New Integral Representations for the Maslov Canonical Operator on an Isotropic Manifold with a Complex Germ

Abstract

The objective of our paper is to generalize the representation obtained in 2017 by S.Yu. Dobrokhotov, V.E. Nazaikinskii, and A.I. Shafarevich to the general case of the canonical operator on an isotropic manifold with complex germ.

A. Stationary phase method in the complex case

Let us present the stationary phase method for a complex phase function (see [35, Th. 7.7.12]).

Theorem 3.

Let $S(x, \psi)$ be a complex-valued function of class $C^\infty$ in some neighborhood of the point $(0,0)$ in $\mathbb{R}^{n+m}$ satisfying the conditions

$$\operatorname{Im}S\geq 0, \qquad \operatorname{Im}S(0,0) =0, \qquad S_\psi(0,0)=0, \qquad \det S_{\psi \psi}(0,0)\neq 0,$$

and let $a(x, \psi)\in C_0^\infty(K)$ , where $K$ is a small neighborhood of $(0,0)$ in $\mathbb{R}^{n+m}$ . Then there is a $C>0$ such that

$$ \Big|\frac{1}{(2\pi h)^{m/2}}\int a \exp(iS/h) d\psi - \frac{a^0 \exp(i S^0/h) }{\sqrt{(\det (-iS_{\psi \psi}))^0}}\Big| \leq C h,$$

(A.1)

where, for functions $F(x, \psi)$ , the symbol $F^0(x)$ stands for an arbitrary function of a single $x$ belonging to the same residue class modulo the ideal generated by the functions $S_{\psi_j}$ , $j=1,\ldots, m$ . For $x=0,$ the square root is defined by the rule described below, and, for small $x\neq 0$ , by continuity.

For real positive definite matrices $A$ we set $\sqrt{\det A} > 0$. According to this condition, a continuous branch $\sqrt{\det A}$ is chosen on the closure of the open convex set of all symmetric $n\times n$-matrices $A$ with positive definite real part $\operatorname{Re}A = (A+A^*)/2$. Note that, for a symmetric matrix $A$ with positive definite real part, the specified definition of $\sqrt{\det A}$ gives

$$\int_{\mathbb{R}^n} \exp(-x^T Ax ) dx = \frac{\pi^{n/2}}{\sqrt{\det A}}.$$

B. Proof of Proposition 5

The function $S_I$, up to an insignificant permutation of the subscripts of the variables $\Psi$, is equal to the function $(S_{I_1})_{I_2\ldots}$ obtained by th successive application of finitely many transformations $S \mapsto S_I$ for $|I|=1$. Therefore, it suffices to prove the assertion in the case of $|I|=1$. Let $I=(1)$ for definiteness. We write $x = (x_1, x')$, $p =(p_1, p')$. Compute the derivatives of $S_I$:

$$(S_I)_x = (-\psi', S_{x'})|_{x_1 = \psi'}, \qquad (S_I)_\Psi = (S_{\psi}, S_{x_1})|_{x_1 = \psi'} - (0, x_1),$$

(B.1)

$$d^2 S_I = \begin{pmatrix} (S_I)_{xx} & (S_I)_{x \Psi} \\ (S_I)_{\Psi x} & (S_I)_{\Psi \Psi} \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 & -1 \\ 0 & S_{x' x'} & S_{x' \psi} & S_{x' x_1} \\ 0 & S_{\psi x'} & S_{\psi \psi} & S_{\psi x_1} \\ -1 & S_{x_1 x'} & S_{x_1 \psi} & S_{x_1 x_1} \end{pmatrix}\Bigg|_{x_1 = \psi'}.$$

(B.2)

Introduce the mapping

$$\begin{aligned} \, g\colon \mathbb{R}^{n+m+1} & \to \mathbb{R}^{n+m} \\ (x_1, x', \psi, \psi') & \mapsto (\psi', x', \psi). \end{aligned}$$

The function $S_I = g^*S - x_1\psi'$ is treated as an element of $C^\infty(g^{-1}(U))$. Since $\operatorname{Im} S_I = \operatorname{Im} g^*S$, it follows that condition (A1) for $S$ implies the validity of condition (A1) for $S_I$. It is clear that $\gamma_{S_I} = g^{-1}(\gamma_S)$. As is well known, the preimage of the manifold under the submersion $g$ is a submanifold, and $\operatorname{codim} \gamma_{S_I} = \operatorname{codim} \gamma_S$. We have proved (A2) for $S_I$. It can be seen from (B.2) that, for $(x, \Psi) \in \gamma_{S_I}$ (and hence, for $g(x, \Psi) \in \gamma_S$), the equality $\operatorname{rank} \operatorname{Im} d^2S_I(x, \Psi) = \operatorname{rank} \operatorname{Im} d^2S(g(x, \Psi))$ holds. By property (A3) for the function $S$, the rank on the right-hand side of the equality is equal to the codimensions of $\gamma_S$ and $\gamma_{S_I}$. This implies the validity of (A3) for $S_I$.

