13
$\begingroup$

The following Wronskian identity can be proved by expanding both sides and checking that two sides are the same. But how to prove it more elegantly?

Let $u_1(x), u_2(x), u_3(x), u_4(x)$ be four functions. Define q-shift Wronskian as follows: $$W(u_1, u_2, u_3)(x)=\det \begin{bmatrix} u_1(x) & u_1(xq^{-2}) & u_1(xq^{-4}) \\ u_2(x) & u_2(xq^{-2}) & u_2(xq^{-4}) \\ u_3(x) & u_3(xq^{-2}) & u_3(xq^{-4}) \end{bmatrix}.$$

Similarly for $W(u_1, u_2)(x)$ and $W(u_1, u_2, u_3, u_4)(x)$. Then we have a Wronskian identity:

$$ W(W(u_1, u_3, u_4)(x), W(u_2, u_3, u_4)(x))(x) = W(u_1, u_2, u_3, u_4)(x) \cdot W(u_3, u_4)(xq^{-2}). $$ Thank you very much.

Edit: The general version of the identity is the following. Let $W_s(i)=W(u_1, \ldots, \hat{u_i}, \ldots, u_{s+1})$, where $\hat{u_i}$ means without $u_i$. Given functions $u_1, \ldots, u_{s+1}$.

$$ W_{k+1}(W_s(s-1)(x), W_s(s-2)(x), \ldots, W_{s}(s-k-1)(x))(x) = \\ \left(\prod_{j=1}^{k} W_{s+1}(u_1, \ldots, u_{s+1})(xq^{-2(j-1)})\right) \cdot W_{s-k}(u_{k+2}, \ldots, u_{s+1})(xq^{-2k}). $$

$\endgroup$
8
  • $\begingroup$ Missing a $\det$. Is there a more general version of the formula that you know about? $\endgroup$
    – anon
    Jan 29, 2012 at 22:36
  • $\begingroup$ @anon, thank you. det has been added. Yes, there is a more general version of the formula. $\endgroup$
    – LJR
    Jan 30, 2012 at 1:26
  • $\begingroup$ user, it might be easier to see a route to proving this if you posted the general form. Also, I'd like to see it, out of curiosity.. $\endgroup$
    – anon
    Jan 30, 2012 at 1:28
  • $\begingroup$ Cool, thanks.$\text{ }$ $\endgroup$
    – anon
    Jan 30, 2012 at 18:40
  • 1
    $\begingroup$ Excuse me, my mathematics is a bit rusty. Why is W called a Wronskian, when no differentiation is involved? $\endgroup$
    – user1551
    Nov 15, 2012 at 6:42

1 Answer 1

1
$\begingroup$

Disclaimer: This is not an answer...just an idea which doesn't fit in a comment.

I can't come up with a nice trick to get a slick proof, but here's a suggestion for organizing the data you're working with.

Let $u_{ij} = u_i(xq^{-2(j-1)})$. Then $u_{21} = u_2(x)$ and $u_{23} = u_2(xq^{-4})$ etc. Then

$$W(u_1,u_2,\dots,u_n)(x) = \mathrm{det} \begin{bmatrix} u_1(x) & u_1(xq^{-2}) & \cdots & u_1(xq^{-2n-2}) \\ u_2(x) & u_2(xq^{-2}) & \cdots & u_2(xq^{-2n-2}) \\ \vdots & \vdots & \vdots & \vdots \\ u_n(x) & u_n(xq^{-2}) & \cdots & u_n(xq^{-2n-2}) \end{bmatrix} $$ $$ = \mathrm{det} \begin{bmatrix} u_{11} & u_{12} & \cdots & u_{1n} \\ u_{21} & u_{22} & \cdots & u_{2n} \\ \vdots & \vdots & \vdots & \vdots \\ u_{n1} & u_{n2} & \cdots & u_{nn} \end{bmatrix} = W(u_{11},u_{21},u_{31},u_{41})$$

Notice that, for example, $u_{23}(x) = u_{21}(xq^{-4}) = u_{22}(xq^{-2})$. So $W(u_3,u_4)(xq^{-2})=$ $W(u_{31},u_{41})(xq^{-2})=$ $W(u_{32},u_{42})(x)$

Again, I know this doesn't solve the problem, but maybe reorganizing the data this way will yield a reasonable proof.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.