Zoran Skoda exp(xF G)

Let $F = F(d/d x), G = G(d/d x)$ be polynomial differential operators. The commutator $[F,x] =: D(F)$ is the derivative of $F$ with respect to its argument, namely $d/d x$.

For convenience, let us introduce a parameter $\lambda$. We want to polarize the operator $exp(\lambda x F + G)$.

Let $A_{s,l}$, be a double sequence defined as follows: $A_{s,l} = 0$ unless $0\leq s\leq l$, $A_{s,s} = F^s$ ($s$-th power) and the basic recursion

$A_{s,l+1} = F (D(A_{s,l}) + A_{s-1,l}) + A_{s,l} G$

holds.

Theorem. $exp(\lambda (x F + G)) = \sum_{s=0}^\infty \frac{x^s \alpha^s}{s!} C$ where $C = \sum_{l = 0}^\infty \frac{\lambda^l A_{0,l}}{l!}$ and $\alpha = \frac{1}{C}\sum_{l = 0}^\infty \frac{\lambda^l A_{1,l}}{l!}$.

Sketch of the proof. We can write $(x F + G)^l = \sum_{s=0}^l x^s B_{s,l}$. Clearly, $B(s,s) = F^s$. Now

$(x F + G)^{l+1} = \sum_{s=0}^l x^s B_{s,l} (x F + G)$

Commuting $B_{s,l}$ and $x$ we get

$= \sum_{s=0}^l (x^{s+1} B_{s,l} F + x^s D (B_{s,l})+ x^s B_{s,l} G),$

and after the renaming of the indices a comparison shows that the recursion relation for $A$ is satisfied by $B$. Extending $B$ by zero to other values of $s,l$ (and recalling the initial conditions) we see that $A=B$ and

$exp(\lambda (x F + G)) = \sum_{l=0}^\infty \frac{\lambda^l}{l!} (x F + G)^l = \sum_{s,l=0}^\infty\frac{\lambda^l}{l!} x^s A_{s,l}$

It is clear that $C \alpha^s/s! = \sum_{l} \lambda^l \frac{A_{l,s}}{l!}$ stands along $x^s$ in the polarized form of
$exp(\lambda (x F + G))$. Then

$s (C\alpha) (C \alpha^{s-1}/s!) = C (C\alpha^s/s!)$

translates into

$s\sum_{r_1+r_2 = l} \frac{A_{1,r_1}}{r_1!}\frac{A_{s-1,r_2}}{r_2!} = \sum_{l_1+l_2 = l} \frac{A_{0,l_1}}{l_1!}\frac{A_{s,l_2}}{l_2!}$

If this is true for some $l$ let us prove it for $l+1$. As this is true identically, we can apply $D$ to the identity, multiply by $F$ and apply the Leibniz rule for $D$, obtaining

$s \sum_{r_1+r_2 = l} \frac{F D(A_{1,r_1}) A_{s-1,r_2} + F A_{1,r_1} D(A_{s-1,r_2})}{r_1! r_2!} =$
$= \sum_{l_1+l_2 = l} \frac{F D(A_{0.l_1}) A_{s,l_2}+ F A_{0.l_1} D(A_{s,l_2})}{l_1! l_2!}$

hence

$s\sum_{r_1+r_2 = l} \frac{(A_{1,r_1+1}- F A_{0,r_1} - A _{1,r_1} G) A_{s-1,r_2} + A_{1,r_1} (A_{s-1,r_2+1}- F A_{s-2,r_2} - A_{s-1,r_2} G)}{r_1! r_2!} =$
$= \sum_{l_1+l_2 = l} \frac{(A_{0.l_1+1} -A_{1.l_1}) A_{s,l_2}+ A_{0.l_1} (A_{s,l_2+1}- F A_{s-1,l_2} - A_{s,l_2} G)}{l_1! l_2!}$

By the induction hypothesis for $l$, all the terms containing $G$ clearly cancel. Similarly, the terms with $G$: $s F_{0,r_1} F_{s-1,r_2}$ splits into the sum of $(s-1)F_{0,r_1} F_{s-1,r_2}$ and $F_{0,r_1} F_{s-1,r_2}$; the first cancels with the corresponding term on the LHS by hypothesis and the other term with a term which is identical after renaming indices. The remaining terms multiply by $l!$. One is left with the identity

$s \sum_{r_1+r_2 = l} \frac{l!}{r_1! r_2!} (A_{1,r_1+1} A_{s-1,r_2} + A_{1,r_1} A_{s-1,r_2+1}) = \sum_{l_1+l_2 = l} \frac{l!}{l_1! l_2!} (A_{0,l_1+1} A_{s,l_2} + A_{0,l_1} A_{s,l_2+1})$

Rename the indices so that the two summands on each side have identical labels:

$s \sum_{r_1+r_2 = l+1} \left(\frac{l!}{(r_1-1)! r_2!} +\frac{l!}{r_1! (r_2-1)!}\right) A_{1,r_1} A_{s-1,r_2} = \sum_{l_1+l_2 = l+1} \left(\frac{l!}{(l_1-1)! l_2!} +\frac{l!}{l_1! (l_2-1)!}\right) A_{0,l_1} A_{s,l_2}$

hence by the Pascal triangle we get the identity we needed to prove.

Last revised on May 12, 2014 at 07:40:09. See the history of this page for a list of all contributions to it.