Local classification of singular hexagonal 3-webs with holomorphic Chern connection form and infinitesimal symmetries

Abstract

Implicit ODE, cubic in derivative, generically has no infinitesimal symmetries even at regular points with distinct roots. Cartan showed that at regular points, ODEs with hexagonal 3-web of solutions have symmetry algebras of the maximal possible dimension 3. At singular points such a web can lose all its symmetries. In this paper we study hexagonal 3-webs having at least one infinitesimal symmetry at singular points. In particular, we establish sufficient conditions for the existence of non-trivial symmetries and show that under natural assumptions such a symmetry is semi-simple, i.e. is a scaling in some coordinates. Using the obtained results, we provide a complete classification of hexagonal singular 3-web germs in the complex plane, satisfying the following two conditions: 1) the Chern connection form is holomorphic at the singular point, 2) the web admits at least one infinitesimal symmetry at this point. As a by-product, a classification of hexagonal weighted homogeneous 3-webs is obtained.

Appendices

Appendix A: solutions to $[12+2t^2+9tU]\frac{\hbox {d}U}{\hbox {d}t}=L(4+27U^2)$

First consider the case of a holomorphic solution $U$. By $T=3\sqrt{3} L\frac{f'(z)}{f(z)},$ $U=-\frac{2 }{3\sqrt{3}}\tan (z)$ we linearize the ODE. The general solution of the linear equation $f''-\frac{\tan (z)}{3L}f'+\frac{2}{9L^2}f=0$ can be expressed in terms of the Legendre functions. In fact, the substitution $f(z)=(1-x^2)^{-\frac{\mu }{2}}g(x)$, $x=\sin z$ transforms the above equation to the Legendre equation

$$\begin{aligned} (1-x^2)g''-2xg'+\left[ \nu (\nu +1)-\frac{\mu ^2}{1-x^2} \right] g=0 \end{aligned}$$

(34)

with $\mu =\frac{1}{2}(1-\frac{1}{3L})$, $\nu =\frac{1}{2}(\frac{1}{L}-1)$. Thus we obtain the normal form 1) with $U$ defined by (26).

It is easily seen that $U(T)$ is allowed to have the pole of order 1, which corresponds to $n_0\ge 1$, or of order 2, which corresponds to $n_0\ge 4$. The substitution $T=3\sqrt{3} L\frac{f'(z)}{f(z)},$ $U=\frac{2}{3\sqrt{3} \tan (z)},$ now brings the equation for $U$ to

$$\begin{aligned} f''+\frac{1}{3L\tan (z)}f'+\frac{2}{9L^2}f=0. \end{aligned}$$

(35)

The singular point $z=0$ is regular. A standard local analysis (see, for example, [15]) gives the following types of solutions.

• 1. If $\rho :=1-\frac{1}{3L}\ne {\mathbb {Z}}$ then $f(z)=c_1A(z)+c_2z^{\rho }B(z)$.

• 2. If $\rho =-n$, $n\in {\mathbb {N}}$ then $f(z)=c_1A(z)+c_2[z^{-n}B(z)+\lambda \ln z A(z)]$.

• 3. If $\rho =0$ then $f(z)=c_1A(z)+c_2[\kappa \ln z A(z)+z \psi (z)]$, where $\kappa \ne 0$ and $\psi$ is analytic.

• 4. If $\rho =n$, $n\in {\mathbb {N}}$ (note that $n>1$) then $f(z)=c_1z^nA(z)+c_2[C(z)+\lambda z^n \ln z B(z)]$.

In the above formulas, the functions $A,B,C$ are analytic and non-vanishing at $z\ne 0$, and $\lambda$ is constant. The function $U$ has pole of order 1 at $z=0$ iff $f(z)$ is analytic with $f(0)\ne 0$, $f'(0)=0$, $f''(0)\ne 0$. If there is an analytic non-vanishing at $z=0$ solution, then it automatically verifies $f'(0)=0$, $f''(0)\ne 0$. A solution of types 1, 2 or 3 suits iff $c_2=0$, thus giving (27), where we use $V=-\frac{1}{U}$. In fact, the substitution $f(z)=(1-x^2)^{\frac{\mu }{2}}g(x)$, $x=\cos z$ transforms Eq. (35) to the Legendre Eq. (34). For the solutions of the type 4, an analysis of series expansions of the functions $P^{\mu }_{\nu }(z), Q^{\mu }_{\nu }(z)$ at $z=1$ (see, for instance, [11]) shows that always holds true $\lambda =0$ for odd $n$, and $\lambda \ne 0$ for even $n$. Therefore a solution with desired properties exists only for odd $n$ and is of the form

$$\begin{aligned} f(z)=\sin ^{\mu } (z)[C_1P^{\mu }_{\nu }(\cos z)+ C_2Q^{\mu }_{\nu }(\cos z)] \end{aligned}$$

