A special-case identity for generalized Umemura polynomials

About 26 years old · traced to 2000

Let $U_m=U_m(t;b_1,b_2)$ be the generalized Umemura polynomials, and let $U_{2,m-1}$ denote the corresponding polynomial appearing in the construction. In the specialization $b_1=0$ , write

U_m:=U_m(0,b_2),\qquad U_{2,m-1}:=U_{2,m-1}(0,b_2).

Special-case Umemura-polynomial identity. If $b_1=0$ , then

U_{m+1}U_{m-1}-(2m+1)^2U_m^2=\frac{1}{4b_2^2}U_{2,m-1}^2.

This identity gives a factorization of the denominator occurring in the expression for the Painlevé VI solution $q_m$ and relates generalized Umemura polynomials to a special case of Umemura's polynomial. The supplied text does not indicate whether the assertion has been proved or remains open.

References

Primary source

Anatol N. Kirillov and Makoto Taneda, “Generalized Umemura polynomials”, arXiv:math/0010279 (2000).

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

A reader has supplied an unverified numerical example that would disprove the identity at its first nontrivial case, so the claim is not settled.

The identity is printed in the original paper and repeated as Proposition 6 in a later paper, but no named proposer or independent proof is identified in the retrieved material.

Posted attempt

A posted attempt claims a complete disproof at $m=1$ , using $b_1=0$ , $b_2=1$ , $z=3/4$ , and $w=5/4$ . It reports $U_2U_0-9U_1^2=-12$ but $U_{2,0}^2/(4b_2^2)=225/4$ ; the calculation has not been independently verified.

Current status (as of August 2026): The printed identity has an unverified claimed counterexample at $m=1$ ; without independent checking, its validity remains unsettled.

Sources

arxiv.org

CounterexampleThis solution needs a summarySee full solutionHide full solution

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