Capelli's Identity - Statement

Statement

Suppose that xij for i,j = 1,...,n are commuting variables. Write Eij for the polarization operator

The Capelli identity states that the following differential operators, expressed as determinants, are equal:

\begin{vmatrix} E_{11}+n-1 & \cdots &E_{1,n-1}& E_{1n} \\ \vdots& \ddots & \vdots&\vdots\\ E_{n-1,1} & \cdots & E_{n-1,n-1}+1&E_{n-1,n} \\ E_{n1} & \cdots & E_{n,n-1}& E_{nn} +0\end{vmatrix} = \begin{vmatrix} x_{11} & \cdots & x_{1n} \\ \vdots& \ddots & \vdots\\ x_{n1} & \cdots & x_{nn} \end{vmatrix} \begin{vmatrix} \frac{\partial}{\partial x_{11}} & \cdots &\frac{\partial}{\partial x_{1n}} \\ \vdots& \ddots & \vdots\\ \frac{\partial}{\partial x_{n1}} & \cdots &\frac{\partial}{\partial x_{nn}} \end{vmatrix}.

Both sides are differential operators. The determinant on the left has non-commuting entries, and is expanded with all terms preserving their "left to right" order. Such a determinant is often called a column-determinant, since it can be obtained by the column expansion of the determinant starting from the first column. It can be formally written as

where in the product first come the elements from the first column, then from the second and so on. The determinant on the far right is Cayley's omega process, and the one on the left is the Capelli determinant.

The operators Eij can be written in a matrix form:

where are matrices with elements Eij, xij, respectively. If all elements in these matrices would be commutative then clearly . The Capelli identity shows that despite noncommutativity there exists a "quantization" of the formula above. The only price for the noncommutivity is a small correction: on the left hand side. For generic noncommutative matrices formulas like

do not exist, and the notion of the 'determinant' itself does not make sense for generic noncommutative matrices. That is why the Capelli identity still holds some mystery, despite many proofs offered for it. A very short proof does not seem to exist. Direct verification of the statement can be given as an exercise for n' = 2, but is already long for n = 3.

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