\section{Introduction }
Today it is well known that strong correlated systems
in condensed matter can be sucessfully described with the help of non-relativistic holography, for review see for example \cite{Hartnoll:2016apf}. This duality is based on the idea that the strongly coupled theory on the boundary
can be described by string theory in the bulk. Further, when the curvature of the space-time is small we can use the
classical gravity instead full string theory machinery. In case of non-relativistic holography the situation is even more interesting since we have basically two possibilities: Either we  use Einstein metric with non-relativistic isometries
\cite{Son:2008ye,Balasubramanian:2008dm,Herzog:2008wg} or we
introduce non-relativivistic gravities in the bulk
\cite{Son:2013rqa,Janiszewski:2012nb}, like Newton-Cartan gravity \cite{Cartan:1923zea}
\footnote{For some recent works, see
	\cite{Bergshoeff:2017dqq,Bergshoeff:2016lwr,Hartong:2016yrf,Hartong:2015zia,
		Afshar:2015aku,	Bergshoeff:2015ija,Bergshoeff:2015uaa,Bergshoeff:2014uea,
		Andringa:2010it}.}
or Ho\v{r}ava gravity \cite{Horava:2009uw}. Then it is certainly very interesting
to study matter coupled to non-relativistic gravity. We can either study
field theories on non-relativistic background as in
\cite{Bergshoeff:2015sic,Bagchi:2015qcw,Festuccia:2016caf,Geracie:2015dea,Jensen:2014aia,Hartong:2014pma} or particles \cite{Kuchar:1980tw,Barducci:2017mse,Bergshoeff:2014gja,Kluson:2017pzr}
or even higher dimensional objects, as for example non-relativistic strings
and p-branes \cite{Andringa:2012uz,Harmark:2017rpg}.

In this work we would like to focus on the  canonical formulation of non-relativistic
string and p-brane in Newton-Cartan background. The starting point of our analysis is
the relativistic string in general background that couples to NSNS two form.
Then we use the limiting procedure that was proposed in
\cite{Bergshoeff:2015sic} and try to find corresponding string action.
Note that this is different limiting procedure than in case of the non-relativistic string in flat background where the non-relativistic limit is performed
on coordinates \cite{Gomis:2000bd,Gomis:2005bj,Gomis:2005pg}
\footnote{For recent work, see for example \cite{Batlle:2016iel,Kluson:2017djw,Kluson:2017ufb,Kluson:2017vwp}.}.
It is important to stress that if we apply this limiting procedure that leads to corank-1 spatial metric and rank one temporal metric of Newton-Cartan gravity to the case of the string action
 we find that there is no way how to ensure that this action is finite. In order
 to resolve this problem we have to select two flat target space   longitudial directions exactly in the same way as in \cite{Andringa:2012uz}. Then we  propose such an ansatz for NSNS two form field that is constructed with
 the help of the fields that define Newton-Cartan geometry and where the divergent contribution from the coupling to NSNS two form exactly cancels the divergent
 contribution coming from Nambu-Gotto part of the action. As a result we obtain
 an action for the string in Newton-Cartan background that was proposed in
 \cite{Andringa:2012uz} using different procedure.
 As the next step we proceed to the canonical formulation of this theory. Then however
 we encounter an obstacle in the form that we are not able to invert relation
 between conjugate momenta and velocity in case of non-zero gauge field $m_\mu^{ \ a}$ whose explicit definition will be given in the next section. For that reason
 we restrict ourselves to the case of the zero gauge field keeping in mind that
 the case of on-zero gauge field deserves further study. Then we find  Hamiltonian for this non-relativistic string that is linear combination of two first class constraints
 which is the manifestation of the fact that two dimensional string action
 is invariant under world-sheet diffeomorphism.  As the next step  we generalize this analysis to the case of $p-$brane.
 We firstly determine well defined action for non-relativistic p-brane
 when we consider specific form of the background $p+1$ form that couples to the world-volume of p-brane. Then we introduce an equivalent form of $p-$brane action
 that allows us to consider canonical analysis of this theory. Finally we determine constraint structure of this theory and we show that there are $p+1$ first class constraints, $p-$ spatial diffeomorphism constraints and one Hamiltonian constraint. We again show that these constraints are the first class constraints.

 This paper is organized as follows. In the next section (\ref{second}) we determine the form of non-relativistic string in Newton-Cartan background and perform its Hamiltonian analysis. Then in section (\ref{third}) we generalize this analysis to the case of
 $p-$brane. Finally in conclusion (\ref{fourth}) we outline our results and suggest possible extension of this work.

\section{Review of the Non-relativistic Limit for Nambu-Gotto String}
\label{second}
In this section we derive non-relativistic form of the string action
in Newton-Cartan background using the limiting procedure developed in
\cite{Bergshoeff:2015uaa}. We start with
the Nambu-Gotto form of the action in the general background
\begin{equation}\label{funstringact}
S=-\tilde{\tau}_F \int d\tau d\sigma\sqrt{-\det (E_\mu^{ \ A}E_\nu^{ \ B}\eta_{AB}
	\partial_\alpha x^\mu\partial_\beta x^\nu)}+\tilde{\tau}
\int d\tau d\sigma B_{\mu\nu}\partial_\tau x^\mu \partial_\sigma x^\nu \ ,
\end{equation}
where $E_\mu^{ \ A}$ is $d-$dimensional vierbein so that the metric components have the form
\begin{equation}
G_{\mu\nu}=E_\mu^{ \ A}E_\nu^{ \ B}\eta_{AB} \ , \eta_{AB}=\mathrm{diag}(-1,1,\dots,1)
\end{equation}
Note that the metric inverse $G^{\mu\nu}$ is defined with the help of the inverse vierbein $E^\mu_{ \ B}$ that obeys the relation
\begin{equation}
E_\mu^{ \ A}E^\mu_{ \ B}=\delta^A_{ \ B} \ ,  \quad E_\mu^{ \ A}E^\nu_{ \ A}=
\delta_\mu^{ \ \nu} \ .
\end{equation}
Further, $B_{\mu\nu}$ is NSNS two form field that plays the crucial role in the
limiting procedure. Finally $x^\mu \ ,\mu=0,\dots,d-1$ are embedding coordinates of the string where the two dimensional world-sheet is parameterised by  $\sigma^\alpha\equiv(\tau,\sigma)$.

Let us now proceed to the brief description of the procedure that
leads to Newton-Cartan background from general background, for more detailed discussion, see the original paper \cite{Bergshoeff:2015uaa}. The starting point
is following ansatz for $d-$dimensional vierbein \cite{Bergshoeff:2015uaa}
\begin{equation}\label{vierbeinomega}
E_\mu^{ \ 0}=\omega\tau_\mu+\frac{1}{2\omega}m_\mu \ , \quad  E_\mu^{ \ a'}= e_\mu^{ \ a'} \ ,
\end{equation}
where $a'=1,\dots,d-1$ and where $\omega$ is free parameter which goes to infinity in the Newton-Cartan limit. Note that in this case the metric has the form
\begin{eqnarray}
& &G_{\mu\nu}=E_\mu^{ \ A}E_\nu^{ \ B}\eta_{AB}
=-\omega^2 \tau_\mu \tau_\nu-\frac{1}{2}\tau_\mu m_\nu-\frac{1}{2}\tau_\nu m_\mu+h_{\mu\nu}+\frac{1}{4\omega^2}m_\mu m_\nu=\nonumber \\
&=&-\omega^2\tau_\mu\tau_\nu
+\bar{h}_{\mu\nu}+\frac{1}{4\omega^2}m_\mu m_\nu \ , \quad
\bar{h}_{\mu\nu}=h_{\mu\nu}-\frac{1}{2}\tau_\mu m_\nu-\frac{1}{2}\tau_\nu m_\mu \ , \quad
h_{\mu\nu}=e_\mu^{ \ a'}e_\nu^{ \ b'}\delta_{a'b'} \ . \nonumber \\
\end{eqnarray}
Inserting this metric into the Nambu-Gotto action and performing expansion
with respect to $\omega$ we obtain
\begin{equation}
S=-\tilde{\tau}_F \omega^2\int d\tau d\sigma
\sqrt{-\det\mathbf{a}} -\frac{\tilde{\tau}_F}{2}\int d\tau d\sigma
\sqrt{-\det\mathbf{a}}\mathbf{a}^{\alpha\beta}\bar{h}_{\alpha\beta} \ ,
\end{equation}
where we defined
\begin{equation}\label{actnaive}
\mathbf{a}_{\alpha\beta}=\tau_{\mu\nu}\partial_\alpha x^\mu \partial_\beta x^\nu \ ,
\quad \mathbf{a}^{\alpha\beta}\mathbf{a}_{\beta\gamma}=\delta^\alpha_\gamma \ , \quad
\bar{h}_{\alpha\beta}=\bar{h}_{\mu\nu}\partial_\alpha x^\mu \partial_\beta x^\nu \ .
\end{equation}
We also used the fact that $\mathbf{a}_{\alpha\beta}$ is $2\times 2$ matrix that is non-singular.
Apparently we see from (\ref{actnaive}) that there is a term proportional to $\omega^2$ that cannot be made
finite by rescaling of $\tilde{\tau}_F$. In case of the string in the flat non-relativistic
background such a term
can be canceled with the suitable form of the background $NSNS$ two form. Further,
, this two form field should be build from the fields  that define  Newton-Cartan theory as $m_\mu,\tau_\nu$. However it turns out that it is not possible to find such a  NSNS two form due to the fact that it has to be antisymmetryc in space-time indicies. In order to find solution of this problem we implement  the generalization of the Newton-Cartan gravity that was introduced in
\cite{Andringa:2012uz}. Explicitly,  we split the target-space indices $A$ into $A=(a',a)$ where now $a=0,1$ label longitudial  and $a'=2,\dots,d-1$ correspond to transverse directions. Then we introduce $\tau_\mu^{ \ a}$ so that we write
\begin{equation}
  \tau_{\mu\nu}=\tau_\mu^{ \ a}\tau_\nu^{ \ b}
\eta_{ab} \ , \quad  a,b=0,1 \ , \quad \eta_{ab}=\mathrm{diag}(-1,1) \ .
\end{equation}
In the same way we introduce vierbein $e_\mu^{ \ a'}, a'=2,\dots,d-1$ and also  we generalize $m_\mu$ into $m_\mu^{ \ a}$. The $\tau_\mu^{ \ a}$ can be interpreted as the gauge fields of the longitudinal translations while $e_\mu^{ \ a'}$  as the gauge fields of the transverse translations
\cite{Andringa:2012uz}. Then we can also introduce their inverses with respect to their longitudinal and transverse subspaces
\begin{eqnarray}
e_\mu^{ \ a'}e^\mu_{ \ b'}=\delta^{a'}_{b'} \ ,  \quad
e_\mu^{ \ a'}e^\nu_{ \ a'}=\delta_\mu^\nu-\tau_\mu^{ \ a}
\tau^\nu_{ \ a} \ , \quad \tau^\mu_{ \ a}\tau_\mu^{ \ b}=\delta_a^b \ , \quad
\tau^\mu_{ \ a}e_\mu^{ \ a'}=0 \ , \quad
\tau_\mu^{ \ a}e^\mu_{ \ a'}=0 \ . \nonumber \\
\end{eqnarray}
Performing this generalization implies following form
of the vierbein
\begin{equation}
E_\mu^{ \ a}=\omega \tau_\mu^{ \ a}+\frac{1}{2\omega}m_\mu^{ \ a} \ , \quad
E_\mu^{ \ a'} =e_\mu^{ \ a'}
\end{equation}
so that we find following form of the metric
\begin{eqnarray}
G_{\mu\nu}&=&E_\mu^{ \ a}E_\nu^{ \ b}\eta_{ab}+E_\mu^{ \ a'}E_\nu^{ \ b'}\delta_{a'b'}
=\nonumber \\
&=&\omega^2 \tau_{\mu\nu}+h_{\mu\nu}+\frac{1}{2}\tau_\mu^{ \ a}m_\nu^{ \ b}\eta_{ab}+
\frac{1}{2}m_\mu^{ \ a}\tau_\nu^{ \ b}\eta_{ab}+\frac{1}{4\omega^2}m_\mu^{ \ a}m_\nu^{ \ b}
\eta_{ab} \ .  \nonumber \\
\end{eqnarray}
It was shown  in \cite{Bergshoeff:2015uaa} that in order to find the right form
of the particle action in Newton-Cartan background we should consider following
ansatz for the background gauge field $A_\mu=\omega \tau_\mu-\frac{1}{2\omega}m_\mu$.
In order to find correct form of the action for the string in Newton-Cartan background
we propose analogue form of the NSNS two form
\begin{eqnarray}
B_{\mu\nu}&=&\left(\omega\tau_\mu^{ \ a}-\frac{1}{2\omega}m_\mu^{ \ a}\right)\left( \tau_\nu^{ \ b}-\frac{1}{2\omega}m_\nu^{ \ b}\right)\epsilon_{ab}=
\nonumber \\
&=&\omega^2\tau_\mu^{ \ a}\tau_\nu^{ \ b}\epsilon_{ab}-
\frac{1}{2\omega}(m_\mu^{ \ a}\tau_{\nu}^{ \ b}+
\tau_\mu^{\  a}m_\nu^{ \ b})\epsilon_{ab}+\frac{1}{4\omega^2}
m_\mu^{ \ a}m_\mu^{ \ b}\epsilon_{ab} \ , \quad
  \epsilon_{ab}=-\epsilon_{ba} \ , \quad
  \epsilon_{01}=1 \ .
\nonumber \\
\end{eqnarray}
It is important that terms proportional to $\omega^{-2}$ and $\omega^{-4}$ vanish
in the limit $\omega\rightarrow \infty$ while the divergent contribution cancel the divergence coming from Nambu-Gotto part of the action since
\begin{eqnarray}
&- &\tilde{\tau}_F\omega^2\int d^2\sigma\sqrt{-\det\mathbf{a}}+\frac{\tilde{\tau}_F}{2} \int d^2\sigma
\epsilon^{\alpha\beta}B_{\mu\nu}\partial_\alpha x^\mu
\partial_\beta x^\nu \nonumber \\
&=&
-\tilde{\tau}_F \omega^2\int d^2\sigma \det \tau_\alpha^{ \ a}+
\omega^2 \frac{\tilde{\tau}_F}{2}\int d^2\sigma \epsilon^{\alpha\beta}\epsilon_{ab}\tau_\alpha^{ \ a}\tau_\beta^{ \ b}=0 \ ,
\nonumber \\
\end{eqnarray}
where we introduced $2\times 2$ matrix $\tau_\alpha^{ \ a}\equiv \tau_\mu^{ \ a}\partial_\alpha x^\mu$ and where we used the fact that $\det \tau_\alpha^{ \ a}=
\frac{1}{2}\epsilon^{\alpha\beta}\epsilon_{ab}\tau_\alpha^{ \ a}\tau_\beta^{ \ b}$ where $\epsilon^{\alpha\beta}=-\epsilon^{\beta\alpha} \ , \epsilon^{01}=1
$ is antisymmetric symbol with upper indices.

In summary we obtain the action for non-relativistic string in Newton-Cartan background in the form
\begin{equation}\label{nonrelstringNC}
S=-\frac{\tau_F}{2}\int d^2\sigma \sqrt{-\det\mathbf{a}}\mathbf{a}^{\alpha\beta}
\bar{h}_{\alpha\beta} \ ,
\end{equation}
where $\tilde{\tau}_F=\tau_F$. Note that the action (\ref{nonrelstringNC})
was derived previously in \cite{Andringa:2012uz}
using slightly different procedure.

