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Let $R$ be a real closed field. An example of Positivstellensatz is that for a real polynomial $f\in R[x_1,\ldots, x_n]$, that is strictly positive on $R^n$, then there are sums of squares $s$ and $t$ such that $sf=1+t$. This was first proved by Stengle.

In the case where $R$ is the real number field, we can view this statement as a Positivstellensatz for the coordinate ring of the real algebraic variety $R^n$.

I'm wondering if there are Positivstellensatz for projective real algebraic varieties, eg $RP^n$. For example, Artin's solution of Hilbert's 17th problem tells us we can find for homogeneous $f$, homogeneous sums of squares $s,t$ such that $sf=t$. This is more like a Nichtnegativstellensatz. Can we modify Stengle into a Positivstellensatz for projective case?

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The answer is yes, because Stengle's positivstellensatz follows from a much more formal statements about rings. What follows is adapted from the book by Bochnak Coste and Roy.

(Prop. 4.4.1 in the French version) Let $A$ be a ring, and $(a_i), (b_j), (c_k)$ arbitrary families of elements of the ring. Let $P=cone(a_i)$, $M=monoid(b_j)$ and $I=Ideal(c_k)$. TFAE:

  1. There is no prime cone $Q$ with $P \subset Q\subset A$ such that the support of $Q$ contains $I$ but none of the elements $b_j$.
  2. There is no homomorphism $\phi: A \to F$ where $F$ is a real closed field, for which $\phi(a_i)\geq 0$, $\phi(b_j)\neq 0$, $\phi(c_k)=0$.
  3. There exists $p\in P, b\in M, c\in I$ with $p+b^2+c=0$.

(The support of $Q$ btw is simply $Q\cap -Q$.) So $P$ represents your $\geq 0$ functions, $M$ your $\neq0$ functions and $I$ your $=0$ functions: if you can write zero as a sum of something nonnegative, something positive and something which is zero, they cannot be functions in, say the coordinate ring of a variety with real points.

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  • $\begingroup$ What does that mean for a real-algebraic variety or basic semi-algebraic set? In which form does a Positivstellensatz hold? $\endgroup$ – Andreas Thom Oct 29 '10 at 5:59
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If I understand correctly, you begin with a homogeneous $f$ of degree $2d$ (say), which is positive except at the origin. You would like a certificate that exhibits this positivity. If you dehomogenize one variable, say $x_i$, apply Stengle's result, then rehomogenize, you get something like $s_i f = x_i^{2k_i} + t_i$, where $t_i$ and $s_i$ are homogeneous sums of squares of degree $2k_i$ and $2k_i - 2d_i$, respectively. The concern is that $x_i$ may divide $t_i$ and so we may not have certified that $f$ is strictly positive if $x_i=0$ (but we are not at the origin).

However, I think we can just do this for each variable $i$ and then piece things together. By multiplying the equation $s_if = x_i^{2k_i} + t_i$ by a suitable even power of $x_i$, we can assume the right hand sides have the same degree $2k$ for all $i$, as do all the $s_i$. Adding these together, we get an expression $s f = t$, where $s$ and $t$ are homogeneous sums of squares and $t$ is strictly positive everywhere but at the origin, because at every such point one of the $x_i$ is nonzero. This means $x_i^{2k}+t_i>0$ and $x_j^{2k}+t_j\geq 0$ for all $j\neq i$, so $t = \sum_i [x_i^{2k}+t_i] > 0$.

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  • $\begingroup$ This is sort of against the spirit of Positivstellensatz in that we cannot see from the form the equation $sf=t$ that $f$ is strictly positive. You put conditions on $t$. $\endgroup$ – user2529 Oct 28 '10 at 15:10
  • $\begingroup$ (I will stop using $t$ lest I get confusing.) Basically Stengle's condition is $sf = 1 + SOS$, whereas I wrote $sf = \sum_i x_i^{2k} + SOS$. It seems to me that $\sum_i x_i^{2k}$ is a reasonable homogeneous analog of Stengle's $1$: a term which makes strict positivity evident. I would imagine one could also use something like $(\sum_i x_i^2)^k + SOS$ to be a little more natural,. I somehow feel like I'm missing the point. Could you give an example of a certificate for homogeneous polynomials which would look more natural to you (whether it works or not)? Or was that the question? $\endgroup$ – Noah Stein Oct 28 '10 at 16:24
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The most general result is - I believe - Schmüdgen's Positivstellensatz, which applies to all compact basic semi-algebraic sets. These are the sets which are described by a finite set of polynomial inequalities. The theorem says that every (strictly) positive polynomial function on a compact basic semi-algebraic set is obtained by addition and multiplication from sums of squares and the defining polynomials of the semi-algebraic set.

Hence, if your set is given by polynomial equalities then it is really a sum of squares in the algebra of functions on the real-algebraic variety.

Konrad Schmüdgen, The $K$-moment problem for compact semi-algebraic sets, Math. Ann. 289 (1991), no. 2, 203–206.

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  • $\begingroup$ That result is not the most general since it requires compactness. Anyway, Schmüdgen's result is related to, but different from Stengle's result. Schmüdgen says that if your set is compact, the result is effective, to the extent that any family of polynomials that do cut out your set can be used to generate the positive cone. If your set is not compact, then there may be positive functions that cannot be expressed from the inequalities that you chose to define your set. $\endgroup$ – Thierry Zell Oct 28 '10 at 21:35
  • $\begingroup$ You are right, for example for $\mathbb R^2$ the result fails. I wrote this answer since I (unfortunatelly) failed to understand the implications of the theorem that you mention in your answer. See the comment there. $\endgroup$ – Andreas Thom Oct 29 '10 at 5:57
  • $\begingroup$ Andreas: I'll flesh out my answer and give you some more details. But I won't be able to do that until Monday at the earliest, so be patient :) $\endgroup$ – Thierry Zell Oct 30 '10 at 17:40
  • $\begingroup$ Does Schmüdgen's Positivstellensatz hold for a general real closed field $R$ or only for the real field $\mathbb{R}$? $\endgroup$ – user2529 Feb 11 '16 at 13:24
  • $\begingroup$ @ColinTan: No, I do not think. This would imply bounds on the degrees of the SOS representation, but these do not exist. If you assume $p \geq1/n$, then it holds. $\endgroup$ – Andreas Thom Feb 11 '16 at 14:40

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