Proof. The term ‘generic’ used in this statement refers to an element of some fixed open dense set. The set of $x \in X$ such that $\sigma$ has constant rank in a neighborhood of $x$ is open and dense, so this reduces to the case of a Poisson vector space $(W, \sigma )$. The proposition then follows from the fact that the set of $k$-dimensional subspaces $V \subseteq W$ such that $V + \sigma (V^{\circ })$ has maximal dimension is open and dense in the Grassmannian of $k$-planes in $W$.

Proof. Let $\varphi : (\ker d\mathsf {t})\vert _X \longrightarrow T^{*}X$ be the isomorphism given by (2.2), so that

\begin{align*} \sigma \circ \varphi = d\mathsf{s} : (\ker d\mathsf{t})\vert_X \longrightarrow TX. \end{align*}

Let $x \in S$. The definition of $\mathcal {H}$ then implies that

\begin{align*} \ker d\mathsf{t}_x \cap T_x\mathcal{H} = \varphi^{-1}(\sigma^{-1}(T_xS) \cap T_xS^{\circ}) = \ker d\mathsf{t}_x \cap d\mathsf{s}^{-1}_x(T_xS) \cap T_xS^{\Omega}. \end{align*}

It follows that

\begin{align*} T_x\mathcal{H} = T_xS \oplus (\ker d\mathsf{t}_x \cap T_x\mathcal{H}) = T_xS \oplus (\ker d\mathsf{t}_x \cap d\mathsf{s}^{-1}_x(T_xS) \cap T_xS^{\Omega}). \end{align*}

However, $T_xS \subseteq T_xS^{\Omega }$ because $X$ is Lagrangian in $\mathcal {G}$, so

\begin{align*} T_x\mathcal{H} = (T_xS \oplus (\ker d\mathsf{t}_x \cap d\mathsf{s}^{-1}_x(T_xS))) \cap T_xS^{\Omega} = d\mathsf{s}_x^{-1}(T_xS) \cap d\mathsf{t}_x^{-1}(T_xS) \cap T_xS^{\Omega}. \end{align*}

Proof. Consider the map

\begin{align*} F : \mathcal{G} \times M \times M^{-} \longrightarrow \mathbb{R}, \quad F(g, p, q) = f(\mathsf{s}(g)) - f(\mu(q)). \end{align*}

The graph $\Gamma$ of the action is coisotropic and $F\vert _\Gamma = 0$, so $X_F$ is tangent to $\Gamma$. However,

\begin{align*} dF = (d(\mathsf{s}^{*}f), 0, -d(\mu^{*}f)), \end{align*}

so $X_F = (X_{\mathsf {s}^{*}f}, 0, X_{\mu ^{*}f})$. Hence,

\begin{align*} (X_{\mathsf{s}^{*}f}(g), 0, X_{\mu^{*}f}(g \cdot p)) = (u, v, d\psi(u, v)) \end{align*}

for some $(u, v) \in T_{(g, p)}(\mathcal {G} \times _X M)$, which implies that $v = 0$ and $u = X_{\mathsf {s}^{*}f}(g)$. We conclude that

\begin{align*} X_{\mu^{*}f}(g \cdot p) = d\psi(u, 0) = d\psi_p(X_{\mathsf{s}^{*}f}(g)). \end{align*}

Proof. By Lemma 2.11, we have

(2.4)\begin{equation} T_x\mathcal{H}_x = \ker d\mathsf{t}_x \cap d\mathsf{s}_x^{-1}(T_xS) \cap (T_xS)^{\Omega}. \end{equation}

Let $v = X_{\mathsf {s}^{*}f}(x)$ be as stated. Then $d\mathsf {t}(v) = 0$ (see [Reference Mikami and WeinsteinMW88, Theorem 1.6(i)]), $d\mathsf {s}(v) = X_f(x) \in T_xS$ because $\mathsf {s}$ is a Poisson map, and

\begin{align*} \Omega(v, T_xS) = df(d\mathsf{s}(T_xS)) = df(T_xS) = 0. \end{align*}

Hence, $v \in T_x\mathcal {H}_x$ by (2.4).