Our goal is to find Hamiltonian formulation of this theory. To do this we
rewrite the Lagrangian density introduced in  (\ref{nonrelstringNC}) into the form
\begin{eqnarray}\label{actionextended}
\mathcal{L}&=&\frac{1}{4\lambda^\tau}(\bar{h}_{\tau\tau}-2\lambda^\sigma
\bar{h}_{\tau\sigma}+(\lambda^\sigma)^2\bar{h}_{\sigma\sigma})-\lambda^\tau
\tau_F^2\bar{h}_{\sigma\sigma}+\nonumber \\
&+&B^\tau \left(\lambda^\tau-\frac{\sqrt{-\det\mathbf{a}}}{2\tau_F\mathbf{a}_{\sigma\sigma}}\right)
+B^\sigma \left(\lambda^\sigma-\frac{\mathbf{a}_{\tau\sigma}}{\mathbf{a}_{\sigma\sigma}}\right) \ .
\nonumber \\
\end{eqnarray}
It is easy to see an equivalence of these two Lagrangians since the
 equations of motion for $B^\tau$ and $B^\sigma$
give
\begin{equation}
\lambda^\tau=\frac{\sqrt{-\det\mathbf{a}}}{2\tau_F\mathbf{a}_{\sigma\sigma}} \ , \quad
\lambda^\sigma=\frac{\mathbf{a}_{\tau\sigma}}{\mathbf{a}_{\sigma\sigma}} \ .
\end{equation}
Inserting this result into (\ref{actionextended}) and using the fact that
\begin{eqnarray}
& &\frac{1}{\lambda^\tau}=
-2\tau_F \mathbf{a}^{\sigma\sigma}\sqrt{-\det\mathbf{a}} \ , \quad
\frac{\lambda^\sigma}{\lambda^\tau}
=2\tau_F \mathbf{a}^{\tau\sigma}
\sqrt{-\det\mathbf{a}} \ ,
\nonumber \\
& & \frac{1}{4\lambda^\tau}
((\lambda^\sigma)^2-4\tau_F^2(\lambda^\tau)^2)=
-\frac{\tau_F}{2}\mathbf{a}^{\sigma\sigma}\sqrt{-\det\mathbf{a}}
\nonumber \\
\end{eqnarray}
we find that (\ref{actionextended}) reduces into (\ref{nonrelstringNC}).
Then  from (\ref{actionextended}) we obtain conjugate momenta
\begin{eqnarray}\label{pmu}
p_\mu&=&\frac{1}{2\lambda^\tau}\bar{h}_{\mu\nu}\partial_\tau x^\nu-
\frac{\lambda^\sigma}{2\lambda^\tau}\bar{h}_{\mu\nu}\partial_\sigma x^\nu
-B^\tau\frac{1}{2\tau_F\mathbf{a}_{\sigma\sigma}}\tau_{\mu\nu}\partial_\alpha x^\nu \mathbf{a}^{\alpha\tau}\sqrt{-\det\mathbf{a}}-\frac{B^\sigma}{\mathbf{a}_{\sigma\sigma}} \tau_{\mu\nu}\partial_\sigma x^\nu \ , \nonumber \\
P^\tau_B&=&\frac{\partial L}{\partial \partial_\tau B^\tau}\approx 0 \ , \quad
P^\sigma_B=\frac{\partial L}{\partial\partial_\tau B^\sigma}\approx 0
\ , \quad  P_\lambda^\tau=\frac{\partial L}{\partial \partial_\tau \lambda^\tau}\approx 0 \ , \quad  P_\lambda^\sigma=\frac{\partial L}{\partial \partial_\tau \lambda^\sigma}\approx 0
\nonumber \\
\end{eqnarray}
Now we come to the most important problem in  our analysis which is an imposibility to invert the relation between $p_\mu$ and $\partial_\tau x^\mu$  in order to express $\partial_\tau x^\mu$ using the canonical variables. The reason why we are not able to do it is in the presence of the vector field $m_\mu^{ \ a}$ so that the contraction of the metric  $\bar{h}_{\mu\nu}$  with $\tau^\mu$ is non-zero. For that reason we restrict to the simpler case when $m_\mu^{\  a }=0$. Despite of this simplification  we will see that even in this case the Hamiltonian formulation of the non-relativistic string in Newton-Cartan background is non-trivial task.
In case when $m_\mu^{ \ a}=0$ we have
 $\bar{h}_{\mu\nu}=h_{\mu\nu}$ and $\bar{h}_{\mu\nu}h^{\nu\rho}
\bar{h}_{\rho\sigma}=h_{\rho\sigma}$ and hence from (\ref{pmu}) we obtain
\begin{eqnarray}
\left(p_\mu+\frac{\lambda^\sigma}{2\lambda^\tau}h_{\mu\rho}
\partial_\sigma x^\rho\right) h^{\mu\nu}\left(p_\nu
+\frac{\lambda^\sigma}{2\lambda^\tau}h_{\nu\sigma}
\partial_\sigma x^\sigma\right)=\frac{1}{4(\lambda^\tau)^2}
\partial_\tau x^\mu h_{\mu\nu}\partial_\tau x^\nu \ .
\end{eqnarray}
On the other hand let us multiply both sides of expression
(\ref{pmu}) with
$\tau^\mu_{ \ a}\eta^{ab}\epsilon_{bc}\tau_\rho^{ \ c}\partial_\sigma x^\rho$ and we obtain
\begin{eqnarray}\label{pmumilt}
& &p_\mu \tau^\mu_{ \ a}\eta^{ab}\epsilon_{bc}\tau_\rho^{ \ c}\partial_\sigma x^\rho
=-B^\tau\frac{1}{2\tau_F \mathbf{a}_{\sigma\sigma}}\partial_\alpha x^\mu \tau_\mu^{ \ a}\epsilon_{ab}
\tau_\rho^{ \ b}\partial_\sigma x^\rho\mathbf{a}^{\alpha\tau}\sqrt{-\det\mathbf{a}}-B^\sigma\frac{1}{\mathbf{a}_{\sigma\sigma}}
\partial_\sigma x^\mu \tau_\mu^{ \ a}\epsilon_{ab}\tau_\nu^{ \ b}\partial_\sigma x^\nu\nonumber \\
\nonumber \\
&=&-B^\tau\frac{1}{2\tau_F\mathbf{a}_{\sigma\sigma}}\partial_\tau x^\mu \tau_\mu^{ \ a}\epsilon_{ab}\tau_\rho^{ \ b}\partial_\sigma x^\rho
\frac{\mathbf{a}_{\sigma\sigma}}{\det\mathbf{a}}\sqrt{-\det\mathbf{a}}=\frac{1}{2\tau_F}B^\tau\nonumber \\
\end{eqnarray}
using
\begin{equation}
\tau^\mu_{ \ a}\tau_{\mu\nu}=\tau_\nu^{ \ b}\eta_{ba}
\end{equation}
and also we used the fact that
\begin{equation}
\sqrt{-\det \mathbf{a}}=\det \tau_\alpha^{ \ a}=
\tau_\tau^a\tau_\sigma^b\epsilon_{ab} \ ,
\end{equation}
where $\tau_\alpha^{ \ a}=\tau_\mu^{ \ a}\partial_\alpha x^\mu$.
Then the relation (\ref{pmumilt})  implies following primary constraint
\begin{equation}\label{prim1}
\Gamma^\tau\equiv
2\tau_Fp_\mu \tau^\mu_{ \ a}\eta^{ab}\epsilon_{bc}\tau_\rho^{ \ c}\partial_\sigma x^\rho-B^\tau \approx 0 \ .
\end{equation}
On the other hand if we multiply the  relation (\ref{pmu}) with $\tau^{\mu\nu}\tau_{\nu\rho}\partial_\sigma x^\rho$ we obtain
\begin{eqnarray}
p_\mu \tau^{\mu\nu}\tau_{\nu\rho}\partial_\sigma x^\rho
=-
B^\tau \frac{1}{2\tau_F \mathbf{a}_{\sigma\sigma}}
\mathbf{a}_{\sigma\alpha}
\mathbf{a}^{\alpha\tau}\sqrt{-\det\mathbf{a}}-B^\sigma =
-B^\sigma \
\nonumber \\
\end{eqnarray}
and hence we obtain second primary constraint
\begin{equation}
\Gamma^\sigma\equiv p_\mu \tau^{\mu\nu}\tau_{\nu\rho}\partial_\sigma x^\rho+B^\sigma\approx 0 \ .
\end{equation}
As a result  the extended
 Hamiltonian with all primary constraints included has the form
\begin{eqnarray}
H_E&=&\int d\sigma\left(\lambda^\tau (p_\mu h^{\mu\nu}p_\nu+\tau_F^2 h_{\sigma\sigma})
 -B^\tau \lambda^\tau-B^\sigma \lambda^\sigma
+\lambda^\sigma p_\mu h^{\mu\nu}h_{\nu\rho}\partial_\sigma x^\rho+\right.
\nonumber \\
&+&\left. U_\tau \Gamma^\tau+U_\sigma \Gamma^\sigma +v
_\tau^B P^\tau_B+v_\sigma^B P_B^\sigma+v^\lambda_\tau P_\lambda^\tau+
v_\sigma^\lambda P_\lambda^\sigma\right) \ .
\nonumber\\
\end{eqnarray}
Let us now analyze the requirement of the preservation of the primary constraints
$P_\lambda^\tau\approx 0 , P_\lambda^\sigma\approx 0$. In case of $P_\lambda^\tau$ we obtain
\begin{eqnarray}
\partial_\tau P_\lambda^\tau&=&\pb{P_\lambda^\tau,H_E}=
-p_\mu h^{\mu\nu}p_\nu-\tau_F^2 h_{\sigma\sigma}+B^\tau=
\nonumber \\
&=&
-p_\mu h^{\mu\nu}p_\nu-\tau_F^2h_{\sigma\sigma}+2\tau_F p_\mu
\tau^\mu_{ \ a}\eta^{ab}\epsilon_{bc}\tau_\rho^{ \ c}\partial_\sigma x^\rho-
2\tau_F \Gamma^\tau \approx \nonumber \\
&\approx &-p_\mu h^{\mu\nu}p_\nu-\tau_F^2h_{\sigma\sigma}+2\tau_F p_\mu
\tau^\mu_{ \ a}\eta^{ab}\epsilon_{bc}\tau_\rho^{ \ c}\partial_\sigma x^\rho\equiv -\mathcal{H}_\tau
\approx 0 \nonumber \\
\end{eqnarray}
and also
\begin{equation}
\partial_\tau P_\lambda^\sigma=\pb{P_\lambda^\sigma,H}=
-p_\mu\partial_\sigma x^\mu+p_\mu \tau^{\mu\nu}\tau_{\nu\rho}\partial_\sigma x^\rho
+B^\sigma=-p_\mu\partial_\sigma x^\mu+\Gamma^\sigma\approx -p_\mu\partial_\sigma x^\mu
\equiv -\mathcal{H}_\sigma\approx 0 \ .
\end{equation}
We see that the requirement of the preservation of the primary constraints
$P_\lambda^\tau\approx 0 \ , P_\lambda^\sigma\approx 0$  implies an existence of two secondary  constraints:
\begin{eqnarray}
\mathcal{H}_\sigma=p_\mu\partial_\sigma x^\mu\approx 0 \ , \quad
\mathcal{H}_\tau=p_\mu h^{\mu\nu}p_\nu+\tau_F^2h_{\sigma\sigma}-2\tau_F p_\mu
\tau^\mu_{ \ a}\eta^{ab}e_{bc}\tau_\rho^{ \ c}\partial_\sigma x^\rho\approx 0 \ .
\nonumber \\
\end{eqnarray}
Further, since $\pb{P_B^\tau(\sigma),\Gamma^\tau(\sigma')}=\delta(\sigma-\sigma'),
\pb{P_B^\sigma(\sigma),\Gamma^\sigma(\sigma')}=-\delta(\sigma-\sigma')$ we see that these constraints are the second class constraints that can be explicitly solved for $B^\tau$ and $B^\sigma$. Then these constraints vanish strongly and hence the
Hamiltonian is linear combination of the constraints
\begin{equation}
\mathcal{H}_E=\lambda^\tau \mathcal{H}_\tau+\lambda^\sigma \mathcal{H}_\sigma+v_\tau^\lambda P_\lambda^\tau+
v_\sigma^\lambda P_\lambda^\sigma  \ .
\end{equation}
As the next step we should check that $\mathcal{H}_\tau\approx 0 \ ,\mathcal{H}_\sigma\approx 0$
are the first class constraints. To do this we introduce the smeared
forms of these constraints
\begin{eqnarray}
\mathbf{T}_\tau(N)=\int d\sigma N \mathcal{H}_\tau \ ,
\mathbf{T}_\sigma(N^\sigma)=\int d\sigma N^\sigma \mathcal{H}_\sigma \ .
\nonumber \\
\end{eqnarray}
First of all we easily find
\begin{eqnarray}
\pb{\mathbf{T}_\sigma(N^\sigma),\mathbf{T}_\sigma(M^\sigma)}=
\mathbf{T}_\sigma(N^\sigma\partial_\sigma M^\sigma-N^\sigma
\partial_\sigma M^\sigma) \ . \nonumber \\
\end{eqnarray}
In case of the Hamiltonian constraints the situation is more involved since the explicit calculation gives
\begin{eqnarray}
& & \pb{\mathbf{T}_\tau(N),\mathbf{T}_\tau(M)}=\int d\sigma
(N\partial_\sigma M-M\partial_\sigma N)4\tau_F^2p_\mu h^{
	\mu\nu}h_{\nu\rho}\partial_\sigma x^\rho
+
\nonumber \\
&+&2\int d\sigma \tau_F (N\partial_\sigma M-M\partial_\sigma N)
p_\mu V^\mu_{ \ \nu}h^{\nu\omega}p_\omega+
\nonumber \\
&+&\int d\sigma (N\partial_\sigma M-M\partial_\sigma N)
p_\rho V^\rho_{ \ \sigma}V^\sigma_{\ \omega}\partial_\sigma x^\omega+
\nonumber \\
&+&\tau_F^2\int d\sigma (N\partial_\sigma M-M\partial_\sigma N)4\tau_F^2V^\mu_{ \ \nu}\partial_\sigma x^\nu h_{\mu\rho}\partial_\sigma x^\rho\ ,
\nonumber \\
\end{eqnarray}
where we defined
\begin{equation}
V^\mu_{ \ \nu}=-2\tau_F\tau^\mu_{ \ a}\eta^{ab}\epsilon_{bc}
\tau_\nu^{\ c} \ .
\end{equation}
To proceed further we calculate
\begin{eqnarray}
& &4p_\mu h^{\mu\nu}h_{\nu\rho}\partial_\sigma x^\rho=
4\tau_F^2p_\mu \partial_\sigma x^\mu-4\tau_F^2 p_\mu\tau^{\mu\rho}\tau_{\rho\nu}
\partial_\sigma x^\nu \ ,   \quad
 \nonumber \\
& &p_\rho V^\rho_{ \ \mu}V^\mu_{ \ \nu}\partial_\sigma x^\nu=4\tau_F^2 p_\mu \tau^\mu_{ \ a}\tau^a_{ \ \nu}\partial_\sigma x^\nu=4\tau_F^2 p_\mu
\tau^{\mu\rho}\tau_{\rho\nu}\partial_\sigma x^\nu \ , \nonumber \\
& & V^\mu_{ \ \nu}\partial_\sigma x^\nu h_{\mu\rho}\partial_\sigma x^\rho=0 \ ,
\quad
p_\mu V^\mu_{ \ \nu}h^{\nu\omega}p_\omega=0 \ .
\nonumber \\
\end{eqnarray}
Collecting all these results together we finally obtain
\begin{equation}
\pb{\mathbf{T}_\tau(N),\mathbf{T}_\tau(M)}=\mathbf{T}_\sigma ((N\partial_\sigma M-M\partial_\sigma N)
4\tau_F^2 ) \ .
\end{equation}
Finally we also calculate  Poisson bracket between $\mathbf{T}_\sigma(N^\sigma)$ and $\mathbf{T}_\tau(M)$ and we find
\begin{equation}
\pb{\mathbf{T}_\sigma(N^\sigma),\mathbf{T}_\tau(M)}=
\mathbf{T}_\tau(\partial_\sigma MN^\sigma-M\partial_\sigma N^\sigma) \ .
\end{equation}
These results show that $\mathcal{H}_\tau\approx 0,\mathcal{H}_\sigma \approx 0 $ are the first class constraints and the non-relativistic string is well defined system from the Hamiltonian point of view.

