Geometric Scattering Monodromy

Abstract

In this paper we give geometric conditions so that the integral mapping of a Liouville integrable Hamiltonian system with a focus-focus equilibrium point has scattering monodromy. Using a complex version of the Morse lemma, we show that scattering monodromy is the same as the scattering monodromy of the standard focus-focus system.

Appendix

In this appendix we prove a complex version of the Morse lemma. Our proof was inspired by the proof of the focus-focus Morse lemma in [6].

Lemma A1

(Morse lemma). Let

$$\begin{aligned} \widehat{{\mathcal {H}}}: [0,2] \times {\mathbb {C}}^2 \rightarrow \mathbb {C}: (t,z) \mapsto {{\mathcal {H}}}_t(z) = Q(z) + t R(z), \end{aligned}$$

(27)

where Q is a nondegenerate homogeneous quadratic polynomial and R is a smooth function, which is flat to second order at the origin (0, 0). Then there is an open neighborhood U of (0, 0) in ${\mathbb {C}}^2$ and a diffeomorphism $\Phi$ of U into itself with $\Phi (0,0) = (0,0)$ such that ${\Phi }^{*}{{\mathcal {H}}}_1 = Q$ on U. Moreover, $\Phi$ is isotopic to ${\textrm{id}}_U$.

Proof

By a complex linear change of coordinates we may assume that $Q(z) = z_1z_2$. We want to find a time dependent vector field $X = X_t + \frac{\partial }{\partial t}$ on $(0,2) \times {\mathbb {C}}^2$ whose flow ${\varphi }^X_t$ satisfies

$$\begin{aligned} ({\varphi }^X_t)^{*}{{\mathcal {H}}}_t = Q, \, \, \, \text{ for } \text{ every } t \in [0,1]. \end{aligned}$$

(28)

Differentiating (28) gives $0 = ({\varphi }^X_t)^{*}\big ( \frac{\partial \widehat{{\mathcal {H}}} }{\partial t} + L_{X_t}{{\mathcal {H}}}_t \big )$. Since $\frac{\partial {\mathcal {H}}}{\partial t} = R$, we need to find a vector field $X_t$ on ${\mathbb {C}}^2$ such that

$$\begin{aligned} \mathop {\! \,\textrm{d} \!}\nolimits {{\mathcal {H}}}_t(z) X_t(z) = - R(z) \, \, \, \text{ for } \text{ all } t \in [0,1]. \end{aligned}$$

(29)

Now $\mathop {\! \,\textrm{d} \!}\nolimits H_t(z) = \big ( z_2 + t \frac{\partial R}{\partial z_1}(z) \big ) \mathop {\! \,\textrm{d} \!}\nolimits z_1 + \big ( z_1 + t \frac{\partial R}{\partial z_2}(z) \big ) \mathop {\! \,\textrm{d} \!}\nolimits z_2$. For some smooth functions A and B on $(0,2) \times {\mathbb {C}}^2$

$$\begin{aligned} X_t(z) = A(t,z) \frac{\partial }{\partial z_1} + B(t,z) \frac{\partial }{\partial z_2}. \end{aligned}$$

(30)

Since $R(0) =0$, by the integral form of Taylor’s theorem $R(z) = G_1(z) z_1 + G_2(z) z_2$, where $G_j$ are smooth functions with $G_j(0) = \frac{\partial R}{\partial z_j}(0)$ for $j=1,2$. Since R is flat to second order at 0, we get $G_j(0) =0$. Thus (29) can be written as

$$\begin{aligned} -G_1(z) z_1 - G_2(z) z_2&= A(t,z) \left( z_2 + t\frac{\partial R}{\partial z_1}(z) \right) + B(t,z) \left( z_1 + \frac{\partial R}{\partial z_2}(z) \right) . \end{aligned}$$

(31)

