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\begin{document}
\maketitle{}
\thispagestyle{empty}
\begin{abstract}
The complete list of Poisson--Lie structures on 4-d Galilei group is presented.
\end{abstract}
\newpage
\section{Introduction \label{s1}}

\indent Quantum groups have emerged in physics in connection with an attempt to
understand the symmetries underlying exact solvability of certain
quantum--mechanical and statistical models; they appeared to be quite powerful
in this respect.

Therefore, it is natural to ask whether their range of applicability as
a~mathematical tool for describing physical symmetries is wider and covers,
in particular, the most important case of space--time symmetries. The
theory of Hopf algebras offers a variety of structures which can be viewed
as deformations of classical space--time symmetry groups. For example,
a number of deformed Poincare groups were considered~\cite{b1}.
They posses many attractive features. However, if one is going to take
seriously the very idea of quantum space time symmetries, many conceptual
problems arise which solution seems to be quite difficult. It is rather obvious
that one should concentrate on deformations of relativistic symmetries because
it is  a high energy/ small distance region where the deviations from
the predictions of ``classical'' theory should occur. However, as far as the
conceptual problems are concerned, the nonrelativistic region provides similar
challenge. On the other hand it seems that the study of nonrelativistic deformed
symmetries is slightly simpler. One is not here faced with some complications
typical for relativistic quantum theory as, for example, the nonexistence
of particle number conserving nontrivial interactions.

In the present paper as a first step towards the understanding of
nonrelativistic quantum space--time symmetries we classify all
nonequivalent Poisson--Lie structures on
the four-dimensional Galilei group.
The method we use (c.f.\cite{b2}) is based on solving directly the cocycle
condition; it has been already used for finding Poisson--Lie structures
on the two-dimensional Galilei group~\cite{b3},\cite{b4}.

We find families of structures which cannot be related to each other by the
automorphisms of the Galilei group; contrary to the case of the
Poincare group \cite{b5}
many of them are not of the coboundary type.

The paper is organized as follows.
In sec.\ref{s2}\ we sketch a~general strategy while the results are presented
in sec.\ref{s3}. All technical details are relegated to the number of appendices.

Let us conclude the introduction which some details concerning
the Galilei\\ group\cite {b6}.

The generic element $g$\ of the
ten--parameter Galilei group $G$\ is denoted by
\be
g&=&(t,\vec{a},\vec{v},R)\label{w1}
\ee
We shall denote by\\
$T=\{(t,\vec{0},\vec{0},I)\}$\\
$S=\{(0,\vec{a},\vec{0},I)\}$\\
$V=\{(0,\vec{0},\vec{v},I)\}$\\
${\cal R}=\{(0,\vec{0},\vec{0},R)\}$\\
the subgroups of time translations, space--translations, pure Galilei transformations (boosts),
rotations, respectivly.

The group law is expressed by
\be
g''=g'\cdot g=(t',\vec{a'},\vec{v'},R')(t,\vec{a},\vec{v},R)=\nn\\
=(t+t',\vec{a'}+R'\vec{a}+\vec{v'}t,\vec{v'}+R'\vec{v},R'R)\label{w2}
\ee
The identity for the group is
\be
e=(0,\vec{0},\vec{0},I)\label{w3}
\ee
and  the inverse of the generic element is given by
\be
g^{-1}=&(t,\vec{a},\vec{v},R)^{-1}&=(-t,-R^{-1}(\vec{a}-t\vec{v}),-R^{-1}\vec{v},R^{-1})\label{w4}
\ee
The generators $H,\vec{P},\vec{K}$\ and $\vec{J}$\ of
the Galilei Lie algebra
${\cal G}$\ are defined with the help of the exponential parametrization
\be
g&=&e^{-itH}e^{i\vec{a}\vec{P}}e^{i\vec{v}\vec{K}}e^{i\vec{\theta}\vec{J}}\label{w5}
\ee
and they obey the
following commutation rules (only the nonvanishing ones are written up)
\be
{}[J_i,J_j]&=&i\varepsilon_{ijk}J_k\nn\\
{}[J_i,K_j]&=&i\varepsilon_{ijk}K_k\label{w6}\\
{}[J_i,P_j]&=&i\varepsilon_{ijk}P_k\nn\\
{}[K_i,H]&=&iP_i\nn\\
\ee
here and in the sequel the summation over repeated indices is understood
$(i,j,k=1,2,3)$.

The automorphism group of $G$\ consists of the inner automorphisms together
with two outer ones generated by space dilations and time dilations~:
\be
(t,\vec{a},\vec{v},R)&\to&(t,a\vec{a},a\vec{v},R)\nn\\
(t,\vec{a},\vec{v},R)&\to&(bt,\vec{a},b^{-1}\vec{v},R)\label{w7}
\ee
In what follows we shall need the right-invariant vector fields on $G$.
Denoting by $\hat{X}$\ the right-invariant vector field corresponding to the
element $X$\ of the Lie algebra ${\cal G}$\ we have
\be
\hat{H}&=&-i{\p\over\p t}\nn\\
\hat{P_i}&=&i{\p\over\p a_i}\label{w8}\\
K_i&=&i(t{\p\over\p a_i}+{\p\over\p v_i})\nn\\
J_i&=&-i\varepsilon_{ijk}a_j{\p\over\p a_k}-i\varepsilon_{ijk}v_j{\p\over\p v_k}
-i\varepsilon_{ijk}R_{jl}{\p\over\p R_{kl}}\nn
\ee
\section{Poisson--Lie structures on Galilei group --- the general strategy\label{s2}}

\indent Let us remind the notion of Poisson--Lie group~\cite{b7},\cite{b8}.
It is a Lie group $\tilde{G}$\ which has a~Poisson structure \hbox{\{ , \}}\
such that the multiplication map \hbox{$m: \tilde{G}\times\tilde{G}\to\tilde{G}$}\
is a~Poisson map (here \hbox{$\tilde{G}\times\tilde{G}$}\ is given
by the product
 Poisson structure).

Poisson--Lie structures can be described explicitly as follows. Let ${\tilde{\cal G}}$\
be the Lie algebra of $\tilde{G}$; denote by $\{X_i\}$\ an arbitrary basis in
${\tilde{\cal G}}$\ and  let $c_{ij}^k$\ be the corresponding structure constants.
One defines a~mapping \hbox{$\eta:\tilde{G}\to\wedge^2\tilde{\cal G}$},
\be
\eta(g)\equiv\eta^{ij}(g)X_i\otimes X_j &,& \eta^{ij}(g)=-\eta^{ji}(g)\label{w9}
\ee
Let $\{X^R_i\}$\ be the realization of $\tilde{\cal G}$\ in terms of right
invariant vector fields on~$\tilde{G}$. The Poisson bracket on $\tilde{G}$\
given by
\be
\{\Phi,\Psi\}&=&\eta^{ij}(X^R_i\Phi)(X_j^R\Psi)\label{w10}
\ee
defines the Poisson--Lie structure on $\tilde{G}$\ provided the following conditions are
obeyed\\
(i) Poisson--Lie property (co-cycle condition)
\be
\eta(g'g)&=&\eta(g')+Ad(g')\eta(g)\label{w11}
\ee
(ii) Jacobi identity
\be
\eta^{il}X^R_l\eta^{jk}+\eta^{kl}X^R_l\eta^{ij}+\eta^{jl}X^R_l\eta^{ki}-\nn\\
{}-c_{lp}^j\eta^{il}\eta^{pk}-c^i_{lp}\eta^{kl}\eta^{pj}-
c^k_{lp}\eta^{jl}\eta^{pi}=0\label{w12}
\ee

The inverse is also true: any Poisson--Lie structure on $\tilde{G}$\ can be
written in the above form. The infinitesimal analogues of Poisson--Lie
groups are Lie bialgebras. For any \hbox{$X\in\tilde{\cal G}$} define
\be
\delta(X)&\equiv&{d\over dt}\eta(e^{itX})|_{t=0}\label{w13}
\ee
Then it can be easily shown that $\delta:\tilde{\cal G}\to \wedge^2\tilde{\cal G}$\
has the following properties which are the infinitesimal counterparts of~(i) and~(ii):\\
(i') co-cycle conditions
\be
\delta([X,Y])&=&[X\otimes I+I\otimes X,\delta(Y)]+[\delta(X),I\otimes Y+Y\otimes I]\label{w14}
\ee
(ii') co-Jacobi identity
\be
\sum\limits_{c.p.}(\delta\otimes id)\circ\delta(X)&=&0\label{w15}
\ee
where $c.p.$\ means the summation over cyclic permutation of the factors in
\hbox{$\tilde{\cal G}\otimes\tilde{\cal G}\otimes\tilde{\cal G}$}.

Every Poisson--Lie structure on $\tilde{G}$\ defines the bialgebra structure
 on~$\tilde{\cal G}$. The inverse is true provided $\tilde{G}$\ is connected and
simply connected\cite{b9}. Two Poisson--Lie structures on~$\tilde{G}$\ will
be called equivalent if there exists an automorphism of~$\tilde{G}$\ which is a
Poisson map.

The main aim of the present paper is to classify, up to equivalence, all
Poisson--Lie structures on the four-dimensional
Galilei group~$G$. We adopt the
following, rather straightforward, strategy.

First, we write out $\eta$\ in the form
\be
\eta(g)&=&\Psi_i(g)H\wedge J_i+\Phi_i(g)H\wedge P_i+\Gamma_i(g) H\wedge K_i\nn\\
&&+\Lambda_i(g)\varepsilon_{ijk}P_j\wedge P_k+\Upsilon_{ij}(g)P_i\wedge K_j\label{w16}\\
&&+\Sigma_{ij}(g)P_i\wedge J_j+\Xi_i(g)\varepsilon_{ijk}K_j\wedge K_k+\nn\\
&&+\Omega_{ij}(g)K_i\wedge J_j+\Pi_i(g)\varepsilon_{ijk}J_j\wedge J_k\nn
\ee
where
\hbox{$g\equiv (t,\vec{a},\vec{v},R)$} is an arbitrary element of
the Galilei group~$G$.