(36)

where $C_1,C_2$ are chosen to guarantee $c_2\ne 0$. By Lemma 6 all corresponding 3-webs are equivalent, thus one can choose $C_2=0$ and get (27).

The function $U$ has pole of order 2 at $z=0$ iff $f(z)$ is of the type 1 with $c_1\ne 0$, $c_2\ne 0$. Therefore $L= -\frac{2}{3}$ and the solution is of the form (36). Due to Lemma 6 all corresponding ODEs are equivalent and we can choose $C_1=C_2=1.$ This gives (28).

Appendix B: tables of invariants

Elliptic case

#	$\alpha ,\beta$	$\mu$	$[w_1:w_2]$	$\Delta$	$[\gamma ]$
1	$\scriptstyle 1-\dfrac{n_0}{2}, 1+\frac{m_0}{2}$	2	$\scriptstyle [m_0+2:2-n_0]$	$\scriptstyle x^{n_0}y^{m_0}$	$\frac{n_0}{2}\frac{\hbox {d}x}{x}$
1	$\scriptstyle 1+\dfrac{n_0}{2}, 1-\frac{m_0}{2}$	3	$\scriptstyle [2-m_0:n_0+2]$	$\scriptstyle x^{n_0}y^{m_0}$	$\frac{n_0}{2}\frac{\hbox {d}x}{x}+\frac{m_0\hbox {d}y}{y}$
2	$\scriptstyle 1+\dfrac{n_0}{3}, 1-\frac{m_0}{2}$	3	$\scriptstyle [2-m_0:n_0+3]$	$\scriptstyle x^{n_0}y^{m_0}(4x^{3+n_0}+27y^{2-m_0})$	$\frac{n_0}{3}\frac{\hbox {d}x}{x}+\frac{m_0\hbox {d}y}{y}+\frac{\hbox {d}\ln (4x^{3+n_0}+27y^{2-m_0})}{2}$
3	$\scriptstyle 1+\dfrac{n_0}{3}, 1-\frac{m_0}{2}$	3	$\scriptstyle [2-m_0:n_0+3]$	$\scriptstyle x^{n_0}y^{m_0}(32x^{3+n_0}+27y^{2-m_0})$	$\frac{n_0}{3}\frac{\hbox {d}x}{x}+\dfrac{m_0\hbox {d}y}{y}$
4	$\scriptstyle 1+\dfrac{n_0}{2}, 1$	3	$\scriptstyle [1:n_0+2]$	$\scriptstyle yx^{n_0}(12y+\lambda x^{2+n_0})$	$\frac{n_0}{2}\frac{\hbox {d}x}{x}+\frac{2}{3}\frac{\hbox {d}y}{y}+\dfrac{\hbox {d}\ln (12y+\lambda x^{2+n_0})}{2}$
5	$\scriptstyle 1+\dfrac{n_0}{2}, 1$	3	$\scriptstyle [1:n_0+2]$	$\scriptstyle yx^{n_0}(3y+\lambda x^{2+n_0})$	$\frac{n_0}{2}\frac{\hbox {d}x}{x}+\frac{2}{3}\frac{\hbox {d}y}{y}$
6	$\scriptstyle 1+\dfrac{n_0}{2}, 1$	3	$\scriptstyle [1:n_0+2]$	$\scriptstyle x(x^{2+n_0}-6\lambda y)(x^{2+n_0}-3\lambda y)$	$\frac{2n_0+1}{3}\frac{\hbox {d}x}{x}+\frac{\hbox {d}\ln (x^{2+n_0}-6\lambda y)}{2}$
7	$\scriptstyle 1+\dfrac{n_0}{2}, 1$	3	$\scriptstyle [1:n_0+2]$	$\scriptstyle x^{n_0}(x^{2+n_0} -12\lambda y)(x^{2+n_0}-3\lambda y)$	$\frac{n_0}{2}\frac{\hbox {d}x}{x}+\frac{\hbox {d}\ln (x^{2+n_0}-12\lambda y)}{2}$
8	$\scriptstyle 1+\dfrac{n_0}{2}, 1$	3	$\scriptstyle [1:n_0+2]$	$\scriptstyle x^{n_0}(x^{2+n_0} -3\lambda y)(2x^{2+n_0}+3\lambda y)$	$\frac{n_0}{2}\frac{\hbox {d}x}{x}$
