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Every separable Banach space is a linear quotient of $\ell_1$, however not every separable Banach algebra is a Banach-algebra quotient of $\ell_1(G)$ for some group $G$ (these are the so called unitary Banach algebras).

Is every separable Banach algebra a quotient of $\ell_1(S)$ for some countable semigroup?

I'd be interested in knowing what could be the Banach-algebra quotients of $\ell_1(\mathbf{N}_0)$ and $\ell_1(\mathbf{Z})$.

Does either of them quotient onto $C(X)$ for some infinite compact metric space $X$?

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  • $\begingroup$ Does ${\bf N}$ mean the strictly positive integers? If so, are you happy with an answer that only addresses $\ell_1({\bf N}_0)$? (I don't think it ends up making much difference, but I find it easier to assume things are unital) $\endgroup$ – Yemon Choi Mar 24 '17 at 17:29
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(What follows is largely the result of digging around online, based on knowing a few more magic words than the OP.)

Answer to the first question (I think).

Let $V$ be a separable Banach space. The standard proof that $V$ is isometrically isomorphic to a linear quotient of $\ell_1$ works by choosing a countable set $X$ that is a dense subset of the unit sphere of $V$, defining $f:\ell_1(X) \to V$ in the obvious way, and then verifying two things:

(i) $f$ is surjective;

(ii) the natural map $f_1: \ell_1(X)/\ker(f) \to V$ is an isometry.

Suppose $V$ is actually a unital Banach algebra. Let $S$ be the multiplicative subsemigroup of $V$ generated by $X$; of course $S$ is countable. Although $S$ might not lie in the unit sphere of $V$, it does lie in the unit ball, and so we do get a norm-decreasing homomorphism $h: \ell_1(S) \to V$.

We may factorize $f$ as $h\circ \iota$ where $\iota: \ell_1(X)\to \ell_1(S)$ is the natural isometric embedding induced from $X\subset S$. Since $f$ is surjective, so is $h$. Moreover, since $f_1$ is an isometry: given $b$ in the unit ball of $V$ and $K>1$ we can find $\xi\in \ell_1(X)$ with $f(\xi)=b$ and $\Vert\xi\Vert_1 < K$. So $b=h(\iota(\xi))$ where $\Vert\iota(\xi)\Vert_1 < K$. This shows that $h$ is actually a quotient map of Banach spaces, as required.

Comment on the intermediate question. I am not sure right now what one can say about the Banach algebra quotients of these algebras in general.

Partial answer to the final question. Via the Gelfand transform we can regard $\ell_1({\bf Z}_+)$ as an algebra of continuous functions on the closed unit disc, and hence for any closed subset $E$ of the unit disc, there is a norm-decreasing homomorphism $r_E: \ell_1({\bf Z}_+)\to C(E)$ given by restriction of these functions. It is known that for some infinite closed $E\subset {\bf T}$ this restriction homomorphism $r_E$ can be surjective: according to remarks in Volume 2 of Hewitt&Ross, these were originally called Carleson sets, but Wik showed that this coincides with the more widely known notion of a Helson subset of ${\bf T}$, which corresponds to $r_E: \ell_1({\bf Z}) \to C(E)$.

MR0125404 (23 #A2707) I. Wik, On linear dependence in closed sets. Ark. Mat. 4 1961 209–218.

The condition that $r_E:\ell_1({\bf Z})\to C(E)$ be a quotient map of Banach spaces corresponds to the classical notion of $E$ being a subset of ${\bf T}$ with Helson constant one. Such sets are known to exist, but I confess I don't know anything about the proofs/constructions. Wik's result shows that Helson sets are Carleson sets, but it is not clear to me from his proof if "1-Helson sets" are automatically "1-Carleson".

So I can't answer to your question in the $\ell_1({\bf Z}_+)$ case, but the answer is positive in the $\ell_1({\bf Z})$-case.

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  • $\begingroup$ Interesting! Do you know if Helson sets are zero-dimensional? $\endgroup$ – user512365 Mar 24 '17 at 18:16
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    $\begingroup$ I'm not sure, but it is known that Helson sets in ${\bf T}$ cannot contain "intervals", so in particular are totally path-disconnected. (There is a notion of Helson set in any LCA group and for e.g. ${\bf T}^2$ one can find path-connected infinite Helson sets.) $\endgroup$ – Yemon Choi Mar 24 '17 at 19:27
  • $\begingroup$ Does $\ell_1(\mathbf{N})$ (without zero) have factorization? Maybe that would help? $\endgroup$ – Tomek Kania Mar 25 '17 at 15:59
  • $\begingroup$ @TomekKania All such things in that ideal $J$ are of the form $zf$ where $f$ is analytic on the open disc; so $J$ is not even square-dense $\endgroup$ – Yemon Choi Mar 25 '17 at 16:14

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