text:
\begin{document}

\title{{An extension of the Marsden-Ratiu reduction for Poisson manifolds}}

\author{Fernando Falceto}
\address{Departamento de F\'{\i}sica Te\'orica and
Instituto de Biocomputaci\'on y F\'{\i}sica de Sistemas Complejos,
Universidad de Zaragoza,
E-50009 Zaragoza (Spain)}
\email{falceto@unizar.es}

\author{Marco Zambon}
\address{Centre de Recerca Matematica, Apartat de correus 50, 08193 Bellaterra
(Spain)}
\email{mzambon@crm.cat}

\thanks{ {\it Keywords:} Poisson manifolds. Reduction.\\
{\it 2000 MSC:} 53D17, 53D20, 53D99.\\}

\begin{abstract}
\vskip 0.3cm
\noindent
We propose a generalization of the reduction of Poisson manifolds by
distributions introduced by Marsden and Ratiu.
Our proposal overcomes some of the restrictions of the original procedure,
and  makes the reduced Poisson structure
effectively dependent on the distribution.
Different applications are discussed, as well as the algebraic
interpretation of the procedure and its formulation in terms
of Dirac structures.
\end{abstract}
\maketitle

\section{Introduction}

Symplectic manifolds model phase spaces of physical systems, and
their theory of reduction   is a classical subject. A case in which
reduction occurs naturally is when a Lie group $G$ acts on a
symplectic manifold $(M,\omega)$ with equivariant moment map
$J:M\rightarrow \g^*$: under regularity assumptions the
Marsden-Weinstein theorem  states that the quotients
$J^{-1}(\mu)/G_{\mu}$ inherit a symplectic form. Another case is
given by submanifolds $C\subset M$ such that $TC^{\omega}\subset TC$
(coisotropic submanifolds), for in that case the quotient $C/
TC^{\omega}$, when smooth, inherits a symplectic form. The theory of
reduction  extends naturally to  Poisson manifolds, which encode
phase spaces of physical systems with symmetry.

The hamiltonian reduction and coisotropic reduction mentioned above
extend in a straightforward way to Poisson manifolds. Further,  both
are recovered as special cases of a reduction theorem stated in 1986
by Marsden and Ratiu  \cite{MarRat}.

The starting data of the Marsden-Ratiu theorem is a pair $(N,B)$
where $N$ is a submanifold  of the Poisson manifold $(M,\Pi)$ and
$B$  a subbundle of $T_NM$, the restriction of $TM$ to $N$. The
role of $B$ is to prescribe how
 to extend certain functions on $N$ to functions on the whole of
$M$, and is needed because the Poisson bracket of $M$ is defined
only for elements of $C^{\infty}(M)$. The conclusion of the theorem
is that, when the assumptions are met, the quotient $N/(B\cap TN)$
inherits a Poisson bracket from the one on  $M$.\\

The aim of this paper is two-fold.  First  we argue that the
assumptions of the Marsden-Ratiu theorem are too strong, in the
sense that the theorem allows to recover only  Poisson structures (on
quotients of $N$) which lose most of the information encoded by the
subbundle $B$. See Thm. \ref{MR2} -- our rephrasing of the original Marsden-Ratiu theorem -- and Prop. \ref{pr:unique}.

Then we set weaker assumptions on the pair $(N,B)$ which  ensure the
existence of  a Poisson structure on $N/(B\cap TN)$   encoding the
subbundle $B$. The main difficulty consists in ensuring that the
bracket of functions on the quotient satisfies the Jacobi identity.
 In
Prop. \ref{onn} we set assumptions  similar in spirit to those of
\cite{MarRat}, whereas in Prop. \ref{thetaD} the assumptions involve
an additional piece of data, namely a foliation on $M$. We apply
these results to the symplectic setting (with and without moment
map) as well.

The paper is organized as follows:
in the next section we review the original reduction of Marsden and Ratiu.
In section 3 we present the most general form of the extension
that we propose, while section 4 is devoted to the application of the previous
results to some special situations and examples.
We collect in the appendix some complementary results,
like the algebraic interpretation of our reduction,
its description in term of Dirac structures and other auxiliary
material necessary for the main body of the paper.

We finish remarking that an
extension of the Marsden-Ratiu reduction
using supergeometry is
being worked out in \cite{CZsup}.\\

\noindent{\bf Acknowledgments:} We thank  J.P. Ortega  for pointing
out a necessary assumption in Prop. \ref{thetaD}.
 Research partially
supported by grants FPA2003-02948 and FPA2006-02315, MEC and
SB2006-0141(Spain) as well as SNF grant 20-113439 (Switzerland).

 \section{Marsden-Ratiu reduction}\label{MRpaper}

We start by recalling the Poisson reduction by distributions as it was stated
by Marsden and Ratiu in \cite{MarRat}, see also \cite{OrtRat}.
The set-up we consider here and in the rest of the paper is the following:
\begin{center}
\fbox{
\parbox[c]{12.6cm}{\begin{center}
$(M,\{\cdot,\cdot\})$ is a Poisson manifold\\
$N$ is a submanifold with embedding $\iota: N\hookrightarrow M$\\
${B}\subset T_N M$ is  a smooth
subbundle of $TM$ restricted to $N$.\\
${{F}}:={B}\cap TN$ is an integrable regular distribution on $N$.
\end{center}
}}
 \end{center}

In this section we shall also assume that $\un{N}:=N/F$ is a smooth manifold so that the projection map is a submersion.
Since we are concerned with  functions on $N$, instead of working with functions on the whole of $M$, we work on  a suitable tubular neighborhood  $M'$ of $N$ in $M$.

\begin{lemma}\label{surj}

There exists an open subset    $M'$ of $M$ containing $N$ with the property that
any $F$-invariant function on $N$ can be extended to a $B$-invariant function on $M'$.
\end{lemma}
\begin{proof} Fix a Riemannian metric on $M$.
Consider the normal\footnote{Here ``normal'' is taken to mean a  complement to $TN$
in $T_NM$, not necessarily the orthogonal bundle to $TN$ w.r.t. the Riemannian metric.}
bundle to $N\subset M$
given by $\pi: \tilde{B}\oplus (TN+B)^{\perp}  \rightarrow N$, where $\tilde{B}$ is the orthogonal complement to $F=B\cap TN$ in $B$, and $(TN+B)^{\perp}$ denotes the orthogonal complement to $TN+B=TN\oplus \tilde{B}$ in the vector bundle $T_NM$.
The exponential map associated to the
Riemannian metric on $M$ identifies a neighborhood of the zero section of this normal bundle
with a tubular neighborhood $M'$ of $N$ in $M$ \cite{MS}\cite{Lee}.  If $f\in C^{\infty}(N)_F$ then
$\pi^*f$ is an extension lying in  $C^{\infty}(M')_B$, for its differential at points of $N$ annihilates both $F$ and $\tilde{B}$.
\end{proof}

The concrete choice of tubular neighborhood $M'$ is immaterial, because taking Poisson brackets is a local operation. To keep the notation simple, in the sequel we will assume that $M$ itself satisfies the above extension property, i.e. we will assume $M'=M$.

\begin{definition}\cite{MarRat}
The subbundle $B\subset T_N M$ is called

{\bf canonical}
if for any elements $f_1, f_2$ of
$C^\infty(M)_{B} \equiv\{f\in C^\infty(M)\mid
\dd f\vert_{B}=0\}$ we have
$\{f_1,f_2\}\in C^\infty(M)_{B}$.
\end{definition}

In other words, $B$ is canonical if  the Poisson bracket of
${B}$-invariant functions is ${B}$-invariant.
Note that in the previous definition,
$\dd f\vert_{B}$ stands for the restriction (not pullback) of $\dd f$ to
$N$ and then to sections of ${B}$.

In the next definition we consider $C^\infty(N)_{{F}} \equiv\{f\in C^\infty(N)\mid
\dd f\vert_{F}=0\}$, which is naturally isomorphic to $C^\infty(\un{N})$.
\begin{definition}\label{def:reducible}\cite{MarRat}

 $(M,\{\cdot,\cdot\}, {N}, B)$ is {\bf Poisson reducible}
if there is a Poisson bracket $\{\cdot,\cdot\}_{\un{N}}$ on $\un{N}$
such that for any
$f_1, f_2\in  C^\infty(\un{N})\cong  C^\infty(N)_{{F}}$ we have:
$$\{f_1,f_2\}_{\un{N}}=\iota^*\{f_1^B,f_2^B\}$$
for all extensions $f_i^B\in C^\infty(M)_{B}$ of $f_i$.
\end{definition}

With the previous definitions we can state the Marsden-Ratiu reduction theorem.

\begin{theorem}\label{th:MarRat}
(Marsden-Ratiu \cite{MarRat}) Assume that ${B}\subset T_NM$ is a
canonical subbundle. Then $(M,\{\cdot,\cdot\}, {N}, B)$  is Poisson
reducible if and only if
$$\sharp {B}^{\circ}\subset TN+{B}.$$

\end{theorem}
In the above theorem
$\sharp: T^*M \rightarrow TM$ denotes
the contraction with the Poisson bivector on $M$, and
${B}^{\circ}={\rm Ann}({B})$ consists of elements of
$T_N^*M$ that kill all vectors in ${B}$.
The proof of the theorem can be found in \cite{MarRat} and
\cite{OrtRat}.\\

 In the rest of this section we shall discuss the
implications of the  assumptions of the Marsden-Ratiu theorem.

The main observation is

that the assumption made in Theorem \ref{th:MarRat}
that ${B}$ is canonical
 is a rather strong requirement.

\begin{lemma}\label{condonMR}

Assume that ${B}\subset T_NM$ is a canonical subbundle.

Then either $\sharp{B}^{\circ}\subset TN$ or ${B}=0$.
\end{lemma}
\begin{proof}

Assume that there is a point $p\in N$
s. t. $(\sharp{B}^{\circ})_p\not\subset T_pN$. Then there is
a $B$-invariant function $h\in C^\infty(M)_{B}$
and a constraint
$g\in \cI\equiv\{f\in C^\infty(M)\ \mbox{s. t.}\ f\vert_N=0\}$
that satisfies $\{g,h\}(p)\not=0$.

It is clear that $g^2$ is ${B}$-invariant, and  the canonicity of
$B$ implies that $\dd\{g^2,h\}\vert_{B}=0$.
In particular one must have $i_{v}(\dd g)_p \{g,h\}(p)=0$
and we then deduce that $i_{v}(\dd g)_p=0$ for any $v\in{B}_p$.

Consider now any other constraint $g'\in \cI $,
we again have that $g\cdot g'$ is ${B}$-invariant
and therefore $i_{v}(\dd g')_p \{g,h\}(p)=0$.
From this we deduce that $i_{v}(\dd g')_p=0$ for any constrain $g'$ and any
$v\in{B}_p$. This is equivalent to saying
$${B}_p\subset T_pN.$$

By the assumption of constant rank for ${B}\cap TN$

we must have ${B}\subset TN$ everywhere.
This   implies that $f\cdot g$
is $B$-invariant

for any $f\in C^\infty(M)$ and therefore
$i_{v}(\dd f)_p \{g,h\}(p)=0$ for any $v\in{B}_p$.
But this is possible only if ${B}_p=0$
which implies ${B}=0$ and the proof is complete.
\end{proof}

\begin{remark}
Consider the familiar situation in which $G$ is a compact Lie group
acting freely on a symplectic manifold $(M,\omega)$ with equivariant
moment map $J:M\rightarrow \g^*$.
Fix $\mu\in \g^*$, let $N=J^{-1}(\mu)$ and $B$ be given by the tangent
spaces to the orbits of the $G$-action at points of $N$.
By Example B of \cite{MarRat} the subbundle $B$ is  canonical,
and the Marsden-Ratiu theorem recovers the familiar symplectic
structure on $J^{-1}(\mu)/G_{\mu}$.

Now    take $N$ as above but $B'\subset TN$ to be given by the
tangent spaces to the $G_{\mu}$-orbits  \emph{at points of $N$}, and
assume that $\mu$ is not a fixed point of the coadjoint action. Then
$B'$ is \emph{not} a canonical subbundle. This fact is consistent
with Lemma \ref{condonMR}, and is of course no contradiction to the
fact that the $G_{\mu}$-invariant functions on $M$ are closed under
the Poisson bracket (i.e. that the tangent spaces to the
$G_{\mu}$-orbits at \emph{all points of $M$} form a canonical
distribution).

\end{remark}

\begin{remark} If the subbundle $B \neq 0$ is canonical then it follows from Lemma \ref{condonMR}
 that $B^\circ \rightarrow N$ is a Lie subalgebroid of $T^*M$ (with the Lie algebroid structure induced by the Poisson structure on $M$).

\end{remark}

In view of Lemma \ref{condonMR}, Theorem \ref{th:MarRat}
becomes:

\begin{theorem}\label{MR2}

Assume that ${B}\subset T_NM$ is a canonical subbundle.
\begin{itemize}
\item  If $B\not=0$ then  $(M,\{\cdot,\cdot\}, {N}, B)$  is Poisson reducible.
\item If $B=0$, then $(M,\{\cdot,\cdot\}, {N}, B)$  is Poisson reducible iff $N$ is a Poisson submanifold.
\end{itemize}

\end{theorem}
\begin{proof}
If $B\not=0$ then by Lemma \ref{condonMR} we have $\sharp B^{\circ}\subset TN$, which
by Thm. \ref{th:MarRat} implies Poisson reducibility. For the case $B=0$ we can apply directly
Thm. \ref{th:MarRat}.
\end{proof}

Further, the induced Poisson structure on $\un{N}$ depends only on $F=B\cap TN$ and not on $B$, as stated by the following proposition.

\begin{proposition}\label{pr:unique}
If $B, B' \subset T_NM$ are non-zero canonical subbundles such that $B\cap TN=B'\cap TN$,
then the induced Poisson structures on $\un{N}=\un{N'}$ agree.
\end{proposition}
\begin{proof}

By Lemma \ref{condonMR} we know that $\sharp B^{\circ},\sharp B'^{\circ}\subset TN$.

Let $f_1,f_2\in C^\infty(M)_B$ and
$f_1', f_2'\in C^\infty(M)_{B'}$ s. t.
$\iota^*f_i=\iota^*f'_i$, $i=1,2$.