9	$\scriptstyle 1+\dfrac{n_0}{2}, 1$	3	$\scriptstyle [1:n_0+2]$	$\scriptstyle x^{n_0}(x^{2+n_0} +6\lambda y)(x^{2+n_0}-3\lambda y)$	$\scriptstyle \frac{n_0}{2}\frac{\hbox {d}x}{x}\,\,+\,\,\frac{\hbox {d}\ln (x^{2+n_0}+6\lambda y)}{2}+{\scriptstyle \hbox {d}\ln (x^{2+n_0}-3\lambda y)}$
10	$\scriptstyle 1+\dfrac{n_0}{2}, 1$	3	$\scriptstyle [1:n_0+2]$	$\scriptstyle x^{n_0}(x^{2+n_0} +\lambda y)(x^{4+2n_0}+2\lambda x^{2+n_0}y+4\lambda ^2 y^2)$	$\frac{n_0}{2}\frac{\hbox {d}x}{x}\,\,+\,\,\frac{\hbox {d}\ln (x^{4+2n_0}+2\lambda x^{2+n_0}y+4\lambda ^2 y^2)}{2}$
11	$\scriptstyle 1\,\,+\,\,\dfrac{n_0}{2}, 1$	3	$\scriptstyle [1:n_0+2]$	$\scriptstyle x^{n_0}(x^{2+n_0} -\lambda y)(x^{2+n_0}\,\,+\,\,\lambda y)$	$\frac{n_0}{2}\frac{\hbox {d}x}{x}\,\,+\,\,\frac{\hbox {d}\ln (x^{2+n_0}+\lambda y)}{3}$
12	$\scriptstyle 1+\dfrac{n_0}{2}, 1$	3	$\scriptstyle [1:n_0+2]$	$\scriptstyle x^{n_0}(x^{2+n_0} +4\lambda y)(x^{2+n_0}+2\lambda y)$	$\scriptstyle \frac{n_0}{2}\frac{\hbox {d}x}{x}\,\,+\,\,\dfrac{\hbox {d}\ln (x^{2+n_0}+4\lambda y)}{2}\,\,+\,\,\frac{\hbox {d}\ln (x^{2+n_0}+2\lambda y)}{3}$
13	$\scriptstyle 1+\dfrac{n_0}{2}, 1$	3	$\scriptstyle [1:n_0+2]$	$\scriptstyle x^{n_0}(x^{2+n_0} -4\lambda y)(x^{4+2n_0}-2\lambda x^{2+n_0}y-2\lambda ^2 y^2)$	$\scriptstyle \frac{n_0}{2}\frac{\hbox {d}x}{x}\,\,+\,\,\frac{\hbox {d}\ln (x^{2+n_0}-4\lambda y)}{2}$
14	$\scriptstyle 1+\dfrac{n_0}{3}, 1$	3	$\scriptstyle [1:n_0+3]$	$\scriptstyle yx^{n_0}(x^{3+n_0} -27\lambda y)$	$\scriptstyle \frac{n_0}{3}\frac{\hbox {d}x}{x}\,\,+\,\,\frac{2}{3}\frac{\hbox {d}y}{y}+\hbox {d}\ln (x^{3+n_0}-27\lambda y)$
15	$\scriptstyle 1+\dfrac{n_0}{3}, 1$	3	$\scriptstyle [1:n_0+3]$	$\scriptstyle yx^{n_0}(2x^{3+n_0} +27\lambda y)(x^{3+n_0} \,\,+\,\,54\lambda y)$	$\scriptstyle \frac{n_0}{3}\frac{\hbox {d}x}{x}\,\,+\,\,\dfrac{2}{3}\frac{\hbox {d}y}{y}+\frac{\hbox {d}\ln (x^{3+n_0}+54\lambda y)}{2}$
16	$\scriptstyle 1+\dfrac{n_0}{3}, 1$	3	$\scriptstyle [1:n_0+3]$	$\scriptstyle yx^{n_0}(8x^{3+n_0} +27\lambda y)$	$\scriptstyle \frac{n_0}{3}\frac{\hbox {d}x}{x}\,\,+\,\,\frac{2}{3}\frac{\hbox {d}y}{y}$
17	$\scriptstyle 1+\dfrac{n_0}{3}, 1$	3	$\scriptstyle [1:n_0+3]$	$\scriptstyle x^{n_0}(2x^{3+n_0}- 9\lambda y)(2x^{3+n_0}-27 \lambda y)$	$\scriptstyle \frac{n_0}{3}\frac{\hbox {d}x}{x}\,\,+\,\,\dfrac{\hbox {d}\ln (2x^{3+n_0}-9\lambda y)}{2}+\frac{\hbox {d}\ln (2x^{3+n_0}-27\lambda y)}{2}$
