A symmetric determinantal lower bound for diagonal power sums via polar degree

Abstract

The symmetric determinantal complexity sdc(f) of a polynomial f is the least m such that f = det(M) for an m x m symmetric matrix M of affine-linear forms. We prove, over the complex numbers, that sdc(sum_{i=1}^n x_i^n) >= (1/(2e) - o(1)) n^2. This is a symmetric companion to the author's non-symmetric polar-degree preprint (arXiv:7680505); the method parallels that work, but the proof here is self-contained and redoes the load-bearing local incidence analysis in the symmetric setting. The general theorem: if X = V(f) in P^{N-1} is a smooth degree-d hypersurface, N >= 3, and f = det(A_0 + sum x_i A_i) with all A_i symmetric of size m, then the top polar degree d(d-1)^{N-2} is at most 2^{N-2} C(m, N-1). The proof uses the symmetric rank-one kernel incidence M(z,x) u = 0. At a genuine polar point M has rank m-1, and a symmetric Schur-complement normal form eliminates the unique kernel line scheme-theoretically; on the resulting local graph the lifted conormal forms u^T A_i u are a common unit multiple of the partials d_i f, so the lifted polar equations cut the ordinary polar slice up to units and each genuine lifted polar point is a zero-dimensional isolated solution. Multihomogeneous Bezout on P^N x P^{m-1} then yields the bound 2^{N-2} C(m, N-1). For F_n = sum x_i^n this gives the constant 1/(2e). More generally, for F_{N,d} = sum_{i=1}^N x_i^d the same theorem gives sdc(F_{N,d}) >= (1/(2e) - o_N(1)) N(d-1) as N -> infinity. We give an explicit symmetric representation of F_{N,d} of size 2N(d+1)+1, so the diagonal bounds are non-vacuous and tight up to a constant. The result is for exact symmetric determinantal complexity in characteristic zero; it is not a border statement and not a uniform positive-characteristic theorem.