A special-case identity for generalized Umemura polynomials

Counterexample

The identity is false already at $m=1$. Take

$$b_1=0,\qquad b_2=1,\qquad z=\frac34,\qquad w=\frac54.$$

These are valid nonsingular values for the source variables, since

$$w^2-z^2=\frac{25-9}{16}=1,$$

and $b_2,z,w$ are all nonzero.

First evaluate the ordinary Umemura polynomials. The source gives

$$U_0=U_1=1.$$

For $U_2$, formula (2.6) is a sum over the two subsets of $[1]=\{1\}$. The representation dimension for $\{1\}$ is $3$, while the empty-subset dimension is $1$. With $b_1=0$ and $b_2=1$, the coefficient factors are

$$c_1=1,\qquad d_1=1-4b_2^2=-3.$$

The source identifies the ordinary arguments with $z^2,w^2$, so

$$U_2=3c_1z^2+d_1w^2 =3\frac9{16}-3\frac{25}{16} =-3.$$

Now evaluate the generalized polynomial $U_{2,0}$. Its indexing set is

$$[2;0]=\{1,2\}.$$

To avoid confusing the generalized coefficient sequences with the external parameters $b_1,b_2$, call them $A_i,B_i$. Here

$$A_1=0,\qquad A_2=1,\qquad B_1=-4,\qquad B_2=-3.$$

Every subset containing $1$ therefore vanishes. Only the empty subset and $\{2\}$ remain. Since

$$d_{2,0}(\{2\}) = \left|\frac{2+1}{2-1}\right| =3,$$

the defining subset sum gives

$$\begin{aligned} U_{2,0} &=B_1B_2w^3+3A_2B_1z^2w\\ &=12w^3-12z^2w\\ &=12w(w^2-z^2)\\ &=15. \end{aligned}$$

Consequently, the two sides of the claimed identity are

$$U_2U_0-3^2U_1^2 =(-3)(1)-9(1)^2 =-12$$

and

$$\frac{U_{2,0}^2}{4b_2^2} = \frac{15^2}{4} = \frac{225}{4}.$$

Thus the displayed identity would require

$$-12=\frac{225}{4},$$

which is impossible. Therefore the conjecture is false as printed.

The same formula is repeated as Proposition 6 in the later paper below, so that proposition is also contradicted by the source definitions. This calculation does not identify the intended corrected formula.

Original source: https://arxiv.org/abs/math/0010279

Later repetition: https://arxiv.org/abs/math/0106025