Finally we also say few words about the gauge fixed theory. We showed above that the Hamiltonian and spatial diffeomorphism constraints are the first class. Standard way how to deal with such a theory is to gauge fix them. For example, we can impose the static gauge when we introduce two gauge fixing functions
\begin{equation}
\mathcal{G}^0=x^0-\tau \approx 0 \ , \quad \mathcal{G}^1=x^1-\sigma \approx 0 \ .
\end{equation}
It is easy to see that
 $\mathcal{G}^a\approx 0$ are the second class constraints together with $\mathcal{H}_\tau\approx 0 , \mathcal{H}_\sigma\approx 0$. Since then these constraints vanish strongly we can identify the
Hamiltonian density on the reduced phase space from the action principle
\begin{equation}
S=\int d\tau d\sigma (p_\mu \partial_\tau x^\mu-H)=
\int d\tau d\sigma (p_i\partial_\tau x^i+p_0)
\end{equation}
so that it is natural to identify $-p_0$ as the Hamiltonian on the reduced phase space
$H_{red}=-p_0$. The explicit form of the Hamiltonian follows from the Hamiltonian constraint that can be solved for $p_0$.  Note also that $\mathcal{H}_\sigma$ can be solved for $p_1$ as $p_1=-p_I\partial_\sigma x^I \ , I=2,\dots,d-1$.
\section{Generalization: Non-relativistic p-brane}\label{third}
In this section we perform generalization of the analysis presented
previously to  the case of non-relativistic $p-$brane.
As the first step we determine an action for non-relativistic p-brane in Newton-Cartan backgorund in the same way as in the string case. Explicitly, we start with the relativistic p-brane action coupled to $C^{p+1}$ form
whose action has the form
\begin{equation}\label{pbraneaction}
S=-\tilde{\tau}_p\int d^{p+1}\xi\sqrt{-\det \mathbf{A}_{\alpha\beta}}
+\tilde{\tau}_p\int C^{(p+1)} \ , \quad \mathbf{A}_{\alpha\beta}=G_{\mu\nu}\partial_\alpha x^\mu
\partial_\beta x^\nu \ ,
\end{equation}
where $\xi^\alpha \ , \alpha=0,\dots,p$ label world-volume of p-brane
and where
\begin{equation}
C^{(p+1)}=C_{\mu_1\dots \mu_{p+1}}dx^{\mu_1}\wedge \dots dx^{\mu_{p+1}}=
\frac{1}{(p+1)!}\epsilon^{\alpha_1\dots \alpha_{p+1}}
C_{\mu_1\dots \mu_{p+1}}\partial_{\alpha_1}x^{\mu_1}\dots \partial_{\alpha_{p+1}}x^{\mu_{p+1}} \ ,
\end{equation}
where again $\epsilon^{\alpha_1\dots \alpha_{p+1}}$ is totally antisymmetric
symbol.

 With the help of the  action (\ref{pbraneaction}) we can proceed to the definition of non-relativistic p-brane in  Newton-Cartan background. As we have seen in case of the non-relativistic string
the requirement that the action for non-relativistic string should be finite
select two longitudial directions. Then we can deduce that in case of non-relativistic p-brane we should select $p+1$ longitudial directions. Explicitly, we presume that in case of the probe p-brane the background metric has the form
\begin{eqnarray}
G_{\mu\nu}&=&E_\mu^{ \ a}E_\nu^{ \ b}\eta_{ab}+E_\mu^{ \ a'}E_\nu^{ \ b'}\delta_{a'b'}
=\nonumber \\
&=&\omega^2 \tau_{\mu\nu}+h_{\mu\nu}+\frac{1}{2}\tau_\mu^{ \ a}m_\nu^{ \ b}\eta_{ab}+
\frac{1}{2}m_\mu^{ \ a}\tau_\nu^{ \ b}\eta_{ab}+\frac{1}{4\omega^2}m_\mu^{ \ a}m_\nu^{ \ b}
\eta_{ab} \ ,  \nonumber \\
\end{eqnarray}
where now $a,b=0,\dots,p$ and $a',b'\dots=(p+1,\dots,d-1)$. Further, $\tau_{\mu\nu}$
and $h_{\mu\nu}$ are defined as
\begin{equation}
\tau_{\mu\nu}=\tau_\mu^{ \ a}\tau_\nu^{ \ b}
\eta_{ab} \ , \quad \eta_{ab}=\mathrm{diag}(-1,\dots,1) \  , \quad h_{\mu\nu}=
e_\mu^{ \ a'}e_\nu^{ \ b'}\delta_{a'b'}  \ .
\end{equation}
We also introduce their inverses with respect to their longitudinal and transverse dimensions
\begin{eqnarray}
e_\mu^{ \ a'}e^\mu_{ \ b'}=\delta^{a'}_{b'} \ ,  \quad
e_\mu^{ \ a'}e^\nu_{ \ a'}=\delta_\mu^\nu-\tau_\mu^{ \ a}
\tau^\nu_{ \ a} \ , \quad \tau^\mu_{ \ a}\tau_\mu^{ \ b}=\delta_a^b \ , \quad
\tau^\mu_{ \ a}e_\mu^{ \ a'}=0 \ , \quad
\tau_\mu^{ \ a}e^\mu_{ \ a'}=0 \ . \nonumber \\
\end{eqnarray}
In case of $p+1$-form $C^{(p+1)}$ we presume, with the analogy with the
string case, that it has the form
\begin{eqnarray}
C_{\mu_1\dots \mu_{p+1}}=\left(\omega\tau_{\mu_1}^{ \ a_1}-\frac{1}{2\omega}m_{\mu_1}^{ \ a_1}\right)\times \dots\times \left( \tau_{\mu_{p+1}}^{ \ a_{p+1}}-
\frac{1}{2\omega}m_{\mu_{p+1}}^{ \ a_{p+1}}\right)\epsilon_{a_1\dots a_{p+1}} \ ,
\end{eqnarray}
where  $\epsilon_{a_1\dots a_{p+1}}$ is  totally antisymmetric symbol.
Now we are ready to define  non-relativistic limit of p-brane action. We start with the kinetic term and we obtain
\begin{eqnarray}
S_{DBI}=-\tilde{\tau}_p\omega^{p+1}\int d^{p+1}\xi\sqrt{-\det\mathbf{a}}-
\frac{\tilde{\tau}_p}{2}\omega^{p-1}\int d^{p+1}\xi\sqrt{-\det\mathbf{a}}\tilde{\mathbf{a}}^{\alpha\beta}
\bar{h}_{\alpha\beta} \ ,
\nonumber \\
\end{eqnarray}
where $\tilde{\mathbf{a}}^{\alpha\beta}$ is inverse to $\mathbf{a}_{\alpha\beta}$. In fact, it is reasonable to presume that $\mathbf{a}_{\alpha\beta}=\partial_\alpha x^\mu \eta_{ab}\partial_\beta x^\nu=\tau_\alpha^{ \ a}\eta_{ab}\tau_\beta^{ \ b}$ since
$\tau_\alpha^{ \ a}$ and $\eta_{ab}$ are  $(p+1)\times (p+1)$ non-singular matrices.
From the requirement that the kinetic term is finite we have to perform following rescaling
\begin{equation}
\tilde{\tau}_p \omega^{p-1}=\tau_p \ .
\end{equation}
Further, the divergent term can be written as
\begin{equation}
\tilde{\tau}_p \omega^{p+1}\int d^{p+1}\xi \sqrt{-\det\mathbf{a}}=
\tau_p \omega^2\int d^{p+1}\xi\det \tau_\alpha^{ \ b} \ , \quad   \tau_\alpha^{ \ a}=
\tau_\mu^{ \ a}\partial_\alpha x^\mu \ .
\end{equation}
Let us now concentrate on the second term in the action
(\ref{pbraneaction}). If we express $\tilde{\tau}_p$ using $\tau_p$ as
$\tilde{\tau}_p=\frac{1}{\omega^{p-1}}\tau_p$ we find that the only non-zero
contribution comes from the product of $\tau_\mu^{ \ a}$'s while remaining terms vanish in the limit $\omega\rightarrow \infty$. Then we obtain
\begin{eqnarray}
S_{WZ}&=&\frac{1}{\omega^{p-1}}\tau_p\int d^{p+1}\epsilon^{\alpha_1\dots \alpha_{p+1}}
\omega\tau_{\mu_1}^{ \ a_1}\partial_{\alpha_1}x^{\mu_1}\dots \omega\tau_{\mu_{p+1}}
^{ \ a_{p+1}}\partial_{\alpha_{p+1}}x^{\mu_{p+1}}\epsilon_{a_1\dots a_{p+1}}
\nonumber \\
&=&\omega^2\tau_p\int d^{p+1}\xi \frac{1}{p!}
\epsilon^{\alpha_1\dots \alpha_{p+1}}\tau_{\alpha_1}^{ \ a_1}
\dots \tau_{\alpha_{p+1}}^{ \ a_{p+1}}=
\omega^2\tau_p \int d^{p+1}\xi \det\tau_\alpha^{ \ a}
\nonumber \\
\end{eqnarray}
and we again see that these two divergent contributions cancel. As a result we obtain well defined action for non-relativistic p-brane in Newton-Cartan background
\begin{equation}\label{nonrelpbraneaction}
S=-\frac{\tau_p}{2}\int d^{p+1}\xi
\sqrt{-\det\mathbf{a}}\tilde{\mathbf{a}}^{\alpha\beta}\bar{h}_{\mu\nu}\partial_\alpha x^\mu
\partial_\beta x^\nu \ .
\end{equation}
Now we proceed to the Hamiltonian formulation of this theory. With the analogy with the string case we write the action as
\begin{eqnarray}\label{pbraneactionextended}
S&=&\int d^{p+1}
\xi \left(\frac{1}{4\lambda^\tau}(\partial_0 x^\mu-\lambda^i\partial_i x^\mu)
h_{\mu\nu}(\partial_0 x^\nu-\lambda^j\partial_j x^\nu)-\lambda^\tau \tau^2_p\det\mathbf{a}_{ij}\mathbf{a}^{ij}h_{ij} \right.
\nonumber \\
& &\left.+B^0\left(\lambda^0-\frac{\sqrt{-\det \mathbf{a}}}{2\tau_p
\det \mathbf{a}_{ij}}\right)+B^i(\lambda_i-\mathbf{a}_{i0})\right) \ ,
\nonumber \\
\end{eqnarray}
where
\begin{equation}
\lambda^i=\mathbf{a}^{ij}\mathbf{a}_{j0} \ ,  \quad
\mathbf{a}_{ij}\mathbf{a}^{jk}=\delta_i^ k \ .
\end{equation}
In order to see
 an equivalence between the action (\ref{pbraneactionextended}) and
(\ref{nonrelpbraneaction})  we note that the inverse matrix $\tilde{\mathbf{a}}^{\alpha\beta}$ to the matrix $\mathbf{a}_{\alpha\beta}$ has  the form
\begin{eqnarray}\label{inversematrix}
\tilde{\mathbf{a}}^{00}&=&\frac{\det\mathbf{a}_{ij}}{\det\mathbf{a}} \ , \quad
\tilde{\mathbf{a}}^{0i}=-\mathbf{a}_{0k}\mathbf{a}^{kj}\frac{\det\mathbf{a}_{ij}}{\det\mathbf{a}} \ ,  \nonumber \\
\tilde{\mathbf{a}}^{i0}&=&-\mathbf{a}^{ik}\mathbf{a}_{k0}\frac{\det\mathbf{a}_{ij}}{\det\mathbf{a}} \ , \quad
\tilde{\mathbf{a}}^{ij}=\mathbf{a}^{ij}+\frac{\det\mathbf{a}_{ij}}{\det\mathbf{a}}
\mathbf{a}^{ik}\mathbf{a}_{k0}\mathbf{a}_{0l}\mathbf{a}^{lj} \ , \nonumber \\
\end{eqnarray}
where $\mathbf{a}^{ij}\mathbf{a}_{jk}=\delta^i_k$.
 Then the equation of motion for
 $B^0$ and $B^i$ imply
\begin{equation}
\lambda^0=-\frac{\sqrt{-\det\mathbf{a}}}{2\tau_p \det\mathbf{a}_{ij}} \ , \quad  \lambda_i=\mathbf{a}_{0i} \ .
\end{equation}
Inserting this result into (\ref{pbraneactionextended}) we obtain that it is equal to the action (\ref{nonrelpbraneaction}).\
Let us now return to the action  (\ref{pbraneactionextended})
and determine conjugate momenta from it
\begin{eqnarray}\label{pmubrane}
& &p_\mu=\frac{\partial \mathcal{L}}{\partial (\partial_0 x^\mu)}\nonumber \\
&=&\frac{1}{2\lambda^\tau}\bar{h}_{\mu\nu}\partial_0 x^\nu-\frac{\lambda^i}{2\lambda^\tau}
\bar{h}_{\mu\nu}\partial_i x^\nu-B^\tau\frac{1}{2\tau_p\det\mathbf{a}_{ij}}
\tau_{\mu\nu}\partial_\alpha x^\nu (\mathbf{a}^{-1})^{\alpha 0}\sqrt{-\det\mathbf{a}}-
B^i\tau_{\mu\nu}\partial_i x^\nu \ , \nonumber \\
\nonumber \\
& &P_0=\frac{\partial \mathcal{L}}{\partial(\partial_0 \lambda^0)}\approx 0 \ , \quad
P_i=\frac{\partial \mathcal{L}}{\partial(\partial_0 \lambda^i)}\approx 0 \ , \quad
P^B_0=\frac{\partial \mathcal{L}}{\partial(\partial_0 B^0)}\approx 0 \ , \quad
P^B_i=\frac{\partial \mathcal{L}}{\partial(\partial_0 B^i)}\approx 0 \ .
\nonumber \\
\end{eqnarray}
From the same reason as in case of the fundamental string we have to restrict
to the case $m_\mu^{ \ a}=0$ so that $\bar{h}_{\mu\nu}=h_{\mu\nu}$. Then if we multiply
(\ref{pmubrane}) with  $h^{\mu\nu}$  we obtain
\begin{eqnarray}
\left(p_\mu+\frac{\lambda^i}{2\lambda^\tau}h_{\mu\rho}\partial_i x^\rho\right)
h^{\mu\nu}\left(p_\nu+\frac{\lambda^j}{2\lambda^\tau}h_{\nu\sigma}\partial_jx^\sigma\right)=
\frac{1}{4(\lambda^\tau)^2}\partial_0 x^\mu h_{\mu\nu}\partial_0 x^\nu \ . \nonumber \\
\end{eqnarray}
On the other hand let us multiply both sides of (\ref{pmubrane}) with $\tau^{\mu\nu}\tau_{\nu\rho}
\partial_i x^\rho$ and  we obtain
\begin{eqnarray}
p_\mu \tau^{\mu\nu}\tau_{\nu\rho}\partial_i x^\rho=
-B^\tau \frac{1}{2\tau_p\det\mathbf{a}_{ij}}
\partial_i x^\mu \tau_{\mu\nu}\partial_\alpha x^\nu \mathbf{a}^{\alpha 0}
\sqrt{-\det\mathbf{a}}-\partial_i x^\mu \tau_{\mu\nu}\partial_j x^\nu B^j=
-\mathbf{a}_{ij}B^j
\nonumber \\
\end{eqnarray}
that imlies $p-$primary constraints
\begin{eqnarray}\label{primpcon}
\Sigma^i\equiv p_\mu \tau^{\mu\nu}\tau_{\nu\sigma}\partial_j x^\sigma \mathbf{a}^{ji}+B^i \approx 0  \ .
\nonumber \\
\end{eqnarray}
On the other hand let us multiply (\ref{pmubrane}) with following expression
\begin{equation}
\frac{1}{p!}\tau^\mu_{ \ a}\eta^{aa_1}\epsilon_{a_1\dots a_{p+1}}
\tau^{\ a_2}_{\nu_2}\dots \tau^{\ a_{p+1}}_{\nu_{p+1}}
\epsilon^{j_2\dots j_{p+1}}\partial_{j_2}x^{\nu_2}\dots \partial_{j_{p+1}}
x^{\nu_{p+1}} \ .
\end{equation}
Then using the fact that
\begin{eqnarray}
& &\partial_i x^\nu\tau_{\nu\mu}\tau^\mu_{ \ a}\eta^{aa_1}
\epsilon_{a_1\dots a_{p+1}}\tau_{\nu_2}^{ \ a_2}
\dots \tau_{\nu_{p+1}}^{\ a_{p+1}}\epsilon^{j_2\dots j_{p+1}}
\partial_{j_2}x^{\nu_2}\dots \partial_{j_{p+1}}x^{\nu_{p+1}}=0  \ ,
\nonumber \\
& &\frac{1}{p!}\mathbf{a}^{0\alpha}\partial_\alpha x^\nu \tau_{\nu\mu}
\tau^\mu_{ \ a}\eta^{aa_1}\epsilon_{a_1\dots a_{p+1}}
\tau_{\nu_2}^{ \ a_2}\dots \tau_{\nu_{p+1}}^{ \ a_{p+1}}
\epsilon^{j_2\dots j_{p+1}}\partial_{j_2}x^{\nu_2}\dots \partial_{j_{p+1}}
x^{\nu_{p+1}}\frac{\sqrt{-\det\mathbf{a}}}{\det\mathbf{a}_{ij}}=\nonumber \\
& &=-\frac{1}{(p+1)!}\epsilon_{a_1\dots a_{p+1}}
\epsilon^{j_1\dots j_{p+1}}\tau_{\nu_1}^{ \ a_1}\partial_{\alpha_1}x^{\nu_1}
\dots \tau_{\nu_{p+1}}^{ \ a_{p+1}}\partial_{\alpha_{p+1}}x^{\nu_{p+1}}
\frac{1}{\det \tau_\alpha^{ \ a}}
=-1  \nonumber \\
\end{eqnarray}
 we obtain second primary constraint
\begin{equation}
\Sigma^0 \equiv
2\tau_p p_\mu \frac{1}{p!}\tau^\mu_{ \ a}\eta^{aa_1}\epsilon_{a_1\dots a_{p+1}}
\tau^{\ a_2}_{\nu_2}\dots \tau^{\ a_{p+1}}_{\nu_{p+1}}
\epsilon^{j_2\dots j_{p+1}}\partial_{j_2}x^{\nu_2}\dots \partial_{j_{p+1}}
x^{\nu_{p+1}}-B^0 \approx 0 \ .
\end{equation}
Using all these results we determine extended Hamiltonian with all primary constraints included in the form
\begin{eqnarray}\label{Hamextendedbrane}
H_E=\int d^p\xi (\lambda^0 p_\mu h^{\mu\nu}p_\nu+\lambda^i p_\mu h^{\mu\nu}h_{\nu\sigma}
\partial_i x^\sigma+\lambda^\tau \tau_p^2\det\mathbf{a}_{ij}\mathbf{a}^{ij}h_{ij}-
\nonumber \\
-B^0\lambda^\tau-B^i\lambda_i+v^0P_0+v^iP_i+v^0_BP_0^B+v_i^B P^i_B
+\Psi_0\Sigma^0+\Psi_i\Sigma^i) \ .
\nonumber \\
\end{eqnarray}
Since $\pb{P_0^B(\xi),\Sigma^0(\xi')}=\delta(\xi-\xi') \ ,
\pb{P^i_B(\xi),\Sigma^j(\xi')}=-\delta^{ij}\delta(\xi-\xi')$ we see that
 that $P_0^B$ together with $\Psi^0$ are the couple of $p+1$ second class constraints. Then we can solve these constraints with respect to $B^0,B^i$ and we
we obtain the Hamiltonian
in the form
\begin{eqnarray}
H_B
=\int d^p\xi (\lambda^0\mathcal{H}_0+\lambda^i\mathcal{H}_i+v^0P_0+v^i P_i) \nonumber \\
\end{eqnarray}
where
\begin{eqnarray}
\mathcal{H}_0&=&p_\mu h^{\mu\nu}p_\nu
+\tau_p^2\det\mathbf{a}_{ij}\mathbf{a}^{ij}h_{ij} -\nonumber \\
&-&2\tau_p p_\mu \frac{1}{p!}\tau^\mu_{ \ a}\eta^{aa_1}\epsilon_{a_1\dots a_{p+1}}
\tau^{\ a_2}_{\nu_2}\dots \tau^{\ a_{p+1}}_{\nu_{p+1}}
\epsilon^{j_2\dots j_{p+1}}\partial_{j_2}x^{\nu_2}\dots \partial_{j_{p+1}}
x^{\nu_{p+1}}\approx 0 \ ,  \nonumber \\
\mathcal{H}_i&=&p_\mu\partial_ix^\mu \approx 0 \ .
\nonumber \\
\end{eqnarray}
Then the requirement of the preservation of the constraint $P_0\approx 0\ ,  P_i\approx 0$ implies $p+1$ secondary constraints
\begin{equation}
\mathcal{H}_0\approx 0 \ , \quad \mathcal{H}_i\approx 0 \ .
\end{equation}
Now we have to check that these constraints are the first class constraints. We
introduce their smeared form
\begin{equation}
\mathbf{T}_T(N)=\int d^p\xi N\mathcal{H}_0 \ , \quad
\mathbf{T}_S(N^i)=\int d^p\xi N^i\mathcal{H}_i \
\end{equation}
and calculate corresponding  Poisson brackets. First of all we have
\begin{eqnarray}
\pb{\mathbf{T}_S(N^i),\mathbf{T}_S(M^j)}=\mathbf{T}_S(N^j\partial_j M^i-M^j\partial_j N^i) \ .
 \nonumber \\
\end{eqnarray}
In case of the calculation of the Poisson brackets between $\mathbf{T}_S(N^i)$ and
 $\mathbf{T}_T(M)$ we have
to be more careful. First of all we have
\begin{equation}
\pb{\mathbf{T}_S(N^ i),\tau_i^{ \ a}}=-N^k\partial_k \tau_i^{ \ a}-
\partial_i N^j \tau_j^{ \ a} \ , \quad \tau_i^{ \ a}\equiv \partial_i x^\mu
\tau_\mu^{ \ a} \ .
\end{equation}
Then we obtain
\begin{eqnarray}
& &\pb{\mathbf{T}_S(N^i),\mathbf{a}_{ij}}=-N^k\partial_k \mathbf{a}_{ij}-\partial_i N^k\mathbf{a}_{kj}-
\mathbf{a}_{ik}\partial_j N^k \ ,  \nonumber \\
& &\pb{\mathbf{T}_S(N^i),\mathbf{a}^{ij}}=-N^k\partial_k \mathbf{a}^{ij}+\partial_k N^i\mathbf{a}^{kj}+
\mathbf{a}^{ik}\partial_k N^j \ , \nonumber \\
& &\pb{\mathbf{T}_S(N^i),\det\mathbf{a}_{ij}}=-N^k\partial_k (\det\mathbf{a}_{ij})-
2\partial_i N^i\det\mathbf{a}_{ij} \ . \nonumber \\
\end{eqnarray}
Using also the fact that
\begin{eqnarray}
& &\pb{\mathbf{T}_S(N^i),\partial_i x^\mu}=-N^k\partial_k(\partial_i x^\mu)-\partial_i
N^k\partial_k x^\mu \ , \nonumber \\
& &\pb{\mathbf{T}_S(N^i),h_{ij}}=-N^k\partial_k h_{ij}-\partial_i N^k h_{kj}
-h_{ik}\partial_j N^k \ \nonumber \\
\end{eqnarray}
we finally obtain
\begin{equation}
\pb{\mathbf{T}_S(N^i),\det\mathbf{a}_{ij}\mathbf{a}^{kl}h_{kl}}=-N^k\partial_k (\det\mathbf{a}_{ij}
\mathbf{a}^{kl}h_{kl})-2\partial_i N^i\det\mathbf{a}_{ij}\mathbf{a}^{kl}h_{kl} \ .
\end{equation}
Let us introduce following vector
\begin{equation}
V^\mu=-2\tau_p \frac{1}{p!}\tau^\mu_{ \ a}\eta^{aa_1}\epsilon_{a_1\dots a_{p+1}}
\tau^{\ a_2}_{\nu_2}\dots \tau^{\ a_{p+1}}_{\nu_{p+1}}
\epsilon^{j_2\dots j_{p+1}}\partial_{j_2}x^{\nu_2}\dots \partial_{j_{p+1}}
x^{\nu_{p+1}} \ .
\end{equation}
Then after some algebra we obtain
\begin{equation}
\pb{\mathbf{T}_S(N^i),V^\mu}=-N^k\partial_k V^\mu-2\partial_k N^k V^\mu \ .
\end{equation}
Collecting all these results together we finally find
\begin{equation}
\pb{\mathbf{T}_S(N^i),\mathbf{T}_S(M)}=\mathbf{T}_T(N^i\partial_iM-\partial_i N^i M) \ .
\end{equation}
Finally we calculate Poisson brackets of the smeared forms of Hamiltonian
constraints
and we obtain
\begin{eqnarray}\label{pbTTpbrane}
& &\pb{\mathbf{T}_T(N),\mathbf{T}_T(M)}=\int d^p\xi (N\partial_iM-M\partial_iN)4\tau_p^2
\det\mathbf{a}_{ij}\mathbf{a}^{ij}p_\mu h^{\mu\nu}h_{\nu\sigma}\partial_j x^\sigma+
\nonumber \\
&+&2\tau_p\int d^p\xi
(N\partial_i M-M\partial_iN)\frac{1}{(p-1)!}p_\nu\tau^\nu_{ \ a}\eta^{aa_1}
\epsilon_{a_1\dots a_{p+1}}\tau_{\mu}^{ \ a_2}\tau_{\nu_3}^{ \ a_3}
\dots \tau_{\nu_{p+1}}^{ \  a_{p+1}}\times \nonumber \\
&\times &
\epsilon^{i j_3\dots j_{p+1}}\partial_{j_3}x^{\nu_3}\dots \partial_{j_{p+1}}
x^{\nu_{p+1}}V^\mu \ .  \nonumber \\
\end{eqnarray}
Then after some lengthy calculations we find
that the last expression is equal to
\begin{equation}
4\tau_p^2 (N\partial_i M-M\partial_i N)\mathbf{a}^{ij}p_\mu \tau^{\mu\nu}
\tau_{\nu\sigma}\partial_j x^\sigma \det \mathbf{a}_{ij} \ .
\end{equation}
Inserting this result into (\ref{pbTTpbrane} we obtain  final result
\begin{equation}
\pb{\mathbf{T}_T(N),\mathbf{T}_T(M)}=\mathbf{T}_S((N\partial_iM-M\partial_iN)4\tau_p^2\mathbf{a}^{ij}
\det\mathbf{a}_{ij}) \ .
\end{equation}
These results show that $\mathcal{H}_0$ and $\mathcal{H}_i$ are the first class constraints
that reflect diffeomorphism invariance of non-relativistic p-brane.

