MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $f$ and $g$ be positive definite forms in the polynomial ring ${\mathbb{R}}[x_0,\ldots, x_n]$ such that $\deg(g)$ divides $\deg(f)$. A generalization of a theorem by Reznick is that $g^N f$ is a sum of squares for large $N$.

A corollary of the Representation Theorem says that if $V$ is an affine real variety with compact set $V({\mathbb{R}})$ of real points, then every $f\in {\mathbb{R}}[V]$ that is strictly positive on $V(\mathbb{R})$ is a sum of squares in ${\mathbb{R}}[V]$.

A proof of the theorem by Reznick using this corollary goes like this: Let $V$ be the complement of the hypersurface $g=0$ in real projective space ${\mathbb{P}}^{n-1}$. This variety $V$ is affine and has a compact set of real points. Let $r=\deg f/\deg g$. By assumption $f/g^r>0$ and thus a sum of squares.

My question is: Why is there a need to remove the hypersurface $g=0$. In the first place, if $g$ is positive-definite, isn't this hypersurface $g=0$ an empty set?

share|cite|improve this question
up vote 1 down vote accepted

The real point of $g = 0$ are the empty set, but $g = 0$ is still a non-trivial hypersurface over $\mathbb R$, just with no real points. (The fact that is has no real points is the reason why the real points of $V$ are compact.)

share|cite|improve this answer

Could you perhaps provide a reference to the generalisation you are talking about? I don't think it is well-known.

share|cite|improve this answer
I found this in "Positivity and sums of squares: A guide to some recent results" by C Scheiderer. – user2529 Dec 25 '10 at 13:54
Thank! It completely escaped my attention somehow... When Schederer writes $\mathbb{P}^{n-1}$ he certainly means the complex projective space---I tried to work with him some years back :-) – Dima Pasechnik Dec 25 '10 at 18:08

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.