# det(A)det(B) = det(AB+correction), Capelli identities, “factorzied” representation of gl_n

Context: some probably know that there are Capelli identities which state det(A)det(B) = det(AB+correction) for some matrices with non-commuting elements, they go back to 19-th century, but also investigated up to present due relations with repesentation theory, quantum integrable systems, etc. In forthcoming paper I will discuss a more general question/conjecture on det(A+correction)det(B+correction) = det(AB+correction). But what just I understood that there is quite a simple to ask question, which seems to be open and let me ask it here.

Notations Consider sets of commuting variables $a_i$ and another set $b_i$, but $[a_i, b_j] \ne 0$, more precisely let require the following commutation relations:

Denote $E_{ij}=a_i b_j$ and require that $[E_{ij}, E_{kl} ] = \delta_{jk}E_{il}-\delta_{li}E_{kj}$, i.e. the same relations as satisfied by the "matrix units" $e_{ij}$, i.e. matrices having 1 at position (i,j) and zeros everywhere else.

Question 1 Is it true that $det^{column}(E+diag(n-1,n-2, ..., 1,0)) = 0$ ? (n>1).

Where $E$ is a matrix with elements $E_{ij}$ at position (ij), $det^{column}$ means we use usual definition of the determinant, but the elements of first column comes first in the products, second column - second, etc.

Motivations: If $a,b$ would be commuting variables then $E$ is matrix of rank 1 and its determinat equal to zero. The correction $+diag(n-1,n-2, ..., 1,0)$ is usual correction for the Capelli type identities.

Checks For 2x2 matrix I have checked it. If commutation relations between $[a_i, b_j]=-\delta_{ij}$ then it is also known to be true as a part of the modern Capelli story. I.e. if $b_j = \partial_{a_j}$ we get standard construction of the $S^*(C^n)$ representation of $gl_n$ by differential operators.

Subquestion: For 3x3 case it is not easy to check by hands, may be it is possible to check by some software ?

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I would suggest to see if this is true even when you consider non-commutative variables $E_{ij}$ that are not necessarily of the form $a_ib_j$. – DamienC Dec 4 '12 at 8:54
Damien, thank you for your comment however I a little disagree - if E_{ij} not of this form, but say of E is "full rank matrix" we will get det(A)det(B) and it is not equal to zero, but conjectural generalization of classical Capelli identity. It should be zero when A,B are of small rank. I will put it as question 2 later... – Alexander Chervov Dec 4 '12 at 9:09