It is clear that the following equality holds for the set $C_{S_I}$ of zeros of the functions $\operatorname{Im}S_I$, $(S_I)_\Psi$:

$$ C_{S_I} = \{(x, \Psi) \in g^{-1}(U) \mid g(x, \Psi) \in C_S, \ x_1 = g^*S_{x_1} \}.$$

(B.3)

This is the graph of the smooth dependence of $x_1$, on the points $(\psi', x', \psi)$ of a smooth submanifold $C_S$ in the domain $U$, expressed by the formula $x_1 = g^*S_{x_1}$, considered in this case with the coordinates $(\psi', x', \psi)$ instead of $(x_1, x', \psi)$. Thu,s $C_{S_I}$ is a submanifold of $g^{-1}(U)$, and $g$ realizes a diffeomorphism between $C_{S_I}$ and $C_S$. The equality $\gamma_{S_I} = g^{-1}(\gamma_S)$ implies the following assertion.

$$ (\delta x, \delta \Psi) \in T_{(x, \Psi)} \gamma_{S_I} \iff [dg(x, \Psi)] (\delta x, \delta \Psi) \in T_{g(x, \Psi)} \gamma_S.$$

(B.4)

Equality (B.3) implies that

$$(\delta x, \delta \Psi) \in T_{(x, \Psi)} C_{S_I} \iff [dg(x, \Psi)] (\delta x, \delta \Psi) \in T_{g(x, \Psi)} C_S\qquad\qquad\qquad\qquad$$

(B.5)

$$\qquad \qquad\qquad\qquad\qquad\qquad\text{and} \ \delta x_1 = (g^*S_{x_1 x'} \delta x' + g^*S_{x_1 \psi} \delta \psi + g^*S_{x_1 x_1} \delta \psi').$$

(B.6)

The families $R_S$ and $R_{S_I}$ of planes are defined as the kernels of the matrices $(S_{\psi x}, S_{\psi \psi})$ and $((S_I)_{\Psi x}, (S_I)_{\Psi \Psi})$. Formula (B.2) implies the assertion which holds for every $(x, \Psi) \in C_{S_I}$:

$$(\delta x, \delta \Psi) \in R_{S_I}(x, \Psi) \iff [dg(x, \Psi)] (\delta x, \delta \Psi) \in R_S(g(x, \Psi)) \qquad\qquad\qquad\qquad $$

(B.7)

$$ \qquad \qquad\qquad\qquad\qquad\qquad\text{and} \ \delta x_1 = (g^*S_{x_1 x'} \delta x' + g^*S_{x_1 \psi} \delta \psi + g^*S_{x_1 x_1} \delta \psi'),$$

(B.8)

where the values of the matrix functions $g^*S_{x_1 x'}$, $g^*S_{x_1 \psi}$, and $g^*S_{x_1 x_1}$ are taken at the point $(x, \Psi)$. We have assumed that $S$ has equivalent pairs of properties (C1)–(C2) and (D1)–(D2). Then, since $C_{S_I}$ is a submanifold, it follows from assertions (B.4), (B.5), (B.7) that (D1) holds for $S_I$. The fact that the rank of $((S_I)_{\Psi x}, (S_I)_{\Psi \Psi})$ is full for $(x, \Psi) \in C_{S_I}$ follows from (B.2) and from the fact that the rank of $(S_{\psi x}, S_{\psi \psi})$ is full for $(x, \psi) \in C_S$. Thus, $S_I$ satisfies (D2). We see that conditions (D1)–(D2) and the equivalent conditions (C1)–(C2) hold for the function $S_I$.

The linear mapping $[dg(x, \Psi)](\delta x, \delta \Psi) = (\delta \psi', \delta x', \delta \psi)$ takes $R_{S_I}(x, \Psi)$ bijectively onto $R_S(g(x, \psi))$. The mapping and its differential $(g, dg)$ implement an isomorphism of the bundles $(C_{S_I}, R_{S_I})$ and $(C_S, R_S)$.

Consider the mappings

$$\rho(x, \psi) = (x, S_x(x, \psi)), \qquad \rho_I(x, \Psi) = (x, (S_I)_x(x, \Psi)),$$

which (see the diagram (6.6)) implement, respectively, the isomorphism of the bundle $(C_S, R_S)$ with the bundle $(\Lambda, r)$ and an isomorphism of the bundle $(C_{S_I}, R_{S_I})$ with the bundle $(\Lambda_I, r_I)$. We claim that the mapping $J^{-1}_I(x_1, x', p_1, p') = (-p_1, x', x_1, p')$ implements an isomorphism of the bundles $(\Lambda_{S_I}, r_{S_I})$ and $(\Lambda_S, r_S)$, and the compositions of the bundle isomorphisms coincide, as shown in the diagramFootnote

The operations of restriction of the domains of the mappings onto the corresponding bundles are not shown to simplify the notation.

(B.9)

The operations of restriction of the domains of the mappings onto the corresponding bundles are not shown to simplify the notation.

The chain of equalities

$$\begin{aligned} \, J^{-1}_I \circ \rho_I(x, \psi) & = J^{-1}_I(x_1, x', -\psi', g^*S_{x'}(x, \Psi)) = (\psi', x', x_1, g^*S_{x'}(x, \Psi)) \\ &= (\psi', x', g^* S_{x_1}(x, \Psi), g^*S_{x'}(x, \Psi)) = \rho \circ g (x, \Psi) \end{aligned}$$

for $(x, \Psi)\in C_{S_I}$ (and hence, $x_1 = g^* S_{x_1}(x, \Psi)$) shows that $\rho \circ g|_{C_{S_I}} = J^{-1}_I \circ \rho_I|_{C_{S_I}}$. Now we present the following calculations at the point $(x, \Psi)\in C_{S_I}$ and $(\delta x, \delta \Psi) \in R_{S_I}(x, \Psi)$ (and hence $x_1 = g^* S_{x_1}(x, \Psi)$ and $\delta x_1 = g^*S_{x_1 x'} \delta x' + g^*S_{x_1 \psi} \delta \psi + g^*S_{x_1 x_1} \delta \psi'$):

$$\begin{aligned} \, & [dJ^{-1}_I(\rho_I(x, \Psi)) \circ d\rho_I(x, \Psi)](\delta x, \delta \Psi) \\ & \quad = [dJ^{-1}_I(\rho_I(x, \Psi))](\delta x_1, \delta x', -\delta \psi', g^* S_{x'x'}\delta x' + g^* S_{x'\psi} \delta \psi + g^* S_{x' x_1} \delta \psi') \\ & \quad = (\delta \psi', \delta x', g^*S_{x_1 x'} \delta x' + g^*S_{x_1 \psi} \delta \psi + g^*S_{x_1 x_1} \delta \psi', g^* S_{x'x'}\delta x' + g^* S_{x'\psi} \delta \psi + g^* S_{x' x_1} \delta \psi'). \end{aligned}$$