Conversely, let $v \in T_x\mathcal {H}_x$. By (2.4), we have $d\mathsf {t}(v) = 0$, $d\mathsf {s}(v) \in T_xS$ and $\Omega (v)\vert _{T_xX} \in T_xS^{\circ }$. The latter implies that $\Omega (v)\vert _{T_xX} = df_x$ for some $f \in C^{\infty }(X)$ such that $f\vert _S = 0$. Then $\Omega (v) = df \circ d\mathsf {s}_x$ because $T_x\mathcal {G} = T_xX \oplus \ker d\mathsf {s}_x$ and $(\ker d\mathsf {s}_x)^{\Omega } = \ker d\mathsf {t}$. It follows that $v = X_{\mathsf {s}^{*}f}(x)$ and $X_f(x) = d\mathsf {s}(v) \in T_xS$.

Proof of Theorem 2.10 To show that

\begin{align*} T_p(\mathcal{H} \cdot p) \subseteq T_pN \cap \tau(T_pN^{\circ}), \end{align*}

let $x := \mu (p)$ and $v \in T_x\mathcal {H}_x$. Note that

\begin{align*} u := d\psi_p(v) \in T_p(\mathcal{H} \cdot p). \end{align*}

Then $u \in T_pN$ because $\mathcal {H}$ acts on $N$. By Lemma 2.13, we have $v = X_{\mathsf {s}^{*}f}(x)$ for some $f \in C^{\infty }(X)$ with $f\vert _S = 0$ and $X_f(x) \in T_xS$, and by Lemma 2.12,

\begin{align*} u = d\psi_p(v) = X_{\mu^{*}f}(p). \end{align*}

However, $d\mu (X_{\mu ^{*}f}(p)) = X_f(x) \in T_xS$ since $\mu$ is a Poisson map; this follows from [Reference Mikami and WeinsteinMW88, Theorem 3.8], as explained in [Reference Bursztyn and CrainicBC05, § 4.2]. Hence, $u \in T_pN$. We also have

\begin{align*} d(\mu^{*}f)(T_pN) = df(d\mu(T_pN)) \subseteq df(T_xS) = 0, \end{align*}

so $u = X_{\mu ^{*}f}(p) \in \tau (T_pN^{\circ })$.

Conversely, let $v \in T_pN \cap \tau (T_pN^{\circ })$. As

\begin{align*} T_pN^{\circ} = (d\mu^{-1}(T_xS))^{\circ} = d\mu^{*}(T_xS^{\circ}), \end{align*}

we have $v = X_{\mu ^{*}f}(p)$ for some $f \in C^{\infty }(X)$ with $f\vert _S = 0$. We also have

\begin{align*} X_f(x) = d\mu(v) \in T_xS. \end{align*}

We thus have $X_{\mathsf {s}^{*}f}(x) \in T_x\mathcal {H}_x$ by Lemma 2.13 and $v = d\psi _p(X_{\mathsf {s}^{*}f}(x))$ by Lemma 2.12, so $v \in T_p(\mathcal {H} \cdot p)$.

Proof. We have $\ker d\pi _p = T_p(\mathcal {H} \cdot p)$ for all $p \in N$ by assumption. Hence, as in standard symplectic reduction [Reference Marsden and WeinsteinMW74, Reference Mikami and WeinsteinMW88], it suffices to show that the distribution $\ker (i^{*}\omega ) = TN \cap TN^{\omega }$ on $N$ coincides with that induced by the $\mathcal {H}$-orbits. This is precisely the content of Theorem 2.10.

Proof. Regarding $\ker (d\varphi _p)_x$ as a subspace of $T_x\mathcal {H}_x \subseteq T_x\mathcal {G}$, the statement can be reformulated as

\begin{align*} T_xS + \operatorname{im} d\mu_p = T_xX \cap (\ker (d\varphi_p)_x)^{\Omega}. \end{align*}

We prove this reformulated version in the following.

