text:
\section{Introduction}
\label{s:intro}

In \cite{pea-rankham}, it was shown that one-point blow up $(\widetilde M,\widetilde\omega_\rho)$ of a
closed 4-dimensional symplectic manifold $(M,\omega)$ has a loop $\{\widetilde \psi_t\}$ of Hamiltonian
diffeomorphisms such that the class $[\widetilde\psi_t]$  has infinite order in $\pi_1(\textup{Ham}
(\widetilde M,\widetilde\omega_\rho)).$  Actually the argument presented in \cite{pea-rankham}
works for any dimension as long as it is greater than two. The restriction to the 4-dimensional case had a
specific purpose;  it was  related to the  study of the abelian group
$\pi_1(\textup{Ham} (\widetilde M,\widetilde\omega_\rho)).$
Namely, for any positive integer $k$
that there exists closed symplectic 4-manifold such that the rank of
 $\pi_1(\textup{Ham} )$ is at least $k$.
This phenomena about  the  rank of $\pi_1(\textup{Ham} (\widetilde M,\widetilde\omega_\rho))$ being
positive was first discovered by D. McDuff
in \cite{mcduff-blowup}; the difference between the results of  \cite{pea-rankham}
and those of \cite{mcduff-blowup}
is that the element of infinite order in $\pi_1(\textup{Ham})$  is written explicitly
in \cite{pea-rankham}.
The goal of this article consists on
extending this result. Instead of blowing up a zero-dimensional
submanifold, we blow up a symplectic submanifold $N\subset (M,\omega)$ of codimension $2k>2$.
In this way we also obtain a loop of Hamiltonian diffeomorphisms on $(M,\omega)$, that induces an element
of infinite order in $\pi_1(\textup{Ham}(\widetilde M,\widetilde\omega_{N,\rho}))$
where $(\widetilde M,\widetilde\omega_{N,\rho})$ stands for the symplectic blow up
along the submanifold $N$.

\begin{thm}
\label{t:mainrank}
Let $(M,\omega)$ be a rational closed symplectic manifold and $N\subset (M,\omega)$
a closed symplectic submanifold of codimension $2k>2$.
Let $(\widetilde M,\widetilde\omega_{N,\rho})$ be the symplectic blow up
of $(M,\omega)$ along $N$. Then for some small values of $\rho$ the rank of
$\pi_1(\textup{Ham}(\widetilde M,\widetilde\omega_{N,\rho}))$
is positive.
\end{thm}

In $(\widetilde M,\widetilde\omega_{N,\rho})$, the symplectic blow
up of   $(M,\omega)$ along $N$,
the parameter $\rho$ in the description of $(\widetilde M,\widetilde\omega_{N,\rho})$
is related to the $\omega^n$-volume of a tubular neighborhood of
$N\subset(M,\omega)$. Therefore, in the case when $N$ is a single point,
it corresponds to the weight of the one-point blow up.

Although, Theorem  \ref{t:mainrank} is  a statement about the topology of the
group $\textup{Ham}(\widetilde M,\widetilde\omega_{N,\rho})$, that is as much
as we know about its topology.
Nonetheless, the study of the homotopy type of $\textup{Ham}(M,\omega)$
has succeeded in some cases, all of which are 4-dimensional.
For instance the homotopy type of
 $(\mathbb{C}P^2,\omega_{FS})$
and $(\mathbb{C}P^1\times\mathbb{C}P^1,\omega_{FS}\oplus \omega_{FS})$
was computed by M. Gromov \cite{gromov-psudo};
 in \cite{abreu-mcduff-topology},
M. Abreu and D. McDuff
settled the problem for the one-point blow up of  $(\mathbb{C}P^2,\omega_\textup{FS})$.
For other examples of  4-dimensional symplectic manifolds see  the works of
S. Anjos and S. Eden \cite{anjos-eden},
S. Anjos and M. Pinsonnault \cite{anjos-pin-homo}, J. Evans \cite{Evans-symplectic-mapping} and
F. Lalonde and M. Pinsonnault  \cite{Lalonde-Pin}.
As for computations of the fundamental group of $\textup{Ham}(M,\omega)$, there are the works of
J. Li and  T.-J. Li  \cite{lili-symp}  and J. Li, T.-J. Li and W. Wu \cite{liliwu}.

The techniques that are used to prove Theorem \ref{t:mainrank} are soft techniques.
Furthermore, we give an explicit description of the loop $\{\widetilde\psi_{N,t}\}$ of Hamiltonian
diffeomorphisms on $(\widetilde M,\widetilde\omega_{N,\rho})$ that induces the element of
infinite order in its fundamental group. In fact, the loop $\{\widetilde\psi_{N,t}\}$ is induced from
a Hamiltonian loop $\{ \psi_t\}$  on $(M,\omega)$ that is supported in a tubular
neighborhood of $N$. Finally, in order determine that  an element on the fundamental group
of $\textup{Ham}(M,\omega)$ has infinite order, we relay on
Weinstein's morphism,
$$
\mathcal{A}:\pi_1(\textup{Ham}(M,\omega) ) \to \mathbb{R}/\mathcal{P}(M,\omega).
$$
In this direction we have the following relation
between Hamiltonian diffeomorphism on $(M,\omega)$ and on
$(\widetilde M,\widetilde\omega_{N,\rho})$ for a particular
class of Hamiltonian loops. In the formula of the above result,
${\Phi(\mathcal{U}_{\rho}(\nu_0))}$ stands for
a particular tubular neighborhood of the symplectic submanifold $N\subset (M,\omega)$.

\begin{thm}
\label{t:weinsrelation}
Let $(M,\omega)$ be a closed symplectic manifold and $N\subset (M,\omega)$
a symplectic submanifold of codimension $2k>2$.
If $\{\psi_t\}$ is a loop of Hamiltonian diffeomorphisms on $(M,\omega)$
that is $N$-liftable to a loop $\{\widetilde\psi_{N,t}\}$
on $(\widetilde M_N,\widetilde \omega_{N,\rho})$, then
\begin{eqnarray*}
\mathcal{A}([\widetilde\psi_{N,t}]) =
\left[\mathcal{A}([\psi_t]) +
\frac{1}{\textup{Vol}(\widetilde M_N,(\widetilde\omega_{N,\rho})^n)}
\int_0^1
\int_{\Phi(\mathcal{U}_{\rho}(\nu_0))}  H_t\, \omega^n dt
 \right]
\end{eqnarray*}
in $\mathbb{R}/\mathcal{P}(\widetilde M_N,\widetilde \omega_{N,\rho})$, where
$H_t$ if the normalized Hamiltonian function of the loop $\{\psi_t\}$.
\end{thm}

Loosely speaking, the reason why the results of \cite{pea-rankham} extend
to the case of a submanifold of positive dimension is because many of the
arguments that are involved in the definition of symplectic blow up are in a
sense $U(k)$-equivariant.
Furthermore, the diffeomorphisms
$\{\psi_t\}$ and $\{\widetilde\psi_{N,t}\}$ that appear in the above result are induced
by matrices that lie in the center of $U(k)$. This condition turns to be crucial,
since in many instances the commutativity of the matrices is fundamental.

The paper is organized as follows.  In Section \ref{sec:blow} we review the
 definition of the symplectic one-point blow up as well as some facts
 of \cite{pedroza-hamiltonian} and \cite{pea-rankham}.
The definition of the symplectic blow up
 a long a submanifold is review in Section
\ref{s:blowN}. In Section \ref{s:hamlif} we give conditions that guarantee
that a Hamiltonian diffeomorphisms can be lifted to the symplectic blow up.
Finally, Sections \ref{s:wein} and \ref {s:main} deal with normalization
of Hamiltonian functions and the proofs of the main results respectively.

\section{The case of a one-point blow up}
\label{sec:blow}

\subsection{Symplectic structure on the one-point blow up.}
In this section we review how the  symplectic structure on the blow up of the
origin in $(\mathbb{C}^k,\omega_0)$ is defined.
Following  \cite[Sec. 2]{pedroza-hamiltonian},
we also review the conditions that a Hamiltonian diffeomorphism
on $(\mathbb{C}^k,\omega_0)$ needs to satisfy in order to induced
 a Hamiltonian diffeomorphism on the one-point blow up.
Later, we will impose similar conditions on a Hamiltonian
diffeomorphism in order to induced a Hamiltonian on  the symplectic manifold
that is obtain by blowing up a compact symplectic submanifold.

As a manifold, the one-point blow up of $\mathbb{C}^k$  at the origin
 is the total space of the tautological line bundle,
\begin{equation*}
	\widetilde{\mathbb{C}}^k:=\{(z,\ell)\in \mathbb{C}^k\times \mathbb{C}P^{k-1}  \mid z\in \ell\}.
\end{equation*}
The projection maps are denoted by $\textup{pr} \colon \widetilde{\mathbb{C}}^k\to \mathbb{C}P^{k-1}$ and
$\pi \colon \widetilde{\mathbb{C}}^k\to \mathbb{C}^k$.
Here, $\pi$ and $E:=\pi^{-1}(0)$ are called
 the blow up map  is  the exceptional divisor respectively.
Let $\omega_\textup{FS}$ be the
Fubini-Study symplectic form on $\mathbb{C}P^{k-1}$ normalized so that
 $\langle \omega_\textup{FS}, \mathbb{C}P^1 \rangle = \pi$.
For each $\rho>0$ define on $\widetilde{\mathbb{C}}^k$     the symplectic form
\begin{equation*}
	\omega_{\rho}:=\pi^*\omega_0+ \rho^2\textup{pr}^*\omega_\textup{FS}.
\end{equation*}
Notice that the $\omega_{\rho}$-area of  a line in the exceptional divisor is $\pi \rho^2$.
Denote by  $B_r\subset \mathbb{C}^k$  the open ball of radius $r$ centered at the origin
and  its preimage under $\pi$  by
$$
	L_r := \pi^{-1}(B_r).
$$

According to \cite[Prop. 7.1.13]{MS-book} for each $\epsilon\in (0,1)$ there exists
a symplectic form $\widetilde{\omega}_{\rho,\epsilon}$ on $\widetilde{\mathbb{C}}^k$ such that
\begin{itemize}
\item $\widetilde{\omega}_{\rho,\epsilon}=\omega_\rho$ on $L_\epsilon$ and
\item $\widetilde{\omega}_{\rho,\epsilon}=\pi^{*}(\omega_0)$ on $\widetilde{\mathbb{C}}^k \setminus
\overline{L}_{\rho+\epsilon}.$
\end{itemize}
Later in this note it will be important to know specific details on how  the symplectic form
$\widetilde\omega_{\rho,\epsilon}$ is defined. To that end, one considers
a diffeomorphism
$F_{\rho,\epsilon}: \mathbb{C}^k\setminus \{0\}\to \mathbb{C}^k\setminus \overline{B}_\rho$
that consists of stretching the punctured space,
defined as
\begin{eqnarray}
\label{e:strech}
F_{\rho,\epsilon}(z):=f_{\rho,\epsilon}(|z|)\frac{z}{|z|}
\end{eqnarray}
where  $f_{\rho,\epsilon}:\mathbb{R}\to \mathbb{R}$ is a smooth function
 such that
$f_{\rho,\epsilon}(x)=\sqrt{\lambda^2+x^2}$ on $(0,\epsilon)$ and
is equal to the identity on
$[\rho+\epsilon,\infty)$.
Then the above symplectic form
$\widetilde\omega_{\rho,\epsilon}$ is defined as
$\widetilde\omega_{\rho,\epsilon}:=(\pi^*\circ F_{\rho,\epsilon}^*)(\omega_0)$
on  $\widetilde{\mathbb{C}}^k \setminus E$.

\begin{comment}
Given $r_\rho>\rho$, there exists a diffeomorphism
 $F_\rho: L_{r_\rho}\setminus E\to B_{r_\rho}\setminus
\overline{B}_\rho$ defined as
$$
F_\rho(z):= \beta (|z|) \frac{z}{|z|}
$$
where $\beta $ is a real-valued function and $F_\rho$
is such that    $F_\rho^*(\omega_0)=\omega_\rho$
near the exceptional divisor
  and  $F_\rho^*(\omega_0)=\omega_0$
  on
$L_{r_\rho}\setminus L_{(r_\rho+\rho)/2}$.
Notice that $F_\rho$ is $U(k)$-equivariant; this mean that
$F_\rho(U(z))=U(F_\rho(z))$. This observation will become important
in the definition of the blow up along a submanifold.
Using $F_\rho$ as a gluing map, the symplectic  blow up
of the origin in $(\mathbb{C}^k,\omega_0)$
of weight $\rho$ is defined as
\begin{equation}
\label{e:symplecticformblow}
	(\widetilde{\mathbb{C}}^k, \widetilde \omega_{\rho}) :=
	(\mathbb{C}^k \setminus \overline{B_\rho}, \omega_0) \cup_{F_\rho} (L_{r_\rho}, \omega_{\rho}).
\end{equation}
The blow up map $\pi \colon (\widetilde{\mathbb{C}}^k, \widetilde{\omega}_\rho)\to
( \mathbb{C}^k,\omega_0)$ is defined as
\begin{eqnarray}
 \pi (z):=
\left\{
	\begin{array}{ll}
		z   & \textup{if } z\in \mathbb{C}^k \setminus \overline{B_\rho}\\
		F_\rho(z) &  \textup{if } z\in L_{r_\rho}\setminus E\\
		0 &  \textup{if } z\in E.
		\end{array}
\right.
\end{eqnarray}
and is a symplectic diffeomorphism on the complement of $L_{r_\rho}$.
(Note that there is an omission of the parameter $\rho$ on the notation of the blow up map.)
It is important to note that in the definition of the blow up a closed ball $\overline{B_\rho}$ of
$\mathbb{C}^k$ was removed; not only a
\end{comment}

In this case $(\widetilde{\mathbb{C}}^k, \widetilde{\omega}_{\rho,\epsilon})$ is defined
as  the symplectic blow up of the origin in $( \mathbb{C}^k,\omega_0)$ of weight $\rho$.
It is important to note that the blow up map $\pi \colon (\widetilde{\mathbb{C}}^k,
\widetilde{\omega}_{\rho,\epsilon})\to ( \mathbb{C}^k,\omega_0)$
is a symplectic diffeomorphism on the complement of
$\overline{L}_{\rho+\epsilon} \subset  \widetilde{\mathbb{C}}^k.$

The complete details of symplectic one-point blow up, including the exact
definitions of $\widetilde \omega_{\rho,\epsilon}$ and $f_{\rho,\epsilon}$
appear in \cite[Ch. 7]{MS-book}.

\subsection{Induced Hamiltonian on the one-point blow up.}
\label{ss:inducedHam}
Next we review some results of \cite{pedroza-hamiltonian} that deal with the problem of lifting a
Hamiltonian diffeomorphism on $( \mathbb{C}^k,\omega_0)$ to a Hamiltonian on
$(\widetilde{\mathbb{C}}^k, \widetilde{\omega}_{\rho,\epsilon})$.

