This curious identity arose from studying reductions of the maximal ideal in certain monomial algebra. It can be proved "by hand", (i.e, using Macaulay 2), but I am seeking a more conceptual understanding and related references if they exist.

Let $R$ be a commutative ring. For two vectors $v=(a,b,c,d), w=(A,B,C,D)\in R^4$, we define $v\star w:= (aA,aB+bA,bB, cC,cD+dC,dD)\in R^6$. Given any 3 vectors $v_1,v_2,v_3\in R^4$, we can form a $6\times 6$ matrix $M$ whose rows are $v_i\star v_j$, $1\leq i,j\leq 3$. Then: $$\det(M)=0$$

It is not clear to me how to explain this. The kernel of $M$ is a column of degree $6$ polynomials, so the relations are quite complicated.

Question: Is there a way to conceptually explain the vanishing of $\det(M)$? Have you seen similar identities?


Three vectors $v_1,v_2,v_3$ lie in a hyperplane $H:\alpha x+\beta y+\gamma z+\delta t=0$, in this plane we have $Q(v,v):=(\alpha x+\beta y)^2-(\gamma z+\delta t)^2=0,\forall v\in H$. Thus by polarization $Q(v,w)=\frac 14 (Q(v+w,v+w)-Q(v-w,v-w))=0$ for all $v,w\in H$ that yields a relation between columns of your matrix: if $v=(a,b,c,d), w=(A,B,C,D)$, then $Q(v,w)=\alpha^2 aA+\beta^2 bB+\alpha\beta(aB+bA)-\gamma^2 cC-\delta^2 dD-\gamma\delta(cD+dC)$.

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    $\begingroup$ So in other words you are saying that the vector $(\alpha^2,\alpha\beta,\beta^2,-\gamma^2,-\gamma\delta,-\delta^2)$ is in the kernel of the matrix. Since the $\alpha,\beta,\gamma,\delta$ are cubic in the vectors $v_1,v_2,v_3$, this explains the degree $6$ polynomials. $\endgroup$ – Zach Teitler Oct 27 '19 at 8:40

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