Proof Since f is a b-map, we have that $W\in \mathfrak {X}(M_2)$ is tangent to $Z_{2}$ , so indeed $W=\rho _{2}(\overline {W})$ for unique $\overline {W}\in \Gamma ({}^{b}TM_{2})$ . Now, first consider $p\in M_{1}\setminus Z_{1}$ . Commutativity of the diagram (1.2) implies that

$$ \begin{align*} \rho_{2}\big(({}^{b}f_{*})_{p}(\overline{Y}_{p})\big)=(\,f_{*})_{p}\big(\rho_{1}(\overline{Y}_{p})\big)=(\,f_{*})_{p}(Y_{p})=W_{f(p)}. \end{align*} $$

But we also have $\rho _{2}(\overline {W}_{f(p)})=W_{f(p)}$ , so that injectivity of $\rho _{2}$ at $f(p)\in M_{2}\setminus Z_{2}$ implies $({}^{b}f_{*})_{p}(\overline {Y}_{p})=\overline {W}_{f(p)}$ . Next, we choose $p\in Z_{1}$ . Since f is a b-map, we can take a (one-dimensional) slice S through p transverse to $Z_{1}$ , such that the restriction $f|_{S}\colon S\rightarrow f(S)$ is a diffeomorphism. Since $({}^{b}f_{*})|_{S}$ is a vector bundle map covering the diffeomorphism $f|_{S}$ , the expression $({}^{b}f_{*})|_{S}(\overline {Y}|_{S})$ is well defined and smooth. Moreover, it is equal to $\overline {W}|_{f(S)}$ on the dense subset $f(S)\setminus (\,f(S)\cap Z_{2})\subset f(S)$ , as we just proved. By continuity, the equality $({}^{b}f_{*})\mid_{S}(\overline {Y}|_{S})=\overline {W}|_{f(S)}$ holds on all of $f(S)$ , so that, in particular, $({}^{b}f_{*})_{p}(\overline {Y}_{p})=\overline {W}_{f(p)}$ . This concludes the proof.▪

Proof Fixing some notation, we have anchor maps $\widetilde {\rho }\colon\; {}^{b}TC\rightarrow TC$ and $\rho \colon\; {}^{b}TM\rightarrow TM$ , and we put $\widetilde {\mathbb {L}}:=\text{Ker}(\widetilde {\rho }|_{C\cap Z})$ and $\mathbb {L}=\text{Ker}(\rho |_{Z})$ as before. If $i\colon (C,C\cap Z)\hookrightarrow (M,Z)$ denotes the inclusion, then we get a commutative diagram with exact rows, for points $p\in C\cap Z$ :

(1.3)

We obtain $({}^{b}i_{*})_{p}(\widetilde {\mathbb {L}}_{p})=\mathbb {L}_{p}$ : the inclusion “ $\subset $ ” holds by the above diagram, and the equality follows by dimension reasons, since ${({}^{b}i_{*})_{p}} $ is injective. In particular, $\mathbb {L}_{p}$ is contained in the image of ${({}^{b}i_{*})_{p}}$ , as we wanted to show.▪

Proof

1. (i) We first note that

   $$ \begin{align*} (\,f|_{C_{1}})^{-1}(C_{2}\cap Z_{2})&=C_{1}\cap f^{-1}(C_{2}\cap Z_{2})=C_{1}\cap f^{-1}(C_{2})\cap f^{-1}(Z_{2})\\ &=C_{1}\cap f^{-1}(C_{2})\cap Z_{1}=C_{1}\cap Z_{1}, \end{align*} $$
   since f is a b-map and $C_{1}\subset f^{-1}(C_{2})$ . Next, choosing $p\in C_{1}\cap Z_{1}$ , we have to show that
   (1.4) $$ \begin{align} (\,f_{*})_{p}(T_{p}C_{1})+T_{f(p)}(C_{2}\cap Z_{2})=T_{f(p)}C_{2}. \end{align} $$
   We clearly have the inclusion “ $\subset $ ”. For the reverse, we choose $v\in T_{f(p)}C_{2}$ . By transversality $f\, \pitchfork\, Z_{2}$ , we know that $(\,f_{*})_{p}(T_{p}M_{1})+ T_{f(p)}Z_{2}=T_{f(p)}M_{2}$ . So we have $v=(\,f_{*})_{p}(x)+y$ for some $x\in T_{p}M_{1}$ and $y\in T_{f(p)}Z_{2}$ . Next, since $C_{1}\, \pitchfork\, Z_{1}$ , we have $T_{p}C_{1}+T_{p}Z_{1}=T_{p}M_{1}$ so that $x=x_{1}+x_{2}$ for some $x_{1}\in T_{p}C_{1}$ and $x_{2}\in T_{p}Z_{1}$ . So we have
   (1.5) $$ \begin{align} v=(\,f_{*})_{p}(x_{1})+\big[(\,f_{*})_{p}(x_{2})+y\big]. \end{align} $$
   The term in square brackets clearly lies in $T_{f(p)}Z_{2}$ , and being equal to $v-(\,f_{*})_{p}(x_{1})$ , it also lies in $T_{f(p)}C_{2}$ . So it lies in $T_{f(p)}(C_{2}\cap Z_{2})$ , using the transversality $C_2\, \pitchfork\, Z_{2}$ . Hence, the decomposition (1.5) is as required in (1.4).

