TITLE: Is it possible to compute if an algebraic variety is a differential manifold?
QUESTION [6 upvotes]: Is it possible for a computer to decide if a given real algebraic (or semi-algebraic) variety is a differential manifold ?
Let $f_1,…,f_p, g_1,…,g_q$ polynomials in $n$ variables with coefficients in $\mathbb{R}$. Let $V$ be the variety defined by $f_1=f_2=…=f_p=0$ and $g_1>0,…,g_q>0$. Is it possible to compute if $V$ is a differential manifold ?
(the topology on $V$ is the topology induced by the distance in $\mathbb{R}^n$)

REPLY [3 votes]: Here are two (related) misconceptions on the problem. 
1) That if a map $F:\mathbb R^n\to \mathbb R^p$ does not have maximal rank, the subset $F^{-1}(0)\subset \mathbb R^n$ will not be a regular submanifold of $\mathbb R^n$.
A counterexample is $F:\mathbb R\to \mathbb R:x\to x^2$.
The map has $0$ as critical value and yet $F^{-1}(0)=\lbrace 0\rbrace$ is a regular submanifold of $\mathbb R$ 
2) That a singular  algebraic variety cannot be a smooth manifold.
Just take the algebraic variety $x^2+y^2=0$ in $\mathbb R^2$  to get a counterexample.  
A very satisfying answer to this conundrum has been given by Robert Bryant on MathOverflow