Inserting the expansion \mref{w16} into the one-cocycle condition \mref{w11} one
obtains the following set of functional equations for the coefficients $\Psi_i,\Phi_i$\ , etc.
\be
\Psi_i(gg')&=&\Psi_i(g)+R_{il}\Psi_l(g')\nn\\
\Phi_i(gg')&=&\Phi_i(g)+R_{il}\Phi_l(g')+\varepsilon_{ink}(v_nt-a_n)R_{kl}\Psi_l(g')-\nn\\
&&-tR_{il}\Gamma_l(g')\nn\\
\Gamma_i(gg')&=&\Gamma_i(g)+R_{il}\Gamma_l(g')-\varepsilon_{ink}v_nR_{kl}\Psi_l(g')\nn\\
\Lambda_i(gg')&=&\Lambda_i(g)+R_{il}\Lambda_l(g')-{1\over 2}\varepsilon_{imn}v_mR_{nl}\Phi_l(g')+\nn\\
&&+\fr [(\vec{v}^2t-a_mv_m)\delta_{in}-v_n(v_it-a_i)]R_{nl}\Psi_l(g')+\nn\\
&&+\fr t\varepsilon_{imn}v_mR_{nl}\Gamma_l(g')-\fr t R_{il}\varepsilon_{lmn}\Upsilon_{mn}(g')+\nn\\
&&+\fr [(tv_i-a_i)\delta_{mp}-R_{im}R_{np}(tv_n-a_n)](\Sigma_{pm}(g')\nn\\
&&-\fr t\Omega_{pm}(g'))+t^2R_{il}\Xi_l(g')+\nn\\
&&+(t^2v_pv_i-v_pa_it-v_ia_pt+a_pa_i)R_{pm}\Pi_m(g')\nn\\
\Upsilon_{ij}(gg')&=&\Upsilon_{ij}(g)+R_{im}R_{jn}\Upsilon_{mn}(g')+v_i\varepsilon_{jnk}v_nR_{kl}\Psi_l(g')\nn\\
&&-v_iR_{jl}\Gamma_l(g')-\varepsilon_{jnl}v_nR_{ip}R_{lk}\Sigma_{pk}\nn\\
&&-2t\varepsilon_{ijs}R_{sl}\Xi_l(g')\nn\\
&&+[\varepsilon_{inl}(a_n-v_nt)R_{jp}+\varepsilon_{jnl}v_ntR_{ip}]R_{lk}\Omega_{pk}(g')\nn\\
&&+2[\varepsilon_{ijn}v_n(a_p-v_pt)R_{pm}-\varepsilon_{njr}a_nv_rR_{im}]\Pi_m(g')\label{w17}\\
\Sigma_{ij}(gg')&=&\Sigma_{ij}(g)+R_{ip}R_{jk}\Sigma_{pk}(g')-v_iR_{jl}\Psi_l(g')\nn\\
&&-tR_{ip}R_{jk}\Omega_{pk}(g')+\nn\\
&&+2[(a_j-tv_j)R_{im}-(a_l-tv_l)R_{lm}\delta_{ij}]\Pi_m(g')\nn\\
\Xi_i(gg')&=&\Xi_i(g)+R_{in}\Xi_n(g')-\fr (v_i\delta_{pm}-R_{im}R_{np}v_n)\Omega_{pm}(g')+\nn\\
&&+v_iv_pR_{pm}\Pi_m(g')\nn\\
\Omega_{ij}(gg')&=&\Omega_{ij}(g)+R_{ip}R_{jk}\Omega_{pk}(g')+2(R_{im}v_j-R_{lm}v_l\delta_{ij})\Pi_m(g')\nn\\
\Pi_i(gg')&=&\Pi_i(g)+R_{im}\Pi_m(g')\nn
\ee

In spite of their complicated structure they can be solved in the following way
(cf.\cite{b2},\cite{b3},\cite{b4}). One decomposes the general element
\hbox{$g=(t,\vec{a},\vec{v},R)$} into the product of four elements belonging
to the subgroups of time--and--space--translations, boosts and rotations,
(see sec.\ref{s1}).
\be
(t,\vec{a},\vec{v},R)&=&(t,\vec{0},\vec{0},I)\cdot(0,\vec{a},\vec{0},I)\cdot(0,\vec{0},\vec{v},I)\cdot
(0,\vec{0},\vec{0},R)\label{w18}
\ee
According to the condition (i) one can successively calculate $\eta(g)$\ using the
above decomposition provided the form of $\eta$\ for all four subgroups is
known. In order to find the latter we specify eqs.\mref{w17} to those
subgroups. The resulting equations can be easily solved; this is done in Appendix~A.
However, in obtaining the final form of $\eta$\ we apply eq.(19) with some definite
order of multiplication
(for example using \hbox{$(t,\vec{a},\vec{v},R)=(t,\vec{0},\vec{0},I)\cdot$}\\
\hbox{$\cdot((0,\vec{a},0,I)((0,0,\vec{v},I)(0,\vec{0},\vec{0},R)))$})
so there could be further constraints on the
parameters entering $\eta$\ following
from associativity. Therefore, we reinsert $\eta$\ into eq.\mref{w11} with arbitrary
$g$\ and $g'$\ to find all missed relations between parameters. In this way we
produce the general solution to eq.\mref{w11} described in Appendix~B (see eq.\mref{wb1}).

There remains to solve (ii) which imposes additional relations between
coefficients of $\eta$. It is very tedious to try to solve eq.\mref{w12} directly
so we adopt a~different method. From our general form of $\eta$\ we calculate
$\delta$\ and impose (ii') which, in this context, is equivalent to (ii). On
the other hand it is well known that (ii') can be restated as the condition
that the dual map $\delta^*$\ defines a Lie algebra structure on ${\cal G}^*$.
Therefore, we first calculate $\delta$\ and the commutators on ${\cal G}^*$\
resulting from it and then we solve the Jacobi identities. This is still
a complicated problem but it can be simplified by using the boost and translation
automorphisms of the Galilei group/algebra. Once this is done there remains only
to use the residual automorphisms to put our solutions in canonical position.
The more detailed discussion is given in Appendix~C.

At the end, having the form of $\eta$\ (eq.\mref{w16}) and the form of the right-invariant
vector fields (eq.\mref{w8}), using eq.\mref{w10} and taking into
account that for all families of solutions $\Pi_i(g)\equiv 0$\ (see eq.\mref{wb1}
and sec.\ref{s3}) one can easily calculate the following fundamental
Poisson--Lie brackets.
\be
\{R_{ab},R_{cd}\}&=&0\nn\\
\{v_a,R_{bc}\}&=&\varepsilon_{bjl}R_{lc}\Omega_{aj}\nn\\
\{a_a,R_{bc}\}&=&\varepsilon_{bjl}R_{lc}(\Sigma_{aj}+t\Omega_{aj})\nn\\
\{t,R_{bc}\}&=&-\varepsilon_{bjl}R_{lc}\Psi_j\nn\\
\{t,v_a\}&=&\Gamma_a-\varepsilon_{ajl}v_l\Psi_j\label{w19}\\
\{v_a,v_b\}&=&-2\varepsilon_{abj}\Xi_j+\varepsilon_{bjl}v_l\Omega_{aj}-\varepsilon_{ajl}v_l\Omega_{bj}\nn\\
\{a_a,v_b\}&=&-\Upsilon_{ab}+\varepsilon_{bjl}v_l\Sigma_{aj}-\nn\\
&&-2t\varepsilon_{abj}\Xi_j+t\varepsilon_{bjl}v_l\Omega_{aj}-\varepsilon_{ajl}a_l\Omega_{bj}\nn\\
\{a_a,a_b\}&=&-2\varepsilon_{abj}\Lambda_j-t\Upsilon_{ab}+t\Upsilon_{ba}-2t^2\varepsilon_{abj}\Xi_j\nn\\
&&+\varepsilon_{bjl}a_l\Sigma_{aj}-\varepsilon_{ajl}a_l\Sigma_{bj}+t\varepsilon_{bjl}a_l\Omega_{aj}\nn\\
&&-t\varepsilon_{ajl}a_l\Omega_{bj}\nn\\
\{t,a_a\}&=& -\varepsilon_{ina}\alpha_n\Psi_i+\Phi_a+\Gamma_a\nn
\ee

