2
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ Consider Fourier series. $\endgroup$ – Daniel Fischer Dec 15 '13 at 21:33
  • $\begingroup$ All this really generalises as far as possible: if $\Gamma$ is a discrete Abelian group, and if $C_r^\ast(\Gamma)$ (the reduced group $C^\ast$-algebra of $\Gamma$) is the completion of $\mathbb{C}[\Gamma]$ in $B(\ell^2(\Gamma))$, then the character space $M(C_r^\ast(\Gamma))$ can be identified with the Pontrjagin dual $\widehat{\Gamma}$ of $\Gamma$, and hence the Gel'fand transform $C_r^\ast(\Gamma) \cong C(M(C_r^\ast(\Gamma)))$ is just the Fourier transform $C_r^\ast(\Gamma) \cong C(\widehat{\Gamma})$ (i.e., $\ell^1(\Gamma) \subset C_r^\ast(\Gamma) \cong C(\widehat{\Gamma})$. $\endgroup$ – Branimir Ćaćić Dec 16 '13 at 21:32
5
$\begingroup$

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: $$ F_\mathbb{Z}:\ell_1(\mathbb{Z})\to C(\mathbb{T}): c_n\mapsto \left(z\mapsto\sum_{n=-\infty}^{+\infty} c_n z^n\right) $$

$\endgroup$
4
  • $\begingroup$ Thank you for your answer. I'm not sure why each character must be on that form, that is, why we cannot have any other bounded linear $l : \mathbb{Z} \rightarrow \mathbb{C}$ in place of $z^n$? $\endgroup$ – guest Dec 16 '13 at 12:53
  • $\begingroup$ So assuming a bounded linear $l_z : \mathbb{Z} \rightarrow \mathbb{C}$, in order for us to have $\chi_z((a_n)*(b_n)) = \chi_z((a_n))\chi_z((b_n))$ we must have $l_z(n) = l_z(n-k)l_z(k)$. This is clearly true for $z \mapsto z^n$, but I can't see why we cannot have $l_z$ on another form. $\endgroup$ – guest Dec 16 '13 at 13:54
  • $\begingroup$ Oh cannot edit now, but I did not mean for $l$ to be linear, just bounded. $\endgroup$ – guest Dec 16 '13 at 14:05
  • $\begingroup$ @guest see edits to my answer $\endgroup$ – Norbert Dec 16 '13 at 20:37
3
$\begingroup$

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.


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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.