Let $\phi$ and $\widetilde \phi$   be  Hamiltonian diffeomorphisms on $(\mathbb{C}^k,\omega_0)$ and
$(\widetilde{\mathbb{C}}^k, \widetilde{\omega}_{\rho,\epsilon})$ respectively.
We say that the diffeomorphism $\widetilde \phi$
{\em lifts} the diffeomorphism $\phi$ if the following diagram commutes
$$
\xymatrix{(\widetilde{\mathbb{C}}^k, \widetilde{\omega}_{\rho,\epsilon})\ar[d]_\pi\ar[r]^{\widetilde{\phi}} &
 (\widetilde{\mathbb{C}}^k, \widetilde{\omega}_{\rho,\epsilon})\ar[d]^\pi \\
({\mathbb{C}}^k, {\omega}_0)\ar[r]^\phi & ({\mathbb{C}}^k, {\omega}_0).
}$$
This means that $	\pi\circ\widetilde{\phi}=\phi\circ\pi$.

Consider $\{\psi_t\}_{0\leq t\leq 1}$ a path of Hamiltonian diffeomorphisms on $(\mathbb{C}^k,\omega_0)$
with Hamiltonian function $H_t$ such that $\psi_0=1_{\mathbb{C}^k}$.
The goal is to define a path of Hamiltonian diffeomorphisms
$\{\widetilde{\psi}_t\}$
on 	$(\widetilde{\mathbb{C}}^k, \widetilde \omega_{\rho,\epsilon})$ that lifts
the path $\{\psi_t\}$; that is
 $	\pi\circ\widetilde{\psi}_t=\psi_t\circ\pi$
  for each
$t\in[0,1]$.
At this point it is important to note that
the Hamiltonian path induced by the function
$H_t\circ \pi:(\widetilde{\mathbb{C}}^k, \widetilde{\omega}_{\rho,\epsilon})\to\mathbb{R}$ fails to
lift the path $\{\psi_t\}$.
Away from  the exceptional divisor is true that the path induced by
$H_t\circ \pi$ is a lift of the original path; the problem occurs  near the
exceptional divisor. Note for instance that $H_t\circ \pi$ is independent of the parameter $\rho$
and such parameter is relevant in the symplectic structure on the one-point blow up.
As we will see, the Hamiltonian function that induces the desired Hamiltonian path
will depend  on the parameter $\rho$.

The required condition for a path of Hamiltonian diffeomorphisms to lift to the one-point blow up is
that for each $t\in[0,1]$
there exists a matrix $U_t\in U(k)$  such that $\psi_t=U_t$ on
${B}_{\rho+\epsilon}\subset\mathbb{C}^k$.

\begin{lem}
\label{l:hamfunc}
Let $\psi_t :(B_{r},\omega_0)\to (B_{r},\omega_0)$
be a Hamiltonian path given by unitary matrices
starting at the identity matrix with Hamiltonian function $H_t$. Then
$$
H_t(z)=H_t(\lambda z)
$$
for $z\in B_r$ and $\lambda \in S^1$.
\end{lem}
\begin{pf}
Denote by $X_t$ the time-dependent vector field of
the path $\{\psi_t\}$. For $\lambda\in S^1$, let $\phi_\lambda:B_r\to B_r$ be matrix multiplication
by $\lambda I$. Since $\phi_\lambda$ is in the center of $U(k)$ it follows that
\begin{eqnarray*}
X_t\circ \phi_\lambda =\left.\frac{d}{ds}\right|_{s=t} \psi_s\circ \psi_t ^{-1}\circ  \phi_\lambda
=\left.\frac{d}{ds}\right|_{s=t} \phi_\lambda \circ \psi_s\circ \psi_t ^{-1} = (\phi_\lambda)_* X_t.
\end{eqnarray*}

Therefore
$$
d (H_t\circ \phi_\lambda)=\omega_0(X_t, (\phi_\lambda)_*( \cdot)) =
(\phi_\lambda) ^*\omega_0 (X_t,\cdot)=d H_t.
$$
Since  both functions   $H_t$ and $H_t\circ \phi_\lambda$ agree at the origin, it follows that
 $H_t(z)=H_t(\lambda z)$.
\end{pf}

Let $\{\psi_t\}$ be a path of Hamiltonian diffeomorphisms on $(\mathbb{C}^k,\omega_0)$
such that for each $t$
the restriction of $\psi_t$  to ${B}_{\rho+\epsilon}\subset\mathbb{C}^k$
is given by a unitary matrix. If $H_t: (\mathbb{C}^k,\omega_0)\to \mathbb{R}$ is the Hamiltonian
function of the path, define the function
$\widetilde H_t:(\widetilde{\mathbb{C}}^k, \widetilde \omega_{\rho,\epsilon})\to \mathbb{R}$
by
\begin{eqnarray}
\label{e:hamblow}
 \widetilde H_t(z,\ell) :=
\left\{
	\begin{array}{ll}
			H_t\circ F_{\rho,\epsilon}\circ \pi (z,\ell)
			& \textup{if } (z,\ell)\in \widetilde{\mathbb{C}}^k\setminus E \\
			H_t\left(  \frac{\rho}{|w|} w\right) & \textup{if $z=0$ and $[w]=\ell  \in \mathbb{C}P^{k-1}$}.\\
	\end{array}
\right.
\end{eqnarray}
It follows from Lemma \ref{l:hamfunc} and the definition of $F_{\rho,\epsilon}$ that $ \widetilde H_t$
is well defined. Note that $ \widetilde H_t$ depends on the diffeomorphism $F_{\rho,\epsilon}$ and hence
on the parameter $\rho$.
The path of Hamiltonian diffeomorphisms on $(\widetilde{\mathbb{C}}^k, \widetilde \omega_{\rho,\epsilon})$
induced by $\widetilde H_t$
is the path that lifts the initial path on $({\mathbb{C}}^k, {\omega}_0)$.
The proof of this fact  as well as the proof of the next proposition appears in
\cite[Sec. 3]{pedroza-hamiltonian}.

\begin{prop}
The time-dependent vector field $\widetilde X_t$ and path of diffeomorphisms
$\{\widetilde \psi_t\}$ induced by the function
$\widetilde H_t:(\widetilde{\mathbb{C}}^k, \widetilde \omega_{\rho,\epsilon})\to \mathbb{R}$,  defined in
(\ref{e:hamblow}), are such that
\begin{enumerate}
\item $\pi_*(\widetilde X_t)=X_t$ for each $t\in[0,1]$ and
\item $\widetilde \psi_t$ is a lift of $\psi_t$ for each $t\in[0,1]$.
\end{enumerate}
In particular, $\{\widetilde \psi_t\}$ is a Hamiltonian path of
$(\widetilde{\mathbb{C}}^k, \widetilde \omega_{\rho,\epsilon})$ that lifts the Hamiltonian path
$\{\psi_t\}$.
\end{prop}

Thus if $\psi$ is a Hamiltonian diffeomorphism on $(\mathbb{C}^k,\omega_0)$
such that it can be joined to the identity by a path of Hamiltonian diffeomorphisms such that
each one of them is equal to a unitary matrix on the ball ${B}_{\rho+\epsilon}$,
then $\psi$ admits a
Hamiltonian lift to the one-point blow up $(\widetilde{\mathbb{C}}^k, \widetilde \omega_{\rho,\epsilon})$.

This definition can be extended to
the one-point blow up of weight $\rho$ of
 an arbitrary symplectic manifold $(M,\omega).$
For the purpose of this note it is not relevant to review such definition.
For further details  see \cite{pedroza-hamiltonian}.

\section{Review of the blow up along a submanifold}
\label{s:blowN}

Following \cite[Ch. 7]{MS-book},
we review the construction of the symplectic blow up along a compact symplectic submanifold.
The complete details of this construction are relevant for the Hamiltonian
diffeomorphism that we will defined on the blown up manifold.

\subsection{The blow up along a compact submanifold.}
Let $N\subset (M,\omega)$ be a compact symplectic submanifold of codimension $2k>2$.
Then
$\pi_N:TN^\omega\to N$ corresponds to its normal bundle which is
a symplectic vector bundle. Since
the symplectic linear group contracts to the  unitary group,
$\pi_N$ is in fact a complex vector bundle.
For simplicity, write $\nu$ for  $TN^\omega$.
Thus there exists a  Hamiltonian linear action of $U(k)$ on $(\mathbb{R}^{2k},\omega_0)
=(\mathbb{C}^{k},\omega_0)$
and  a principal
$U(k)$-bundle $P\to N$  such that the associated fibration coincides with the normal bundle,
$\nu\simeq P\times_{U(k)}\mathbb{C}^{k}$.
Let $\mu:(\mathbb{C}^{k},\omega_0)\to \mathfrak{u}(k)^*$ be the
corresponding moment map
and  fix  a connection 1-form $A\in\Omega^1(P, \mathfrak{u}(k))$ on  $P\to N$.
On
$P\times\mathbb{C}^k$ define the 1-form
$$
\langle \mu,A \rangle_{(p,z)} (u+v):=
\langle \mu(z),A_p(u) \rangle
$$
where $ u\in T_pP $ and $ v\in  T_z \mathbb{C}^k$.
Here $\langle\cdot ,\cdot\rangle$
stands for the pairing between  $\mathfrak{u}(k)^*$ and $\mathfrak{u}(k)$.
Notice that the 1-form $\langle \mu,A \rangle$ is independent of the vectors that lie in $T\mathbb{C}^k$.
For $\xi\in \mathfrak{u}(k)$, let $X_\xi$ be the vector
on $\mathbb{C}^k$ induced by the action of $U(k)$.
 Then  on  $P\times\mathbb{C}^{k}$ define  the 2-form
\begin{eqnarray*}
\omega_{A}(u_1+v_1,u_2+v_2 )&:=&\omega_0(v_1+X_{A(u_1)},v_2+X_{A(u_2)})-\\
& & d\langle \mu, A\rangle(u_1+v_1,u_2+v_2 ).
\end{eqnarray*}
So defined, the 2-form $\omega_A$ is closed. Moreover, according to  \cite[Thm. 6.6.3]{MS-book}
$\omega_A$   is $U(k)$-invariant and horizontal.
Henceforth, it induces a  closed  2-form on $\nu\simeq P\times_{U(k)}\mathbb{C}^{k}$.
In order to avoid cumbersome notation, we will make no distinction between
a basic form and the induced form on the orbit space.
Note that on each fibre of $\pi_N:\nu\to N$ the form
$\omega_A$ restricts  to $\omega_0$.
Finally, define the 2-form
\begin{equation}
\label{e:formonN}
\omega_{\nu,A}:=\omega_A  + \pi_N^*(\omega)
\end{equation}
on $\nu.$ Notice that $\omega_{\nu,A}$  also restricts to $\omega_0$ on each fibre.
Furthermore, if  $\nu_0$ stands for the zero-section of $\nu$ then  $\omega_{\nu,A}$
restricted to $\nu_0$ is equal to $\pi_N^*(\omega)$.
These observations imply the following lemma. First, for $\epsilon >0$  denote by
$\mathcal{U}_{\epsilon}(\nu_0):=
P\times_{U(k)} B_{\epsilon}$ the $\epsilon$-disk subbundle of
$\nu\to N$.

\begin{lem}
\label{l:sympformnormal}
Let $N\subset (M,\omega)$ be a compact symplectic submanifold of codimension $2k$.
Then there exist
$\epsilon_0>0$ such that the 2-form
$$
\omega_{\nu,A}=\omega_A  + \pi_N^*(\omega)
$$
is symplectic on the
disk subbundle $\mathcal{U}_{\epsilon_0}(\nu_0)$.  Moreover $\omega_{\nu,A}$
restricts to $\omega_0$ on each fibre and to
$\pi_N^*(\omega)$ on the zero-section.
\end{lem}

\begin{comment}

Hence, the 2-form
\begin{eqnarray}
\omega_{\nu,A}:=\omega_A + \pi_N^*(\omega)
\end{eqnarray}
that restricts to $\omega_0$  on each fibre
and to $\pi_N^*(\omega)$ on the zero section of the vector bundle
$\pi_N:\nu\to N$.
\end{comment}

It follows by the symplectic neighborhood theorem there exists a symplectic diffeomorphism
\begin{eqnarray}
\label{e:symplecticdiffeo}
\Phi: (\mathcal{U}_{\epsilon_0}(\nu_0),\omega_{\nu,A}) \to (\mathcal{U}(N), \omega)
\end{eqnarray}
where $\mathcal{U}(N)\subset (M,\omega)$ is some neighborhood
of $N$.

The outline of the construction of the symplectic blow up of $(M,\omega)$ along the symplectic
submanifold
$N$ is as follows:
The  neighborhood  $ (\mathcal{U}(N), \omega)$ of $N\subset (M,\omega)$ is identified with
 the open set $(\mathcal{U}_{\epsilon_0}(\nu_0),\omega_{\nu,A})$
 via the symplectic diffeomorphism $\Phi$ and thus the zero-section $\nu_0$ in $\nu$
is identified with the submanifold $N$ is blown up. Hence we give a description
of the blow up of the zero-section in $\nu$.
By the blow up of $\nu_0$  we mean the total space of the associated bundle
that is  obtained by $\widetilde{\mathbb{C}}^{k}$ and $P\to N$ which is  denoted by
$\widetilde\nu$.
Finally, $\widetilde\nu$ carries a symplectic form $\widetilde\omega_{\widetilde\nu,A,\rho}$
and such model
is glued back to $ (M\setminus N,\omega)$.

The complex-linear action of $U(k)$ on $\mathbb{C}^k$ has a canonical lift
 to $\widetilde{\mathbb{C}}^{k}$
defined as
$$
U.(z,\ell)\mapsto (U(z),U(\ell)) \hspace{1.5cm} \textup{ for }  U\in U(k) \textup{ and }(z,\ell)\in
\widetilde{\mathbb{C}}^{k}.
$$
This $U(k)$-space together with the $U(k)$-principal bundle $P\to N$
induce the associated fibration
$\widetilde\pi_N:\widetilde\nu\to N$ where $\widetilde \nu:=P\times_{U(k)}\widetilde{\mathbb{C}}^{k}$.
Since the blow up map $\pi:\widetilde{\mathbb{C}}^{k}\to\mathbb{C}^{k}$ is $U(k)$-equivariant,
 it induces the map
$$
\widetilde\pi: P\times_{U(k)}\widetilde{\mathbb{C}}^{k}\to P\times_{U(k)}{\mathbb{C}}^{k}
$$
defined as $\widetilde\pi[p,(z,\ell)]:=[p,\pi(z,\ell)]=[p,z].$ Notice that $\widetilde\pi:\widetilde\nu\to \nu$
corresponds to the blow up of the zero-section $\nu_0$ in $\nu$.
As in the case of the one-point blow up, $\widetilde\pi$ is a diffeomorphism on the complement of
$\widetilde{\pi}^{-1}(\nu_0)\subset \widetilde\nu$.