On the other hand,

$$\begin{gathered} \, [d\rho(g(x, \Psi)) \circ dg(x, \Psi)](\delta x, \delta \Psi) = [d\rho(g(x, \Psi))](\delta \psi', \delta x', \delta \psi) \\= (\delta \psi', \delta x', g^*S_{x_1 x'} \delta x' + g^*S_{x_1 \psi} \delta \psi + g^*S_{x_1 x_1} \delta \psi', g^* S_{x'x'}\delta x' + g^* S_{x'\psi} \delta \psi + g^* S_{x' x_1} \delta \psi'). \end{gathered}$$

The equality of these expressions means that the equality shown on diagram (B.9) holds,

$$(\rho, d\rho) \circ (g, dg)|_{(C_{S_I}, R_{S_I})} = (J^{-1}_I, dJ^{-1}_I) \circ (\rho_I, d\rho_I)|_{(C_{S_I}, R_{S_I})}.$$

We have proved the commutativity of the mappings on the diagram (B.9). The mapping $(J^{-1}_I, dJ^{-1}_I)$ defines an isomorphism of bundles, which follows from the commutativity of the mappings on the diagram and from the fact that the other three maps on it are bundle isomorphisms.

It follows from the diagram (B.9) and from the definitions of $A_{(S, a)}$ and $A_{(S_I, a_I)}$ that $A_{(S_I, a_I)}(\lambda) = A_{(S, a)}(J^{-1}_I \lambda)$ for $\lambda \in \Lambda_{S_I}$. It remains to prove the assertion for the measures. By analogy with the definition of $d\sigma$, we can assign to the function $S_I$ (not uniquely) an $n$-form $d\sigma_I$. By definition, $d\sigma$ and $d\sigma_I$ must satisfy the equalities

$$d\sigma \wedge d(-S_{\psi}) = dx \wedge d\psi, \qquad d\sigma_I \wedge d(-(S_I)_\Psi) = dx \wedge d\Psi.$$

It is clear that $d(-(S_I)_\Psi) = [g^*d(-S_{\psi})]\wedge d(-g^*S_{x_1}+x_1)$. Let us show that the form $d\sigma_I = -g^*d\sigma$ satisfies the definition. Indeed,

$$\begin{aligned} \, -[g^*d\sigma]\wedge d(-S_I)_\Psi & = -[g^*(d\sigma \wedge d(-S_\psi))]\wedge d(-g^*S_{x_1}+x_1) \\&= -[g^*(dx\wedge d\psi)]\wedge d(-g^*S_{x_1}+x_1) \\&= -d\psi'\wedge dx' \wedge d\psi \wedge d(-g^*S_{x_1}+x_1) \\&= -d\psi'\wedge dx' \wedge d\psi\wedge dx_1 \\&= dx\wedge d\Psi. \end{aligned}$$

Here the validity of the penultimate equality follows from the fact that $(g^*S_{x_1})_{x_1} = 0$. The measures $d\mu_S$ and $d\mu_{S_I}$ are defined as restrictions of $d\sigma$ and $d\sigma_I$ to $(C_S, R_S)$ and $(C_{S_I}, R_{S_I})$, respectively, with the subsequent transfer, by $(\rho, d\rho)$ and $(\rho_I, d\rho_I)$, to $(\Lambda_S, r_S)$ and $(\Lambda_{S_I}, r_{S_I})$ respectively. Here we established that the mapping $(g, dg)$ connects the forms $d\sigma_I$ and $-d\sigma$. According to the diagram (B.9), we have $d\mu_{S_I}(\lambda) = -(J^{-1}_I)^* d\mu_S(J^{-1}_I \lambda)$ for $\lambda \in \Lambda_{S_I}$. This completes the proof of the proposition.

C. Proof of Proposition 7

Lemma 9.

In some neighborhood $U'$ , part $(1)$ of Proposition $7$ holds.

Proof.

By Theorem 7.5.7 of [35], in some neighborhood $U'$ of the point $(x_0, \psi_0)$, for every function $f(x, \psi)\in C^\infty(U')$, there is a function $f^0(x) \in C^\infty(\pi_x(U'))$ such that the representation $f^0 = f+S_\psi q_0$ is possible for some smooth $m$-dimensional function $q_0(x, \psi)$. The neighborhood $U'$ can be chosen to be independent of $f$ (see the discussion before Theorem 7.5.7 in [35]). Thus, the existence of the desired objects $S^0$, $(\det S_{\psi \psi})^0$, and $a^0$ is proved with the exception that the function $a^0$ existing by the cited theorem need not necessarily have a compact support in $\pi_x(U')$. However, the possibility of choosing $a^0$ with compact support is proved by multiplying the equality $a^0=a+S_\psi q_2$ by some smooth function $g(x)$, with compact support, which is equal to $1$ in the neighborhood $\pi_x(\operatorname{supp} a)$, and taking $g a^0$ instead of $a^0$ and $g q_2$ instead $q_2$.

The following lemma implies part (2) of Proposition 7.

Lemma 10.