2. (ii) Denoting the inclusions $i_{1}\colon\; (C_1,C_1\cap Z_{1})\hookrightarrow (M_{1},Z_{1})$ and $i_{2}\colon\; (C_2,C_2\cap Z_{2})\hookrightarrow (M_{2},Z_{2})$ , we have $f\circ i_{1}=i_{2}\circ f|_{C_1}$ . Hence by functoriality, ${}^{b}f_{*}\circ {}^{b}(i_{1})_{*}={}^{b}(i_{2})_{*}\circ {}^{b}(\,f|_{C_1})_{*}$ , which implies the claim.▪

Proof

1. (i) One checks that the inclusion $\Gamma (D)\subset \Gamma ({}^{b}TM)$ induces a well-defined vector bundle map

   $$ \begin{align*} \sigma\colon D\longrightarrow{}^{b}TM\colon\; v\longmapsto X_{p}, \end{align*} $$
   where $X\in \Gamma (D)$ is any extension of $v\in D_{p}$ . This map $\sigma $ satisfies $\rho \circ \sigma = \text{Id} _{D}$ , so in particular $\rho (\sigma (D))=D$ .

2. (ii) We only have to show that if $D^{\prime }$ is a subbundle of ${}^{b}TM$ intersecting trivially $\ker (\rho )$ , then $\sigma (\rho (D^{\prime }))=D^{\prime }$ . Denote $D:=\rho (D^{\prime })$ , a distribution on M tangent to Z. The canonical splitting $\sigma \colon D\rightarrow {}^{b}TM$ is injective, and D and $D^{\prime }$ have the same rank, hence it suffices to show that $\sigma (D)\subset D^{\prime }$ . If X is a section of D, then $X=\rho (Y)$ for unique $Y\in \Gamma (D^{\prime })$ . We get

   $$ \begin{align*} \rho\big(\sigma(X)\big)=X=\rho(Y), \end{align*} $$
   and since the anchor $\rho $ is injective on sections, this implies that $\sigma (X)=Y$ .▪

Proof Under the bijection of Lemma 1.10(ii), $\text{Ker}(\,f_{*})$ corresponds to $\text{Ker}({}^{b}f_{*})$ , as a consequence of Lemma 1.5.▪

Proof Denote by $VE:=\text{Ker}(\pi _{*})$ the vertical bundle. By Corollary 1.11 there is a canonical lift $\sigma \colon VE\hookrightarrow {}^{b}TE$ such that $\sigma (VE)=\text{Ker}({}^{b}\pi _{*})$ . So we get a short exact sequence of vector bundles over E:

(1.6)

Here,

$$\begin{align*}\pi^{*}({}^{b}TM)=\big\{(e,v)\in E\times{}^{b}TM:\pi(e)=pr(v)\big\} \end{align*}$$

is the pullback of the vector bundle $pr\colon {}^{b}TM\rightarrow M$ by $\pi $ , and the surjective vector bundle map

$$ \begin{align*} \widetilde{{}^{b}\pi_{*}}\colon {}^{b}TE\longrightarrow \pi^{*}({}^{b}TM),\quad(e,v)\longmapsto \left[e,({}^{b}\pi_{*})_{e}(v)\right] \end{align*} $$

is induced by the b-map $\pi \colon (E,E|_{Z})\rightarrow (M,Z)$ .

Restricting (1.6) to the zero section $M\subset E$ gives a short exact sequence of vector bundles over M:

$$ \begin{align*} 0\longrightarrow E \hookrightarrow{}^{b}TE|_{M}\overset{{{}^{b}\pi_{*}}}{\longrightarrow}{}^{b}TM\longrightarrow 0. \end{align*} $$

This sequence splits canonically through the map ${}^{b}i_{*}\colon {}^{b}TM\rightarrow {}^{b}TE|_{M}$ induced by the inclusion $i\colon (M,Z)\hookrightarrow (E,E|_{Z})$ .▪

Proof

1. (i) Since M is a b-submanifold of $(E,E|_{Z})$ , we have that ${}^{b}TM$ is a Lie subalgebroid of ${}^{b}TE$ . In particular, $\widetilde {\rho }$ and $\rho $ agree on ${}^{b}TM$ . Next, we know that $\widetilde {\rho }$ takes $E\subset {}^{b}TE|_{M}$ isomorphically to $E\subset TE|_{M}$ , thanks to Lemma 1.5 applied to $\pi $ . To see that $\widetilde {\rho }|_{E}= \text{Id} _{E}$ , we choose $v\in E_{p}$ and extend it to $V\in \Gamma (VE)$ . Denote by $\sigma \colon VE\hookrightarrow {}^{b}TE$ the canonical splitting of $\widetilde {\rho }$ , as in the proof of Lemma 1.12. Then $\widetilde {\rho }(v)=[\widetilde {\rho }(\sigma (V))]_{p}=V_{p}=v$ .