\section{Conclusion}\label{fourth}
In this paper we formulated non-relativistic actions for string and p-brane in Newton-Cartan background. Then we found their Hamiltonian formulations and we determined structure of constraints in the special case where the gauge field $m_\mu^{ \ a}$ is zero. We argued that we restricted to this case since we were not able to express time derivatives of $x^\mu$ or their combinations as functions of canonical variables in the case when $m_\mu^{ \ a}\neq 0$. Certainly this more general case deserves further study.
One possibility is to  address this problem from different point of view when we
start with the Hamiltonian formulation string in general background, then we
perform limiting procedure on the background metric and NSNS two form field
and derive corresponding Hamiltonian. This problem is currently under study and we
hope to report on this analysis in near future.

\acknowledgments{This  work  was
	supported by the Grant Agency of the Czech Republic under the grant
	P201/12/G028. }


\section{Introduction and outline}

\subsection{Background}
A hypergraph  $H$ consists of a set $V(H)$ together with a family of subsets $E(H)$ of $V(H)$
called hyperedges. Two vertices $u,v \in V(H)$ are adjacent if there is a hyperedge $e \in E(H)$
such that both $u \in e$ and $v \in e$.
$H$ is $r$-uniform if every hyperedge in $E(H)$ has exactly $r$ elements.
Two hypergraphs $H_1, H_2$ are isomorphic,
denoted by $H_1 \simeq H_2$,
if there is a bijective map $h: V(H_1) \rightarrow V(H_2)$
such that for any two vertices $u, v \in V(H_1)$  we have $u$ and $v$ are adjacent iff
$h(u)$ and $h(v)$ are adjacent in $H_2$.

We denote by
$\frak{H}$ ($\frak{H}^r$)
the class of all ($r$-uniform) hypergraphs,
and by
$\frak{H}_n$ ($\frak{H}_n^r$)
the set of all ($r$-uniform) hypergraphs $H$ with $V(H) = [n]$.
Here $[n] = \{ 1, 2, \ldots, n\}$.
We note that graphs are $2$-uniform hypergraphs.

A (univariate) hypergraph polynomial $P(H;X)$ is a function
$ P: \frak{H} \rightarrow \mathbb{Z}[X]$
which preserves hypergraph isomorphisms.
Let $P(H;X)$ be a univariate hypergraph polynomial.
A hypergraph $H$ is $P$-unique if for every hypergraph $H_1$ with $P(H_1:X) = P(H;X)$
we have that $H_1$ is isomorphic to $H$.
Similarly,
a $r$-uniform hypergraph $H$ is $r$-$P$-unique if for every  $r$-uniform hypergraph $H_1$ with $P(H_1:X) = P(H;X)$
we have that $H_1$ is isomorphic to $H$.

A hypergraph polynomial $P(H;X)$ is {\em complete
(for $r$-uniform hypergraphs)
}, if all
($r$-uniform) hypergraphs are $P$-unique.
Let $\mathcal{H}(n)$ be the number of non-isomorphic hypergraphs on $n$ vertices, and let
and $\mathcal{H}^r(n)$ be the number of non-isomorphic $r$-uniform hypergraphs on $n$ vertices.
Furthermore, let
$\mathcal{U}_P(n)$
($\mathcal{U}_P^r(n)$)
be the number of non-isomorphic $P$-unique ($r$-uniform)  hypergraphs on $n$ vertices.

$P$ is {\em almost complete}
if
$$
\lim_{n \rightarrow \infty} \frac{\mathcal{U}_P(n)}{\mathcal{H}(n)} = 1.
$$
It is almost complete on
on $r$-uniform hypergraphs  if
$$
\lim_{n \rightarrow \infty} \frac{\mathcal{U}_P^r(n)}{\mathcal{H}^r(n)} = 1.
$$
$P$ is {\em weakly distinguishing} if
$$
\lim_{n \rightarrow \infty} \frac{\mathcal{U}_P(n)}{\mathcal{H}(n)} = 0.
$$
and analogously for $r$-uniform hypergraphs.

No known
univariate hypergraph polynomial
in the literature
is complete, although one can construct such polynomials using some clever encoding of the isomorphism types
of finite hypergraphs.

\subsection{Three hypergraph polynomials}
Hypergraph polynomials studied in the literature are the chromatic polynomial $\chi(H,X)$
\cite{ar:Tomescu-2004,ar:Tomescu-2007,ar:Tomescu-2014,ar:BorowieckiLazuka-2000,ar:BorowieckiLazuka-2007},
the independence polynomial $\mathrm{Ind}(H;X))$,
\cite{ar:Trinks2016},
and the matching polynomial $\mathrm{M}(H;X)$,
\cite{ar:GuoZhaoMao2017}.
There where also attempts to extend spectral graph theory to hypergraphs, cf.
\cite{ar:CooperDutle-2012,ar:PearsonZhang-2014} and the references therein.
The monograph \cite{bk:BrouwerHaemers2012}
summarizes what is known about graphs unique
for the characteristic and the Laplacian polynomial.
In \cite{bk:BrouwerHaemers2012}  the authors also
suggest that the characteristic polynomial
is almost complete on graphs.
However, we do not discuss here which hypergraphs are unique for these polynomials.