18	$\scriptstyle 1+\dfrac{n_0}{3}, 1$	3	$\scriptstyle [1:n_0+3]$	$\scriptstyle x^{n_0}(25x^{3+n_0}- 18\lambda y)(25x^{3+n_0}+27 \lambda y)$	$\scriptstyle \frac{n_0}{3}\frac{\hbox {d}x}{x}\,\,+\,\,\dfrac{\hbox {d}\ln (25x^{3+n_0}-18\lambda y)}{2}$
19	$\scriptstyle 1+\dfrac{n_0}{3}, 1$	3	$\scriptstyle [1:n_0+3]$	$\scriptstyle x^{n_0}(x^{3+n_0}- 3\lambda y)(2x^{3+n_0}+3 \lambda y)$	$\scriptstyle \frac{n_0}{3}\frac{\hbox {d}x}{x}\,\,+\,\,\dfrac{\hbox {d}\ln (x^{3+n_0}-3\lambda y)}{3}$
20	$\scriptstyle 1+\dfrac{n_0}{3}, 1$	3	$\scriptstyle [1:n_0+3]$	$\scriptstyle x^{n_0}(4x^{3+n_0}- 75\lambda y)(4x^{3+n_0}+15\lambda y)(x^{3+n_0}-30\lambda y)$	$\scriptstyle \dfrac{n_0}{3}\frac{\hbox {d}x}{x}+\frac{\hbox {d}\ln (x^{3+n_0}-30\lambda y)}{2}+\dfrac{\hbox {d}\ln (4x^{3+n_0}-75\lambda y)}{3}$
21	$\scriptstyle 1+\dfrac{n_0}{3}, 1$	3	$\scriptstyle [1:n_0+3]$	$\scriptstyle x^{n_0}(x^{3+n_0}+27\lambda y)(x^{3+n_0}-9\lambda y)$	$\scriptstyle \frac{n_0}{3}\frac{\hbox {d}x}{x}+\dfrac{\hbox {d}\ln (x^{3+n_0}+27\lambda y)}{2}+\frac{3}{2}\hbox {d}\ln (x^{3+n_0}-9\lambda y)$
22	$\scriptstyle 1+\dfrac{n_0}{3}, 1$	3	$\scriptstyle [1:n_0+3]$	$\scriptstyle x^{n_0}(x^{3+n_0}-18\lambda y)(2x^{3+n_0}-27\lambda y)(4x^{3+n_0}+9\lambda y)$	$\scriptstyle \dfrac{n_0}{3}\frac{\hbox {d}x}{x}+\frac{\hbox {d}\ln (x^{3+n_0}-18\lambda y)}{2}$
23	$\scriptstyle 1+\dfrac{n_0}{3}, 1$	3	$\scriptstyle [1:n_0+3]$	$\scriptstyle x^{n_0}(8x^{3+n_0}-27\lambda y)(8x^{3+n_0}+9\lambda y)$	$\scriptstyle \frac{n_0}{3}\frac{\hbox {d}x}{x}$
24	$\scriptstyle 1+\dfrac{n_0}{3}, 1$	3	$\scriptstyle [1:n_0+3]$	$\scriptstyle x^{n_0}(x^{3+n_0}+9\lambda y)(x^{6+2n_0}+36\lambda x^{3+n_0}y+972\lambda ^2 y^2)$	$\scriptstyle \dfrac{n_0}{3}\frac{\hbox {d}x}{x}+\frac{\hbox {d}\ln (x^{6+2n_0}+36\lambda x^{3+n_0}y+972\lambda ^2 y^2)}{2}$
25	$\scriptstyle 1+\dfrac{n_0}{3}, 1$	3	$\scriptstyle [1:n_0+3]$	$\scriptstyle x^{n_0}(x^{3+n_0}-216\lambda y)(x^{3+n_0}-144\lambda y)$	$\scriptstyle \frac{n_0}{3}\frac{\hbox {d}x}{x}+\dfrac{3}{2}\hbox {d}\ln (x^{3+n_0}-144\lambda y)$
26	$\scriptstyle 1+\dfrac{n_0}{3}, 1$	3	$\scriptstyle [1:n_0+3]$	$\scriptstyle x^{n_0}(25x^{3+n_0}+432\lambda y)(25x^{3+n_0}+72\lambda y)$	$\scriptstyle \frac{n_0}{3}\frac{\hbox {d}x}{x}+\dfrac{\hbox {d}\ln (25x^{3+n_0}+432\lambda y)}{2}$