\section{Poisson--Lie structures on Galilei group --- the results\label{s3}}

\indent By applying the procedure outlined above we  solve the relevant
Jacobi identities for dual algebra (making use of boost and translation automorphisms) and we
arrive at the following families of Poisson--Lie structures (for all cases
\hbox{$\vec{n}=0$}).
\begin{description}
\item[I.] $\vec{\alpha}$--arbitrary,\ $\beta\not=0,\;\;\vec{\gamma}=0,\;\;\vec{\phi}=0,\;\;
v=0,\;\;\vec{\xi}=0,\;\;\theta=0,\;\;\rho=0,\;\;\sigma_{ij}=0,\;\;\chi_{ij}=0,\;\;\omega_{ij}=0$\\
free parameters: $\vec{\alpha},\beta\not=0$
\item[II.] $\vec{\alpha}\not=0,\;\;\beta=0,\;\;\vec{\gamma}=0,\;\;\vec{\phi}=F\vec{\alpha},\;\;
\vec{\lambda}=L\vec{\alpha},\;\;v$--arbitrary,\ $\vec{\xi}=0,\;\;\theta=0,\;\;\rho=0,\;\;\sigma_{ij}=0,\;\;
\omega_{ij}=W(\vec{\alpha^2}\delta_{ij}-\alpha_i\alpha_j),\;\; \chi_{ij}=B(\alpha_i\alpha_j-{1\over 3}\vec{\alpha^2}\delta_{ij})+2Wv\varepsilon_{ijk}\alpha_k$\\
free parameters: $F,L,v,W\not=0,B$
\item[III.] $\vec{\alpha}=0,\;\;\beta=0,\;\;\vec{\gamma}=0,\;\;\vec{\phi}=F\vec{\mu},\;\;
\vec{\lambda}=L\vec{\mu},\;\;v=0,\;\;\vec{\xi}=0,\;\;\theta=0,\;\;\rho=0,\;\;\sigma_{ij}=0,\;\;
\omega_{ij}=W(\delta_{ij}-\mu_i\mu_j), \;\;\chi_{ij}=B(\mu_i\mu_j-{1\over 3}\delta_{ij})+C\varepsilon_{ijk}\mu_k$\\
free parameters: $F\not=0,L,B,C,W\not=0,\vec{\mu},||\vec{\mu}||=1$
\item[IV.] $\vec{\alpha}=0,\;\;\beta=0,\;\;\vec{\gamma}=0,\;\;\vec{\phi}=0,\;\;
\vec{\lambda}=L\vec{\mu},\;\;v=0,\;\;\vec{\xi}=X\vec{\mu},\;\;\theta$--arbitrary,\ $\rho=0,\;\;\sigma_{ij}=0,\;\;
\omega_{ij}=W(\delta_{ij}-\mu_i\mu_j), \;\;\chi_{ij}=B(\mu_i\mu_j-{1\over 3}\delta_{ij})+C\varepsilon_{ijk}\mu_k$\\
free parameters: $L,X,\theta,W\not=0,B,C,\vec{\mu},||\vec{\mu}||=1$
\item[V.] $\vec{\alpha}\not=0,\;\;\beta=0,\;\;\vec{\gamma}=0,\;\;\vec{\phi}=F\vec{\alpha},\;\;
\vec{\lambda}=L\vec{\alpha},\;\;v$--arbitrary,\ $\vec{\xi}=0,\;\;\theta=0,\;\;\rho=0,\;\;\sigma_{ij}=0,\;\;
\omega_{ij}=0,\;\; \chi_{ij}=B(\alpha_i\alpha_j-{1\over 3}\vec{\alpha^2}\delta_{ij})$\\
free parameters: $\vec{\alpha}\not=0,F,L,v,B$
\item[VI.] $\vec{\alpha}=0,\;\;\beta=0,\;\;\vec{\gamma}=0,\;\;\vec{\phi}=0,\;\;v=0,\;\;,\vec{\xi}=0\;\;
\theta$--arbitrary, $\;\; \rho=0,\;\;\sigma_{ij}=0,\;\;
\omega_{ij}=0,$\\
$\vec{\lambda}$\ and $\chi_{ij}$-- arbitrary except that $\varepsilon_{abc}\chi_{ab}=0$\\
free parameters: $\vec{\lambda},\chi_{ab},\theta$
\item[VII.] $\vec{\alpha}=0,\;\;\beta=0,\;\;\vec{\gamma}=0,\;\;\phi_i=F\varepsilon_{imn}\chi_{mn},\;\;
\vec{\lambda}=0,\;\;v=0,\;\;\vec{\xi}=0,\;\;\theta=0,\;\;\rho=0,\;\;\sigma_{ij}=0,\;\;
\omega_{ij}=0,$\\
$\chi_{ij}$-- arbitrary except that $\varepsilon_{imn}\chi_{mn}\not=0$\\
free parameters: $F\not=0,\chi_{mn}$
\item[VIII.] $\vec{\alpha}=0,\;\;\beta=0,\;\;\vec{\gamma}=0,\;\;\phi_i\not=F\varepsilon_{imn}\chi_{mn}\;\; \hbox{and}\;\; \vec{\phi}\not=0,\;\;
\vec{\lambda}=L\vec{\phi},\;\;v=0,\;\;\vec{\xi}=0,\;\;\theta=0,\;\;\rho=0,\;\;\sigma_{ij}=0,\;\;
\omega_{ij}=0,$\\
$\chi_{ij}$-- arbitrary\\
free parameters: $\vec{\phi}\not=0,L,\chi_{ij}$
\item[IX.] $\vec{\alpha}=0,\;\;\beta=0,\;\;\vec{\gamma}=0,\;\;\vec{\phi}\hbox{--arbitrary },\;\;
\vec{\lambda}=0,\;\;v\not=0,\;\;\vec{\xi}=0,\;\;\theta=0,\;\;\rho=0,\;\;\sigma_{ij}=0,\;\;
\omega_{ij}=0,\;\;\chi_{ij}\hbox{--arbitrary }$\\
free parameters: $\vec{\phi},v\not=0,\chi_{ij}$
\item[X.] $\vec{\alpha}=0,\;\;\beta=0,\;\;\vec{\gamma}=0,\;\;\vec{\phi}=0,\;\;
\vec{\lambda}\hbox{--arbitrary},\;\;v=0,\;\;\vec{\xi}\not=0,\;\;\theta\hbox{--arbitrary},\;\;\rho=0,\;\;\sigma_{ij}=0,\;\;
\omega_{ij}=0,\;\;\chi_{ij}\hbox{--arbitrary except that }\\ \varepsilon_{abc}\chi_{ab}\xi_c=0$\\
free parameters: $\vec{\lambda},\vec{\xi}\not=0,\theta,\chi_{ij}$
\item[XI.] $\vec{\alpha}=0,\;\;\beta=0,\;\;\vec{\gamma}=0,\;\;\vec{\phi}=0,\;\;
\vec{\lambda}=L\vec{\mu},\;\;v=0,\vec{\xi}=0,\;\;\theta\hbox{--arbitrary},\;\;\rho=-{1\over 3}S,\;\;\sigma_{ij}=S(\mu_i\mu_j-{1\over 3}\delta_{ij}),\;\;
\omega_{ij}=0,\;\;\chi_{ij}=B(\mu_i\mu_j-{1\over 3}\delta_{ij})$\\
free parameters: $S\not=0,L,B,\theta,\vec{\mu},||\vec{\mu}||=1$
\item[XII.] $\vec{\alpha}=0,\;\;\beta=0,\;\;\vec{\gamma}=0,\;\;\vec{\phi}=0,\;\;
\vec{\lambda}=0,\;\;v=0,\;\;\vec{\xi}=0,\;\;\theta\hbox{--arbitrary},\;\;\rho=-{1\over 3}S,\;\;\sigma_{ij}=S(\mu_i\mu_j-{1\over 3}\delta_{ij}),\;\;
\omega_{ij}=0,\;\;\chi_{ij}=B(\mu_i\mu_j-{1\over 3}\delta_{ij})+C\varepsilon_{ijk}\mu_k$\\
free parameters: $S\not=0,B,C\not=0,\theta,\vec{\mu},||\vec{\mu}||=1$
\item[XIII.] $\vec{\alpha}=0,\;\;\beta=0,\;\;\vec{\gamma}=0,\;\;\vec{\phi}=0,\;\;
\vec{\lambda}=0,\;\;v\not=0,\;\;\vec{\xi}=0,\;\;\theta=0,\;\;\rho=-{1\over 3}S,\;\;\sigma_{ij}=S(\mu_i\mu_j-{1\over 3}\delta_{ij}),\;\;
\omega_{ij}=0,\;\;\chi_{ij}=B(\mu_i\mu_j-{1\over 3}\delta_{ij})+C\varepsilon_{ijk}\mu_k$\\
free parameters: $S\not=0,v=0,B,C,\vec{\mu},||\vec{\mu}||=1$
\item[XIV.] $\vec{\alpha}=0,\;\;\beta=0,\;\;\vec{\gamma}=0,\;\;\vec{\phi}=0,\;\;
\vec{\lambda}=L\vec{\mu},\;\;v=0,\;\;\vec{\xi}=X\vec{\mu},\;\;\theta\hbox{--arbitrary },\;\;\rho=-{1\over 3}S,\;\;\sigma_{ij}=S(\mu_i\mu_j-{1\over 3}\delta_{ij}),\;\;
\omega_{ij}=0,\;\;\chi_{ij}=B(\mu_i\mu_j-{1\over 3}\delta_{ij})$\\
free parameters: $S\not=0,X\not=0,L,\theta,B,\vec{\mu},||\vec{\mu}||=1$
\item[XV.] $\vec{\alpha}=0,\;\;\beta=0,\;\;\vec{\gamma}=0,\;\;\vec{\phi}=F\vec{\mu},\;\;
\vec{\lambda}=0,\;\;v\not=0,\;\;\vec{\xi}=0,\;\;\theta=0,\;\;\rho=-{1\over 3}S,\;\;\sigma_{ij}=S(\mu_i\mu_j-{1\over 3}\delta_{ij}),\;\;
\omega_{ij}=0,\;\;\chi_{ij}=B(\mu_i\mu_j-{1\over 3}\delta_{ij})+C\varepsilon_{ijk}\mu_k$\\
free parameters: $S\not=0,F\not=0,v\not=0,B,C,\vec{\mu},||\vec{\mu}||=1$
\item[XVI.] $\vec{\alpha}=0,\;\;\beta=0,\;\;\vec{\gamma}=0,\;\;\vec{\phi}=F\vec{\mu},\;\;
\vec{\lambda}=0,\;\;v=0,\;\;\vec{\xi}=0,\;\;\theta=0,\;\;\rho=-{1\over 3}S,\;\;\sigma_{ij}=S(\mu_i\mu_j-{1\over 3}\delta_{ij}),\;\;
\omega_{ij}=0,\;\;\chi_{ij}=B(\mu_i\mu_j-{1\over 3}\delta_{ij})+C\varepsilon_{ijk}\mu_k$\\
free parameters: $S\not=0,F\not=0,B,C,\vec{\mu},||\vec{\mu}||=1$
\item[XVII.] $\vec{\alpha}=0,\;\;\beta=0,\;\;\vec{\gamma}=0,\;\;\vec{\phi}=F\vec{\mu},\;\;
\vec{\lambda}=L\vec{\mu},\;\;v=0,\;\;\vec{\xi}=0,\;\;\theta=0,\;\;\rho=-{1\over 3}S,\;\;\sigma_{ij}=S(\mu_i\mu_j-{1\over 3}\delta_{ij}),\;\;
\omega_{ij}=0,\;\;\chi_{ij}=B(\mu_i\mu_j-{1\over 3}\delta_{ij})$\\
free parameters: $S\not=0,F\not=0,B,L,\vec{\mu},||\vec{\mu}||=1$
\item[XVIII.] $\vec{\alpha}=0,\;\;\beta=0,\;\;\vec{\gamma}\not=0,\;\;\vec{\phi}=0,\;\;
\vec{\lambda}=L\vec{\gamma},\;\;v=0,\;\;\vec{\xi}=X\vec{\gamma},\;\;\theta=0,\;\;\rho=0,\;\;\sigma_{ij}=-\varepsilon_{ijk}\gamma_k,\;\;
\omega_{ij}=0,\;\;\chi_{ij}=0$\\
free parameters: $X,L$.
\end{description}

Let us note that all Poisson--Lie structures with $\beta=0,\;v=0$\ and $\theta=0$\
are coboundary and the corresponding r--matrix reads
\be
r&=&i\phi_kH\wedge P_k+i\gamma_kH\wedge K_k+i\alpha_k H\wedge J_k+\nn\\
&&+i\varepsilon_{ijk}\lambda_k P_i\wedge P_j+i(\sigma_{ij}-\rho\delta_{ij})P_i\wedge J_j\label{w20}\\
&&+i\chi_{ij} P_i\wedge K_j-i(2\omega_{ij}-\omega_{nn}\delta_{ij})J_i\wedge K_j\nn\\
&&+i\varepsilon_{ijk}\xi_kK_i\wedge K_j\nn
\ee

Now there remains only to classify the orbits under the action of residual
automorphisms corresponding to the rotations and scaling and put our structure in the canonical
form. This is a~straightforward although very tedious task. The result can be
summarized as follows. There are 69 families of inequivalent Poisson--Lie
structures which have been grouped for convenience into eight groups. Each
group is described by an appropriate tables which are given below. They
provide the main result of our paper.

Let us note that for all groups $\vec{n}=0$. In the last column
(labelled by \#) we indicate the number of essential parameters.

I. $\rho=0,\sigma_{ij}=0,\omega_{ij}=0,\chi_{ij}=0,\vec{\gamma}=0$
\vskip0.4cm
\begin{tabular}{|c|c|c|c|c|c|c|c|c|}
\hline
N&$\vec{\alpha}$&$\beta$&$\vec{\phi}$&$\vec{\lambda}$&$v$&$\vec{\xi}$&$\theta$&\#\\
\hline
1&$(0,0,\alpha)$&1&0&0&0&0&0&1\\
&$\alpha\geq0$&&&&&&&\\
2&(0,0,1)&0&(0,0,1)&(0,0,$L$)&$v$&0&0&2\\
3&(0,0,1)&0&0&(0,0,$\pm$1)&$v$&0&0&1\\
4&(0,0,1)&0&0&0&1&0&0&0\\
5&(0,0,1)&0&0&0&0&0&0&0\\
6&0&0&(0,0,1)&(0,0,$\pm$1)&0&0&0&0\\
7&0&0&(0,0,1)&0&0&0&0&0\\
8&0&0&(0,0,1)&0&1&0&0&0\\
9&0&0&0&$\lambda_1=0$&0&(0,0,1)&$\theta$&2\\
&&&&$\lambda_2^2+\lambda_3^2=1$&&&&\\
10&0&0&0&(0,0,1)&0&0&$\pm1$&0\\
11&0&0&0&(0,0,1)&0&0&0&0\\
12&0&0&0&0&0&(0,0,1)&$\theta$&1\\
13&0&0&0&0&0&0&$\pm1$&0\\
14&0&0&0&0&0&0&0&0\\
\hline
\end{tabular}
\vskip0.3cm
II. $\beta=0,\vec{\gamma}=0,\rho=0,\sigma_{ij}=0,\omega_{ij}=W(\delta_{ij}-\delta_{i3}\delta_{j3}),
\chi_{ij}=B(\delta_{i3}\delta_{j3}-{1\over 3}\delta_{ij})+C\varepsilon_{ij3}$
\vskip0.4cm
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|}
\hline
N&$\vec{\alpha}$&$\vec{\phi}$&$\vec{\lambda}$&$v$&$\vec{\xi}$&$\theta$&$B$&$C$&$W$&\#\\
\hline
15&(0,0,1)&(0,0,$F$)&(0,0,$L$)&$v$&0&0&$B$&$2v$&1&4\\
16&(0,0,1)&(0,0,$F$)&(0,0,$L$)&$v$&0&0&1&0&0&3\\
17&0&(0,0,$\pm1$)&(0,0,$L$)&0&0&0&$B$&$C$&1&3\\
18&0&$\vec{\phi}$&0&1&0&0&0&1&0&3\\
19&0&0&(0,0,$L$)&0&(0,0,$X$)&$\theta$&\multicolumn{2}{c|}{${\scriptstyle 2B^2+6C^2=3}$}&1&4\\
20&0&0&(0,0,$\pm1$)&0&(0,0,$X$)&$\theta$&0&0&1&2\\
21&0&0&0&0&(0,0,$X$)&$\theta$&0&0&1&2\\
\hline
\end{tabular}
\vskip0.5cm
III. $\vec{\alpha}=0,\beta=0,\vec{\gamma}=0,\rho=-{1\over 3},
\sigma_{ij}=(\delta_{i3}\delta_{j3}-{1\over 3}\delta_{ij}),\omega_{ij}=0,
\chi_{ij}=B(\delta_{i3}\delta_{j3}-{1\over 3}\delta_{ij})+C\varepsilon_{ij3}$
\vskip0.4cm
\begin{tabular}{|c|c|c|c|c|c|c|c|c|}
\hline
N&$\vec{\phi}$&$\vec{\lambda}$&$v$&$\vec{\xi}$&$\theta$&$B$&$C$&\#\\
\hline
22&(0,0,1)&(0,0,$L$)&0&0&0&$B$&0&2\\
23&(0,0,1)&0&$v\not=0$&0&0&$B$&0&2\\
24&(0,0,1)&0&$v$&0&0&$B$&$C\not=0$&3\\
25&0&(0,0,$L$)&0&(0,0,$\pm1$)&$\theta$&$B$&0&3\\
26&0&(0,0,$L$)&0&0&$\pm1$&$B$&0&2\\
27&0&(0,0,$L$)&0&0&0&1&0&1\\
28&0&(0,0,$L$)&0&0&0&0&0&1\\
29&0&0&1&0&0&$B$&$C$&2\\
30&0&0&0&0&$\theta$&\multicolumn{2}{l|}{$\scriptstyle 2B^2+6C^2=3$}&2\\
&&&&&&\multicolumn{2}{c|}{$\scriptstyle C\not=0$}&\\
\hline
\end{tabular}