For any $\epsilon>0$, the ball $B_\epsilon\subset\mathbb{C}^k$ is  $U(k)$-invariant
and hence  $L_\epsilon\subset\widetilde{\mathbb{C}}^{k}$ is also $U(k)$-invariant.
Write ${\mathcal{U}}_\epsilon(\widetilde{\pi}^{-1}(\nu_0))$ for the subbundle
$P\times_{U(k)}L_\epsilon$. Notice that
 $\widetilde{\pi}$ maps
${\mathcal{U}}_\epsilon(\widetilde{\pi}^{-1}(\nu_0))$
onto $\mathcal{U}_{\epsilon}(\nu_0)$. Furthermore, $\widetilde \pi$
maps  ${\mathcal{U}}_\epsilon(\widetilde{\pi}^{-1}(\nu_0))\setminus \widetilde{\pi}^{-1}(\nu_0)$
diffeomorphically onto   $\mathcal{U}_{\epsilon}(\nu_0)\setminus \nu_0$. Summarizing, we have the
following commutative diagram of maps
$$
\xymatrix{{\mathcal{U}}_\epsilon(\widetilde{\pi}^{-1}(\nu_0)) \ar[r]^{\widetilde \pi}\ar@{->}
[dr]_{\widetilde \pi_N} &
\mathcal{U}_\epsilon(\nu_0) \ar[d]^{\pi_N}  \ar@{^(->}[r] \ar@{->}
[dr]^{\Phi}  & \nu \\
&N \ar@{^(->}[r] & \mathcal{U}(N)\ar@{^(->}[r]  & M.
}$$

We are now ready to define the manifold that results by blowing up  $(M,\omega)$ along the submanifold $N$.
If $\epsilon_0>0$ is as before,
the blow up of $M$ along the compact symplectic submanifold $N$ is defined as
\begin{eqnarray}
\label{e:mblow}
\widetilde M_N:=M\setminus N \ \
\cup_{\Phi\circ \widetilde\pi } \ \ {\mathcal{U}}_{\epsilon_0}(\widetilde{\pi}^{-1}(\nu_0))
\end{eqnarray}
where
${\mathcal{U}}_{\epsilon_0}(\widetilde{\pi}^{-1}(\nu_0))\setminus
\widetilde{\pi}^{-1}(\nu_0)$ is identified with $\mathcal{U}(N)\setminus N$
 via the diffeomorphism $\Phi\circ \widetilde\pi$.

As in the case of the one-point blow up, there is a projection map
$\widetilde\pi_{(M,N)}: \widetilde M_N\to M$, also called the blow up map,
defined as
\begin{eqnarray}
\label{e:hamblowmap}
 \widetilde\pi_{(M,N)}(p) :=
\left\{
	\begin{array}{ll}
	p & \textup{if } p \in M\setminus N \\
	\Phi\circ \widetilde\pi(p) & \textup{if }
	p=[p^\prime, (z,\ell)]\in {\mathcal{U}}_{\epsilon_0}(\widetilde{\pi}^{-1}(\nu_0)).
	\end{array}
\right.
\end{eqnarray}
Note that  $\widetilde\pi_{(M,N)}$ depends on the disk subbundle
 ${\mathcal{U}}_{\epsilon_0}(\nu_0)$. We omit such dependency from the notation of the map.
The preimage of the blown up submanifold $E_N:={\widetilde\pi}_{(M,N)}^{-1}(N)$ is called the exceptional
divisor and ${\widetilde\pi}_{(M,N)}$ restricted to the exceptional divisor,
$E_N\to N$, is a fibration with fiber $\mathbb{C}P^{k-1}$ that corresponds
to the projectivization of the normal bundle $\nu\to M$.

It remains to define a symplectic structure on $\widetilde M_N$.
In the definition of $\widetilde M_N$,
the parameter $\epsilon_0>0$ of the disk subbundle was
irrelevant.
Its importance will become apparent
in the description of the symplectic structure on it.

\subsection{The symplectic structure on $\widetilde M_N$.}
\label{sec:sympblow}
Let $\widetilde \mu_\rho: (\widetilde{\mathbb{C}}^{k},\widetilde\omega_{\rho,\epsilon})
\to\mathfrak{u}(k)^*$ be the  moment map of the induced action of $U(k)$
on $(\widetilde{\mathbb{C}}^{k},\widetilde\omega_\rho)$ described above.
Using the same connection 1-form $A$
on the principal bundle $P\to N$, we have the analog
of the form $\omega_A$ defined before, but now on
$P\times\widetilde{\mathbb{C}}^{k}$.
That is, on $P\times\widetilde{\mathbb{C}}^{k}$ we have the
following 2-form
\begin{eqnarray*}
\widetilde\omega_{A,\rho}(u_1+v_1,u_2+v_2 )&:=&
\widetilde\omega_{\rho,\epsilon}(v_1+X_{A(u_1)},v_2+X_{A(u_2)})-  \\
& & d\langle \widetilde\mu_\rho, A\rangle(u_1+v_1,u_2+v_2 )
\end{eqnarray*}
that is closed and  descends to the orbit space
$\widetilde\nu:=P\times_{U(k)} \widetilde{\mathbb{C}}^{k}$. Moreover it
 restricts to  $\widetilde\omega_{\rho,\epsilon}$ on each fibre of $\widetilde\pi_N:\widetilde\nu\to N$.
As before, on  $\widetilde\nu$ we define the 2-form
\begin{equation}
\label{e:symplecformblow}
\widetilde\omega_{\widetilde\nu,A,\rho}:=\widetilde\omega_{A,\rho} +\widetilde\pi_N^*(\omega).
\end{equation}

Recall that $\pi: (\widetilde{\mathbb{C}}^{k},\widetilde\omega_{\rho,\epsilon})\to (\mathbb{C}^{k},\omega_0)$
is  a symplectic diffeomorphism on the complement of $\overline{L}_{\rho+\epsilon}$.

\begin{lem}
\label{l:symprel}
Let $\rho$ and $\epsilon\in(0,1)$ such that
$\rho+\epsilon<\epsilon_0$. Then under the map
$\widetilde \pi: \widetilde\nu\to \nu$ we have that
$$
\widetilde\pi^*(\omega_{\nu,A})= \widetilde\omega_{\widetilde\nu,A,\rho}
$$
on
$P\times_{U(k)}(\widetilde{\mathbb{C}}^k\setminus \overline{L}_{\rho+\epsilon})$.
\end{lem}
\begin{proof}
In order to prove the claim, consider the forms on $P\times {\mathbb{C}}^k$
instead of $P\times_{U(k)} {\mathbb{C}}^k$. Similarly for  the forms on
$P\times_{U(k)} \widetilde{\mathbb{C}}^k$. Hence we must show that
$$
(1\times\pi)^*(\omega_{\nu,A})= \widetilde\omega_{\widetilde\nu,A,\rho}.
$$

First note that since $\pi_N$ and $\widetilde\pi_N$ are fibrations over $N$ and $\widetilde\pi$
is a bundle map, it follows that $\widetilde\pi_N^*(\omega)=   \widetilde\pi^*
\circ\pi_N^*(\omega)$ on all $\widetilde\nu$.
$$
\xymatrix{
P\times \widetilde{\mathbb{C}}^k \ar[r]\ar[d]_{1\times \pi} & \widetilde\nu
\ar[dr]^{\widetilde\pi_N} \ar[d]_{\widetilde\pi}& \\
P\times {\mathbb{C}}^k \ar[r]                         &   \nu \ar[r]_{\pi_N} & N
}$$
Therefore
$(1\times \pi)^*(   \pi_N^*(\omega)  )=   \widetilde\pi_N^*(\omega)$.
It remains to show that
$
(1\times\pi)^*(\omega_{A})= \widetilde\omega_{A,\rho}.
$ In this step the parameter $\rho$ will be relevant. Since
 $\pi: (\widetilde{\mathbb{C}}^{k},\widetilde\omega_{\rho,\epsilon})\to (\mathbb{C}^{k},\omega_0)$
is  a symplectic diffeomorphism on the complement of $\overline{L}_{\rho+\epsilon}$ and is $U(k)$-equivariant,
the moment maps satisfy
$\widetilde\mu_\rho=\mu\circ\pi$    on
$P\times(\widetilde{\mathbb{C}}^k\setminus \overline{L}_{\rho+\epsilon})$.
This in turn imply  that
$(1\times\pi)^*\langle \mu,A  \rangle =\langle \widetilde\mu_\rho,A  \rangle$
also on
$P\times(\widetilde{\mathbb{C}}^k\setminus \overline{L}_{\rho+\epsilon})$.

Once again, using the fact that the blow up map is a symplectic diffeomorphism
on the complement of $\overline{L}_{\rho+\epsilon}$ and the fact that $\pi_*(\widetilde X_{\xi})
=X_{\xi}$ it follows that
 \begin{eqnarray*}
\omega_0(\pi_*(v_1)+X_{A(u_1)},  \pi_*(v_2)+X_{A(u_2)}  )
&=& \omega_0(\pi_*(v_1)+\pi_*(\widetilde{X}_{A(u_1)}),\\
& & \hspace{.6cm}\pi_*(v_2)+\pi_*(\widetilde{X}_{A(u_2)})  )\\
&=& \pi^*(\omega_0)(v_1+\widetilde{X}_{A(u_1)}, v_2+\widetilde{X}_{A(u_2)}  )\\
&=& \omega_{\rho,\epsilon}(v_1+\widetilde{X}_{A(u_1)},  v_2+\widetilde{X}_{A(u_2)}  )
\end{eqnarray*}
where $u_j\in TP$ and $v_j\in T\widetilde{\mathbb{C}}^k$.

\begin{comment}
Recall from Sec. \ref{sec:blow} that there is a diffeomorphism $F_{\rho,\epsilon}: \mathbb{C}^k\setminus \{0\}
\to \mathbb{C}^k\setminus \overline{B}_\rho$
such that $F_{\rho,\epsilon}(z)=f_{\rho,\epsilon}(|z|)z/|z|$ and
$\pi^*\circ F_{\rho,\epsilon}^*(\omega_0)=\widetilde\omega_{\rho,\epsilon}$.
So defined,
$ F_{\rho,\epsilon}\circ\pi:\widetilde{\mathbb{C}}^k\setminus E\to \mathbb{C}^k\setminus \overline{B}_\rho
$ is $U(k)$-equivariant. Moreover, the induced map
$$
\widetilde F_{\rho,\epsilon}\circ\widetilde\pi :\widetilde\nu\setminus \pi^{-1}(\nu_0)  \to\nu
$$
is such that
$\widetilde \pi^*\circ \widetilde F_{\rho,\epsilon}^*(\omega_0)=\widetilde\omega_{\rho,\epsilon}$.
\end{comment}

All these facts  show that
$(1\times \pi)^*(\omega_{\nu,A})= \widetilde\omega_{\widetilde\nu,A,\rho}$ on
$P\times(\widetilde{\mathbb{C}}^k\setminus \overline{L}_{\rho+\epsilon})$. Henceforth,
$
\widetilde\pi^*(\omega_{\nu,A})= \widetilde\omega_{\widetilde\nu,A,\rho}
$
on
$P\times_{U(k)}(\widetilde{\mathbb{C}}^k\setminus \overline{L}_{\rho+\epsilon})$.
\end{proof}

For the rest of the article we fix the parameters $\rho>0$ and $\epsilon\in(0,1)$ such that
$\rho+\epsilon<\epsilon_0$.
Remember that
$
\Phi: (\mathcal{U}_{\epsilon_0}(\nu_0),\omega_{\nu,A}) \to (\mathcal{U}(N), \omega)
$ is a symplectic diffeomorphism
and by Lemma \ref{l:symprel}
that
$
\widetilde\pi^*(\omega_{\nu,A})= \widetilde\omega_{\widetilde\nu,A,\rho}
$
on $P\times_{U(k)}(\widetilde{\mathbb{C}}^k\setminus \overline{L}_{\rho+\epsilon})$.
Therefore on $P\times_{U(k)}(\widetilde{\mathbb{C}}^k\setminus \overline{L}_{\rho+\epsilon})$
we have that $  \widetilde\pi^* \circ \Phi^* (\omega) = \widetilde\omega_{\widetilde\nu,A,\rho}$.
Hence it follows that $\widetilde M_N$ carries a symplectic form
$\widetilde \omega_{N,\rho}$ defined as
\begin{eqnarray}
\label{e:symplcformblow}
\widetilde \omega_{N,\rho}:=
\left\{
	\begin{array}{ll}
	\omega & \textup{on }M\setminus\Phi(\overline{ \mathcal{U}_{\epsilon^\prime}(\nu_0)  })  \\
	\widetilde\omega_{\widetilde\nu,A,\rho}& \textup{on }\mathcal{U}_{\epsilon_0}(\widetilde \pi^{-1}(\nu_0))
	\end{array}
\right.
\end{eqnarray}
where $\epsilon^\prime$  is such that $\rho+\epsilon <\epsilon^\prime<\epsilon_0$.
Note that $\widetilde\pi_{(M,N)}$ is a symplectic diffeomorphism on
$M\setminus\Phi(\overline{ \mathcal{U}_{\epsilon^\prime}(\nu_0)  })$.

\section{Induced Hamiltonians on the blown up manifold}
\label{s:hamlif}

In this section we give conditions  on a Hamiltonian diffeomorphism on
 $(M,\omega)$
in order to ensure that it lifts to  a Hamiltonian diffeomorphism
on the blown up manifold $(\widetilde M_N, \widetilde\omega_{N,\rho})$.
As stated in the previous section, a diffeomorphism $\widetilde \phi$
on $\widetilde M_N$ is said to be a lift
of the diffeomorphism $\phi$ on $M$ if  the following
diagram commutes
$$
\xymatrix{
\widetilde{M}_N\ar[d]_{\widetilde\pi_{(M,N)}}\ar[r]^{\widetilde\phi} &
\widetilde{M}_N\ar[d]^{\widetilde{\pi}_{(M,N)}}\\
M\ar[r]^\phi & M.
}$$
By the nature of the blow up construction, these conditions will only take place in a
neighborhood of the symplectic submanifold $N\subset M$.
To that end,
 fix a tubular neighborhood $\mathcal{U}(N)$ of $N$  and a symplectic
diffeomorphism
$
\Phi: (\mathcal{U}_{\epsilon_0}(\nu_0),\omega_{\nu,A}) \to (\mathcal{U}(N), \omega)
$ as before.

\begin{defi}
Let $N\subset (M,\omega)$ be a compact symplectic submanifold
of codimension $2k$.
A symplectic diffeomorphism $\psi: (M,\omega)\to (M,\omega)$
is called {\em $N$-liftable} if
\begin{itemize}
\item[a)] $\psi(\mathcal{U}(N))\subset \mathcal{U}(N)$  and
\item[b)] the map  $\Phi^{-1}\circ \psi\circ\Phi:
(\mathcal{U}_{\epsilon_0}(\nu_0),\omega_{\nu,A}) \to
 (\mathcal{U}_{\epsilon_0}(\nu_0),\omega_{\nu,A})$
 takes the form
$$
 [p,z]\mapsto [p,U(z)]
 $$
where
$U$ lies in the center of $ U(k)$.
\end{itemize}
\end{defi}

\begin{rema}
\begin{itemize}
\item
The above definition depends on the tubular neighborhood $\mathcal{U}(N)$
and  the diffeomorphism $\Phi$.
In order to avoid cumbersome notation we omit
such dependencies from the definition.

