# Zeros of polynomials as a finite union of manifolds

Consider a polynomial in $d$ variables, $p:\mathbb{R}^{d}\rightarrow\mathbb{R}$. Denote by $\mathcal{C}$ its set of zeros, i.e. $$\mathcal{C}=\{x\in\mathbb{R}^{d}\ |\ p(x)=0\}.$$

Q. Is it possible to find finitely many (not necessarily disjoint) manifolds $M_{1},\dots, M_{n}\subset\mathbb{R}^{d}$, with possibly different dimensions, such that $$\mathcal{C}=\bigcup_{k=1}^{n}M_{k}?$$

My question arises in the context of degenerate real matrices, namely, the case where $\mathcal{C}$ consists on the set of symmetric real matrices with at least one repeated eigenvalue (in this context, $p(x)$ is the discriminant of the symmetric real matrix $x$). Any suggestion on how to approach this problem or a reference related to the problem will be greatly appreciated.

Your set $\mathcal{C}$ is by definition a (semi)algebraic subset of $\mathbb{R}^d$, and any such a subset has a Whitney stratification with finitely many (semi)algebraic strata.
M. Tib$\check{\rm u}$ar, Polynomials and vanishing cycles, Cambridge Tracts in Mathematics 170 (2007), ZBL1126.32026, page 219 and the references given therein.
• Note however that the wikipedia reference that you provide contains a big mistake. If $X$ is a singular variety, the collection $Sing(X), Sing(Sing(X)),...$ is not a Whitney stratification in general. Apr 27, 2017 at 18:16
• Please, what happens if we replace $\mathbb{R}$ by $\mathbb{C}$? Apr 28, 2017 at 9:34