\vskip0.5cm
IV. $\vec{\alpha}=0,\beta=0,\vec{\phi}=0,v=0,,\theta=0,\rho=0,
\omega_{ij}=0,
\chi_{ij}=0,\sigma_{ij}=-\varepsilon_{ij3}$
\vskip0.4cm
\begin{tabular}{|c|c|c|c|c|}
\hline
N&$\vec{\gamma}$&$\vec{\lambda}$&$\vec{\xi}$&\#\\
\hline
31&(0,0,1)&(0,0,$L$)&(0,0,$\pm1$)&1\\
32&(0,0,1)&(0,0,$L$)&0&1\\
\hline
\end{tabular}
\vskip0.5cm
V.$\vec{\alpha}=0,\beta=0,\vec{\gamma}=0,v=0,\rho=0,\sigma_{ij}=0,
\omega_{ij}=0,tr\chi=0,\sum\limits^3_{ij}\chi_{ij}^2=1,\chi_{ij}=diag(\chi_{11},\chi_{22},\chi_{33})$

$\chi_{11}\not=\chi_{22}\not=\chi_{33}$
\vskip0.4cm
\begin{tabular}{|c|c|c|c|c|c|}
\hline
N&$\vec{\phi}$&$\vec{\lambda}$&$\vec{\xi}$&$\theta$&\#\\
\hline
33&0&$||\vec{\lambda}||=1$&0&$\theta$&4\\
34&0&$\vec{\lambda}$&$||\vec{\xi}||=1$&$\theta$&7\\
35&$||\vec{\phi}||=1$&$\vec{\lambda}=L\vec{\phi}$&0&0&4\\
\hline
\end{tabular}
\vskip0.4cm
(Vb)\hskip0.4cm$\chi_{11}=\chi_{22}\not=\chi_{33}$
\vskip0.3cm
\begin{tabular}{|c|c|c|c|c|c|}
\hline
N&$\vec{\phi}$&$\vec{\lambda}$&$\vec{\xi}$&$\theta$&\#\\
\hline
36&0&$\lambda_1=0$&0&$\theta$&2\\
&&$\lambda_2^2+\lambda_3^2=1$&&&\\
37&0&$\vec{\lambda}$&$\xi_1=0$&$\theta$&5\\
&&&$\xi_2^2+\xi_3^2=1$&&\\
38&$\phi_1=0$&$\vec{\lambda}=L\vec{\phi}$&0&0&2\\
&$\phi_2^2+\phi_3^2=1$&&&&\\
\hline
\end{tabular}
\vskip0.4cm

VI.$\vec{\alpha}=0,\beta=0,\vec{\gamma}=0,v=0,\rho=0,\sigma_{ij}=0,
\omega_{ij}=0,tr\chi=0,\sum\limits^3_{ij}\chi_{ij}^2=1,\chi_{32}=-\chi_{23}\not=0,
\chi_{12}=\chi_{21},\chi_{13}=\chi_{32}$

(VIa)\hskip0.4cm$\chi_{22}\not=\chi_{33}$
\vskip0.3cm
\begin{tabular}{|c|c|c|c|c|c|}
\hline
N&$\vec{\phi}$&$\vec{\lambda}$&$\vec{\xi}$&$\theta$&\#\\
\hline
39&0&$\lambda_1=0$&0&$\theta$&6\\
&&$\lambda_2^2+\lambda_3^2=1$&&&\\
40&0&$\vec{\lambda}$&$\xi_1=0$&$\theta$&9\\
&&&$\xi_2^2+\xi_3^2=1$&&\\
41&$||\vec{\phi}||=1$&$\vec{\lambda}=L\vec{\phi}$&0&0&7\\
&$\phi_2^2+\phi_3^2=1$&&&&\\
\hline
\end{tabular}
\vskip0.5cm
(VIb)\hskip0.4cm$\chi_{22}=\chi_{33}$
\vskip0.3cm
\begin{tabular}{|c|c|c|c|c|c|}
\hline
N&$\vec{\phi}$&$\vec{\lambda}$&$\vec{\xi}$&$\theta$&\#\\
\hline
42&0&$\lambda_1=\lambda_2=0$&0&$\theta$&4\\
&&$\lambda_2=\pm1$&&&\\
43&0&$\vec{\lambda}$&$\xi_1=\xi_2=0$&$\theta$&7\\
&&&$\xi_3=\pm1$&&\\
44&($F$,0,1)&$\vec{\lambda}=L\vec{\phi}$&0&0&5\\
\hline
\end{tabular}
\vskip0.5cm
VII.$\vec{\alpha}=0,\beta=0,\vec{\gamma}=0,\vec{\lambda}=0,\vec{\xi}=0,
\rho=0,\sigma_{ij}=0,
\omega_{ij}=0,tr\chi=0,\sum\limits^3_{ij}\chi_{ij}^2=1,$
\vskip0.4cm
(VIIa)\hskip0.4cm$\chi_{11}\not=\chi_{22}\not=\chi_{33},\chi_{13}=-\chi_{31},\chi_{23}=-\chi_{32},\chi_{12}=-\chi_{21}$
\vskip0.3cm
\begin{tabular}{|c|c|c|c|c|}
\hline
N&$\vec{\phi}$&$v$&$\theta$&\#\\
\hline
45&0&0&$\pm1,0$&4\\
46&$\vec{\phi}$&1&0&7\\
\hline
\end{tabular}
\vskip0.4cm
(VIIb)\hskip0.4cm$\chi_{11}=\chi_{22}\not=\chi_{33},\chi_{13}=-\chi_{31}\not=0,\chi_{23}=\chi_{32}=0,\chi_{12}=-\chi_{21}$
\vskip0.3cm
\begin{tabular}{|c|c|c|c|c|}
\hline
N&$\vec{\phi}$&$v$&$\theta$&\#\\
\hline
47&0&0&$\pm1,0$&2\\
48&$\vec{\phi}$&1&0&5\\
\hline
\end{tabular}
\vskip0.4cm
(VIIc)\hskip0.4cm$\chi_{11}=\chi_{22},\chi_{13}=\chi_{31}=\chi_{23}=\chi_{32}=0,\chi_{12}=-\chi_{21}$
\vskip0.3cm
\begin{tabular}{|c|c|c|c|c|}
\hline
N&$\vec{\phi}$&$v$&$\theta$&\#\\
\hline
49&0&0&$\pm1,0$&1\\
50&$(0,\phi_2,\phi_3)$&1&0&3\\
\hline
\end{tabular}
\vskip0.5cm
VIII.$\vec{\alpha}=0,\beta=0,\vec{\gamma}=0,\vec{\phi}=(0,0,1),
\vec{\lambda}=0,v=0,\vec{\xi}=0,\theta=0,
\rho=0,\sigma_{ij}=0,
\omega_{ij}=0,tr\chi=0,\sum\limits^3_{ij}\chi_{ij}^2=1,\chi_{12}=-\chi_{21}={1\over F}$
\vskip0.3cm
(VIIIa)\hskip0.4cm$\chi_{11}\not=\chi_{22},\chi_{13}=\chi_{31},\chi_{23}=\chi_{32}$

(VIIIb)\hskip0.4cm$\chi_{11}=\chi_{22},\chi_{13}=\chi_{31}=0,\chi_{23}=\chi_{32}$
\\

Now, inserting the appropriate values of the
parameters (listed above) to eq.\mref{wb1}
and using eq.\mref{w19} one can easily calculate the fundamental Poisson brackets
for all nonequivalent structures.

\section{Summary}

We have obtained all Lie--Poisson structures
on the four-dimensional Galilei group and classified
them up to the equivalence implied by group
automorphisms. The resulting set of structures appears
to be quite rich; in particular, it includes many
non--coboundary structures, to be contrasted with the
Poincare group case~\cite{b5}. In spite of that, part of them
can surely be obtained from those on the Poincare group
by a~contraction procedure.

The next step to be done is to quantize the
Lie-Poisson structures. In general, the consistent
quantization is not an easy task. However, the preliminary
study already done by us shows, that most of the
cases described here are quantization friendly.