\item The reason that the matrix $U$ must lie in the center
of $U(k)$ is to ensure that
 $\widetilde\mu_\rho(U(z))=\widetilde\mu_\rho(z)$,
 where $\widetilde \mu_\rho: (\widetilde{\mathbb{C}}^{k},\widetilde\omega_{\rho,\epsilon})
\to\mathfrak{u}(k)^*$ is the moment map of the induced action
used in the definition of the symplectic form on
described in Sec. \ref{sec:sympblow}.
It will also be useful when proving that the induced diffeomorphisms on
the blown up manifold by a Hamiltonian is also Hamiltonian.

\item If  $\phi$ is a gauge transformation of $P\to N$, then the map
$
 [p,z]\mapsto [\phi(p),U(z)]
$ is a diffeomorphism of $\mathcal{U}_{\epsilon_0}(\nu_0)$. However,
if $\phi\neq 1_P$, such diffeomorphism
on  $(\mathcal{U}_{\epsilon_0}(\nu_0),\omega_{\nu,A})$
will not be symplectic.
\end{itemize}
\end{rema}

\medskip

If $\psi: (M,\omega)\to (M,\omega)$ is $N$-liftable, denote by $\psi_\nu$
the diffeomorphism $\Phi^{-1}\circ \psi\circ\Phi$ defined on
$(\mathcal{U}_{\epsilon_0}(\nu_0),\omega_{\nu,A})$.
Hence
 $\psi_\nu$ induces the diffeomorphism
$\widetilde\psi_\nu  :
\mathcal{U}_{\epsilon_0}(\widetilde \pi^{-1}(\nu_0)) \to \mathcal{U}_{\epsilon_0}(\widetilde \pi^{-1}(\nu_0)) $
given  by
$$
\widetilde\psi_\nu [p, (z, \ell)]:=[p, (U(z), U(\ell))].
$$
Since $\psi$ is $N$-liftable it  maps the complement of  $\; \mathcal{U}(N)\subset M$ to itself.
Therefore $\widetilde\psi_\nu$ together  with $\psi$ give rise to
a well-defined diffeomorphism on the
blown up manifold $\widetilde M_N$ that we denote by  $\widetilde\psi_N$ .
So defined $\widetilde\psi_N$ is a lift of $\psi$, that is
 $\widetilde\pi_{(M,N)}\circ\widetilde\psi_N= \psi \circ\widetilde\pi_{(M,N)}$.
It remains to show that it is a symplectic diffeomorphism.

\begin{prop}
\label{p:liftsymplectic}
If $\psi: (M,\omega)\to (M,\omega)$ is a symplectic diffeomorphism that is
$N$-liftable, then $\widetilde\psi_N$ is a symplectic diffeomorphism
of $(\widetilde M_N,\widetilde\omega_{N,\rho}).$
\end{prop}
\begin{proof}
By definition of the symplectic form  $\widetilde\omega_{N,\rho}$ in Eq.
(\ref{e:symplcformblow})
and of the diffeomorphism $\widetilde\psi_N$, if follows that $\widetilde\psi_N$ is a
symplectic diffeomorphism on
$M\setminus\Phi(\overline{ \mathcal{U}_{\epsilon^\prime}(\nu_0)  })$. It only remains to
show that $\widetilde\psi_N$ is a
symplectic diffeomorphism on
$ \mathcal{U}_{\epsilon_0}(\widetilde \pi^{-1}(\nu_0))$.

Since $\psi$ is $N$-liftable, on $\mathcal{U}_{\epsilon_0}(\widetilde \pi^{-1}(\nu_0))$ the diffeomorphism
$\widetilde\psi_N$ is given by $ \widetilde \psi_\nu$. That is
$$
\widetilde \psi_\nu[p,(z,\ell)]=[p,(U(z),U(\ell))]
$$
where $U$ lies in the center of $U(k)$.
In order to show that $\widetilde\psi_\nu^*(\widetilde\omega_{\widetilde\nu,A,\rho})=
\widetilde\omega_{\widetilde\nu,A,\rho}$, we regard  the symplectic form
on $P\times L_{\epsilon_0}$ instead of
$P\times_{U(k)}L_{\epsilon_0}$. In particular, $\widetilde\omega_{\widetilde\nu,A,\rho}$
is a basic form.
On $P\times L_{\epsilon_0}$ the diffeomorphism
$\widetilde\psi_\nu$ is given by
$1_P\times U$, where $U$ lies in the
center of $U(k)$. Thus
$ (1\times U)^*  \widetilde\omega_{A,\rho}=  \widetilde\omega_{A,\rho}$
and $\widetilde\psi_N$ is a symplectic diffeomorphism
of $(\widetilde M_N,\widetilde\omega_{N,\rho}).$

\begin{comment}
On $P\times L_{\epsilon_0}$ the diffeomorphism
$\widetilde\psi_\nu$ is given by
$1_P\times U$.  Since the matrix $U$ lies in the
center of $U(k)$ and $\widetilde\omega_{\rho,\epsilon}$ is invariant under the action
of $U(k)$ it follows that
\begin{eqnarray*}
(1\times U)^*\widetilde\omega_{\rho,\epsilon}(v_1+X_{A(u_1)},v_2+X_{A(u_2)})
&=& \widetilde\omega_{\rho,\epsilon}(v_1+X_{A(u_1)},v_2+X_{A(u_2)})
\end{eqnarray*}
for $u_j\in TP$ and $v_j\in T\widetilde{\mathbb{C}}^{k}$.
Now,
the 1-form $\langle\widetilde\mu_\rho,A\rangle$ only depends
on the vectors in $TP$.
Since the matrix $U$ satisfies
 $\widetilde\mu_\rho(U)=U.\widetilde\mu_\rho=\widetilde\mu_\rho$,
therefore
$$
(1\times U)^* \langle\widetilde\mu_\rho,A\rangle=
 \langle\widetilde\mu_\rho,A\rangle.
$$
From these two observations we conclude that
$ (1\times U)^*  \widetilde\omega_{A,\rho}=  \widetilde\omega_{A,\rho}$.
\end{comment}

\begin{comment}
Since the action of $U(k)$ on $(\widetilde{\mathbb{C}}^{k},\widetilde{\omega}_\rho)$
is Hamiltonian
\begin{eqnarray*}
\widetilde\omega_{A,\rho}(u_1+v_1,u_2+v_2 ):&=&
\widetilde\omega_\rho(v_1-X_{A_p(u_1)},v_2-X_{A_p(u_2)})-  \\
& & d\langle \widetilde\mu_\rho, A\rangle(u_1+v_1,u_2+v_2 )
\end{eqnarray*}
 it follows that
$
(\phi\times U)^*(\widetilde\omega_\rho)=\widetilde\omega_\rho.
$
\end{comment}

\begin{comment}
Finally, note that  $\widetilde \pi \circ(1\times U)=\widetilde \pi$.
Thereafter
$
(1\times U)^*(\widetilde\pi^*(\omega))=\widetilde\pi^*(\omega)$ and
the lift $\widetilde\psi_N$ is a symplectic diffeomorphism on
$(\widetilde M_N,\widetilde\omega_{N,\rho}).$
\end{comment}

\end{proof}

Next we prove that the lift of a Hamiltonian diffeomorphism, under some constraints, is again a
Hamiltonian diffeomorphism on the blown up manifold.
To that end, consider $\psi$  a Hamiltonian diffeomorphism on $(M,\omega)$
such that there exists a path $\{\psi_t\}_{0\leq t\leq 1}$ that start at the identity,
ends at $\psi$ and  for each $t$ the diffeomorphism $\psi_t$ is $N$-liftable.
Hence,  $\{\psi_{\nu,t}\}$  is a Hamiltonian path on
$ (\mathcal{U}_{\epsilon_0}(\nu_0),\omega_{\nu,A})$.
Moreover for each $t\in [0,1]$
there exist a matrix $U_t$ in the center of $U(k)$
such that
$$
\psi_{\nu,t}[p,z]=[p,U_t(z)].
$$
Since each matrix $U_t$ is of the form $\lambda_t \cdot 1_{k\times k}$ for
$\lambda_t\in \mathbb{C}$ and $|\lambda_t |=1$, it follows that the induced
Hamiltonian function of the path of unitary matrices is
autonomous.
Moreover, it is given by
$H(z):=c \sum_j  (x_j^2+y_j^2)=c| z|^2$ for some $c\in\mathbb{R}$.
Recall from Section \ref{ss:inducedHam}  that such Hamiltonian function
$H$ induces the Hamiltonian
$\widetilde H:L_{\epsilon_0}\to \mathbb{R}$ defined in Eq. (\ref{e:hamblow}).
Further it also satisfies the relation $\widetilde H(U(z),U(\ell))=\widetilde H(z,\ell)$.
Henceforth, we define the function
$\widetilde H_{t,N}:(\widetilde M_N,\widetilde\omega_{N,\rho}) \to \mathbb{R}$ by
\begin{eqnarray}
\label{e:hamblowN}
\widetilde H_{N,t}(q) :=
\left\{
	\begin{array}{ll}
		H_t(\widetilde\pi_{(M,N)}(q))   &  \textup{if }q\in
					M\setminus\Phi(\overline{ \mathcal{U}_{\epsilon^\prime}(\nu_0)  })  \\
	\widetilde H(z,\ell) &   \textup{if }  q=[p,(z,\ell)] \in\mathcal{U}_{\epsilon_0}(\widetilde \pi^{-1}(\nu_0)).
	\end{array}
\right.
\end{eqnarray}

Next we show that $\widetilde H_{N,t}$
 is the Hamiltonian function of the lifted path of symplectic diffeomorphisms.

\begin{prop}
Let  $\psi_t: (M,\omega)\to (M,\omega)$ be a path of Hamiltonian diffeomorphisms
such that each $\psi_t$ is $N$-liftable and $\psi_0=1$.
Then $\widetilde \psi_{N,t}: (\widetilde M_N,\widetilde\omega_{N,\rho})\to
 (\widetilde M_N,\widetilde\omega_{N,\rho})$ is a Hamiltonian path
induced by $\widetilde H_{N,t}$.
\end{prop}
\begin{proof}
Since the Hamiltonian path $\{ \psi_t \} $ is $N$-liftable, by Prop. \ref{p:liftsymplectic}
the path  $\{\widetilde\psi_{N,t}\}$ consists of symplectic diffeomorphisms on
$(\widetilde M_N,\widetilde\omega_{N,\rho})$ and satisfies
   $\psi_t\circ\widetilde\pi_{(M,N)}=\widetilde\pi_{(M,N)}\circ \widetilde\psi_{N,t}$
for each $t$. Thus if $\widetilde X_t$ is the time-dependent vector field
induced by $\{\widetilde\psi_{N,t}\}$, then  $(\widetilde\pi_{(M,N)})_*\widetilde X_t=X_t$.
Thus by the definition of the function $\widetilde H_{N,t}$ on (\ref{e:hamblowN}) and the fact that
$\widetilde\pi_{(M,N)}$ is a symplectic diffeomorphism on
$M\setminus\Phi(\overline{ \mathcal{U}_{\epsilon^\prime}(\nu_0)  }) $
it follows that
$$
\widetilde\omega_N(\widetilde X_t ,\cdot)=d\widetilde H_{t,N}
$$
holds on $\widetilde M_N	\setminus \mathcal{U}_{\epsilon_0}(\widetilde \pi^{-1}(\nu_0))$.

Now on $ \mathcal{U}_{\epsilon_0}(\widetilde \pi^{-1}(\nu_0))\simeq P\times_{U(k)}L_{\epsilon_0}$
the path   $\{\widetilde \psi_{N,t}\}$ is given by $[p,(z,\ell)]\mapsto [p,(U_t(z),U_t(\ell))]$.
Thus the time-dependent vector field is actually independent of time and takes the form
$0+\widetilde X$ where $\widetilde X$ is a vector field on $L_{\epsilon_0}$.
Also, on
$ \mathcal{U}_{\epsilon_0}(\widetilde \pi^{-1}(\nu_0))$
the symplectic form $\widetilde\omega_N$ is equal to
$\widetilde\omega_{\widetilde \nu, A,\rho}=\widetilde\omega_{A,\rho}+\widetilde\pi_N^*(\omega)$.
Therefore,
\begin{eqnarray*}
\widetilde\omega_{\widetilde \nu, A,\rho}  (\widetilde X_t,\cdot)
 &=& \widetilde\omega_{\widetilde \nu, A,\rho}(0+\widetilde X,\cdot)\\
&=& \widetilde\omega_{A,\rho}(0+\widetilde X,\cdot)+\widetilde\pi_N^*(\omega)(0+\widetilde X,\cdot) \\
&=&
\widetilde\omega_{\rho,\epsilon}(\widetilde X+0,\cdot)-  d\langle \widetilde\mu_\rho, A\rangle(0+\widetilde X,\cdot )
+\widetilde\pi_N^*(\omega)(0+\widetilde X,\cdot) \\
&=&
\widetilde\omega_{\rho,\epsilon}(\widetilde X+0,\cdot)
\end{eqnarray*}
By
\cite[Prop. 3.7]{pedroza-hamiltonian}, on $L_{\epsilon_0}$ we have that
$\widetilde\omega_{\rho,\epsilon}(\widetilde X,\cdot)=d\widetilde H$
where the function $\widetilde H$ was
defined in Eq. (\ref{e:hamblow}).
Thus the proposition
follows.
\end{proof}

Notice that if $\{\psi_t\}$ is a loop of Hamiltonian diffeomorphisms on $(M,\omega)$
that is $N$-liftable, then the lift  $\{\widetilde \psi_{N,t}\}$ is also loop of Hamiltonian diffeomorphisms.
We are interested in understanding the
the behavior of the loop $\{\widetilde \psi_{N,t}\}$ in the fundamental group of $\textup{Ham}(\widetilde M_N,
\widetilde\omega_{N,\rho}).$

\section{Weinstein's morphism}
\label{s:wein}

The period group $\mathcal{P}(M,\omega)$ of $(M,\omega)$
is defined as the image of the pairing $\langle\omega,\cdot \rangle\colon H_2(M;\mathbb{Z})
\to \mathbb{R}$. Then Weinstein's morphism \cite{weinstein-coho},
$$
\mathcal{A}:\pi_1 (\textup{Ham}(M,\omega)) \to \mathbb{R}/ \mathcal{P}( M,\omega)
$$
is defined via the action functional on a loop $\psi=[\{\psi_t\}]$  as
$$
\mathcal{A}(\psi)=
\left[
-\int_{D} u^*( \omega) +\int_{0}^1 H_t(\psi_t(x_0)) dt \right].
$$
Here $D$ is the unit closed disk and  $u:D\to M$ is a smooth function such that
$u(\partial D)$ agrees with  the loop $\{\psi_t(x_0)\}$, $x_0\in M$  a base point and
$H_t$ is the 1-periodic Hamiltonian function
induced by  the Hamiltonian  loop $\{\psi_t\}$ subject to the normalized condition
$$
\int_M H_t \, \omega^n=0
$$
for every $t\in [0,1]$.

