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Update: see below

Let $M$ be an $n\times n$ matrix that's constructed as follows. Construct the right-most column of $M$ as $[\alpha_1(x_1),\cdots,\alpha_n(x_n)]^T$ for some class of fixed functions $\alpha_i(x)$. Now, moving left, each column of $M$ is the derivative of the last. So the next column is $[\alpha_1'(x_1),\cdots,\alpha'_n(x_n)]^T$, and the very first column is $[\alpha_1^{(n-1)}(x_1),\cdots,\alpha^{(n-1)}_n(x_n)]^T$. I'm curious if there's a name for such matrices. Specifically, I'm wondering about the determinant of such matrices: $G(x_1,\cdots,x_n)=\det(M(x_1,\cdots,x_n))$. As Jose rightfully pointed out when all variables are set equal we get the usual Wronskian.

I'm particularly curious about $\alpha_i(x)= x^{d_i}/(d_i)!$ for some decreasing positive integer sequence $d_i$. A similar example would be when $M_{ij}=\frac{x_i^{d_i+j-i}}{(d_i+j-i)!}$, with the understanding that $M_{ij}$ is zero once we are taking negative factorials. This is starting to look like an offset Vandermonde matrix, or a confluent Vandermonde. The determinant isn't quite a Schur polynomial but perhaps a known generalization? Is there a connection of these determinants to some well studied class of polynomials (it seems unlikely they will be symmetric)?

Note that when written as $M_{ij}=\frac{x_i^{d_i+j-i}}{(d_i+j-i)!}$, we have that if $|d|=\sum_{i=1}^nd_i$, then when $x_1=x_2=\cdots=x_n=1$, $|d|!\det(M)$ is equal to the number of standard Young tableau of shape $(d_1,d_2,\cdots,d_n)$.

Alternatively, is there a way to write $\det(M)=C\prod_{i=1}^dx_{i}^{d_i}\prod_{i\neq j}(x_i-x_j)^{a_ib_j}?$ (or possibly more complicated). Unfortunately numerical evidence suggests that fixing a variable, say $x_1$ and thinking of the determinant as a polynomial in $x_1$ gives wildly complex expressions for its roots. So the $(x_i-x_j)$ factors aren't quite right either. I've seen "generalized Vandermondes" here but unfortunately it looks like the structure in the link is simpler.

As an example, take $n=4$, and $d_i=9-2i$. Then the resulting matrix is:

\begin{pmatrix} \frac{x_1^4}{4!} & \frac{x_1^5}{5!} & \frac{x_1^6}{6!} & \frac{x_1^7}{7!}\\ \frac{x_2^2}{2!} & \frac{x_2^3}{3!} & \frac{x_2^4}{4!} & \frac{x_2^5}{5!}\\ 1 & x_3 & \frac{x_3^2}{2!} & \frac{x_3^3}{3!}\\ 0 & 0 & 1 & x_4 \end{pmatrix}

The determinant equals:

$$\frac{1}{302400}x_1^4 x_2^2 \left(10 x_1^3 x_2-30 x_1^3 x_3-70 x_1^2 x_2 x_4+210 x_1^2 x_3 x_4-21 x_1 x_2^3+105 x_1 x_2^2 x_4+210 x_1 x_3^3-630 x_1 x_3^2 x_4+105 x_2^3 x_3-525 x_2^2 x_3 x_4-350 x_2 x_3^3+1050 x_2 x_3^2 x_4\right)$$

and it looks like if we single out $x_1$ to make a polynomial in $x_1$, we get complex roots involving $x_2,x_3,x_4$. So, if there is a way to write the determinant as the sum over something nice (not involving signs of permutations), I would also be very interested in see that.

Context: These kinds of determinants come up when counting Young Tableau via integrals over sub-simplices, (specifically (4.1),(4.2)). It's a surprisingly powerful way for calculating the number of non-standard Young tableau shapes. I'm just wondering if there's a neater (representation-theoretic?) interpretation of where these determinants arise from.

Edit:

Consider the matrix $\mathcal{M}=\left[\frac{t_i^{x+j-2}}{(x+j-2)!}\right]_{i,j=1^d}$. It's determinant is:

$$\mbox{det}(\mathcal{M})=\frac{1}{\prod_{j=1}^n(x+j-2)}\prod_{i=1}^d t_i^{x-1} \prod_{1\leq i< j\leq d}(t_j-t_i),$$

where we recognize the vandermonde determinant. Then my matrix $M$, and it's determinant can be calculated from this by taking an appropriate amount of derivatives of $t_i$ in each row.

So I think this does have something to do with Schur functions afterall. The Jacobi Trudi identity states that:

$s_{\lambda}=\det(h_{\lambda_i+j-i})_{i,j=1}^d.$

The exponential specialization of $h_k$ to parameter $t$ gives $t^k/k!$. With $t=1$, one gets that the exponential specialization the Schur polynomial above gives the identity:

$$f^{\lambda}=\det\left[\frac{1}{(\lambda_i+j-i)!}\right]_{i,j=1}^d.$$

So the question is, is it sensible to generalize exponential specialization to get an identity of the above form for $t_1,t_2,\cdots,t_d$ instead of just $t_1=\cdots=t_n=t=1$?