2. (ii) It is routine to check that $\varphi $ is a b-map, so we only prove the second statement. Taking the b-derivative of both sides of the equality $\pi _2\circ \varphi =f\circ \pi _1$ at a point $p\in M_{1}$ , we know that $({}^{b}\pi _{2})_{*}({}^{b}\varphi _{*}(E_{1})_{p})={}^{b}f_{*}(({}^{b}\pi _{1})_{*}(E_{1})_{p})=0$ , since $(E_{1})_{p}=\text{Ker}[({}^{b}\pi _{1})_{*}]_{p}$ . Hence, ${}^{b}\varphi _{*}(E_{1})_{p}\subset \text{Ker}[({}^{b}\pi _{2})_{*}]_{f(p)}=(E_{2})_{f(p)}$ by the proof of Lemma 1.12. Using (i) and the diagram (1.2), we have a commutative diagram

   (1.8)

   

   It implies that

   $$ \begin{align*} {}^{b}\varphi_{*}|_{(E_{1})_{p}}= \varphi_{*}|_{(E_{1})_{p}}= \varphi|_{(E_{1})_{p}}. \end{align*} $$
   Finally, $ {}^{b}\varphi _{*}|_{{}^{b}TM_{1}}={}^{b}f_{*}$ holds by Lemma 1.9(ii).▪

Proof We adapt the proof of [Reference Cannas da Silva3, Proposition 6.8]. We first choose a suitable tubular neighborhood of C that is compatible with the hypersurface Z. Due to transversality $C\, \pitchfork\, Z$ , we can pick a complement V to $TC$ in $TM|_{C}$ such that $V_{p}\subset T_{p}Z$ for all $p\in C\cap Z$ . Fix a Riemannian metric g for which $Z\subset (M,g)$ is totally geodesic (e.g.,[Reference Milnor19, Lemma 6.8]). The associated exponential map then establishes a b-diffeomorphism between a neighborhood of C in $(V,V|_{C\cap Z})$ and a neighborhood of C in $(M,Z)$ .

So we can work instead on the total space of $\pi \colon (V,V|_{C\cap Z})\rightarrow (C,C\cap Z)$ . Consider the retraction of V onto C given by ${r}\colon V\times [0,1]\rightarrow V\colon (p,v,t)\mapsto (p,tv)$ , and notice that the ${r}_{t}$ are b-maps. The associated time-dependent vector field $X_{t}$ is given by $X_{t}(p,v)={\frac {1}{t}v}$ , which is a b-vector field that vanishes along C. It follows that we get a well-defined b-de Rham homotopy operator

$$ \begin{align*} I\colon {}^{b}\Omega^{k}(V)\longrightarrow{}^{b}\Omega^{k-1}(V)\colon \alpha\longmapsto\int_{0}^{1}{}^{b}{r}_{t}^{*}(\iota_{X_{t}}\alpha)dt, \end{align*} $$

which satisfies

(1.10) $$ \begin{align} {}^{b}{r}_{1}^{*}\alpha-{}^{b}{r}_{0}^{*}\alpha={}^{b}d I(\alpha)+I({}^{b}d\alpha). \end{align} $$

Since ${r}_{1}= \text{Id} $ and ${r}_{0}=i\circ \pi $ , formula (1.10) gives $\beta ={}^{b}d I(\beta )$ . Now set $\eta :=I(\beta )$ .▪

Proof Consider the convex combination $\omega _{t}:=\omega _{0}+t(\omega _{1}-\omega _{0})$ for $t\in [0,1]$ . There exists a neighborhood U of C such that $\omega _{t}$ is non-degenerate on U for all $t\in [0,1]$ . Shrinking U if necessary, Lemma 1.18 yields $\eta \in {}^{b}\Omega ^{1}(U)$ such that $\omega _{1}-\omega _{0}={}^{b}d\eta $ and $\eta |_{C}=0$ . As in the usual Moser trick, it now suffices to solve the equation

$$ \begin{align*} \iota_{X_{t}}\omega_{t}+\eta=0 \end{align*} $$

for $X_{t}\in {}^{b}\mathfrak {X}(U)$ , which is possible by non-degeneracy of $\omega _{t}$ . The b-vector fields $X_{t}$ thus obtained vanish along C, since $\eta |_{C}=0$ . Further shrinking U if necessary, we can integrate the $X_{t}$ to an isotopy $\{{\phi _{t}}\}_{t\in [0,1]}$ defined on U. Note that the ${\phi _{t}}$ are b-diffeomorphisms that restrict to the identity on C. By the usual Moser argument, we have ${}^{b}{\phi _{1}}^{*}\omega _{1}=\omega _{0}$ , so setting $\varphi :={\phi _{1}}$ finishes the proof.▪

Proof If C is coisotropic, then at points of $C \cap (M\setminus Z)$ , we have that $TC^\omega \subset TC$ , i.e., $({}^{b}TC)^\omega \subset {}^{b}TC$ . By continuity, this inclusion of subbundles holds at all points of C. Conversely, if this inclusion holds on C, it follows that $C \cap (M\setminus Z)$ is coisotropic in $M\setminus Z$ , and using characterization (ii) in Definition 2.1, we see that C is coisotropic in M.▪

Proof At points $p\in C\setminus (C\cap Z)$ , equality (2.1) holds by symplectic linear algebra. So let $p\in C\cap Z$ . Denote by ${}^{b}\Pi :={-}\omega ^{-1}\in \Gamma (\wedge ^{2}({}^{b}TM))$ the lift of $\Pi $ as a b-bivector field. Note that

(2.2) $$ \begin{align} ({}^{b}T_{p}C)^{\omega_{p}}=(\omega_{p}^{\flat})^{-1}\big(({}^{b}T_{p}C)^{0}\big)={}^{b}\Pi^{\sharp}\big(({}^{b}T_{p}C)^{0}\big), \end{align} $$

where the annihilator is taken in ${}^{b}T^{*}_{p}M$ . We now assert:

Claim. $({}^{b}T_{p}C)^{0}=\rho _{p}^{*}(T_{p} C^{0}).$

To prove the claim, we first note that the dimensions of both sides agree, since

$$ \begin{align*} \text{Ker}(\rho_{p}^{*})\cap T_{p}C^{0}= Im(\rho_{p})^{0}\cap T_{p}C^{0}=T_{p}Z^{0}\cap T_{p}C^{0}=(T_{p}Z+T_{p}C)^{0}=\{0\}, \end{align*} $$

where the last equality holds by transversality $C\, \pitchfork\, Z$ . Now it is enough to show that the inclusion “ $\supset $ ” holds, which is clearly the case, since $\rho _{p}({}^{b}T_{p}C)\subset T_{p}C$ . This proves the claim.