The aim is to relate  Weinstein's morphism   on $\pi_1(\textup{Ham}(M,\omega))$ and
$\pi_1(\textup{Ham} (\widetilde M_N,\widetilde\omega_{N,\rho})      )$ on a particular class of loops.
To that end, recall that we must consider a Hamiltonian diffeomorphisms  on $(M,\omega)$
that are $N$-liftable. As seen above, such Hamiltonians  are
 induced by Hamiltonians functions on $(\mathbb{C}^n,\omega)$ that on a neighborhood of
the origin take the form
 $c \sum_j  (x_j^2+y_j^2)=c| z|^2$ for some $c\in\mathbb{R}$.

\begin{lem}
\label{l:hamfuncrel}
Let $H:M\to\mathbb{R}$ be a smooth function that is induced by
a Hamiltonian diffeomorphism that is $N$-liftable, $\rho$ as before and
$\widetilde H_N:\widetilde M_N\to\mathbb{R}$ the induced function defined in
(\ref{e:hamblowN}). Then
$$
\int_{\widetilde M_N} \widetilde H_N \, \widetilde\omega_{N,\rho}^n
=\int_{M}  H\, \omega^n
-\int_{\Phi(\mathcal{U}_{\rho}(\nu_0))}  H\, \omega^n.
$$
\end{lem}
\begin{proof}
First note that $\widetilde M_N$ is the union of the disjoint sets
$M\setminus\Phi(\overline{ \mathcal{U}_{\epsilon^\prime}(\nu_0)  })$
and  $\mathcal{U}_{\epsilon^\prime}  (\widetilde\pi^{-1}(\nu_0)) $. Recall that
$\epsilon^\prime$ is such that
 $\rho+\epsilon<\epsilon^\prime < \epsilon_0.$
Then from the definition
of $\widetilde H_N$ in (\ref{e:hamblowN}) and of $\widetilde H$ in (\ref{e:hamblow}) it follows that
\begin{eqnarray*}
\int_{\widetilde M_N} \widetilde H_N \,\widetilde\omega_{N,\rho}^n
&=&
\int_{M\setminus\Phi(\overline{ \mathcal{U}_{\epsilon^\prime}(\nu_0)  }) }
 \hspace{-.6cm} \widetilde H_N \,\widetilde\omega_{N,\rho}^n
+\int_{\mathcal{U}_{\epsilon^\prime  }
(\widetilde\pi^{-1}(\nu_0))} \hspace{-.6cm} \widetilde H_N \,\widetilde\omega_{N,\rho}^n\\
&=&
\int_{M\setminus\Phi(\overline{ \mathcal{U}_{\epsilon^\prime}(\nu_0)  }) }
 \hspace{-.6cm} H\circ \widetilde\pi_{(M,N)} \,\widetilde\omega_{N,\rho}^n
+\int_{\mathcal{U}_{\epsilon^\prime  }
(\widetilde\pi^{-1}(\nu_0))} \hspace{-.6cm} \widetilde H \,\widetilde\omega_{N,\rho}^n\\
&=&
\int_{M\setminus\Phi(\overline{ \mathcal{U}_{\epsilon^\prime}(\nu_0)  }) }
 \hspace{-.6cm} H \,\omega^n
+\int_{\mathcal{U}_{\epsilon^\prime  }
(\widetilde\pi^{-1}(\nu_0))} \hspace{-.6cm} \widetilde H \,\widetilde\omega_{\widetilde\nu,A,\rho}^n\\
&=&
\int_{M\setminus\Phi(\overline{ \mathcal{U}_{\epsilon^\prime}(\nu_0)  }) }
 \hspace{-.6cm} H \,\omega^n
+\int_{(P\times_{U(k)}L_{\epsilon^\prime})\setminus \nu_0}
 \hspace{-.6cm}   \widetilde H \,\widetilde\omega_{\widetilde\nu,A,\rho}^n\\
\end{eqnarray*}

Recall that the diffeomorphism
 $F_{\rho,\epsilon}: \mathbb{C}^k\setminus \{0\}\to \mathbb{C}^k\setminus \overline{B}_\rho$
defined in
(\ref{e:strech}) is $U(k)$-equivariant.
Hence it induces a diffeomorphism
${\bf F}_{\rho,\epsilon}: P\times_{U(k)}(\mathbb{C}^k\setminus \{0\})
\to P\times_{U(k)}(\mathbb{C}^k\setminus \overline{B}_\rho)$ and
\begin{eqnarray*}
{\bf F}_{\rho,\epsilon} ^*(\omega_{\nu,A})
&=&
{\bf F}_{\rho,\epsilon} ^*(\omega_A)  + {\bf F}_{\rho,\epsilon} ^*(\pi_N^*(\omega))\\
&=&
 \widetilde\omega_{A,\rho} + \widetilde\pi_N^*(\omega)\\
 &=&
\widetilde\omega_{\widetilde\nu,A,\rho}.
\end{eqnarray*}
It then follows that
\begin{eqnarray*}
\int_{(P\times_{U(k)}L_{\epsilon^\prime})\setminus \nu_0}

\hspace{-.6cm} \widetilde H \,\widetilde\omega_{\widetilde\nu,A,\rho}^n
&=&
\int_{\mathcal{U}_{\epsilon^\prime  }
(\widetilde\pi^{-1}(\nu_0))} \hspace{-.6cm} \widetilde H \,
{\bf F}_{\rho,\epsilon} ^*\omega_{\nu,A}^n \\
&=&
\int_{P\times_{U(k)}(B_{\epsilon^\prime}\setminus \overline{B_\rho})}
 \hspace{-.6cm}  H \,
\omega_{\nu,A}^n.
\end{eqnarray*}

Hence
\begin{eqnarray*}
\int_{\widetilde M_N} \widetilde H_N \,\widetilde\omega_{N,\rho}^n
&=&
\int_{M\setminus\Phi(\overline{ \mathcal{U}_{\epsilon^\prime}(\nu_0)  }) }
 \hspace{-.6cm} H \,\omega^n
+\int_{P\times_{U(k)}(B_{\epsilon^\prime}\setminus \overline{B_\rho})}
 \hspace{-.6cm} H   \,\omega_{\nu,A}^n\\
 &=&
 \int_{M}  H\, \omega^n
-\int_{\Phi(\mathcal{U}_{\rho}(\nu_0))}  H\, \omega^n.
\end{eqnarray*}
\end{proof}

\section{Proof of Theorems
 \ref{t:mainrank} and \ref{t:weinsrelation}}
 \label{s:main}

Now we prove the main results of this article that appeared at the Introduction.

\smallskip
\noindent {\bf Theorem \ref{t:weinsrelation}.} {\em
Let $(M,\omega)$ be a closed symplectic manifold and $N\subset (M,\omega)$
a symplectic submanifold of codimension $2k>2$.
If $\{\psi_t\}$ is a loop of Hamiltonian diffeomorphisms on $(M,\omega)$
that is $N$-liftable to a loop $\{\widetilde\psi_{N,t}\}$
on $(\widetilde M_N,\widetilde \omega_{N,\rho})$, then
\begin{eqnarray*}
\mathcal{A}([\widetilde\psi_{N,t}]) =
\left[\mathcal{A}([\psi_t]) +
\frac{1}{\textup{Vol}(\widetilde M_N,(\widetilde\omega_{N,\rho})^n)}
\int_0^1
\int_{\Phi(\mathcal{U}_{\rho}(\nu_0))}  H_t\, \omega^n dt
 \right]
\end{eqnarray*}
in $\mathbb{R}/\mathcal{P}(\widetilde M_N,\widetilde \omega_{N,\rho})$, where
$H_t$ if the normalized Hamiltonian function of the loop $\{\psi_t\}$. }

\begin{proof
Let $\{\psi_t\}$ be a loop of Hamiltonian diffeomorphisms on $(M,\omega)$ that is
$N$-liftable with normalized Hamiltonian function $H_t$. If $\{\widetilde\psi_{N,t}\}$
is the induced loop on $(\widetilde M_N ,\widetilde\omega_{N,\rho})$, denote by
 $\widetilde H_{N,t}$ its Hamiltonian function defined by Eq. (\ref{e:hamblowN}).

It follows from Lemma \ref{l:hamfuncrel} that for each $t$,
\begin{eqnarray}
\label{e:intloc}
\int_{\widetilde M_N} \widetilde H_{N,t} \,\widetilde\omega_{N,\rho}^n
=
-\int_{\Phi(\mathcal{U}_{\rho}(\nu_0))}  H_t\, \omega^n.
\end{eqnarray}
Denote such number by $c(\rho,t)\in\mathbb{R}$. Hence the normalized Hamiltonian
function of the loop $\{\widetilde\psi_{N,t}\}$ is
$$
\widetilde{\bf H}_{N,t}:=\widetilde H_{N,t} -\frac{1}{\textup{Vol}(\widetilde M_N,
\widetilde\omega_{N,\rho}^n)} c(\rho,t).
$$

Thus, taking a disk $D$ and a point $x_0$ in $\widetilde M_N	\setminus \mathcal{U}_{\epsilon_0}(\widetilde \pi^{-1}(\nu_0))$,
\begin{eqnarray*}
\mathcal{A}([\widetilde\psi_{N,t}])
&=&
\left[ -\int_D u^*(\widetilde\omega_{N,\rho}) + \int_0^1
\widetilde{\bf H}_{N,t} (\widetilde\psi_{N,t}(x_0)) dt\right]\\
 &=&
\left[ -\int_D u^*(\omega) + \int_0^1
\widetilde H_{N,t} (\widetilde\psi_{N,t}(x_0)) dt \right. \\
& &\left. - \frac{1}{\textup{Vol}(\widetilde M_N,
\widetilde\omega_{N,\rho}^n)}    \int_0^1  c(\rho,t) dt \right].
\end{eqnarray*}

Hence from Eq. (\ref{e:intloc}) we get
\begin{eqnarray*}
\mathcal{A}([\widetilde\psi_{N,t}]) =
\left[\mathcal{A}([\psi_t]) +
\frac{1}{\textup{Vol}(\widetilde M_N,\widetilde\omega_{N,\rho}^n)}
\int_0^1
\int_{\Phi(\mathcal{U}_{\rho}(\nu_0))}  H_t\, \omega^n dt
 \right].
\end{eqnarray*}
\end{proof}

In order to give meaning to the expression that was proven  above, now we relate the periods
groups of $(\widetilde M_N,\widetilde\omega_{N,\rho})$ and $( M,\omega)$.
Recall that the blow up map $\widetilde\pi_{(M,N)} : \widetilde M_N\to M $ is a diffeomorphism
on the complement of $\widetilde\pi_{(M,N)}^{-1}(N)$. Moreover,  $\widetilde\pi_{(M,N)}$
induces a symplectic diffeomorphism on the complement of a neighborhood of
$\widetilde\pi_{(M,N)}^{-1}(N)$. Therefore, the maps
$$
\langle \widetilde\omega_{N,\rho}, \cdot \rangle: H_2(\widetilde M_N \setminus \widetilde\pi_{(M,N)}^{-1}(N);\mathbb{Z})
\to \mathbb{R}
\textup{ and }
\langle \omega, \cdot \rangle: H_2( M_N;\mathbb{Z})
\to \mathbb{R}
$$
have the same image. It remains to determine the value of
$\langle \widetilde\omega_{N,\rho}, \cdot \rangle$ on the kernel of $\widetilde\pi_{(M,N)*}:
H_2(\widetilde M_N;\mathbb{Z})\to H_2( M;\mathbb{Z})$.
From the  commutative diagram
{\vskip .4cm}
$$
\xymatrix{
  & H_1(\nu\setminus\nu_0;\mathbb{Z}) \ar[d]_1\ar[l] &  H_2(\widetilde M_N;\mathbb{Z}) \ar[l] \ar[d]_{\widetilde\pi_{(M,N)*}}
&  H_2(M\setminus N;\mathbb{Z})\oplus H_2(\widetilde\nu;\mathbb{Z})\ar[d]_{1\oplus \widetilde\pi_*}\ar[l] & \ar[l]    \\
    & H_1(\nu\setminus\nu_0;\mathbb{Z}) \ar[l]        &  H_2(M;\mathbb{Z}) \ar[l]
&  H_2(M\setminus N;\mathbb{Z})\oplus H_2(\nu;\mathbb{Z}) \ar[l] & \ar[l]
}
$$
it suffices to look at the map $\widetilde\pi_* : H_2(\widetilde\nu;\mathbb{Z})
\to H_2(\nu;\mathbb{Z})$.

Recall that the closed 2-form $\widetilde\omega_{\widetilde\nu,A,\rho}$ on the
total space of the fibration    $\widetilde\pi_N:\widetilde\nu\to N$
restricts to  $\widetilde\omega_{\rho,\epsilon}$ on each fibre.

The relation between the homology of $\widetilde{M}_N$ and $M$ is as follows
$$
H_2(\widetilde M_N,\mathbb{Z})\simeq H_2(M,\mathbb{Z})\oplus \textup{ker}\{
{\widetilde\pi}_{N,*} :H_2(\widetilde \nu,\mathbb{Z})\to H_2(N,\mathbb{Z})\}.
$$
Moreover, if $\alpha\in H_2(\widetilde M_N;\mathbb{Z})$ takes the form
$\widetilde\pi_{(M,N)*}(\alpha)+ 0$ in the above decomposition, then
$$
\langle \widetilde\omega_{N,\rho},  \alpha \rangle =
\langle \omega,  \widetilde\pi_{(M,N)*}(\alpha)   \rangle.
$$
Since $\widetilde\pi_{(M,N)*}:
H_2(\widetilde M_N;\mathbb{Z})\to H_2( M;\mathbb{Z})$
is surjective, we have that
$$
\mathcal{P}( M,\omega)\subset \mathcal{P}(\widetilde M_N,\widetilde\omega_{N,\rho}).
$$
Finally,
$\textup{ker}\{
{\widetilde\pi}_{N,*} :H_2(\widetilde \nu;\mathbb{Z})\to H_2(N;\mathbb{Z})\}$
is generated by a single class that its restriction to the fiber
is the generator of $H_2(\mathbb{C}P^{k-1};\mathbb{Z})$.
Hence we have the following result.

\begin{lem}
\label{l:relperiod}
Let $(M,\omega)$ and $(\widetilde M_N,\widetilde\omega_{N,\rho})$ as above.
Assume that $N$ is compact,
then
\begin{eqnarray*}
\mathcal{P}(\widetilde M_N,\widetilde\omega_{N,\rho})=
\mathcal{P}(M,\omega)+
\mathbb{Z}\langle\pi \rho^2 \rangle.
\end{eqnarray*}
\end{lem}

Recall  from Lemma \ref{l:sympformnormal} that on a neighborhood
$P\times_{U(k)} B_{\epsilon_0}$
of the zero-section of
$\nu=P\times_{U(k)}\mathbb{C}^k$ the symplectic form is given by
\begin{eqnarray}
\label{e:expan}
\omega_{\nu,A}(u_1+v_1,u_2+v_2 ) &=&
\omega_0(v_1+X_{A(u_1)},v_2+X_{A(u_2)})- \label{omega0} \\
& & d\langle \mu, A\rangle(u_1+v_1,u_2+v_2 )+ \nonumber\\
& & \pi_N^*(\omega)(u_1+v_1,u_2+v_2 ) \nonumber
\end{eqnarray}
for $u_j\in TP$ and $v_j\in T\mathbb{C}^k$. Hence on the normal bundle $\nu$ we have that
\begin{eqnarray}
\omega_{\nu,A}^n &=&   \sum_{j=0}^n
\omega_0^j   \; \pi_N^*(\omega)^{n-j}   + d\alpha
\end{eqnarray}
for some $(2n-1)$-form $\alpha$. (Keep in mind that in this equation  the 2-form
$\omega_0$ must be evaluated as in Eq. (\ref{e:expan})).

\begin{lem}
\label{l:integralfunction}
Let $H:\nu\to \mathbb{R}$ be a smooth function.
Then,
\begin{eqnarray*}
\int_{  \nu } H\; \omega_{\nu,A}^n
&=& \int_{N}      (\pi_N)_*(   H\; \omega_0^k ) \;\omega^{n-k}.
\end{eqnarray*}
\end{lem}
\begin{proof}
First we integrate along the fiber with respect to $\pi_N: \nu\to N$ and then
use the commutativity of the diagram
$$
\xymatrix{
\nu \ar[r]^{\pi_N}
\ar[dr]&  N\ar[d]\\
     & \{pt\}.
}$$
Therefore by the projection formula we get that
\begin{eqnarray*}
(\pi_N)_*( H\; \omega_{\nu,A}^n ) &=&   \sum_{j=0}^n (\pi_N)_*\left(
   H\; \omega_0^j   \; \pi_N^*(\omega)^{n-j}  \right )\\
&=&  \sum_{j=0}^n (\pi_N)_*(
   H\; \omega_0^j ) \;\omega^{n-j}. \\
\end{eqnarray*}
Note that in this expression, the symplectic form $\omega$ is restricted to the submanifold $N$.

