Banach algebra structure of $\ell^p$ sequence space

Question

Long intro: questions after examples.

The space of absolutely convergent sequences $\ell^1 = \{ \vec{a} = (a_1, a_2, a_3, ...) \in \Bbb{R}^\Bbb{N}: \sum_{n} |a_n| < \infty \}$ is closed under vector multiplication and scalar addition, and has a natural norm (though not inner product) given by $|| \vec{a} || = \sum_n |a_n|$. While this space is typically considered as a complete normed vector space, I'm curious about its structure as a commutative ring endowed with various types of permutation-invariant multiplication. A multiplication $\otimes$ will be permutation invariant if for every permutation $\sigma: \Bbb{N} \rightarrow \Bbb{N}$, we have $$||(|a_{\sigma(n)}|) \otimes (|b_{\sigma(n)}|)|| = ||(|a_n|) \otimes (|b_n|)||.$$

There are three "natural"-seeming, permutation-invariant multiplications to define, which are each continuous with respect to the norm:

• $\underline{\text{Componentwise.}}$ $\ell^1$ is closed under componentwise multiplication: if $( a_n ), ( b_n ) \in \ell^1$, then the componentwise product $$( a_n ) \cdot ( b_n ) := ( a_n b_n ) \in \ell^1,$$ and $||\vec{a} \cdot \vec{b}|| \leq ||\vec{a}|| ||\vec{b}||$, so this makes $\ell^1$ into a Banach algebra. Every ideal is generated by one or more ideals of the form $$\mathfrak{a}_I = \{ (a_n) \in \ell^1 : a_k = 0, \forall k \in I \},$$ for some (possibly empty) index set $I \subseteq \Bbb{N}$. The ideal $\mathfrak{a}_I$ is all of $\ell^1$ only if $I = \emptyset$ and a prime or maximal ideal only if $I = \{ k \}$ is a single element.

• $\underline{\text{Cauchy product.}}$ Since the Cauchy product of two absolutely convergent sequences is absolutely convergent, $\ell_1$ is closed under Cauchy products: $$( a_n ) * ( b_n ) := ( c_n ) \in \ell^1,$$ where $c_n := \sum_{k=1}^{n} a_k b_{n-k+1}$ for $n = 1, 2, 3, ...$ and it only takes a little algebra (omitted) to show that $|| \vec{a} * \vec{b} || \leq ||\vec{a}|| ||\vec{b}||$. So this is a Banach algebra again, and because elements of $\ell^1$ are in 1-1 correspondence with power series $\sum_n a_n x^{n-1}$ that converge absolutely on the closed interval $[-1, 1]$, the maximal ideals of $\ell^1$ are in 1-1 correspondence with points $x \in [-1, 1]$ via the correspondence $$x \mapsto \mathfrak{m}_x := \{ (a_n): \sum_{n} a_n x^{n-1} = 0 \}.$$ Every other ideal is an intersection of some (finite number of) the $\mathfrak{m}_x$, a power of one of the $\mathfrak{m}_x$, or a combination of the two.

• $\underline{\text{Dirichlet convolution.}}$ For two elements $(a_n), (b_n) \in \ell^1$, the Dirichlet convolution of these sequences is also in $\ell^1:$ $$(a_n) \star (b_n) = (d_n) \in \ell^1 ,$$ where $d_n = \sum_{k | n} a_k b_{n/k}$. We can show that $(d_n) \in \ell^1$ similarly to showing that the Cauchy product converges:

\begin{align*} \sum_{n=1}^N |d_n| &= \sum_{n=1}^N \left| \sum_{k | n} a_k b_{n/k} \right| \\ &\leq \sum_{n=1}^N \sum_{k | N} |a_k| |b_{n/k}| \\ &= \sum_{\substack{1 \leq j, k \leq N \\ jk \leq N}} |a_j| |b_k| \\ &\leq \sum_{j=1}^N \sum_{k=1}^N |a_j| |b_k| \\ &= \left(\sum_{j=1}^N |a_j|\right)\left(\sum_{k=1}^N |b_k|\right), \ \end{align*}

and by letting $N \to \infty$ we see that $||\vec{d}|| = ||\vec{a} \star \vec{b}||$ is finite, and $||\vec{a} \star \vec{b}|| \leq ||\vec{a}|| ||\vec{b}||$, so $\ell^1$ is once again a Banach algebra under this multiplication. In this case, elements of $\ell^1$ are in 1-1 correspondence with Dirichlet series $\sum_n a_n n^{-s}$ that converge absolutely on $[0, \infty)$, and so maximal ideals of $\ell^1$ are in 1-1 correspondence with points $s \in [0, \infty)$ via $$s \mapsto \mathfrak{m}_s := \{ (a_n): \sum_{n} a_n n^{-s} = 0 \},$$ and every other ideal is an intersection of finitely many $\mathfrak{m}_s$, a power of one of the $\mathfrak{m}_s$, or some combination of the two.

Question 1: Is there a "canonical" way to create a Banach algebra by equipping $\ell^1$ with a permutation-invariant multiplication? If not, does every permutation-invariant Banach algebra multiplication on $\ell^1$ arise in a manner similar to the second and third examples, and can the first example also be related to this format?

Question 2: In the Cauchy example, the maximal ideals $\mathfrak{m}_1$ and $\mathfrak{m}_{-1}$ have a different structure than other maximal ideals $\mathfrak{m}_x$ for $x \in (-1, 1)$, as $\mathfrak{m}_{\pm 1}^p = \mathfrak{m}_{\pm 1}$ for all integers $p > 0$, but for other values of $x$, $\mathfrak{m}_x \supsetneq \mathfrak{m}_x^2 \supsetneq \mathfrak{m}_x^3 \supsetneq...$ Something similar happens in the Dirichlet example, where $\mathfrak{m}_0^p = \mathfrak{m}_0$ for all integers $p > 0$ but for $s > 0$, $\mathfrak{m}_s \supsetneq \mathfrak{m}_s^2 \supsetneq \mathfrak{m}_s^3 \supsetneq ...$ Do these special ideals tell us anything significant about the algebra overall, and is any Banach algebra constructed from $\ell^1$ uniquely characterized by the number and nature of such special ideals?

Question 3: How do the answers to Q1 and Q2 change for $\ell^p$ with $p > 1$? In particular, what are the corresponding answers to Q1 and Q2 for $\ell^2$ (which has an inner product) and $\ell^\infty$? Do they have a canonical permutation-invariant multiplication that turns them into a Banach algebra, and if not, can every Banach algebra over $\ell^2$ or $\ell^\infty$ be characterized by a set of special maximal ideals?