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Consider the ring of polynomials $R:=\mathbb{Z}[x_1,x_2,x_3]$. Define the operators $E, I:R\rightarrow R$ by $Ef(x_1,x_2,x_3)=f(x_1-1,x_2,x_3)$ and the identity $If=f$.

Let $\mathcal{L}:R\rightarrow R$ be the operator given by $$\mathcal{L}f=[(x_1+x_2)(x_1+x_3)E-x_1^2I]f.$$

Let $1$ stand for the constant function $f(x_1,x_2,x_3)=1$ and $\mathcal{L}^2f=\mathcal{L}(\mathcal{L}f)$, etc.

CLAIM. Experiments suggest that $\mathcal{L}^n1$ is always a symmetric polynomial in $R$. Any proof?

EDIT. This has found a resolution (see Pietro Majer's answer).

For example, $\mathcal{L} 1=e_2$ and $\mathcal{L}^21=e_2^2-e_1e_2+e_3$ where $e_1=x_1+x_2+x_3, e_2=x_1x_2+x_1x_3+x_2x_3, e_3=x_1x_2x_3$ are the standard elementary symmetric polynomials.

QUESTIONS. (EDIT) These did not find a definitive answer (apart from Brendan McKay's evidence and argument).

(1) Are there other orbits of symmetric polynomials under $\mathcal{L}$?

(2) Are there other non-trivial operators with similar property over $R$?

(3) What about over rings of many more variables?

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    $\begingroup$ Hm, would be interesting to see what it does to the classical bases .. the elementary, the power-sums, Schurs, etc... $\endgroup$ Commented Dec 1, 2016 at 4:14
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    $\begingroup$ If you start with $e_1$ then $\mathcal{L}e_1$ is not symmetric, for instance. $\endgroup$ Commented Dec 1, 2016 at 4:18
  • $\begingroup$ I think it may be useful to reexpress $E$ using Taylor series, so that $E$ becomes an entirely local operator: $Ef = f - \frac{\partial f}{\partial x_1} + \frac{\partial^2 f}{\partial x_1^2} + ... + (-1)^i \frac{\partial^i f}{\partial x_1^i} + ...$. $\endgroup$ Commented Dec 1, 2016 at 5:37
  • $\begingroup$ Is it still true that $\mathcal{L}^m(e_2^n)$ is generally symmetric? $\endgroup$ Commented Dec 1, 2016 at 5:38
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    $\begingroup$ You can start with a linear combination of polynomials in the orbit you have and that will form a different orbit. But that's cheating. $\endgroup$ Commented Dec 1, 2016 at 6:12

2 Answers 2

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Suppose $f(x,y,z)$ is symmetric (in the following, symmetric tout court always means "symmetric w.r.to the three variables $(x,y,z)$") . Then $\mathcal{L}f(x,y,z):=(x+y)(x+z)f(x-1,y,z)-x^2f(x,y,z)$ is already symmetric w.r.to $(y,z)$, so it is symmetric if and only if it is symmetric w.r.to $(x,y)$, that is, after simplifications, if and only if $f$ satisfies the functional equation $$(x-y)f(x,y,z)-(x+z)f(x-1,y,z)+(y+z)f(x,y-1,z)=0.$$

Claim: if both $f$ and $\mathcal{L}f$ are symmetric, so is $\mathcal{L}^2f$. In other words, the space of all symmetric solutions to the above functional equation is $\mathcal{L}$-invariant.

Proof: According to the observation above, in order to prove the claim, we need to check that the following is identically zero: $$(x-y)\mathcal{L}f(x,y,z)-(x+z)\mathcal{L}f(x-1,y,z)+(y+z)\mathcal{L}f(x,y-1,z) $$ and we exploit the symmetry of $\mathcal{L}f$ writing it $$(x-y)\mathcal{L}f(x,y,z)-(x+z)\mathcal{L}f(y,x-1,z)+(y+z)\mathcal{L}f(x,y-1,z) $$ namely $$(x-y)\big[(x+y)(x+z)f(x-1,y,z)-x^2f(x,y,z)\big]$$$$ -(x+z)\big[(y-1+x)(y+z)f(y-1,x-1,z)-y^2f(y,x-1,z)\big]$$$$ +(y+z)\big[(x+y-1)(x+z)f(x-1,y-1,z)-x^2f(x,y-1,z)\big]$$