Next we integrate on the submanifold $N$,
\begin{eqnarray*}
\int_{  \nu } H\; \omega_{\nu,A}^n
&=& \sum_{j=0}^n  \int_{N}      (\pi_N)_*(
   H\; \omega_0^j ) \;\omega^{n-j} \\
&=& \sum_{j=k}^n  \int_{N}      (\pi_N)_*(
   H\; \omega_0^j ) \;\omega^{n-j} \\
&=& \int_{N}      (\pi_N)_*(
   H\; \omega_0^k ) \;\omega^{n-k}.
\end{eqnarray*}
The last equality follows from the fact that the form $\omega_0$ comes from the
fiber $\mathbb{C}^k$ of $\pi_N$.
\end{proof}

Notice that  from this result the evaluation of $\omega_0$ as in
Eq. (\ref{e:expan}) becomes irrelevant.

In particular, restrict the above integral to  ${\mathcal{U}_{\rho}(\nu_0)}$
and take
$H\equiv 1$. Then are able to compute the volume of ${\mathcal{U}_{\rho}(\nu_0)}$,
\begin{eqnarray*}
\textup{Vol}({\mathcal{U}_{\rho}(\nu_0)} , \omega_{\nu,A}^n )&=&
\int_{  {\mathcal{U}_{\rho}(\nu_0)} }  \omega_{\nu,A}^n \\
&=& \int_{N}      (\pi_N|_{\mathcal{U}_{\rho}(\nu_0)})_*(  \omega_0^k ) \;\omega^{n-k}\\
&=& \textup{Vol}(B_\rho , \omega_0^k )\cdot \textup{Vol}(N , \omega^{n-k} )\\
&=& \frac{\pi^{k} \rho^{2k}}{k!} \cdot \textup{Vol}(N , \omega^{n-k} ).
\end{eqnarray*}

\subsection{Description of the loop.}

In order to prove Thm. \ref{t:mainrank}, we need to defined a loop of Hamiltonian diffeomorphisms
on $(M,\omega)$ supported on $P\times_{U(k)}B_{\epsilon_0} $, a neighborhood of $N$.
The definition of such loop follows the same line of thought as
the Hamiltonian loop constructed in \cite[Sec. 2]{pea-rankham}.

Let $\alpha:\mathbb{R}\to\mathbb{R}$ such that $\alpha(0)=0$ and $\alpha(1)\neq 0$. Then consider
the path of diagonal  matrices $A_t:=\exp{(2\pi i\alpha(t))}\cdot 1_{k\times k}$. Hence $A_t$ lies
in the center of $U(k)$ and it induces on  $(\mathbb{C}^k,\omega_0)$
 a path $\{\psi^\alpha_t\}_{0\leq t\leq 1}$ of Hamiltonian diffeomorphisms
starting at the identity and
  with Hamiltonian function
$$
H^\alpha_t(z_1,\ldots , z_k):= \pi\; \alpha^\prime(t)\; \sum_{j=1}^k |z_j |^2.
$$
As before we have the positive  parameters $\rho,\epsilon_0$ and $\epsilon\in (0,1)$ such that
$\rho+\epsilon<\epsilon_0$, that are used in the definition of $(\widetilde M_N,\widetilde
\omega_{N,\rho})$.
Now fix a $U(k)$-invariant smooth function
 $g:\mathbb{C}^k\to \mathbb{R}$ supported in $B_{\epsilon_0}$ such that
 $g\equiv 1$ on  $B_{\rho}$.
Consider the Hamiltonian function
$$
H^{\alpha,g}_t: = g\cdot H^{\alpha}_t
$$
and let $\{\psi^{\alpha,g}_t\}_{0\leq t\leq 1}$ be the induced Hamiltonian path. Since $\omega_0$
and $H^{\alpha,g}_t$ are $U(k)$-equivariant, it follows that  each $\psi^{\alpha,g}_t$
is also $U(k)$-equivariant. Furthermore, on $B_\rho$ we have that
$\psi^{\alpha,g}_t \equiv \psi^{\alpha}_t$ and $H^{\alpha,g}_t\equiv
H^{\alpha}_t$.
Note that as in \cite[Lemma 2.4]{pea-rankham},
\begin{eqnarray*}
\int_0^1 \int_{B_\rho}  \;H^{\alpha,g}_t\;\omega_0^k\;dt
&= & (\alpha(1)-\alpha(0)) \;\pi
\int_{B_\rho} \sum_{j=1}^k x_j^2 + y_j^2 \frac{1}{k!}dx_1\cdots dy_k \\
&=&\alpha(1) \cdot \frac{\pi}{k!} \cdot \frac{\pi^{k} \rho^{2k+2}}{k+1}.
\end{eqnarray*}

Since each  $\psi^{\alpha,g}_t$ is $U(k)$-equivariant, it induces
a path of diffeomorphisms $1\times \psi^{\alpha,g}_t$ on $\nu=P\times_{U(k)}
\mathbb{C}^k.$ Furthermore, each diffeomorphism is
supported on $\mathcal{U}_{\epsilon}(\nu_0)=P\times_{U(k)} B_{\epsilon_0}$.

\begin{lem}
\label{l:hamnormal}
The path $1\times \psi^{\alpha,g}_t$ restricted to $\mathcal{U}_{\epsilon_0}(\nu_0)$
 is a  Hamiltonian path with Hamiltonian function $1\times H^{\alpha,g}_t.$
\end{lem}
\begin{proof}
Recall that the 2-form $\omega_{\nu,A}$ is
a basic form on $P\times B_{\epsilon_0}$, in particular is $U(k)$-invariant.
Therefore, $1\times \psi^{\alpha,g}_t$ preserves the form $\omega_{\nu,A}$
 and is a symplectic diffeomorphism
on  $\mathcal{U}_{\epsilon_0}(\nu_0)$.

It remains to prove that is Hamiltonian. Note that by definition on $(B_{\epsilon_0},
\omega_0)$  we have that $\{\psi^{\alpha,g}_t\}$ is Hamiltonian with Hamiltonian
function $H^{\alpha,g}_t$. Let $X^{\alpha,g}_t$ be the associated Hamiltonian vector
field, $\omega_0( X^{\alpha,g}_t, \cdot ) = dH^{\alpha,g}_t$. Recall from Eq.
(\ref{e:expan}) that
\begin{eqnarray*}
\omega_{\nu,A}(u_1+v_1,u_2+v_2 ) &=&
\omega_0(v_1+X_{A(u_1)},v_2+X_{A(u_2)})- \label{omega0} \\
& & d\langle \mu, A\rangle(u_1+v_1,u_2+v_2 )+ \nonumber\\
& & \pi_N^*(\omega)(u_1+v_1,u_2+v_2 ) \nonumber
\end{eqnarray*}
for $u_j\in TP$ and $v_j\in TB_{\epsilon_0}$.
If $X_{\xi}$ is the vector field on $B_{\epsilon_0}$ induced
by $\xi\in \mathfrak{u}(k)$ we have that
$\omega_0(X_{\xi_1},X_{\xi_2})=\langle \mu , [{\xi_1}, {\xi_2}]\rangle $
for any $\xi_1,\xi_2\in \mathfrak{u}(k)$. In particular, the vector field $X^{\alpha,g}_t$
is of the form $X_{\xi}$ for ${\xi}$ in the center of $ \mathfrak{u}(k)$ and therefore
$\omega_0(X^{\alpha,g}_t ,X_{\xi})=0$ for any ${\xi}$. Then for
$u\in TP$ and   $v\in TB_{\epsilon_0} $
\begin{eqnarray*}
\omega_{\nu,A}( X^{\alpha,g}_t,  u+v )
&=&\omega_0(X^{\alpha,g}_t ,v+X_{A(u)})  \\
&=&\omega_0(X^{\alpha,g}_t ,v)  \\
&=&  dH^{\alpha,g}_t  (    v  ) \\
&=& d(1\times H^{\alpha,g}_t)  (u,v)
\end{eqnarray*}
where $\langle \mu, A\rangle$ vanishes since $v_1=0$.
\end{proof}

Now we will consider another path of Hamiltonian diffeomorphisms in order
to create a loop by concatenating it with the path
$\{\psi^{\alpha,g}_t\}_{0\leq  t\leq 1}$. Hence, consider  $\beta:\mathbb{R}\to\mathbb{R}$
such that $\beta(0)=0$ and  $\alpha(t)-\beta(t)\equiv1$ in a neighborhood of 1.
As above denote
by  $\{\psi^{\beta,g}_t\}_{0\leq t\leq 1}$ the induced the path of Hamiltonian diffeomorphisms
 with Hamiltonian function $
H^{\beta,g}_t$ supported on $B_{\epsilon_0}$. Then the desired loop of Hamiltonian
 diffeomorphisms  $\{
\psi^{\mathbb{C}^k}_t  \}_{0\leq t\leq 2}$    is defined as
\begin{eqnarray}
\label{e:loop}
\psi^{\mathbb{C}^k}_t :=
\left\{
	\begin{array}{ll}
		\psi^{\alpha,g}_t  & t\in[0,1] \\
		\psi^{\beta,g}_{2-t}  & t\in[1,2]
	\end{array}
\right.
\end{eqnarray}
with Hamiltonian function
\begin{eqnarray}
\label{e:Hloop}
H^{\mathbb{C}^k}_t :=
\left\{
	\begin{array}{ll}
		H^{\alpha,g}_t  & t\in[0,1] \\
		H^{\beta,g}_{2-t}  & t\in[1,2].
	\end{array}
\right.
\end{eqnarray}

Furthermore, we have that
\begin{eqnarray*}
\int_0^2 \int_{B_\rho}  \; H^{\mathbb{C}^k}_t   \;\omega_0^k\;dt
= (\alpha(1)-\beta(1)) \; \frac{\pi^{k+1} \rho^{2k+2}}{(k+1)!}
= \frac{  (\pi \rho^2)^{k+1}}{(k+1)!}.
\end{eqnarray*}

We have all the prerequisites to proceed with the proof of the main theorem.

\medskip
\noindent {\bf Theorem \ref{t:mainrank}.} {\em
Let $(M,\omega)$ be a rational closed symplectic manifold and $N\subset (M,\omega)$
a closed symplectic submanifold of codimension $2k>2$.
Let $(\widetilde M,\widetilde\omega_{N,\rho})$ be the symplectic blow up
of $(M,\omega)$ along $N$. Then for some small values of $\rho$ the rank of
$\pi_1(\textup{Ham}(\widetilde M,\widetilde\omega_{N,\rho}))$
is positive.}

\begin{proof
Consider the loop  of Hamiltonian diffeomorphisms $\{  \psi^{\mathbb{C}^k}_t  \}_{0\leq t\leq 2}$
defined above.
Since each  $\psi^{\mathbb{C}^k}_t$ is $U(k)$-equivariant, it induces
a loop of diffeomorphisms $\{\psi^{\nu}_t\}$ on $\nu=P\times_{U(k)}
\mathbb{C}^k.$ Furthermore, each diffeomorphism is
supported on $P\times_{U(k)} B_{\epsilon_0}$.

From Lemma \ref{l:hamnormal}, we have that $\{1\times \psi^{\nu}_t\}$
is a loop of Hamiltonian diffeomorphisms on $P\times_{U(k)} B_{\epsilon_0}$
with Hamiltonian function $H^\nu_t=1\times H^{\mathbb{C}^k}_t.$
Since $P\times_{U(k)} B_{\epsilon_0}$ is
symplectomorphic to a neighborhood $\mathcal{U}(N)$, we consider
$\{\psi^{\nu}_t\}$ as a loop of Hamiltonian diffeomorphisms on $(M,\omega)$.
Denote such loop by $\{\psi^M_t\}$ with Hamiltonian function $H^M_t$.
Notice that such loop is contractible in $\textup{Ham}(M,\omega)$ and the
Hamiltonian function is not normalized. Thus consider the function
\begin{eqnarray}
\label{e:normalconstant}
c_t:= \int_M  H^M_t\; \omega^n
= \int_{\mathcal{U}(N)} H^M_t\; \omega^n.
\end{eqnarray}
Then the normalized Hamiltonian function of $\{\psi^M_t\}$ is defined as
\begin{eqnarray*}
H^{M,\textup{norm}}_t:=  H^M_t -  \frac{c_t}{\textup{Vol}(M,\omega^n)}.
\end{eqnarray*}

From its definition, the loop  $\{\psi^M_t\}_{0\leq t\leq 2}$ is $N$-liftable.
Denote by $\{\widetilde\psi_{N,t}\}_{0\leq t\leq 2}$ the induced loop of
Hamiltonian diffeomorphisms.
Hence  by Thm \ref{t:weinsrelation} we have that
\begin{eqnarray*}
\mathcal{A}([\widetilde\psi_{N,t}]) =
\left[
\frac{1}{\textup{Vol}(\widetilde M_N,\widetilde\omega_{N,\rho}^n)}
\int_0^2
\int_{\Phi(\mathcal{U}_{\rho}(\nu_0))}  H^{M,\textup{norm}}_t\, \omega^n dt
 \right]
\end{eqnarray*}
in $\mathcal{P}(M,\omega)+
\mathbb{Z}\langle\pi \rho^2 \rangle.$
From the definition of the symplectic blow up we know that
$  \textup{Vol}(\widetilde M_N,\widetilde\omega_{N,\rho}^n)$ is equal to
$\textup{Vol}(M,\omega^n) -
\textup{Vol}  ( \mathcal{U}_{\rho}(\nu_0), \omega_{\nu,A}^n )$.

