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.


1 Answer 1


The answer is yes.

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.

See A. Dimca, Singularities and Topology of Hypersurfaces, ZBL07535700, Chapter 1, Corollary 1.12 for more details.

Edit. It is also true that any closed analytic (or subanalytc) set of an analytic manifold admits a Whitney stratification whose strata are analytic submanifolds. In the complex setting, this is called a complex Whitney stratification. See

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.

  • $\begingroup$ 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. $\endgroup$
    – Libli
    Apr 27, 2017 at 18:16
  • $\begingroup$ @Libli: you are right. This does not define in general a Whitney stratification, but some refinement will, see J. L. Verdier, Stratifications de Whitney et theoreme de Bertini-Sard, Invent. Math. 36 (1976), 295-312. Thank you for the remark. $\endgroup$ Apr 27, 2017 at 18:32
  • $\begingroup$ Please, what happens if we replace $\mathbb{R}$ by $\mathbb{C}$? $\endgroup$
    – user237522
    Apr 28, 2017 at 9:34
  • $\begingroup$ Whitney stratification still exist in the complex setting. See my edit to the answer. $\endgroup$ Apr 28, 2017 at 9:50

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