restriction to $\mathbb C[\mathbb Z]$, and so it might help to determine the homomorphisms $\mathbb C[\mathbb Z] \to \mathbb C$ first.

[Aside: Actually, all homomorphisms $\ell^1(\mathbb Z) \to \mathbb C$ are continuous, because maximal ideals are automatically closed; but it is helpful to remember the continuity property, since it allows one to interpolate from the pure algebra of $\mathbb C[\mathbb Z]$ to the analytic context of $\ell^1$.]

Now these latter homomorphisms are determined by where the element $1 \in \mathbb Z$ goes; once we know that, everything else is determined by the homomorphism condition. (In more representation-theoretic terms, homomorphisms $\mathbb C[\mathbb Z] \to \mathbb C$ are the same as one-dimensional representations, or characters of $\mathbb Z$, and these are determined by their value on the generator $1$ of $\mathbb Z$.)

Thus the homomorphisms $\mathbb C[\mathbb Z] \to \mathbb C$ are in correspondence with the possible images of $1 \in \mathbb Z$, which can be any non-zero complex number, say $z \in \mathbb C \setminus \{0\}.$

But not all of these will extend to continuous homomorphisms $\ell^1(\mathbb Z) \to \mathbb C$.

Slightly informally, any such homomorphism would (by continuity) have to map the sequence $(a_n)_{n \in \mathbb Z}$ to the complex number $\sum_n a_n z^n$, but this latter series won't converge in general.

Note that since $(a_n)$ is assumed to be in $\ell^1$, this series will converge if $|z| = 1$; this shows you that of $|z| = 1$, our homomorphism does extend from $\mathbb C[\mathbb Z]$ to $\ell^1$. A little more argument shows that if $|z| \neq 1$, then there is no such continuous extension.

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General philosophy: think of $\ell^1$ of a group as a completion of the group ring. First see what you can work out by pure algebra (manipulating the group ring), and then see what additional constraints the analytic conditions impose.