MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 2 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. – Colin Tan 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.

Not the answer you're looking for? Browse other questions tagged or ask your own question.