Let $(\widetilde{x}, \widetilde{\psi}) \in C_S \cap U'$ . Then

• 1. $S^0(\widetilde{x}) = S(\widetilde{x}, \widetilde{\psi})$ , $a^0(\widetilde{x}) = a(\widetilde{x}, \widetilde{\psi})$ , $(\det (-iS_{\psi \psi}))^0(\widetilde{x}) = \det (-iS_{\psi \psi})(\widetilde{x}, \widetilde{\psi})$ .

• 2. $q(\widetilde{x}, \widetilde{\psi}) =0$ .

• 3. $(S^0)_x(\widetilde{x}) = S_x(\widetilde{x}, \widetilde{\psi})$ .

• 4. $(S^0)_{xx}(\widetilde{x}) = [S_{xx} - S_{x\psi}S_{\psi \psi}^{-1} S_{\psi x}](\widetilde{x}, \widetilde{\psi})$ .

Proof.

• 1. Substitute $(\widetilde{x}, \widetilde{\psi})$ to (7.5), (7.6) and use the equality $S_\psi(\widetilde{x}, \widetilde{\psi}) =0$, which follows from the definition of $C_S$.

• 2. Take the derivative with respect to $\psi$ of both sides of the equality $S^0=S+S_\psi q$ and substitute $(\widetilde{x}, \widetilde{\psi})$. This gives $0 = [q^T S_{\psi \psi}](\widetilde{x}, \widetilde{\psi})$. Use the invertibility of $S_{\psi \psi}(\widetilde{x}, \widetilde{\psi})$.

• 3. Take the derivative with respect to $x$ of both sides of the equality $S^0=S+S_\psi q$ and substitute $(\widetilde{x}, \widetilde{\psi})$. This point obviously follows from the equalities $S_\psi(\widetilde{x}, \widetilde{\psi}) =q(\widetilde{x}, \widetilde{\psi}) =0$.

• 4. Take the derivative with respect to $x\psi$ of both parts of the equality $S^0=S+S_\psi q$, multiply by $S_{\psi \psi}^{-1}S_{\psi x}$ from the right, and substitute $(\widetilde{x}, \widetilde{\psi})$:

  $$0 = [S_{x \psi} S_{\psi \psi}^{-1} S_{\psi x} + q_x^T S_{\psi x} + S_{x \psi} q_\psi S_{\psi \psi}^{-1} S_{\psi x}](\widetilde{x}, \widetilde{\psi}).$$

  Let us sum this equality with its transposition:

  $$ 0 = [S_{x \psi} (2 S_{\psi \psi}^{-1} + q_\psi S_{\psi \psi}^{-1} + S_{\psi \psi}^{-1} q_\psi^T) S_{\psi x} + q_x^T S_{\psi x} + S_{x \psi}q_x](\widetilde{x}, \widetilde{\psi}).$$

  (C.1)

  Take the derivative with respect to $\psi \psi$ of both sides of the equality $S^0=S+S_\psi q$, multiply from the left and right by $S_{\psi \psi}^{-1}$, and substitute $(\widetilde{x}, \widetilde{\psi})$:

  $$ 0 = [S_{\psi \psi}^{-1} + q_\psi S_{\psi \psi}^{-1} + S_{\psi \psi}^{-1} q_\psi^T](\widetilde{x}, \widetilde{\psi}).$$

  (C.2)

  Take the derivative with respect to $x x$ of both sides of the equality $S^0=S+S_\psi q$, and substitute $(\widetilde{x}, \widetilde{\psi})$:

  $$ [(S^0)_{xx}](\widetilde{x}) = [S_{xx} + q_x^T S_{\psi x} + S_{x \psi} q_x](\widetilde{x}, \widetilde{\psi}).$$

  (C.3)

  The equality at this part of the lemma is obtained by subtracting (C.1) from (C.3) and summing with equality (C.2) multiplied from the left by $S_{x\psi}$ and from the right by $S_{\psi x}$.

It is clear that parts (1), (2) of Proposition 7 hold also for all possible neighborhoods of $(x_0, \psi_0)$ contained in $U'$.

Lemma 11.

Reducing the neighborhood $U'$ if necessary, we can achieve the validity of part $(3)$ of Proposition $7.$

Proof.

The restriction $\pi_x|_{C_S}$ is an immersion, and hence an embedding in a sufficiently small neighborhood $U'$ of the point $(x_0, \psi_0)$. Indeed, at $(x, \psi) \in C_S$, by property (D1), we have $T^\mathbb{C}_{(x, \psi)} C_S \in R_S(x, \psi)$, and $R_S(x, \psi)$, for $(x,\psi)$ close to $(x_0, \psi_0)$, has the form of a graph over the $x$-plane (see formula (7.2)). Consider the submanifold $\gamma = \pi_x(C_S \cap U')$.

Let us show that conditions (B1)–(B2) hold for $S^0$ and $\gamma$. Indeed, for $(x, \psi) \in C_S \cap U'$, the equalities $S^0(x) = S(x, \psi)$ and $(S^0)_x(x) = S_x(x, \psi)$ hold, whose imaginary components of the right-hand sides vanish by the definition of $C_S$. We claim now that, for $S^0$ and $\gamma$, condition (B3) holds. Indeed, for $(x, \psi) \in C_S \cap U'$, we have $(S^0)_{xx}(x) = [S_{xx} - S_{x\psi}S_{\psi \psi}^{-1} S_{\psi x}](x, \psi)$. On the other hand, by (7.3), for $(x, \psi) \in C_S \cap U'$, we have the following expression for the fiber of the complex germ:

$$r_S(\rho(x, \psi)) = \{(\delta x, \delta p_x) = (\delta x, [(S^0)_{xx}(x)]\delta x) \mid \delta x \in \mathbb{C}^n_x\}.$$