Lemma 2.11 tells us that

\[ \ker (d\varphi_p)_x \subseteq T_x\mathcal{H} \subseteq (T_xS)^{\Omega}, \]

so

\[ T_xS \subseteq T_xX \cap (\ker (d\varphi_p)_x)^{\Omega}. \]

To show that $\operatorname {im} d\mu _p \subseteq (\ker (d\varphi _p)_x)^{\Omega }$, let $v \in T_pM$ and $w \in \ker (d\varphi _p)_x$. By Lemma 2.13, $w = X_{\mathsf {s}^{*}f}(x)$ for some $f \in C^{\infty }(X)$ such that $f\vert _S = 0$ and $X_f(x) \in T_xS$. Lemma 2.12 then shows that $X_{\mu ^{*}f}(p) = d\varphi _p(w) = 0$, so $d(\mu ^{*}f)_p = 0$. It follows that

\begin{align*} \Omega(d\mu(v), w) =-df(d\mathsf{s}(d\mu(v))) = -df(d\mu(v)) = 0. \end{align*}

It remains only to show that

\begin{align*} T_xS^{\circ} \cap (\operatorname{im} d\mu_p)^{\circ} \subseteq T_xX^{\circ} + \Omega(\ker (d \varphi_p)_x), \end{align*}

where all annihilators are taken in $T^{*}_x\mathcal {G}$. To this end, let $\xi \in T_xS^{\circ } \cap (\operatorname {im} d\mu _p)^{\circ }$ and write $\xi \vert _{T_xX} = \Omega (v)\vert _{T_xX}$ with $v \in \ker d\mathsf {t}_x$. It then suffices to prove that $v \in T_x\mathcal {H}_x$ and $d\varphi _p(v) = 0$. We begin by verifying that $v \in T_x\mathcal {H}_x$. First note that $v \in \ker d\mathsf {t}_x \cap (T_xS)^{\Omega }$. By Lemma 2.11, showing that $d\mathsf {s}(v) \in T_xS$ would suffice to prove that $v \in T_x\mathcal {H}_x$. As $\mu$ is Poisson, we have

\begin{align*} d\mathsf{s}(v) = \sigma(\xi\vert_{T_xX}) = d\mu(\omega^{-1}(d\mu^{*}(\xi\vert_{T_xX}))), \end{align*}

and the latter is $0$ because $\xi \in (\operatorname {im} d\mu _p)^{\circ }$. It follows that $v \in T_x\mathcal {H}_x$. To show that $d\varphi _p(v) = 0$, first note that Lemma 2.13 implies that $v = X_{\mathsf {s}^{*}f}(x)$ for some $f \in C^{\infty }(X)$ such that $f\vert _S = 0$ and $X_f(x) \in T_xX$. However,

\begin{align*} df_x =\Omega(v)\vert_{T_xX} = \xi\vert_{T_xX} \in (\operatorname{im} d\mu_p)^{\circ}, \end{align*}

so $d(\mu ^{*}f)_p = 0$ and hence $d\varphi _p(v) = X_{\mu ^{*}f}(p) = 0$ by Lemma 2.12.

Proof. This follows from Proposition 2.19 using the fact that $\ker (d\varphi _p)_x = 0$ if $\mathcal {H}$ acts freely.

Proof. As $\mathcal {H}$ is closed in $\mathcal {G}\vert _S$, it suffices to show that the action of $\mathcal {G}\vert _S$ on $N$ is proper. By assumption, the map $\theta : \mathcal {G}_{\mathsf {t}}{\times }_\mu M \longrightarrow M \times M$ defined by (2.1) is proper.

Consider the pullback of $\theta$ by the inclusion $\iota : N \times N \hookrightarrow M \times M$, i.e.

