## A sum involving derivatives of Vandermonde

Consider the standard Vandermonde $V(x_1, \ldots, x_n) = \prod_{i < j} (x_i - x_j)$. I am intersted in the calculation of the following expression for fixed $k$: $$\sum_i (x_i)^k (d/dx_i)^k V(x_1 , \ldots , x_n).$$ My guess is that it equals $c \cdot V(x_1, \ldots, x_n)$ where $c$ is an expression depending on $k$ and $n$ but not on the $x_i$'s. Is it true ? If yes, what is this constant $c$?

I think, if it is true, then it is pretty well-known. Would you be so kind to provide with the answer and/or proof and/or references? What can be the context people study it? Symmetric functions? Quantum Calogero-Moser?

-
Those question marks scare me. – Andrej Bauer Oct 19 2011 at 6:27
Andrej, you checked this is wrong ? I see yours answer in my mail box, but not on forum. So I am puzzled, are you sure it is wrong. For k=1 it is right if we did not make mistake. – Alexander Chervov Oct 19 2011 at 7:10
@Alexander: at first I put into Mathematica the Vandermonde matrix, I was confused. You probably received my confused answer by email while I was fixing it. – Andrej Bauer Oct 19 2011 at 7:35

Perhaps a little Mathematica program will help us form a conjecture. For $k \geq n$ the answer is $0$, so we list your $c$ for $k < n$, as follows:

In[1]:= V[x_] := Product[x[[i]] - x[[j]], {i,1,Length[x]},{j,1,i-1}]
In[2]:= V[x/@Range[3]]
Out[2]= (-x[1]+x[2]) (-x[1]+x[3]) (-x[2]+x[3])
In[3]:= s[k_,x_] := Sum[x[[i]]^k*D[V[x],{x[[i]],k}],{i,1,Length[x]}]
In[4]:= Table[s[k,x/@Range[n]]/V[x/@Range[n]], {n,1,6},{k,0,n-1}]//Simplify//TableForm
Out[4]//TableForm=
1
2       1
3       3       2
4       6       8       6
5       10      20      30      24
6       15      40      90      144     120


Each row is showing a fixed $n$, and each column a fixed $k$. The first column is easy. The second one is triangular number $n(n+1)/2$, OEIS tells me the third one is $2 {n \choose 3}$ and the fourth one $n(n+1)(n+2)(n+3)/4$. After a bit of experimentation:

In[19]:= c[n_,k_] := Product[(n-k+i),{i,0,k}]/(k+1)
In[20]:= Table[c[n,k],{n,1,6},{k,0,n-1}]//TableForm
Out[20]//TableForm=
1
2       1
3       3       2
4       6       8       6
5       10      20      30      24
6       15      40      90      144     120


There we have it, your constant seems to be $n (n-1) \cdots (n-k)/(1+k)$. A true combinatorist should now be able to prove that this is really so. I already believe it.

-
 Great Thanks ! So if it is true it should be known. I believe should be some papers playing with such things. It is not the only calculation which I need... That is why I am interested in what people already know. – Alexander Chervov Oct 19 2011 at 7:19 Denis's answer provide a derivation, then should be some simple formula like: n + n-1 + n-2 + ... + 1 = n(n-1) /2 – Alexander Chervov Oct 19 2011 at 12:22 @Chernov This table matches OEIS-A111492: a(n,k) = (k-1)!C(n,k) For k > 1, a(n,k) = the number of permutations of the symmetric group S_n that are pure k-cycles. – Tom Copeland Jun 28 at 7:31

Your expression is a polynomial $V^k(x_1,\ldots,x_n)$ that is still skew-symmetric. Therefore it is divisible by $V$. In addition, $V^k$ has the same degree as $V$. Thus the quotient $V^k/V$ is a constant. Hence the answer to your question is Yes.

By looking at the coefficient of the monomial $x_1^{n-1}x_2^{n-2}\cdots x_{n-1}$, one finds the constant $$c=\sum_{r=k}^{n-1}\frac{r!}{(r-k)!}.$$

-
 Thank You ! Any refrences where it might be discussed ? (Origin of the problem is this "the Harish-Chandra" map for invariant differential operators (from e.g. Etingof Ginzburg paper). I.e. obtaining Calogero as Hamiltonian reduction from D(gl_n), Vandermonde appears as volume of the group. Also it appears in matrix models by the same reason... – Alexander Chervov Oct 19 2011 at 7:22 So, which one is right, Denis's formula or mine? They don't seem to be equal. – Andrej Bauer Oct 19 2011 at 7:34 Good question :) I will try to check in an hour. – Alexander Chervov Oct 19 2011 at 7:50 I checked Denis's answer. It seems it is correct. – Alexander Chervov Oct 19 2011 at 9:46 It seems Denis's answers coincide with Yours ! I checked with WolframAlpha for several cases. It should be true in general. – Alexander Chervov Oct 19 2011 at 12:21
show 1 more comment

I think your guess is true, and using the formula for the derivative of a determinant, one can compute $c = (n-1)^{k-1}\prod_{j=k}^{j=n}\left(n-1 + \frac{(j-1)!}{(j-1-k)!}\right)$. The proof goes as follows: We have

$$x_m^k \frac{d^k}{dx_1^k}V = \mathrm{det}(V^m),$$ where $V_{ij}^m = x_i^{j-1}$ for $i \neq m$ and $V^m_{mj} = \frac{(j-1)!}{(j-1-k)!}x_m^j$ (this is $0$ by convention when $j < k$). Now notice that when $x_m= x_n$, $V_m$ and $V_n$ differ by an interchange of rows, and hence $det(V_m) + det(V_n) = 0$, while for $s\neq n,m$, $det(V_s) = 0$ since rows $m$ and $n$ are equal. This shows that $x_m - x_n$ is a factor of $S \triangleq \sum_{i}det(V_i)$, and hence $V$ is a factor too. By degree considerations, each of these terms is a factor with multiplicity at most $1$, and hence $S = cV$ for some constant $c$. To compute $c$, one can put in $x_i = i$ on both sides and compute. See the answer above by Denis Serre for a much simpler method to do this.

-

Here is a direct way to obtain Denis Serre's formula: Just note that $x_i^k\frac{\partial^k}{\partial x_i^k}$ multiplies a monomial in the determinant by $\frac{r!}{(r-k)!}$ where $r$ is the power of $x_i$ in that monomial, provided that $r\geq k$. Otherwise, it acts by 0. That's all you need, since in each monomial, all powers between 1 and $n-1$ occur exactly once.

-