We have
$$\iota^*\{f_1,f_2\}-\iota^*\{f_1',f_2'\}
=\iota^*\{f_1-f_1',f_2\}+\iota^*\{f_1',f_2-f_2'\},$$
and as for any $p\in N$ we have $d(f_i-f_i')_p\in TN^\circ$ and
$\sharp(df_i)_p\in TN$, the right hand side vanishes. Hence
$$\iota^*\{f_1,f_2\}=\iota^*\{f_1',f_2'\}.$$
\end{proof}

\begin{remark} To every submanifold $N$ of the Poisson manifold $M$
  is   canonically associated a Poisson algebra, as
follows\footnote{This is an algebraic version of Example $D$ in
 \cite{MarRat}; the latter holds when $\sharp TN^{\circ}$ and $\sharp TN^{\circ}\cap TN$
  have constant rank.}. Let  $\cI$ be the ideal of functions on
  $M$ vanishing on $N$. Its Poisson normalizer $\cN\equiv\{f\in C^\infty(M)\mid
\{f,\cI\}\subset \cI\}$ is a Poisson subalgebra, so the
quotient $\cN/(\cN\cap \cI)$ is a Poisson algebra (see also \cite{glmv}).
Notice that $\cN$ consists of functions whose differentials
annihilates all vectors in  $\sharp TN^{\circ}$.

Now let $B$ be a nonzero canonical subbundle. Then
$C^{\infty}(\un{N})$, with the Poisson bracket induced as in Thm.
\ref{MR2}, is a \emph{Poisson subalgebra} of  $\cN/(\cN \cap \cI)$. Indeed by Lemma \ref{condonMR} we have $B \supset
\sharp TN^{\circ}$, so $C^{\infty}(M)_B\subset \cN$, hence
$C^{\infty}(\un{N})=C^{\infty}(M)_B/(C^{\infty}(M)_B\cap \cI)$
sits inside $\cN/(\cN\cap \cI)$ and is a Poisson
subalgebra. Notice that $\cN/(\cN\cap \cI)$ does not
``see'' the subbundle $B$, in agreement with Prop. \ref{pr:unique}
above. \end{remark}

\begin{remark} We complete
Prop. \ref{pr:unique} by dealing with  the trivial case $B=0$ (which is clearly canonical). As we saw above,
$(M,\{\cdot,\cdot\}, {N}, B=0)$ is Poisson reducible iff $N$ is a Poisson submanifold.
If $B'$ is some canonical subbundle with $B'\cap TN=0$ then the Poisson structures induced by $B'$ and $B=0$ on $N$ agree, as $N$ is a Poisson submanifold.
\end{remark}

The conclusion of Prop. \ref{pr:unique} is that, when the Marsden-Ratiu reduction endows
 $\un{N}$ with an
induced Poisson structure, this structure depends only on $F$.

This result is against the original idea of reduction by distributions,
where the role played by $B$ is expected to be more prominent.
In order to accomplish this objective we will proceed, in the coming section,
to relax the condition of canonicity of the distribution while maintaining
the requirement of having a Poisson structure induced on  $\un{N}$.

\section{Extension of the Marsden-Ratiu reduction}\label{extMR}

The set-up of this section consists of  the geometric data of the
Mardsen-Ratiu theorem; we will set various conditions on these data
which guarantee Poisson reducibility.
So   let $(M,\Pi)$ be a Poisson manifold,
$N\subset M$ a submanifold and $B \subset T_NM$ a   subbundle
with $F:=B\cap TN$ a regular, integrable distribution.  Without loss of generality (see
Lemma \ref{surj}), here and in the rest of the paper, we
 assume that the restriction map
$\iota^*: C^{\infty}(M)_B \rightarrow  C^{\infty}(N)_F$ is
surjective.
We do not need to assume that
$\un{N}:=N/F$
is a smooth manifold, even though this is of course the case of interest. In that case $C^{\infty}(\un{N})\cong
C^{\infty}(N)_F$.

We would like to  define a bilinear operation $\{\cdot,\cdot\}_{\un{N}}$ on $C^{\infty}(N)_F$ by the following rule:
 \begin{equation}\label{defbr}
\{f,g\}_{\un{N}}:=\iota^*\{f^B,g^B\}
\end{equation}
where  $f^B$, $g^B$ are arbitrary extensions to elements of
$C^{\infty}(M)_B$.
 As the restriction map
$\iota^*: C^{\infty}(M)_B \rightarrow  C^{\infty}(N)_F$ is
surjective there is at most one bilinear
operation $\{\cdot,\cdot\}_{\un{N}}$. Our task is to determine when
$\{\cdot,\cdot\}_{\un{N}}$ is well-defined and when it is a Poisson
bracket.\\

The r.h.s. of eq. \eqref{defbr} is independent of the chosen extensions
(for all $f,g \in  C^{\infty}(N)_F$) iff
\begin{equation}\label{tnb}\sharp B^{\circ}\subset
TN+B \end{equation}
 (see the proof of the Marsden-Ratiu theorem or the proof of Thm.
\ref{gen} below). If this is the case, the r.h.s. of \eqref{defbr}
lies in  $C^{\infty}(N)_F$ iff for one choice of  extensions
$f^B,g^B$ we have $\iota^*\{f^B,g^B\} \in C^{\infty}(N)_F$, or
equivalently if
\begin{equation}\label{desc} \{ {C^{\infty}(M)_B},
{C^{\infty}(M)_B}\}\subset C^{\infty}(M)_F. \end{equation}

In this case clearly $\{\cdot,\cdot\}_{\un{N}}$ will be a
skew-symmetric operation on $C^{\infty}(N)_F$ which is a
biderivation w.r.t. the product; if $\un{N}$ is smooth, this means
that $\{\cdot,\cdot\}_{\un{N}}$ defines a bivector field on it.

Now we want to determine conditions under which
$\{\cdot,\cdot\}_{\un{N}}$ satisfies the Jacobi identity, for when this is the case
$(M,\{\cdot,\cdot\}, {N}, B)$ is Poisson reducible.
 Checking
the Jacobi identity suggests to require that  for any $f,g \in
C^{\infty}(N)_F$ there \emph{exist} extensions $f^B,g^B$ whose bracket
annihilate not only $F$ but actually a larger subbundle  (not necessarily tangent to $N$). This leads us to  a condition that
  involves two pieces of data: an additional subbundle $D$ of
$T_NM$ and a subspace $\cB$ of $C^{\infty}(M)_B$ which contains the above
extensions.

In the Appendix we give an algebraic interpretation of these data,
and at the end of Subsection \ref{appldis} we give a geometric
interpretation.

 \begin{theorem}\label{gen}
Let $(M,\{\cdot,\cdot\})$ be a Poisson manifold, $N \subset M$ a submanifold and
$B \subset T_NM$ a   subbundle with $F:=B\cap TN$ a regular,
integrable distribution.
Let $D$ be a subbundle of $T_NM$ satisfying\footnote{Equivalently
$D\subset B$ and $B\cap TN=D\cap TN$.} $F\subset D \subset B$ and
\begin{equation}\label{sha}
\sharp B^{\circ}\subset D+TN.
\end{equation}
 Let $\cB \subset C^{\infty}(M)_B$ be a
multiplicative subalgebra such that the restriction map $\iota^*:
\cB  \rightarrow  C^{\infty}(N)_F$

is surjective. Assume that
\begin{equation}\label{bra}
\{\cB ,\cB \}\subset
C^{\infty}(M)_D.\end{equation}

Then $(M,\{\cdot,\cdot\}, {N}, B)$ is Poisson reducible.
\end{theorem}

\begin{proof}
Consider functions $f,g \in C^{\infty}(N)_F$ and extensions
$f^B,g^B$ in $\cB $. If we choose a different
extension ${f^B}'$ for $f$, the differential of  $f^B-{f^B}'$
annihilates $TN+B$, so because of $\sharp(TN+B)^{\circ}\subset
\sharp(D+TN)^{\circ}\subset B$ (eq. \eqref{sha}) we have $\iota^* \{f^B-{f^B}',g^B\}=0$.
Hence the expression for $\{f,g\}_{\un{N}}$ is independent of the
choice of extensions. By eq. \eqref{bra} it actually lies in
$C^{\infty}(N)_F$.

Now $\{f^B,g^B\}$ and $(\{f,g\}_{\un{N}})^B$ by definition agree on
$N$, and are elements respectively of $C^{\infty}(M)_D$ (by eq.
\eqref{bra}) and $\cB $. So their difference
annihilates $D+TN$ and by eq. \eqref{sha} the Poisson bracket of their difference with any
element of $\cB $ vanishes on $N$. This explains the second equality in the identity
$$\{\{f,g\}_{\un{N}},h\}_{\un{N}}=\iota^* \{(\{f,g\}_{\un{N}})^B,h^B\}=
\iota^* \{\{f^B,g^B\},h^B\}.$$ From this is clear that the Jacobi
identity for $\{\cdot,\cdot\}_{\un{N}}$ holds as a consequence of
that for $\{\cdot,\cdot\}$.
\end{proof}

\begin{remark} Enlarging $D$ makes the constraint  \eqref{bra} more
severe, so in applications one should choose $D$ satisfying
\eqref{sha} to have dimension as small as possible. In general there
is no  unique minimal choice of $D$.
\end{remark}

\section{Applications and examples} In this section we consider special cases
of Thm. \ref{gen}. As usual  $(M,\Pi)$ is a Poisson
manifold, $N\subset M$ a submanifold and $B \subset T_NM$ a
subbundle with $F:=B\cap TN$ a regular, integrable distribution.

\subsection{A straightforward application}
 Setting $D=F$ and
$\cB =C^{\infty}(M)_B$ in Thm. \ref{gen}. we
obtain a minor improvement of the Marsden-Ratiu theorem (Thm. \ref{th:MarRat}), where the
condition on the canonicity of $B$ is weakened:
\begin{proposition}\label{onn}
If
\begin{equation} \label{BBF} \{C^{\infty}(M)_B,C^{\infty}(M)_B\}\subset C^{\infty}(M)_F
\end{equation} and $\sharp B^{\circ}\subset TN$ then
$(M,\{\cdot,\cdot\}, {N}, B)$ is Poisson reducible.

\end{proposition}

\begin{remark}\label{remliederF}
In the above proposition condition \eqref{BBF} is equivalent to
the following, which is more suited for computations: locally
  there exists a frame of sections $X_i$ of $F$ and extensions thereof to
  vector fields on $M$ such that
 \begin{equation}\label{liederF} (\cL_{X_i} \Pi)|_N \subset B \wedge
 T_NM.\end{equation}
This can be shown  using formula \eqref{Xfg} below and  $\sharp
B^{\circ}\subset TN$.
\end{remark}

We present an example where the assumptions of Prop. \ref{BBF} are
satisfied but $B$ is not canonical.
\begin{example}\label{zxy}
Let $(M,\Pi)$ be $(\RR^3, z\frac{\partial}{\partial x}\wedge
\frac{\partial}{\partial y})$ and $N$ the plane given by $z=0$. Let
$B=\RR \frac{\partial}{\partial z}$. The conditions of Prop.
\ref{BBF} are satisfied because $\Pi$ vanishes at points of $N$ and
because $F=B\cap TN=\{0\}$. However $C^{\infty}(M)_B$ is not closed
w.r.t. the Poisson bracket: for instance $x,y$ lie in
$C^{\infty}(M)_B$ but $\{x,y\}=z$ does not.
\end{example}

If $\sharp B^{\circ}\subset TN$ then necessarily $\sharp
TN^{\circ}\cap TN \subset F$. When this last inclusion is an
equality eq. \eqref{BBF} holds automatically, so the interesting
case is when the inclusion is strict, as in the following example,
in which we use Remark \ref{remliederF}  to check condition
\eqref{BBF}.
\begin{example}
Let $(M,\Pi)=(\RR^6, \sum_i \frac{\partial}{\partial x_i}\wedge
\frac{\partial}{\partial y_i})$ and $N$ be the (coisotropic)
hyperplane $\{y_3=0\}$.  Let $B={\rm span}\{\frac{\partial}{\partial x_3},
\frac{\partial}{\partial x_1}+ \alpha\frac{\partial}{\partial
y_2}\}\subset TN$ where $\alpha \in C^{\infty}(N)$. $F=B$ is integrable iff
$\alpha$ is independent of $x_3$. We have $\sharp B^{\circ}\subset
TN$  since $B$ contains the characteristic distribution of $N$.