(2.5)\begin{equation} (\mathcal{G}_{\mathsf{t}}{\times}_\mu M) \times_{M \times M} (N \times N) \longrightarrow N \times N. \end{equation}

As $(\mathcal {G}_{\mathsf {t}}{\times }_\mu M) \times _{M \times M} (N \times N)$ is closed in $(\mathcal {G}_{\mathsf {t}}{\times }_\mu M) \times N \times N$ and $\theta$ is proper, (2.5) is proper. Note that

\begin{align*} (\mathcal{G}_{\mathsf{t}}{\times}_\mu M) \times_{M \times M} (N \times N) = (\mathcal{G}\vert_N)_{\mathsf{t}}{\times}_\mu N \end{align*}

and that (2.5) is the action map $(\mathcal {G}\vert _N)_{\mathsf {t}}{\times }_\mu N \longrightarrow N \times N$. This shows that the action of $\mathcal {G}\vert _S$ on $N$ is proper, as desired.

Proof. Recall that if $\mathcal {H}$ is source-connected, then $S / \mathcal {H}$ is the Marsden–Ratiu reduction of the triple $(X, S, \sigma (TS^{\circ }))$ (see [Reference Marsden and RatiuMR86, Example D]). We are thereby reduced to proving the following claim: if $f, g, F, G$ are as in the statement, then $\{F, G\}_X$ is $\mathcal {H}$-invariant. In other words, we want to show that $j^{*}\mathsf {s}^{*}\{F, G\}_X = j^{*}\mathsf {t}^{*}\{F, G\}_X$, where $j : \mathcal {H} \longrightarrow \mathcal {G}$ is the immersion of $\mathcal {H}$ in $\mathcal {G}$.

Recall from § 2.2 that there is a Poisson transversal $Y$ in $X$ containing $S$ as a coisotropic submanifold. The restriction $\mathcal {G}\vert _Y := \mathsf {s}^{-1}(Y) \cap \mathsf {t}^{-1}(Y)$ is then a symplectic subgroupoid of $\mathcal {G}$ (see [Reference Cattaneo and ZambonCZ09, Lemma 7.1]) and $\mathcal {H}$ is an isotropic Lie subgroupoid of $\mathcal {G}\vert _Y$. By [Reference CattaneoCat04, § 5], $\mathcal {H}$ is Lagrangian in $\mathcal {G}\vert _Y$. We therefore have

(2.6)\begin{equation} T_h\mathcal{H} = T_h\mathcal{H}^{\Omega} \cap T_{j(h)}(\mathcal{G}\vert_Y) \end{equation}

for all $h \in \mathcal {H}$, where $\Omega$ is the symplectic form on $\mathcal {G}$ and $T_h\mathcal {H}$ is identified as subspace of $T_{j(h)}\mathcal {G}$ via the immersion $j : \mathcal {H} \longrightarrow \mathcal {G}$.

As $TX\vert _Y = TY \oplus \sigma (TY^{\circ })$, there is a locally defined smooth function $\tilde {F}$ on $X$ such that $\rho ^{*}f = i^{*}\tilde {F}$ and $d\tilde {F}(\sigma (TY^{\circ })) = 0$. Note that $\sigma (T_xS^{\circ }) \cap T_xY \subseteq T_xS$ for all $x \in S$ because $S$ is coisotropic in $Y$. We therefore have

\begin{align*} \sigma(T_xS^{\circ}) = (\sigma(T_xS^{\circ}) \cap T_xS) \oplus \sigma(T_xY^{\circ}), \end{align*}

so $d\tilde {F}(\sigma (TS^{\circ })) = 0$. On the other hand, we have $i^{*}\{\tilde {F}, G\}_X = i^{*}\{F, G\}_X$ as $i^{*}\tilde {F} = i^{*}F$. We may then assume without loss of generality that $dF(\sigma (TY^{\circ })) = 0$, i.e. $X_F$ is tangent to $Y$.

