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I wish to find the coefficients of the Laurent expansion of the inverse of the Vandermonde determinant, that is, the Laurent expansion at 0 of $$\prod_{1\leq i<j \leq n}(x_j-x_i)^{-1}.$$

We can iteratively do a series expansion in $x_1,x_2,$ and so on, (the full expansion depends on the order of expansion).

An answer I would be satisfied with is if I can say that the coefficient $c_a(k)$ of $x_1^{a_1+k}x_1^{a_2+k}\cdots x_{n-1}^{a_{n-1}+k} x_n^{a_n-k}$ in the Laurent expansion of $$\prod_{1\leq i<j \leq n}(x_j-x_i)^{-d}.$$ is a polynomial in $k,$ for every fixed positive integer $d$.

A paper stating this, or a short argument of this fact would be terrific.

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Hm, I believe I can actually prove the thing I seek, however, if there is a short argument, or if this is done in a nice manner somewhere, then I'd prefer to reference it. –  Per Alexandersson Jul 3 '13 at 15:42

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up vote 11 down vote accepted

In your question, the exponent of $x_2$ is $a_2+k$. I am assuming that is a typo, and you intended $c_a(k)$ to be the coefficient of $$x_1^{a_1+k} x_2^{a_2} x_3^{a_3} \cdots x_{n-2}^{a_{n-2}} x_{n-1}^{a_{n-1}} x_n^{a_n-k}.$$ If so, I can answer all of your questions.

It will be convenient to switch the sign in your Vandermonde determinant, so I'll set $\Delta = \prod_{1 \leq i < j \leq n} (x_j-x_i)$. Let $\rho$ be the vector $(0,1,2,\ldots, n-1)$. So $$\Delta = x^{\rho} \prod_{1 \leq i < j \leq n} ( 1- x_i/x_j)$$ and $$\Delta^{-1} = x^{-\rho} \prod_{1 \leq i < j \leq n} \frac{1}{1-x_i/x_j} =x^{-\rho} \prod_{1 \leq i < j \leq n} \sum_{c_{ij}=0}^{\infty} \left( \frac{x_i}{x_j} \right)^{c_{ij}}.$$

So the coefficient of $x^a$ in $\Delta^{-1}$ is the number of ways to write $a+\rho$ in the form $\sum_{1 \leq i < j \leq n} c_{ij} (e_i-e_j)$, where $e_k$ is the $k$-th basis vector and the $c_{ij}$ are nonnegative integers. This is called the Kostant partition function of $a+\rho$; I'll denote it by $K(a+\rho)$.

If you do the same argument for $\Delta^{-d}$, you are counting the number of ways to write $$a+\rho = \sum_{1 \leq i < j \leq n} \sum_{1 \leq k \leq d} c_{ij}^k (e_i-e_j)$$ where $c_{ij}^{k}$ are nonnegative integers. Let's call this $K^d(a+\rho)$.

As I will explain below, the values of $K^d(\beta)$ are piecewise polynomial, with domains of polynomiality being convex polyhedral cones. The line $a+\rho + k (e_1-e_n)$ will, for $k$ sufficiently large, eventually lie in just one cone, so it will be polynomial for $k$ sufficiently large. I'll discuss the switch between "eventually polynomial" and "polynomial" below.

The piecewise polynomiality of $K^d(\beta)$ is a special case of the following theorem of Blakley; I like the presentation of it in this paper of Sturmfels

Let $v_1$, $v_2$, ..., $v_N$ be a finite list of vectors in $\mathbb{Z}^M$, all lying in an open half space. Let $\Lambda$ be the lattice generated by the $v_i$. For $\lambda \in \Lambda$, let $c(\lambda)$ be the number of ways to write $\lambda$ as $\sum c_r \lambda_r$, with $c_r \in \mathbb{Z}_{\geq 0}$. Then can partition $\mathbb{R} \Lambda$ into finitely many convex polyhedral cones $K_j$, so that $c$ is a quasi-polynomial on $\Lambda \cap K_j$.

