Hilbert's seventeenth problem asks to prove that every positive semidefinite form can be written as the sum of squares of rational functions. Currently we don't seem to have a good understanding of how many such squares are sufficient, or how ugly the denominators will eventually get.

One theorem that bears some psychological value in this direction is the following result of Pólya (discussed somewhere in "Inequalities" by Hardy, Littlewood and Pólya):

If a homogeneous polynomial $p$ in $n$ variables $x_1,x_2,\dots,x_n$ is positive in the standard simplex then there exists an $N=N(p)\in \mathbb N$ so that $(x_1+x_2+\cdots+x_n)^N p$ has all coefficients positive.

This means that every even positive definite form can be multiplied by a high enough power of $x_1^2+\cdots+x_n^2$ to have all coefficients positive and thus be a sum of squares of polynomials. In fact it is true that after multiplying any positive definite form with a high enough power of $x_1^2+\cdots+x_n^2$ one gets a sum of squares of polynomials.

Now, let $S_{n,d}$ be the set of positive semidefinite forms in $n$ variables of degree $d$. Let $\Sigma$ be the set of all forms that are a sum of squares of polynomials.

Question: Can we always choose a countable collection of forms $Q=\{q_1,q_2\dots\}$ so that for each $p\in S_{n,d}$ we have $q_i^2p\in \Sigma$ for some $i\in \mathbb N$?

I have proven that a finite $Q$ does not suffice, however I can't seem to prove that one can get a way with countably many denominators. It would be nice if this was true and moreover one could find a $Q$ with some simple structure like in Polya's theorem. On the other hand, if it wasn't true it would be interesting to know if one can put any condition at all.


I passed this question on to Bruce Reznick, who points to http://www.math.uiuc.edu/~reznick/aud.pdf,

MR2159759 (2006d:12001)
Reznick, Bruce(1-IL)
On the absence of uniform denominators in Hilbert's 17th problem. (English summary) Proc. Amer. Math. Soc. 133 (2005), no. 10, 2829--2834 (electronic).

Bruce adds, "I already proved that finitely many won't work."

  • $\begingroup$ Thanks! Now I want to take a look at the articles by Delzell referenced in there as they also seem to be relevant. $\endgroup$ Jul 29 '10 at 4:20

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