3
$\begingroup$

Let $M$ be a $4\times 4$ symmetric matrix whose entries $m_{i,j}$ for $i,j =1,\dots,4$ are homogeneous polynomials of degree $2$ in $3$ variables. Assume that $m_{1,1} = 0$.

Does there exist a decomposition of the determinant of $M$ as: $$ \det(M) = A^2 + BC $$ where $A$ is a non zero homogeneous polynomial of degree $4$, and $B,C$ are non constant homogeneous polynomials?

The determinant of $M$ could be written as follows:

$$ (m_{0,1}m_{2,3}+m_{0,2}m_{1,3}+m_{0,3}m_{1,2})^2-2m_{0,1}m_{0,2}m_{2,3}m_{1,3}-4m_{0,1}m_{0,3}m_{1,2}m_{1,3}-4m_{0,2}m_{0,3}m_{1,2}m_{1,3}+2m_{0,1}m_{0,2}m_{1,2}m_{3,3}+2m_{0,2}m_{0,3}m_{1,1}m_{2,3}-m_{0,1}^2m_{2,2}m_{3,3}-m_{0,2}^2m_{0,3}m_{1,1}-m_{0,3}^2m_{1,1}m_{2,2}. $$

So, if for instance $m_{0,1}=0$, the answer is positive since we have $$ \det(M) = (m_{0,2}m_{1,3}+m_{0,3}m_{1,2})^2 +m_{0,3}(-4m_{0,2}m_{1,2}m_{1,3}+2m_{0,2}m_{1,1}m_{2,3}-m_{0,2}^2m_{1,1}-m_{0,3}m_{1,1}m_{2,2}). $$

$\endgroup$
4
  • 2
    $\begingroup$ no idea. Can you show us any examples where your idea does work? $\endgroup$
    – Will Jagy
    Dec 28, 2022 at 18:57
  • 2
    $\begingroup$ I added a class of examples for which the formula works. $\endgroup$
    – Puzzled
    Dec 28, 2022 at 22:20
  • 2
    $\begingroup$ There is some index problem --- you first assumed $m_{1,1} = 0$, then $m_{0,1} = 0$, but in your example both are present. $\endgroup$
    – Sasha
    Dec 29, 2022 at 4:55
  • 1
    $\begingroup$ Just to connect to some body of literature which may be useful for this question: a positive answer would imply that the {\it strength} of the determinant is at most two. There may be results giving lower bounds on the notion of strength studied in the article by Ananyan and Hochster arxiv.org/abs/1610.09268 $\endgroup$ Dec 29, 2022 at 15:30

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.