Next we compute the integrals. To that end, by the definition
of the Hamiltonian function on the neighborhood
$\Phi(\mathcal{U}_{\rho}(\nu_0))$
of $N$ we
have that
\begin{eqnarray*}
\int_0^2
\int_{\Phi(\mathcal{U}_{\rho}(\nu_0))}  H^{M,\textup{norm}}_t\, \omega^n \, dt
&=&
\int_0^2
\int_{\Phi(\mathcal{U}_{\rho}(\nu_0))}   H^M_t -   \frac{c_t}{\textup{Vol}(M,\omega^n)}   \,   \omega^n  \,dt \\
&=&
\int_0^2
\int_{\mathcal{U}_{\rho}(\nu_0)}       1\times H^{\mathbb{C}^k}_t     \,  \omega_{\nu,A}^n  \,dt    \\
 &  &   -
    \frac{ \textup{Vol}  ( \mathcal{U}_{\rho}(\nu_0), \omega_{\nu,A}^n )  }{\textup{Vol}(M,\omega^n)} \int_0^2
c_t   \, dt.
\end{eqnarray*}
By replacing the normal bundle $\nu$ by the disk bundle $\mathcal{U}_{\rho}(\nu_0)$ on the
first sumand, we get
by Lemma \ref{l:integralfunction} that
\begin{eqnarray*}
\int_{\mathcal{U}_{\rho}(\nu_0)}       1\times H^{\mathbb{C}^k}_t      \omega_{\nu,A}^n
&=&
\int_{N}      (\pi_N  |_{\mathcal{U}_{\rho}(\nu_0)} )_*(    1\times H^{\mathbb{C}^k}_t   \; \omega_0^k ) \;\omega^{n-k}.
\end{eqnarray*}
Since
\begin{eqnarray*}
 (\pi_N  |_{\mathcal{U}_{\rho}(\nu_0)} )_*(    1\times H^{\mathbb{C}^k}_t   \; \omega_0^k )
&=&
 \int_{B_\rho}  \; H^{\mathbb{C}^k}_t   \;\omega_0^k,
\end{eqnarray*}
it follows that
\begin{eqnarray*}
\int_0^2 \int_{\mathcal{U}_{\rho}(\nu_0)}       1\times H^{\mathbb{C}^k}_t      \omega_{\nu,A}^n  \,dt
&=&
 \frac{  (\pi \rho^2)^{k+1}}{(k+1)!}\textup{Vol}(N,\omega^{n-k})
\end{eqnarray*}
since $\alpha(1)-\beta(1)=1.$

Notice that $c_t$ can be rewritten as
\begin{eqnarray*}
 {c_t}
&=&
\int_{M} H^M_t\; \omega^n\\\
&=&
\int_{\mathcal{U}(N)} H^M_t\; \omega^n\\
&=&
\int_{\nu}  1\times H^{\mathbb{C}^k}_t
 \; \omega_{\nu,A}^n \\
&=&
\int_{N}      (\pi_N   )_*(    1\times H^{\mathbb{C}^k}_t   \; \omega_0^k ) \;\omega^{n-k}.
 \end{eqnarray*}
Before the fiber was $(B_\rho,\omega_0)$, in this case the fiber is $(\mathbb{C}^k,\omega_0)$
and  $ H^{\mathbb{C}^k}_t $ has  compact support.
But on  $(\mathbb{C}^k,\omega_0)$, we know that the
Calabi morphism vanishes. Therefore we have that
$$
\int_0^2    {c_t} \; dt=
\int_0^2 (\pi_N   )_*(    1\times H^{\mathbb{C}^k}_t   \; \omega_0^k )\; dt =0
$$
and
\begin{eqnarray*}
\mathcal{A}([\widetilde\psi_{N,t}]) =
\left[
\frac{1}{\textup{Vol}(M,\omega^n) -
  \frac{\pi^{k} \rho^{2k}}{k!} \cdot \textup{Vol}(N , \omega^{n-k} )
 } \cdot
 \frac{  (\pi \rho^2)^{k+1}}{(k+1)!}\textup{Vol}(N,\omega^{n-k})
 \right]
\end{eqnarray*}
in $\mathcal{P}(M,\omega)+\mathbb{Z}\langle\pi \rho^2 \rangle.$ By hypothesis,
$\textup{Vol}(M,\omega^n),  \textup{Vol}(N,\omega^{n-k})$ and
 $\mathcal{P}(M,\omega)$  are in $\mathbb{Q}$.
Therefore, it follows from the above  expression that if $\rho$ is such that $\pi\rho^2$ is a
transcendental number
then $\mathcal{A}([\widetilde\psi_{N,t}]^m)\neq 0$ in
$\mathcal{P}(M,\omega)+\mathbb{Z}\langle\pi \rho^2 \rangle$
for all non zero $m\in\mathbb{Z}$. Henceforth, $\pi_1(\textup{Ham}(\widetilde M,\widetilde\omega_{N,\rho}))$
has positive rank.
\end{proof}

\bibliographystyle{acm}

redpajama_set_name: RedPajamaArXiv


text:
Celles-sur-Plaine és un municipi francès, situat al departament dels Vosges i a la regió del Gran Est. L'any 2007 tenia 851 habitants.

Demografia

Població
El 2007 la població de fet de Celles-sur-Plaine era de 851 persones. Hi havia 384 famílies, de les quals 132 eren unipersonals (52 homes vivint sols i 80 dones vivint soles), 116 parelles sense fills, 100 parelles amb fills i 36 famílies monoparentals amb fills.

La població ha evolucionat segons el següent gràfic:

Habitants censats

Habitatges
El 2007 hi havia 574 habitatges, 385 eren l'habitatge principal de la família, 125 eren segones residències i 64 estaven desocupats. 472 eren cases i 93 eren apartaments. Dels 385 habitatges principals, 278 estaven ocupats pels seus propietaris, 100 estaven llogats i ocupats pels llogaters i 7 estaven cedits a títol gratuït; 20 tenien dues cambres, 53 en tenien tres, 82 en tenien quatre i 230 en tenien cinc o més. 276 habitatges disposaven pel capbaix d'una plaça de pàrquing. A 191 habitatges hi havia un automòbil i a 127 n'hi havia dos o més.

Piràmide de població
La piràmide de població per edats i sexe el 2009 era:

Economia
El 2007 la població en edat de treballar era de 519 persones, 366 eren actives i 153 eren inactives. De les 366 persones actives 306 estaven ocupades (175 homes i 131 dones) i 60 estaven aturades (23 homes i 37 dones). De les 153 persones inactives 58 estaven jubilades, 41 estaven estudiant i 54 estaven classificades com a «altres inactius».

Ingressos
El 2009 a Celles-sur-Plaine hi havia 382 unitats fiscals que integraven 871,5 persones, la mediana anual d'ingressos fiscals per persona era de 15.143 €.

Activitats econòmiques
Dels 24 establiments que hi havia el 2007, 1 era d'una empresa de fabricació d'elements pel transport, 1 d'una empresa de fabricació d'altres productes industrials, 4 d'empreses de construcció, 7 d'empreses de comerç i reparació d'automòbils, 1 d'una empresa de transport, 2 d'empreses d'hostatgeria i restauració, 1 d'una empresa de serveis, 5 d'entitats de l'administració pública i 2 d'empreses classificades com a «altres activitats de serveis».

Dels 9 establiments de servei als particulars que hi havia el 2009, 1 era una oficina de correu, 1 taller de reparació d'automòbils i de material agrícola, 1 guixaire pintor, 1 fusteria, 3 lampisteries, 1 perruqueria i 1 restaurant.

Dels 2 establiments comercials que hi havia el 2009, 1 era una botiga de més de 120 m² i 1 una fleca.

Equipaments sanitaris i escolars
L'únic equipament sanitari que hi havia el 2009 era una farmàcia.

El 2009 hi havia una escola elemental.

Poblacions més properes
El següent diagrama mostra les poblacions més properes.

Referències

 Résumé statistique Fitxa resum de dades estadístiques de Celles-sur-Plaine a l'INSEE.
 Évolution et structure de la population Fitxa amb el detall de dades de Celles-sur-Plaine a l'INSEE
 France par commune  Dades detallades de tots els municipis de França accessibles a partir del mapa.

Municipis dels Vosges

redpajama_set_name: RedPajamaWikipedia


text:
Q: heroku deploy failed using nodejs I am trying to push local files to heroku and getting below error.
I have my code in github
Can someone help me in this. Thanks
$ heroku buildpacks:set heroku/nodejs
Buildpack set. Next release on haz will use heroku/nodejs.
Run git push heroku master to create a new release using this buildpack.
$ git push heroku master
Counting objects: 693, done.
Delta compression using up to 8 threads.
Compressing objects: 100% (500/500), done.
Writing objects: 100% (693/693), 63.10 MiB | 2.54 MiB/s, done.
Total 693 (delta 220), reused 639 (delta 171)
remote: Compressing source files... done.
remote: Building source:
remote:
remote: -----> Using set buildpack heroku/nodejs
remote:
remote:  !     Push rejected, failed to detect set buildpack heroku/nodejs
remote: More info: https://devcenter.heroku.com/articles/buildpacks#detection-failure
remote:
remote: Verifying deploy....
remote:
remote: !   Push rejected to haz.
remote:
To https://git.heroku.com/haz.git
 ! [remote rejected] master -> master (pre-receive hook declined)
error: failed to push some refs to 'https://git.heroku.com/haz.git'
$

for package.json
{
  "name": "haz",
  "version": "1.0.0",
  "description": "Hazzir: An Ionic project",
  "private": "true",
  "dependencies": {
    "express": "^4.13.3"
  },
  "main": "serve.js",
  "scripts": {
    "start": "node serve.js",
    "postinstall": "bower install && grunt build",
    "test": "echo \"Error: no test specified\" && exit 1"
  },
  "engines": {
      "node": "4.1.2",
      "npm": "3.4.0"
    },

  "keywords": [
    "Haz",
    "product"
  ],
  "author": "Asim Khan",
  "license": "ISC",
  "bugs": {
    "url": "https://github.com/asimkh/apps/issues"
  },
  "homepage": "https://github.com/asimkh/apps#readme"
}

I tested locally, app is runnning at port 5000 using express
var express = require('express'),
    app = express();

app.use(express.static('www'));

// CORS (Cross-Origin Resource Sharing) headers to support Cross-site HTTP requests
app.all('*', function(req, res, next) {
    res.header("Access-Control-Allow-Origin", "*");
    res.header("Access-Control-Allow-Headers", "X-Requested-With");
    next();
});

// API Routes
// app.get('/blah', routeHandler);

app.set('port', process.env.PORT || 5000);

app.listen(app.get('port'), function () {
    console.log('Express server listening on port ' + app.get('port'));
});

A: I see a few issues here
1) You need a procfile - https://devcenter.heroku.com/articles/getting-started-with-nodejs#define-a-procfile
2)
You don't specify the engines in your package.json
"engines": {
  "node": "0.10.x"
},

See here- https://discussion.heroku.com/t/the-official-node-js-buildpack-is-going-on-a-diet/100
3) You have comitted your node_modules directory to git. You should be able to download packages using npm install (that's what your packages.json file is for)
Delete this directory, and commit the delete to git.
Then, create a .gitignore file. Add the following line to your .gitignore file:
node_modules

commit this .gitignore file to your repository.
Git will now ignore your node_modules_ directory

A: Try: git add -f package.json
It worked for me.

A: git init and then readding heroku remote worked
heroku git:remote -a yourappname

redpajama_set_name: RedPajamaStackExchange


Christmas in Brisbane just got better with the New Biggest Water Park for corporate Christmas parties being held in the first two weeks of December!
Choose the party that suits you and your staff and share the cost and the fun with other companies. All you need to do is get to the Rocklea Show Grounds on your booked day. For a small outlay per person we are offering a fully catered Christmas/Xmas party, complete with the following catering, Water Slides and pools with entertainers. Whether you employ 3 staff or 300 staff, we can offer you a hassle free party with all the fun of a Summer Event! We have limited numbers per event so be quick book your place.
We have removed the stress and time out of organising an affordable End of Year Staff Christmas Party in Brisbane by dealing with the one person to book this in.
Lifeguards and staff to run event!
For as little as $35 per person, you can add a sit down meal, including dessert, in a private room, or General Public under covered area.
Available ON-THE-DAY CASH PURCHASES AND ORDER IN ADVANCE!
A variety of hot food and drinks are available for purchase by individuals including burgers, and chips, chicken strips, Dagwood Dogs and more. Click on the tabs below to view individual product prices.
Staff alcoholic drink coupons can be purchase prior to the day, and also on the day, by the company or individuals. PLUS Bar Tabs can be arranged — staff are presented with an armband to wear that represents your company, and you will be kept informed of the running balance.