We thus obtain

$$ \begin{align*} \rho_{p}\big(({}^{b}T_{p}C)^{\omega_{p}}\big)=(\rho_{p}\circ{}^{b}\Pi_{p}^{\sharp}\circ\rho_{p}^{*})(T_{p}C^{0}) =\Pi_{p}^{\sharp}(T_{p}C^{0}), \end{align*} $$

where in the first equality we used (2.2), and the claim just proved, and in the second, we used the diagram (1.9).▪

Proof Diagram (2.3) is a diagram of vector subbundles of diagram (1.9), by the claim in the proof of Lemma 2.5 and by equation (2.2).

For (i), since the morphism ${}^{b}\Pi ^{\sharp }$ in diagram (1.9) is an isomorphism of Lie algebroids, it suffices to show that $({}^{b}TC)^{\omega }$ is a Lie subalgebroid of ${}^{b}TM$ . Since $({}^{b}TC)^{\omega }$ is the kernel of the closed b-2-form ${}^{b}i^* \omega $ , a standard Cartan calculus computation shows that this is indeed the case. It is well known that $TC^{\circ }$ and $TC$ are also Lie subalgebroids, proving (ii). ▪

Proof The intersection $graph(\,f)\cap Z_1\times Z_2$ is clean, since it coincides with $graph(\,f|_{Z_1})$ thanks to the assumption $f(Z_1)\subset Z_2$ . Hence, $graph(\,f)$ can be “lifted” to X.

The resulting submanifold $\overline {\overline {graph(\,f)}}$ is coisotropic: $graph(\,f)$ is coisotropic in $M_1\times {\widehat {M}_2}$ , because f is a Poisson map, so $p^{-1}(graph(\,f)\setminus Z_1\times Z_2)$ is coisotropic in X (since p is a Poisson diffeomorphism away from the exceptional divisor), and the same holds for its closure.

To finish the proof, we have to show that $\overline {\overline {graph(\,f)}}$ is transverse to the exceptional divisor $E:=p^{-1}(Z_1\times Z_2)$ . Let $(x^{(1)}_{i})$ be local coordinates on $M_1$ such that $Z_1=\{x^{(1)}_{1}=0\}$ , and similarly, let $(x^{(2)}_{j})$ be local coordinates on $M_2$ such that $Z_2=\{x^{(2)}_{1}=0\}$ . Then

$$ \begin{align*} \big(x^{(1)}_{i},f^*(x^{(2)}_{j})-x^{(2)}_{j}\big) \end{align*} $$

are local coordinates on $M_1\times M_2$ that are adapted $graph(\,f)$ , but also to

$$ \begin{align*}(Z_1\times Z_2)=\big\{x^{(1)}_{1}=0,f^*(x^{(2)}_{1})-x^{(2)}_{1}=0\big\}, \end{align*} $$

due to the hypothesis $f(Z_1)\subset Z_2$ . Hence, we can apply Lemma 2.9, which yields the desired transversality.▪

Proof We have

$$ \begin{align*}Bl_Z({\mathbb R}^{n+m})=\overline{\Big\{\Big((x,y),[x]\Big): x\in {\mathbb R}^{n}\setminus\{0\}, y\in {\mathbb R}^{m}\Big\}}\;\subset {\mathbb R}^{n+m}\times {\mathbb R} P^{n-1},\end{align*} $$

where $[\cdot ]$ denotes the class in projective space. Notice that by taking the closure, we are adding exactly the exceptional divisor

$$ \begin{align*}E=\{0\}\times {\mathbb R}^{m}\times {\mathbb R} P^{n-1}.\end{align*} $$

We have

$$ \begin{align*} \overline{\overline{S}}=\overline{\Big\{\Big((x_1,0,y_1,0),[(x_1,0)]\Big): x_1\in {\mathbb R}^{k}\setminus\{0\}, y_1\in {\mathbb R}^{l}\Big\}}. \end{align*} $$

By taking the closure we are adding exactly

$$ \begin{align*}\{0\}\times \big({\mathbb R}^{l}\times\{0\}\big)\times {\mathbb R} P^{k-1}=\overline{\overline{S}}\cap E.\end{align*} $$

For every point $p\in \overline {\overline {S}}\cap E$ there is a curve of the form

$$ \begin{align*}\gamma\colon t\longmapsto \big((tx_1,0,y_1,0),[(x_1,0)]\big)\end{align*} $$

lying in $\overline {\overline {S}}$ with $\gamma (0)=p$ , and clearly $\frac {d}{dt}|_0\gamma (t)\notin T_pE$ . Since $\frac {d}{dt}|_0\gamma (t)\in T_p\overline {\overline {S}}$ and E has codimension $1$ , we obtain $T_pE+T_p\overline {\overline {S}}=T_pBl_Z({\mathbb R}^{n+m})$ .▪

Proof Fix a complement G to E in ${}^{b}TC$ , and let $j\colon E^{*}\hookrightarrow {}^{b}T^{*}C$ be the induced inclusion. It is clear that $j(E^{*})=G^{0}$ . Since both the bundle projection $\pi \colon (E^{*},E^{*}|_{Z_{C}})\rightarrow (C,Z_{C})$ and the inclusion $j\colon (E^{*},E^{*}|_{Z_{C}})\rightarrow ({}^{b}T^{*}C,{}^{b}T^{*}C|_{Z_{C}})$ are b-maps, we can define a b-two-form $\Omega _{G}$ on $(E^{*},E^{*}|_{Z_{C}})$ by

(2.6) $$ \begin{align} \Omega_{G}:={}^{b}\pi^{*}\omega_{C}+{}^{b}j^{*}\omega_{{can}}. \end{align} $$

Here, $\omega _{{can}}$ denotes the canonical b-symplectic form on ${}^{b}T^{*}C$ as in Example 1.17, and the subscript G is used to stress that the definition depends on the choice of complement G.