We claim that $X_{\mathsf {s}^{*}F} - X_{\mathsf {t}^{*}F}$ is tangent to $\mathcal {H}$. A first observation is that $j^{*}\mathsf {s}^{*}F = j^{*} \mathsf {t}^{*}F$, because

\begin{align*} j^{*}\mathsf{s}^{*}F = \mathsf{s}^{*} i^{*}F = \mathsf{s}^{*} \rho^{*} f = \mathsf{t}^{*} \rho^{*}f = \mathsf{t}^{*} i^{*}F = j^{*} \mathsf{t}^{*} F. \end{align*}

This implies that

\begin{align*} \Omega(X_{\mathsf{s}^{*}F} - X_{\mathsf{t}^{*}F}, T\mathcal{H}) = (d(\mathsf{s}^{*}F) - d(\mathsf{t}^{*}F))(T\mathcal{H}) = 0, \end{align*}

so $X_{\mathsf {s}^{*}F} - X_{\mathsf {t}^{*}F}$ takes values in $T\mathcal {H}^{\Omega }$. Now recall that $d\mathsf {s}(X_{\mathsf {t}^{*}F}) = 0 = d\mathsf {t}(X_{\mathsf {s}^{*}F})$ (see e.g. [Reference Mikami and WeinsteinMW88, Theorem 1.6(i)]), so $d\mathsf {s}(X_{\mathsf {s}^{*}F} - X_{\mathsf {t}^{*}F}) = X_F$ and $d\mathsf {t}(X_{\mathsf {s}^{*}F} - X_{\mathsf {t}^{*}F}) = X_F$ are tangent to $Y$. By combining the last two sentences with (2.6), we obtain

\begin{align*} (X_{\mathsf{s}^{*}F} - X_{\mathsf{t}^{*}F})\vert_{j(h)} \in (T_h\mathcal{H})^{\Omega} \cap T_{j(h)}(\mathcal{G}\vert_Y) = T_h\mathcal{H} \end{align*}

for all $h \in \mathcal {H}$.

The identity $j^{*} \mathsf {s}^{*} G = j^{*} \mathsf {t}^{*} G$ therefore implies that for all $h \in \mathcal {H}$ and $g := j(h)$,

\begin{align*} (j^{*}\mathsf{s}^{*}\{F, G\}_X)(h) &= \{\mathsf{s}^{*}F, \mathsf{s}^{*}G\}_X(g) = d(\mathsf{s}^{*}G)(X_{\mathsf{s}^{*}F}|_g) = d(\mathsf{s}^{*}G)(X_{\mathsf{s}^{*}F}|_g - X_{\mathsf{t}^{*}F}|_g) \\ &= d(\mathsf{t}^{*}G)(X_{\mathsf{s}^{*}F}|_g - X_{\mathsf{t}^{*}F}|_g) = -d(\mathsf{t}^{*}G)(X_{\mathsf{t}^{*}F}|_g) = -\{\mathsf{t}^{*}F, \mathsf{t}^{*}G\}(g) \\ &= (j^{*}\mathsf{t}^{*}\{F, G\}_X)(h).\end{align*}

Proof. Let $N := \mu ^{-1}(S)$ and consider the following commutative diagram.

Let $f, g$ be locally defined smooth functions on $S / \mathcal {H}$ and $F, G$ local extensions of $\rho ^{*}f, \rho ^{*}g$ as in Proposition 2.23. We want to show that $\bar {\mu }^{*}\{f, g\} = \{\bar {\mu }^{*}f, \bar {\mu }^{*}g\}$ or, equivalently,

\[ \pi^{*}\bar{\mu}^{*}\{f, g\} = \pi^{*}\{\bar{\mu}^{*}f, \bar{\mu}^{*}g\}. \]