\paragraph{The chromatic polynomial.}
The chromatic polynomial for hypergraphs defined below generalizes the chromatic polynomial for graphs,
but also show distinctly different behaviour in the case of hypergraphs, cf. \cite{ar:ZhangDong-2017}.

Let $k \in \mathbb{N}$ and
$f: V(H) \rightarrow [k]$.
$f$ is a {\em proper coloring of $H$ with at most $k$ colors}
if every $e \in E(H)$ contains two vertices $u,v \in e$ with $f(u) \neq f(v)$.
We denote by $\chi(H;k)$ the number of proper colorings of $H$ with at most $k$ colors.

A set $I \subseteq V(H)$ is {\em independent} of there is no edge $e \subseteq I$.
For $i \in \mathbb{N}$
let $b_i(H)$ be number of partitions of $V(H)$ into $i$ independent sets.
For $X$ we denote by $X_{(i)}$ the polynomial
$$
X \cdot (X_1) \cdot \ldots \cdot (X-i+1).
$$

\begin{prop}[\cite{ar:BorowieckiLazuka-2000,Zhang}]
\label{pr:chromatic}
\begin{enumerate}[(i)]
\item
$\chi(H;k)$ is a polynomial in $k$, hence can be extended to a polynomial $\chi(H;X) \in \mathbb{Z}[X]$.
\item
$\chi(H;X) = \sum_i^n b_i(H) \cdot X_{(i)}$.
\end{enumerate}
\end{prop}

$\chi$-unique hypergraphs were presented compactly  in \cite{Zhang}
which we summarize here.
The definitions of hypercycles, hyperpaths and sunflower hypergraphs
are standard in the hypergraph literature, cf.
the books by
C. Berge \cite{Berge73}.
V. I. Voloshin \cite{bk:Voloshin-2009},
and
A. Bretto \cite{bk:Bretto-2013}.

\begin{prop}[\cite{ar:Tomescu-2004,ar:Tomescu-2007,ar:Tomescu-2014,Zhang}]
\begin{enumerate}[(i)]
\item
For $r\geq 3$, $r$-uniform hypercycle $\mathcal{C}_{m}^{r}$ is $r$-$\chi$-unique but it is not $\chi$-unique;
\item
For every $p, r\geq 3$, $B_{p, p}^{2, 2}$ is $\chi$-unique, where $B_{p,p}^{2, 2}$
denote the hypergraphs obtained by identifying extremities $x$ and $y$ of
$\mathcal{P}_{p-1}^{r}$ with distinct vertices of degree $2$ in the same edge of $\mathcal{C}_{p}^{r}$;
\item
The sunflower hypergraph $SH(n, 1, r)$ is $\chi$-unique;
\item
Let $n=r+(k-1)p$, where $r\geq 3, k\geq 1$ and $1\leq p\leq r-1$.
Then $SH(n, p, r)$ is $r$-$\chi$-unique for every $1\leq p\leq r-2$; for $p=r-1$,
$SH(n, r-1, r)$ is $r$-$\chi$-unique for $k=1$ or $k=2$ but it has not this property for $k\geq 3$.
\end{enumerate}
\end{prop}

In \cite{ar:BollobasPebodyRiordan2000} it is conjectured that the chromatic polynomial
is almost complete on graphs.
Our first result shows that the conjecture is not true for hypergraphs.
\begin{thm}
\label{th:main-chrom}
The chromatic polynomial
$\chi(H;X)$ is weakly distinguishing
\begin{enumerate}[(i)]
\item
on hypergraphs:
$$\lim_{n \rightarrow \infty} \frac{\mathcal{U}_{\chi}(n)}{\mathcal{H}(n)} = 0.$$
\\
\item
on $r$-uniform hypergraphs:
For every $r \geq 3$
$$\lim_{n \rightarrow \infty} \frac{\mathcal{U}_{\chi}^r(n)}{\mathcal{H}^r(n)} = 0.$$
\end{enumerate}
\end{thm}

\paragraph{The independence polynomial.}
The independence polynomial for hypergraphs is defined as
$$
\mathrm{Ind}(H:X) =
\sum_i \mathrm{ind}_i(H) \cdot X^i.
$$
where  $\mathrm{ind}_i(H)$ is the number of independent sets $I \subseteq V(H)$ with $|I|=i$.

The independence polynomial for hypergraphs was studied in
\cite{ar:Trinks2016}.

\begin{thm}
\label{th:main-indep}
The independence polynomial $\mathrm{Ind}(H;X)$ is weakly distinguishing
\begin{enumerate}[(i)]
\item
on hypergraphs:
$$
\lim_{n \rightarrow \infty} \frac{\mathcal{U}_{Ind}(n)}{\mathcal{H}(n)} = 0.
$$
\item
on $r$-uniform hypergraphs:
For every $r \geq 2$
$$
\lim_{n \rightarrow \infty} \frac{\mathcal{U}_{Ind}^r(n)}{\mathcal{H}^r(n)} = 0.
$$
\end{enumerate}
\end{thm}

The case $r=2$ was shown in \cite{ar:JAMRakita2017}.
The proof for $r \geq 3$ is given in Section \ref{se:indep}.

\paragraph{The matching polynomial.}
A {\em $k$-matching $m$ of a hypergraph $H$} is a set $m \subseteq E(H)$ of $k$ disjoint
hyperedges.
Let $\mu_k(H)$ be the number of $k$-matchings of $H$.
The {\em matching polynomial $\mathrm{M}(H;X)$} of a hypergraph $H$ is defined by
$$
\mathrm{M}(H;X) = \sum_{k=1}^{\lfloor\frac{n}{k_H}\rfloor}\mu_k(H) \cdot X^k.
$$
where $k_H$ is the minimum size of the edges in $E(H)$.
The matching polynomial for hypergraphs was studied \cite{ar:GuoZhaoMao2017}.
M. Noy  \cite{ar:Noy03} studied $\mathrm{M}$-unique graphs.

\begin{thm}
\label{th:main-match}
The matching polynomial
$\mathrm{M}(H;X)$ is weakly distinguishing
\begin{enumerate}[(i)]
\item
on hypergraphs:
$$ \lim_{n \rightarrow \infty} \frac{\mathcal{U}_{M}(n)}{\mathcal{H}(n)} = 0.  $$
\item
on $r$-uniform hypergraphs:
For every $r \geq 3$
$$ \lim_{n \rightarrow \infty} \frac{\mathcal{U}_{M}^r(n)}{\mathcal{H}^r(n)} = 0.
$$
\end{enumerate}
\end{thm}
The case $r=2$ is still open.
The proof is also given in Section \ref{se:indep}.

To the best of our knowledge
no explicite description of
$\mathrm{Ind}$-unique
and
$\mathrm{M}$-unique
hypergraphs is given in the literature.

\section{General strategy of the proofs}
\label{se:strategy}

The general strategy of our proofs has been motivated by the first author's work with V. Rakita
\cite{ar:JAMRakita2017}.

Let $\overline{\mathcal{H}}(n)$ and $\overline{\mathcal{H}}^r(n)$ denote the number of {\em labeled} hypergraphs and
$r$-uniform hypergraphs of order $n$.

\begin{lem}
\label{le:labeled}
For any positive integer $n$,
\begin{enumerate}[(i)]
\item
$\overline{\mathcal{H}}(n) \leq 2^{2^n}$.
\item
$ \overline{\mathcal{H}}(n) \leq \mathcal{H}(n) \cdot n!$
\item
$\overline{\mathcal{H}}^r(n) \leq 2^{{n \choose r}}$.
\item
$ \overline{\mathcal{H}}^r(n) \leq \mathcal{H}^r(n) \cdot n!$
\end{enumerate}
\end{lem}
\begin{proof}
(i) and (iii):
are obvious.
\\
(ii) and (iv):
Not every ($r$-uniform) hypergraph has automorphisms.
\end{proof}

Let $P(H;X)$ be a hypergraph polynomial.
We denote by
$\mathcal{B}_{P}(n)$
($\mathcal{B}_{P}^r(n)$)
the number of polynomials $p(X)$ such that there is a  ($r$-uniform) hypergraph $H$
of order $n$ with $p(X)=P(H;X)$.

We shall use the following observations:

\begin{lem}
\label{le:strategy}
\begin{enumerate}[(i)]
\item
$\mathcal{B}_{P}^r(n) \leq \mathcal{B}_{P}(n)$.
\item
$U_{P}(n)\leq \mathcal{B}_{P}(n)\leq \mathcal{H}(n)$.
\item
$U_{P}^r(n)\leq \mathcal{B}_{P}^r(n)\leq \mathcal{H}^r(n)$.
\end{enumerate}
\end{lem}
\begin{proof}
(i): There are more candidate polynomials for
$\mathcal{B}_{P}(n)$ than  for
$\mathcal{B}_{P}^r(n)$.
\\
(ii) and (iii):
There cannot be more $P$-unique hypergraphs than polynomials in $\mathcal{B}_{P}(n)$ ($\mathcal{B}_{P}^r(n)$).
There cannot be more
polynomials in $\mathcal{B}_{P}(n)$ ($\mathcal{B}_{P}^r(n)$)
than there are hypergraphs in  $\mathcal{H}(n)$ ($\mathcal{H}^r(n)$).
\end{proof}

To prove our Theorems we will use Lemmas  \ref{le:labeled} and \ref{le:strategy} and estimate the numbers
$\mathcal{B}_{P}^r(n)$ or
$\mathcal{B}_{P}(n)$.

We also use an observation from pre-calculus:

\begin{prop}
\label{pr:precalculus}
Let $f, g$ be two non-decreasing real functions.
\\
If $\lim_{n \rightarrow \infty} \frac{\log(f(n))}{\log(g(n))} =0$ then
$\lim_{n \rightarrow \infty} \frac{f(n)}{g(n)} =0$.
\end{prop}

\section{Proof of Theorem \ref{th:main-chrom}}
\label{se:chromatic}
By Proposition \ref{pr:chromatic}(ii)
the chromatic polynomial of hypergraphs can be written as
$$\chi(H;X)=\sum_{i=1}^{n}b_i(H) \cdot X_{(i)}.$$

where $b_i(H)$
is the number of partitions of $V(H)$
into $i$ non-empty independent subsets.
Let
$S(n,i)$ denote the Stirling number of the second kind, which counts the number of partitions of $[n]$
into $i$ distinct subsets.

\begin{lem}
$\mathcal{B}_{\chi}(n)\leq \prod_{i=1}^{n}S(n,i).$
\end{lem}
\begin{proof}
Clearly, $0\leq  b_i(H) \leq S(n, i)$.
\end{proof}

\begin{thm}
\label{th:1}
$$\lim_{n\rightarrow \infty}\frac{\mathcal{B}_{\chi}(n)}{\mathcal{H}(n)}=0.$$
\end{thm}

We first recall some facts from \cite{bk:GrahamKnuthPatashnik94}.
\begin{facts}
\begin{enumerate}[(i)]
\item
The Stirling approximation for the factorial:
\begin{gather}
n!\sim \sqrt{2\pi n}(\frac{n}{e})^{n}.
\notag
\end{gather}
\item
For fixed $n$, $S(n,i)$ is unimodal in $i$, i.e.,
it has a single maximum, which is attained for at most two consecutive values of $i$.
That is, there is an integer $K_{n}$ such that
\begin{gather}
S(n, 1)<S(n, 2)<\cdots<S(n, K_{n}) \notag \\
S(n, K_{n})\geq S(n, K_{n}+1)>\cdots>S(n, n).
\notag
\end{gather}
\item
When $n$ is large enough, $K_{n}\sim \frac{n}{\log n}$.
\end{enumerate}
\end{facts}

Now we  prove Theorem \ref{th:1}.

\begin{proof}
\begin{gather}
\frac{\mathcal{B}_{\chi}(n)}{\mathcal{H}(n)} \leq
\frac{\mathcal{B}_{\chi}(n)\cdot n!}{\overline{\mathcal{H}}(n)}=
\frac{\mathcal{B}_{\chi}(n)\cdot n!}{2^{2^{n}}}
\notag
\end{gather}

\begin{gather}
\frac{\mathcal{B}_{\chi}(n)\cdot n!}{2^{2^{n}}}
\leq  \frac{\prod_{i=1}^{n}S(n, i)\cdot n!}{2^{2^{n}}}
\leq \frac{(S(n, K_{n}))^{n}\cdot n!}{2^{2^{n}}}
\notag \\
\sim \frac{(S(n, \frac{n}{\log n}))^{n}\cdot n!}{2^{2^{n}}}.
\notag
\end{gather}
Let us define
$\mathcal{B}_{\chi}^{*}(n)=S(n,\frac{n}{\log n}).$

We estimate
$\mathcal{B}_{\chi}^{*}(n)$.
\begin{gather}
\mathcal{B}_{\chi}^{*}(n)
\sim
\frac{(\frac{n}{\log n})^{n}}{(\frac{n}{log n})!}
=
\frac{n^{n}}{(\log n)^{n}\cdot (\frac{n}{\log n})!}
\notag \\
\sim \frac{n^{n}\cdot (e\cdot \log n)^{\frac{n}{\log n}}}{(\log n)^{n}
\cdot \sqrt{2\pi\cdot \frac{n}{\log n}}\cdot n^{\frac{n}{\log n}}}
\notag \\
=
\frac{n^{n-\frac{n}{\log n}}\cdot e^{\frac{n}{\log n}}\cdot (\log n)^{\frac{n}{\log n}-n}}{\sqrt{2\pi\cdot \frac{n}{\log n}}}
\leq n^{n}\cdot e^{n}.
\notag
\end{gather}

Therefore,
\begin{gather}
\frac{\mathcal{B}_{\chi}(n)}{\mathcal{H}(n)}\leq  \frac{(n^{n}\cdot e^{n})^{n}\cdot n!}{2^{2^{n}}}
\leq  \frac{n^{n^{2}}\cdot e^{n^{2}}\cdot n^{n}}{2^{2^{n}}}=\frac{n^{n^{2}+n}\cdot e^{n^{2}}}{2^{2^{n}}}.
\notag
\end{gather}

Now we use Proposition \ref{pr:precalculus} and
take
base $2$
logarithms
of the numerator and denominator, and we get:
\begin{gather}
\label{eq:endofproof-chi}
\lim_{n\rightarrow \infty}\frac{\log_{2} \mathcal{B}_{\chi}(n)}{\log_{2} \mathcal{H}(n)}
\leq
\lim_{n\rightarrow \infty}\frac{(n^{2}+n)\log_{2}n+n^{2}\log_{2}e}{2^{n}}=0,
\end{gather}
which implies that $\lim_{n\rightarrow \infty}\frac{\mathcal{B}_{\chi}(n)}{\mathcal{H}(n)}=0$.
\end{proof}

Now Theorem \ref{th:main-chrom}(i) follows using Lemma \ref{le:strategy}(ii).
Theorem \ref{th:main-chrom}(ii) follows
using Lemma \ref{le:strategy}(i) and (iii)
by replacing
$\mathcal{H}(n)$ by $\mathcal{H}^r(n) \leq 2^{{n \choose r}}$ in
the proof of Theorem  \ref{th:1}.
We get instead of Equation (\ref{eq:endofproof-chi})
\begin{gather}
\label{eq:r-uniform}
\lim_{n\rightarrow \infty}\frac{\log_{2} \mathcal{B}_{\chi}^r(n)}{\log_{2} \mathcal{H}^r(n)}
\leq
\lim_{n\rightarrow \infty}\frac{(n^{2}+n)\log_{2}n+n^{2}\log_{2}e}{{n \choose r}}=0,
\notag
\end{gather}
which still holds for $r \geq 3$.

