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

I would appreciate if someone can point out to the literature related to characterizing the set of all different ways to write real quartic diagonal $\sum \limits_{k=1}^n x_k^4, x \in \mathbb{R^n}$ as a sum of squares of real quadratic forms. Murray Marshall in his book "Positive polynomials and sums of squares" show that quartic diagonal is in the interior of the cone of sum of squares. Does anyone knows details about such representation.

In particular, suppose $\sum \limits_{k=1}^n x_k^4= \sum_p (x^T A_p x)^2$ ($x$ is a column vector, and $x^T$ its transpose, $A_p \in \mathbb{R^n \times R^n}$) then what $Q\in \mathbb{R^n \times R^n}$ can be represented by the expression $\sum_p (x^T A_p x - {x^*}^T A_p x^*)^2= \sum \limits_{k=1}^n x_k^4 - 2x^T Q x+ const$, $x^*$ is a point in $\mathbb{R^n}$. More specifically, whether $Q$ is dense around identity matrix (in a small neighborhood).

share|cite|improve this question
up vote 2 down vote accepted

I passed your question on to a friend who knows about these things, and he replied,

The theorem that $\sum_{k=1}^m x_k^{2r}$ is interior to the sum of squares of appropriate degree, can be found, with proof, in a paper of R. M. Robinson: Some definite polynomials which are not sums of squares of real polynomials, Izdat. "Nauka" Sibirsk. Otdel. Novosibirsk (1973), 264-282, Selected questions of algebra and logic (a collection dedicated to the memory of A. I. Mal'cev), Abstract in Notices Amer. Math. Soc. 16 (1969), p. 554.

My friend also recommended the Memoir by Bruce Reznick, which can be found, scanned, on Bruce's webpage at UIUC.

share|cite|improve this answer
Thanks a lot. Nevertheless this answer does not provide any specific information about quartic polynomials of interest. As far as I know no one was studying them in details. – mkatkov Oct 17 '11 at 7:52

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.