\section{Appendix A: The cocycle condition for subgroups}
\renewcommand{\theequation}{A.\arabic{equation}}
\zero
\indent Let us specify the equations \mref{w17} for the subgroups of
rotations, boosts, space and time translations. They read, respectively:
\be
\Psi_i(RR')&=&\Psi_i(R)+R_{il}\Psi_l(R')\nn\\
\Phi_i(RR')&=&\Phi_i(R)+R_{il}\Phi_l(R')\nn\\
\Gamma_i(RR')&=&\Gamma_l(R)+R_{il}\Gamma_l(R')\nn\\
\Lambda_i(RR')&=&\Lambda_i(R)+R_{il}\Lambda_l(R')\nn\\
\Upsilon_{ij}(RR')&=&\Upsilon_{ij}(R)+R_{im}R_{jl}\Upsilon_{ml}(R')\label{wa1a}\\
\Sigma_{ij}(RR')&=&\Sigma_{ij}(R)+R_{im}R_{jl}\Sigma_{ml}(R')\nn\\
\Xi_{i}(RR')&=&\Xi_{i}(R)+R_{im}\Xi_{m}(R')\nn\\
\Omega_{ij}(RR')&=&\Omega_{ij}(R)+R_{im}R_{jl}\Omega_{ml}(R')\nn\\
\Pi_i(RR')&=&\Pi_i(R)+R_{il}\Pi_l(R')\nn\\
&&\nn\\
\Psi_i(\vec{v}+\vec{v'})&=&\Psi_i(\vec{v})+\Psi_i(\vec{v'})\nn\\
\Phi_i(\vec{v}+\vec{v'})&=&\Phi_i(\vec{v})+\Phi_i(\vec{v'})\nn\\
\Gamma_i(\vec{v}+\vec{v'})&=&\Gamma_i(\vec{v})+\Gamma_i(\vec{v'})-\varepsilon_{ink}v_n\Psi_k(\vec{v'})\nn\\
\Lambda_i(\vec{v}+\vec{v'})&=&\Lambda_i(\vec{v})+\Lambda_i(\vec{v'})-\fr\varepsilon_{imn}v_m\Phi_n(\vec{v'})\label{wa1b}\\
\Upsilon_{ij}(\vec{v}+\vec{v'})&=
&\Upsilon_{ij}(\vec{v})+\varepsilon_{jnk}v_n\Psi_k(\vec{v'})v_i-
\Gamma_j(\vec{v'})v_i-\varepsilon_{jnk}v_n\Sigma_{ik}(\vec{v'})\nn\\
\Sigma_{ij}(\vec{v}+\vec{v'})&=&\Sigma_{ij}(\vec{v})+\Sigma_{ij}(\vec{v'})-v_i\Psi_j(\vec{v'})\nn\\
\Xi_i(\vec{v}+\vec{v'})&=&\Xi_i(\vec{v})+\Xi_i(\vec{v'})-\fr(v_i\Omega_{mm}(\vec{v'})-\nn\\
&&-v_p\Omega_{pi}(\vec{v'}))+\Pi_m(\vec{v'})v_mv_i\nn\\
\Omega_{ij}(\vec{v}+\vec{v'})&=&\Omega_{ij}(\vec{v})+\Omega_{ij}(\vec{v'})+2\Pi_i(\vec{v'})v_j-2\Pi_m(\vec{v'})v_m\delta_{ij}\nn\\
\Pi_i(\vec{v}+\vec{v'})&=&\Pi_i(\vec{v})+\Pi_i(\vec{v'})\nn\\
&&\nn\\
\Psi_i(\vec{a}+\vec{a'})&=&\Psi_i(\vec{a})+\Psi_i(\vec{a'})\nn\\
\Phi_i(\vec{a}+\vec{a'})&=&\Phi_i(\vec{a})+\Phi_i(\vec{a'})-\varepsilon_{ink}a_n\Psi_k(\vec{a'})\nn\\
\Gamma_i(\vec{a}+\vec{a'})&=&\Gamma_i(\vec{a})+\Gamma_i(\vec{a'})\nn\\
\Lambda_i(\vec{a}+\vec{a'})&=&\Lambda_i(\vec{a})+\Lambda_i(\vec{a'})+\Pi_m(\vec{a'})a_ma_i-\fr(a_i\delta_{lm}-a_l\delta_{im})\Sigma_{lm}(\vec{a'})\nn\\
\Upsilon_{ij}(\vec{a}+\vec{a'})&=&\Upsilon_{ij}(\vec{a})+\Upsilon_{ij}(\vec{a'})+\varepsilon_{ink}a_n\Omega_{jk}(\vec{a'})\label{wa1c}\\
\Sigma_{ij}(\vec{a}+\vec{a'})&=&\Sigma_{ij}(\vec{a})+\Sigma_{ij}(\vec{a'})+2(\Pi_i(\vec{a'})a_j-\Pi_l(\vec{a'})a_l\delta_{ij})\nn\\
\Xi_{i}(\vec{a}+\vec{a'})&=&\Xi_i(a)+\Xi_i(a)\nn\\
\Omega_{ij}(\vec{a}+\vec{a'})&=&\Omega_{ij}(\vec{a})+\Omega_{ij}(\vec{a'})\nn\\
\Pi_i(\vec{a}+\vec{a'})&=&\Pi_i(\vec{a})+\Pi_i(\vec{a'})\nn\\
&&\nn\\
\Psi_i({t}+{t'})&=&\Psi_i({t})+\Psi_i({t'})\nn\\
\Phi_i({t}+{t'})&=&\Phi_i({t})+\Phi_i({t'})-t\Gamma_i(t')\nn\\
\Gamma_i({t}+{t'})&=&\Gamma_i({t})+\Gamma_i({t'})\label{wa1d}\\
\Lambda_i({t}+{t'})&=&\Lambda_i({t})+\Lambda_i({t'})-\fr t\varepsilon_{imn}\Upsilon_{mn}(t')+t^2\Xi_i(t')\nn\\
\Upsilon_{ij}(t+t')&=&\Upsilon_{ij}(t)+\Upsilon_{ij}(t')-2t\varepsilon_{ijl}\Xi_l(t')\nn\\
\Sigma_{ij}(t+t')&=&\Sigma_{ij}(t)+\Sigma_{ij}(t')-t\Omega_{ij}(t')\nn\\
\Xi_i(t+t')&=&\Xi_i(t)+\Xi_i(t')\nn\\
\Omega_{ij}(t+t')&=&\Omega_{ij}(t)+\Omega_{ij}(t')\nn\\
\Pi_i(t+t')&=&\Pi_i(t)+\Pi_i(t')\nn
\ee
Note that all eqs\mref{wa1a} have the same structure:
\be
T_{i_1\ldots i_k}(RR')&=&T_{i_1\ldots i_k}(R)+
R_{i_1j_1}\ldots R_{i_kj_k}T_{j_1\ldots j_k}(R')\label{wa2}
\ee
They can be solved by integrating over $R'$\ with respect to the
Haar measure on SO(3):
\be
T_{i_1\ldots i_k}(R)&=&(R_{i_1j_1}\ldots R_{i_kj_k}-\delta_{i_1 j_1}\ldots\delta_{i_kj_k})c_{j_1\ldots j_k}\label{wa3a}
\ee
where $c_{j_1\ldots j_k}$ are constants.
This result agrees with general theorem that all cocycles on semisimple groups
are coboundaries. On the other hand it follows immediately from \hbox{eqs.(\ref{wa1b}--\ref{wa1d})}
that all functions entering there are polynomials in the relevant parameters.
This allows us to write out explicitly the general solutions.
\be
\Psi_i(\vec{v})&=&0\nn\\
\Phi_i(\vec{v})&=&0\nn\\
\Gamma_i(\vec{v})&=&a_{ij}v_j\nn\\
\Lambda_i(\vec{v})&=&b_{ij}v_j\label{wa3b}\\
\Upsilon_{ij}(\vec{v})&=&c_{ijk}v_k-\fr(\delta_{ik}a_{jl}+\delta_{ji}a_{lk}-a_{lj}\delta_{ki}+\fr\varepsilon_{jkl}\varepsilon_{inm}a_{nm})v_kv_l\nn\\
\Sigma_{ij}(\vec{v})&=&(\fr\delta_{jk}\varepsilon_{inm}a_{nm}-\varepsilon_{ijn}a_{kn})v_k\nn\\
\Xi_i(\vec{v})&=&d_{ik}v_k+({1\over 4}e_{jik}-{1\over 8}(e_j\delta_{ik}+e_k\delta_{ij}))v_jv_k\nn\\
\Omega_{ij}(\vec{v})&=&(e_{ijk}+\fr(e_k\delta_{ji}-e_i\delta_{jk}))v_k\nn\\
\Pi_i(\vec{v})&=&0,\nn
\ee
where $e_{ijk}=e_{kji},\;e_{iik}=0$;
\be
\Psi_i(\vec{a})&=&0\nn\\
\Phi_i(\vec{a})&=&f_{ij}a_j\nn\\
\Gamma_i(\vec{a})&=&g_{ij}a_j\nn\\
\Lambda_i(\vec{a})&=&h_{ij}a_j+{1\over 4}(h_{jik}-\fr(h_j\delta_{ik}+h_k\delta_{ij}))a_ja_k\nn\\
\Upsilon_{ij}(\vec{a})&=&k_{ijk}a_k\label{wa3c}\\
\Sigma_{ij}(\vec{a})&=&(h_{ijk}+\fr(h_k\delta_{ij}-h_i\delta_{jk}))a_k\nn\\
\Xi_i(\vec{a})&=&l_{ij}a_j\nn\\
\Omega_{ij}(\vec{a})&=&0\nn\\
\Pi_i(\vec{a})&=&0,\nn
\ee
where $h_{ijk}=h_{kji},\;h_{iik}=0$;
\be
\Psi(t)&=&p_it\nn\\
\Phi_i(t)&=&r_it-\fr s_it^2\nn\\
\Gamma_i(t)&=&s_it\nn\\
\Lambda_i(t)&=&u_it-{1\over 4}\varepsilon_{imn}x_{mn}t^2+{1\over 3}w_it^3\nn\\
\Upsilon_{ij}(t)&=&x_{ij}t-\varepsilon_{ijk}w_kt^2\label{wa3d}\\
\Sigma_{ij}(t)&=&y_{ij}t-\fr z_{ij}t^2\nn\\
\Xi_i(t)&=&w_it\nn\\
\Omega_{ij}(t)&=&z_{ij}t\nn\\
\Pi_i(t)&=&m_it.\nn
\ee
\section{Appendix B: The general solution to the cocycle condition}
\renewcommand{\theequation}{B.\arabic{equation}}
\zero
\indent According to the procedure outlined in sec.\ref{s2}, we use
the cocycle condition(i) together with the decomposition \mref{w18} and
the expressions written out in Appendix~A to produce the Ansatz
for \hbox{$\eta(g)$.} Inserting it back into \mref{w17} we find the
general solution for $\eta$\ of the form.
\be
\Psi_i(g)&=&(R_{ij}-\delta_{ij})\alpha_j\nn\\
\Phi_i(g)&=&(R_{ij}-\delta_{ij})\phi_j+\beta(a_i-v_it)-\gamma_jR_{ij}t+\varepsilon_{ijk}
\alpha_lR_{jl}(a_k-v_kt)\nn\\
\Gamma_i(g)&=&(R_{ij}-\delta_{ij})\gamma_j+\beta v_i+\varepsilon_{ijk}\alpha_lR_{jl}v_k\nn\\
\Lambda_i(g)&=&(R_{ij}-\delta_{ij})\lambda_j+(\rho-\fr\sigma_{nn})(a_i-v_i t)+\fr\beta\varepsilon_{ijk}a_jv_k\label{wb1}\\
&&+\fr\varepsilon_{ijk}\phi_lR_{jl}v_k-\fr\alpha_jR_{ij}(a_kv_k-\vec{v}^2t)+\nn\\
&&+\fr\alpha_jR_{kj}v_k(a_i-v_it)+\xi_jR_{ij}t^2-\fr\varepsilon_{ijk}\gamma_lR_{jl}v_kt+\nn\\
&&-\fr\varepsilon_{jkl}\chi_{kl}R_{ij}t-\omega_{jl}R_{ij}R_{kl}a_kt+\nu v_i+\nn\\
&&+\fr\sigma_{ij}R_{ij}R_{kl}(a_k-v_kt)+n_kR_{mk}(a_ia_m+2v_mv_it)-\nn\\
&&-n_l(a_kR_{kl}v_i+a_iR_{kl}v_k)t\nn\\
\Upsilon_{ij}(g)&=&(R_{ik}R_{jl}-\delta_{ik}\delta_{jl})\chi_{kl}+\delta_{ij}\theta t-
\fr\delta_{ij}\beta\vec{v}^2+\nn\\
&&+\varepsilon_{jkl}\alpha_nR_{ln}v_kv_i-\gamma_kR_{jk}v_i-2\varepsilon_{ijk}\xi_lR_{kl}t\nn\\
&&-\varepsilon_{jkl}\sigma_{ns}R_{in}R_{ls}v_k+\rho\varepsilon_{ijk}v_k+\omega_{nn}\varepsilon_{ijk}a_k\nn\\
&&+2\omega_{ns}(\varepsilon_{ikl}R_{js}R_{ln}a_k-\varepsilon_{ijl}R_{ks}R_{ln}v_kt)\nn\\
&&+2n_l(R_{jl}v_k-R_{ml}v_m\delta_{ik})\varepsilon_{kin}a_n+\nn\\
&&-2n_kv_sv_mR_{mk}\varepsilon_{sij}t+\omega_{lk}R_{nk}R_{il}v_nt^2\nn\\
\Sigma_{ij}(g)&=&(R_{ik}R_{jl}-\delta_{ik}\delta_{jl})\sigma_{kl}-\beta\varepsilon_{ijk}v_k-\alpha_kR_{jk}v_i\nn\\
&&-2\omega_{lk}R_{ik}R_{jl}t+\omega_{nn}\delta_{ij}t+\nn\\
&&+2n_k(R_{ik}a_j-R_{mk}a_m\delta_{ij}-R_{ik}v_jt+R_{mk}v_m\delta_{ij}t)\nn\\
\Xi_i(g)&=&(R_{ij}-\delta_{ij})\xi_j+\omega_{jl}R_{ij}R_{kl}v_k+n_kR_{mk}v_mv_i\nn\\
\Omega_{ij}(g)&=&2(R_{ik}R_{jl}-\delta_{ik}\delta_{jl})\omega_{lk}+2n_k(R_{ik}v_j-R_{mk}v_m\delta_{ij})\nn\\
\Pi_i(g)&=&(R_{ij}-\delta_{ij})n_j\nn
\ee
\section{Appendix C: Jacobi identities}
\renewcommand{\theequation}{C.\arabic{equation}}
\zero
Using the general form of $\eta$\ described in Appendix~B we find from eq.\mref{w13}:
\be
\delta(H)&=&\gamma_iH\wedge P_i+\fr(\chi_{kj}-\chi_{jk})P_j\wedge P_k+(2\varepsilon_{ijk}\xi_k-\theta\delta_{ij})P_i\wedge K_j\nn\\
&&(2\omega_{ji}-\omega_{nn}\delta_{ji})P_i\wedge J_j\nn\\
\delta(P_s)&=&(\beta\delta_{is}+\varepsilon_{iks}\alpha_k)H\wedge P_i+\varepsilon_{ijk}(\rho\delta_{is}-\fr\sigma_{nn}\delta_{is}+\nn\\
&&+\fr\sigma_{si})P_j\wedge P_k+(2\varepsilon_{lis}\omega_{lj}-\varepsilon_{jis}\omega_{nn})P_i\wedge K_j\nn\\
&&+2(n_i\delta_{sj}-n_s\delta_{ij})P_i\wedge J_j\label{wc1}\\
\delta(K_s)&=&(\beta\delta_{is}+\varepsilon_{iks}\alpha_k)H\wedge P_i+(v\varepsilon_{sjk}-\fr(\phi_k\delta_{js}-\phi_j\delta_{ks}))P_j\wedge P_k\nn\\
&&+(\rho\varepsilon_{ijs}-\varepsilon_{kjs}\sigma_{ik}-\gamma_j\delta_{is})P_i\wedge K_j+\nn\\
&&-(\beta\varepsilon_{ijs}+\alpha_j\delta_{is})P_i\wedge J_j+\varepsilon_{ijk}\omega_{is}K_j\wedge K_k+\nn\\
&&+2(n_j\delta_{sk}-n_s\delta_{jk})K_j\wedge J_k\nn\\
\delta(J_s)&=&\varepsilon_{sij}\alpha_jH\wedge J_i+\varepsilon_{sij}\phi_jH\wedge P_i+\varepsilon_{ijs}\gamma_j H\wedge K_i\nn\\
&&+(\lambda_j\delta_{ks}-\lambda_k\delta_{js})P_j\wedge P_k+(\varepsilon_{sik}\lambda_{kj}+\varepsilon_{sjk}\chi_{ik})P_i\wedge K_j\nn\\
&&+(\varepsilon_{sik}\sigma_{kj}+\varepsilon_{sjk}\sigma_{ik})P_i\wedge J_j+(\xi_i\delta_{ks}-\xi_k\delta_{js})K_j\wedge K_k\nn\\
&&+2(\varepsilon_{sik}\omega_{jk}+\varepsilon_{sjk}\omega_{ki})K_i\wedge J_j+2n_kJ_k\wedge J_s\nn
\ee
Let $\tilde{X_i}$\ denote a basis in ${\cal G}^*$\ defined by $<\tilde{X_i},X_j>=\delta_{ij}$.
Then $\delta$\ imposes the following commutator structure in ${\cal G}^*$.
\be
{}[\tilde{H},\tilde{J_k}]&=&\varepsilon_{ikl}\alpha_l\tilde{J_i}\nn\\
{}[\tilde{H},\tilde{P_k}]&=&\gamma_k\tilde{H}+(\beta\delta_{ik}+\varepsilon_{ikl}\alpha_l)\tilde{P_i}+\varepsilon_{ikl}\phi_l\tilde{J_i}\nn\\
{}[\tilde{H},\tilde{K_k}]&=&(\beta\delta_{ik}+\varepsilon_{ikl}\alpha_l)\tilde{K_i}+\varepsilon_{ikl}\gamma_l\tilde{J_i}\nn\\
{}[\tilde{K_k},\tilde{J_l}]&=&2(n_k\delta_{li}-n_i\delta_{kl})\tilde{K_i}+2(\varepsilon_{ikn}\omega_{ln}+\varepsilon_{iln}\omega_{nk})\tilde{J_i}\nn\\
{}[\tilde{P_l},\tilde{P_m}]&=&(\chi_{lm}-\chi_{ml})\tilde{H}+2\varepsilon_{klm}[\rho\delta_{ki}+\fr(\sigma_{ik}-\sigma_{nn}\delta_{ik})]\tilde{P_i}\nn\\
&&+2[v\varepsilon_{ilm}+\fr(\phi_l\delta_{im}-\phi_m\delta_{li})]\tilde{K_i}+\label{wc2}\\
&&+2(\lambda_l\delta_{im}-\lambda_m\delta_{ik})\tilde{J_i}\nn\\
{}[\tilde{P_k},\tilde{K_l}]&=&(2\xi_n\varepsilon_{nkl}-\theta\delta_{kl})\tilde{H}+(2\varepsilon_{nki}\omega_{nl}-\omega_{nn}\varepsilon_{lki})\tilde{P_i}\nn\\
&&+(\rho\varepsilon_{kli}-\varepsilon_{lin}\sigma_{kn}-\delta_{ki}\gamma_l)\tilde{K_i}+\nn\\
&&+(\varepsilon_{ikn}\chi_{nl}+\varepsilon_{iln}\chi_{kn})\tilde{J_i}\nn\\
{}[\tilde{P_k},\tilde{J_l}]&=&(2\omega_{ik}-\omega_{nn}\delta_{lk})\tilde{H}+2(n_k\delta_{li}-n_i\delta_{kl})\tilde{P_i}\nn\\
&&-(\beta\varepsilon_{kli}+\alpha_l\delta_{ki})\tilde{K_i}+(\varepsilon_{ikn}\sigma_{nl}+\varepsilon_{iln}\sigma_{kn})\tilde{J_i}\nn\\
{}[\tilde{K_m},\tilde{K_n}]&=&2\varepsilon_{kmn}\omega_{ki}\tilde{K_i}+2(\xi_m\delta_{ni}-\xi_n\delta_{mi})\tilde{J_i} \nn\\
{}[\tilde{J_k},\tilde{J_l}]&=&2(n_k\delta_{li}-n_l\delta_{ki})\tilde{J_i}\nn
\ee
Now we have to solve the Jacobi identities for the structure described in \mref{wc2}.
This is still a~complicated task so we apply a~mixed procedure consisting in
solving part of Jacobi identities directly and applying the group of automorphisms
to simplify the remaining ones.