\ifskip
\else
\begin{thm}
\label{th:main-i}
For every $r \geq 3$
$ \lim_{n \rightarrow \infty} \frac{\mathcal{U}_{\chi}^r(n)}{\mathcal{H}^r(n)} = 0 $
\end{thm}
\begin{proof}
We replace $\mathcal{H}(n)$ by $\mathcal{H}^r(n) = 2^{{n \choose r}}$.
and
$\mathcal{B}_{\chi}(n)$
by
$\mathcal{B}_{\chi}^r(n)$.
We get instead of Equation (\ref{eq:endofproof})
\begin{gather}
\label{eq:r-uniform}
\lim_{n\rightarrow \infty}\frac{\log_{2} \mathcal{B}_{\chi}(n)}{\log_{2} \mathcal{H}(n)}
\leq
\lim_{n\rightarrow \infty}\frac{(n^{2}+n)\log_{2}n+n^{2}\log_{2}e}{{n \choose r}}=0,
\end{gather}
which still holds for $r \geq 3$.
\end{proof}
\fi

\section{Proof of Theorems \ref{th:main-indep} and \ref{th:main-match}}
\label{se:indep}
Let $\mathcal{B}_{\mathrm{Ind}}(n)$ be the number of polynomials $p(X)$ such that there is a
hypergraph $H$ of order $n$ with $p(X)=\mathrm{Ind}(H;X)$.

\begin{thm}
\label{th:2}
\begin{enumerate}[(i)]
\item
$
\lim_{n\rightarrow \infty}
\frac{{\mathcal{B}_{\mathrm{Ind}}(n)}}
{\mathcal{H}^(n)}=0.
$
\item
$
\lim_{n\rightarrow \infty}
\frac{{\mathcal{B}^{r}_{\mathrm{Ind}}(n)}}
{\mathcal{H}^{r}(n)}=0
$.
\end{enumerate}
\end{thm}
\begin{proof}
(i):
As $0\leq \mathrm{ind}_{i}(H)\leq \binom{n}{i}$, then
$$
\mathcal{B}_{\mathrm{Ind}}(n)
\leq
\prod_{i=1}^{n}\binom{n}{i}
<
\prod_{i=1}^{n}(\frac{n\cdot e}{i})^{i}
<
\prod_{i=1}^{n}(n\cdot e)^{i}.
$$

From Lemma \ref{le:labeled}, we have that
\begin{gather}
\label{eq 1}
\frac{\mathcal{B}_{\mathrm{Ind}}(n)}{\mathcal{H}(n)}
\leq
\frac{\mathcal{B}_{\mathrm{Ind}}(n)\cdot n!}{\overline{\mathcal{H}}(n)}
<
\frac{\prod_{i=1}^{n}(n\cdot e)^{i}\cdot n!}{2^{2^{n}}}
<
\frac{\prod_{i=1}^{n}(n\cdot e)^{i}\cdot n^{n}}{2^{2^{n}}}.
\end{gather}

Now we use Proposition \ref{pr:precalculus} and
take
base $2$
logarithms
of the numerator and denominator, and we get in
Equation (\ref{eq 1}).

\begin{gather}
\frac{\log\mathcal{B}_{\mathrm{Ind}}(n)}{\log \overline{H}(n)}
<
\frac{\frac{n\cdot (n+1)}{2}\cdot \log(n\cdot e)+n\cdot \log n}{2^{n}\cdot log 2}.
\notag
\end{gather}
Therefore,
\begin{gather}
\label{eq 2}
\lim_{n\rightarrow\infty}\frac{\log\mathcal{B}_{\mathrm{Ind}}(n)}{\log \overline{H}(n)}
<
\lim_{n\rightarrow\infty}\frac{\frac{n\cdot (n+1)}{2}\cdot \log(n\cdot e)+n\cdot \log n}{2^{n}\cdot log 2}
=0,
\end{gather}
which implies that
$$
\lim_{n\rightarrow \infty}\frac{\mathcal{B}_{\mathrm{Ind}}(n)}{\mathcal{H}(n)}=0.
$$
\\
(ii):
As in the case of the chromatic polynomial
we replace $\mathcal{H}(n)$ by $\mathcal{H}^r(n) \leq 2^{{n \choose r}}$
and
$\mathcal{B}_{\mathrm{Ind}}(n)$
by
$\mathcal{B}_{\mathrm{Ind}}^r(n)$
in
Equation (\ref{eq 2}) and use
Lemma \ref{le:strategy}(i) and (iii).
\hfill
\end{proof}

Now Theorem \ref{th:main-indep}(i) and (ii) follow.

\ifskip
\else
2. A matching in $\mathcal{H}$ is a set of disjoint edges in $\mathcal{E}(\mathcal{H})$.

3. Matching polynomial of a hypergraph:

$$\mathrm{M}(H;X)=\sum_{i=0}^{\lfloor\frac{n}{k}\rfloor}\omega_{i}x^{i},$$
where $k$ is the minimum size of edge in $\mathcal{H}$ and $\omega_{i}$ is the number of matchings of size $i$ in $\mathcal{H}$.
\fi

Let $\mathcal{B}_{\mathrm{M}}(n)$ be the number of polynomials $p(X)$ such that there is a
hypergraph $H$ of order $n$ with $p(X)=\mathrm{M}(H;X)$.

\ifskip
\else
A hypergraph $\mathcal{H}$ is called an \textbf{matching-unique hypergraph} if the equality
$\mathcal{M}(\mathcal{H}, x)=\mathcal{M}(\mathcal{G}, x)$
implies that $\mathcal{H}$ and $\mathcal{G}$ are isomorphic.
\begin{thm}
Almost all hypergraphs are not matching-unique, i.e.,
$\lim_{n\rightarrow \infty}\frac{\mathcal{B}_{\mathrm{M}}(n)}{\mathcal{H}(n)}=0$.
\end{thm}
\fi

\begin{thm}
\begin{enumerate}[(i)]
\item
$
\lim_{n\rightarrow \infty}\frac{\mathcal{B}_{\mathrm{M}}(n)}{\mathcal{H}(n)}=0.
$
\item
$
\lim_{n\rightarrow \infty}\frac{\mathcal{B}^{r}_{\mathrm{M}}(n)}{\mathcal{H}^{r}(n)}=0.
$
\end{enumerate}
\end{thm}
\begin{proof}
(i):
As $0\leq \omega_{i}(\mathcal{H})\leq \binom{\lfloor\frac{n}{k}\rfloor}{i}$, then $$\mathcal{B}_{\mathrm{M}}(n)
\leq
\prod_{i=1}^{n}\binom{\lfloor\frac{n}{k}\rfloor}{i}
<
\prod_{i=1}^{n}(\frac{\lfloor\frac{n}{k}\rfloor\cdot e}{i})^{i}
<
\prod_{i=1}^{n}(\lfloor\frac{n}{k}\rfloor\cdot e)^{i}.$$

From Lemma \ref{le:labeled}, we have that
\begin{gather}\label{eq 4}
\frac{\mathcal{B}_{\mathrm{M}}(n)}{\mathcal{H}(n)}
\leq
\frac{\mathcal{B}_{\mathrm{M}}(n)\cdot n!}{\overline{\mathcal{H}}(n)}
<
\frac{\prod_{i=1}^{n}(\lfloor\frac{n}{k}\rfloor\cdot e)^{i}\cdot n!}{2^{2^{n}}}
<
\frac{\prod_{i=1}^{n}(\lfloor\frac{n}{k}\rfloor\cdot e)^{i}\cdot n^{n}}{2^{2^{n}}}.
\end{gather}

Now we use Proposition \ref{pr:precalculus} and
take
base $2$
logarithms
of the numerator and denominator, and we get in
Equation (\ref{eq 4}):
\begin{gather}
\frac{\log\mathcal{B}_{\mathrm{M}}(n)}{\log \overline{\mathcal{H}}(n)}
<
\frac{\frac{\lfloor\frac{n}{k}\rfloor\cdot (\lfloor\frac{n}{k}\rfloor+1)}{2}\cdot \log(n\cdot e)+n\cdot \log n}{2^{n}\cdot log 2}.
\notag
\end{gather}
Therefore,
\begin{gather}
\label{eq 5}
\lim_{n\rightarrow\infty}\frac{\log\mathcal{B}_{\mathrm{M}}(n)}{\log \overline{\mathcal{H}}(n)}
<
\lim_{n\rightarrow\infty}\frac{\frac{\lfloor\frac{n}{k}\rfloor\cdot (\lfloor\frac{n}{k}\rfloor+1)}{2}\cdot \log(n\cdot e)+n\cdot \log n}{2^{n}\cdot log 2}
=0,
\end{gather}
which implies that $\lim_{n\rightarrow \infty}\frac{\mathcal{B}_{\mathrm{M}}(n)}{\mathcal{H}(n)}=0$.
\\
(ii):
As in the case of the chromatic polynomial
we replace $\mathcal{H}(n)$ by $\mathcal{H}^r(n) \leq 2^{{n \choose r}}$
and
$\mathcal{B}_{\mathrm{M}}(n)$
by
$\mathcal{B}_{\mathrm{M}}^r(n)$
in
Equation (\ref{eq 5}) and use
Lemma \ref{le:strategy}(i) and (iii).
\hfill
\end{proof}

Now Theorem \ref{th:main-match}(i) and (ii) follow.

\section{Conclusions and further research}

We have shown that for $r$-uniform hypergraphs with $r \geq 3$, and hypergraphs in general,
there are very few hypergraphs which are unique for $\chi$, $\mathrm{Ind}$ and $\mathrm{M}$.
This is not so surprising as there are many more $r$-uniform hypergraphs of order $n$ than graphs.
Still, it is interesting to search for such graphs.

\begin{problem}
Find more $P$-unique $r$-uniform hypergraphs for $\chi$, $\mathrm{Ind}$ and $\mathrm{M}$.
\end{problem}
For $\mathrm{Ind}$ and $\mathrm{M}$ it seems this has not been properly investigated.

In \cite{ar:White2011} the Tutte polynomial $T(G;X,Y)$ and the most general edge elimnation polynomial $\xi(G;X,Y,Z)$
of \cite{ar:AverbouchGodlinMakowsky10} are generalized to hypergraphs.
$T(H;X,Y)$ is a substitution instance of $\xi(H;X,Y,Z)$ both on graphs and hypergraphs.
In \cite{ar:Kalman-2013} another version of a Tutte polynomial for hypergraphs is proposed.

\begin{problem}
Is $T(H;X,Y)$ almost complete for $r$-uniform hypergraphs.
\end{problem}
Note that the original BPR-conjecture asserts this for graphs, and is still open.
\begin{problem}
Is $\xi(H;X,Y,Z)$ almost complete for $r$-uniform hypergraphs.
\end{problem}

\subsection*{Acknowledgements}

This paper was conceived and written during
the {\em Tutte Centenary retreat 2017}, held at the MATRIX,
(https://www.matrix-inst.org.au/events/tutte-centenary-retreat/)
Creswick Campus of the University of Melbourne November 25- December 1, 2017.
We would like to thank the organizers
for inviting both authors and
for the pleasent working atmosphere at the retreat.
We would like to thank Graham Farr for his encouragement and
Vsevolod Rakita for insightful discussions.

\bibliographystyle{alpha}


\section{Introduction}
\label{sec:intro}
There has been a dramatic decrease in the extent and thickness of Arctic sea ice over the past few decades, with multi-year ice giving way to mostly first-year ice in the present-day Arctic \citep{Stroeve:2007aa, Comiso:2008aa,OneWatt}. While aspects of this decline are simulated in climate models, it is useful to develop heuristic understanding of the mechanisms causing it with simpler, analytic models. The relatively thin Arctic sea ice is much more vulnerable to external forcing than the Antarctic polar cap, which sits on land and is a few kilometers thick. To leading order, the balance of heat fluxes across the ice to the atmosphere and the ocean controls its thickness, with the ice being much more sensitive to the variation in oceanic heat flux ($O$(1) $Wm^{-2}$) than the atmospheric heat flux ($O$(100) $Wm^{-2}$) \citep{Maykut:1971aa}.

The sea-ice cover in the Arctic, though perennial, has been thinning rapidly in the past few decades. Because this perennial ice cover requires the presence of Multi-Year Ice (MYI), i.e. ice that persists for more than one melt season, there have been discussions regarding when and if the Arctic can transition to a seasonal ice cover \citep{EW09, Tietsche:2011, SerrezeBarry:2011, MW:2011, MW:2012, Toppaladoddi:2017aa}. The debate { whether or not the Arctic has surpassed a \emph{tipping point}} has mainly arisen because of the decay of MYI cover that covered nearly two-thirds of the basin in the 1970s to less than one-third in the past decade \citep{OneWatt}, leading to a state where the Arctic is mostly covered by first-year ice. Since the first-year ice retreats rapidly, knowing the location of the ice edge is an important step in determining how external forcings influence the basin-wide sea-ice properties.

Bering Strait, the Canadian Archipelago and Fram Strait are some of the gateways through which the Arctic Ocean interacts with the rest of the world oceans \citep{Manley:1995aa, Woodgate:2005aa, Kwok:2004aa, Meredith:2001aa}. While all these gateways are important, Fram Strait is the largest, deepest and most significant \citep{Untersteiner:1988aa, Vinje:1986aa, Vinje:1998aa}. The amount of fresh water exchange and deep water formation in these regions influence global ocean circulations \citep{Aagaard:1968aa, Aagaard:1968ab, Aagaard:1989aa, Rudels:1991aa, Steur:2009aa}. {The sea-ice export  through the Fram Strait constitutes nearly 10\% of all the sea ice in the Arctic, with 60\% of this occuring in the winter and the rest during the summer \citep{Smedsrud:2017aa}.} \citet{Colony:1985}, used a simple stochastic model to study the statistics of residence time of sea ice in different sectors of the Arctic. They demonstrated that, while it takes nearly 5 years for sea ice in the Beaufort sector to either melt or exit through the Fram Strait, the time for termination in the marginal ice zones is only 1-2 years. They also showed that the ice at the North Pole has a 35\% probability that it will exit in 1 year through the Strait and 96\% that it will exit through the Strait sometime. The sea-ice edge in the Fram Strait forms the northern boundary of the Atlantic Meridional Overturning Circulation (AMOC) \citep{Mauritzen:1997aa, Jungclaus:2005aa, Koenigk:2006aa, Miles:2014aa}. The AMOC circulates the cold and fresh waters from the Arctic to the Antarctic and back, forming a circulation that regulates global climate.

\citet{Untersteiner:1988aa} studied the interaction of southwards-flowing ice with the northward-bound warm and salty waters of the West Spitsbergen Current \citep{Vowinckel:1962aa, Aagaard:1987aa}. The interaction of these two currents gives rise to an ice wedge, which can be imagined as follows: A rectangular slab of ice starts to melt as it comes in contact with warm water underneath. This melt water in turn is fresh and cold and starts to cool the incoming warm water (see Fig. \ref{fig:IceWedge}(a)), giving rise to an ice wedge, which we refer to as \emph{Untersteiner's Sea-ice Wedge}. In steady state, the heat energy required to melt the ice is comparable to the heat lost from the warm water current. \citet{Untersteiner:1988aa} computed several integral quantities over this wedge, such as the depth, temperature and salinity of the mixed layer in steady state, without calculating the length of the wedge itself or its detrended profile. { In reality the North Atlantic Warm Waters (NAWW) divide upon entering the Fram Strait into multiple branches, with a part circulating back in the EGC, another entering west of Spitsbergen, another branch dividing near the Barents sea and circulating cyclonically as a boundary current underneath the Arctic Ocean and returning under the EGC. In this study, as a simplification and as comparison to \citet{Untersteiner:1988aa}, we treat the WSC as the branch of the NAWW entering the Arctic west of Svalbard and ignore the rest of the branches. There are a multitude processes that happen in the Arctic such as formation of polynyas, river runoff from the Eurasian and American basins, cooling, freezing and melting of ice. Polynyas are one of most significant sources of saline rejection to mixed layer and deep waters in the Arctic, while river runoff brings fresh water to the Arctic. These water transformations in the complete Arctic basin lead to the boundary current of NAWW that enters the basin to exit near the Fram Strait as cool and fresh water, as the EGC. While the EGC can be thought of as an output of this complex non-linear large-scale dynamical system that is the whole Arctic, we treat the EGC and WSC as two individual currents and study their influence on the sea-ice formation and edge dynamics in the North Atlantic. }

\begin{figure*}[htbp]
    \centering
    \includegraphics[trim = 0 0 0 0, clip, width = \textwidth]{Figure-1.pdf}
    \caption{(a) Schematic of an ice wedge, redrawn from Fig. 3 of \citep{Untersteiner:1988aa}. The warm and salty West Spitsbergen Current (WSC) flows northwards, which melts the southwards-flowing ice to create a wedge shape, whereas the relatively cold and fresh East Greenland Current (EGC) flows southwards with the ice and therefore does not generate such a wedge. (b) Schematic of our simple longitudinal model used to study the dynamics of the sea-ice edge in detail.}
    \label{fig:IceWedge}
\end{figure*}

Here, we study this phenomenon in more detail { by building a simple model for sea-ice export in the Fram Strait. This model allows us to analyze the individual physical mechanisms and their effect in much more detail than a complicated Global Circulation Model.} We look at the sea-ice-thickness distributions and mixed-layer properties along a longitude from a central basin in the Arctic to the sea-ice edge. Some of the key questions we ask are: What are the various factors responsible for formation of this ice wedge; Can we determine the ice edge based on external parameters, to know the location of the wedge; what is the typical length of the \emph{Untersteiner's Sea-ice Wedge} and what controls it; How do these ice-ocean interactions impact the dynamics of ice extent in the Arctic?