We check  condition \eqref{liederF}, which is easier than checking
directly  condition \eqref{BBF}. We have $\cL_{\frac{\partial}{\partial x_3}}\Pi=0$,
and $(\cL_{ \frac{\partial}{\partial x_1}+
\alpha\frac{\partial}{\partial y_2}}\Pi)|_N=-\sharp \dd\alpha\wedge
\frac{\partial}{\partial y_2}$ surely lies in $B \wedge T_NM$ if
$\alpha$ depends only on the coordinates $y_1$ and $x_2$. In this
case by Prop.  \ref{onn} the quotient $\un{N}\cong \RR^3$ has an
induced Poisson structure, which in suitable coordinates is given by
$\{y_1,y_2\}=\alpha$, $\{x_2,y_2\}=1$ and $\{y_1,x_2\}=0$.
\end{example}

\subsection{An application involving distributions}\label{appldis}

 If $M$ is endowed with a suitable distribution we can
weaken the condition $\sharp B^{\circ}\subset TN$ (which, as seen in
Lemma \ref{condonMR}, is an assumption of the Marsden-Ratiu theorem
for $B\not=0)$.

\begin{definition}\label{compat}
Let  $\theta_D$ be an integrable distribution  on $M$ such that
$F\subset {\theta_D}|_N \subset B$. We say that  $\theta_D$ and
$B$ are {\bf compatible} if  $\iota^*:C^{\infty}(M)_B\cap
C^{\infty}(M)_{\theta_D}\rightarrow C^{\infty}(N)_F$ is surjective.
\end{definition}

The above compatibility is satisfied for instance when ${\theta_D}|_N = B$ or
$F:=B\cap TN=\{0\}$.
In the appendix (Prop. \ref{equichar})
we shall give an equivalent characterization of
Def. \ref{compat}.

\begin{proposition}\label{thetaD}
Suppose that on $M$  there is an integrable distribution $\theta_D$
 such that $F\subset D:={\theta_D}|_N \subset B$ and so that
 $\theta_D$ is compatible with $B$.
 Assume that
 \begin{equation}\label{shathetaD}\sharp B^{\circ}\subset
 D+TN\end{equation} and that, for any section $X$ of $\theta_D$,
 \begin{equation}\label{lieder} (\cL_X \Pi)|_N \subset B \wedge T_NM.
 \end{equation}
 Then $(M,\{\cdot,\cdot\}, {N}, B)$ is Poisson reducible.
\end{proposition}
\begin{proof} Set
$\cB =C^{\infty}(M)_B\cap C^{\infty}(M)_{\theta_D}$ in Thm.
\ref{gen}. By assumption  $\iota^*:\cB\rightarrow
C^{\infty}(N)_F$ is surjective.
 Condition \eqref{bra}  reads $$

\{C^{\infty}(M)_B \cap
C^{\infty}(M)_{\theta_D},
 C^{\infty}(M)_B \cap C^{\infty}(M)_{\theta_D}\}\subset
C^{\infty}(M)_D.$$ This is equivalent to \eqref{lieder}, as one can see
evaluating at points of $N$ the following equation: for $X \in
\Gamma(\theta_D)$ and $f,g\in C^{\infty}(M)_B \cap
C^{\infty}(M)_{\theta_D}$,
\begin{equation}\label{Xfg} X\{f,g\}=

(\cL_X
\Pi)(df,dg)+\Pi(d(Xf),dg)+\Pi(df,d(Xg)).\end{equation}
 \end{proof}

\begin{remark}\label{remlocal}
 It is sufficient to apply Prop. \ref{thetaD} locally.
More precisely: let $\{U^{\alpha}\}$ be an open cover
of a tubular neighborhood of $N$, and suppose that on each
$\{U^{\alpha}\}$
there exists an integrable distribution $\theta^{\alpha}_D$
as in the proposition. Then in particular eq. \eqref{tnb} is
satisfied, so eq. \eqref{defbr} determines a well-defined map
$C^{\infty}(N)_F\times  C^{\infty}(N)_F \rightarrow  C^{\infty}(N)$.
Applying Prop. \ref{thetaD} on each open set $U^{\alpha}$ ensures that this map
defines a Poisson bracket on $C^{\infty}(N)_F$.

Further it is sufficient to check condition \eqref{lieder}
locally on a frame $\{X_i\}$ of sections of $\theta_D$.
\end{remark}

To further illustrate the properties of the reduction discussed in
Prop. \ref{thetaD} we provide some concrete examples that highlight
different aspects of the reduction. The first examples are
particularly simple, since  there $B\oplus TN=T_NM$, so that
formula \eqref{defbr} defines a bivector field on $\un{N}=N$.
\begin{example}\label{poissonex}
Consider the symplectic manifold $(\RR^4, \sum_i dx_i\wedge dy_i)$,
 let $N$ be given by the constraints $x_1=x_2=0$ and let
$B={\theta_D}|_N$ where $\theta_D$ is the distribution on $\RR^4$
given by
  $\theta_D:=span\{\frac{\partial}{\partial x_1},
\frac{\partial}{\partial x_2}- \lambda\frac{\partial}{\partial
y_1}\}$ (with $\lambda \in \RR$).  $C^{\infty}(M)_{\theta_D}$ is
closed under the bracket and $\sharp (TN+B)^{\circ}=0$, so the
assumptions of
 Prop. \ref{thetaD} are met. The quotient $\un{N}$ is $\RR^2$ with
 natural coordinates $y_1,y_2$ and Poisson bivector
 $\lambda\frac{\partial}{\partial y_1}\wedge \frac{\partial}{\partial
 y_2}$.

 Notice that in this example the condition $\sharp B^{\circ}\subset
 TN$ is violated. The final Poisson structure depends on $B$
(while $B\cap TN=\{0\}$ is independent of $\lambda$). As shown in Prop.
\ref{pr:unique} this  can not happen in the Marsden-Ratiu reduction
(Thm. \ref{th:MarRat}).
\end{example}

The next example illustrates the fact that,
even if we have a well-defined smooth bivector on $\un{N}$,
we need extra conditions to satisfy the Jacobi identity.
\begin{example}\label{exbsb}
 Let $(M,\Pi)$ be the Poisson manifold
$(\RR^4, \sum_i \frac{\partial}{\partial x_i}\wedge
\frac{\partial}{\partial y_i})$, consider the hyperplane
$N=\{y_2=0\}$ and the subbundle $B$ of $T_NM$ spanned by
$\frac{\partial}{\partial
y_2}+ \alpha \frac{\partial}{\partial x_1}$,
 where $\alpha
\in C^{\infty}(N)$.

The bivector field induced by eq. \eqref{defbr} on $N$ is
$\frac{\partial}{\partial x_1}\wedge (\frac{\partial}{\partial
y_1}+\alpha \frac{\partial}{\partial x_2})$, hence it is Poisson iff
$\alpha$ is independent of $x_1$.

All the Poisson structures obtained above can be obtained using
Prop. \ref{thetaD}. Indeed, if we extend $B$ to the distribution
$\theta_D:=\RR ( \frac{\partial}{\partial y_2}+ \alpha
\frac{\partial}{\partial x_1})$, eq. \eqref{lieder} is satisfied iff
$\frac{\partial}{\partial x_1}\alpha=0$.

\end{example}

In the previous example we have seen an obstruction to obtaining a
Poisson structure after the reduction, namely eq. \eqref{lieder}. In
the following example the distribution $F$ on $N$ is non-trivial,
and we shall also exhibit an obstruction to have a well defined
bivector field on $\un N$ in the first place.

\begin{example}\label{exr6} Let $(M,\Pi)$ be the Poisson manifold
$(\RR^6, \sum_i \frac{\partial}{\partial x_i}\wedge
\frac{\partial}{\partial y_i})$, consider the hyperplane
$N=\{y_2=0\}$ and the subbundle of $T_NM$ given by
$B={\rm span}\{\frac{\partial}{\partial y_1}, \frac{\partial}{\partial
y_2}+ \alpha \frac{\partial}{\partial x_1}\}$, where
$\alpha\in C^{\infty}(N)$. Clearly $\sharp B^{\circ} \subset TN+B$, and
$F:=B \cap TN=\RR \frac{\partial}{\partial y_1}$.

Now the bracket of the $B$-invariant extensions
of the coordinate functions $x_1,x_2$ is:
$\{x_1^B, x_2^B\}\vert_N=\alpha$, which is well defined on $\un N$
iff $\alpha$ does not depend on $y_1$.  This condition ensures that we have
a bivector field on $\un N$ but still is not enough
to guarantee reducibility.

Prop. \ref{thetaD} can be applied to determine when the bracket
$\{\cdot, \cdot\}_{\un{N}}$ is a well-defined Poisson bracket. We

extend $B$ constantly in the $y_2$ direction to obtain the
distribution $\theta_D=span\{\frac{\partial}{\partial y_1},
\frac{\partial}{\partial y_2}+ \alpha\frac{\partial}{\partial
x_1}\}$ on $M$.  The distribution $\theta_D$ is integrable iff
$\alpha$ does not depend on $y_1$.  Now
$\cL_{\frac{\partial}{\partial y_1}}\Pi=0$, and
$\cL_{\frac{\partial}{\partial y_2}+ \alpha\frac{\partial}{\partial
x_1}}\Pi= -X_{\alpha}\wedge \frac{\partial}{\partial x_1} \subset
B\wedge T_NM$
iff $\alpha$ does not depend on the coordinates $x_3$ and $y_3$.
Hence Prop. \ref{thetaD} allows us to conclude that, when $\alpha$
depends only on the coordinates $x_1$ and $x_2$, we obtain a Poisson
bivector on $\un{N}\cong \RR^4$. In the natural coordinates, the
induced Poisson bivector is $\alpha \frac{\partial}{\partial
x_1}\wedge \frac{\partial}{\partial x_2} +\frac{\partial}{\partial
x_3}\wedge \frac{\partial}{\partial y_3}$.
\end{example}

The following is a simple example in which $M$ is a linear Poisson
manifold.
\begin{example}\label{dual}
Let $\g$ be a Lie algebra, $V\subset \g$ a subspace and $\h\subset
\g$ a Lie subalgebra satisfying $[\h, V\cap \h]\subset V$. We set
$M:=\g^*$, $N:=V^{\circ}$, and $B_x:=\h^{\circ}\subset T_xM$ at all
$x\in N$. Using  Lemma 5.4 of \cite{CZbis} and the assumptions,  we
see $\sharp B_x^{\circ}=\{ad_h^*(x): h\in \h\}\subset T_xN+B_x$ at
all $x\in N$. Extending $B=\h^{\circ}$ by translation to a
distribution $\theta_D$ on $M$ and noticing that the projection
$\g^*\rightarrow \g^*/\h^{\circ}\cong \h^*$ is a Poisson map we see
that eq. \eqref{lieder} is satisfied. By Prop. \ref{thetaD} we
conclude that there is an induced (linear) Poisson structure on
$\un{N}=\frac{V^{\circ}}{V^{\circ}\cap \h^{\circ}}\cong
(\frac{V+\h}{V})^*$. It corresponds to the Lie algebra structure on
$\frac{\h}{\h\cap V}$, which as a vector space is canonically
isomorphic to $\frac{V+\h}{V}$.
\end{example}

Our last example  shows that conditions of Prop. \ref{thetaD} are
not necessary

in order to obtain a Poisson structure after the reduction.
\begin{example}
 Let $(M,\Pi)$ be $(\RR^3, z\frac{\partial}{\partial x}\wedge
\frac{\partial}{\partial y})$,  $N$ the plane given by $z-x=0$ and
$B=\RR \frac{\partial}{\partial z}$. Formula \eqref{defbr} defines
the Poisson structure $\{x,y\}=x$ on $N$, however
  Prop.
\ref{thetaD} can not be applied   because a distribution $\theta_D$
as in the proposition does not exist. Indeed $\theta_D$ has to be
one-dimensional because of eq. \eqref{shathetaD}.
  For any vector field
$X$ which restricts to $\frac{\partial}{\partial z}$ on $N$, we have
$(\cL_X \Pi)|_p=X(z)|_p \cdot \frac{\partial}{\partial x}\wedge
\frac{\partial}{\partial y}=\frac{\partial}{\partial x}\wedge
\frac{\partial}{\partial y}$ at any point   $p\in N$ of the form
$(0,y,0)$, so eq. \eqref{lieder} is not satisfied.
\end{example}

We conclude the subsection giving a geometric interpretation
of Prop. \ref{thetaD}. Assume
that the quotient $\un{M}:=M/\theta_D$ is smooth  and that
$C^{\infty}(M)_{\theta_D}$ is closed under the Poisson bracket, so
that $\un{M}$ has a Poisson structure  for which the projection $M
\rightarrow \un{M}$ is a Poisson map.  Assume $\un{N} \subset
\un{M}$ is a { Poisson-Dirac submanifold} \cite{CrF}, so that it has
an induced Poisson structure. Then the Poisson bracket of functions
on $\un{N}$ is computed by lifting to functions in
$C^{\infty}(M)_{B}$ where $B$ is  a subbundle   as in Prop.
\ref{thetaD}. In this interpretation the case $D=B$ corresponds to
the case where $\un{N}$ is actually a
\emph{Poisson} submanifold of $\un{M}$.\\

\subsection{An application to hamiltonian actions}

Here is an instance where the assumptions of Prop. \ref{thetaD} are
naturally met. Given an action of a Lie group on a manifold $M$ we
denote by $\g_M(p)$ the span at $p\in M$ of the vector fields
generating the action (i.e. the tangent space of the $G$-orbit
through $p$).
\begin{proposition}\label{action}
Let the Lie group $G$ act on the symplectic manifold $(M,\omega)$ so
that $\g_M$ has constant rank and  with equivariant moment map
$J:M\rightarrow \g^*$. Let $m\in J^{-1}(0)$ and $N$ be a submanifold through
$m$ so that
\begin{equation}\label{dirsum}
T_mN\oplus ker(d_mJ)=T_mM.
\end{equation}
Then $N$,  after shrinking it to a smaller neighborhood of $m$ if
necessary, has an induced Poisson structure, obtained extending
functions from $N$ to $M$ so that they annihilate
$[\g_M+(TN+\g_M)^{\omega}]|_N$

\end{proposition}
\begin{proof} Consider $B:=[\g_M+(TN+\g_M)^{\omega}]|_N\subset T_NM$ and the
distribution $\theta_D:=\g_M$. We now check that the assumptions of
Prop. \ref{thetaD} are automatically satisfied; we will make use
repeatedly of $\g_M(m)\subset ker(d_mJ)=\g_M(m)^\omega$, which
holds by the equivariance of $J$.

First of all $B$ has constant rank, at least near $m$. Indeed the
sum of $TN$ and $\g_M$ has constant rank because their intersection
at $m$ is trivial. Further
 $TN+\g_M$ is a symplectic subbundle of $T_NM$.
 To this aim we check that at the point $m$ we have
\begin{equation}
\begin{aligned}
T_mN^{\omega}\cap \left[\g_M(m)^{\omega}\cap (T_mN+\g_M(m))\right]&=
T_mN^{\omega}\cap
\g_M(m)=\hfill\cr&=\left(T_mN+\g_M(m)^\omega\right)^\omega=\{0\}.
\end{aligned}
\end{equation}
We conclude that $B=\g_M\oplus (TN\oplus \g_M)^{\omega}$ has
constant rank near $m$. Further we have $F=B\cap TN=\{0\}$ since
$B_m\subset ker(d_mJ)$.

Compatibility of $\theta_D$ and $B$ holds because $F=\{0\}$.

Condition \eqref{shathetaD} as well as $D:={\theta_D}|_N
\subset B$ are trivially satisfied. Condition \eqref{lieder} is
satisfied since the $G$-action preserves $\omega$.
\end{proof}

\begin{remark}\label{remaction}
1) The geometric interpretation of Prop. \ref{thetaD} applied to the
special case of  Prop. \ref{action} is the following: if the $G$
action is free and proper it is known that $M/G$ is a Poisson
manifold, whose symplectic leaves  are given by $J^{-1}(\cO)/G$ as
$\cO\subset \g^*$ ranges over all coadjoint orbits. Therefore
$\un{N}\cong N$ is a submanifold of $M/G$ which intersects
transversely the symplectic leaf $J^{-1}(0)/G$, and has such it has
a Poisson structure induced from $M/G$. This Poisson structure agree
with the one that Prop. \ref{action} induces on $N$.