We want to show that $\Omega _{G}$ is b-symplectic on a neighborhood of $C\subset (E^{*},E^{*}|_{Z_{C}})$ . As $\Omega _{G}$ is clearly b-closed, it suffices to prove that $\Omega _{G}$ is non-degenerate at points $p\in C$ .

We claim that under the decomposition

$$\begin{align*}{}^{b}T_{p}({}^{b}T^{*}C)\cong{}^{b}T_{p}C\oplus{}^{b}T_{p}^{*}C, \end{align*}$$

of Lemma 1.12, the canonical b-symplectic form is the usual pairing

(2.7) $$ \begin{align} (\omega_{{can}})_{p}(v+\alpha,w+\beta)=\langle v,\beta\rangle - \langle w,\alpha\rangle. \end{align} $$

This claim can be checked writing in cotangent coordinates $ \omega _{{can}}=\frac {dx_{1}}{x_{1}}\wedge dy_{1}+\sum _{i=2}^{n}dx_{i}\wedge dy_{i}, $ and noticing that $y_{i}$ is a linear coordinate on each fiber ${}^{b}T_{p}^{*}C$ , i.e., $y_{i}\in ({}^{b}T_{p}^{*}C)^{*}\cong {}^{b}T_{p}C$ .

Consider now the decomposition

(2.8) $$ \begin{align} {}^{b}T_{p}E^{*}\cong{}^{b}T_{p}C\oplus E_{p}^{*}= E_{p}\oplus G_{p}\oplus E_{p}^{*} \end{align} $$

given by Lemma 1.12. Using Lemma 1.13(ii), we have $({}^{b}j_{*})_{p}= \text{Id} _{ {}^{b}T_{p}C}\oplus j|_{E_{p}^{*}}$ . Hence, under decomposition (2.8), we have

$$ \begin{align*} ({}^{b}j^{*}\omega_{{can}})_{p}(v+w+\alpha,x+y+\beta)&=\big(\omega_{{can}})_{p}(v+w+j(\alpha),x+y+j(\beta)\big)\\ &=\langle v+w,j(\beta)\rangle - \langle x+y,j(\alpha)\rangle\\ &=\langle v,j(\beta)\rangle-\langle x,j(\alpha)\rangle, \end{align*} $$

using the above claim and recalling that $j(E_{p}^{*})=G_{p}^{0}$ . In matrix notation,

(2.9)

for some matrix A of full rank. Similarly, we have $({}^{b}\pi _{*})_{p}= \text{Id} _{ {}^{b}T_{p}C}\oplus 0$ , applying Lemma 1.13(ii) to $\pi $ (regarded as a vector bundle map). Therefore, under (2.8), we get

$$ \begin{align*} ({}^{b}\pi^{*}\omega_{C})_{p}(v+w+\alpha,x+y+\beta)=(\omega_{C})_{p}(v+w,x+y) \end{align*} $$

so that we get a matrix representation of the form

(2.10)

where we also use that $E=\text{Ker}(\omega _{C})$ . Note that the matrix B in (2.10) is of full rank, since the restriction of $(\omega _{C})_{p}$ to $G_{p}$ is non-degenerate. Combining (2.9) and (2.10), we have that

(2.11)

which is of maximal rank. Therefore, $\Omega _{G}$ is non-degenerate at points $p\in C\subset E^{*}$ .▪

Proof Proof of Theorem 2.13(a)

We show that the inclusion $(C,Z_{C},\omega _{C})\overset {i}\hookrightarrow (E^{*},E^{*}|_{Z_{C}},\Omega _{G})$ is indeed a b-coisotropic embedding; i.e.,

1. (i) ${}^{b}i^{*}\Omega _{G}=\omega _{C}$ ;

2. (ii) ${}^{b}TC^{\Omega _{G}}\subset {}^{b}TC$ .

We have ${}^{b}i^{*}\Omega _{G}={}^{b}(\pi \circ i)^{*}\omega _{C}+{}^{b}(j\circ i)^{*}\omega _{\text{can}}=\omega _{C}+{}^{b}(j\circ i)^{*}\omega _{\text{can}}$ . Note that $j\circ i$ is the inclusion of C into ${}^{b}T^{*}C$ , so that ${}^{b}(j\circ i)^{*}\omega _{\text{can}}=0$ , since C is b-Lagrangian in $({}^{b}T^{*}C,\omega _{\text{can}})$ . This proves (i).