Therefore, we first give the action of automorphism group on parameters
of $\delta$. Let us start with  inner automorphisms. They  read
\be
\vec{\alpha_k}=\vec{\alpha_i}&&\gamma'_k=\gamma_k-\beta v_k+\varepsilon_{kil}v_i\alpha_l\nn\\
\beta'=\beta&&\lambda_k'=\lambda_k-vv_k-\fr\varepsilon_{kil}\phi_iv_\alpha\nn\\
\vec{\phi'}=\vec{\phi}&&\xi'_k=\xi_k-\omega_{kn}v_n-v_k(v_nn_n)\label{wc3a}\\
\theta'=\theta&&\rho'=\rho-{1\over 3}\alpha_nv_n\nn\\
v'=v&&\omega'_{ij}=\omega_{ij}+v_in_j+(v_nn_n)\delta_{ij}\nn\\
\vec{n'}=\vec{n}&&\sigma'_{ia}=\sigma_{ia}+\beta\varepsilon_{ial}v_l+\alpha_av_i-{1\over 3}\alpha_nv_n\delta_{ai}\nn\\
&&\chi'_{ab}=\chi_{ab}+v_a\gamma_b+\varepsilon_{bnm}v_n\alpha_mv_a-\fr\varepsilon_{abk}\sigma_{nk}v_n\nn\\
&&-\rho\varepsilon_{abn}v_n-\fr\varepsilon_{alm}\sigma_{bl}v_m-\fr\varepsilon_{blm}\sigma_{al}v_m\nn\\
&&-{1\over 3}(v_n\gamma_n+\varepsilon_{lnm}\sigma_{nl}v_m)\delta_{ab}\nn
\ee
for boosts,
\be
\vec{n'}=\vec{n},&\vec{\alpha'}={\alpha},&\beta'=\beta\nn\\
v'=v,&\vec{\xi'}=\vec{\xi},&\theta'=\theta\nn
\ee
\be
\phi_i'&=&\phi_i-\beta a_i-\varepsilon_{ink}\alpha_na_k\nn\\
\gamma'_k&=&\gamma_k-2\varepsilon_{knm}a_mn_n\label{wc3b}\\
\lambda'_k&=&\lambda_k-\fr\sigma_{nk}a_n-\rho a_k-\vec{a^2}n_k-(n_ia_i)n_k\nn\\
\rho'&=&\rho-{4\over 3}a_kn_k\;\;\sigma'_{ab}=\sigma_{ab}+{2\over 3}(a_kn_k)\delta_{ab}-2n_aa_b\nn\\
\chi'_{ab}&=&\chi_{ab}-\varepsilon_{abn}\omega_{nk}a_k+\varepsilon_{amk}\omega_{mb}a_k+\nn\\
&&+\varepsilon_{bmk}\omega_{ma}a_k+{2\over 3}\varepsilon_{mnk}\omega_{mn}a_k\delta_{ab}\nn\\
\omega'_{ab}&=&\omega_{ab}\nn
\ee
for space translations and
\be
\vec{n'}=\vec{n},&\vec{\alpha}=\alpha,&\beta'=\beta,\;\;\;\;\vec{\gamma'}=\vec{\gamma}\nn\\
v'=v,&\vec{\xi'}=\vec{\xi},&\theta'=\theta,\;\;\;\;\omega'_{ij}=\omega_{ij}\nn\\
\vec{\phi'}=\vec{\phi}+t\vec{\gamma}&,&\lambda'_i=\lambda_i-\fr t\varepsilon_{imn}\chi_{nm}+t^2\xi_i\nn\\
\rho'=\rho+{1\over 3}t\omega_{nn}&,&\sigma'_{ab}=\sigma_{ab}+2t\omega_{ba}-{2\over 3}t\omega_{nn}\delta_{ab}\nn\\
\chi'_{ab}=\chi_{ab}+2t\varepsilon_{abk}\xi_k\nn
\ee
for time translations, respectively.

