## Friday, April 19, 2013

### Wronski, Vandermonde, and Moore!

This post is based on a recent letter by Matt Papanikolas outlining some results he has discovered whilst writing a (highly anticipated!) monograph on $L$-values in finite characteristic. In staring at Matt's letter, I realized that he allowed one to relate the big 3 matrices (Wronski, Vandermonde and Moore) in one simple formula which I will present below and then pose a related question.

I apologize if this is challenging to read; I am having a devil of a time getting my tex to compile correctly.

Matt's perspicacious insight was based on a paper by Felipe Voloch which was published in the Journal of Number Theory  in 1998. (Full disclosure: I had previously played with Voloch's paper and totally missed these ideas...)

Let me begin with some standard notation. As is now becoming standard, following Greg Anderson, let us put $A:= F$_q$[\theta]$. For each nonnegative integer $i$, let's put $A(i):= \{a\in A$ $\mid$ $\deg (a)\leq i\}$. Moreover, for such $i$, following Carlitz, we set $[i]:=\theta$^{q^i}-$\theta$. These are basic building blocks of $A$; for instance, one easily sees that $[i]$ is the product of all monic primes of degree dividing $i$. Various types of factorials are products of these elements....

Now recall that the Wronskian matrix is made up out of the derivatives of a function. Obviously in finite characteristic one can only differentiate a little bit before ending up with identically zero functions. As such we modify the Wronskian slightly here and use the "divided derivatives" $\partial$_i (i.e., those elements actually giving the coefficients of Taylor series etc.) in order to end up with nontrivial matrices.

Let $x$_0,..., $x$_i be i indeterminates. We define the Vandermonde matrix $V(i)(x$_0,..,$x$_i) as usual by having the $i$-th row consist of $(1$,$a$ ,..., $a$^i$)$, where $a=x$_{i-1}, and so on.

Finally of course the Moore matrix is made up out of the $q$^i powers of elements in a very similar
fashion to both the Wronski and Vandermonde matrices.

Let $g\in F[[\theta]]$. Notice that $g$^{q^i} is simply $g(\theta$^{q^i}) as the elements of $F$ are
fixed.  Thus, as Felipe observed, a bit of calculus (e.g., Taylor expansions) immediately gives $g$^{q^i}=$\sum$_j $\partial$_j(g) [i]^j.

Let $a\in A(i)$. Let M(i)=(a, a^q,..., a^{q^i})^t be the column vector consisting of a and its q^j-th powers and let W(i):=(a, $\partial a$, ..., $\partial$_ia)^t be the column vector consisting of the divided derivatives of a. Matt realized that Felipe's result implies
$$M(i)=V(i)([0],[1],...,[i])W(i).$$
Now simply choose a bunch more elements in A(i) and sling the above equations together. We immediately deduce an equality of matrices of the form M=VW where M is a Moore matrix, V is the Vandermonde and W the Wronskian. Thus taking determinants allows us, in this case, to calculate the Wronskian determinant in terms of the Moore and Vandermonde determinants.

It is very wild that one can relate the q-th power mapping, which is a field embedding, in this fashion with the divided derivatives which satisfy the Leibniz identity and I wonder if there is much more here.

In any case, the Moore determinant is intimately related to having elements linearly independent over F_q.  In classical theory, the Wronskian of analytic functions determines whether they are linearly independent (a result evidently due to Peano); see arXiv:1301.6598 for a modern approach. In characteristic p, the p-th power mapping clearly wreaks some havoc as usual (think of the Wronskian of x^p and x^{p^2}). But I wonder if there might not be some clear statement about linear dependence involving both the Wronskian and the p-th power mapping...?

(Updated 4-22-13: If you choose as your elements $a\in A(i)$ the set $\{1,\theta,...,\theta$^i}\$, you
obtain a Wronskian with determinant 1 thus giving an equality between the determinant of Matt's Vandermonde matrix and the determinant of the Moore matrix. As in Carlitz's original paper back in the 1930's, this Moore determinant is then easily computed to be a Carlitz factorial... It should also be possible to compute the determinant of Matt's matrix directly using the well-known formula for the Vandermonde determinant.)

(Updated 4-23-13: I thank Sangtae Jeong for pointing out the work of F.K. Schmidt, in 1939, and the work of Felipe Voloch and Arnaldo Garcia, in 1987, that goes very far with Wronskians in finite characteristic. See "Wronskians and linear independence in fields of prime characteristic", Manuscripta Math., 59, 1987, 457-469.)

(:1301.6598arXiv:1301.6598

#### 1 comment:

Felipe Voloch said...

Hi. Just saw this. Thanks! Matt will be here in a couple of weeks so I can ask him more.