# 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 '11 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 '11 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 '11 at 7:35
Just a very minor comment, since you asked in what context this result appears. I happended to use it in a recent paper (Eq. (3.30) in arXiv:1312.5879). The context there is symmetric polynomials related to exactly solvable models of statistical mechanics. But anyone working with symmetric functions would view it as a standard result, and I am sure it has been used very often. – Hjalmar Rosengren Dec 9 '14 at 9:04

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 '11 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 '11 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 '12 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 '11 at 7:22
So, which one is right, Denis's formula or mine? They don't seem to be equal. – Andrej Bauer Oct 19 '11 at 7:34
Good question :) I will try to check in an hour. – Alexander Chervov Oct 19 '11 at 7:50
I checked Denis's answer. It seems it is correct. – Alexander Chervov Oct 19 '11 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 '11 at 12:21

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.

-

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.

-

Dividing the $k$th column of the lower triangular matrix $T$ (OEIS A111492) in Andrej Bauer's answer by $(k-1)!$ for each column generates A135278 (the $f$-vectors, or face-vectors for the $n$-simplexes). Then ignoring the first column gives A104712, so $T$ acting on the column vector $(-0,d,-d^2/2!,d^3/3!,...)$ gives the Euler classes for hypersurfaces of degree $d$ in $CP^n$. (See Daniel Dugger, A Geometric Introduction to K-Theory, pg. 168.)

$T$ also has relations to the number of permutations of the symmetric group $S_n$ that are pure $k$-cycles, colored forests of "naturally-grown" trees, disposition of flags on flagpoles, the colorings of the vertices of the complete graphs $K_n$, encoded in their chromatic polynomials (see A130534), and the commutator $[log(D), x^nD^n]=d(x^nD^n)/d(xD)$ for $D=d/dx$ (cf. A238363).

Update (Apr 26 and May 20 2014):

The Vandermonde matrix $V_n$ is intimately connected to the $(n-1)$-simplex and its edge projection onto a plane, the complete graphs $K_n$. There are several definitions in use, so to be definite let

$$V_n=V_n(x_1,x_2,...,x_n) = \left[ \begin{array}{} 1 & 1 & \cdots & 1\\ x_1 & x_2 & \cdots & x_n \\ \vdots & \vdots & \ddots & \vdots\\ x^{n-1}_1 & x^{n-1}_2 & \cdots & x^{n-1}_{n}\end{array} \right]$$

and determinant

$$|V_n|=|V_n(x_1, \ldots, x_n)| = \prod_{1 \leq i < j \leq n} (x_j - x_i).$$

To obtain a generating function for the rows of $T$ from Chervov's operator, first note the action of the generalized shift/dilation operator $exp(t:xd/dx:)f(x)=f((1+t)x)$ (a generalization of $e^{td/dx}f(x)=f(x+t))$, where $(:x_i\frac{d}{dx_i}:)^n=x_i^n(\frac{d}{dx_i})^n$, i.e., the power distributes over the expressions between colons. Also let $p_k(x_1,...,x_n)$ be the power sum symmetric polynomial. Then act on $|V_n|$ with a sum of $exp(t:x_id/dx_i:)$ obtaining

$$W_n(x_1 , \ldots , x_n;t)=\sum_{k\geq0} \frac{t^k}{k!}\sum_{i=1}^{n} x_i^k \frac{d^k}{dx_i^k} |V_n|$$

$$=exp[t \cdot p.(:x_1\frac{d}{dx_1}:,...,:x_n\frac{d}{dx_n}:)]|V_n|$$

$$=\sum_{i=1}^{n}exp(t :x_i\frac{d}{dx_i}:) |V_n(x_1 , \ldots , x_n)|$$

$$= |V_n((1+t)x_1 ,x_2, \ldots , x_n)|+|V_n(x_1 ,(1+t)x_2, \ldots , x_n)|+ \ldots+|V_n(x_1 ,x_2,\ldots,(1+t) x_n)|$$

