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The first Painlevé equation $$P_I:y''=6y^2-x $$ has the symmetries $$x \mapsto \omega x \\y \mapsto \omega^3 y$$ for any fifth root of unity $\omega$. At the same time, the near-infinity asymptotics of $P_I$ involve the five rays $$\Gamma_k= \left\{x:\arg x= \frac{2 \pi i k}{5} \right\}, \quad k=0,1,2,3,4.$$

Looking at the second Painlevé equation $$P_{II}: y''=2y^3+xy+\alpha $$ the scaling symmetries are $$x \mapsto \omega x \\ y \mapsto \omega^2 x $$ for any third root of unity $\omega$. However, the near-infinity asymptotics of $P_{II}$ actually involve six rays $$\Gamma_k= \left\{x:\arg x= \frac{2 \pi i k}{6} \right\}, \quad k=0,1,2,3,4,5.$$ My question is: is there a direct relationship between the discrete symmetries of the second Painlevé equation, and its near-infinity asymptotics, as in the case of the first Painlevé equation? If so, how come I only found three symmetries, which should correspond to the rays $$\left\{x: \arg x=\frac{2 \pi i k}{3} \right\}, \quad k=0,1,2 $$

Thanks!

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Yes, there are relationships, they depend on the weights chosen to compactify the system at infinity. First, write the equation as the companion autonomous vector field in $\mathbb C^3$ $$X(x,y_1,y_2)=\partial_x+y_2\partial_{y_1}-6z_1^2\partial_{z_2}$$ For $P_1$, you use the Boutroux compactification given by the chart transition mapping $(t,u_1,u_2)\mapsto(x,y_1,y_2)$ with $$\begin{cases}{x=t^{-4/5}}\\{y_1=(u_1+\frac{i}{\sqrt6})t^{-2/5}}\\{y_2=u_2t^{-3/5}}\end{cases}$$The new vector field, after division by $-\frac{5}{4t^{1/5}}$, writes $$W(t,u_1,u_2)=t^2\partial_t+(\cdots)\partial_{u_1}+(\cdots)\partial_{u_2}$$which has an irregular singularity at $(0,0,0)$ with Poincaré rank-$1$.

The lines $\Gamma_k$ are the preimages in the starting chart of the Stokes line of the previous vector field. There are five of them because of the order-$5$ ramification in the $x$-variable.

In the case of $P_2$ after compactification and reduction, you obtain two irregular singularities at infinity (Poincaré rank-$1$), each one bringing a Stokes line in. Hence the $6=2\times3$ lines in the original chart. (I need to check that precisely, though, I'll update the answer).

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