Are there projective version of real Positivstellensatz?

Question

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?

Answer 1

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.

Answer 2

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 $t = \sum_i [x_i^{2k}+t_i] > 0$.

Answer 3

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.