4
$\begingroup$

Consider a binary shuffle algebra $\mathcal{W}$ of two letters $a, b$. As usual the concatination of two words $u = u_1 \dots u_m$, $v = v_1 \dots v_n$ is defined as: $$u \bullet v := u_1 \dots u_m v_1 \dots v_n$$ and the shuffle product is defined recursively as: $$(k \bullet u) * (l \bullet v) := k \bullet (u * (l \bullet v) ) + l \bullet ( (k \bullet u) * v)$$ where $k,l \in \{a,b\}$ are some letters... If necessary, the coproduct of $w \in \mathcal{W}$ is: $$\Delta (w) = \sum_{u \bullet v = w} u \otimes v.$$

An element $c \in \mathcal{W}$ is said to be in echalon form of weight $N$, if it is the concatination exponent of some linear combination of $a$ and $b$, i.e.: $$c = (\alpha a + \beta b)^N = \sum_{k=0}^N \alpha^{N-k} \beta^{k} (a^{N-k} * b^{k}).$$

The problem I have is to prove (relatively easy) that any shuffle element $a^m * b^n$ can be represented as a combination of echalon elements of wieght $m+n$ and to find a formula (the hard part) for this, e.g. something like: $$a^m * b^n = \sum_{k=1}^{m+n} \gamma_k^{(m,n)} (\alpha_k^{(m,n)} a + \beta_k^{(m,n)} b)^{m+n} \, ?$$ Examples: $$a*b = ab + ba = \frac{1}{2} \left( (a+b)^2 - (a-b)^2 \right),$$ $$a^2 * b = a^2b + aba + ba^2 = \frac{1}{2} \left( (a+b)^3 - (a-b)^3 - 2 b^3 \right)...$$

$\endgroup$
4
  • $\begingroup$ Your definition of the shuffle product looks incorrect - presumably you want some combination of $k$ and $l$ on the right, not $a$ and $b$? As you've written it, the shuffle product won't actually depend on its arguments at all... $\endgroup$ Commented Aug 12, 2015 at 19:32
  • $\begingroup$ Yep, sorry about that. I wrote the difinition at first with $a$ and $b$ as letters, but then realised that these are the primary letters of the algebra, so went to change them with $k$ and $l$ and it seems I hadn't changed them everywhere. Now it should be fine. $\endgroup$
    – Newbie
    Commented Aug 12, 2015 at 19:44
  • $\begingroup$ You can get an answer by inverting the Vandermonde matrix that appears in your binomial formula. It is another question how explicit this answer can get... Probably it will contain Stirling numbers of one or the other kind. What answer do you expect, or want, to get? $\endgroup$ Commented Aug 12, 2015 at 20:25
  • $\begingroup$ Anything, which can be written as an explicite formula... I have proven the existence recursivly and simply would like to illustrate it with something, which can be computed given the input $(m,n)$. Regarding what I expect - something like $\sum_{\alpha} c [ (\alpha a + b)^{m+n} \pm (\alpha a - b)^{m+n} ] \pm Ca^{m+n}$ /all explicit results that I got so far are of this symmetric form/. $\endgroup$
    – Newbie
    Commented Aug 12, 2015 at 20:50

1 Answer 1

0
$\begingroup$

I've found by applying the difference formula:

$$\Delta^n f(x) = \sum_{k=0}^n (-1)^{n-k} \binom{n}{k} f(x + k),$$

where $\Delta f(x) = f(x+1) - f(x)$, on $f(x) = x^{n+1}$:

$$ \frac{n}{2} + x = \sum_{k=0}^n \frac{(-1)^{n-k}}{(n+1)!} \binom{n}{k} (x + k)^{n+1}, \\ \Rightarrow a * b^n = \sum_{k=0}^n \frac{(-1)^{n-k}}{n!} \binom{n}{k} (a + k \, b)^{n+1} - \binom{n+1}{2} b^{n+1},$$

which provides a recursive step that solves the given problem.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .