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The formulation of the Capelli's identity is very elementary; it has important applications in invariant theory and representation theory, see http://en.wikipedia.org/wiki/Capelli%27s_identity

To remind it, it requires some notation. Let $x_{ij}$, $1\leq i,j\leq n$, be commuting variables. Define differential operators on the space of functions on $n\times n$ matrices: $$E_{ij}=\sum_{a=1}^n x_{ia}\frac{\partial}{\partial x_{aj}},\, 1\leq i,j\leq n.$$

The Capelli identity states that $$\det\left[\begin{array}{cccc} E_{11}+n-1&\dots&E_{1,n-1}&E_{1n}\\ \vdots&\vdots&\dots&\vdots\\ E_{n-1,1}&\dots&E_{n-1,n-1}+1&E_{n-1,n}\\ E_{n,1}&\dots&E_{n,n-1}&E_{n,n}+0 \end{array}\right]=\\\det\left[\begin{array}{ccc} x_{11}&\dots&x_{1n}\\ \dots&\dots&\dots\\ x_{n1}&\dots&x_{nn} \end{array} \right]\cdot \det\left[\begin{array}{ccc} \frac{\partial}{\partial x_{11}}&\dots&\frac{\partial}{\partial x_{1n}}\\ \dots&\dots&\dots&\\ \frac{\partial}{\partial x_{n1}}&\dots&\frac{\partial}{\partial x_{nn}} \end{array}\right]. $$

Note that in the right hand side of the equality the two matrices have commuting entries, while in the left hand side the entries do not commute. Hence one has to be careful to define the determinant. The convention for the determinant of such matrices is $$\det(a_{ij})=\sum_{\sigma\in S_n}sgn(\sigma) a_{1\sigma(1)}a_{2\sigma(2)}\dots a_{n\sigma(n)},$$ where the order of terms is important.

This order of terms in the determinant is the most mysterious point for me. Is there any reason for it? What happens if one chooses some other ordering of terms: will the Capelli identity be modified somehow or it will not work at all?

There are more recent generalizations of the Capelli identity, see e.g. the above link to Wikipedia. However unfortunately they do not clarify to me the original Capelli idenity, but rather use it as a basic inspiration for further extensions (may be I am missing something).

UPDATE. So far the most conceptual approach to the Capelli identity I was able to find in the literature is due to I. Gelfand and V. Retakh, Funktsional. Anal. i Prilozhen. 25 (1991), no. 2, 13--25 (Section 3.4). This is their first paper on their general theory of non-commutative determinants. They use it to rewrite the Capelli identity in their language. I have not studied their proof in detail, but apparently they do not use this strange and seemingly arbitrary definition of det as a sum over all permutation of terms with prescribed ordering. I still have to understood how difficult and natural it is to pass from their language to the classical one.

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This order of terms in the determinant is the most mysterious point for me. Is there any reason for it? What happens if one chooses some other ordering of terms: will the Capelli identity be modified somehow or it will not work at all?

I think the answers to these questions can be found in the paper http://arxiv.org/abs/0809.3516 (Noncommutative determinants, Cauchy-Binet formulae, and Capelli-type identities. I. Generalizations of the Capelli and Turnbull identities, by S. Caracciolo, A. Sportiello and A.D. Sokal).

As first indicated in http://arxiv.org/abs/math/9309212 (Combinatorial Proofs of Capelli's and Turnbull's Identities from Classical Invariant Theory, by D. Foata and D. Zeilberger), the reason for using this particular order of terms in the l.h.s. determinant is to kill the "bad guys" which otherwise will spoil the identity (see the paper for how these "bad guys" are identified).

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  • $\begingroup$ Thank you for the references! I have had a look at the first paper. There it is explained, among other things, what happens if one choses a column ordering of det. $\endgroup$ – MKO Jan 31 '15 at 7:55
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I am somewhat late for a year, sorry, but since the wiki-article is mainly written by me and I somehow worked on the subject, it is difficult to resist writing an answer.

Currently there is some understanding about non-commutative determinants which makes identity less mysterious:

Point 1. There are determinants (and whole linear algebra) for very special classes of matrices with noncommutative elements. Although for generic matrices with non-commutative elements there is NO distingueshed determinant, however for some special classes of matrices with noncommuting entries which are in certain sense not very far from commutativity, there are natural definitions of the determinant and moreover many theorems of linear algebra can be naturally extended to such matrices. Probably q-determinant for quantum groups is the most well-known example.

Point 2. Capelli matrix "E" is of that special type. The matrix "E" actually of this type, i.e.
it can be related to "Manin matrices", although this relation is not unique. For Manin matrices everything is perfect: determinant, linear algebra ... everything ... - clear concept with no mysteries.

Point 3. In particular your questions:

the determinant is the most mysterious point for me. Is there any reason for it? 

For Manin matrices there are clear and simple reasons, for Capelli matrix "E" we can obtain the determinant above from determinant for Manin. (If you find relation Capelli <-> Manin "natural", you might not - look below).

What happens if one chooses some other ordering of terms: 
will the Capelli identity be modified somehow or it will not work at all?  

For Manin matrices defining determinant you use arbitrary ordering. It implies that there is control on what happens when you change ordering in the determinant for "E".


Let me sketch how to relate Capelli matrix $E$ with Manin matrices. We must introduce auxiliary formal variable "u" And consider: $$ E(u) = E + u Id$$ So we can restore $E$ by taking $u=0$. That is not a Manin matrix, but it is almost Manin matrix.

The idea that is that sometimes matrix $T_0$ may not be Manin, but it is possible to find corrections $T_i$, such that $T(u) = T_0+ u T_1 + u^2 T_2 + ...$, would be (almost) Manin matrix. The Capelli case is the most simple where correction is the most simple $T_1 = uId$.

Finally I need to explain what I mean saying "almost Manin matrix", short way of saying it is that if one introduces new element $K = exp(-d/du)$, which commute with $u$ as a shift operator $K f(u) = f(u+1) K$, then the matrix $K E(u)$ would be Manin matrix.

Longer, but more explicit way is look on the defining relations for Manin matrices, for example column-commutativity: $ac = ca$ are substituted by $a(u+1) c(u) = c(u+1) a(u) $ i.e. shift in $u$ to $u+1$ is introduced.

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