Under the rotations the parameters transform as tensors of appropriate rank.

Besides, there are two outer automorphisms, which correspond to rescaling
of space and time unit, \hbox{($\vec{a}\to a\vec{a},\;t\to bt$)}. They read:
\be
n'=\vec{n},&\alpha'={1\over b}\vec{\alpha},&\beta'={1\over b}\beta\nn\\
\vec{\gamma}={1\over a}\gamma,&\vec{\phi}={1\over ab}\vec{\phi},&\vec{\lambda}={1\over a^2}\vec{\lambda}\nn\\
v'={1\over ab}v,&\vec{\xi}={b^2\over a^2}\vec{\xi},&\theta'={b^2\over a^2}\theta\label{wc4}\\
\rho'={1\over a}\rho,&\sigma'_{ab}={1\over a}\sigma_{ab},&\chi'_{ab}={b\over a^2}\chi_{ab}\nn\\
\omega'_{ab}={b\over a}\omega_{ab}\nn
\ee
We shall not enter into all details. Let us rather give a sketch of the procedure.

By solving Jacobi identities for the subalgebra generated by $\tilde{H},\tilde{J_l},\tilde{K_n}$\
we find the following six families of constraints on parameters $\vec{n},\beta,\vec{\alpha},
\omega_{ij},\vec{\gamma},\hbox{\ and\ }\vec{\xi}$.\\
a)$\vec{n}=0,\vec{\alpha}\hbox{--arbitrary\ },\beta\not=0,\vec{\gamma}\hbox{--arbitrary,\ }\vec{\xi}=0,
\omega_{ij}=0$\\
b)$\vec{n}=0,\vec{\alpha}\not=0,\beta=0,\vec{\gamma}\hbox{--arbitrary,\ }\vec{\xi}=\xi\vec{\alpha}+W(\vec{\alpha}\times\vec{\gamma}),
\omega_{ij}=W(\vec{\alpha^2}\delta_{ij}-\alpha_i\alpha_j), W\not=0$\\
c)$\vec{n}=0,\vec{\alpha}=0,\beta=0,\vec{\xi}\hbox{--arbitrary,\ }\omega_{ij}=W(\delta_{ij}-\mu_i\mu_j)+V\varepsilon_{ijk}\mu_k,
||\vec{\mu}||=1$\\
$\vec{\gamma}=\left\{
\begin{array}{c}
\gamma\vec{\mu}\;\;\;\;\; \hbox{if\ } |W|+|V|\not=0\\
\hbox{arbitrary\ }\;\hbox{if\ } W=V=0
\end{array}\right.$\\
d)$\vec{n}=0,\vec{\alpha}\not=0,\beta=0,\omega_{ij}=W\delta_{ij},\vec{\gamma}={1\over W}(\vec{\alpha}\times\vec{\xi}),
W\not=0,\vec{\xi}\hbox{--arbitrary}$\\
e)$\vec{n}=0,\vec{\alpha}\not=0,\beta=0,\omega_{ij}=0,\vec{\xi}||\vec{\alpha},\vec{\gamma}\bot\vec{\alpha}$\\
f)$\vec{n}=0,\vec{\alpha}=0,\beta=0,\omega_{ij}=W\delta_{ij},\vec{\gamma}=0,\vec{\xi}\hbox{--arbitrary}$\\

Now we have to solve the remaining Jacobi identities. First of all let us
note that there are ambiguities in determining the matrices $\sigma_{ij}$\ and
$\chi_{ij}$. Namely, both can be redefined by adding the arbitrary multiplies
of unit matrix. In order to remove this ambiguity we put
\hbox{$tr\sigma=tr\chi=0$}. Now  we use the automorphisms generated by
boosts, space and time translations to simplify the Jacobi identities. For
example in the  cases (a),(d) and~(e) we may use the boost to put
$\vec{\gamma}=0$\ from the very beginning. On the other hand in the case
(b) by solving Jacobi identity for $\tilde{H},\tilde{P_i},\tilde{P_k}$\
we find $\vec{\gamma}\vec{\alpha}=0$\ and again the boost can be used to put
$\vec{\gamma}=0$. Following this way we obtain the eighteen families of solutions
described in sec.\ref{s3}.

{\Large\bf Acknowledgment}

The authors acknowledge Profs. P.Kosi\'nski  and S.Giller for many helpful discussions.


TITLE: Order statistics and indices
QUESTION [1 upvotes]: Let $X_1,\dots, X_n$ be iid continuous random variables with cdf $F$ with support $[a,b]$. Let $x,y\in [a,b]$ such that $x<y$. Let $k\in\{0,\dots,n\}$.
Fix some $i\in\{1,\dots,n\}$.
I would like to compute $$\mathbb{P}\Big(\text{There are exactly } k \text{ random variables }X_j \text{ other than } i  \text{ such that either } (X_j>y)  \text{ and }  (j < i) \text{ or } (X_j>x) \text{ and }(j>i)\Big).$$
And from this compute
$\mathbb{P}\Big(\text{There are less than } k \text{ random variables }X_j \text{ other than }i  \text{ such that either } X_j>y \text{ and } j<i \text{ or } X_j>x \text{ and }j>i\Big)$
(which I would do buy summing the above over $l<k$).
If simpler to solve (which I doubt), the uniform distribution case would be a start.

REPLY [3 votes]: There are $n - 1$ random variables other than $X_i$, and we partition them into two sets: $\mathcal{X}_1 = \{X_1, X_2, \ldots, X_{i-1}\}$ with $i - 1$ random variables, and $\mathcal{X}_2 = \{X_{i+1}, X_{i+2}, \ldots, X_n\}$ with $n - i$ random variables.
When $k \leq i -1$ and $k \leq n - i$, we have
$$ \sum_{u=0}^k \binom {i - 1} {u} [1 - F(y)]^uF(y)^{i-1-u}
\binom {n - i} {k - u} [1 - F(x)]^{k-u}F(x)^{n-i-k+u}$$
The first half of the summand is the probability of having exactly $u$ of $X_j \in \mathcal{X}_1$ with $X_j > y$, and the second half is similar.
When $k > i - 1$ or $k > n - i$ we just need to adjust the bounds of the summation.


TITLE: Finding the amount of possibilities to split two parties in pairs
QUESTION [1 upvotes]: I have two parties. Each party contains $3$ people. I'm trying to figure out how many possibilities there are to divide those two parties into pairs so each pair contains a person from party A and a person from party B.
I think the answer is $6$. But I'm trying to prove it. First of all, I put the three people from party A in a row in $3!$ options. I also put the three people from party B in a separate row in $3!$ options. Each pair will be in the people by indexes (first pair contains the first person from the first row and the first person from the second row and so on).
But now we need to divide by something. How should I continue this idea?

REPLY [0 votes]: You are correct that there are six ways to match people from party $A$ with people from party $B$ in pairs.
Line up the people in party $A$ in some order, say alphabetically.  There are three ways to match a person from party $B$ with the first person in line, two ways to match one of the remaining people from party $B$ with the second person in line, and one way to match the remaining person from party $B$ with the third person in line.  Hence, there are $3! = 6$ ways to pair people from party $A$ with people from party $B$.
To continue with your idea, match the person from party $A$ with the corresponding person in party $B$.  Notice that doing so counts each possible pairing $3!$ times, once for each of the $3!$ ways we could arrange the same set of pairs in the two lines.  Hence, there are
$$\frac{3!3!}{3!} = 3!$$
ways to pair people from party $A$ with people from party $B$.


TITLE: What happens to the acceleration of a body undergoing constant force?
QUESTION [0 upvotes]: Here's the problem I have, specifically relating to a rocket in deep space:
A rocket burns fuel at a constant rate. Assuming its mass remains constant, what happens to its acceleration? I seem to get two different answers depending on how I solve it (using conservation of energy or conservation of momentum):
Conservation of Energy
$\frac12mv^2=E_{exhaust}$, where $E_{exhaust}$ varies with time.
$v=\sqrt{\frac2mt}$, meaning that velocity varies with $\sqrt t$, meaning that acceleration decreases over time.
This also means that depending on your frame of reference's velocity relative to the rocket, the ship's acceleration must change...which makes about 0 sense,
Conservation of Momentum
$P_{exhaust}=mv$
Meaning that as more fuel is consumed, the velocity of the ship increases too (as $P_{exhaust}$ increases linearly with time, $v$ must also increase linearly with time).
I am assuming that the mass of the spaceship is constant. While not quite true, the effect would be negligible and the paradox I'm running into would remain true anyways.
Can anyone tell me what I'm doing wrong? The acceleration cannot possibly change depending on where you observe it from, but the acceleration also cannot remain constant as that violates conservation of energy.
Thanks in advance!

REPLY [2 votes]: Your conception of Conservation of Energy is wrong.  First of all, what do you mean by $E_{exhaust}$ ?   The kinetic energy of the exhaust?  If so, you have the KE of the exhaust, and the KE of the rocket ... and where did all of this KE come from?
To begin applying conservation of energy, you'd have to say something like: the sum of the KE of the rocket plus the KE of the exhaust is equal to the chemical energy released during the burning of the fuel".
Your conservation of momentum argument is more or less correct, but you would have to explain why the exhaust momentum increases linearly.  Won't exhaust be expelled at different velocities as the speed of the rocket increases?  A more complete analysis is needed.
But your original question has a simple answer.   If you assume, as you do, that the mass of the rocket+fuel does not change significantly, and also that the fuel is burned at a constant rate, then the thrust (force) will be constant, and you can apply Newton's second law.


TITLE: Maps from $V\otimes V^*$ to the underlying field
QUESTION [1 upvotes]: Let $V\in \operatorname{Vect}(k)$ I was asked to consider if the evaluation map was the only map from  $V^*\otimes V$ to $k$.  My question is, is it correct for me to rephrase this question to be, can you find a map $\phi:V^*\otimes V \to k$ that is independent of any choice of basis?  An answer to my rephrased question is something I think I can solve.

REPLY [2 votes]: Maps $V^*\otimes V \to k$ are the same thing as elements of $(V^*\otimes V)^*$. Since this is not the zero vector space (assuming $V \neq 0$), it has more than one element. If your field is infinite, say $k = \mathbb R$ or $k = \mathbb C$, then $(V^*\otimes V)^*$ is also infinite. This proves that the evaluation map is not the only map.


TITLE: Can we have chaotic motion due to the finite precision of our calculations?
QUESTION [5 upvotes]: I understand chaotic motion to mean that very small perturbations in the initial starting condition can lead to very different trajectories in phase space. For this reason, we can never predict the motion accurately, because we can never have 100% accurate initial conditions.
Can we look at the inability to predict the future states in a different way, related to the precision of our calculations (done on a computer)? Are there situations where we may know the initial conditions with 100% accuracy, but still cannot trust any of the predicted motions, because the motion depends on the accuracy of intermediate calculations, which, being done on a computer, are finite and therefore not perfectly precise?
For example, if I needed to calculate a numeric integral as a step towards a final answer, if my integral were computer to 16 floating point vs 32 floating point accuracy, this would correspond to a difference at the sixteenth significant digit, which could then be sufficient to induce wildly different behavior in the subsequent trajectories.
We could imagine a case that no matter how accurate your calculations were, additional accuracy in the calculations would cause the trajectory to diverge chaotically. Is this phenomenon known to exist, and are there examples of it?