$$= [1+(1+t)+(1+t)^2+\;...+(1+t)^{n-1}]\; |V_n(x_1 , \ldots , x_n)|$$

$$= \frac{(1+t)^n-1}{t}\; |V_n(x_1 , \ldots , x_n)|.$$

Therefore, $$G_n(t)=\frac{W_n(x_1 , \ldots , x_n;t)}{|V_n(x_1 , \ldots , x_n)|}=\frac{(1+t)^n-1}{t},$$

which gives an exponential generating function for the rows of the matrix $T$ (OEIS A111492, A238363) in Bauer's guess that is in agreement with Serre's answer, and an ordinary generating function for the f-polynomials (f-vectors) of the number of k-faces of the $(n-1)$-simplex (OEIS A135278).

For example,

$$G_4(t)=\frac{(1+t)^4-1}{t}=4+6t+8 \frac{t^2}{2!}+6 \frac{t^3}{3!}=4+6t+4t^2+t^3.$$

$V_n,K_n,G_n$ are associated to the $(n−1)$-simplex, and the $3$-simplex is the tetrahedron with $4$ vertices, $6$ edges, $4$ triangles, $1$ polyhedron. (The number of factors in the product formula for $|V_n|$ is given by the number of edges of $K_n$ (OEIS A000217). See also the MO-Qs Cyclotomic Polynomials in Combinatorics and my notes on The Vandermonde Matrix and Goin' with the Flow at my website.)

There is another expression for $G_n(t)$:

$G_n(t)=\left[ \begin{array}{} 1 & t & \frac{t^2}{2!} & \cdots & \frac{t^{n-1}}{(n-1)!}\end{array} \right]|V_n|^{-1}V_n(:x_1 \frac{d}{dx_1}:,...,:x_n \frac{d}{dx_n}:) |V_n| \left[ \begin{array}{} 1 \\ 1 \\ \; \vdots\\ 1\end{array} \right]$

since

$\left[ \begin{array}{} 1 & t & \frac{t^2}{2!} & \cdots & \frac{t^{n-1}}{(n-1)!}\end{array} \right]V_n\left[ \begin{array}{} 1 \\ 1 \\ \; \vdots\\ 1\end{array} \right]=\sum_{k<n} \frac{t^k}{k!}\sum_{i=1}^{n} x_i^k=\sum_{k<n} \frac{t^k}{k!}p_k(x_1,...,x_n),$

and when acting on a polynomial of degree $\leq (n-1)$, the finite operator sum $\sum_{k<n} \frac{t^k}{k!}p_k(:x_1d/dx_1:,...,:x_nd/dx_n:)$ is equivalent to $exp[t \cdot p.(:x_1\frac{d}{dx_1}:,...,:x_n\frac{d}{dx_n}:)]$ above.

The row polynomials of $T$ are given by replacing $t^j/j!$ by $t^j$ in the operator expression.

(Edit 6/2014)

Derivatives of the generating function $W_n(x_1,...,x_n;t)$ generate the inverse of $V_n$:

Note that $D^k_{t=-1}f[(1+t)x]=x^k f^{(k)}(0)=x^k D^k_{x=0}f(x)$, so acting on the two different expressions for $W_n$ gives, for $k=0,...,n-1$,

$$D^k_{t=-1} W_n(x_1,...,x_n;t)= \sum_{i=1}^n x_i^k D^k_{x_i=0} |V_n(x_1,...,x_n) |= k! |V_n(x_1,...,x_n)|.$$

Writing out the determinants in matrix form, you can identify the coefficients with the Cramer's rule soln. to the elements of the inverse of $V_n$. Each equation is then proportional to the inner product of a column covector of the adjugate matrix with a row vector of $V_n$.

-
One way to get a relation to the elementary symmetric polynomials $e_k$ is to express the power sums in terms of them. Also, the determinant can be written in terms of $e_k$, e.g., $$|V_3|=-|\left[ \begin{array}{} 1 & 1 & 1\\ e_1(x_2,x_3) & e_1(x_1,x_3) & e_1(x_1,x_2) \\ e_2(x_2,x_3) & e_2(x_1,x_3) & e_2(x_1,x_2)\end{array} \right]|$$. – Tom Copeland May 20 '14 at 6:57