2) In  the case that the $G$-action in  Prop.
\ref{action}  is free and proper

one has a dual pair $M/G\leftarrow M \rightarrow \g^*$, and Thm 8.1
of \cite{AW} says that the Poisson structure on $\un{N}$ (as in part 1)
above) is isomorphic up to sign to the one on the open subset $J(N)$
of  $\g^*$. However the   identification  $\un{N}\cong N \cong J(N)$
given by the dual pair does not preserve the Poisson structures in
general. (A sufficient condition is that $N$ is isotropic.)
\end{remark}

The following is an example for Prop. \ref{action}.
\begin{example}
Consider the action of $G=U(2)$ on $M=GL(2,\CC)$ by left multiplication, and endow $M$ with the symplectic form induced by the natural embedding in $\CC^4$. This action is Hamiltonian with moment map $J:GL(2,\CC)\rightarrow \uu^*(2)\cong \uu(2)$ given by $J(A)=\frac{1}{2i}(AA^*-I)$ \cite{Ca}.
A slice transverse to $J^{-1}(0)$ at the identity is given by
$$N :=\left\{ \left(
 \begin{smallmatrix} {x_1} & x_2+ix_3 \\ 0 & x_4 \end{smallmatrix}
\right) \right \}$$ where $x_1,x_4$ are real numbers close to $1$
and $x_2,x_3$ are close to $0$. A straightforward computation shows
that extending the coordinates $x_i$ on $N$ so that they annihilate
$[\g_M+(TN+\g_M)^{\omega}]|_N= [\g_M]|_N$ delivers the following
bracket on $N$:

\begin{eqnarray*}
 \{x_1,x_2\}=\frac{x_3}{x_1},& \{x_1,x_3\}=-\frac{x_2}{x_1},&
\{x_1,x_4\}=0\\
 \{x_2,x_3\}=1-\frac{x_4^2}{x_1^2},&
 \{x_2,x_4\}=\frac{x_3x_4}{x_1^2},&
\{x_3,x_4\}=-\frac{x_2x_4}{x_1^2}.
\end{eqnarray*}
Prop. \ref{action} states that this is a Poisson bracket.

In the new coordinates $\xi_1=\frac12x_1x_2, \xi_2=\frac12x_1x_3,
\xi_3=\frac14(x_1^2-x_2^2-x_3^2-x_4^2),
\eta=\frac14(x_1^2+x_2^2+x_3^2+x_4^2)$ the Poisson bracket is linear
and coincides  with that of $\uu^*(2)$, in agreement with Remark
\ref{remaction}.
\end{example}

\subsection{The symplectic case}

We end this section  asking when eq. \eqref{defbr} defines a \emph{symplectic} structure on the quotient $\un{N}$. We focus on the case where $M$ has a symplectic (not just Poisson) structure $\omega$.

\begin{lemma}\label{sympl}
Assume that eq. \eqref{defbr} endows $\un{N}$ with a well-defined bivector field $\un{\Pi}$.  $\un{\Pi}$ corresponds to a non-degenerate 2-form   iff
\begin{equation}\label{BBperp}
TN+B=B^{\omega}+B.
\end{equation}
In this case the 2-form on $\un{N}$ is obtained pushing down  $\omega^B\in \Omega^2(N)$ given by $\omega^B(X_1,X_2)
=\omega(X_1+b_1,X_2+b_2)$, where $b_i\in B$ are such that $X_i+b_i\in  B^{\omega}$.
\end{lemma}
\begin{proof}
From Lemma \ref{pushpull} in the Appendix it follows that $\un{\Pi}$ is invertible iff the almost Dirac structure $\iota^*(L_{\Pi}^B)$ on $N$ is the graph of a 2-form with kernel $F$. Writing out explicitly $\iota^*(L_{\Pi}^B)$ one sees that it is the graph of a 2-form iff $TN\subset B^{\omega}+B$, which in turn is equivalent to eq. \eqref{BBperp} since   eq. \eqref{tnb} holds. In this case the kernel of the 2-form is automatically $F$. This shows the equivalence claimed in the lemma.

A  computation shows that $\iota^*(L_{\Pi}^B)$ is the graph of the 2-form $\omega^B$ defined above.
\end{proof}

A simple instance of Lemma \ref{sympl}  is the case when $N$ is a symplectic submanifold of $(M,\omega)$ and $B$ is small perturbation of $TN^{\omega}$.
Then $N$ is endowed with a
non-degenerate 2-form $\omega^B$, which is intertwined with
$\iota^*\omega$ by the bundle isomorphism $TN\cong B^{\omega}$ (given
by projection along $B$).

Suppose that  $B$ can be extended locally to an integrable distribution $\theta$ on $M$ so that the $\theta$-invariant functions are closed w.r.t. the Poisson bracket. Then $\omega^B$ is a closed form, for it is just the pullback to $N$ of the symplectic form on the quotient $M/\theta$ (this is an instance of Prop. \ref{thetaD}).
In general, writing $B$ as
the graph of a bundle map $A: TN^{\circ}\cong TN^{\omega} \rightarrow TN$,
it would be interesting to spell out in terms on $A$ when $\omega^B$ is a \emph{symplectic}
structure.

\appendix

\section{}
\subsection{Algebraic interpretations}

We  provide an algebraic interpretation of Thm. \ref{gen}.
\begin{proposition}\label{alg}
Let $\cM$ be a Poisson algebra, $\cB \subset \cD$ multiplicative
subalgebras of $\cM$ and $\cI$ a multiplicative ideal of $\cM$.
Assume that the images of $\cB$ and $\cD$ under the projection $\cM
\rightarrow \cM/\cI$ are equal   and that
\begin{equation}\label{shaalg}
    \{\cB,\cI\cap \cD\}\subset \cI
\end{equation}
and
\begin{equation}\label{braalg}
 \{\cB,\cB\}\subset \cD.
\end{equation}
Then there is an induced Poisson algebra structure on
$\frac{\cB}{\cB\cap \cI}$, whose bracket is determined by the
commutative diagram
\[
\xymatrix{
 \cB \times \cB \ar[d] \ar[r]^{\{\cdot,\cdot\}}& \cD \ar[d]
\\
\frac{\cB}{\cB\cap \cI} \times \frac{\cB}{\cB\cap \cI}  \ar[r]&
\frac{\cD}{\cD\cap \cI}=\frac{\cB}{\cB\cap \cI}\\
 }.
\]
\end{proposition}

Thm. \ref{gen} is recovered setting $\cM=C^{\infty}(M)$,

$\cD=C^{\infty}(M)_D$ and $\cI=\{f\in C^{\infty}(M): \iota^*f=0\}$.
Conditions  \eqref{shaalg} and \eqref{braalg} become conditions
\eqref{sha} and \eqref{bra}
 respectively.

The proof of Prop. \ref{alg} is similar to that of Prop. \ref{gen}
and will not be given here. We just mention that condition
\eqref{shaalg} can be interpreted as ``$\cI$ behaves like an ideal
in $\cB$'', and condition \eqref{braalg} as ``$\cB$ behaves like a
Poisson subalgebra'', showing that one has  a well-defined almost
Poisson bracket on $\frac{\cB}{\cB\cap \cI}$. To show that it
satisfies the Jacobi identity one needs to use
once more both conditions.\\

\subsection{Descriptions in terms of Dirac structures}\label{dirac}

In the next proposition we interpret in terms of Dirac
structures the operation $\{\cdot,\cdot\}_{\un{N}}$ given by eq.
\eqref{defbr}.
Let $(M,\Pi)$ be a Poisson manifold,
$N\subset M$ a submanifold and $B \subset T_NM$ a   subbundle with
$F:=B\cap TN$ a regular, integrable distribution.

\begin{proposition}\label{pushpull}
Assume  that $\un{N}:=N/F$ is   smooth and
 that the prescription \eqref{defbr} gives a well-defined
bivector field on $\un{N}$, and denote by $L_{\un{N}}$ its graph.
Then the pullback of the almost Dirac structure  $L_{\un{N}}$ under
$p: N \rightarrow \un{N}$ is $\iota^*(L_{\Pi}^B)$.

 Here $L_{\Pi}^B$
is the stretching \cite{CFZ} of $L_{\Pi}=graph(\Pi)$ in direction of
$B$, defined as $[L_{\Pi}|_N\cap (T_NM \oplus B^{\circ})]+(B\oplus
0)$.

  \end{proposition}
\begin{proof}
We will show that the Poisson algebras of admissible functions for
$\iota^*(L_{\Pi}^B)$ and $p^*(L_{\un{N}})$ match, hence the
subbundles have to agree too. Short computations using
$\sharp(TN+B)^{\circ}\subset B$ (which holds since we assume that
eq. \eqref{defbr} gives a well-defined expression) show that
$\iota^*(L_{\Pi}^B)$ is a smooth almost Dirac structure on $N$ and
  that its kernel is exactly $F$. Hence its set of admissible
  functions is $ C^{\infty}(N)_F$.
  If $f,g \in C^{\infty}(N)_F$ their $\iota^*(L_{\Pi}^B)$-bracket is
  $\langle X_{f^B}+b, dg^B \rangle$ (where
  $f^B,g^B \in C^{\infty}(M)_B$ are extensions and $b\in \Gamma(B)$ is such
  that $X_{f^B}+b\in TN$), which is equal to $\{f^B,g^B\}$.

The kernel of $p^*(L_{\un{N}})$ is clearly also $F$, and if $f,g \in
C^{\infty}(N)_F$ their $p^*(L_{\un{N}})$-bracket is
$\{f,g\}_{\un{N}}$. Using eq. \eqref{defbr} this  concludes the
proof.
\end{proof}

\begin{remark}
The following statements complement Proposition \ref{pushpull} and are
proved similarly. Assume that  $\sharp(TN+B)^{\circ}\subset B$. Then
eq.  \eqref{defbr} defines a bivector field on $\un{N}$  iff
$\iota^*(L_{\Pi}^B)$ pushes forward under $p: N \rightarrow \un{N}$
(to the graph of eq. \eqref{defbr}).  If   $\iota^*(L_{\Pi}^B)$ is
integrable (i.e. if it is a Dirac structure on $N$) then it
automatically pushes forward, and therefore eq.  \eqref{defbr}
defines a Poisson structure on $\un{N}$.

Hence, assuming $\sharp(TN+B)^{\circ}\subset B$,  eq.  \eqref{defbr} defines a Poisson structure on $\un{N}$ iff $\iota^*(L_{\Pi}^B)$ is integrable. Unfortunately we were not able to express the integrability of the latter in simple
terms.
\end{remark}

\subsection{Compatibility with foliations}
We now address the question of compatibility
stated in Def. \ref{compat} and give an equivalent characterization.
By Remark \ref{remlocal} we can work locally, so in the following
we will assume that $\un{N}$ and $\un{M}:=M/\theta_D$ are smooth.

\begin{proposition}\label{equichar}
$\theta_D$ and $B$ as in Def. \ref{compat}
are compatible if and only if
\begin{eqnarray}\label{thetadescequiv}
&&\text{There exists a subbundle }
\hat{B} \text{ with }
B\subset\hat{B}\subset T_NM
\text { and }\hat{B}\cap TN=F\notag \\
&&\text{ such that }
pr: M  \rightarrow \un{M} \text{ maps }
\hat{B}\text{ to a well-defined subbundle of }T_{\un{N}}\un{M}.
\end{eqnarray}
\end{proposition}

\begin{proof}
To show the ``if'' part notice that $pr_*{\hat{B}}$
intersects trivially $T\un{N}$ (since $F \subset D$), hence any
function on $\un{N}$ can be extended to an element of
$C^{\infty}(\un{M})_{pr_*\hat{B}}$, and the pullback under $pr$ is then an
element of $C^\infty(M)_B\cap C^\infty(M)_{\theta_D}$.
Conversely, if $\theta_D$ and $B$ are compatible,
we can extend a set of coordinates on $\un{N}$ to
functions  $x_i$ on $\un{M}$ so that $pr^* x_i \in
C^{\infty}(M)_B$, and
$\hat{B}:=pr_*^{-1}(\cap{\rm ker }\dd x_i)\subset T_NM$
will satisfy the condition above.
\end{proof}

In general it is not trivial to check whether the conditions of the previous
proposition are satisfied. One can however compute easily a sufficient
condition for the compatibility in the case one can take $\hat B=B$.

To state the result we introduce
$\tilde{\Gamma}(B):= \{X\in \Gamma(TM): X|_N\subset B\}$ and
$\Gamma'(\theta_D):=\Gamma(\theta_D)\cap \tilde{\Gamma}(F)$.
Then one can prove that (\ref{thetadescequiv}) holds with
$\hat B=B$ if and only if
\begin{equation}\label{thetadescsec}
[\Gamma'(\theta_D),\tilde{\Gamma}(B)]
\subset \tilde{\Gamma}(B),
\end{equation}
which implies the compatibility of $\theta_D$ and $B$.

We conclude remarking that, given a subbundle $D$ with $F\subset D\subset B$,
locally one can always find an extension of $D$ to an involutive
distribution $\theta_D$ compatible with $B$.

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\providecommand{\MRhref}[2]{
  \href{http://www.ams.org/mathscinet-getitem?mr=#1}{#2}
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TITLE: What does $H^n(L)$ means in algebraic geometry?
QUESTION [1 upvotes]: I encountered the notation $H^n(L)$ where $L$ is a line bundle over some projective variety $X$ in the book by Le Potier and some other papers. I've only seen notations like $H^n(X,L)$, so I am wondering if the notation $H^n(L)$ has any special meaning.