$$=\big[(x-y) (x+y)(x+z) +(x+z) y^2 \big]f(x-1,y,z)$$$$-(x-y)x^2f(x,y,z) -(y+z) x^2f(x,y-1,z) =$$

$$=-x^2\big[(x-y)f(x,y,z)-(x+z)f(x-1,y,z)+(y+z)f(x,y-1,z)\big]$$

which is indeed zero, according to the initial remark, since $\mathcal{L}f$ was assumed symmetric.

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    $\begingroup$ This is quite interesting because it suffices to check only $f, \mathcal{L}f$ are symmetric. Cool. I'll check the details. $\endgroup$ Commented Dec 2, 2016 at 0:44
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    $\begingroup$ Based on Pietro's proof, we may ask: (a) is it possible for a non-symmetric $f$ we get $\mathcal{L}f$ symmetric? (b) can we identify all those symmetric polynomials $g$ such that $\mathcal{L}g$ is symmetric? $\endgroup$ Commented Dec 2, 2016 at 1:03
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    $\begingroup$ your proof seems correct, up-voted. $\endgroup$ Commented Dec 2, 2016 at 1:04
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    $\begingroup$ @T.Amdeberhan Haglund et al uses a similar proof in Lemma 4.3.1: math.berkeley.edu/~mhaiman/ftp/jim-conjecture/nsmac.pdf to characterize the image of an operator using symmetric properties of the input and output. $\endgroup$ Commented Dec 2, 2016 at 1:16
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    $\begingroup$ The next step is to understand the restriction of $\cal L$ to this subspace. $\endgroup$ Commented Dec 2, 2016 at 7:41
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Noting Pietro's finite check, I can report that all symmetric polynomials $p(x_1,x_2,x_3)$ up to degree 18 inclusive such that $\mathcal{L}p$ is symmetric are linear combinations of $1,\mathcal{L}1,\mathcal{L}^21,\ldots\,$.

This strongly suggests that there are no others.

An observation that might lead to an elementary proof is that, up to degree 18, all polynomials $p$ such that both $p$ and $\mathcal{L}p$ are symmetric are uniquely determined by the coefficients of the powers of $e_2$ (i.e. the terms in the representation in the base $\{e_1,e_2,e_3\}$ which have the form $c e_2^k$).

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  • $\begingroup$ This is quite an intensive search and I tend to believe your observation "no others". Voting up! On the other hand, what do you mean when you said "... determined by the coefficients..."? From my comment above, $\mathcal{L}(e_2^2)$ is not symmetric. $\endgroup$ Commented Dec 2, 2016 at 2:57
  • $\begingroup$ @T.Amdeberhan : For example, consider all degree 4 symmetric polynomials: $p=c_0 + c_1e_1 + c_2e_2 + c_{11}e_1^2 + c_{111}e_1^3 + c_{1111}e_1^4 + c_{13}e_1e_3 + c_{112}e_1^2e_2 + c_{22}e_2^2$. Then the equation $p=\mathcal{L}p$ has solutions $c_0 + c_2 \mathcal{L}1 + c_{22}\mathcal{L}^21$ only. $\endgroup$ Commented Dec 2, 2016 at 8:12
  • $\begingroup$ But, why do you equate $p=\mathcal{L}p$? Are you looking for eigenfunctions? Otherwise, we're trying to determine: given $q$ symmetric, is $\mathcal{L}q=p$ where $p, q$ are of the form you wrote. Are we not? $\endgroup$ Commented Dec 2, 2016 at 13:33
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    $\begingroup$ @T.Amdeberhan : Sorry I mistyped. I'm not solving $p=\mathcal{L}p$. What I intended to write is that $\mathcal{L}p$ is symmetric iff $p = c_0 + c_2\mathcal{L}1 + c_{22}\mathcal{L}^21$. $\endgroup$ Commented Dec 3, 2016 at 3:13

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