# Is it possible to compute if an algebraic variety is a differential manifold?

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 $\mathbb{R}^n$)

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The algebraic case is explained on page 104 of Shafarevich I. In particular, you only need that the variety is nonsingular. – Andrew Oct 1 '12 at 16:36
Thanks, Andrew. – francis-jamet Oct 1 '12 at 17:08

Yes it is. For algebraic varieities, you need only consider the implicit function theorem which tells you when the level sets are regular submanifolds. Say you have $f_i : X \to \mathbb{R}$ and you consider the set $x \in X$ such that $f_i(x) = 0$ for all $1 \le i \le n.$ The set is a regular submanifold if the map $F : X \to \mathbb{R}^n$ given by $x\mapsto (f_1(x),\ldots,f_n(x))$ has zero as a regular value.

As an example, consider $f_i : \mathbb{R}^3 \to \mathbb{R}$ given by $f_1(x,y,z) = x^2 + y^2 + z^2 -1$ and $f_2(x,y,z) = z.$ We are interested in the variety given by $(x,y,z) \in \mathbb{R}^3$ such that $f_1(x,y,z) = f_2(x,y,z)=0.$ This is actually the intersection of the unit sphere with the $xy$-plane and so is a circle in the $xy$-plane. To verify this, let $F:\mathbb{R}^3 \to \mathbb{R}$ be given by $F(x,y,z) := (f_1(x,y,z),f_2(x,y,z)).$ To verify that $(0,0) \in \mathbb{R}^2$ is a regular value, we consider the Jacobian matrix:

$$J_F = \left[\begin{array}{ccc} 2x & 2y & 2z \\ 0 & 0 & 1 \end{array}\right] .$$ The critical points are given by $x=y=0$ and so the critical values are $(-1,z).$ It follows that $(0,0)$ is regular value of $F$ and so $F^{-1}(0,0)$ is a regular submanifold.

In the case of semi-algebraic sets, we have to use the idea of transversality and, in particular, Thom's Transversality Theorem. Transversality generalises the idea of critical points/values. Instead of a mapping having a point in the image as a regular value, we talk about a mapping being transverse to a submanifold in the image.

If you had some functions $g_i : X \to \mathbb{R}$ and you wanted the set of $x \in X$ such that $g_i(x) > 0$ for all $1 \le i \le n$ then you first consider the map $G : X \to \mathbb{R}^n$ given by $x \mapsto (g_1(x),\ldots,g_n(x)).$ Then you need to prove that $G$ is tranverse to the set $S := \{(y_1,\ldots,y_n) \in \mathbb{R}^n : y_i > 0\}.$ If $G$ is transverse to $S$ then $G^{-1}(S)$ will be a regular submanifold of $X$.

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Thanks a lot. With the idea of transversality, have we a sufficient condition or a necessary and sufficient condition ? – francis-jamet Oct 1 '12 at 18:01
The map $G : X \to \mathbb{R}^n$ is transverse to $S \subseteq \mathbb{R}^n$ if $G_*(T_xX) + T_{G(x)}S = T_{G(x)}\mathbb{R}^n$ for all $x \in X,$ where $G_*$ is the differential, $T_xX$ is the tangent space to $X$ at $x$ and $T_{G(x)}S$ is the tangent space to $S$ at $G(x)$, etc. It's basically an exercise in linear algebra. – Fly by Night Oct 1 '12 at 18:07
Dear Fly by Night, the implicit function theorem mentioned in your first section gives a sufficient condition, but not not a necessary one: cf. my answer. – Georges Elencwajg Oct 1 '12 at 18:44
@GeorgesElencwajg Thanks Georges. I've changed "...if and only if..." to "...if...". My answer still answers the OP's question: Is it possible? Yes it is. It is possible if... – Fly by Night Oct 1 '12 at 19:08
Dear FlybyNight, yes your answer perfectly answers the OP's question and I have just upvoted it to prove I believe that! – Georges Elencwajg Oct 1 '12 at 19:13

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

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Point 1 is a nice one. Is this usually dealt with by some irreducibility assumption in algebraic geometry? Perhaps this is worded in terms of divisors? – Fly by Night Oct 1 '12 at 19:24