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Suppose $m \oplus n$ is a commutative and associative binary relation $\oplus: \Bbb{N} \times \Bbb{N} \to \Bbb{N}$, and that $1$ is an identity element for this operation. In other words, $(\oplus, \Bbb{N}, 1)$ is a commutative monoid. Let us also assume that $\oplus(m, n) \geq \max(m, n)$. How many such binary relations can there be? And, is there a nice characterization of such binary relations?

Background: In a previous question I was thinking about ways to give $\ell^1(\Bbb{R})$, the space of absolutely convergent sequences of reals, a multiplication that would turn it into a Banach algebra. The Cauchy product and the Dirichlet convolution are two examples of multiplications that are both:

-Norm-preserving on nonnegative sequences- Let $(p_n)$ and $(q_n)$ be two nonnegative sequences such that $\sum_n p_n < \infty$, $\sum_n q_n \leq \infty$. Then both the Cauchy product and the Dirichlet convolution are absolutely convergent, and in both cases, the norm of the product is the product of the norm.

-Permutation-invariant on nonnegative sequences- If any permutation $\sigma: \Bbb{N} \rightarrow \Bbb{N}$ of the natural numbers is given, and $(p_n)$ and $(q_n)$ are two nonnegative sequences which converge in $\ell^1$, then the norm of the Cauchy product is equal to the Cauchy product of the norm, and vice versa.

Both multiplications also come from thinking of them as coefficients in a series: either $\sum_n p_n x^{n-1}$ for the Cauchy product, or $\sum_n p_n n^{-s}$ for the Dirichlet convolution. So we can generalize. Let's say we have some sequence of functions $f(x, n)$ for each $n \geq 1$, such that $\sum_n p_n f(x, n)$ converges absolutely whenever $(p_n) \in \ell^1$ and for all $x \in (a, b)$, where $(a, b) \subseteq \Bbb{R}$ is some open interval. Let's also say that $\sum_n p_n f(x, n)$ converges absolutely for $x = a$ and $x = b$, and that at these points, $|f(a, n)| = |f(b, n)| = 1$ for all $n \in \Bbb{N}$. Then any product on $\ell^1$ which is norm-preserving and permutation-invariant on nonnegative sequences, comes from a functional relation $$f(x, m)f(x, n) = f(x, m \oplus n),$$ where $m \oplus n$ is a binary relation on $\Bbb{N}$ turning it into a commutative monoid with identity $1$.

Rivers McForge
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  • Seems that $m \oplus n=a(m-1)(n-1)+m+n-1$ works for any integer $a\geq 0$ fixed, so there are infinitely many. The question is are there any other? – Sil Aug 23 '20 at 02:02
  • The free monoid on $\aleph_0$ generators has cardinality $\aleph_0$, and any bijection from it to $\mathbb N$ induces an isomorphic monoid structure on $\mathbb N$. Even with your additional constraints that $1$ is to serve as the identity and that $\oplus(x,y)\geq\max(x,y)$, there still seem to be $2^{\aleph_0}$ possibilities. And you obviously can't have more than that. – Andreas Blass Aug 23 '20 at 02:08
  • @Sil, $m \oplus n = \max(m, n)$ is a commutative and associative operator on the naturals with identity element $1$ that cannot be put in the form $a(m-1)(n-1) + m + n -1$ for fixed $a \geq 0$. I'm impressed how quickly you came up with that family of examples, though. – Rivers McForge Aug 23 '20 at 03:18
  • @AndreasBlass I don't know if that's a lower bound, because it's possible the condition $a \oplus b \geq \max(a, b)$ narrows the possibilities. But I also don't know if it's an upper bound, because not every monoid operation on $\Bbb{N}$ could arise from such a bijection. E.g. $a \oplus b = a+b-1$ or $\max(a, b)$ cannot result from such a bijection ($n$ large enough, corresponding to $n = g_i g_j$ for some generators $g_i, g_j$ of the free monoid on $\aleph_0$ generators, will be equal to $a \oplus b$ for "too many" pairs $(a, b)$). – Rivers McForge Aug 23 '20 at 03:48
  • @AndreasBlass Although we can clearly get multiplication, $a \oplus b = ab$, via the means you outline: if the generators of the free monoid on $\aleph_0$ generators are $g1,g2,g3,...$ just send $g_n \mapsto p_n$, the nth prime. (This is a bijection only on the free commutative monoid on $\aleph_0$ generators, of course....) – Rivers McForge Aug 23 '20 at 05:36
  • @RiversMcForge True, I have actually encountered $\max(m,n)$ but then forgot about it when dealing with $mn$ and $m+n-1$ :) (this is anyway how the generalization above arose). – Sil Aug 23 '20 at 08:18
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    As I said earlier, $2^{\aleph_0}$ is obviously an upper bound; the point is that there are only $2^{\aleph_0}$ functions $\mathbb N\times\mathbb N\to\mathbb N$ altogether. I don't claim that every monoid operation on $\mathbb N$ arises in the way I described. There are obviously others (like the commutative one you mentioned) but they don't increase the cardinality. – Andreas Blass Aug 23 '20 at 12:55

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