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Informal context

In $\mathbb{C}[X]$, all polynomials are split, so monic polynomials are in bijection with the multisets of their roots, i.e. the finite sub-multisets of $\mathbb{C}$.
Polynomial multiplication corresponds to multiset union.

To define a polynomial, we need a ring (and to get always-split polynomials we need an algebraically closed field). But to define multisets, we need nothing at all except a set! So we can define monic "polynomials" with their multiplication in any set we want.

How far can we get this? We'd like to have addition and derivation, how can we define them with multisets?

First steps

We need to get all the polynomials, not merely monic ones, because addition and derivation are not stable inside monic polynomials.

So we need a scaling factor. Looking at the way addition and derivation behave, the scalar set could be $\mathbb{Z}$ or anything bigger. Let's take $\mathbb{R}$.

Let's consider weighted multisets, i.e. if $E$ is the set of values, $M$ the set of multisets of $E$, and $\mathbb{R}^* $ is the multiplicative group $\mathbb{R} - \{0\}$, we define weighted multisets on $E$ as $W = \mathbb{R}^* \times M \cup (0, \emptyset)$.

  • A polynomial $\lambda (X-a_1)(X-a_2)...(X-a_n)$ with $\lambda \in \mathbb{R}$ is represented by $(\lambda, \{a_1, a_2, ..., a_n\})$.
  • The constant $\lambda$ is represented by $(\lambda, \emptyset)$.

Multiplication of polynomials is union of weighted multisets, with the coefficients being multiplied, i.e. $(\lambda, \{a_1, ..., a_n\}) \cup (\mu, \{b_1, ..., b_m\}) = (\lambda \mu, \{a_1, ..., a_n, b_1, ..., b_m\})$.

Question

How can we use these weighted multisets to provide the equivalent of polynomial addition, multiplication and derivation, for spaces such as $\mathbb{R}^n, n>2$, while continuously extending the polynomial operations on $\mathbb{R}^2$ (being identified with $\mathbb{C}$)?

  • addition and multiplication shall have commutative ring properties;
  • derivation of an $n$ multiset shall be an $n-1$ multiset;
  • derivation shall verify the usual sum and product derivation rules;
  • when all roots are in a (2-dim) plane of $\mathbb{R}^n$, the results of addition or derivation shall be the same as in $\mathbb{C}$ (*);
  • continuity is for a distance on same-size multisets, such as: if $A=(a_i)$ and $B=(b_i)$ are $2$ multisets with $n$ elements, $d(A,B)=\min_{\sigma \in S_n} \sum_{i=1}^n \lVert a_i-b_{\sigma(i)} \rVert$.

(*) Note that in $\mathbb{C}$, derivation commutes with all geometric similarities: if polynomial $Q$ is polynomial $P$ translated or rotated or symmetrized or scaled, roots of $Q'$ are roots of $P'$ transformed by the same translation/rotation/symmetry/scaling. So when considering derivation in $\mathbb{R}^n$, if we take a multiset which is entirely contained in a (2-dim) plane, the derivation operation has only one possible result, whatever basis we choose to identify the plane with $\mathbb{C}$. The same can be said about addition. So "extending the polynomial operations on $\mathbb{R}^2$" above is well defined.

Getting addition seems difficult: there is no simple way to deduce roots of $P+Q$ from roots of $P$ and roots of $Q$, except in the most simple of cases, e.g. if $P=\alpha(X-a), Q=\beta(X-b),$
$P+Q=(\alpha+\beta)X-(\alpha a + \beta b) = (\alpha+\beta)(X-\frac {\alpha a + \beta b} {\alpha+\beta})$
In terms of weighted multisets, this is a barycenter operation. Unfortunately this does not extend to degree $2$ and above polynomials.

Getting derivation may use Bôcher's theorem: zeros of $P'$ which are not also zeros of $P$, are the equilibrium points of an $\frac 1 r$ force field created by particles of same mass on zeros of $P$. This could be extended to $\mathbb{R}^n$, but do we have the property that $p$ distinct zeros of $P$ create $p-1$ equilibrium points?

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  • $\begingroup$ I wish I had the knowledge to understand this question :c $\endgroup$
    – Babu
    Commented Oct 8, 2022 at 22:20
  • $\begingroup$ @TrystwithFreedom That's less a question of knowledge, than a question of the OP providing good explanations, and I fear I may have failed at that, sorry. $\endgroup$ Commented Oct 8, 2022 at 22:26

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