\section{Model}
\label{sec:CM}

Consider a slab of ice moving southwards from the north pole with velocity $U_i$ and an ocean current moving northwards with velocity $U_w$. The temperature and salinity of the oncoming ocean are considered to be constant at $T_w = 2^\circ{C}$ and $S_w = 36$\textperthousand, respectively. The relatively fresh, though cold melt water is less dense than the salty ocean water and forms a mixed layer beneath the melting ice (see Fig. \ref{fig:IceWedge}(b)).  The heat flux from the mixed layer to the underside of the ice $F_{ml} = \lambda_{ml} (T_{ml} - T_f)$, where $T_{ml}$ and $S_{ml}$ are the temperature and salinity of the mixed layer, $\lambda_{ml} = \rho c_p St |\bf{U_i} - \bf{U_w}|$ is the heat transfer coefficient \citep{McPhee:2008aa}, $c_p = 4186$ $J/kg/^\circ{C}$ is the specific heat capacity of water, and $St = 1.7 \times 10^{-4}$ is the Stanton number \citep{Omstedt:1992aa}. We assume that there is an abyssal heat flux from the ocean to the mixed layer $F_b$. In the presence of salt, the liquidus temperature $T_f$, which is decreased due to the presence of salt, can be approximated as $T_f = -\gamma S_{ml}$, where $\gamma \approx 0.055 ^\circ{C}/psu$. The net outgoing atmospheric heat flux at the surface of the ice is parameterized by $F_a = \lambda_a (T_s - T_a)$, where $T_s$ is the surface temperature at the ice surface, $T_a$ is the prescribed atmospheric temperature, and $\lambda_a$ is the heat-transfer coefficient, assumed to be a constant. Here, we take $T_a = -40 \cos \frac{\pi x}{2000}$ as a simple atmospheric forcing proxy to explore the role of latitudinally varying heat flux, where $x$ is distance in kilometers, measured southwards from the pole. The density of ice is assumed to be the same as the ocean ($\rho_{ice} = \rho_{ocean} = \rho = 1000kg/m^3$). For large \emph{Stefan number} ($= L/c_p \Delta T$), { where $\Delta T$ is the temperature difference between the top and bottom ice surfaces}, the heat-flux balance across the interfaces of ice of thickness $h$ and mixed layer of depth $H$ can be written as
\begin{eqnarray}
&F_{b}& - F_a = -\rho L \frac{D^i}{D t} h + \rho c_p\frac{D^w}{D t} H(T_{ml} - T_w)
\label{eq:FbFa_NS}
\end{eqnarray}
for the total heat flux balance and
\begin{equation}
F_b - F_{ml} = \rho c_p \frac{D^w}{D t}H(T_{ml} - T_w)
\label{eq:FbFml_NS}
\end{equation}
for the balance across the mixed layer. Here, $L = 334 kJ/kg$ is the latent heat of fusion of ice, and $\frac{D^i}{D t}$ and $\frac{D^w}{D t}$ are material derivatives with respect to $U_i$ and $U_w$, respectively.

The fresh water from the ice melt is entrained by the mixed layer, thereby reducing the mixed-layer salinity. In other words, the salt content is conserved in the sea-ice -- mixed-layer system. Writing this salt flux balance for the whole system gives
\begin{eqnarray}
\frac{D^i}{Dt}(S_i - S_w)h + \frac{D^w}{Dt} H(S_{ml} - S_w) = 0,
\label{eq:Salt}
\end{eqnarray}
where, $S_i$ is the salinity of ice. Assuming that the ice is fresh for simplicity, we take $S_i = 0$.

The mixed layer is assumed to deepen until it attains neutral buoyancy with respect to the underlying ocean to maintain a continuous density profile. This implies that the mixed layer has constant buoyancy $b$, which with an assumption of a linear equation of state gives
\begin{eqnarray}
\frac{D^w}{Dt}b &\equiv& \frac{D^w}{Dt}\left [ b_0 + (\alpha(T_{ml} - T_w) - \beta(S_{ml} - S_w) \right ] = 0. \nonumber \\
\Rightarrow
\alpha\frac{D^w}{Dt}T_{ml} &=& \beta\frac{D^w}{Dt}S_{ml}.
\label{eq:buoyancy}
\end{eqnarray}
where $\alpha$ and $\beta$ are coefficients of thermal expansion and solutal buoyancy, respectively.

The system of Eqns. \ref{eq:FbFa_NS}, \ref{eq:FbFml_NS}, \ref{eq:Salt} and \ref{eq:buoyancy} forms the complete system to be solved to determine sea-ice thickness $h$, mixed-layer depth $H$, mixed-layer temperature $T_{ml}$ and mixed-layer salinity $S_{ml}$.

\subsection{Steady State}
\label{sec:SS}
In steady state, the coupled system of Eqns. \ref{eq:FbFa_NS} -- \ref{eq:buoyancy} can be written as
\begin{eqnarray}
\frac{\partial}{\partial x}H(T_w - T_{ml}) &=& \frac{\beta S_w(F_a - F_{ml}) }{\alpha L\rho U_w}, \nonumber \\
\frac{\partial}{\partial x} h &=& \frac{F_a - F_{ml}}{L \rho U_i}, \nonumber \\
\mathcal{F} = \frac{\alpha L (F_b - F_{ml})}{\beta c_p S_w (F_a - F_{ml})} &=& 1, \nonumber \\
\mathcal{B} = \frac{\alpha (T_w - T_{ml})}{\beta (S_w - S_{ml})} &=& 1.
\label{eq:solve2}
\end{eqnarray}
Here, $\mathcal{F}$ is the energy conservation parameter and $\mathcal{B}$ is the density ratio between the mixed layer and the ocean underneath. In the second equation above, the sea-ice melt occurs due to a mismatch between its bottom and top fluxes and hence our model is agnostic with regard to the location of the melt. Using the conditions that $\mathcal{F} = 1$ and $\mathcal{B} = 1$ (from Eqns. \ref{eq:FbFa_NS}, \ref{eq:FbFml_NS}, \ref{eq:Salt}, \ref{eq:buoyancy} and applying $L'H\hat{o}pital$'s rule, see Supporting Information (SI)) over the domain, we can obtain $T_{ml}$, substitute it back in the solution for the first equation above and obtain the mixed-layer depth $H$. Thus we find
\begin{eqnarray}
T_{ml} &=& \frac{K F_b - F_a}{\lambda_{ml}(K-1)} - \gamma S_{ml}, \qquad \textnormal{where } K = \frac{\alpha L}{\beta c_p S_w}.
\label{eq:Tml}
\end{eqnarray}
Hence,
\begin{eqnarray}
S_{ml} &=& S_w - \frac{\alpha}{\beta}(T_w - T_{ml}).
\label{eq:Sml}
\end{eqnarray}

This formulation reduces the coupled system of four equations to two equations for sea-ice thickness $h$ and heat content of the mixed layer $H(T_w - T_{ml})$, with the last two equations in (\ref{eq:solve2}) providing the conditions to determine the other properties.

\subsection{Time-dependent Periodic Forcing}
\label{sec:time}
We would also like to understand how these dynamics play a role in the sea-ice export in the Fram Strait, under an ideal periodic forcing of atmospheric temperature. A seasonal cycle is included by varying the atmospheric temperature $T_a$ by $10^\circ{C}$ over 365 days, i.e. $T_a(x,t) = (T_a(x) - 5) + 5 \cos \frac{2 \pi t}{365}$.
The fully coupled time-dependent system of equations can be written as,
\begin{eqnarray}
\left(\frac{\partial}{\partial t} + {\bf{U_w}} \frac{\partial}{\partial x}\right)H(T_w - T_{ml}) &=& \frac{\beta S_w(F_a - F_{ml}) }{\alpha L\rho} \nonumber \\ \nonumber \\
\left(\frac{\partial}{\partial t} + {\bf{U_i}} \frac{\partial}{\partial x}\right) h &=& \frac{F_a - F_{ml}}{L \rho }\nonumber \\ \nonumber \\
\mathcal{F} = \frac{\alpha L (F_b - F_{ml})}{\beta c_p S_w (F_a - F_{ml})} &=& 1 \nonumber \\ \nonumber \\
\frac{\alpha (T_w - T_{ml})}{\beta (S_w - S_{ml})} &=& 1 \nonumber\\ \nonumber \\
sgn({\bf{U_w}}) &=& \left\{
     \begin{array}{@{}l@{\thinspace}l}
       -1  &: WSC\\
        1 &: EGC\\
     \end{array}
   \right \}
\label{eq:FullSystem}
\end{eqnarray}

The system of Eqs. \ref{eq:FullSystem} make up a simple model for mixed-layer properties and sea-ice thickness in the Fram Strait. We solve this hyperbolic system numerically. At first glance, one can see that this is a moving-boundary problem. Since the ice edge evolves in time, we map the domain onto a fixed rectangle using a new variable $\xi = x / x_e(t)$, where $0<x<x_e$ and $x_e$ is the ice edge at any time $t$. Substitution in to Eqs. \ref{eq:FullSystem} gives the equation governing the sea-ice thickness as
\begin{equation}
\frac{\partial}{\partial t}h - \frac{\dot{x_e}}{x_e} \xi \frac{\partial}{\partial \xi}h +  \frac{U_i}{x_e}\frac{\partial}{\partial \xi}h = \frac{F_a - F_{ml}}{L \rho}.
\label{eq:hscale}
\end{equation}
and the dynamics of the ice edge $x_e$ is given by
\begin{eqnarray}
\dot{x_e} &=& U_i, \qquad h_e > 0  \nonumber \\
\dot{x_e} &=& U_i - x_e \frac{\Delta F}{\frac{\partial h}{\partial \xi}}\big|_{\xi = 1}, \qquad h_e = 0.
\label{eq:xm}
\end{eqnarray}
where $\Delta F = \frac{F_a - F_{ml}}{L \rho}$ and $h_e$ is the ice thickness at the edge.

We solved the coupled system of Eqs. \ref{eq:FullSystem}, \ref{eq:hscale} and \ref{eq:xm} numerically using an upwind scheme. We prescribe a single ocean velocity profile across the Fram Strait based approximately on data from \citep{Aagaard:1968aa, Aagaard:1968ab, Foldvik:1988aa, Woodgate:1999aa, Cokelet:2008aa, Sutherland:2008aa, Steur:2009aa}  as Eq. \ref{eq:FS_Velocity} with distance measured west to east (with grid points $n$ every 21km), also depicted in Fig. \ref{fig:FS_VXM}(a), and keep the ice velocity fixed at $U_i = 0.1$ $ms^{-1}$. It has been shown that the waters in the Fram Strait do not interact significantly longitudinally in comparison to their latitudinal interaction, i.e. the measured east-west velocity is much smaller than the north-south component \citep{Boyd:1994aa, Woodgate:1999aa}. This allows us to use the model above to compute individual sea-ice profiles for each ($U_i,U_w$) velocity pair and stack them together. We limit the mixed-layer velocity in the East Greenland Current $U_w \le 0.075$ $ms^{-1}$ by assuming that the flow of the mixed layer is controlled by the flow of sea ice above it.

\begin{eqnarray}
{\hspace{-15mm}
U_w = \left\{
     \begin{array}{@{}l@{}ll}
     	\text{ EGC}&&\\
       0.075 \text{ ms}^{-1} &: 0km - 105km, &1 \le n \le 6,\\

	\text{EGC to WSC}&&\\
       0.075\cos \left(\frac{\pi (n-6)}{20} \right) \text{ ms}^{-1} &: 105km - 525km, &7 \le n \le 25,\\

	\text{ WSC}&&\\
       -0.257 \text{ ms}^{-1} &: 525km - 630km, &26 \le n \le 31,\\
     \end{array}
   \right.
   \label{eq:FS_Velocity}
   }
\end{eqnarray}

\section{Discussion and Results}
\label{sec:results}
\subsection{Steady State Results}
{Steady-state dynamics help in explaining how different components of the model interact with each other under a constant forcing. In addition to the longitudinal profiles for sea-ice thickness, and mixed layer properties, our simple model also allows us to compute the location of the sea-ice edge, the basin properties, and a wedge length scale analogous to \emph{Untersteiner's sea-ice wedge} analytically. }

\subsubsection{Analytic Solution for the Location of Ice Edge}
This simple model allows us to compute the location of the ice edge analytically, based on external parameters, which can be more easily observed.

From  Eq. \ref{eq:solve2}, $\mathcal{F} = 1$ at all places wherever there is ice, and hence $\mathcal{F}_{edge} = 1$ ($\mathcal{F}$ at edge). Therefore using $\mathcal{F}_{edge} = 1$ we can obtain the location of the steady-state ice edge $x_e$ from
\begin{eqnarray}
F_{a_{edge}}(x_e) &=& \frac{\alpha L}{\beta c_p S_w} (F_b - F_{ml_{edge}}) + F_{ml_{edge}},
\label{eq:Edge}
\end{eqnarray}
where $ F_{a_{edge}}$ and $F_{ml_{edge}}$ are the atmospheric and mixed-layer heat fluxes at the ice edge, respectively. Because the boundary conditions for temperature and salinity at the ice edge are known externally ($F_{a_{edge}}(x) = -\lambda_a (\gamma S_w + T_a(x))$,  $F_{ml_{edge}} = \lambda_{ml}(T_w + \gamma S_w)$), we can substitute those back in Eq. \ref{eq:Edge} and solve for $T_{a_{edge}}$. Since $T_a(x)$ is prescribed, the location of the ice edge $x_e$ is the root of the equation
\begin{equation}
T_a(x_e) = T_{a_{edge}}.
\label{eq:TaEdge}
\end{equation}

The sea-ice edge in Fig. \ref{fig:TmlH_h} and Figs. S1-S2 in the SI were computed using this analysis and then the system was integrated northwards to get the sea-ice profile and mixed-layer properties for the complete domain.