To check (ii), we let $p\in C$ and choose $v+w+\alpha \in E_{p}\oplus G_{p}\oplus E_{p}^{*}\cong {}^{b}T_{p}E^{*}$ lying in ${}^{b}T_{p}C^{\Omega _{G}}$ . Let $x\in E_{p}\subset {}^{b}T_{p}C$ be arbitrary. Thanks to (2.11), we then have

$$ \begin{align*} 0=(\Omega_{G})_{p}(x,v+w+\alpha)=(\Omega_{G})_{p}(x,\alpha), \end{align*} $$

which forces that $\alpha =0$ due to non-degeneracy of $(\Omega _{G})_{p}$ on $E_{p}\times E_{p}^{*}$ . Hence, $v+w+\alpha =v+w$ lies in $E_{p}\oplus G_{p}={}^{b}T_{p}C$ , as desired.▪

Proof Since $\omega |_{G\times G}$ is non-degenerate, we have a decomposition ${}^{b}TM|_{C}=G\oplus G^{\omega }$ as symplectic vector bundles. Note that E is a Lagrangian subbundle of $(G^{\omega },\omega )$ , since

(2.12) $$ \begin{align} E^{\omega}\cap G^{\omega}=(E\oplus G)^{\omega}={}^{b}TC^{\omega}={}^{b}TC^{\omega}\cap{}^{b}TC=E. \end{align} $$

We fix a Lagrangian complement V to E in $(G^{\omega },\omega )$ , i.e., $G^{\omega }=E\oplus V$ .

The idea of the proof is to construct a b-diffeomorphism between neighborhoods of C in M and $E^*$ —obtained as a composition of b-diffeomorphisms to a neighborhood in V— whose b-derivative at points of C pulls back $\Omega _{G}$ to $\omega $ , and then apply a Moser argument.

We start by establishing a b-geometry version of the tubular neighborhood theorem, in which V plays the role of the normal bundle to C.

We will construct this map in two steps:

$$\begin{align*}V\xrightarrow[(1)]{} \rho(V)\xrightarrow[(2)]{} M. \end{align*}$$

Step 1. Let $\rho \colon {}^{b}TM\rightarrow TM$ denote the anchor map of ${}^{b}TM$ and notice that its restriction to V is injective. To see this, recall the decomposition

(2.13) $$ \begin{align} {}^{b}TM|_{C}=G\oplus G^{\omega}=G\oplus E\oplus V={}^{b}TC\oplus V \end{align} $$

and the fact that $\text{Ker}(\rho |_{C})\subset {}^{b}TC$ by Lemma 1.8, so that $\text{Ker}(\rho )$ intersects V trivially. As such, we get a b-diffeomorphism $\rho \colon (V,V|_{C\cap Z})\rightarrow (\rho (V),\rho (V)|_{C\cap Z})$ .Step 2. The distribution $\rho (V)$ is complementary to $TC$ , i.e.,

$$ \begin{align*} TM|_{C}=TC\oplus\rho(V). \end{align*} $$

Indeed, by Step 1, we have at any point $p\in C$ ,

$$\begin{align*}\dim(T_{p}M)=\dim(T_{p}C)+\dim(V_{p}) =\dim(T_{p}C)+\dim(\rho(V_{p})), \end{align*}$$

and moreover, if $v\in V_{p}$ is such that $\rho (v)\in T_{p}C$ , then $v\in {}^{b}T_{p}C\cap V_{p}=\{0\}$ . Now fix a Riemannian metric g on M such that $Z\subset (M,g)$ is totally geodesic (e.g., [Reference Milnor19, Lemma 6.8]). The corresponding exponential map $\exp _{g}$ takes a neighborhood of $C\subset \rho (V)$ diffeomorphically onto a neighborhood of $C\subset M$ . Moreover, the fibers of $\rho (V)$ over $C\cap Z$ are mapped into Z, since $\rho (V_{p})\in T_{p}Z$ for $p\in C\cap Z$ and Z is totally geodesic. Therefore, the mapFootnote ² $\exp _{g}\colon (\rho (V),\rho (V)|_{C\cap Z})\rightarrow (M,Z)$ is a b-diffeomorphism between neighborhoods of C.

We now show that $\phi :=\exp _{g}\circ \rho \colon V\to M$ has the claimed property. That is, we show that $[{}^{b}(\exp _{g}\circ \rho )_{*}]|_{C}$ is the identity map on ${}^{b}TV|_{C}\cong {}^{b}TC\oplus V={}^{b}TM|_{C}$ , by checking that it acts as the identity on sections. We will need the commutative diagram

(2.14)

which implicitly uses Lemma 1.13(i). We will also use that for all $q\in C$ the ordinary derivative reads

(2.15) $$ \begin{align} \big[(\exp_{g}\circ\rho)_{\ast}\big]_{q}\colon T_{q}V\cong T_{q}C\oplus V_{q} & \longrightarrow T_{q}M = T_{q}C\oplus\rho(V_{q})\nonumber \\ w+v & \longmapsto w+\rho(v). \end{align} $$

For a section $X+Y\in \Gamma ({}^{b}TC\oplus V)$ , we now compute

$$ \begin{align*} \rho[{}^{b}(\exp_{g}\circ\rho)_{*}(X+Y)]&=(\exp_{g}\circ\rho)_{*}(\rho(X)+Y)\\ &=\rho(X)+\rho(Y)\\ &=\rho(X+Y), \end{align*} $$

using (2.14) in the first equality and (2.15) in the second. Since the anchor $\rho $ is injective on sections, this implies that ${}^{b}(\exp _{g}\circ \rho )_{*}(X+Y)=X+Y$ , as desired. Claim 2.16 is proved.