We have

\begin{align*} \pi^{*}\bar{\mu}^{*}\{f, g\} = \tilde{\mu}^{*}\rho^{*}\{f, g\} = \tilde{\mu}^{*}j^{*}\{F, G\} = i^{*} \mu^{*}\{F, G\} = i^{*}\{\mu^{*}F, \mu^{*}G\}, \end{align*}

as $\mu$ is a Poisson map. As the Poisson structure on $N / \mathcal {H}$ is also obtained from Poisson reduction, it suffices to show the following:

1. (1) $\mu ^{*}F$ and $\mu ^{*}G$ vanish on $TN^{\omega }$;

2. (2) $\mu ^{*}F$ and $\mu ^{*}G$ restrict to $\pi ^{*}\bar {\mu }^{*}f$ and $\pi ^{*}\bar {\mu }^{*}g$ on $N$.

To verify part (1), suppose that $v \in TN^{\omega }$. We have $v = \omega ^{-1}(d\mu ^{*}(\xi ))$ for $\xi \in TS^{\circ }$. As $\mu$ is Poisson, we obtain $d\mu (v) = \sigma (\xi ) \in \sigma (TS^{\circ })$. It follows that $d(\mu ^{*}F)(v) = dF(d\mu (v)) = 0$. To show part (2), note that $i^{*}\mu ^{*}F = \tilde {\mu }^{*}j^{*}F = \tilde {\mu }^{*}\rho ^{*}f = \pi ^{*}\bar {\mu }^{*}f$. One has analogous identities for $g$ and $G$.

Proof. To see that (3.1) does not depend on the choice of $F$ and $G$, we need to show the following: if $p \in N$ and $F, G \in \mathcal {O}_{M, p}^{E}$ are such that $F\vert _N = 0$, then $\{F, G\}\vert _N = 0$. Note that $X_F\vert _N$ takes values in $\tau (TN^{\circ } \cap E^{\circ })$, whereas condition (2) implies that

\begin{align*} TN^{\circ} \cap E^{\circ} = (TN + E)^{\circ} \subseteq \tau(E^{\circ})^{\circ} = \tau^{-1}(E). \end{align*}

It follows that $\tau (TN^{\circ } \cap E^{\circ }) \subseteq E$ and, hence, $\{F, G\}\vert _N = dG(X_F\vert _N) = 0$.

Let $f, g \in \mathcal {O}_N^{D}(U)$ for some open set $U \subseteq N$. We then get a well-defined element $\{f, g\}' \in \mathcal {O}_N(U)$ defined by the collection of germs $\{F, G\}\vert _N \in \mathcal {O}_{N, p}$ for all $p \in U$ and $F, G \in \mathcal {O}_{M, p}^{E}$ satisfying $F\vert _N = f_p$ and $G\vert _N = g_p$. We also have $\{f, g\}' \in \mathcal {O}_N^{D}(U)$, because

\begin{align*} d\{f, g\}'(D) = d\{F, G\}(D) \subseteq d\{F, G\}(E) = 0 \end{align*}

by condition (1).

The Leibniz identity follows from the fact that $\{f, gh\}' = \{F, GH\}\vert _N$ if $f, g, h \in \mathcal {O}_{N, p}$ are related to $F, G, H \in \mathcal {O}_{M, p}$ as in condition (3), which holds because $d(GH)$ also vanishes on $E$. For the Jacobi identity, note that $\{f, \{g, h\}'\}' = \{F, \{G, H\}\}\vert _N$, because $\{g, h\}' = \{G, H\}\vert _N$ by definition and $d\{G, H\}(E) = 0$ by condition (1).

Proof. Note that this is a local statement. We may therefore assume that $E = D \oplus R$ for some holomorphic subbundle $R$ of $TM\vert _N$, and then extend $R$ to a holomorphic subbundle $S$ containing $R$ such that $TM\vert _N = TN \oplus S$. Then there exists a tubular neighborhood of $N$ in $S$ that is biholomorphic to a neighborhood of $p$ in $M$. We may define $F = f \circ \pi$, where $\pi : S \longrightarrow N$ is the bundle map.