\subsubsection{Basin Properties}

The basin properties in this idealized model can be computed analytically as follows. The Arctic basin in steady state can be assumed to have $F_b = F_{ml}$ and $F_b = F_a$, i.e. the abyssal heat flux from the ocean is fully transmitted through the mixed layer and ice to the atmosphere, which gives
\begin{eqnarray}
F_b = \lambda_{ml} (T_{ml} + \gamma S_{ml}), \textnormal{ and}
\label{eq:Basin2}
\end{eqnarray}
\begin{eqnarray}
F_b &=& \lambda_a (T_s - T_a) \nonumber \\
\Rightarrow T_s &=& \frac{F_b}{\lambda_a} + T_a.
\label{eq:Ts2}
\end{eqnarray}

Substituting $S_{ml}$ from Eq. \ref{eq:Sml} in Eq. \ref{eq:Basin2} gives us an equation for $T_{ml_{Basin}}$
\begin{equation}
T_{ml_{Basin}} = \frac{F_b - \gamma\lambda_{ml} \left ( S_w - \frac{\alpha}{\beta}T_w \right )}{\lambda_{ml} \left ( 1 + \gamma\frac{\alpha}{\beta} \right )}.
\label{eq:Basin3}
\end{equation}
Substitution of $T_{ml_{Basin}}$ from Eq. \ref{eq:Basin3} in Eq. \ref{eq:Sml} gives us $S_{ml_{Basin}}$. For large \emph{Stefan number} ($= L/c_p \Delta T$), the temperature profile in the ice can be assumed to be linear, which gives
\begin{equation}
T_s = \frac{-k\gamma S_{ml} + \lambda_a h T_a}{k + \lambda_a h}
\label{eq:Ts1}
\end{equation}
where $k = 2.2 W m^{-1} K^{-1}$ is the heat conductivity of ice. The thickness of sea ice in the basin is obtained by equating Eqs. \ref{eq:Ts1} and \ref{eq:Ts2}, which gives
\begin{equation}
h_{Basin} = \frac{-k \left ( \gamma S_{ml} + T_a(0) + \frac{F_b}{\lambda_a}\right )}{F_b}
\label{eq:hBasin}
\end{equation}

The assumption of $F_b = F_{ml}$ and $F_b = F_a$ in the Arctic basin is similar to the assumption by \citet{Colony:1985} of no ice formation in the Central Arctic. In other words, since ice is always present in the basin, it requires that the rate of formation of ice must always be greater than the ice velocity $U_i$. This strict assumption holds in Figs. S1 and S2, where we show both $F_a = F_{ml} = F_b = 20 Wm^{-2}$. $T_{ml_{Basin}}$, $S_{ml_{Basin}}$ and $h_{Basin}$ are marked in Figs. S1 and S2, which show very good agreement with this analysis.

\subsubsection{Untersteiner's Sea-ice Wedge}

Imagine a rectangular block of sea ice moving southwards. As it starts to interact with a northward-bound warm and salty ocean current, it starts to melt \citep{Untersteiner:1988aa}. In steady state, it takes a wedge shape, with the heat required to melt the ice being equal to heat released from cooling of the incoming warm waters (see Fig. \ref{fig:IceWedge}(a)). A length scale $\eta$ can be ascribed to this wedge related to this transfer of heat. The maximum amount of heat being advected with respect to ice is

\begin{equation}
F_{Advect} = \rho c_p |U_w | H_o (T_{ml} - T_f)
\label{eq:F1}
\end{equation}
where $H_o$ is the mixed layer depth in the basin, calculated numerically. The amount of heat transferred from the mixed layer to ice that results in the melting of ice is
\begin{equation}
F_{Transfer} = \rho c_p St |U_i - U_w| \eta (T_{ml} - T_f).
\label{eq:F2}
\end{equation}
Here, $\eta$ is the length of the wedge over which $F_{Advect} = F_{Transfer}$. Equating Eqs. \ref{eq:F1} and \ref{eq:F2}, we get
\begin{equation}
\eta = \frac{|U_w| H_o}{|U_i - U_w| St}.
\label{eq:wedge}
\end{equation}
We show this wedge length scale in the solutions in Fig. \ref{fig:TmlH_h}.

We plot the solutions to Eqns. \ref{eq:solve2} in Fig. \ref{fig:TmlH_h} for $U_i = 10^{-1}m/s, U_w = 2.5\times10^{-1}m/s$. Fig. \ref{fig:TmlH_h} shows that the \emph{Untersteiner Sea-ice Wedge} is more apparent in the mixed-layer properties than in the sea-ice profile itself whose shape is determined dominantly by the latitudinal variation in atmospheric forcing. This wedge forms a boundary layer between the incoming ocean current and the mixed layer generated by the melting of ice. The wedge length scale in \citep{Untersteiner:1988aa} was approximated as 150 km, by arguing that melting ice is still apparent at 100 km from the ice edge, whereas at 200 km there is no sign of further melting. Using similar velocity profiles and external parameters, we calculate the wedge to have a length of approximately 105 km (see Fig. \ref{fig:TmlH_h}(c)). Untersteiner assumed $T_w = 2 ^\circ{C}$, $S_w = 35$\textperthousand, $U_i/U_w = 1/2$, and the sea-ice thickness at the end of the wedge to be 2.5m which gave a mixed-layer thickness at the end of the wedge of 24m, with mixed-layer temperature $T_{ml}$ of $-1.74^\circ{C}$. Given approximately the same parameters, we calculate $T_{ml}$ in this model as $-1.56^\circ{C}$. The average heat flux from the mixed layer to ice $F_{ml}$ was approximated as $300$ $Wm^{-2}$ in \citep{Untersteiner:1988aa}, while we calculate this quantity as $\sim 400$ $Wm^{-2}$. { To compare our model with observations, we used the mean atmospheric temperature profile over the Arctic from the \emph{Berkeley Earth Temperature} database \citep{Rohde:2013aa} over the period 1951-1980. This gives us $h = 2.43m$, $T_{ml} = -1.73^\circ{C}$ and $H = 37m$ in the Central Arctic, in agreement with the observed quantities during that period. }

\begin{figure}[htbp!]
	\includegraphics[trim = 0 0 0 0, clip, width = 0.5\textwidth]{Figure-2.pdf}
    \caption{With ice velocity $10^{-1}m/s$ and ocean velocity $2.5\times10^{-1}m/s$.(a)Ice thickness, $h$;(b)mixed-layer (ML) depth, $H$;(c)ML Temp., $T_{ml}$, Thick black line is the \emph{Untersteiner's Sea-ice Wedge}, $\eta = 105.5km$;(d)ML Salinity., $S_{ml}$; (e)mixed-layer heat flux $F_{ml}$; (f) Atmospheric heat flux $F_a$}
    \label{fig:TmlH_h}
\end{figure}

We vary the ocean current and sea-ice velocities to study the effect of $U_i$ and $U_w$ on these profiles. As the ice velocity is decreased, keeping the ocean current velocity fixed, the ice wedge becomes sharper (see Fig. S1 in SI, $U_i = 10^{-3}m/s, U_w = 2.5\times10^{-1}m/s$). As both ice and ocean currents are slowed down, the ice wedge becomes extremely narrow, behaving more like the rectangular ice slab that flows south (see Fig. S2 in SI,  $U_i = 10^{-5}m/s, U_w = 2.5\times10^{-2}m/s$). Fig. S4 in SI compares four velocity pairs here with the situation if there were only atmospheric forcing. This shows the sensitivity of sea-ice thickness to the heat flux from the mixed layer as well as the formation of the \emph{Untersteiner Sea-ice Wedge.}

\begin{figure}[htbp]
    \centering
    \includegraphics[trim = 0 0 0 0, clip, width = 0.5\textwidth]{Figure-3.pdf}
    \caption{(a)Velocity profile of the mixed layer in the Fram Strait from Eq. \ref{eq:FS_Velocity}. (b) Location of the ice edge in the Fram Strait as a function of time. The ice edge extends for more than 200km in the East Greenland Current compared to the West Spitsbergen Current.}
    \label{fig:FS_VXM}
\end{figure}

\subsection{Time-dependent Forcing}

To understand how a changing climate affects the sea-ice profile and its associated properties, one needs to look at the fully time-dependent dynamics because, depending on the rate of change of climate forcing, the system may never reach a steady state and will always be in a state of evolution. In Fig. \ref{fig:FS_VXM}(b) we plot the location of the ice edge in the Fram Strait as a function of time, where day number denotes the time from the start of the simulation. Fig. \ref{fig:FS_VXM}(b) shows the asymmetry in the slow growth rate of ice as it moves southwards vs the fast melt rate as it retreats northwards. It takes more than 45 days to grow approximately 150km while it takes less than 30 days to retreat back more than 250km. The difference in extent between the East Greenland Current vs. the West Spitsbergen Current is more than 200km, determined only by the variation in the ocean-current velocity. Another point to note is the difference in time when the respective longitudes reach the maximum ice-extent. The ice-extent in EGC keeps increasing even when the ice-edge in the WSC has started to retreat. This lag produces a nonlinear response to an almost linear change in the relative velocity. \citet{Agarwal:2017aa} showed that sea-ice velocity may be the dominant physical process in explaining the variability in the ice-extent dynamics in the Arctic on timescales up to a few years. This independent modeling study shows that only a small change in the relative velocity between ice and ocean currents can cause a very large variation in the sea-ice extent. We show the time series of the sea-ice edge for two velocity profiles in Fig. S5. These time series also show the transient behavior that the system exhibits to come to a steady seasonal cycle. The entire simulation showing the daily sea-ice edge for two seasonal cycles is shown in Movie S1. { To compare this model with the observations, similar to the steady state case, we used the \emph{Berkeley Earth Temperature} database \citep{Rohde:2013aa}. Varying the ocean velocity from 0.095$ms^{-1}$ in the EGC to 0.25$ms^{-1}$ in the WSC produces a sea-ice tongue in excess of 600km, showing the robustness of this model in studying the individual mechanisms of sea-ice export.}

\subsection{Implications for the Atlantic Meridional Overturning Circulation}
Sea ice is a major component of the Atlantic Meridional Overturning Circulation (AMOC), which regulates global climate by transporting heat and salt across the global oceans. Southwards flowing sea ice  and the relatively fresh mixed layer waters from the EGC provide a net freshwater flux to the AMOC, thus modulating its strength on timescales of decades and more \citep{Aagaard:1989aa, Timmermann:2003aa,  Sevellec:2017aa}. The location of the Arctic sea-ice edge can have huge influence on the strength of the AMOC \citep{Sevellec:2017aa}. Figs. \ref{fig:SS_V}(a,b) show the steady state location of the ice edge and basal sea-ice thickness as the temperature and salinity of the incoming Atlantic waters through the WSC are varied. These calculations show that the basal ice thickness does not vary strongly with the temperature of WSC, but only with its salinity. On the other hand, the location of the sea-ice edge has a strong dependence on both the temperature and salinity of the WSC. These are in agreement with the recent warming of the incoming Atlantic Waters through the WSC \citep{Walczowski:2007aa,Schauer:2008aa}  accompanied by the migration of the sea-ice edge northwards in the Greenland Sea leading to a decrease in the sea-ice extent.
\begin{figure*}[htbp!]
	\includegraphics[trim = 0 0 0 0, clip, width = \textwidth]{Figure-4.pdf}
    \caption{Steady state location of the sea-ice edge and basal sea-ice thickness as a function of $T_w$ and $S_w$ of the incoming Atlantic waters through the WSC, with $F_b = 20 Wm^{-2}$.}
    \label{fig:SS_V}
\end{figure*}
Movie S2 shows the simulation of sea-ice flow across the Fram Strait, with the thickness variation shown in colors. To calculate the annual sea-ice flux across the Fram Strait, we use the thickness profile across the north-most ice edge, which is a consequence of the WSC entering the Arctic near Spitsbergen. The volume flux $F_v$ is calculated as $F_v = \int_{n=1}^{n=31} U_n h_n dx$, where $U_n = 0.1ms^{-1}$ is the ice velocity, $h_n$ is the ice thickness at grid point $n$ and $dx = 21km$ is the spacing between each grid point. We compute this integral using the trapezoidal rule, which gives a value for $F_v = 1234km^3/year$ or about $0.04 Sv$, comparable to the noise amplitude used in simulations for determining the stability of the thermohaline circulations \citep{Timmermann:2003aa}. Figure S6 shows the daily sea-ice flux across the Fram Strait. \citet{Aagaard:1989aa} give a value of $2790km^3/year$ as an estimate of the sea-ice flux to the Greenland-Iceland-Norwegian Seas. While our value for the idealized forcings is significantly lower than their value, there are a few important reasons for this difference: (a) we use a constant sea-ice velocity profile, with $U_i = 0.1m/s$, while the realistic velocities may be larger in the EGC; (b) we use a very simplified ocean velocity profile across the Fram Strait (see Fig. \ref{fig:FS_VXM}(a)); (c) these computations are only looking at the first-year ice, which is considerably thinner than multiyear ice; {(d) the atmospheric temperature profile we assume is much simpler than observed. }
Retreat of the Arctic sea ice has led to increased warming and freshening of the open ocean in the North Atlantic with the warming having a greater impact than the freshening, leading to a destabilizing feedback to the Arctic sea ice. This retreat in turn gives rise to buoyancy anomalies in the North Atlantic causing a slowdown of the AMOC  \citep{Sevellec:2017aa} and hence affecting the global climate dynamics.

\section{Conclusions}
\label{sec:conc}
Fram Strait is the largest gateway in the Arctic through which the polar ocean interacts with the world oceans \citep{Untersteiner:1988aa}, and the sea-ice edge in the Fram Strait forms the northern boundary for the formation of Deep Water in the North Atlantic that drives the Atlantic MOC. Therefore, the location of the sea-ice edge is a factor influencing the strength and dynamics of this circulation, which in turn influences the global climate through ocean circulation, with time scales ranging from decades to centuries.

\citet{Thorndike:1982} have shown that more than $70\%$ of the variability in the sea-ice velocity fields can be explained by the geostrophic winds. \citet{Agarwal:2017aa} have shown further that this variability in the sea-ice velocity fields explains the dynamics of the observed sea-ice extent on time scales ranging from a few days up to a couple of years. Using a simple model, we have shown how the interaction between sea-ice currents flowing southwards and  ocean currents flowing southwards/northwards (East Greenland Current/WestSpitsbergen Current) influences the sea-ice edge and therefore sea-ice extent in the Fram Strait. { While the idealized forcings produce a sea-ice tongue of approximately 200km, the climatological temperature profile obtained from observations leads to a sea-ice tongue on the order of 600km, in agreement with the observations.}

{ The Atlantic waters entering the Arctic have become warmer as compared to the past few decades \citep{Walczowski:2007aa,Schauer:2008aa}, leading to an increased melt of sea ice and the thinner ice, being more susceptible to wind forcing, has become faster when flowing out of the Arctic. Our simple model has allowed us to vary parameters such as the ice and ocean current velocities, Atlantic temperature and salinity as well as atmospheric temperature to study their effect on the dynamics of sea-ice edge and accompanying mixed layer properties in the Fram Strait.} We show how the interaction between the oncoming sea-ice and ocean currents form a boundary-layer structure with a length scale associated with the \emph{Untersteiner Sea-ice Wedge}, which is more apparent in the mixed-layer properties than the sea ice itself. Our simple model also shows how to calculate values of key variables such as the sea-ice edge only from the external forcing, without integrating the complete model. This allows us to study the effect of climate change on the Arctic sea-ice extent directly. The dynamics of sea-ice export in the Fram Strait is a highly nonlinear coupled complex dynamical system, which produces a nonlinear response given a nearly linear velocity change, which is exhibited by our time-dependent solutions that show asymmetrical growth and melt times.

Our simple model is able to capture an ice tongue along Eastern Greenland that extends more than 200km further southwards than the Spitsbergen side of the system. It should be borne in mind that our numerical solutions only involve the export of first-year ice from the Arctic basin, which is then melted as it flows southwards, which produces a \emph{shorter} ice tongue than observed.  A key physical ocean phenomenon prevalent near the sea-ice edge are the eddies. While our simple model does not include these highly complicated flows explicitly, their overall physics is subsumed by the imposition of sea-ice / ocean velocity or the strength of the ocean current which governs the heat flux at the boundaries. There are many other physical mechanisms that have not been included in our simple model above, such as the Coriolis effect on the western boundary current and fresh-water runoff from Greenland, to name a few. However, this simple model illustrates important effects of heat and mass transfer between ice and ocean that contribute dominantly to the formation of the Arctic mixed layer and influence the principal physics of the marginal ice zone and the extent of exported sea ice. It also provides a framework to study the interaction of ocean and sea-ice currents in greater detail.

\acknowledgments
MGW is grateful for support from the Yale Institute for Biospheric Studies. This work started at the 2016 Geophysical Fluid Dynamics Summer Program, Woods Hole Oceanographic Institution, which is supported by the National Science Foundation and the Office of Naval Research. We thank all the Fellows and staff for their useful discussions. We also thank J.S. Wettlaufer for his comments. SA acknowledges support from the David Crighton Fellowship at University of Cambridge where this work was continued and Yale University for financial support.