REPLY [3 votes]: The title question is a bit different from the one in the body of the post, so let's look at them separetelly:

Can we have chaotic motion due to the finite precision of our calculations?

Yes, Lorenz himself described the phenomenon, calling it computational chaos [Lorenz 1989]:

When one seeks approximate solutions of a set of differential equations by stepwise numerical integration, the choice of a time increment $\tau$ [...] may yield chaotic solutions, even when the true solutions approach limit cycles or fixed points.

cannot trust any of the predicted motions [?]

At least for hyperbolic systems actually yes, you can trust them. What gets you covered is the so-called shadowing theorem, which guarantees that, even though you're indeed not simulating the true trajectory of the initial condition you picked, there's a slightly different initial point whose trajectory remains arbitrarily close to the computer-generated trajectory. Check also this answer.
[Lorenz 1989] Computational chaos-a prelude to computational instability, Physica D 35 (3), 1989, Pages 299-317.


TITLE: Proving projection map is onto
QUESTION [1 upvotes]: Let $X$ be a nonempty set and let $R$ be an equivalence relation on $X.$ Given $X/R={[a]:a \in X}$. Prove that there is a map called the projection where $p_x:X\to X/R$ given by $p_x(t)=[t].$ Then this map is onto (surjective).

I know that by definition: Let $X$ be a nonempty set and let $R$ be an equivalence relation on $X$. The set of all equivalence classes, ${[a]:a\in X},$ is called the Quotient Set of $X$ modulo $R,$ written as $X/R.$
I am trying to teach this to myself. I spoke to a friend of mine who made me want to learn some of this. It seemed good to know. How would I prove that the projection map is onto given this nonempty set and equivalence relation. Can someone please help me? It would be nice if I can have a conversation with my friend showing him what I learned on this.

REPLY [2 votes]: Take any equivalence class (and denote it $\mathscr c$) of $X/R$ so by the definition of the equivalence relation $\mathscr c$ isn't empty. Let $a\in\mathscr c$ then we have
$$p_X(a)=\mathscr c$$
hence $p_X$ is onto.


TITLE: Matrix with all eigenvalues $0$ but not triangular?
QUESTION [3 upvotes]: Is the situation described in the title achievable? I am looking for a $3\times 3$ case specifically.

REPLY [2 votes]: To construct examples of this, it's useful to take the $2\times 2$ case and extend it to larger sizes
First observe that there are no diagonalizable matrices with this property.
Thus, such matrices, if they exist are similar to either
$\begin{bmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0\end{bmatrix}$ or $\begin{bmatrix}0 & 1 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}$.
The matrix $\begin{bmatrix} 2 & 0 & -1\\ -4 & -1 & 1\\ 4 & 1 & -1\end{bmatrix}$ is an example of a matrix similar to the first of these. To find it I merely multiplied on the left by an (almost) random matrix and on the right by the inverse of this matrix.
Getting back to the $2\times 2$ case, it's useful to look at matrices similar to $\begin{bmatrix} 0 & 1\\ 0 & 0\end{bmatrix}$. Call this matrix $J_1$.
Take an arbitrary invertible two by two matrix $\begin{bmatrix} a & b\\c & d\end{bmatrix}$, say $P_1$, (its inverse is well known) and find the product $P_1J_1\left(P_1\right)^{-1}$. Now just force this product to not be triangular (it's possible). Finally, to extend it to the $n\times n$ (with $n\ge 3$) case just consider $$J_2=\begin{bmatrix} 0 & 1 & 0 &\ldots  & 0\\ 0 & 0 & 0 & \ldots & 0\\\vdots & \vdots &\ddots & \ldots & \vdots\\ 0 & 0 & 0 & \ldots & 0\end{bmatrix}$$ and for $P_2$ take $$\begin{bmatrix} a & b & 0 & 0 &\ldots  & 0\\ c & d & 0 & 0 &\ldots & 0\\ 0 & 0 & 1 & 0 &\ldots & 0\\ 0 & 0 & 0 & 1 & \ldots & 0\\\vdots & \vdots & \vdots & \vdots &  \ddots & \vdots\\ 0 & 0 & 0 & 0 &\ldots & 1\end{bmatrix}_.$$
The matrix $P_2J_2\left(P_2\right)^{-1}$ is a matrix with the desired property.


TITLE: Find $\lim_\limits{(x,y)\to (0,0)}\frac{x^\alpha y^4}{x^2+y^4}$ where $\alpha > 0$
QUESTION [4 upvotes]: How can we find the following limit? $$\lim_{(x,y)\to(0,0)}\frac{x^\alpha y^4}{x^2+y^4}\qquad \alpha>0$$
By using the polar coordinate, we get
$$\lim_{r\to 0}r^{\alpha+2}\frac{\cos^\alpha\theta \sin^4\theta}{\cos^2\theta+r^2\sin^4{\theta}}=0$$ if $\theta\notin\{\frac{\pi}{2}+\pi k:k\in\Bbb Z\}$.
Now, if $\theta = \frac{\pi}{2}+\pi k$ for some $k\in\Bbb Z$, then we get $$\lim_{(0,y)\to (0,0)}\frac{x^\alpha y^4}{x^2+y^4}=0.$$
Can we conclude that $$\lim_{(x,y)\to(0,0)}\frac{x^\alpha y^4}{x^2+y^4}=0?$$

REPLY [1 votes]: From the AM-GM inequality we have $$\left|\frac{x^ay^b}{x^2+y^4}\right|\le \frac12 |x|^{a-1}|y|^{b-2}$$
where we assume that either $a$ is such that $x^a\in \mathbb{R}$ for $x$ in a neighborhood of $0$ or that the limit is taken as $(x,y)\to (0^+,0)$.
And equality holds when $x=\pm y^2$.  Hence, we find
$$\lim_{(x,y)\to(0,0)}\frac{x^ay^b}{x^2+y^4}=0$$
whenever we have $2a+b>4$. The limit fails to exist otherwise.
In the case at hand, $a=\alpha$ and $b=4$ and we see that the limit is $0$ for $\alpha>0$ and fails to exist otherwise.


TITLE: Will Heat transfer through a metal be faster, if an electrical frequency is applied to the metal
QUESTION [2 upvotes]: Will a metal conduct heat faster if an electrical current is applied to the that metal ? For instance if a frequency current such as the same that would go to a speaker - so like 30 hz or 50hz or 1khz  -  has there ever been studies about such a thing to prove or disprove this ?  Or if metal is vibrating - will it transfer heat better if it is physically vibrating ?  -  So the next thing it this what if the metal is vibrating at 50hz and the metal also has a current of electricity going through it at the same interval - would that metal then be able to transfer heat faster under these conditions. ? Has this been studied ? Have experiments been done ?

REPLY [0 votes]: I am not sure whether experiments have been done on this topic. But my hypothesis for this question is as follows:

The electric field applied across the ends of a metal conductor causes the free electrons to flow in the opposite direction of the field. Here the metal ions do not move but vibrate a little bit due to the collision of free electrons.
If we heat one end of the rod, then the metal atoms vibrate continuously and vibrate the neighboring atoms also, making it to flow through the metal rod.

Here if we apply the electric field across the metal conductor, then there will be a flow of free electrons. But this will not affect the heat transfer in the rod as heat transfer is based on the vibration of the metal ions.
But the inverse is applicable, a heated metal rod will have more electrical resistance to the electric field applied. Because if the temperature of a metal conductor increases, the ions of the metal vibrate more vigorously. This increases the number of collisions between the free electrons and the ions.


TITLE: Clarification on statement of a decomposition for graded ring
QUESTION [0 upvotes]: Above is a decomposition of an element $f \in R$ for $R$ a graded ring. I am confused by the statement $f_j = 0$ for $j.$ Did Professor Eisenbud mean for all but finitely many $j$ as would be the case for infinite direct sums?

More context.

REPLY [1 votes]: This is just a typo, and it should read

... and $f_j=0$ for all but finitely many $j$;


TITLE: Nonrenormalizable but quantizable theory: gravity?
QUESTION [5 upvotes]: In p.8 of Michio Kaku book Introduction to Superstrings and M-Theory-Springer (1998), he said

The gravitational force. Gravity research was totally uncoupled from research in the other interactions. Classical relativists continued to find more and more classical solutions in isolation from particle research. Attempts to canonically quantize the theory were frustrated by the presence of the tremendous redundancy of the theory. There was also the discouraging realization that even if the theory could be successfully quantized, it WOULD ... still be nonrenormalizable.

My question is that

what does Kaku mean for nonrenormalizable but quantizable theory?

what are the criteria to be nonrenormalizable?
what are the criteria to be quantizable?

What are examples of nonrenormalizable but quantizable theories?

Fermi weak interaction theory? Gravity? and how come?

REPLY [9 votes]: The most basic definition of the terms (and I think the one Kaku is using) is
Quantizable theory - A classical theory that can be quantized in a way where the UV divergences are all cancelled by introducing counterterms.
Renormalizable theory - A quantizable theory which needs a finite number of counterterms.
Non-renormalizable theory - A quantizable theory which needs an infinite number of counterterms.
A theory is renormalizable as long as the classical coupling constants have non-negative mass dimension. For instance,

QED has coupling constant $e$ (electric charge) which has mass dimension 0 so it is renormalizable.

Fermi theory has coupling constant $G_F$ (Fermi coupling constant) which has mass dimension $-2$ so it is non-renormalizable.

Gravity has coupling $G$ (Newton's constant) which has dimension $-2$ so it is non-renormalizable.

Mass dimension of a quantity is the total power of mass dimension in a quantity when working in natural units where $[M]=[L]^{-1}=[T]^{-1}$. For instance, the dimension of Newton's constant are $[G] = [L]^3 [M]^{-1} [T]^{-2}$ (because $G = 6.674 \times 10^{-11} m^3 kg^{-1} s^{-2}$). In natural units, we have $[G] = [M]^{-2}$. This implies that $G$ has mass dimension $-2$.


TITLE: Puzzle on the triangle.
QUESTION [5 upvotes]: In triangle top four figures that have to be repositioned to form the "triangle" without a unit square.
How to explain this?

Thank's.

REPLY [3 votes]: Note how the Fibonacci sequence provides a nice way of generating more such triangles, since for any four consecutive elements $S_n, S_{n+1}, S_{n+2},S_{n+3}$ we have:
$S_n \cdot S_{n+3} = S_{n+1} \cdot S_{n+2} \pm 1$ (sign alternates for even and odd values of n).
Then make one of the two triangles on the "not-quite" hypotenuse have dimensions (width x height) $S_{n+2} \times S_n$ and the other $S_{n+3} \times S_{n+1}$. Then in one way of arranging, the rectangular vacancy is $S_{n+2} \times S_{n+1}$ and in the other it is $S_{n+3} \times S_n$, so there is exactly one square missing from the yellow/green rectangle in one of the two arrangements.
For example, to get a "one size smaller" arrangement, make the blue triangle $5 \times 2$ and the red triangle $3 \times 1$, to leave a $5 \times 1$ rectangle in the top half of the original diagram and a $3 \times 2$ rectangle in the bottom half of the diagram. Then use the middle row from the yellow and green pieces in the top half of the diagram, i.e. $2 \times 1$ and $3 \times 1$.