REPLY [8 votes]: If there's only one space under consideration when we're talking about sheaf cohomology, it is not uncommon to omit the space when writing $H^n(X,\mathcal{F})$ and just write $H^n(\mathcal{F})$.


text:
\section{Probability tools}

\begin{lemma}[Matrix Bernstein, Theorem 6.1.1 in \cite{t15}]\label{lem:matrix_bernstein}
Consider a finite sequence $\{ X_1, \cdots, X_m \} \subset \R^{n_1 \times n_2}$ of independent, random matrices with common dimension $n_1 \times n_2$. Assume that
\begin{align*}
\E[ X_i ] = 0, \forall i \in [m] ~~~ \mathrm{and}~~~ \| X_i \| \leq M, \forall i \in [m] .
\end{align*}
Let $Z = \sum_{i=1}^m X_i$. Let $\mathrm{Var}[Z]$ be the matrix variance statistic of sum:
\begin{align*}
\mathrm{Var} [Z] = \max \left\{ \Big\| \sum_{i=1}^m \E[ X_i X_i^\top ] \Big\| , \Big\| \sum_{i=1}^m \E [ X_i^\top X_i ] \Big\| \right\}.
\end{align*}
Then
{\small
\begin{align*}
\E[ \| Z \| ] \leq ( 2 \mathrm{Var} [Z] \cdot \log (n_1 + n_2) )^{1/2} +  M \cdot \log (n_1 + n_2) / 3.
\end{align*}
}
Furthermore, for all $t \geq 0$,
\begin{align*}
\Pr[ \| Z \| \geq t ] \leq (n_1 + n_2) \cdot \exp \left( - \frac{t^2/2}{ \mathrm{Var} [Z] + M t /3 }  \right)  .
\end{align*}
\end{lemma}

\section{Proof of Main Result}
\label{sec:proof_main}
We state the complete version of the proofs shown in Section~\ref{sec:proof_idea} in this section. Parts of Section~\ref{subsec:parapprox}, \ref{subsec:actdiff} and \ref{subsec:rewbound} have been stated in Section~\ref{sec:proof_idea}, we restate here for completeness. In Section~\ref{subsec:parapprox}, we give a concentration bound on the particle approximation of the latent state. In Section~\ref{subsec:actdiff}, we study how the error of inference in each round accumulates through the sequential planning process. In Section~\ref{subsec:rewbound}, we put the pieces together to give the upper bound on the number of particles needed so that the long-run rewards of the two processes are close.  We show the proofs of most of the lemmas in this section to Section~\ref{sec:defproof}.
\subsection{Particle  Concentration}
\label{subsec:parapprox}
We first note that at time $t$, since we know the initial state $x_0$, the transition matrices $A_{0:t-1}$ and $B_{0:t-1}$ and the past actions $\hat{u}_0,...,\hat{u}_{t-1}$, estimating the state $x_t$ is equivalent to estimating $\xi_0,...,\xi_{t-1}$. We show in Lemma~\ref{lem:decomposition} that we can write the states as a function of the initial state, past transformation noise and actions, which follows straightly from our definitions of the processes.
\decomposition*

Recall from Section~\ref{subsec: probsetup} that the estimation $\hat{y}_t$ is given by a weighted average of the states of the simulated particles,
\begin{align}\label{eq:weighted_state}
\hat{y}_t = \frac{\sum_{i=1}^N w_{t}^{(i)} x_t^{(i)}}{\sum_{i=1}^N w_{t}^{(i)}}.
\end{align}
and the estimation $\tilde{y}_t$ is given by the posterior mean of $x^*_t$ given observations $o_{0:t}$,
\begin{align}\label{eq:posterior_state}
\tilde{y}_t = \frac{\int_{x'_{1:t}\in \mathcal{X}^t}\prod_{s=1}^t \P\left[o^*_s~|~x'_s\right] x'_t\d \rho_t(x'_{1:t})}{\int_{x'_{1:t}\in \mathcal{X}^t}\prod_{s=1}^t \P\left[o^*_s~|~x'_s\right] \d \rho_t(x'_{1:t})}.
\end{align}
By Lemma~\ref{lem:decomposition}, we know that to estimate $x_t$ and $x_t^*$, it is enough to estimate $\xi_{0:t-1}$. Surprisingly, we can further show that the estimators $\hat{y}_t$ and $\tilde{y}_t$ can be written as a function of estimators $\hat{\xi}_{t,0:t-1}$ and $\tilde{\xi}_{t,0:t-1}$, past actions $\hat{u}_{0:t-1}$ and $u^*_{0:t-1}$, and the initial state $x_0$. The estimator $\hat{\xi}_{t,0:t-1}$ is given by a weighted average of the noise of the particles, $\xi_{0:t-1}^{(1)},...,\xi_{0:t-1}^{(N)}$, similar to \eqref{eq:weighted_state}. The estimator $\tilde{\xi}_{t,0:t-1}$ is given by the posterior mean of the noise given observations, similar to \eqref{eq:posterior_state}. We show this formally in Lemma~\ref{lem:error_est}.

\begin{restatable}{lemmma}{errorest}
	\label{lem:error_est}
	At time $t \in [T]$, for any $s=0,..,t-1$, if we estimate $\xi_s$ as $\hat{\xi}_{t,s}$, given by,
	\begin{align*}
	\hat{\xi}_{t,s} = \frac{\sum_{i=1}^N w_{t}^{(i)} \xi_s^{(i)}}{\sum_{i=1}^N w_{t}^{(i)}},
	\end{align*}
	$\hat{y}_t $ can be written as
	\begin{align*}
	\hat{y}_t =  \sum_{s=0}^{t-1} \prod_{s'=s+1}^{t-1} A_{s'} \cdot \left(\hat{\xi}_{t,s} + B_{s}\cdot \hat{u}_{s}\right) + \prod_{s=0}^{t-1} A_{s}\cdot x_0.
	\end{align*}
	Moreover, let $\pi_t(\xi_{0:t-1})$ be the distribution of $\xi_{0:t-1}$, given by the density $\prod_{s=0}^{t-1}\eta_s(\xi_s)$. If we estimate $\xi_s$ as $\tilde{\xi}_{t,s}$, given by,
	\begin{align*}
	\tilde{\xi}_{t,s} = \frac{  \int_{\xi'_{0:t-1}} \P\left[o^*_{1:t} ~|~ \xi'_{0:t-1},u^*_{0:t-1},x_0\right]\xi'_{s}\d\pi_t(\xi'_{0:t-1}) }{ \int_{\xi'_{0:t-1}} \P\left[o^*_{1:t} ~|~ \xi'_{0:t-1},u^*_{0:t-1},x_0\right]\d\pi_t(\xi'_{0:t-1}) },
	\end{align*}
	$\tilde{y}_t$ can be written as
	\begin{align*}
	\tilde{y}_t =  \sum_{s=0}^{t-1} \prod_{s'=s+1}^{t-1} A_{s'} \cdot \left(\tilde{\xi}_{t,s} + B_{s}\cdot u^*_{s}\right) + \prod_{s=0}^{t-1} A_{s}\cdot x_0.
	\end{align*}
\end{restatable}
By Lemma~\ref{lem:error_est}, since $\hat{u}_{0:t-1}$ and ${u}_{0:t-1}$ are determined by $\hat{y}_{0:t-1}$ and $\tilde{y}_{t-1}$, to show that $\hat{y}_t$ is close to $\tilde{y}_t$, it is enough to show that $\hat{\xi}_{s,0:s-1}$ is close to $\tilde{\xi}_{s, 0:s-1}$ in all rounds $s = 1,...,t-1$. In this section, we focus on showing how accurately $\xi_{t,0:t-1}$ can approximate $\tilde{\xi}_{t, 0:t-1}$ in one time step. We postpone the discussion of how the error of this approximation accumulates through the process to Section~\ref{subsec:actdiff}.

To see how $\hat{\xi}_{t,s}$ can approximate $\tilde{\xi}_{t,s} $, we study the numerator and the denominator of $\hat{\xi}_{t,s}$ and $\tilde{\xi}_{t,s}$ seperately. To simplify our notation, we make the following definition.

\begin{definition}
	We define the scalar $\gamma_t\in R$ and vector $\Gamma_{t,s} \in \R^d$ for any time $0\leq s< t\leq T$  as follows:
	\begin{align*}
	\gamma_t = & ~  \int_{\xi'_{0:t-1}} \P\left[o^*_{1:t} ~|~ \xi'_{0:t-1},u^*_{0:t-1},x_0\right]\d\pi_t(\xi'_{0:t-1}), \\
	\Gamma_{t,s} = & ~ \int_{\xi'_{0:t-1}} \P\left[o^*_{1:t} ~|~ \xi'_{0:t-1},u^*_{0:t-1},x_0\right]\xi'_{s}\d\pi_t(\xi'_{0:t-1}) .
	\end{align*}
\end{definition}

We can further show that since we couple our two processes using the same noise, the posterior mean of the noise given observations, $o_{1:t}$, and actions $\hat{u}_{0:t-1}$ in the approximate process is the same as that given observations, $o^*_{1:t}$, and actions $u^*_{0:t-1}$ in the ideal process.
\begin{restatable}{lemmma}{comparison}
	\label{lem:comparison}
	For any time $0\leq s< t\leq T$ ,
	\begin{align*}
	\gamma_t = & ~  \int_{\xi'_{0:t-1}} \P\left[o_{1:t} ~|~ \xi'_{0:t-1},\hat{u}_{0:t-1},x_0\right]\d\pi_t(\xi'_{0:t-1}), \\
	\Gamma_{t,s} = & ~ \int_{\xi'_{0:t-1}} \P\left[o_{1:t} ~|~ \xi'_{0:t-1},\hat{u}_{0:t-1},x_0\right]\xi'_{s}\d\pi_t(\xi'_{0:t-1}) .
	\end{align*}
\end{restatable}
Then, to show that the particle approximation, $\hat{\xi}_{t,s}$, is close to $\tilde{\xi}_{t,s} $, it is enough to show that $\hat{\xi}_{t,s}$ concentrates around the posterior mean of $\xi_{t,s}$. We show the relationship between the accuracy of particle approximation and the number of particles, $N$, in the following lemma.
\concentration*
We defer the proof of Lemma~\ref{lem:error_est}, Lemma~\ref{lem:comparison} and Lemma~\ref{lem:concentration} to Appendix~\ref{sec:defproof}.

\subsection{Error Accumulation}
\label{subsec:actdiff}

In Section~\ref{subsec:parapprox}, we studied how the particle approximation concentrates in one time step. In this Section, we discuss how the error of approximation in one time step can affect the actions in the future and further affect the long-run reward of the process.

Lemma~\ref{lem:decomposition} shows that the states of the two processes, $x_t$ and $x_t^*$, at time step $t$, can be written as
\begin{align*}
x_t = \sum_{s=0}^{t-1} \prod_{s'=s+1}^{t-1} A_{s'} \cdot \left(\xi_{s} + B_{s}\cdot \hat{u}_{s}\right) + \prod_{s=0}^{t-1} A_{s}\cdot x_0,\\
x^*_t = \sum_{s=0}^{t-1} \prod_{s'=s+1}^{t-1} A_{s'} \cdot \left(\xi_{s} + B_{s}\cdot {u}^*_{s}\right) + \prod_{s=0}^{t-1} A_{s}\cdot x_0.
\end{align*}
It is easy to see that the distance between $x_t$ and $x_t^*$ is determined by the distance between actions in the past time steps, $\hat{u}_{0:t-1}$ and $u^*_{0:t-1}$.
\statedist*
The actions $\hat{u}_s$ and $u^*_s$ at $s=1,...,t-1$ is determined by the state estimations, $\hat{y}_s$ and $\tilde{y}_s$,
\begin{align*}
\hat{u}_{s} =g(\hat{y}_t) \quad \text{and} \quad u^*_s = g(\tilde{y}_s).
\end{align*}
Moreover, by Lemma~\ref{lem:error_est}, at time step $t$,
\begin{align*}
\hat{y}_t =  \sum_{s=0}^{t-1} \prod_{s'=s+1}^{t-1} A_{s'} \cdot \left(\hat{\xi}_{t,s} + B_{s}\cdot \hat{u}_{s}\right) + \prod_{s=0}^{t-1} A_{s}\cdot x_0,\\
\tilde{y}_t =  \sum_{s=0}^{t-1} \prod_{s'=s+1}^{t-1} A_{s'} \cdot \left(\tilde{\xi}_{t,s} + B_{s}\cdot u^*_{s}\right) + \prod_{s=0}^{t-1} A_{s}\cdot x_0.
\end{align*}
which shows that  $\hat{y}_t$ and $\tilde{y}_s$ are in turn determined by the past actions.
Thus, the key step of bounding the error accumulation is bounding the distance between the actions in the two processes. We show the upper bound on the action distance in the following lemma.
\actiondist*

We defer the proof of Lemma~\ref{lem:action_dist2} to Appendix~\ref{sec:defproof}. Lemma~\ref{lem:state_dist} and Lemma~\ref{lem:action_dist2} together show that we can bound the distance between the states and the action of the two processes in terms of the accuracy of the particle approximation of transformation noise $\xi_{t}$.

