Dear all,

Consider the $(n+1)\times (n+1)$ matrix $A$ with indeterminates $X_i, Y_i$, $0\leq i\leq n$ such that the $(i,j)$-th entry is given by $X_i^jY_i^{n-j}$. The $i$-th row is $(X_i^n,X_i^{n-1}Y_i,\dots,Y_i^n)$ that can be thought as given by the Veronese embedding $\mathbb{P}^1\to\mathbb{P}^n$. The determinant of $A$ is multi-homogeneous and hence defines a variety in $\prod_{i=0}^n\mathbb{P}^1$. It can be checked that $\det(A)$ has the following nice factorization $$ \det(A)=\prod_{0\leq i<j\leq n}\det\left(\begin{smallmatrix}X_j & Y_j\\X_i & Y_i\end{smallmatrix}\right) $$ which gives the decomposition of the corresponding variety. Actually, setting $Y_i=1$, $0\leq i\leq n$, the matrix $A$ simply becomes the Vandermonde matrix and the above factorization becomes $\det(A)=\prod_{0\leq i<j\leq n}(X_j-X_i)$, as is well known.

We may also consider its higher dimensional generalization: for $n,d\geq 1$, let $N={n+d\choose d}$. Consider the $N\times N$ matrix $A$ whose $i$-th row is given by the $n$-uple Veronese embedding $\mathbb{P}^d\to\mathbb{P}^N$ that sents $(X_{i,0},\dots,X_{i,d})$ to the $N$-tuple consisting of all monomials in $X_{i,0},\dots,X_{i,d}$ with total degree $n$.

Question: for $d>1$, does $\det(A)$ admit similar factorization? Or is it simply irreducible?

Setting $X_{i,0}=1$ for all $i$ gives the higher dimensional analogue of the Vandermonde matrix which arises naturally in, for example, the problem of multivariate interpolation. For the multivariate case, I have never seen a simple explicit form of $\det(A)$ as the univariate one, though. So I guess it must be complicated. And my understanding in this case is very limited. I tried to calculate the determinant in a naive way (by mimicing the calculation in the 1-dimension case) but the amount of calculation blows up quickly. But maybe there are structures that I didn't recognize which help simplifying the calculation. So I am also looking for any suggestion for better calculation methods (grobner basis, tools from algebraic combinatorics, representation theory, etc).

Thanks a lot!

  • 3
    $\begingroup$ This paper: arxiv.org/abs/1405.0993 considers higher dimensional analogues of Vandermonde determinants. $\endgroup$
    – pavel
    Aug 7 '18 at 9:03

Looking at the same issue i found this paper : D’Andrea, Carlos, and Luis Tabera. "Tropicalization and irreducibility of generalized Vandermonde determinants." Proceedings of the American Mathematical Society 137.11 (2009): 3647-3656.‏ It's a bit above my algebra skills for now but maybe it could help.


Viewing $X_{i,d}$ as an $N \times d+1$ matrix, $det(A)$ can be written as a polynomial in the determinants of the $d+1\times d+1$ minors. This is just because it only depends on $X_{i,d}$ up to right-multiplication by $SL_{d+1}$, and the invariants of that action are generated by the determinants of the minors. (This is just the system of Plucker coordinates of the Grassmanian.)

But it is not the product of all those determinants, because that would have too high a degree! There are $\left(\begin{array}{c}N \\ d+1 \end{array}\right)$ such minors, and each is a polynomial of degree $d+1$. The determinant is a polynomial of degree $nN$. As long as $d\geq 2$ and $n\geq 2$, the first is larger.

Indeed, while a linear relation among $n+1$ of the rows of $X$ does force the corresponding $n+1$ rows of $A$ to lie in a linear subspace, that linear subspace has dimension larger than $N$.

I don't see any reason for there to be a diffrent sort of factorization.

Suppose there were another type of factorization (into irreducible factors, or otherwise a canonical factorization). Then $S_N$ would act on the factors. It would divide them into orbits of size $1, 2, N, \left(\begin{array}{c}N \\ 2 \end{array}\right),\dots$. For large $N$, The only viable orbits are of size $1$, $2$, and $N$. (I'm not completely sure of this group-theoretic fact.). Orbits of size $1$ and $2$ are uninteresting, so $N$ is the only interesting one. Thus we would need factors of size $n$, or something smaller than $n$, corresponding to rows of $X$. Furthermore since $SL_{d+1}$ is a connected group, the factors themselves must be $SL_{d+1}$ - invariants. It's pretty clear that no such factors exist.

  • $\begingroup$ do you know if there is any paper in the literature which tries to find explicit formulas in terms of these SL invariants? $\endgroup$ Feb 2 '14 at 22:13

My guess is, no. The reason is that there is a positive answer to the following slightly different question, which I blieve is the right one [at least, it is the right one for my project...]

Given a matrix $X = (X_{i,j})$, where $i < m$ and $j \leq n$, can one express the product of the minors of $X$ of order $n+1$ as a single determinant in which the columns correspond to monomials of some fixed degree $d$ in $n+1$ variables?

If $m = \binom{n+d}{n}$ and you apply the Veronese embedding to $X$ this does not work. What does work is to take $m = n+d$, let $Y$ be the matrix for minors of $X$ of order $n$, and apply the Veronese embedding to $Y$. You get a square matrix of order $\binom{n+d}{n}$ whose determinant is the product of the minors of order $n+1$ to the power $n$.

See http://math.univ-lyon1.fr/~begnac/articles/MVVD.pdf


FWIW, for $n=d=2$, this polynomial is irreducible, as I just checked in Magma. The naive code for this (which even the online calculator http://magma.maths.usyd.edu.au/calc/ can handle) is

S<x1,x2,x3,x4,x5,x6,y1,y2,y3,y4,y5,y6,z1,z2,z3,z4,z5,z6> := PolynomialRing(RationalField(),18);
X:=Matrix(S,[[x1^2,y1^2,z1^2,x1*y1,x1*z1,y1*z1], [x2^2,y2^2,z2^2,x2*y2,x2*z2,y2*z2], [x3^2,y3^2,z3^2,x3*y3,x3*z3,y3*z3], [x4^2,y4^2,z4^2,x4*y4,x4*z4,y4*z4], [x5^2,y5^2,z5^2,x5*y5,x5*z5,y5*z5], [x6^2,y6^2,z6^2,x6*y6,x6*z6,y6*z6]]);


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