3
$\begingroup$

I've got a question that I've become curious about as a result of supervising some undergraduate research. Let's suppose we have some sequence of polynomials $f_0, f_1, f_2, \cdots \in \mathbb{Z}[X]$, where $f_0=1$. Now, define the following sequence of symmetric polynomials on variables $x_1, \ldots x_n$:

$$P_m(x_1, \ldots, x_n)=\sum\limits_{m_1+\cdots+m_n=m}f_{m_1}(x_1) \cdots f_{m_n}(x_n)$$

If you want, you can think of this as the coefficient of $y^m$ in the two-variable generating function $\prod\limits_{i=1}^{n}\left(\sum\limits_{j \ge 0}f_j(x_i)y^j\right)$. Now my question is, if you know $x_1, \ldots, x_n$ are positive integers, and you know the values of $P_m(x_1, \ldots, x_n)$ for $m=0, 1, \ldots$, what conditions need to be true on $f_1, f_2, \ldots$ in order for the values of of $x_1, \ldots, x_n$ to be completely determined (up to ordering)? I think it's key that we need to work with inputs in positive integers - if we work over $\mathbb{C}$, the answer may be different (see at the bottom).

For example, if $f_1(x)=x$, and $f_2=f_3=\cdots=0$, then $P_m$ is just the $m$-th elementary symmetric polynomial $\sigma_m$, and then obviously, $x_1, \ldots, x_n$ are determined by $\sigma_1, \sigma_2, \ldots$, being the roots of the polynomial with coefficients $(-1)^i\sigma_i$. If $f_i(x)=x^i$ for each $i$, then it's a little less straightforward, but still not hard: $P_m=\sum\limits_{j=1}^{m}(-1)^{j-1}\sigma_j P_{m-j}$, and so we can show by induction on $m$ that $P_1, \ldots, P_m$ together determine $\sigma_1, \ldots, \sigma_m$, and hence, $x_1, \ldots, x_n$ are again determined. For the general case, we can equivalently ask whether the values of $P_1, P_2, \ldots$ uniquely determine the values of $\sigma_1, \sigma_2, \ldots$ (and this is perhaps a more natural question).

As long as the polynomials $f_1$ aren't all constants, I don't, off the top of my head, know any sequences of polynomials $f_1, f_2, \ldots$ for which the values of $P_1, P_2, \ldots$ don't determine $x_1, \ldots, x_n$. So I'm wondering if it's true given that at least one of the $f_i$'s is nontrivial.

I feel it's worth pointing out that what I'm asking isn't the same thing as asking that the $P_i$ generate the ring of symmetric polynomials. For example, if the values of the $P_i$ were to determine the values of $\sigma_1^2, \sigma_2^2, \ldots$, that would determine the values of $\sigma_1, \sigma_2, \ldots$, even if the $P_i$ don't generate the ring. This is why the condition that the $x_i$'s are positive integers is important. That said, I don't know the answer over $\mathbb{C}$ either.

Anyone have any ideas? If it's of any relevance, in the student's research, the polynomial $f_i$ has degree $2i$ for each $i$.

$\endgroup$

1 Answer 1

1
$\begingroup$

Taking the $\log$ of the two-variable generating function might help. It converts it to $$ \sum_{i=1}^n \log(1 + F(x_i)) = \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k} \sum_{i=1}^n F(x_i)^k,$$ where $F(x)= f_1(x)y + f_2(x)y^2 + \dots $. So for instance, the coefficient of $y$ tells you $f_1(x_1) + \dots + f_1(x_n)$, the coefficient of $y^2$ tells you $f_2(x_1) + \dots + f_2(x_n)- (f_1(x_1)^2 + \dots + f_1(x_n)^2)/2$, etc.

$\endgroup$
1
  • $\begingroup$ Thanks for the comment! It turns out there's a counterexample to the statement I gave (courtesy of Bjorn Poonen), but this statement is stronger than what I needed. I started fiddling around with something else involving symmetric polynomials, and I started getting a calculation quite similar to what you've got here, which I think may help out the student. $\endgroup$ Jul 31, 2013 at 5:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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