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    $\begingroup$ If all $x_i$ are equal, then this is the wronskian of the functions $\alpha_i$. (en.wikipedia.org/wiki/Wronskian) $\endgroup$ – José Figueroa-O'Farrill Sep 15 '14 at 22:26
  • $\begingroup$ Thanks for pointing that out! I've added a small edit mentioning this and am now wondering if there's a Schubert calculus interpretation of such determinants. $\endgroup$ – Alex R. Sep 16 '14 at 2:03
  • $\begingroup$ Another comment, disguised as an answer for the usual reasons. If all the functions coincide and are power functions, then one essentially gets the vandermonde determinants, or slight variants thereof. Thus this concept is a common generalization of the wronskian and the vandermonde determinant, a fact which might give a hint towards a natural interpretation. $\endgroup$ – holloway Sep 16 '14 at 7:48
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Your general determinant has to do with the Wronskian isomorphism for $SL_2$ representations namely a map $$ \wedge^m({\rm Sym}^n(\mathbb{C}^2))\rightarrow {\rm Sym}^m({\rm Sym}^{n-m+1}(\mathbb{C}^2)))\ . $$ If you follow it by the natural map $$ {\rm Sym}^m({\rm Sym}^{n-m+1}(\mathbb{C}^2))\rightarrow {\rm Sym}^{m(n-m+1)}(\mathbb{C}^2) $$ which corresponds to restricting to the diagonal, then you get the usual Wronskian or rather a homogenized version of it where polynomials in one variable are seen as binary forms. You can find more details in this paper and that one.

Great question by the way and thank you for pointing out the article by Ping Sun.

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  • $\begingroup$ Thanks for your answer! It's funny I was adding an edit as you just submitted your answer. I was hoping to clarify something. In the section under edit, my matrix $M$ becomes derivatives of $\mathcal{M}$ in my notation above. From briefly perusing the second link, it looks like I might be able to write the determinant of $M$ in a nice way using Schur polynomials? Or it looks like in 1.4, we get an exact product formula with a Vandermonde term? $\endgroup$ – Alex R. Oct 19 '14 at 21:39
  • $\begingroup$ 1.4 of your second linked paper, that is. $\endgroup$ – Alex R. Oct 19 '14 at 21:46
  • $\begingroup$ the formula in 1.4 is still about the specialization to the diagonal so I don't know the answer to the question in your first comment. $\endgroup$ – Abdelmalek Abdesselam Oct 22 '14 at 20:36
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Couple of quick observations for $\alpha_i(x)=x^{(d_i)}$ ($x^{(d)}=x^d/d!$ as usual).

Note that if $d_i>d_{i+1}$, for all $i$, then $G(x_1,\ldots,x_n)=0$ unless $d_i-d_{i+1}=1$. In particular, if the sequence $\mathbf{d}=(d_i)_{i=1}^n$ is strictly decreasing, then $G\neq 0$ only if $\mathbf{d}=(n-1,\ldots,0)$. In this case $G=1$.

Also, note that if $\mathbf{d}$ is constant, then $G=0$ unless $\mathbf{d}=(n+k-1,n+k-1,\ldots,n+k-1)$, in which case $$G=((n-1)!)^{-n}\prod_{i<j}(x_i-x_j)=((n-1)!)^{-n}(x_1\cdots x_n)^kV(x_1,\ldots,x_n)$$ (the rescaling from the Vandermonde is constant down columns/along rows).

Okay, write $\mathbf{d}=(d_1^{k_1},d_2^{k_2},\ldots)$ where $d_1>d_2>\cdots$ (the $k$'s in the exponent denote multipicity).

Write $G_\mathbf{d}(x_1,\ldots,x_n)$ for the determinant you would get for the associated sequence. Note that if $d_1-d_2>k_1$, then $G_\mathbf{b}=0$ and if $d_1-d_2=k_1$ Then $$G_\mathbf{d}=((k_1-1)!)^{-k_1}V(x_1,\ldots,x_{k_1})G_{\mathbf{e}}(x_{k_1+1},\ldots,x_n)$$ where $\mathbf{e}=(d_2^{k_2},d_3^{k_3},\ldots)$.

This leaves the hard case where $d_1-d_2=b_1<k_1$. This is more difficult to describe, so I will have to think more.

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  • $\begingroup$ Thanks for your answer! I added an example determinant calculating in the body. It unfortunately looks like the $G_\mathbf{e}$ is quite involved, perhaps being a sum of Vandermondes or something similar. $\endgroup$ – Alex R. Oct 8 '14 at 16:10
  • $\begingroup$ @Alex R. I seem to have added the assumption that $d_1=n-1$ in my answer. As you point out, it is much more complicated if this is not the case. I think that if $d_1=n+k-1$, then something like I wrote holds if $d_1-d_2=k_1+k$ (with $G=0$ for $>k_1+k)$. $\endgroup$ – David Hill Oct 8 '14 at 21:29
  • $\begingroup$ are you specifically interested in equation (4.2) in the link? This seems much more tractable than the general problem. $\endgroup$ – David Hill Oct 8 '14 at 21:36
  • $\begingroup$ 4.1 is really the one that I'm after. I agree that 4.2 is much easier. $\endgroup$ – Alex R. Oct 8 '14 at 21:46
  • $\begingroup$ Thanks for the pointers. I've added an edit where I think I've worked out the structure in terms of Vandermonde's. It's basically going to be in terms of a sum of derivatives of Vandermondes. $\endgroup$ – Alex R. Oct 19 '14 at 21:14
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You could consider it as the Wronskian of translates of the functions. That is, if $(\tau_x f)(t) = f(x + t)$, then $G(x_1, \ldots, x_n) = W(\tau_{x_1} f_1, \ldots, \tau_{x_n} f_n)(0)$.

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  • $\begingroup$ Well, yes, but the more-or-less original use of the Wronskian relies on it being restricted to the diagonal. $\endgroup$ – Jesse C. McKeown Oct 6 '14 at 18:53
  • $\begingroup$ Does this Wronskian of translates come up anywhere else? $\endgroup$ – Alex R. Oct 6 '14 at 21:20

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