\subsection{Bound on Reward Difference}
\label{subsec:rewbound}
In this section, we combine the results from Section~\ref{subsec:parapprox} and Section~\ref{subsec:actdiff} to show an upper bound on the number of particle needed so that the rewards of the two processes are close. Lemma~\ref{lem:concentration} upper bounds the number of particles needed so that the particle approximation of the noise $\xi_{t}$ is accurate. Lemma~\ref{lem:state_dist} and Lemma~\ref{lem:action_dist2}  show that the actions and the states of the two processes are close if the particle approximation is accurate. Then, for reward function that depends on states and actions, we can combine these results to upper bound the number of particles that can guarantee the rewards of the two processes are close. We state our main result in Theorem~\ref{thm:main_linear} and show the proof below.
\begin{proof}[Proof of Theorem~\ref{thm:nonlinear} and Theorem~\ref{thm:main_linear}]
	We state the proof for the Lipschitz $g$ case here. The proof for linear $g$ follows the same steps.
	We first show the number of particles needed so that the estimation of the noise, $\xi_{t}$, in a single round is accurate.
	If
	\begin{align}
	\label{eq:num_particle}
	N = \Omega ( \beta^{-2}p^{-1} \log (dT /\delta) ),
	\end{align}
	then
	\begin{align*}
	(d+1) \cdot \exp(-N\beta^2\gamma_t/3 ) \leq & ~ (d+1) \cdot \exp(-N\beta^2p/3 )
	\leq  ~ \delta / (2T^2 ) ,
	\end{align*}
	where the first inequality follows from $$\gamma_t = \P_{O^*_{1:t}}[o_{1:t}^*|u_{0:t}^*,x_0] = \P_{O_{1:t}}[o_{1:t}|\hat{u}_{0:t},x_0] \geq p.$$
	Let $M:=\sqrt{ \frac{d}{m}(1+2\sqrt{\log\beta'/d} +2\log\beta'/d )}$. If we choose $\beta' =  \log (2T^2N/\delta)$ and $\beta = \epsilon /(4MT)$, by Lemma~\ref{lem:concentration}, with success probability at least
	\begin{align*}
	1 - \sum_{t=1}^T\sum_{s=0}^{t-1} \delta/(2T^2) - \sum_{t=1}^T\sum_{s=0}^{t-1} \delta/(2T^2)  \geq 1-\delta,
	\end{align*}
	we have for all time step $t = 1,...,T$ and $s = 0,..,t-1$,
	\[
	\| \hat{\xi}_{t,s}-\tilde{\xi}_{t,s} \| \leq 4\beta M=  \epsilon /T .
	\]

	Next, we bound the distance between actions in the two processes. By Lemma $\ref{lem:action_dist2}$, for any $t = 1,...,T$,
	\begin{align}
	\label{eq:u_dist}
	\left\| \hat{u}_{t} - u_{t}^{*} \right\|   \leq & ~
	 L_g\Sigma_a^{(t)} \left( 1 +  L_g  C_b \Sigma_{ab}^{(t-1)}  \right)  \cdot  \frac \epsilon T.
	\end{align}
	The second inequality follows from our assumption.
	By Lemma~\ref{lem:state_dist} and our assumptions, we can further bound the distance between the states of the two processes as
	\begin{align}
	\label{eq:x_dist}
	\| x_{t} - x_{t}^{*} \|
	= & ~\left \|\sum_{s=0}^{t-1}\left( \prod_{s'=s+1}^{t-1} A_{s'}\right) B_s ( \hat{u}_{s}  - u^*_s ) \right\| \notag
	\leq  ~ C_b  \left( \left\| \hat{u}_{t-1}  - u^*_{t-1}  \right\| + C_a \sum_{s=0}^{t-2} \rho_a^s  \left\| \hat{u}_{s}  - u^*_s  \right\|\right) \notag  \\
	\leq & ~ C_b \Sigma_a^{(t)}\cdot L_g \Sigma_a^{(t)} \left( 1 +  L_g  C_b \Sigma_{ab}^{(t-1)}  \right)  \cdot \frac \epsilon T,
	\end{align}
    Thus, combining \eqref{eq:u_dist} and \eqref{eq:x_dist}, we can get for any $L_r$-Lipschitz reward function $r_T$,
	\begin{align*}
	&~ r_T(x_{1:T}, \hat{u}_{0:T-1}) - r_T(x^{*}_{1:T}, u^{*}_{0:T-1})
	\leq ~ \sum_{t=1}^T  L_r \| x_{t}-x_{t}^{*} \| + \sum_{t=1}^T  L_r \| \hat{u}_{t-1} - u_{t-1}^{*} \| \\
	\leq & ~ L_rL_g \Sigma_a^{(T)} \left( 1 + C_b\Sigma_a^{(T)}\right)\left( 1 +L_g C_b\Sigma_{ab}^{(T-1)}\right)    \epsilon ,
	\end{align*}
	where the first step follows from $r_T$ is $L_r$-Lipschitz and the second step follows from \eqref{eq:u_dist} and \eqref{eq:x_dist}.

	Plugging $\beta^2 =  \epsilon^2 /(16T^2M^2) = \tilde{\Theta}(\epsilon^2 T^{-2}d^{-1}m)$ into \eqref{eq:num_particle}, the number of particles needed is
	\begin{align*}
	N = & ~ \tilde{O} ( \beta^{-2}p^{-1}  )
	=  \tilde{O} ( T^2dm^{-1} \epsilon^{-2} p^{-1}  ),
	\end{align*}
	which completes the proof.
	Similarly, we can also show that the number of particles needed for linear $g$ so that
	\begin{align*}
	&~ r_T(x_{1:T}, \hat{u}_{0:T-1}) - r_T(x^{*}_{1:T}, u^{*}_{0:T-1})  \leq  ~   L_rL_g \Sigma_a^{(T)} \left( 1 + C_b\Sigma_a^{(T)}\right)\left( 1 +C_{bg}\bar{\Sigma}_{ab}^{(T-1)}\right)    \epsilon
	\end{align*}
	is
	$$N =  \tilde{O} ( T^2dm^{-1} \epsilon^{-2} p^{-1}  ).$$
\end{proof}

\subsection{Deferred Proofs}
\label{sec:defproof}

\errorest*

\begin{proof}
	By Lemma~\ref{lem:decomposition}, for every particle $i\in[N]$,
	\begin{align*}
	x_t^{(i)}  = \sum_{s=0}^{t-1} \prod_{s'=s+1}^{t-1} A_{s'} \left(\xi_{s}^{(i)}+ B_{s}\cdot \hat{u}_{s}\right) + \prod_{s=0}^{t-1} A_{s}\cdot x_0.
	\end{align*}
	Then,
	\begin{align*}
	\hat{y}_t & = \frac{\sum_{i=1}^N w_{t}^{(i)} x_t^{(i)}}{\sum_{i=1}^N w_{t}^{(i)}}, \\
	& =  \sum_{s=0}^{t-1} \prod_{s'=s+1}^{t-1} A_{s'} \cdot \left(\frac{\sum_{i=1}^N w_{t}^{(i)} \xi_t^{(i)}}{\sum_{i=1}^N w_{t}^{(i)}}+ B_{s}\cdot \hat{u}_{s}\right) + \prod_{s=0}^{t-1} A_{s}\cdot x_0\\
	& =  \sum_{s=0}^{t-1} \prod_{s'=s+1}^{t-1} A_{s'} \cdot \left(\hat{\xi}_{s} + B_{s}\cdot \hat{u}_{s}\right) + \prod_{s=0}^{t-1} A_{s}\cdot x_0.
	\end{align*}
	Similarly, by Lemma~\ref{lem:decomposition} and the definition of $\rho_t$,
	\begin{align*}
	\tilde{y}_t & = ~\frac{\int_{x'_{1:t}\in \mathcal{X}^t}\prod_{s=0}^t \P\left[o^*_s~|~x'_s\right] x'_t\d \rho_t(x'_{1:t})}{\int_{x'_{1:t}\in \mathcal{X}^t}\prod_{s=0}^t \P\left[o^*_s~|~x'_s\right] \d \rho_t(x'_{1:t})} \\
	& = ~\frac{  \int_{\xi'_{0:t-1}} \P\left[o^*_{1:t} ~|~ \xi'_{0:t-1},u^*_{0:t-1},x_0\right] }{ \int_{\xi'_{0:t-1}} \P\left[o^*_{1:t} ~|~ \xi'_{0:t-1},u^*_{0:t-1},x_0\right]\d\pi_t(\xi'_{0:t-1}) } \left[ \sum_{s=0}^{t-1} \prod_{s'=s+1}^{t-1} A_{s'} \cdot \left({\xi}'_{s} + B_{s}\cdot {u}^*_{s}\right) + \prod_{s=0}^{t-1} A_{s}\cdot x_0\right] \d\pi_t(\xi'_{0:t-1}) \\
	& =   \sum_{s=0}^{t-1} \prod_{s'=s+1}^{t-1} A_{s'} \cdot \left(\tilde{\xi}_{t,s} + B_{s}\cdot u^*_{s}\right) + \prod_{s=0}^{t-1} A_{s}\cdot x_0.
	\end{align*}
\end{proof}
\comparison*
\begin{proof}
	For any $t\in[T]$, we have
	\begin{align*}
	& ~ \P_{O^*_{1:t}}\left[o_{1:t}^{*} ~|~ \xi'_{0:t-1},u_{0:t-1}^{*},x_0\right]\\
	= & ~ \prod_{t'=1}^{t}\P_{O^*_{1:t}}\left[o_{t'}^{*} ~|~ \xi'_{0:t'-1},u_{0:t'-1}^{*},x_0\right]\\
	= & ~ \prod_{t'=1}^{t}\eta_{t'}\left(\left[ \sum_{s=0}^{t'-1} \prod_{s'=s+1}^{t'-1} A_{s'} \cdot \left({\xi}_{s} + B_{s}\cdot \hat{u}_{s}\right) + \prod_{s=0}^{t'-1} A_{s}\cdot x_0 +\zeta_{t'}\right]-\left[ \sum_{s=0}^{t'-1} \prod_{s'=s+1}^{t'-1} A_{s'} \cdot \left(\hat{\xi'}_{s} + B_{s}\cdot \hat{u}_{s}\right) + \prod_{s=0}^{t'-1} A_{s}\cdot x_0\right]\right)\\
	= & ~  \prod_{t'=1}^{t}\eta_{t'}\left(\sum_{s=0}^{t'-1} \prod_{s'=s+1}^{t'-1} A_{s'} \cdot \left({\xi}_{s} -\xi'_s\right) +\zeta_{t'}\right)\\
	= & ~ \P_{O_{1:t}}\left[o_{1:t} ~|~ \xi'_{0:t-1},\hat{u}_{0:t-1},x_0\right],
	\end{align*}
	where the third step follows from  Lemma~\ref{lem:decomposition}, so
	\begin{align*}
	\int_{\xi'_{0:t-1}} \P_{O_{1:t}}\left[o_{1:t} ~|~ \xi'_{0:t-1},\hat{u}_{0:t-1},x_0\right]\d\pi_t(\xi'_{0:t-1}) = & ~  \int_{\xi'_{0:t-1}} \P_{O^*_{1:t}}\left[o^*_{1:t} ~|~ \xi'_{0:t-1},u^*_{0:t-1},x_0\right]\d\pi_t(\xi'_{0:t-1}), \\
	\int_{\xi'_{0:t-1}} \P_{O_{1:t}}\left[o_{1:t} ~|~ \xi'_{0:t-1},\hat{u}_{0:t-1},x_0\right]\xi'_{s}\d\pi_t(\xi'_{0:t-1}) = & ~ \int_{\xi'_{0:t-1}} \P_{O^*_{1:t}}\left[o^*_{1:t} ~|~ \xi'_{0:t-1},u^*_{0:t-1},x_0\right]\xi'_{s}\d\pi_t(\xi'_{0:t-1}) .
	\end{align*}
\end{proof}