It follows from the properties of the complex germ that $\operatorname{rank}\operatorname{Im} (S^0)_{xx}(x) = n-\dim \Lambda_S$. This number is equal to $\operatorname{codim}\gamma$, because $\Lambda_S$, $C_S$, and $\gamma$ have the same dimensions. Thus, the validity of condition (B3) is proved. It follows from the validity of conditions (B1)–(B3) that, when choosing a sufficiently small neighborhood $U'$, conditions (A1)–(A3) are satisfied for the function $S^0$, and $\gamma = \gamma_{S^0} \equiv \{ x\in \pi_x(U') \mid \operatorname{Im} S^0(x) = 0\}$. Conditions (C1)–(C2) for the $\psi$-independent function $S^0$ hold trivially.

Consider the diagram

(C.4)

For a sufficiently small neighborhood $U'$, the restrictions of $(dS, d^2S)$, $(\pi_{xp_x}, d\pi_{xp_x})$ to the corresponding bundles implement the bundle isomorphisms shown in the diagram. Similarly, in the case of a function $S^0$ independent of $\psi$, we obtain similar isomorphisms with the exception that, for an analogueof $\pi_{xp_x}$, we take the identity mapping, which we do not show on the diagram. Note that $C_{S^0} = \gamma_{S^0}$, and the planes $R_{S^0}$ coincide with the $x$-plane. Here the planes $R_S$ have the form of a graph over the $x$-plane. Therefore, the mapping $(\pi_x, d\pi_x)$ is also an isomorphism of bundles in a small neighborhood $U'$. To complete the proof of the lemma, it remains to prove the equality shown in the diagram. To this end, we trace the images of an arbitrary element $(x, \psi, \delta x, \delta \psi) \in (C_S, R_S)$ along two different paths in the diagram. It is necessary to prove the equality indicated in the following diagram (the derivatives of $S$ are taken at a given point $(x, \psi)$, and the derivatives of $S^0$ at the given point $x$):

For $(x, \psi, \delta x, \delta \psi) \in (C_S, R_S)$, by (7.2), we have $\delta \psi = -S_{\psi \psi}^{-1} S_{\psi x}\delta x$; therefore, on the left-hand side of the equality, we have $(S_x, (S_{xx} - S_{x\psi}S_{\psi \psi}^{-1} S_{\psi x})\delta x)$. By part (2) of Proposition 7, which has already been proved, this expression is equal to the right-hand side of the equality in the diagram.

Lemma 12.

Part $(4)$ of Proposition $7$ holds in the neighborhood $U'$ .

Proof.

The equality $d\mu_{S^0} = dx_1 \wedge \ldots \wedge dx_n$ follows trivially from the definition of $d\mu_{S^0}$ for the $\psi$-independent function $S^0$. The other equality of this part follows from (7.4) and from part (2) of Proposition 7 which has already been proved.

Lemma 13.

Part $(5)$ of Proposition $7$ holds in the neighborhood $U'$ .

Proof.

By the equality $a^0(x) = a(x, \psi)$, for $(x, \psi) \in C_S$, the restrictions $a|_{C_S}$, $a_0|_{C_{S^0}}$ coincide if we mean the identification of $C_S$ and $C_{S^0}$ using the mapping $\pi_x$. It remains to use the definitions of $A_{(S^0, a^0)}$ and $A_{(S, a)}$ and the diagram (C.4).

Thus, Proposition 7 is proved.

D. Proof of Proposition 8

Let us first prove the equality of the squares of both parts of (7.8).

Lemma 14.

Let $(\Lambda_{S}, r_{S}) = (\Lambda, r)$ and $d\mu_S= \mathscr{M}^2 d\mu$ . Then, for every $I$ , at every $I$ -nonsingular point $\tau_0\in \Lambda$ , we have

$$\frac{\exp(\frac{i \pi}{2} (m-|I|))}{\det (-iS_I)_{\Psi \Psi}} \Big|_{(\widehat{x}_{0}, \Psi_{0})} = \frac{\mathscr{M}^2}{\mathfrak{m}^{I}_{+0}}\Big|_{\tau_{0}}.$$

Proof.

Let $I=\emptyset$ first. In this case, we are to prove

$$\frac{i^m}{\det (-iS_{\psi \psi})}\Big|_{(x_{0}, \psi_{0})} = \frac{\mathscr{M}^2}{\mathfrak{m}^{\emptyset}_{+0}}\Big|_{\tau_{0}}.$$

Multiply both parts by the form $dx$ that does not vanish on $(\Lambda, r)$. By the definition given in (2.5) for the functions $\mathfrak{m}^{I}_{\varepsilon}$ we can write $(\mathscr{M}^2/\mathfrak{m}^{\emptyset}_{+0}) dx = \mathscr{M}^2 d\mu$ in a neighborhood of $\tau_0$. On the other hand, by parts (4) and (2) of Proposition 7, we have $i^mdx /\det (-iS_{\psi \psi}(x_{0}, \psi_{0})) = d\mu_S(x_{0}, \psi_{0})$. It remains to use the condition $d\mu_S= \mathscr{M}^2 d\mu$.

