Computing symmetric form of certain polynomials

Question

I would like to know the best way to calculate the following.

Let $P(u,v,s,t)$ be a polynomial such that is invariant under permutation of $u\leftrightarrow s$ AND $v\leftrightarrow t$, that is $P(u,v,s,t)=P(s,t,u,v)$

This polynomial can be written as an algebraic combination of these elementary symmetric polynomials: $\{u+s,us,v+t,vt\}$.

An easy example is: $u^2 + s^2 = (u+s)^2 - 2us$.

But if I have $s^n t+u^n v$ things get more complicated (even for $n=1$).

I need to compute this "symmetric form" of a bunch of polynomials, so I would like to know how can I do it with software.

I tried with Maple but it cannot handle $u\leftrightarrow s$ AND $v\leftrightarrow t$ (at the same time).

In case it is relevant to the context, these polynomials are functions on the Jacobian $J$ of a hyperelliptic curve of genus $2$. So these are functions in the function field $k(J)$ where the generic point of $J$ is $\{(u,v),(s,t)\}$.

Thanks

Answer 1

Not every polynomial $P$ satisfying $P\left( u,v,s,t\right) =P\left( s,t,u,v\right) $ can be written as a polynomial in the $u+s,us,v+t,vt$. And I am not sure if every such polynomial can be written as a rational function in the $u+s,us,\dfrac{t-v}{s-u},\dfrac{sv-ut}{s-u}$ either. But your polynomials $s^{n}t+u^{n}v$ can be written in the latter form. Here is how:

Set $a=u+s$, $b=us$, $c=\dfrac{t-v}{s-u}$ and $d=\dfrac{sv-ut}{s-u}$ in the field $\mathbb{Q}\left(s,t,u,v\right)$. Set $p_{n}=s^{n}t+u^{n}v$ for each $n\in\mathbb{N}$. Then, it is easy to check (by straightforward computation) that \begin{align*} p_{0} & =2d+ac;\\ p_{1} & =a^{2}c+ad-2bc. \end{align*} Next, I claim that \begin{equation} p_{n}=ap_{n-1}-bp_{n-2} \label{1} \tag{1} \end{equation} for each $n\geq2$.

[Proof of \eqref{1}: Let $n\geq2$. Then, $p_{n}=s^{n}t+u^{n}v$ and similarly $p_{n-1}=s^{n-1}t+u^{n-1}v$ and $p_{n-2}=s^{n-2}t+u^{n-2}v$. But straightforward computation shows that $as-b=s^{2}$ and $au-b=u^{2}$. Now, \begin{align*} & a\underbrace{p_{n-1}}_{=s^{n-1}t+u^{n-1}v}-b\underbrace{p_{n-2}} _{=s^{n-2}t+u^{n-2}v}\\ & =a\left( \underbrace{s^{n-1}}_{=ss^{n-2}}t+\underbrace{u^{n-1}}_{=uu^{n-2} }v\right) -b\left( s^{n-2}t+u^{n-2}v\right) \\ & =a\left( ss^{n-2}t+uu^{n-2}v\right) -b\left( s^{n-2}t+u^{n-2}v\right) \\ & =ass^{n-2}t+auu^{n-2}v-bs^{n-2}t-bu^{n-2}v\\ & =\underbrace{\left( ass^{n-2}t-bs^{n-2}t\right) }_{=s^{n-2}\left( as-b\right) t}+\underbrace{\left( auu^{n-2}v-bu^{n-2}v\right) } _{=u^{n-2}\left( au-b\right) v}\\ & =s^{n-2}\underbrace{\left( as-b\right) }_{=s^{2}}t+u^{n-2} \underbrace{\left( au-b\right) }_{=u^{2}}v=\underbrace{s^{n-2}s^{2}} _{=s^{n}}t+\underbrace{u^{n-2}u^{2}}_{=u^{n}}v\\ & =s^{n}t+u^{n}v=p_{n}. \end{align*} This proves \eqref{1}.]

From $p_{0}=2d+ac$ and $p_{1}=a^{2}c+ad-2bc$, we conclude that $p_{0}$ and $p_{1}$ are polynomials in $a,b,c,d$. Therefore, by strong induction, we can conclude that each $p_{n}$ (for $n\in\mathbb{N}$) is a polynomial in $a,b,c,d$. (Indeed, \eqref{1} shows that if $p_{n-1}$ and $p_{n-2}$ are polynomials in $a,b,c,d$, then so is $p_{n}$.)

Answer 2

This answer rests on a misreading of the question; see comments to see why. I'm leaving it because I think it may nonetheless prove useful to someone at some point.

First step: symmetrize.,

You've got $P(u, v, s, t) = P(s, v, u, t)$, so apply that and rewrite $P$ as the average of those two. Applying this to the case you asked about, this turns $$ st + uv $$ into $$ \frac{1}{2} \left([st + uv] + [ut + sv] \right). $$ Then do that again with the $v$-to-$t$ symmetry: $$ \frac{1}{4} \left([st + uv] + [ut + sv] + [sv + ut] + [uv + st] \right). $$

At this point, everywhere you have a term $u^k s^\ell t$, you've also got $u^k s^\ell v$. So gather up each $u^k s^\ell$ term, and its coefficient will be some expression that's symmetric in $v$ and $t$.

Then (using the method outlined below), rewrite each $u^k s^\ell$ term (leaving its coefficient in $v$ and $t$ untouched); then, using the same method, rewrite each $v^k t^\ell$ term similarly, leaving the $s$-and-$u$ part untouched. Then you're done!

You can write out the relevant items like $1, us + us, u^2 + s^2, u^2s + s^2 u, u^3s + s^3 u, u^2 s^2 + s^2 u^2, \ldots$ in terms of the given functions, and do this once and for all. $$ 1 = (u+s)^0 \\ us + su = 2 * us \\ u^2 + s^2 = (u+s)^2 - us u^2s + s^2 u = us(s + u) \\ u^3 s + s^3 u =usu ( u^2 + s^2) = su( (u+s)^2 - us ) $$ and pretty soon you start to notice some patterns. If we denote by $P(n, k)$ the polynomial $u^n s^k + s^n u^k$, then there's a recurrence: $$ P(n, k) = \begin{cases} (us)^r \cdot P(n-r, k-r) & \text{for $r = min(n, k) > 0$}\\ P(n, k) = (u+s)^n - Q(n) & \text{when $k= 0$}\\ P(n, k) = (u+s)^n - Q(k) & \text{when $n= 0$}\\ \end{cases} $$ where $Q(n)$ is a polynomial of degree $n-1$ defined as $$ Q(n, k) = \sum_{i = 1}^{n-1} {n \choose i} u^i k^{n-i}. $$ i.e., it's just the binomial expansion of $(u+s)^n$ without the $u^n$ and $s^n$ terms.

Applying this recurrence to each symmetric pair of terms in your polynomial will rewrite the $P(n, k)$ as an algebraic combination of powers of $u+s$ and $us$ (or similarly for $v+t$ and $vt$).

Post-comment addition I've already showed that $st + uv$ is the same as $$ \frac{1}{4} \left([st + uv] + [ut + sv] + [sv + ut] + [uv + st] \right). $$ Gathering powers of $u$ and $s$ we have $$ 4(st + uv) = s(2t + 2v) + u(2t + 2v) = 2(s+u)(t+v). $$ so that $$ st + uv = \frac{1}{2} \left ( (s+u)(t+v) \right). $$