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Let $f_1(x,c_1),\ldots,f_n(x,c_n)$ be $n$ real polynomials in $n$ variables $x=(x_1,\ldots,x_n)$ of degree at most $d$ with coefficients $c=(c_1,\ldots,c_n)$. Thus, for each $i=1,\ldots,n$, we have $c_i\in\mathbb{R}^{{{n+d+1}\choose{d}}} $.

Let $$ \Gamma(c)=\{x\in\mathbb{R}^n\mid f_1(x,c_1)=0,\ldots,f_n(x,c_n)=0\} $$ denote the set of real solutions of the given system of polynomial equations with coefficients $c=(c_1,\ldots,c_n)$. Moreover, define $$ C=\left\{c\in\mathbb{R}^{n\times{{n+d+1}\choose{d}}} \ \middle|\ \Gamma(c)\neq \emptyset\right\} $$ to be the set of coefficients for which there exists a real solution to the associated system of equations.

Let us understand the solution set $\Gamma(c)$ as a set-valued function $\Gamma: C \to 2^{\mathbb{R}^n}$ over $C$. A (single-valued) continuous function $\gamma:C \to\mathbb{R^n}$ is said to be a continuous selection of the set-valued function $\Gamma: C \to 2^{\mathbb{R}^n}$ if $\gamma(c)\in\Gamma(c)$ for all $c\in C$.

Question. Suppose $C$ is ``nice'', say open and connected (if more topological structure is needed, please feel free to assume so). Does $\Gamma$ admit a continuous selection? (A reference is much appreciated.)

Thoughts: Of course, if there is the same number of real solutions over $C$, then this follows from an implicit function theorem. However, if solution paths intersect or the polynomials are not in general position, then I am not sure how to formally proceed. Related questions are here and here, but they only concern a single polynomial and the question is not quite the same.

axelniemeyer
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    Minor criticism : your definition of $C$ (of "possible coefficients") looks a little bit unclear to me. I think what you really mean is that you define $\Gamma(c)$ for any $c\in{\mathbb R}^n$, and then $C$ is simply $C=\lbrace c | \Gamma(c)\neq\emptyset \rbrace$. Is my understanding correct ? – Ewan Delanoy Jan 26 '22 at 19:22
  • This is a good point; let me try to be exact. Say we consider $n$ real polynomials of degree at most $d$ in $n$ variables. Then, the space of coefficients is $\mathbb{R}^{n\times{{n+d+1}\choose{d}}} $. We may then set $ C={c|\Gamma(c)\neq\emptyset} $ as you suggest. In the particular applications, I have in mind, this set $C$ is ``nice'', say open or connected or something like that. I'll edit the original post accordingly. – axelniemeyer Jan 27 '22 at 10:11

1 Answers1

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Kato's theorem is for a continuous choice of complex solutions. This is different from what you are asking. Also, you need more than a constant number of solutions to apply the implicit function theorem. Here is a counterexample.

A continuous selection will remain continuous when restricted to, say, a segment in $C$. Take $n=1$, $d=4$, $f(x, c(t))= ((x-1)^2-t)((x+1)^2+t)$. For $t=0$ we have 2 roots, $1$ and $-1$. For small $|t|$, if $t>0$ the roots are near $1$, and if $t<0$ they are near $-1$. There is no continuous selection $\gamma$. (You can also plot the solutions as function of $t$ to see that.)

Max
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  • Thanks for the counterexample. Do you have a sense of whether there is any hope for making the statement work? Say, I only consider $C$'s such that each coefficient is either $0$ over $C$ or non-zero but doesn't change sign over $C$? Would I "always" be able to construct such counterexamples? I'll accept your answer once the bounty expires because it answers the question but of course, I would be very glad about a more positive answer with further assumptions on $C$. – axelniemeyer Jan 29 '22 at 18:58
  • The zero-non-zero thing won't help, since one can always shift the variables $x\to x+s$ to make things non-zero. I think counterexamples are plentiful. Over the complex numbers things are better, but over the reals I doubt there is anything reasonable to say (though I may be wrong). I think you may be looking for some version of "flat families" - https://mathoverflow.net/questions/6789/why-are-flat-morphisms-flat - but I'm not sure how this plays out over the reals. – Max Jan 31 '22 at 07:19