Reduce the case of an arbitrary $I$ to the case of $I=\emptyset$. Consider an isotropic manifold with complex germ $(J_I\Lambda, J_I r)$ with measure $d\widetilde{\mu} = (-1)^{|I|} (J_I^{-1})^*d\mu$ and the function $\widetilde{\mathscr{M}} = (J_I^{-1})^* \mathscr{M}$ on the manifold. For every $I'$, formula (2.5) defines a function on $J_I\Lambda$, which we denote by $\widetilde{\mathfrak{m}}^{I'}_{+0}$. The connection of the measures $d\widetilde{\mu}$, $d\mu$ implies the equality $\widetilde{\mathfrak{m}}^\emptyset_{+0} = (-1)^{|I|}(J_I^{-1})^*\mathfrak{m}^I_{+0}$. By Proposition 5 and the condition of the lemma, we have $(J_I\Lambda, J_Ir) = (\Lambda_{S_I}, r_{S_I})$ and $d\mu_{S_I} = \widetilde{\mathscr{M}}^2 d\widetilde{\mu}$. The point $(\widehat{x}_{0}, \Psi_{0}) \in C_{S_I}$ corresponds to the point $J_I(\tau_{0})$ with respect to the diffeomorphism $C_{S_I}\cong \Lambda_{S_I}$. Apply the present lemma proved in the case of $I=\emptyset$ to the function $S_I$ and the isotropic manifold with complex germ $(J_I\Lambda, J_Ir)$ with measure $d\widetilde{\mu}$: we have

$$\frac{\exp(\frac{i \pi}{2} (m+|I|))}{\det (-iS_I)_{\Psi \Psi}} \Big|_{(\widehat{x}_{0}, \Psi_{0})} = \frac{\widetilde{\mathscr{M}}^2}{\widetilde{\mathfrak{m}}^\emptyset_{+0}} \Big|_{J_I(\tau_{0})}.$$

Substituting $(\widetilde{\mathscr{M}}^2/\widetilde{\mathfrak{m}}^\emptyset_{+0})|_{J_I(\tau_0)} = (-1)^{|I|} (\mathscr{M}^2/\mathfrak{m}^I_{+0})|_{\tau_0}$ into the right-hand side, we obtain the equality to be proved in the lemma.

Let us introduce some notation for the case in which, in addition to $I$, another list $I'$ is given. For this list, the function $S_{I'}(x, \widetilde{\Psi})$, the coordinates $(\check{x}, \check{p}) = J_{I'} (x, p)$, and the point $(\check{x}_{0}, \widetilde{\Psi}_{0})$, which corresponds to the point $\tau_0$ on $\Lambda$, are defined.

Lemma 15.

Let $(\Lambda_{S}, r_{S}) = (\Lambda, r)$ and $d\mu_S= \mathscr{M}^2 d\mu$ . Then at an $I$ - and $I'$ -nonsingular point $\tau_0\in \Lambda$ , the following equalities hold or fail to hold simultaneously $:$

$$\frac{\exp(\frac{i \pi}{4} (m-|I|))}{\sqrt{\det (-iS_I)_{\Psi \Psi}}}\Big|_{(\widehat{x}_{0}, \Psi_{0})} = \frac{\mathscr{M}}{\sqrt{\mathfrak{m}^{I}_{+0}}}\Big|_{\tau_{0}} \iff \frac{\exp(\frac{i \pi}{4} (m-|I'|)) }{\sqrt{\det (-iS_{I'})_{\widetilde{\Psi} \widetilde{\Psi}}}}\Big|_{(\check{x}_{0}, \widetilde{\Psi}_{0})} = \frac{\mathscr{M}}{\sqrt{\mathfrak{m}^{I'}_{+0}}}\Big|_{\tau_{0}}.$$

Proof.

It suffices to prove the equality

$$ \frac{\sqrt{\det (-iS_I)_{\Psi \Psi}(\widehat{x}_{0}, \Psi_{0})} \exp(\frac{i \pi}{4} |I|)}{\sqrt{\mathfrak{m}^I_{+0}(\tau_0)}} = \frac{\sqrt{\det (-iS_{I'})_{\widetilde{\Psi} \widetilde{\Psi}}(\check{x}_{0}, \widetilde{\Psi}_{0})} \exp(\frac{i \pi}{4} |I'|)}{\sqrt{\mathfrak{m}^{I'}_{+0}(\tau_0)}}.$$

(D.1)

We write

$$\begin{aligned} \, I_0 & = I \cup I', \qquad I_1 = I\setminus I', \qquad I_2 = I'\setminus I, \qquad I_3 = (1,\ldots,n)\setminus (I_1 \cup I_2), \\ \overline{I} & = (1,\ldots,n)\setminus I, \qquad \overline{I'} = (1,\ldots,n)\setminus I'. \end{aligned}$$

The $I$- and $I'$-nonsingularity of $\tau_0$ implies the $I_0$-nonsingularity of $\tau_0$. Denote the arguments of the functions $(S_I)_{I_2}$, $(S_{I'})_{I_1}$by $\Phi = (\Psi, \Psi')$, $\widetilde{\Phi} = (\widetilde{\Psi}, \widetilde{\Psi}')$, respectively. The functions $S_{I_0}$, $(S_I)_{I_2}$, $(S_{I'})_{I_1}$ coincide up to an insignificant permutation of variables and, therefore, the following equality holds:

$$ \sqrt{\det ((-iS_I)_{I_2})_{\Phi \Phi}} = \sqrt{\det ((-iS_{I'})_{I_1})_{\widetilde{\Phi} \widetilde{\Phi}}}.$$

(D.2)

Apply the well-known formula for the block matrix determinant,

$$\det ((-iS_I)_{I_2})_{\Phi \Phi} = \det ((-iS_I)_{I_2})_{\Psi \Psi} \det (-i[((S_I)_{I_2})_{\Psi' \Psi'} - ((S_I)_{I_2})_{\Psi' \Psi} ((S_I)_{I_2})_{\Psi \Psi}^{-1} ((S_I)_{I_2})_{\Psi \Psi'} ]).$$