Next, the map

$$ \begin{align*} \psi\colon V\longrightarrow E^{*},\quad v\longmapsto -\iota_{v}\omega|_{E} \end{align*} $$

is an isomorphism of vector bundles covering $ \text{Id} _{C}$ , whence a b-diffeomorphism between the total spaces (\,For the injectivity, note that $ \iota _{v}\omega |_{E}=0$ implies that $v\in E^{\omega }\cap G^{\omega }=E$ as in (2.12), so that $v\in V\cap E=\{0\}$ ). The composition $\psi \circ \phi ^{-1}\colon (M,Z)\rightarrow (E^{*},E^{*}|_{C\cap Z})$ is a b-diffeomorphism between neighborhoods of C, with $(\psi \circ \phi ^{-1})|_{C}= \text{Id} _{C}$ .

As before, let $\pi \colon E^{*}\rightarrow C$ denote the bundle projection, and let $j\colon E^{*}\hookrightarrow {}^{b}T^{*}C$ be the inclusion induced by the splitting ${}^{b}TC=E\oplus G$ . Since $\psi \colon V\rightarrow E^{*}$ is a vector bundle morphism covering $ \text{Id} _C$ , by Lemma 1.13(ii), we have that

$$ \begin{align*} {}^{b}\psi_{*}|_C\colon {}^{b}TV|_C\cong{}^{b}TC\oplus V\longrightarrow {}^{b}TE^*|_C\cong{}^{b}TC\oplus E^* \end{align*} $$

equals $ \text{Id}_{^{b}TC} \oplus \psi $ . Furthermore, ${}^{b}\phi _{*}|_{C}= \text{Id} _{{}^{b}TM|_{C}}$ by Claim 2.16. Therefore, for $p\in C$ and $x_i+v_i\in {}^{b}T_pC \oplus V_{p}={}^{b}T_{p}M$ , we have

(2.16) $$ \begin{align} \big[{}^{b}(\psi\circ\phi^{-1})^{*}\Omega_{G}\big]_{p}(x_{1}+v_{1},x_{2}+v_{2})=(\Omega_{G})_{p}\big(x_{1}+\psi(v_{1}),x_{2}+\psi(v_{2})\big). \end{align} $$

Recalling equation (2.6) and applying Lemma 1.13(ii) as in the proof of Proposition 2.14, we expand the right-hand side of (2.16) as follows:

$$ \begin{align*} (\Omega_{G})_{p}\big(x_{1}+\psi(v_{1}),x_{2}+\psi(v_{2})\big) & =\omega_{p}(x_{1},x_{2})+(\omega_{{can}})_{p}(x_{1}+j(\psi(v_{1})),x_{2}+j(\psi(v_{2})))\\ &=\omega_{p}(x_{1},x_{2})+\left\langle x_{1},j(\psi(v_{2}))\right\rangle-\left\langle x_{2},j(\psi(v_{1}))\right\rangle\\ &=\omega_{p}(x_{1},x_{2})+\langle e_{1},\psi(v_{2})\rangle - \langle e_{2},\psi(v_{1})\rangle\\ &=\omega_{p}(x_{1},x_{2})+\omega_{p}(e_{1},v_{2})+\omega_{p}(v_{1},e_{2})\\ &=\omega_{p}(x_{1}+v_{1},x_{2}+v_{2}), \end{align*} $$

using equation (2.7) in the second equality, writing $x_i=e_i+g_i\in E_{p}\oplus G_{p}={}^{b}T_{p}C$ , and using in the last equality that V is a Lagrangian subbundle of $(G^{\omega },\omega )$ . This finishes the proof of Claim 2.17.

Applying Proposition 1.19 (relative b-Moser) yields a b-diffeomorphism $\sigma $ , defined on a neighborhood of $C\subset (M,Z)$ , such that ${}^{b}\sigma ^{*}({}^{b}(\psi \circ \phi ^{-1})^{*}\Omega _{G})=\omega $ and $\sigma |_{C}= \text{Id} _{C}$ . So setting $\tau :=\psi \circ \phi ^{-1}\circ \sigma $ finishes the proof.▪

Proof We know that D has constant rank, by Proposition 3.2. As for involutivity, first note that D is generated by Hamiltonians $\left .X_{h}\right |{}_{C}$ of functions $h\in \mathcal {I}_{C}$ . On such generators, we have

$$ \begin{align*} [X_{h_{1}}|_{C},X_{h_{2}}|_{C}]=[X_{h_{1}},X_{h_{2}}]|_{C}= X_{\{h_{1},h_{2}\}} |_{C}, \end{align*} $$

where $\{h_{1},h_{2}\}\in \mathcal {I}_{C}$ due to coisotropicity of C. Hence, D is involutive.

The quotient $\underline {C\cap Z}:=(C\cap Z)/D$ is a smooth submanifold of $\underline {C}$ , since for every slice S in C transverse to D, the intersection $S\cap Z$ is a smooth slice in $C\cap Z$ transverse to D. The leaf space $(\underline {C},\underline {C\cap Z})$ is a b-manifold, and the projection $q:(C,C\cap Z)\rightarrow (\underline {C},\underline {C\cap Z})$ is a b-map. For $p\in C$ , we have an exact sequence

$$\begin{align*}0\longrightarrow D_{p}\hookrightarrow T_{p}C\overset{(q_{*})_{p}}{\longrightarrow} T_{q(p)}\underline{C}\longrightarrow 0, \end{align*}$$

which corresponds with an exact sequence on the level of b-tangent spaces

(3.3) $$ \begin{align} 0\longrightarrow({}^{b}T_{p}C)^{\omega_{p}}\hookrightarrow{}^{b}T_{p}C\overset{({}^{b}q_{*})_{p}}{\longrightarrow}{}^{b}T_{q(p)}\underline{C}\longrightarrow 0. \end{align} $$

To see this, consider the canonical splitting $\sigma \colon D\rightarrow {}^{b}TC$ of the anchor $\rho \colon {}^{b}TC\rightarrow TC$ , as constructed in Lemma 1.10(i), and notice that

$$ \begin{align*} \text{Ker}\big(({}^b q_{*})\big)_p=\sigma(\text{Ker}(q_{*})_{p})=\sigma(D_{p})=({}^{b}T_{p}C)^{\omega_{p}}, \end{align*} $$

where the first equality holds by Corollary 1.11 and the third by Corollary 3.3.