Proof. Note that because $E^{\circ } = \tau (TN^{\circ })^{\circ } = \tau ^{-1}(TN),$ a local holomorphic function $F$ on $M$ satisfies $dF(E) = 0$ if and only if $X_F$ is tangent to $N$. Condition (1) therefore follows from the Jacobi identity of $\tau$, which is equivalent to $X_{\{F, G\}} = [X_F, X_G]$ for all holomorphic functions $F$ and $G$. Condition (2) also holds because $\tau (E^{\circ }) = \tau (\tau ^{-1}(TN)) \subseteq TN$. Part (ii) holds by Lemma 3.2.

Proof. We simply redo the proof of [Reference Cattaneo and ZambonCZ09, Theorem 3.3] locally. Let $S_x$ be the germ of $S$ at $x$. Choose a subbundle $R$ of $TX\vert _{S_x}$ such that

\begin{align*} TX\vert_{S_x} = (TS_x + \sigma(TS_x^{\circ})) \oplus R, \end{align*}

and let $Y$ be a submanifold germ of $X$ containing $S_x$ with $TY\vert _{S_x} = TS_x \oplus R$. Note that the latter is possible because we can choose a tubular neighborhood of $S$ near $x$.

Proof. Conditions (1) and (2) hold by Proposition 3.3(i), so it suffices to verify condition (3). We first observe that this third condition is local. Proposition 3.4 therefore allows us to assume that $S \subseteq Y \subseteq X$, where $Y$ is Poisson transversal in $X$ and $S$ is coiostropic in $Y$. By [Reference Frejlich and MarcutFM17, Lemma 7], $Y$ is transverse to $\mu$, $P := \mu ^{-1}(Y)$ is a Poisson transversal submanifold of $M$, and the restriction map $P \longrightarrow Y$ is Poisson.Footnote ⁴ As the preimage of a coisotropic subspace under a Poisson map is coisotropic, $N=\mu ^{-1}(S)$ is coisotropic in $P$.

Now let $p \in N$ and $f \in \mathcal {O}_{N, p}^{D}$. Extend $f$ to $\tilde {f} \in \mathcal {O}_{P, p}$. As $P$ is smooth and $TP \cap \tau (TP^{\circ }) = 0$, Lemma 3.2 implies that there exists $F \in \mathcal {O}_{M, p}$ such that $F\vert _P = \tilde {f}\vert _P$ and $dF(\tau (TP^{\circ })) = 0$. We also have that $TP\vert _N \cap E \subseteq TN$ as $N$ is coisotropic in $P$, so $E = D \oplus \tau (TP^{\circ })\vert _N$. It follows that

\begin{align*} dF(E) = dF(D \oplus \tau(TP^{\circ})\vert_N) = dF(D) = df(D) = 0.\end{align*}

Proof. By the definition of $N$, we have $T_pN = d\mu _p^{-1}(T_{\mu (p)}S)$. This allows us to repeat the proof of Theorem 2.10 verbatim in the present context.

Proof. Proposition 3.7 implies that $\mathcal {O}_N(U)^{\mathcal{H}} \subseteq \mathcal {O}_N^{D}(U)$, where $D := TN \cap TN^{\omega }$. As the moment map $\mu : M \longrightarrow X$ is Poisson, Proposition 3.6 shows that the complex analytic version of the Marsden–Ratiu theorem (Theorem 3.1) holds, producing a Poisson bracket

\begin{align*} \{{\cdot}\,{,}\,{\cdot}\}' : \mathcal{O}_N^{D}(U) \times \mathcal{O}_N^{D}(U) \longrightarrow \mathcal{O}_N^{D}(U). \end{align*}

It suffices to show that $\mathcal {O}_N(U)^{\mathcal{H}}$ is closed under $\{{\cdot }\,{,}\,{\cdot }\}'$.