Parabolic case

#	$\mu$	$[w_1:w_2]$	$\Delta$	$[\gamma ]$	$[i^3:j^2]$
1	3	$[0:1]$	$yx^N=0$	$\frac{N}{3}\frac{\hbox {d}x}{x} + 2\frac{\hbox {d}y}{y}$	$[0:1]$
2	3	$[0:1]$	$yx^N=0$	$\frac{N}{2}\frac{\hbox {d}x}{x}+\left( 2-\frac{L_0(N+2)}{2\sqrt{3}}\right) \frac{\hbox {d}y}{y}$	$[1:\frac{\tan ^2(L_1)}{-27}]$
3	3	$[0:1]$	$yx^N=0$	$\frac{N}{3}\frac{\hbox {d}x}{x} + \left( 2-\frac{L_0(N+3)}{2\sqrt{3}}\right) \frac{\hbox {d}y}{y}$	$[0:1]$
4	2	$[0:1]$	$x^2y^{N}=0$	$\frac{\hbox {d}x}{x}$

Hyperbolic case

#	$\mu$	$[w_1:w_2]$	$\Delta$	$[\gamma ]$	$[i^3:j^2]$
1	3	$[-(m_0+1):n_0+2]$	$yx^{n_0}=0$	$\frac{n_0}{2}\frac{\hbox {d}x}{x}+\left\{ 2+\left( \frac{L}{2}+1\right) (1+m_0)\right\} \frac{\hbox {d}y}{y}$	$[1:\frac{U^2(0)}{-4}]$
2	3	$[-(m_0+1):n_0+3]$	$yx^{n_0}=0$	$\frac{n_0}{3}\frac{\hbox {d}x}{x}+\left\{ 2+\left( \frac{L}{2}+\frac{5}{6}\right) (1+m_0)\right\} \frac{\hbox {d}y}{y}$	$[0:1]$
2.2	3	$[-(m_0+1):n_0+3]$	$yx^{n_0-3}=0$	$\frac{n_0-3}{6}\frac{\hbox {d}x}{x}+\frac{m_0+7}{3}\frac{\hbox {d}y}{y}$	$[0:1]$
3	2	$[-(m_0+2):n_0+1]$	$xy^{m_0}=0$	$\frac{n_0+3}{2}\frac{\hbox {d}x}{x}$
4	2	$[-(m_0+1):n_0+1]$	$xy=0$	$\left( 1+\frac{(n_0+1)(l_0+3)}{2}\right) \frac{\hbox {d}x}{x}+(1+l_0)(1+m_0)\frac{\hbox {d}y}{y}$

Comments on the hyperbolic case: for $L=0$ one has $U\equiv 0$, the case 2.2 corresponds to $L=-\frac{2}{3}$.

In the table for elliptic case we add the exponents $\alpha ,\beta$ generating normal forms from the corresponding ”basic” equation of Theorem 6. Some comments:

• 1) If $n_0$ or $m_0$ comes with the negative sign in the formulas for $\alpha$ or $\beta$ then $n_0\le 1$ or $m_0\le 1$ respectively.

• 2) $\lambda$ is a non-vanishing constant (its value can be easily computed).

• 3) The invariant $[i^3:j^2]$ is used only once to distinguish between the forms 18) and 26).