It follows from the definition of $(S_I)_{I_2}$ that

$$ \det ((-iS_I)_{I_2})_{\Phi \Phi} = \det (-iS_I)_{\Psi \Psi} \det (-i[(S_I)_{x_{I_2} x_{I_2}} - (S_I)_{x_{I_2} \Psi} (S_I)_{\Psi \Psi}^{-1} (S_I)_{\Psi x_{I_2}} ]),$$

(D.3)

and, in turn, it follows from part (2) of Proposition 7 that

$$ \det ((-iS_I)_{I_2})_{\Phi \Phi} = \det (-iS_I)_{\Psi \Psi} \det ((-iS_I)^0)_{x_{I_2} x_{I_2}} = \det (-i(S_I)_{\Psi \Psi}) \det (-iG_I)_{I_2I_2},$$

(D.4)

where $(-iG_I)_{I_2I_2}$ stands for the matrix of the size $|I_2|\times |I_2|$ composed of the elements of the matrix $-iG_I$ whose row number and the column number is contained in $I_2$. Similarly,

$$ \det ((-iS_{I'})_{I_1})_{\widetilde{\Phi} \widetilde{\Phi}} = \det (-iS_{I'})_{\widetilde{\Psi} \widetilde{\Psi}} \det ((-iS_{I'})^0)_{x_{I_1} x_{I_1}} = \det (-iS_{I'})_{\widetilde{\Psi} \widetilde{\Psi}} \det (-iG_{I'})_{I_1I_1}.$$

(D.5)

By (D.2), (D.4), (D.5), equality (D.1) can be transformed to the equality

$$ \frac{\exp(\frac{i \pi}{4} |I|)}{\sqrt{\mathfrak{m}^I_{+0}} \sqrt{\det (-iG_I)_{I_2I_2}}}\Big|_{\tau_0} = \frac{\exp(\frac{i \pi}{4} |I'|)}{\sqrt{\mathfrak{m}^{I'}_{+0}} \sqrt{\det (-iG_{I'})_{I_1I_1}}}\Big|_{\tau_0}.$$

(D.6)

It follows from the definitions of $\mathfrak{m}^I_{\varepsilon}$ and $G_I$ that $\mathfrak{m}^I_{\varepsilon} = \mathfrak{m}^I_{+0}\det (E_n - i\varepsilon G_I)$. Therefore, $\sqrt{\mathfrak{m}^I_{\varepsilon}} = \sqrt{\mathfrak{m}^I_{+0}} \sqrt{\det (E_n - i\varepsilon G_I)}$; in a similar way, this equality holds for $I'$ instead of $I$. Substitute these equalities for $\varepsilon =1$ into condition (2.6):

$$\sqrt{\det (E_n - i G_I)} \sqrt{\mathfrak{m}^I_{+0}} \exp\left(-\frac{i \pi}{4} |I|\right) = \sqrt{\det (E_n - i G_{I'})} \sqrt{\mathfrak{m}^{I'}_{+0}} \exp\left(-\frac{i \pi}{4} |I'|\right)$$

Thus, equality (D.6) is equivalent to the equality

$$ \sqrt{\det (E_n - i G_I)} \Big(\sqrt{\det (-iG_I)_{I_2I_2}}\Big)^{-1} \Big|_{\tau_0} = \sqrt{\det (E_n - i G_{I'})} \Big( \sqrt{\det (-iG_{I'})_{I_1I_1}} \Big)^{-1} \Big|_{\tau_0}.$$

(D.7)

Introduce a smooth basis $(u_1(\tau), \ldots, u_n(\tau))$ of the fibers $r(\tau)$. To this basis, matrices $C(\tau)$ and $B(\tau)$ correspond. Write

$$\begin{pmatrix} \widehat{C} \\ \widehat{B} \end{pmatrix} = \frac{\partial J_I}{\partial (x,p)} \begin{pmatrix} C \\ B \end{pmatrix}, \qquad \begin{pmatrix} \check{C} \\ \check{B} \end{pmatrix} = \frac{\partial J_{I'}}{\partial (x,p)} \begin{pmatrix} C \\ B \end{pmatrix}.$$

Note the equalities $G_I = \widehat{B} \widehat{C}^{-1}$, $G_{I'} = \check{B} \check{C}^{-1}$. Denote by $\mathscr{E}_I$ the diagonal square matrix of size $n$ whose diagonal entries with numbers in $I$ are $1$ and the other elements are zero. Then it follows from the definitions of $J_I$ and $J_{I'}$ that

$$\widehat{C} = \mathscr{E}_{\overline{I}} C + \mathscr{E}_I B, \qquad \widehat{B} = \mathscr{E}_{\overline{I}} B - \mathscr{E}_I C, \qquad \check{C} = \mathscr{E}_{\overline{I'}} C + \mathscr{E}_{I'} B, \qquad \check{B} = \mathscr{E}_{\overline{I'}} B - \mathscr{E}_{I'} C.$$

We can express

$$\check{C} = \mathscr{E}_{I_3}\widehat{C} + \mathscr{E}_{I_2} \widehat{B} - \mathscr{E}_{I_1} \widehat{B}, \qquad \check{B} = \mathscr{E}_{I_3} \widehat{B} - \mathscr{E}_{I_2} \widehat{C} + \mathscr{E}_{I_1} \widehat{C}.$$

We can also express the matrix

$$G_{I'} = (\mathscr{E}_{I_3} G_I - \mathscr{E}_{I_2} + \mathscr{E}_{I_1}) (\mathscr{E}_{I_3} + \mathscr{E}_{I_2} G_I - \mathscr{E}_{I_1} G_I)^{-1}.$$