text:
LANCE KING Reprograms From The Brink Of Death – "I Only Had Five Minutes"
Nick Balazs
feature heavy metal lance king
It's been a tumultuous eight years for Lance King. The former Pyramaze singer released his solo debut, A Moment In Chiros, in 2011 and was planning to release another album in '13-'14, but then an onslaught of health issues plagued him and his family. "I started writing again in 2013 and '14 for a new album, and I had contacted Jacob Hansen (producer) and got some ideas in the pot, and then a lot of stuff started happening in my life."
From his mother contracting breast cancer and suffering a heart attack to his wife having colon cancer and his mother-in-law having dementia that threw his home life upside down; King was dealt a hand difficult to cope with and he drew a card that would catapult his urgency to record another album.
"I have trained in martial arts for the last 25 years," reveals the singer. "I needed a knee surgery to clean up my meniscus, no big deal. The doctor didn't really give me any feedback on or what I should look for (when in recovery), but I was good to go so whatever. I went home and did what he said, which was to keep it up, even though I was walking around pretty good after surgery, and I'm like 'man, you know, no big deal.'"
Recovery seemed normal until a blood clot developed in his leg. After being given a prescription for blood thinners, the following morning, he was on the eve of death. "I wake up in the morning and I can't breathe; what had happened was the blood clot broke up in my leg and went into my lungs and it gave me a triple pulmonary embolism," recalls the singer. "So not just one, but three different blockages and my lungs, right? So I'm sitting there every time I breathe, my body is automatically having, what I would call a chain reaction. So every time you'd breathe it would hurt so much that your body contraction doesn't allow you to breathe. I can understand why 75% of people die from these things!"
His martial arts training helped him remain calm through this tortuous ordeal. "I was comfortable in a space where it was hard to breathe. I could remain relaxed and not start freaking out, which I imagine a lot of people would do it, would they hyperventilate and then they would just go unconscious and they die," he confirms.
King was rushed to the hospital by his wife that was thankfully close by because the medical personnel told him he was on the brink of death. "The ER people said I had about five minutes left, I would've been gone," says King. After three days in the hospital and a long recovery process, the impetus to record a new album came and it led to ReProgram, recently released through his label Nightmare Records.
"All of a sudden you realize, 'oh yeah, I could have been gone yesterday.' Was I done doing what I wanted to do?" King reflects. "What is it that I want to do now? I want to make music. I don't want to run a record company (laughs). I want to make music."
Not only is there this new album King wrote, but he in fact created a whole new genre, which King calls "celestial metal." So what is celestial metal?
The singer explains, "With my first album, I was trying to look for a definition in my music that, you know, because prog/power or progressive now or all this stuff has been so overused and it's not very definitive anymore. It's kind of cliché almost. When it was cool, everybody started calling music progressive and they weren't a lot of times, so I'm kind of like, well, alright, what is my music? I tried to define it in my own life, figure out what it was because I'd like to set it apart. I would like to create some curiosity and to be accurate with a description. And that's why I called it 'celestial metal'. I mean, it could've been mystical metal, which sounds a little corny (laughs). But it kind of fits the themes and the philosophy that I put in my lyrics and evokes a little question in people's minds like, 'what the heck is that? Let's check that out'."
King utilized the talents of his former Pyramaze bandmates as well as members of Darkwater and Annihilator bassist Rich Hinks to play and help write the record.
"It was very organized with the first album," says the singer. "This one I just contacted a bunch of people individually. I said, 'hey you want to make songs?' And fortunately out of maybe the six that I asked to have, four had the time and the interest to do it and a couple of them were new people that I hadn't worked with before, but I really dug what they did and their talents."
The music all ties nicely into this prog…I mean celestial metal affair that sees King take lyrical criticisms against modern day society, social media, mainstream media, and the overuse of technology. With him taking this lyrical approaches however, the music doesn't suffer for it.
In "Stand Your Ground" for instance, he asks, "What's the use in speaking truth when everyone believes the lies?"
The celestial man explains, "I guess my statement is why fight with everyone when you think you know the truth of something because there's a lot of misleading information that a lot of people believe on all kinds of issues. So what's the point in trying to bash people over the heads and be angry about it instead? Nobody's going to believe the truth because they're so caught up in the programming of the norm going with the flow, you know what I mean? You don't have to go with the flow."
A very interesting perspective, and then in the catchy, simply titled "Technology", King laments in the song, "Never thought I'd see the day when things happened so quickly, getting kind of bored, there's nothing really hard."
When asking him about this, he turns the tables and asks how I interpret it. I told him that I saw it as him lamenting that technology has made everything so easy and that people don't have to work for anything and that it breeds laziness.
Approving of the answer, King states, "Exactly. I like to be working toward things like goal setting and working towards your passions and to bring them to fruition. I feel like a lot of people are so satisfied with the buzz they get from social media and connecting with so many people and being able to just broadcast and get a response and they get this tick out of it that they forget that there's the other things to do, like their passions, and to actually see results and goal setting and getting to those points. Technology makes everything easier, but there's a learning curve with that too, so you can get sucked into constantly learning the technology and not actually doing what you want to do, you know?"
The trembling "Wide Open" takes a stand at today's American political climate. King relates, "If you look at my artwork for both covers that I put on the limited edition of the regular one, they both kind of scream loudly. There's a duality there and you need to find balance in the middle. You don't need to buy everything on a certain platform, there's a lot of other things going on you might not agree with."
The record than finishes with the epic "Mind At War" that serves as King's triumph from darkness and features an interesting monologue in the middle. "It's a eulogy," he explains. "I look at that song kind of like a chunk of my life where I was in a pretty dark place, you know, some of it was really beautiful there. Then in the middle I kind had an epiphany and I would liken that to my near death experience. And then you'll notice that the tone of the song changes in the last part of it. It feels lighter, it feels more joyous. I wouldn't say it's completely from darkness to light. I'd say it's from dim to neutral."
From dim to neutral it is as King went through hell, but you can hear it in his voice as he is calm, reflective, and at peace with what's happened as he related these hardships from his Minnesota home.
If there's a theme to the album, it's simple. He states, "Don't worry about running with the herd. You don't have to be like everybody else."
With his health problems affecting his lungs, did he have any trouble singing again? Was he worried that he wouldn't be the same?
"Mostly it had nothing to do with my health problems and everything to do with the fact that I hadn't sang very much in the last eight years", he reveals. "So the voice was not strong, and I forgotten some simple things like you got to be super hydrated to get the vocal abilities to do everything that they want to do. Because I'm a huge coffee lover and I drink coffee all day, it dehydrates you so you get to certain ranges. It's like, 'man, I can't get past this.' (laughs). So I had to relearn some things and practice. But fortunately during the studio time, that's not really a big deal because it's my house."
On the live front, King has nothing planned at the moment, but is putting something together. "I actually have been talking to a bunch of local musicians that I enjoy playing with over the last long periods of time that I think can actually reproduce the songs and we're getting together, we're going to start rehearsing and for the live show and get out there and do some; I don't know to what extent it's going to be yet because I haven't really broadcast that I want to do that. I haven't talked to any promoters yet or car I guess I wanted to get a feel what it's going to be first before I get too excited and put the cart before the horse."
The best part is King is rejuvenated with a new sense of life and isn't going to wait to start making new music for a third album.
"I'm not waiting, I am going to start writing immediately!"

redpajama_set_name: RedPajamaCommonCrawl


for the 2017 kanthari leadership training program, we received more than 300 applications from all over the world. Out of these 24 participants were selected who started the course in the beginning of May. They come from 12 different countries which are: Cameroon, Ethiopia, India, Kenya, Latvia, Nigeria, Serbia, Sri Lanka, Tanzania, Thailand, Uganda and Zimbabwe. The start was very vibrant and it was inspiring to see how over the past few months, they have become one group that very well understands the importance of peer support and peer-learning.
Act one of the Journey in Five Acts unique kanthari Curriculum started very off very well and currently is moving ahead in great speed.
Yurt-on-wheels, founded by kanthari John Peter is an organisation that works with Gypsy community children, called Narikuravar. Currently 20 children, who now also attend mainstream school, are coming to the center and 10 parents received livelihood support for their families. Learn more about the great work of Yurt on Wheels at: Yurton Wheels Foundaton.
Rajendra lives a life with certain challenges. Rajendra is blind and comes from Nepal. He is the founder of REAL NEPAL. Rajendra has a vision to improve the lives of the blind in his country.. Watch his dream speech to learn more about his plans for a better tomorrow!
You can watch all dream speeches from the 2016 kanthari graduates via links on this webpage.
"Imagine if trees gave off Wifi signals, we would be planting so many trees and we'd probably save the planet too. Too bad they only produce the oxygen we breathe." – Tarun Sarathe – First GREEN TECHNOLOGY CENTRE inaugurated in Kerala. Dr.T.N.Seema, Ex MP, Vice Chair Person, Haritha Keralam Mission, Govt of Kerala inaugurated the first GTC at kanthari. The Green Technology Centre is an institutional set up to provide hands on experience, knowledge and support for implementing and popularizing green technologies. The first GTC was set up by Shyam Kumar of Innovation eXperience in partnership with kanthari. During the inauguration the speakers emphasized the importance of a mindset shift towards the production and use of water, food, energy, housing and social welfare.
On the 21st of April, the 360 Degree Model United Nations Assembly 2017 was inaugurated at the kanthari campus. Honorable Guest was Dr Shashi Tharoor who shared his experiences with the UN with students who came from all over India. Sabriye Tenberken, co-founder of #kanthari welcomed the participants and wished them a fruitful model Assembly.
– SPECIAL VISITOR at kanthari: Parvathi, a 38 year old lady that doesn't live too far away enjoyed her stay at the kanthari campus and all our participants and our colleagues enjoyed meeting her! Many had never met a lady of this size and shape! Parvathi is an Elephant!
– For a special week during which the participants at kanthari focused on development / environment and climate change, we had a special visitor. Dr Nandu, a climate change expert brought a lot of insight and expertise! Several sessions and debates led to a better understanding how all different aspects of life are inter-dependent and how they affect our world.
– The opening of the Green Technology Park at the kanthari campus was covered in the Times of India.
– Rahel Zageye from Ethiopia, Sherin Noordheen from India and Sabriye Tenberken represented kanthari in an interview on Trivandrum's Club FM radio station.
– On the 14th of June 2017, Paul Kronenberg spoke at the Alanus University for Arts and Social Sciences in Bonn.
– A presentation about kanthari was given by Paul Kronenberg at the Europaschool in Bornheim.
– On the 25th of June 2017, Paul Kronenberg spoke and screened the documentary film "kanthari – change from within" was presented at the Indienweek in Cologne.
On behalf of all of our participants and colleagues, we say THANK YOU for your ongoing interest and support. Also we thank you for sharing this newsletter within your network.
We wish you a spicy and wonderful summer.


text:
Pseudomimetis is een geslacht van vlinders van de familie spanners (Geometridae), uit de onderfamilie Larentiinae.

Soorten
P. picta Warren, 1901
P. sylvatica Turner, 1922
P. vailima Prout, 1958


text:
Home » 90 Day Fiance: [SHOCKING] Are Kara Bass & Guillermo Roger Married? Kara's Friend Reveals!
90 Day Fiance: [SHOCKING] Are Kara Bass & Guillermo Roger Married? Kara's Friend Reveals!
By Duskey April 20, 2022
The 90 Day Fiance couple Kara and Guillermo's relationship has been over many turmoils. Guillermo has had some issues with Kara's past. More so, they had many other things to need to be taken care of. However, Kara fell unexpectedly for the handsome and cute waiter she had never thought of. Meanwhile, speculations have been going on that the couple is already married. So, tune in to find out.
90 Day Fiance: Kara Bass's Friend Might Have REVEALED The Biggest Spoiler About The New Couple!
Kara Bass and Guillermo Rojer have faced many issues before entirely getting together. First, the couple lived separately for nine months due to Guillermo's expired passport. Meanwhile, the 90 Day Fiance Season 9 preview shows that Guillermo will have issues with Kara's actions in America. In fact, the trailer shows that the celeb will bring up Kara's past and accuse her of cheating on her exes.
Though, as of now, fans believe that the couple is dating and engaged. But, according to the latest revelation, it isn't so. Turns out one of Kara's friends filled in tea for the followers. Kara's friend posted a picture after a watching party of the reality TV debut. In the caption, she mentioned that she had a lot of fun. Then added, "@90dayfiancé's premier of Kara and her husbands' journey of their engagement to wedding."
Hence the fans have speculated that Kara and Guillermo have already tied the knot. Fans took to the comment section to assume that the husband's passport issues have been resolved as well. Therefore, he is now able to fly to America with ease. Meanwhile, some other fans feel it's all part of the show. And the fight between the couple is scripted. Kara and Guillermo's storyline has already been leaked as the fans know what to expect.
90 Day Fiance: Will Kara And Guillermo End Up Leaving Each Other Due To Age Difference?
The 90 Day fiance couple Kara and Guillermo have made headlines. And the fans love the new lovebirds on the show. Guillermo's innocent and cute looks have the fans lured in more specifically. Turns out, the 21-year-old has also been a protestor when Venezuela was going through turmoil. One of the fans wrote, "A protester? With that face? Who isn't afraid to be a little cheesy?"
Meanwhile, some people feel Guillermo is too young for Kara. And as he looks way younger than her, fans feel it will be a concern for Kara in the future. Likewise, another feels that Kara would try to mother Guillermo and might try to rule over him. How do you like the couple? Let us know in the comments below. And stay tuned to TV Seasons & Spoilers.


text:
January 6, 2021 by Matthew Bennett
Spacebase Startopia enters early access on Xbox
Space station management sim Spacebase Startopia is now available on Xbox Series X/S and Xbox One via Xbox Game Preview, developer Realmforge Studios and publisher Kalypso Media have announced.
The game puts players in charge of a space station filled with a colorful cast of aliens and tasks them with creating the "most impressive trade and tourism operation that infinite space has ever seen." In addition to managing their own station, players will need to keep opposing stations in check through economic sabotage and attacks with mechs.
There's also a Battle Mode and support for both competitive and cooperative multiplayer for up to four players.
Spacebase Startopia is scheduled to officially launch in spring 2021 for Xbox Series X/S, PlayStation 5, Xbox One, PlayStation 4, Nintendo Switch, and PC.
Post tagsKalypso MediaRealmforge StudiosSpacebase Startopia


Ruby Loftus screwing a Breech-ring, 1943, Dame Laura Knight RA. Oil on canvas. Ruby Loftus was something of a celebrity during the Second World War, having mastered a skill in a matter of months that usually took a factory worker several years. As well as being an iconic image of female accomplishment, this image was used heavily to promote female involvement on the home front.
Witness: Women War Artists is the third in the Witness series – which revolve around first person experiences of subjects – to take place at the Imperial War Museum's sister gallery, Imperial War Museum North. It's located in Daniel Liebeskind's specially commissioned building in Manchester.
German photographer Frauke Eigen, whose Kosovo series is a recent addition to the Imperial War Museum's collection.
The show is the first time in 50 years that the Imperial War Museum has held an exhibition devoted to female war artists. The works cover the period from the First World War right up until the conflicts in Kosovo and Iraq and draw on its own collection.
It should be said that from the beginning the aim of the exhibition is to raise the profile of these artists, rather than to make a case for the challenges they faced as female war artists. Frontline warfare, especially the muddy, bloody, unrelenting warfare of the First and Second World Wars, is, unsurprisingly, male-dominated. There are reasons that the output of official women war artists from the first half of the twentieth century is both less than that of their male counterparts, and narrower in scope.
That said, the curators of this exhibition have taken some delight in blowing apart one or two myths. The show opens up with a section titled 'In Action' for example, which features a series of works by female artists with direct experience of working on the front line. Here are the stark and apocalyptic works by Olive Mudie-Cooke, who, having been deployed to France in 1916 with the First Aid Nursing Yeomanry, was commissioned to produce a body of work in 1919 by the Imperial War Museum's Women's Work Sub-Committee, which had been set up in order to produce a record of the important work done by women during the First World War. The section also features the sketches of the 1982 Falklands conflict by Linda Kitson, who was the first female artist to be embedded with the army while it was deployed on active service.
There are also one or two well-known figures here who, like their male counterparts, were pioneers of their profession. The most notable example is Dame Laura Knight, who was already a fully-established artist in her own right by 1938, having become the first female artist to be elected as a Royal Academician for over 160 years. As well as her iconic portraits of war heroines, such as Ruby Loftus and Corporal Daphne Pearson, whose images were commissioned in part to encourage women into the war effort, the curators have also included her unusual and poignant1946 depiction of the Nuremberg trial.
In the latter years of the 20th century, the museum committee responsible for commissioning war artists was resurrected, though the decision was taken to move away from purely archival material to more avant garde, reactionary art works . A fairly recent addition to the collection is Frauke Eigen's Frundstücke Kosovo 2000 series, which is included within the section called 'Parting and Loss'. Taking the form of single, enlarged photographs of half-decayed clothes removed for identification purposes from bodies from a mass grave in Kosovo, the stark images of the clothes, rather than a documentary-style photograph of the bodies themselves, is a haunting evocation of the individuals to whom they once belonged.


Mr. Quail sends news from home and it's not always good. William Tweed, who Jake and Jerrod personally knew to be the leader of a gang of thieves and extortionists (the man who forced Jake to leave Brooklyn), has become very powerful. He is now in the State Legislature and members of his gang have key positions in Brooklyn. Tweed's political corruption has allowed his gang's crime to go unchecked since the police are also on his payroll.
Jake and Jerrod decide to sell their mine and travel home on the newly constructed Transcontinental Railroad. The brothers have evidence that should send Tweed to prison for the rest of his life, but Tweed has heard they are coming and plans to stop them from ever returning to Brooklyn. A contest of wits has begun with "Boss" Tweed, a man with political power, wealth and a gang of criminals at his command.