Again by Taylor’s theorem, $\frac{\partial R}{\partial z_j}(z) = F_j(z) z_1 + E_j(z) z_2$ for $j =1,2$, where $F_j(0) = \frac{{\partial }^2R}{\partial z_1 \, \partial z_j}(0)$ and $E_j(0) = \frac{{\partial }^2R}{\partial z_2 \, \partial z_j}(0)$. Since R is flat to second order at 0, it follows that $F_j(0) =0$ and $E_j(0) =0$. Thus equation (31) becomes

$$\begin{aligned} -G_1(z) z_1 - G_2(z) z_2&= A(t,z) \big ( z_2 +t(F_1(z) z_1 + E_1(z) z_2) \big ) \\&\quad + B(t,z) \big ( z_1 + t (F_2(z)z_1 + E_2(z) z_2) \big ) . \end{aligned}$$

Equating the coefficients of $z_1$ and $z_2$ in the equation above, we get

$$\begin{aligned} -{\begin{pmatrix} G_1(z) \\ G_2(z) \end{pmatrix}} = {\begin{pmatrix} t F_1(z) &{} \quad 1+t F_2(z) \\ 1+t E_1(z) &{} \quad tE_2(z) \end{pmatrix} \, \begin{pmatrix} A(t,z) \\ B(t,z) \end{pmatrix} } = {\mathcal {A}}(t, z){\begin{pmatrix} A(t,z) \\ B(t,z) \end{pmatrix}.} \end{aligned}$$

So

$$\begin{aligned} |\det {\mathcal {A}}(t,z)|&= |1 +t\big ( E_1(z) + F_2(z) \big ) + t^2 \big ( E_1(z)F_2(z)-E_2(z)F_2(z) \big ) | \\&\ge 1 - |t| \, |\big ( E_1(z) + F_2(z) \big ) + t^2 \big ( E_1(z)F_2(z)-E_2(z)F_2(z) \big ) | . \end{aligned}$$

Let U be an open neighborhood of $0 \in {\mathbb {C}}^2$ such that for $i=1,2$

$$\begin{aligned} |E_i(z)|< \frac{{\scriptstyle 1}}{{\scriptstyle 16}} \, \, \, \textrm{and} \, \, \, |F_i(z)| < \frac{{\scriptstyle 1}}{{\scriptstyle 16}}. \end{aligned}$$

(32)

Then

$$\begin{aligned}&|t| \, |\big ( E_1(z) + F_2(z) \big ) + t^2 \big ( E_1(z)F_2(z)-E_2(z)F_2(z) \big ) | \\&\quad \le |t| \big ( |E_1(z)| + |F_2(z)| + |t| \big [ |E_1(z)| \, |F_2(z)| + |E_2(z)|\, |F_1(z)| \big ] \big ) \\&\quad < 2\left[ \frac{{\scriptstyle 1}}{{\scriptstyle 16}} + \frac{{\scriptstyle 1}}{{\scriptstyle 16}} + 2\left( \frac{{\scriptstyle 1}}{{\scriptstyle 16}}\cdot \frac{{\scriptstyle 1}}{{\scriptstyle 16}} + \frac{{\scriptstyle 1}}{{\scriptstyle 16}}\cdot \frac{{\scriptstyle 1}}{{\scriptstyle 16}}\right) \right] , \, \, \text{ using } (31)\hbox { and }t\in [0,2] \\&\quad = \frac{{\scriptstyle 17}}{{\scriptstyle 64}} . \end{aligned}$$

Thus the matrix ${\mathcal {A}}(t,z)$ is invertible for all $z \in U$ and all $t \in [0,2]$. With $\begin{pmatrix} A(t,z) \\ B(t,z) \end{pmatrix}$$= -{{\mathcal {A}}(t,z)}^{-1}$ $\begin{pmatrix} G_1(z) \\ G_2(z) \end{pmatrix}$ we have determined the vector field $X_t$ (30) on $(0,2) \times {\mathbb {C}}^2$ which solves equation (29).

Because $X_t(0,0) = (0,0)$ we can shrink U if necessary so that the flow ${\varphi }^X_t$ of the vector field X sends $[0,1] \times U$ to U. Set $\Phi = {\varphi }^X_1$. Then ${\Phi }^{*}{{\mathcal {H}}}_1 =({\varphi }^X_1)^{*}{{\mathcal {H}}}_t = Q$ on U. The diffeomorphism $\Phi$ is isotopic to ${\textrm{id}}_U$, since it is the time 1 map of a flow. $\square$