Let us present a chain of equalities

$$\begin{aligned} \, E_n - iG_{I'} & = E_n - i (\mathscr{E}_{I_3} G_I - \mathscr{E}_{I_2} + \mathscr{E}_{I_1}) (\mathscr{E}_{I_3} + \mathscr{E}_{I_2} G_I - \mathscr{E}_{I_1} G_I)^{-1} \\ & = (\mathscr{E}_{I_3} + \mathscr{E}_{I_2} G_I - \mathscr{E}_{I_1} G_I -i\mathscr{E}_{I_3} G_I +i\mathscr{E}_{I_2} -i \mathscr{E}_{I_1}) (\mathscr{E}_{I_3} + \mathscr{E}_{I_2} G_I - \mathscr{E}_{I_1} G_I)^{-1} \\ & = (\mathscr{E}_{I_3} + i\mathscr{E}_{I_2} - i \mathscr{E}_{I_1}) (E_n - i G_I) (\mathscr{E}_{I_3} -i(\mathscr{E}_{I_2} + \mathscr{E}_{I_1}) G_I)^{-1} (\mathscr{E}_{I_3} + i\mathscr{E}_{I_2} - i \mathscr{E}_{I_1})^{-1}. \end{aligned}$$

This implies that

$$ \sqrt{\det(E_n - iG_{I'})} = \sqrt{\det(E_n - i G_I)} \sqrt{\det (\mathscr{E}_{I_3} -i(\mathscr{E}_{I_2} + \mathscr{E}_{I_1}) G_I)^{-1}}.$$

(D.8)

Further, note the equalities

$$\sqrt{\det(-i G_{I'})_{I_1 I_1}} = \sqrt{\det (\mathscr{E}_{I_3} + \mathscr{E}_{I_2} - i\mathscr{E}_{I_1} G_{I'})}, \qquad \sqrt{\det (-i G_I)_{I_2 I_2}} = \sqrt{ \det( \mathscr{E}_{I_3} -i \mathscr{E}_{I_2} G_I +\mathscr{E}_{I_1} )}.$$

It follows from these equalities and from the cain of equalities

$$\begin{aligned} \, \mathscr{E}_{I_3} + & \mathscr{E}_{I_2} - i\mathscr{E}_{I_1} G_{I'} = \mathscr{E}_{I_3} + \mathscr{E}_{I_2} - i\mathscr{E}_{I_1} (\mathscr{E}_{I_3} G_I - \mathscr{E}_{I_2} + \mathscr{E}_{I_1}) (\mathscr{E}_{I_3} + \mathscr{E}_{I_2} G_I - \mathscr{E}_{I_1} G_I)^{-1} \\ & = \mathscr{E}_{I_3} + \mathscr{E}_{I_2} - i\mathscr{E}_{I_1} (\mathscr{E}_{I_3} + \mathscr{E}_{I_2} G_I - \mathscr{E}_{I_1} G_I)^{-1} \\ & = ( \mathscr{E}_{I_3} + \mathscr{E}_{I_2} G_I - i\mathscr{E}_{I_1} ) (\mathscr{E}_{I_3} + \mathscr{E}_{I_2} G_I - \mathscr{E}_{I_1} G_I)^{-1} \\ & = (\mathscr{E}_{I_3} + i\mathscr{E}_{I_2} - i \mathscr{E}_{I_1}) ( \mathscr{E}_{I_3} -i \mathscr{E}_{I_2} G_I +\mathscr{E}_{I_1} ) (\mathscr{E}_{I_3} -i(\mathscr{E}_{I_2} + \mathscr{E}_{I_1}) G_I)^{-1} (\mathscr{E}_{I_3} + i\mathscr{E}_{I_2} - i \mathscr{E}_{I_1})^{-1}. \end{aligned}$$

that

$$ \det(-i G_{I'})_{I_1 I_1} = \sqrt{\det (-i G_I)_{I_2 I_2}} \sqrt{\det (\mathscr{E}_{I_3} -i(\mathscr{E}_{I_2} + \mathscr{E}_{I_1}) G_I)^{-1}}.$$

(D.9)

The substitution of (D.8) and (D.9) into (D.7) completes the proof.

Let us complete the proof of Proposition 8. Consider the set

$$\{ \tau_0 \in \Lambda \mid \forall I \ (\tau_0 \text{ is } I\text{-nonsingular} \implies (7.8) \text{holds}) \}.$$

We are to prove that this set coincides with the entire $\Lambda$.

We claim that this set is closed. Let $I$ and an $I$-nonsingular $\tau_0$ be given for which (7.8) fails to hold. Then, due to the continuity of both parts of (7.8) and to the openness of the set of $I$-nonsingular points, there is a neighborhood of $\tau_0$ consisting of $I$-nonsingular points in which equality (7.8) fails to hold.

We claim that the set in question is open. If $\tau_0$ belongs to this set, then, for some $I$, equality (7.8) holds at the point $\tau_0$. Since the squares of both sides of this equality are equal, it follows that this equality holds for given $I$ in a neighborhood of $\tau_0$. By Lemma 15, this equality holds for all $I$ at all $I$-nonsingular points of the given neighborhood. Thus, the openness of the set under consideration is proved.

By the definition of the connectivity of a space, an open-and-closed subset must coincide with either $\Lambda$ or $\emptyset$. Let us show that the last case is impossible. By the condition of the Proposition 8, equality (7.8) holds for some $I$ and $\tau_0$. Then, by Lemma 15 it is satisfied at this point for all $I$ for which $\tau_0$ is $I$-nonsingular. Thus, the set in question is nonempty, since the point $\tau_0$ belongs to it. This completes the proof of Proposition 8.