## Discriminant on boundary of semi-algebraic surface

Let $P(t)$ be a polynomial in $t$ of degree $n$, with some contiguous coefficients (not the first or last) being $x_1,\dots,x_k$ and the rest of the coefficients are fixed.

(E.g. $p(t)=1+2t+x_1t^2+x_2t^3+(5i+7)t^6$ is ok).

Consider the set $S$ defined as $(x_1,\dots,x_k) \in \mathbb{C}^k$ with the property that the roots $t_1,\dots,t_{n}$ of $P(t)=0$ may be ordered increasingly w.r.t modulus, and such that $$|t_j|=|t_{j+1}| = \dots = |t_{k+j}|,$$ for some fixed $j.$

This set is real $k$-dimensional.

Is it true, (in general), that the discriminant $D(x_1,\dots,x_n) = Discr_t(P)$ satisfy $D(x_1,\dots,x_n)=0$ on the $k-1$-dimensional boundary of $S$?

-

Let $P(t)=(t-t_1)(t-t_2)\dots(t-t_k)(t-a)$ where $(-1)^{k+1}t_1t_2\dots t_k a = b$ which has leading coefficient 1 and constant coefficient $b$. If you fix $b$, then $a$ is a rational function of the $t_i$. Treat the case where some $t_i=0$ with care. The other coefficients are $$x_i = (-1)^i \sigma_i(t_1,\dots,t_k,a) = (-1)^i\sum_{1\le j_1<\dots j_i\le k}t_{j_1}\dots t_{j_i} + (-1)^i a\sum_{1\le j_1<\dots j_{i-1}\le k}t_{j_1}\dots t_{j_{i-1}}$$ and the discriminant is $$D=\prod_{i < j} (t_i-t_j)\cdot\prod_i (x_i-a).$$ Also let $$\sigma(t_1,\dots,t_k)=(\sigma_1(t_1,\dots,t_k,a),\dots\sigma_k(t_1,\dots,t_k,a)).$$ Denote by $s_i$ the Newton polynomials $\sum_{j}t_j^i+a^i$ which are related to the elementary symmetric function by $$s_\ell-s_{\ell-1}\sigma_1+s_{k-2}\sigma_2+\dots+(-1)^{\ell-1}s_1\sigma_{\ell-1}+ (-1)^\ell\ell\sigma_\ell=0 \quad (\ell\leq k+1),$$ where $\sigma_{k+1}=b$. The corresponding mappings are related by a polynomial diffeomorphism $\psi$ (treating $a$ as a variable), given by: $$\sigma:=(\sigma_1,\dots,\sigma_k, b):\Bbb R^{k+1}\to \Bbb R^{k+1}$$ $$s:=(s_1,\dots,s_k, s_{k+1}):\Bbb R^{k+1}\to \Bbb R^{k+1}$$
$$s:=\psi^n\circ\sigma^n$$ Note that the Jacobian (the determinant of the derivative) of $s$ is $(k+1)!$ times the Vandermonde determinant: $$\det(ds(t,a))=(k+1)!\,\prod_{i>j}(t_i-t_j)\cdot\prod_i(t_i-a),$$ and even the derivative itself $d(s)(x)$ equals the Vandermonde matrix up to factors $i$ in the $i$-th row.
 Thank you! I'll read your answer and I will probably figure it out! – Per Alexandersson Feb 2 at 15:48 I was a little inprecise about the role of $a$ and edited it somewhat. – Peter Michor Feb 2 at 16:08