\concentration*
\begin{proof}
	We first consider the random variables
	\begin{align*}
	\P\left[ o_{1:t} ~|~ \xi_{0:t-1}^{(i)},\hat{u}_{0:t-1},x_0 \right]= \prod_{t'=1}^{t}\eta_{t'}\left(\sum_{s=0}^{t'-1} \prod_{s'=s+1}^{t'-1} A_{s'} \cdot \left({\xi}_{s} -\xi'_s\right) +\zeta_{t'}\right),
	\end{align*}
	for $i=1,...,N$.
	By the way we generate $\xi_{0:t}^{(1)}, \xi_{0:t}^{(2)},...,\xi_{0:t}^{(N)}$,
	\begin{align*}
	\P\left[o_{1:t} ~|~ \xi_{0:t-1}^{(1)},\hat{u}_{0:t-1},x_0\right], \P\left[o_{1:t} ~|~ \xi_{0:t}^{(2)},\hat{u}_{0:t},x_0\right], ...,\P\left[o_{1:t} ~|~ \xi_{0:t-1}^{(N)},\hat{u}_{0:t-1},x_0\right]
	\end{align*}
	are independent. Also, for $i=1,...,N$, by Lemma~\ref{lem:comparison}
	\begin{align*}
	\E\left[\P\left[o_{1:t} ~|~ \xi_{0:t-1}^{(i)},\hat{u}_{0:t-1},x_0\right]\right]= & ~\int_{\xi'_{0:t-1}}\P\left[o_{1:t} ~|~ \xi'_{0:t-1},\hat{u}_{0:t-1},x_0\right]\d\pi_t(\xi'_{0:t-1})
	=  ~ \gamma_t.
	\end{align*}
	and
	\begin{align*}
	\E\left[\P\left[o_{1:t} ~|~ \xi_{0:t-1}^{(i)},\hat{u}_{0:t-1},x_0\right]\xi'_{s}\right]= & ~\int_{\xi'_{0:t-1}}\P\left[o_{1:t} ~|~ \xi'_{0:t-1},\hat{u}_{0:t-1},x_0\right]\xi'_{s}\d\pi_t(\xi'_{0:t-1})
	=  ~ \Gamma_{t,s}.
	\end{align*}
	By Lemma~\ref{lem:matrix_bernstein},
	\begin{align*}
	& ~ \Pr\left[\left|\frac{1}{N}\sum_{i=1}^{N}\P\left[o_{1:t} ~|~ \xi_{0:t-1}^{(i)},\hat{u}_{0:t-1},x_0\right]-\gamma_t\right|\geq \beta\gamma_t\right]\\
	\leq & ~ \exp\left(-\frac{N\beta^{2}\gamma_t^2}{2\text{Var}\left[\P\left[o_{1:t} ~|~ \xi_{0:t-1}^{(i)},\hat{u}_{0:t-1},x_0\right]\right]+\frac{2}{3}\max \left|\P\left[o_{1:t} ~|~ \xi_{0:t-1}^{(i)},\hat{u}_{0:t-1},x_0\right]\right|\beta\gamma_t}\right)\\
	\leq & ~ \exp ( -  N\beta^{2}\gamma_t / 3 ) .
	\end{align*}
	where the third step follows from
	\begin{align*}
	\text{Var}\left[\P\left[o_{1:t} ~|~ \xi_{0:t-1}^{(i)},\hat{u}_{0:t-1},x_0\right]\right] \leq & ~ \E\left[\P^2\left[o_{1:t} ~|~ \xi_{0:t-1}^{(i)},\hat{u}_{0:t-1},x_0\right]\right] \\
	= & ~ \int_{\xi'_{0:t-1}}\P^2\left[o_{1:t} ~|~ \xi'_{0:t-1},\hat{u}_{0:t-1},x_0\right]\d\pi_t(\xi'_{0:t-1}) \\
	\leq & ~ \gamma_t,
	\end{align*}
	and
	\begin{align*}
	\max\left|\P\left[o_{1:t} ~|~ \xi_{0:t-1}^{(i)},\hat{u}_{0:t-1},x_0\right]\right| \leq 1.
	\end{align*}
	Without loss of generality, we assume the noise $\xi$ has mean zero. Since the noise $\| \xi^{(i)}_{s} \|$ is sub-gaussian, with probability at least $ 1 - N\exp(-\beta')$, for all $i\in[N]$, $$\| \xi^{(i)}_{s} \|^2 \leq M^2 = \frac{d}{m}(1+2\sqrt{\log\beta'/d} +2\log\beta'/d ).$$
	Similarly, by Lemma~\ref{lem:matrix_bernstein}, since the noise $\| \xi^{(i)}_{s} \| \leq M$ for all $i \in [N]$,
	\begin{align*}
	& ~ \Pr\left[\left\| \frac{1}{N}\sum_{i=1}^{N}\P\left[o_{1:t} ~|~ \xi_{0:t-1}^{(i)},\hat{u}_{0:t-1}, x_0\right]\xi_{s}^{(i)}-\Gamma_t\right\| \geq \beta \gamma_t M\right]\\
	\leq & ~ d \cdot \exp\left(-\frac{N\beta^{2}\gamma_t^2M^2}{2\text{Var}\left[\P\left[o_{1:t} ~|~ \xi_{0:t-1}^{(i)},\hat{u}_{0:t-1},x_0\right]\xi_{s}^{(i)}\right]+\frac{2}{3}\max\left\| \P\left[o_{1:t} ~|~ \xi_{0:t-1}^{(i)},\hat{u}_{0:t-1},x_0\right]\xi_{s}^{(i)}\right\| \beta\gamma_t M}\right)\\
	\leq & ~ d \cdot \exp  ( -  N \beta^{2} \gamma_t /  3 ).
	\end{align*}
	where the third step follows from
	\begin{align*}
	\text{Var}\left[\P\left[o_{1:t} ~|~ \xi_{0:t-1}^{(i)},\hat{u}_{0:t-1},x_0\right]\xi_{s}^{(i)}\right] \leq \gamma_t M^2,
	\end{align*}
	and
	\begin{align*}
	\max\left\| \P\left[o_{1:t} ~|~ \xi_{0:t-1}^{(i)},\hat{u}_{0:t-1},x_0\right]\xi_{s}^{(i)}\right\| \leq M.
	\end{align*}
	Then,  with probability at least
	\begin{align*}
	1 - (d+1) \exp ( -  N\beta^{2}\gamma_t / 3 ) - N \exp(-\beta'),
	\end{align*}
	we have
	\begin{align*}
	& ~ \| \hat{\xi}_{t,s}-\tilde{\xi}_{t,s} \|  =  \left\| \hat{\xi}_{t,s} - \frac{\Gamma_{t,s}}{\gamma_t} \right\| \\
	\leq & ~ \max\left\{ \left\| \frac{1}{\left(1-\beta\right)\gamma_t} \sum_{i=1}^{N}w_{t}^{(i)}\xi_{t'}^{(i)} -\frac{\Gamma_{t,s}}{\gamma_t}\right\| ,\left\| \frac{1}{\left(1+\beta\right)\gamma_t } \sum_{i=1}^{N}w_{t}^{(i)}\xi_{t'}^{(i)} -\frac{\Gamma_{t,s}}{\gamma_t}\right\| \right\} \\
	\leq & ~ \max\Bigg\{ \frac{1}{1-\beta}\left\| \frac{\sum_{i=1}^{N}w_{t}^{(i)}\xi_{t'}^{(i)}}{\gamma_t}  -\frac{\Gamma_{t,s}}{\gamma_t}\right\| + \left( \frac{1}{1-\beta}-1\right)\left\|  \frac{\Gamma_{t,s}}{\gamma_t} \right\| ,  \\
	& \frac{1}{1+\beta}\left\| \frac{\sum_{i=1}^{N}w_{t}^{(i)}\xi_{t'}^{(i)} }{\gamma_t} - \frac{\Gamma_{t,s}}{\gamma_t} \right\| + \left(1- \frac{1}{1+\beta}\right) \left\| \frac{\Gamma_{t,s}}{\gamma_t} \right\| \Bigg\} \\
	\leq  & ~ \max \left\{  \frac{\beta M}{1-\beta}+\frac{\beta M}{1-\beta}  , \frac{\beta M}{1+\beta} +\frac{\beta M}{1+\beta}  \right\} \\
	\leq & ~ 4\beta M
	\end{align*}
	where the first step follows from $| \sum_{i=1}^{N} w_{t}^{(i)} - \gamma_t | \leq \beta \gamma_t$, the second step follows from triangle inequality, the third step follows from $\left\|\sum_{i=1}^{N}w_{t}^{(i)}\xi_{t'}^{(i)}-\Gamma_{t,s}\right\|\leq \beta \gamma_t M$, and
	\begin{align*}
	\left\| \frac{\Gamma_{t,s}}{\gamma_t}\right\|
	= & ~ \left\| \frac{\int_{\xi'_{0:t-1}} \P\left[o_{1:t} ~|~ \xi'_{0:t-1},\hat{u}_{0:t-1},x_0\right]\xi'_{s}\d\pi_t(\xi'_{0:t-1}) }{\int_{\xi'_{0:t-1}} \P\left[o_{1:t} ~|~ \xi'_{0:t-1},\hat{u}_{0:t-1},x_0\right]\d\pi_t(\xi'_{0:t-1}) }\right\|
	\leq  ~ M,
	\end{align*}
	and the last step follows from $\beta\leq \frac{1}{2}$.
\end{proof}

\actiondist*
\begin{proof}
	When $t=0$, we have $\hat{u}_{0}=u_{0}^{*}=g(x_{0})$, so $\left\| \hat{u}_{0}-u_{0}^{*} \right\|=0$. For $t > 0$, we study the two cases separately.

	In the first case, for $L_g$-Lipschitz $g$, at time $t>0$,
		\begin{align*}
	&~ \| \hat{u}_{t} - u_{t}^{*} \|
	= ~ \| g( \hat{y}_{t} ) - g ( \tilde{y}_{t} ) \| \\
	= &~ \left\| g\left(  \sum_{s=0}^{t-1} \prod_{s'=s+1}^{t-1} A_{s'} \cdot \left(\hat{\xi}_{t,s} + B_{s}\cdot \hat{u}_{s}\right) + \prod_{s=0}^{t-1} A_{s}\cdot x_0\right) - g\left(  \sum_{s=0}^{t-1} \prod_{s'=s+1}^{t-1} A_{s'} \cdot \left(\tilde{\xi}_{t,s} + B_{s}\cdot u^*_{s}\right) + \prod_{s=0}^{t-1} A_{s}\cdot x_0\right)\right\| \\
	\leq & ~ L_g \cdot \left\| \left(  \sum_{s=0}^{t-1} \prod_{s'=s+1}^{t-1} A_{s'} \cdot \left(\hat{\xi}_{t,s} + B_{s}\cdot \hat{u}_{s}\right) + \prod_{s=0}^{t-1} A_{s}\cdot x_0\right) - \left(  \sum_{s=0}^{t-1} \prod_{s'=s+1}^{t-1} A_{s'} \cdot \left(\tilde{\xi}_{t,s} + B_{s}\cdot u^*_{s}\right) + \prod_{s=0}^{t-1} A_{s}\cdot x_0\right)\right\|  \\
	\leq & L_g \cdot \left\|  \sum_{s=0}^{t-1} \prod_{s'=s+1}^{t-1} A_{s'}  \cdot (\hat{\xi}_{t,s}-\tilde{\xi}_{t,s}) \right\| + L_g \cdot  \left\| \sum_{s=0}^{t-1} \prod_{s'=s+1}^{t-1} A_{s'} B_s \cdot (\hat{u}_{s}-u_{s}^{*})\right\|.
	\end{align*}
	where the first step follows from definitions of $\hat{u}_t$ and $u^*_t$, the second step follows from Lemma~\ref{lem:error_est}, the third step follows from $g$ is $L_g$-Lipschitz and the last step follows from triangle inequality.

	We define $f_t$ and $h_t$ as follows:
	\begin{align*}
	f_{t} = & ~ \left\| \sum_{s=0}^{t-1} \prod_{s'=s+1}^{t-1} A_{s'}  \cdot \left( \hat{\xi}_{t,s}-\tilde{\xi}_{t,s}\right)  \right\|, \\
	h_{t} = & ~ \left\| \sum_{s=0}^{t-1} \prod_{s'=s+1}^{t-1} A_{s'} B_s \cdot \left( \hat{u}_{s}-u_{s}^{*}\right)  \right\|.
	\end{align*}
	Then,
	\begin{align*}
	h_{t}
	\leq & ~ C_ah_{t-1}+ C_b \| \hat{u}_{t-1}-u_{t-1}^{*}\| \\
	\leq & ~ C_a h_{t-1}+ C_b L_g (f_{t-1}+h_{t-1})\\
	\leq & ~  C_b L_g f_{t-1} +(C_a+ C_bL_g)\left(C_a h_{t-2}+ C_b L_g (f_{t-2}+h_{t-2})\right)  \\
	\leq & ~ \cdots \\
	\leq & ~C_b L_g \sum_{s=1}^{t-1}(C_a +C_bL_g)^{t-s-1}f_{s},
	\end{align*}
	The first step follows from definition. The second step follows from $\| \hat{u}_{t-1} - u_{t-1}^{*} \| \leq L_g(f_{t-1} + h_{t-1}).$ The third step and the last step follow from induction.
	Thus, for $L_g$-Lipschitz $g$,
	\begin{align*}
	\| \hat{u}_{t} - u_{t}^{*} \| \leq & ~L_g\cdot f_t + L_g \cdot C_b L_g \sum_{s=1}^{t-1}(C_a+C_bL_g)^{t-s-1}f_{s} \\
	\leq  & ~  L_g\left(  1 + C_a\sum_{s = 0}^{t-2}\rho_a^{s}\right) \epsilon+ L_g \cdot C_b L_g \sum_{s=1}^{t-1}(C_a+C_bL_g)^{t-s-1}\cdot\left( 1+  C_a\sum_{s' = 0}^{s-2}\rho_a^{s'} \right) \cdot \epsilon  \\
	\leq & ~  L_g\Sigma_a^{(t)} \cdot \epsilon + L_g \cdot C_b L_g \Sigma_{ab}^{(t-1)} \cdot \Sigma_a^{(t)}  \cdot \epsilon
	= ~ L_g\Sigma_a^{(t)} \left( 1 +  L_g  C_b \Sigma_{ab}^{(t-1)}  \right)  \cdot \epsilon  .
	\end{align*}
	The second step follows from our assumption and the last two steps follow from our definitions of $\Sigma_a^{(t)} $ and $ \Sigma_{ab}^{(t)} $.

	In the second case, for linear $g = G$, similarly, we have
	\begin{align*}
	&~  \hat{u}_{t} - u_{t}^{*}  = ~ g( \hat{y}_{t} ) - g ( \tilde{y}_{t} )  \\
	= & ~ G\cdot   \sum_{s=0}^{t-1} \prod_{s'=s+1}^{t-1} A_{s'}  \cdot (\hat{\xi}_{t,s}-\tilde{\xi}_{t,s}) + G \cdot   \sum_{s=0}^{t-1} \prod_{s'=s+1}^{t-1} A_{s'} B_s \cdot (\hat{u}_{s}-u_{s}^{*}).
	\end{align*}
	We define $f_t$ and $h_t$ as follows:
	\begin{align*}
	f_{t} = & ~ \sum_{s=0}^{t-1} \prod_{s'=s+1}^{t-1} A_{s'}  \cdot \left( \hat{\xi}_{t,s}-\tilde{\xi}_{t,s}\right) , \\
	h_{t} = & ~  \sum_{s=0}^{t-1} \prod_{s'=s+1}^{t-1} A_{s'} B_s \cdot \left( \hat{u}_{s}-u_{s}^{*}\right).
	\end{align*}
	then by induction,
	\begin{align*}
	 h_{t}
	= & ~ A_{t-1}h_{t-1} + B_{t-1} \left(\hat{u}_{t-1}-u_{t-1}^{*} \right)\\
	\leq & ~ A_{t-1} h_{t-1}+ B_{t-1}G (f_{t-1}+h_{t-1})\\
	\leq & ~  B_{t-1} G f_{t-1} + (A_{t-1} +B_{t-1} G)\left(A_{t-2} h_{t-2}+ B_{t-2} G (f_{t-2}+h_{t-2})\right)  \\
	\leq & ~ \cdots \\
	\leq & ~\sum_{s=1}^{t-1} \prod_{s' = s+1}^{t-1} (A_{s'} + B_{s'}G) B_{s}G f_{s},
	\end{align*}
	Thus, by our assumptions and the definitions of $\Sigma_a^{(t)} $ and $ \bar{\Sigma}_{ab}^{(t)} $.
	\begin{align*}
	\| \hat{u}_{t} - u_{t}^{*} \| = & ~\left\Vert G\cdot  f_t + G \cdot  \sum_{s=1}^{t-1} \prod_{s' = s+1}^{t-1} (A_{s'} + B_{s'}G) B_{s}G f_{s} \right\Vert  \\
	\leq & ~L_g\left( 1+ C_a\sum_{s = 0}^{t-2}\rho_a^{s}  \right) \cdot \epsilon  + L_g \cdot C_{bg} \left(1+ C_{ab} \sum_{s=0}^{t-3}{\rho}_{ab}^s\right)\cdot \left( 1+ C_a\sum_{s' = 0}^{s-2}\rho_a^{s'} \right) \cdot \epsilon  \\
    \leq & ~ L_g\Sigma_a^{(t)}\left( 1 + C_{bg}\bar{\Sigma}_{ab}^{(t-1)}\right) \cdot \epsilon.
	\end{align*}
\end{proof}

\section{Experiment}
\label{sec:exp}

\begin{figure}[b]
	\begin{center}
		\includegraphics[scale=0.58]{exp.png}
	\end{center}
		\caption{Relationship between the regret and the number of particles}
	\label{fig}
\end{figure}

In this section, we use simulations to show the error of particle filtering can accumulate and be amplified through sequential planning. We run a process with a maximum time step $T = 40$ and $d = 1$. Since our bound shows the number of particles needed is insensitive to the dimension $d$, we mainly show how the number of time steps can affect the accuracy of particle filtering.

