Let $S$ be solutions of a system of quadratic polynomials on $\mathbb{R}^n$.

Suppose $q$ is another quadratic polynomial such that $q|_S\geqslant 0$.

Is it possible to find a polynomial $\tilde q$ such that $\tilde q\geqslant0$, $\deg \tilde q\leqslant 2$, and $\tilde q\ |_S=q|_S$?


  • If the system contains only one polynomial, then the answer is yes.

2 Answers 2


No. The main idea is to find a $q$ on $S$ that is the sum of squares, but where the expressions being squared are not linear combinations of the coordinates in ${\bf R}^n$.

Consider the space curve \begin{align*} S &:= \{ (x,x^2,x^4): x \in {\bf R} \} \\ &= \{ (x,y,z) \in {\bf R}^3: y-x^2 = z - y^2 = 0 \} \end{align*} (an incomplete moment curve). The quadratic polynomial $$ q(x,y,z) := y - 2z + yz $$ is non-negative on $S$, since we have a representation as a square $$ q(x,x^2,x^4) = x^2 - 2x^4 + x^6 = (x-x^3)^2.$$ The point here is that the expression $x-x^3$ inside the square is not a linear function of the three coordinates $x,x^2,x^4$, preventing any obvious way to extend $q$ as a sum of squares of linear polynomials on all of ${\bf R}^3$.

Indeed, if $\tilde q$ is a quadratic polynomial that agrees with $q$ on $S$, then $$(\tilde q-q)(x,x^2,x^4) \equiv 0$$ which on expanding the quadratic polynomial $\tilde q-q$ into coefficients reveals that $\tilde q-q$ must take the form $$ (\tilde q-q)(x,y,z) = a (y-x^2) + b(z-y^2)$$ for some scalars $a,b$. Hence $$ \tilde q\ (x,y,z) = a(y-x^2) + b(z-y^2) + y-2z+yz.$$ But this polynomial is linear in $z$ and not independent of $z$, and thus cannot be non-negative on ${\bf R}^3$ regardless of what values one assigns to $a$ and $b$.


This is a bit too long for a comment ...

Forty years ago, I wrote a paper Formes quadratiques et calcul des variations, published in J. Maths. Pures & Appl. 62 (1983), p 177-196. It dealt with such a problem, in the following situation: $S$ is the subset of ${\bf M}_{p\times q}(\mathbb R)$ consisting in the rank-one matrices. It is thus defined by quadratic identities $$a_{ij}a_{k\ell}-a_{i\ell}a_{kj}=0.$$ I showed that

  • if $p\le2$ or $q\le2$, then the answer is positive: for every quadratic form $Q$ satisfying $Q|_S\ge0$, there exists a positive semi-definite quadratic form $\tilde Q$ that coincides with $Q$ over $S$.
  • on the contrary, if both $p,q\ge3$, then there exists such a $Q$, so that no $\tilde Q$ exists.
  • $\begingroup$ Does not it contradict mathoverflow.net/a/455619 for $n=3$? $\endgroup$ Oct 6 at 22:16
  • 1
    $\begingroup$ @AntonPetrunin. It doesn't, because your MO question was about quadratic forms over the space of quadratic forms (say, over the space of symmetric matrices), while my work was about quadratic form over general $p\times q$ matrices. $\endgroup$ Oct 7 at 6:21

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