Start with variables $(a_1, a_2, a_3, … a_n)$ and transform it to the system $(x_1, x_2, x_3, … x_n)$ where the xi’s are the solutions to $x^n + a_1x^{n-1} + a_2x^{n-2} + a_3x^{n-3} +…+ a_n$. The Jacobian transformation seems to be $da_1 da_2 da_3 … da_n = J' dx_1 dx_2 dx_3 … dx_n$ where $J'$ is the square root of the negative of the determinant of $x^n + a_1x^{n-1} + a_2x^{n-2} + a_3x^{n-3} +…+ a_n$. Is this true? Is it well known?
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1$\begingroup$ What is the determinant of a polynomial? $\endgroup$– Igor RivinOct 15, 2012 at 2:30
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1$\begingroup$ "discriminant" must be the intended word (up to $\pm 1$). $\endgroup$– Noam D. ElkiesOct 15, 2012 at 3:08
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2 Answers
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Let the polynomial be $p(x).$ If you write down the Jacobian of the map which maps the $x_i$ to the $a_i$ (which are symmetric functions of the $x_i$), you will see that the columns just have the coefficients of $p(x)/(x-x_i)$ (in other words, the symmetric functions of all but the $i$-th variable). This determinant, as a polynomial in the $x_i$ will have degree $n(n-1)/2,$ and will vanish whenever two of the $x_i$ are equal, so it is a constant multiple of the product of $(x_i - x_j),$ for $i>j$ (which is a square root of the discriminant). Computing the constant is easy by induction.
answered Oct 15, 2012 at 3:20
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You find a lot of information in: Dmitri Alekseevky, Andreas Kriegl, Mark Losik, Peter W. Michor: Choosing roots of polynomials smoothly, Israel J. Math 105 (1998), p. 203-233. (pdf), and in related later papers. The formula that you seek (or the inverse) is on page 7.
answered Oct 15, 2012 at 7:56