Lipschitz regular complex algebraic sets are smooth (Q2790174)

A closed subset \(X\subset\mathbb{R}^n\) is called topologically regular (resp. Lipschitz regular) at \(x_0\in X\) if there exists a homeomorphism (resp. a subanalytic bi-Lipschitz homeomorphism) \(\Phi:B\to U\) where \(B\) is a Euclidean open ball and \(U\) is an open neighbourhood of \(x_0\) in \(X\). \(X\) is called topologically (resp. Lipschitz) regular if \(X\) is topologically (resp. Lipschitz) regular at any \(x\in X\). Mumford proved that a topologically regular complex algebraic surface in \(\mathbb{C}^3\) with an isolated singular point is smooth ([\textit{D. Mumford}, Publ. Math., Inst. Hautes Étud. Sci. 9, 5--22 (1961; Zbl 0108.16801)]).NEWLINENEWLINEIt is proved that any Lipschitz regular complex algebraic set is smooth. Note that the assumptions \(\dim(X)=2\) and isolated singularities are not needed.