Here a function $\phi$ is called quasi-polynomial means that we can find a finite index sublattice $M$ of $\Lambda$ so that $\phi$ is polynomial on cosets of $\Lambda/M$. However, you don't have to worry about the word "quasi" because of a refinement of this result, also stated in Sturmfels' paper: Suppose that, whenever $v_{i_1}$, $v_{i_2}$, ..., $v_{i_D}$ is a $\mathbb{Q}$-basis for $\mathbb{Q} \Lambda$, then $v_{i_1}$, $v_{i_2}$, ..., $v_{i_D}$ is a $\mathbb{Z}$-basis for $\Lambda$. Then we get honest polynomials instead of quasi-polynomials. In this case, we say that the $v$'s are unimodular. The $v$'s in our setting are unimodular, so we get honest polynomials. See the top of page 305 in Sturmfels' paper for more on this.

For discussion of practical computation of these polynomials for the Kostant partition function, see de Loera and Sturmfels and de Loera's website.


For notational simplicity, set $\beta = a+\rho$, so $\sum \beta_i=0$. You asked for $K^d(\beta+k(e_1-e_n))$ to be polynomial, not just eventually polynomial, in $k$. In a dumb sense, this is false. If $\beta = (-10, 0,0,\cdots, 0, 10)$, then $K(\beta + k(e_i-e_j))$ is $0$ for $k < 10$, and then becomes $\geq 0$. However, I'm guessing you might have meant the following:

Suppose that $\beta$ is in the positive integer span of $e_i-e_j$ for $i<j$ (so that $x^{\beta-\rho}$ is an exponent in the Vandermonde inverse). Then $K^d(\beta+k (e_1-e_n))$ is polynomial in $k$ for $k \geq 0$.

I expect this is false for $n=4$ and $d=1$, with $\beta$ of the form $a (e_1-e_2) + b (e_2-e_3) + c (e_3-e_4)$, and $b \gg a > c$. Take a look at Figure 1 in the de Loera-Sturmfels paper, where they draw the chamber complex for $n=4$. I am describing a point in the region they label $3$. For $k \gg 0$, the point $\beta + k (e_1-e_4)$ will be in the region they label $4$. The polynomials on regions $3$ and $4$ are different. Now, different polynomials can become equal when restricted to a line, and I haven't had time to check that this doesn't happen in your case, but I know of no reason it should happen. I'll try to do some examples this evening.

UPDATE (algebra errors in the earlier counter-example now corrected) I now have a specific counterexample. Let $n=4$ and let $\beta = (0,m,-m,0)$. I get that (for $k$, $m \geq 0$) $$K(k,m,-m,-k) = (k+1)(k+2)(2k+3)/6 - \min(0, (k-m+1)(k-m)(k-m-1)/6).$$ So, for $m \geq 2$, this is not polynomial in $k \geq 0$.

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Ah, I had a typo in my question, but it was on another exponent, but I believe this will answer the question none the less. Thank you! –  Per Alexandersson Jul 4 '13 at 6:16
    
In your edited version, the exponent is $(a_1+k, a_2+k, a_3+k, \ldots, a_{n-1}+k, a_n-k)$? But then the total degree varies with $k$, whereas the total degree of any term in $\Delta^{-1}$ is $\binom{k}{2}$. Maybe you still have a typo? –  David Speyer Jul 4 '13 at 10:51
    
No, it is intentional, the total degree is supposed to vary; I mean, in the Laurent expansion you'll get arbitrary large degrees, right? I mean, $1/(1-x) = 1+x+x^2+\cdots$ has all degrees in the right hand side... –  Per Alexandersson Jul 4 '13 at 12:27
    
But $\Delta$ is homogenous of degree $\binom{k}{2}$ (unlike $1-x$). Look at $n=2$. Then $\Delta = x_2-x_1$, so $\Delta^{-1} = x_2^{-1} (1-x_1/x_2)^{-1} = \sum x_1^k x_2^{-k-1}$ which is homogenous of total degree $-1$. –  David Speyer Jul 4 '13 at 12:47
    
Right, silly me! I must have jumped over a step or two when I simplified my problem; but your answer at least gives some insight in how I can expect my computations to be. –  Per Alexandersson Jul 4 '13 at 14:19

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