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?

  • 1
    $\begingroup$ Hm, would be interesting to see what it does to the classical bases .. the elementary, the power-sums, Schurs, etc... $\endgroup$ Dec 1 '16 at 4:14
  • 1
    $\begingroup$ If you start with $e_1$ then $\mathcal{L}e_1$ is not symmetric, for instance. $\endgroup$ Dec 1 '16 at 4:18
  • $\begingroup$ I think it may be useful to reexpress $E$ using Taylor series, so that $E$ becomes an entirely local operator: $Ef = f - \frac{\partial f}{\partial x_1} + \frac{\partial^2 f}{\partial x_1^2} + ... + (-1)^i \frac{\partial^i f}{\partial x_1^i} + ...$. $\endgroup$
    – user44191
    Dec 1 '16 at 5:37
  • $\begingroup$ Is it still true that $\mathcal{L}^m(e_2^n)$ is generally symmetric? $\endgroup$
    – user44191
    Dec 1 '16 at 5:38
  • 1
    $\begingroup$ 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. $\endgroup$ Dec 1 '16 at 6:12

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

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


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

  • 1
    $\begingroup$ This is quite interesting because it suffices to check only $f, \mathcal{L}f$ are symmetric. Cool. I'll check the details. $\endgroup$ Dec 2 '16 at 0:44
  • 1
    $\begingroup$ 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? $\endgroup$ Dec 2 '16 at 1:03
  • 2
    $\begingroup$ your proof seems correct, up-voted. $\endgroup$ Dec 2 '16 at 1:04
  • 1
    $\begingroup$ @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. $\endgroup$ Dec 2 '16 at 1:16
  • 1
    $\begingroup$ The next step is to understand the restriction of $\cal L$ to this subspace. $\endgroup$ Dec 2 '16 at 7:41

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 $c e_2^k$).

  • $\begingroup$ 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. $\endgroup$ Dec 2 '16 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. $\endgroup$ Dec 2 '16 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? $\endgroup$ Dec 2 '16 at 13:33
  • 1
    $\begingroup$ @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 $p = c_0 + c_2\mathcal{L}1 + c_{22}\mathcal{L}^21$. $\endgroup$ Dec 3 '16 at 3:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.