A polynomial implicitization

Question

This is a cross-post from MSE where it did not receive a response.

For integer $n\geq 2$, consider a parameterization of the coordinates $(x_1, x_2, ..., x_{n})$ in terms of the parameters $(s_{1},s_{2}, ..., s_{n-1})$ given by $$x_{j} = \displaystyle\sum_{i=1}^{n-1}(-1)^{i+1}s_{i}^{n-j+1}, \qquad j=1, ..., n.$$ We want to find the implicitization $p(x_1, x_2, ..., x_{n}) = 0$ by eliminating $s_{1},s_{2}, ..., s_{n-1}$.

For example, when $n=2$, the parameterization becomes $$x_{1} = s_{1}^{2},\\ x_{2} = s_{1},$$ and the implicit equation is $p(x_1,x_2) \equiv x_{2}^{2}-x_{1} = 0$.

When $n=3$, the parameterization becomes $$x_{1} = s_{1}^{3} - s_{2}^{3},\\ x_{2} = s_{1}^{2} - s_{2}^{2},\\ x_{3} = s_{1} - s_{2},$$ and elementary algebra gives the implicit equation $p(x_1,x_2,x_3) \equiv x_{3}^{4}-4x_{3}x_{1}+3x_{2}^{2} = 0$.

When $n=4$, the parameterization becomes $$x_{1} = s_{1}^{4} - s_{2}^{4} + s_{3}^{4},\\ x_{2} = s_{1}^{3} - s_{2}^{3} + s_{3}^{3},\\ x_{3} = s_{1}^{2} - s_{2}^{2} + s_{3}^{2},\\ x_{4} = s_{1} - s_{2} + s_{3}.$$

Question: How to systematically derive a general formula for $p(x_1,x_2,...,x_n)$?

Two comments:

(i) I have tried looking into the Macaulay resultant but it is not clear to me if this helps to get a general formula in terms of $n$.

(ii) If we had all plus signs in the $s$ monomials, then the parameterization would reduce to power sums, and we could use Newton's identities to get the implicitization. But I am not sure if that can be generalized here.

Answer 1

I think many places will call these functions supersymmetric power sums. They appear in the work of Kerov, Okounkov and others on the asymptotic representation theory of the symmetric group (via Thoma's simplex etc.). Many of the usual identities on symmetric functions have supersymmetric analogs, including the Newton identities, but the relation you ask for here is of a slightly different flavor.

Let me slightly change your notation so that we have $$p_k(s_1,s_2,s_3,\dots, s_{n-1})=s_1^k-s_2^k+s_3^k+\cdots.$$ and we are interested in the relation between $p_1,p_2,\dots,p_{n}$. Let us introduce the sequence $A_k(s_1,s_2,\dots)$ via the generating function $$F(t)=\sum_{k\geq 0} A_k t^k=\frac{(1-s_1t)(1-s_3t)\cdots}{(1-s_2t)(1-s_4t)\cdots}.$$ Since the generating functions is a rational function with denominator of degree $m=\lfloor \frac{n-1}{2}\rfloor$ then the following Hankel determinant of vanishes $$\det[A_{n-2m+i+j}]_{i,j=0}^{m}=0 \tag{*}$$ (This is a famous relation for the coefficients of a rational function. See this answer for an explanation.) Moreover, it is easy to see that the logarithmic derivative of $F(t)$ is the appropriate generating function of the $p_k$'s, so that $$F(t)=e^{-\left(\sum_{k\geq 1}\frac{p_k}{k}t^k\right)}$$ which implies that every $A_k$ can be written as a polynomial of degree $k$ in the $p$'s (considering $p_i$ of degree i). Substituting each such polynomial expression into $(*)$ gives the desired relation. This will be a relation of degree $(m+1)(n-m)$, which matches your examples. For $n=4$ this produces a relation of degree $6$ and for $n=5$ a relation of degree $9$.