We consider the following process, for all $t \in [T] $,
\begin{align*}
 x_{t} =  x_{t-1}  + u_{t-1} + \xi_{t-1}, \; \text{ and }o_t = x_t + \zeta_t.
\end{align*}
The process can suffer a random shift of size 1, i.e., $\xi_t$ follows a uniform distribution on set $\{0, 1\}$.  $\zeta_t$ follows the standard normal distribution $\mathcal{N}(0, 1)$. The regret is defined as the average $\ell_1$ norm of the states, i.e., $r(x_{1:t}) = \sum_{i=1}^t |x_t|/t$. The policy function is $g(x)=-x$.
We show in Figure~\ref{fig} the regret and its standard deviation of the estimation using different number of particles. The result shows the number of particles needed for an accurate estimation can increase fast as the number of time step increases due to error accumulation. This experiment corroborates the importance of our theoretical results.

config: arxiv


TITLE: $\iint_{S} z+y+\sqrt{(a^2-x^2)} \,ds$ , surface integral
QUESTION [0 upvotes]: $$\iint_{S} z+y+\sqrt{(a^2-x^2)}   \,ds$$
$$ S: x^2+y^2=a^2,0\leqslant z
\leqslant       c   $$
$$ a,c>0 $$
Evaluate surface integral.
I wanted to express x(y),then with it evaluate $ dS $, then project this on yOz. The limits of y would be from - a to a, z from 0 to c.
I have seen  a solution in which $ dS $ was calculated from y. And the integral  was divide into 2 parts for positive and negative y.
Is my idea okay?

REPLY [1 votes]: Sounds right to me. We can then express the integral as following$$I=\iint_Sz+y+|y|ds=\iint_{y>0}z+2yds+\iint_{y<0}zds$$where$$ds=ad\phi dz$$therefore $$I=\int_{0}^{a}\int_{0}^{\pi}z+2a\sin\phi ad\phi dz+2a\pi\int_{0}^{a}zdz$$


TITLE: How to get from heat equation with final condition to one with initial condition?
QUESTION [1 upvotes]: How do I get from the heat equation with end condition
$$\frac{d}{dt}u(x,t) + \Delta u(x,t) = f(x,t)$$
$$u(x,T) = u_0(x)$$
where $t \in (0,T)$ and $x \in \Omega$, to a normal heat equation with initial condition?
if I substitute $s= T-t$, I dont know how to treat the time derivative.

REPLY [3 votes]: If you pose $\tau=T-t$ you have:
$$
\frac{\partial}{\partial \tau}=\frac{\partial}{\partial (T-t)}=-\frac{\partial}{\partial t},
$$
then everything should flow.


TITLE: Showing Lebesgue integral over set of measure zero is in fact zero zero.
QUESTION [4 upvotes]: Question: Let $E\subset\mathbb{R}$ be a Lebesgue measurable set for which $\operatorname{m}(E)=0$, where $\operatorname{m}$ denotes the Lebesgue measure. I'm trying to show that this implies
$$\int_{E}e^{-x^2/2}dx=0.\tag{1}$$

(Attempted) Solution: Let $f:=e^{-x^2/2}$. Since $0 < f \leq 1$ over $\mathbb{R}$, and thus over any $E\subset\mathbb{R}$, it follows
$$0\leq\int_{E}e^{-x^2/2}dx\leq\int_{E}1dx = 1\cdot\operatorname{m}(E) =0.$$
That is,
$$0\leq\int_{E}e^{-x^2/2}dx\leq0.$$
Therefore, indeed we have that $(1)$ holds.

My question is, is this solution correct? If not, can it be repaired? And if not, any help would be appreciated!

REPLY [3 votes]: This works. Your argument shows that if $f$ is bounded on a set $E$ of Lebesgue measure zero, then $\int_E f = 0$. In fact the stronger claim that $\int_E f = 0$ without any hypothesis on $f$ holds! Indeed the integral is, by definition, the supremum over $g$ with $g$ simple and $g \leq f$ of $\int_E g$. But any such $g$ has integral zero.


TITLE: How do I find a function that acts as the zero vector in a vector space
QUESTION [0 upvotes]: Question
I understand that based on the axioms of vector spaces there needs to be a unique member the zero vector in V such that for all v element of V, v+0=v, but how do I find the appropriate function in the bove case?

REPLY [0 votes]: For all $f \in V$ we have $f+z=f$, so for all $x \in \mathbb{R}$: $(f+z)(x) = f(x)+z(x) = f(x)$. For example, $1 = f_1(x) = (f_1+z)(x) = f_1(x)+z(x) = 1 + z(x)$ and so $z(x) = 0$ for all $x \in \mathbb{R}$. The only function $z \in V$ with $z(x)=0$ for all $x \in \mathbb{R}$ is the one with $z(x) = 0 + 0x + 0x^2$. The coordinate vector of the zero-element for either ordered basis must be $(0, 0, 0)^T$.


TITLE: Proof of $a \equiv b \mod n$ implies $a^k \equiv b^k \mod n$
QUESTION [3 upvotes]: Prove that for $n$ in the set of natural numbers, with $n \geq 2$:
For all $a, b \in \mathbb{N}$, $a \equiv b \mod n$ implies that $a^2 \equiv b^2 \mod n$.
also what about this
Prove by induction that for $n$ in the set of natural numbers, $n \geq 2$
For all $a,b \in \mathbb{N}$, $a \equiv b \mod n$ implies that $a^k \equiv b^k \mod n$.

REPLY [0 votes]: Okay, so first let's make a preposition for a^k ≡ b^k, and let's call it P(k). Once we have that, we can move on to our base case. Our base case would be P(0), or when a^0 ≡ b^0 (mod m). That would equal 1 ≡ 1 (mod m). That is true, since, based off of definition, m | (1 - 1), and that equals m | 0, which is true, since m * 0 = 0, so our base case is true. Now it is time for the inductive step. For whenever k ≥ 0, we would have to assume that P(k) in order to prove that P(k + 1) is true. So then assume that a^k ≡ b^k (mod m) is true. When we combine these assumptions and the fact that if a ≡ b (mod m) and c ≡ d (mod m) are true, then a + c ≡ b + d (mod m) would also be true, we would get a^(k + 1) ≡ b^(k + 1) (mod m), so this would therefore be proven by using proof by induction.


TITLE: How evaluate $ \int_0^\infty \frac{\ln^2\;\left|\;\tan\left(\frac{ax}{2}-\frac{\pi}{4}\right)\;\right|}{1+x^2} \;dx$
QUESTION [9 upvotes]: How evaluate (without using Complex analysis)
$$ \int_0^\infty \frac{\ln^2\;\left|\;\tan\left(\frac{ax}{2}-\frac{\pi}{4}\right)\;\right|}{1+x^2}\; dx\quad (a\gt0)$$
My Attempt:
I used the expansion of the following functions: $$ \ln\left|2\sin(x)\right| \text{ and }\ln\left|2\cos(x)\right| $$
To get the following expansion: $$ \ln\;\left|\;\tan\left(\frac{ax}{2}-\frac{\pi}{4}\right)\;\right|=2\sum_{n=1}^\infty (-1)^n \frac{\sin[(2n-1)ax]}{2n-1} $$
Then I expressed the square of the logarithmic function as follows:$$ \ln^2\;\left|\;\tan\left(\frac{ax}{2}-\frac{\pi}{4}\right)\;\right|=4\sum_{m=1}^\infty\sum_{n=1}^\infty (-1)^{m+n} \frac{\sin[(2m-1)ax]\sin[(2n-1)ax]}{(2m-1)(2n-1)} $$
And used the formula of the product of two sines, then integrated the following well known integral under the summation sign:
$$ \int_0^\infty \frac{\cos(bx)}{1+x^2} \;dx$$
And expressed the final result in terms of $\; \arctan\;\left(\;e^{-a}\;\right)\; $
but this was not the right answer according to the one evaluated by Complex analysis, which is
$$ \frac{{\pi}^3}{8}-2\pi\; \arctan^2\;\left(\;e^{-a}\;\right) $$
Any hint for another method or idea?

REPLY [2 votes]: part(1)
$$first\ part\\
I=\int_{0}^{\infty }\frac{ln^2(tan(\frac{ax}{2}-\frac{\pi }{4}))}{x^2+1}dx=\int_{0}^{\infty }\frac{ln^2(cot(\frac{ax}{2}-\frac{\pi }{4}))}{x^2+1}dx\\
\\
we\ have\ ln^2(cot(\frac{ax}{2}-\frac{\pi }{4})=-2\sum_{n,m=1}^{\infty }(-1)^{m+n}\frac{sin(2m-1)ax.sin(2n-1)ax}{(2m-1)(2n-1))}\\
\\
 =2\sum_{n,m=1}^{\infty }\frac{(-1)^{m+n}[cos(2m-2n)ax-cos(2m+2nax))]}{(2m-1)(2n-1)}\\
\\
\therefore I=\frac{\pi }{2}\sum_{m,n=1}^{\infty }\frac{(-1)^{m+n}}{(2m-1)(2n-1)}[e^{-2a(2m-2n)}-e^{-2a(2m+2n-2)}]\\
\\
note\ \sum_{m,n=1}^{\infty }\frac{(-1)^{m+n}}{(2m-1)(2n-1)}e^{-2a(2n+2m-2)}\\
\\
=\sum_{m=1}^{\infty }\frac{(-1)^{m-1}}{(2m-1)}e^{-2a(2m-1)}.\sum_{n=1}^{\infty }\frac{(-1)^{n-1}}{(2n-1)}e^{-2a(2n-1)}=arctan^2(e^{-2a})$$
second  part
$$second\ part \\
\because \sum_{m,n=1}^{\infty }\frac{(-1)^{m+n}}{(2m-1)(2n-1)}e^{-2a(2m-2n)}=\sum_{n=1}^{\infty }\frac{1}{(2n-1)^2}+\sum_{n,m=1}^{\infty }\frac{(-1)^{m-n}}{(2m-1)(2n-1)}e^{-2a(2m-2n)}\\
\\
=\frac{\pi ^{3}}{8}+\sum_{m,n=1}^{\infty }\frac{(-1)^{m-n}}{(m-n)(2n-1)}e^{-2a(2m-2n)}=\frac{\pi ^{3}}{8}+\sum_{n>=1}^{\infty }\frac{1}{(2n-1)}.\sum_{n,m>=1}^{\infty }\frac{(-1)^{m-n}}{(m-n)}e^{-2a|2m-2a}\\
\\
=\frac{\pi ^{3}}{8}+\sum_{n=1}^{\infty }\frac{1}{2n-1}.\sum_{k=n}^{\infty }e^{-4a|k|}\Rightarrow k=m-n\\
\\
=\frac{\pi ^{3}}{8}+\sum_{k=n}^{\infty }\sum_{n=1}^{\infty }\frac{1}{2n-1}\frac{(-1)^{k}}{k}e^{-4ak}=\frac{\pi ^{3}}{8}+\sum_{k=1}^{\infty }[\sum_{n=1}^{k}\frac{1}{2n-1}]\frac{(-1)^{k}}{k}e^{-4ak}$$
part(3)
$$third \ part\\
but\ 2\sum_{n=1}^{k}\frac{1}{2n-1}=\sum_{i=1}^{k}\frac{1}{2i-1}+\sum_{g=1}^{k}\frac{1}{2g-1}\\
\\
=\sum _{i+g=k+1}(\frac{1}{2i-1}+\frac{1}{2g-1})=\sum _{i+g=k+1}\frac{2(i+g-1)}{(2i-1)(2g-1))}\\
\therefore \sum_{n=1}^{k}\frac{1}{2n-1}=\sum _{i+g=k+1}\frac{k}{(2i-1)(2g-1))}\\
\therefore \sum _{k>=1}[\sum _{i+g=k+1}\frac{k}{(2g-1)(2i-1)}]e^{-2a(2m-2n)}=-\sum _{i,g>=1}\frac{(-1)^{i-g}}{(2i-1)(2g-1)}e^{-2a(2i+2g-2)}\\
\\
=-arctan^2(e^{-2a})\ \ \ \ \ \ so \ we \ have\  I=\int_{0}^{\infty }\frac{ln^2(cot(\frac{ax}{2}))-\frac{\pi }{4})}{x^2+1}dx\\
\\
=\frac{\pi ^{3}}{16}-\pi arctan^2(e^{-2a})\\
\\
\therefore I=\int_{0}^{\infty }\frac{ln^2(cot(\frac{ax}{2})-\frac{\pi }{4})}{x^2+4}dx =\frac{\pi ^{3}}{8}-2\pi arctan^2 (e^{-2a})\\
\\
\ \ \ \ \ \ \ \ \ \ ahmed\ hegazi$$


TITLE: Models of infinite groups and 'Group-like' objects
QUESTION [4 upvotes]: Let $G$ be an infinite group, and for simplicity, we will assume that $G$ is also countable. Now, with $G$ in mind, we construct a new language $L_G=\{f_{a_i\_},f_{\_a_i}:a_i\in G\}$ where $f_{a_i\_}(x)=a_ix$ and $f_{\_a_i}(x)=xa_i$ (Note that if $G$ is abelian, we can can restrict the language to $\{f_{\_a_i}:a_i \in G\}$) . Let $T=Th_{L_G}(G)$ (i.e. $T$ is the theory of G in the language $L_G$). It is clear that since $G$ is countable, $L$ is also countable. Therefore, $T$ is a countable, complete theory in a countable language. We say that $M$ is 'group-like' if $M\models T$ for some countably infinite group $G$.
Question 1: If $G$ is countably categorical in $L=\{1,\cdot\}$, then is $T$ countably categorical in $L_G$? (My suspicion is no).
Question 2: What do uncountable 'groups-like' models look like?
Question 3: Can anyone give me an example (countable or uncountable) of a model $M$ such that $M \models T$ but $M$ is "radically not a group" (I know that "radically not a group" is not well defined, but I think the intuition is clear).

REPLY [0 votes]: So, by Levon Haykazyan's first comment, question 1 is answered in the negative.
From this paper, on page 14, we can answer questions 2 and 3.
Example 1.38 states, "Let $G$ be an infinite group of cardinality $\kappa$ say. Let $L = \{\lambda_g : g \in G\}$ where the $\lambda_g$ are unary function symbols. Let $T$ be the theory of free $G$-sets. Then $T$ is categorical in every cardinality $> \kappa$, and complete".
I think this is the fullest answer this question can possibly receive.