Let $f, g \in \mathcal {O}_N(U)^{\mathcal{H}}$, let $p \in U$, and let $h \in \mathcal {H}$ be such that $\mathsf {t}(h) = \mu (p)$. We want to show that

\begin{align*} \{f, g\}(h \cdot p) = \{f, g\}(p). \end{align*}

In other words, we want to show that for any $F, G \in \mathcal {O}_{M, p}^{TN^{\omega }}$ satisfying $F\vert _N = f_p$, $G\vert _N = g_p$, and any $\tilde {F}, \tilde {G} \in \mathcal {O}_{M, h \cdot p}^{TN^{\omega }}$ satisfying $\tilde {F}\vert _N = f_{h \cdot p}$, $\tilde {G}\vert _N = g_{h \cdot p}$, we have

(3.2)\begin{equation} \{F, G\}(p) = \{\tilde{F}, \tilde{G}\}(h \cdot p). \end{equation}

As $\mathsf {t}$ is a submersion, we can find a local holomorphic section

\begin{align*} \rho : V \subseteq S \longrightarrow \mathcal{H} \end{align*}

of $\mathsf {t}$ passing through $h$, i.e. $V$ is a neighborhood of $\mathsf {t}(h)$ in $S$, $\mathsf {t} \circ \rho = \mathrm {Id}$, and $\rho (\mathsf {t}(h)) = h$. We then get a biholomorphism

\begin{align*} \psi_\rho : N \cap \mu^{-1}(V) \longrightarrow N, \quad \psi_\rho(q) = \rho(\mu(q)) \cdot q, \end{align*}

such that $\psi _\rho (p) = h \cdot p$.

To show (3.2), we first claim that

(3.3)\begin{equation} \psi_\rho^{*}i^{*}\omega = i^{*}\omega, \end{equation}

where $i : N \longrightarrow M$ is the inclusion map. Note that for all $q \in N \cap \mu ^{-1}(V)$, $(\rho (\mu (q)), q, \psi _\rho (q))$ is in the graph $\Gamma$ of the action of $\mathcal {G}$ on $M$, which is Lagrangian in $\mathcal {G} \times M \times M^{-}$ by definition.Footnote ⁵ Hence, for all $u, v \in T_qN$, we have

\begin{align*} (d\rho(d\mu(u)), u, d\psi_\rho(u)) \in T\Gamma, \end{align*}

and the analogous statement is true for $v$. It follows that

\begin{align*} \Omega(d\rho(d\mu(u)), d\rho(d\mu(v))) + \omega(u, v) - \omega(d\psi_\rho(u), d\psi_\rho(v)) = 0. \end{align*}

However, $\mathcal {H}$ is isotropic in $\mathcal {G}$, so the first term vanishes. This proves (3.3).

Now note that $X_F$ and $X_G$ are tangent to $N$, as one has

\begin{align*} \omega(X_F, TN^{\omega}) = dF(TN^{\omega}) = 0 \end{align*}

and an analogous statement for $X_G$. In particular, we may evaluate $d\psi _\rho$ at $X_F|_p$. We claim that

(3.4)\begin{equation} d\psi_\rho(X_F|_p) - X_{\tilde{F}}|_{\psi_\rho(p)} \in T_{\psi_\rho(p)}N^{\omega}. \end{equation}

To see this, let $v \in T_{\psi _\rho (p)}N$ and write $v = d\psi _\rho (u)$ for $u \in T_pN$. Then by (3.3), we have

\begin{align*} \omega(d\psi_\rho(X_F|_p), v) = \psi_\rho^{*}\omega(X_F|_p, u) = \omega(X_F, u) = dF(u) = df(u). \end{align*}

However, $f$ is $\mathcal {H}$-invariant, so

\begin{align*} df(u) = d(\psi_\rho^{*}f)(u) = df(v) = d\tilde{F}(v) = \omega(X_{\tilde{F}}|_{\psi_\rho(p)}, v). \end{align*}

This proves (3.4).

At the same time, because $d\tilde {G}(TN^{\omega }) = 0$, (3.4) shows that