Since q is a surjective submersion, it admits sections; hence, for every sufficiently small open subset $U\subset \underline {C}$ , there is a submanifold $S\subset C$ transverse to D such that $q|_{S}\colon S\to U$ is a diffeomorphism. At points $p\in S$ , we have

$$ \begin{align*} {}^{b}T_{p}C =({}^{b}T_{p}C)^{\omega_{p}}\oplus{}^{b}T_{p}S \end{align*} $$

due to sequence (3.3). This implies that ${}^{b}i_{S}^{*}\omega _{C}$ is a b-symplectic form on S, where $i_{S}\colon S\hookrightarrow C$ is the inclusion and $\omega _{C}$ is the restriction of $\omega $ to C. Denote by $\tau \colon U\rightarrow S$ the inverse of $q|_{S}\colon S\rightarrow U$ . Then $\underline {\Omega }:={}^{b}\tau ^{*}({}^{b}i_{S}^{*}\omega _{C})$ is b-symplectic on U. Away from $\underline {C\cap Z}$ , this b-2-form agrees with the symplectic form obtained by symplectic coisotropic reduction from $\omega |_{M\setminus Z}$ . Denote by $-\underline {\Omega }^{-1}$ the non-degenerate b-bivector on U corresponding to $\underline {\Omega }$ . Away from $\underline {C\cap Z}$ , the log-symplectic structure $\underline {\rho }(-\underline {\Omega }^{-1})$ agrees with the reduced Poisson structure, by Remark 3.5. By continuity, the same is true on the whole of U. As U was arbitrary, the reduced Poisson structure on $\underline {C}$ is log-symplectic, and the above reasoning shows that the corresponding b-symplectic form satisfies equation (3.2).▪

Proof

1. (i) We show that $0$ is a regular value of J. Choosing $p\in J^{-1}(0)$ , it is enough to prove that the restriction $d_{p}J\; \colon\; Im (\Pi _{p}^{\sharp })\subset T_{p}M\rightarrow \mathfrak {g}^{*}$ is surjective. To this end, assume that $\xi \in \mathfrak {g}$ annihilates $d_{p}J(Im (\Pi _{p}^{\sharp }))$ . We then get for all $\alpha \in T_{p}^{*}M$ that

   $$ \begin{align*} \left\langle\alpha,(v_{\xi})_{p}\right\rangle=\left\langle\alpha,\Pi_{p}^{\sharp}(d_{p}J^{\xi})\right\rangle=-\left\langle d_{p}J^{\xi},\Pi_{p}^{\sharp}(\alpha)\right\rangle=-\left\langle d_{p}J(\Pi_{p}^{\sharp}(\alpha)),\xi\right\rangle=0, \end{align*} $$
   and therefore $(v_{\xi })_{p}=0$ . Since the action $G\curvearrowright J^{-1}(0)$ is free, this implies that $\xi =0$ . It follows that $d_{p}J(Im (\Pi _{p}^{\sharp }))=\mathfrak {g}^{*}$ , so $0$ is indeed a regular value of J. In particular, $J^{-1}(0)$ is a submanifold of M. The coisotropicity of $J^{-1}(0)$ follows since it is the preimage of a symplectic leaf $\{0\}\subset \mathfrak {g}^{*}$ under a Poisson map.

2. (ii) Let $\mathcal {O}$ denote the symplectic leaf through $p\in J^{-1}(0)$ . By the computation (3.1), it suffices to prove that $\Pi _{p}^{\sharp }|_{[T_{p}J^{-1}(0)]^{0}}$ is injective. Since $0$ is a regular value, this annihilator is given by $[T_{p}J^{-1}(0)]^{0}\nonumber =\{ d_{p}J^{x}: x\in \mathfrak {g}\}.$ We now have a composition of maps

   $$ \begin{align*} &\mathfrak{g}\longrightarrow \big[T_{p}J^{-1}(0)\big]^{0} \longrightarrow\Pi_{p}^{\sharp} \big(\big[T_{p}J^{-1}(0)\big]^{0}\big)\\ &x\longmapsto d_{p}J^{x}\longmapsto\Pi_{p}^{\sharp}(d_{p}J^{x})=(v_{x})_{p}, \end{align*} $$
   that is injective by freeness of $G\curvearrowright J^{-1}(0)$ . In particular, $\Pi _{p}^{\sharp }|_{[T_{p}J^{-1}(0)]^{0}}$ is injective.

3. (iii) We have

   $$ \begin{align*} \Pi_{p}^{\sharp}\big([T_{p}J^{-1}(0)]^{0}\big)=\big\{ \Pi_{p}^{\sharp}(d_{p}J^{x}): x\in\mathfrak{g}\big\}=\big\{ (v_{x})_{p}: x\in\mathfrak{g}\big\}, \end{align*} $$
   which is exactly the tangent space of the G-orbit through p.▪