Let take a real multivariate polynomial $P(x_1, \ldots, x_n, y)$ such as the degree of P relatively to the variable $y$ is odd. Thus, for each $X = x_1,\ldots,x_n \in\mathbb{R}^n$, the univariate polynomial $P(X, y) \in \mathbb{R}[y]$ has at least one real zero.

Given a compact connected subset $A$ of $\mathbb{R}^n$, Is it possible to construct a continuous map $\phi: A \rightarrow \mathbb{R} $ such that for all $X \in A, P(X, \phi(X)) = 0$ (so namely the map $\phi$ choose one of the real roots of the polynomial $P(X,.)$ )?


The continuous dependence of the (complex) roots to the coefficients made me thought that it may be possible.

If we have no singularity then we can apply the implicit functions theorem to obtain a non-empty primary set where this function is defined, then take the closure of that set (and extend the definition of $\phi$ on the cluse) to get a close(by construction) and open( by applying the IFT) subset of $A$ and thus be good by connection.

But what append when singularities occur? I tend to think that if there are not that much, we can get rid of them by extending by continuity the function constructed on the non-singular points.


This won't work. Consider $P(x,y)=y^3-y+x$ and let $A$ be an interval $[-R,R]$, with $R\gg 1$. Then initially, at $x=0$, our polynomial has three zeros, but the second and third will disappear as we increase $x$. So we'd have to take the smallest zero, but this doesn't work either (when we make $x$ small), for the same reason.

  • $\begingroup$ Indeed. But what is appearing is that we can have a decomposition in three connected where such a map exists. (it is in that case a trivial consequence of the point made by Liviu Nicolaescu). And moreover the values taken by the maps are bounded, is it a general property? $\endgroup$ – TomTom May 21 '15 at 9:13

The set $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\bx}{\boldsymbol{x}}$

$$ Z:=\bigl\{ \;(\bx,y)\in\bR^n\times \bR;\;\; P(\bx,y)=0\,\bigr\}, $$

is semialgebraic and the natural projection $\pi: Z\to\bR^n$, $(\bx,y)\mapsto \bx$ is a continuous semialgebraic map.

You are asking whether this map admis sections defined over arbitrary compact subsets $A\subset \bR^n$. As Christian Remling shows in his example, this may not be possible. Still there is something you can say in general.

Using a bit of real algebraic geometry you can show that for any $R>0$ the cube $C_R=[-R,R]^n\subset \bR^n$ admits a (semialgebraic) triangulation with the property that the fibration defined by $\pi$ is locally trivial over the (relative) interior of any face of this triangulation. In particular, it admits continuous semialgebraic sections defined over the (relative) interiors of the faces of the triangulation.

  • $\begingroup$ Bouncing on my previous remark, is it possible to continuously extend it to the closure of these interiors? Thus we could obtain a bounded property by compactness ? $\endgroup$ – TomTom May 21 '15 at 9:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.