TITLE: Proton to neutron mass ratio while free versus when bound inside the nucleus?
QUESTION [1 upvotes]: Is the proton to neutron mass ratio different while they are free versus when they are bound inside a nucleus?

REPLY [0 votes]: However, it is hard to say anything about the masses of individual protons/neutrons in the nucleus, because you can only determine the total mass of the nucleus, but not the mass of a single proton/neutron inside the nucleus.

The above quote states the truth, there is no way to measure individual nucleons in the nucleus. But there is the mathematics developed to describe particles in those dimensions, special relativity and quantum mechanics. One can model mathematically the nucleus as a collective bound state of protons and neutrons.
Special relativity tells us that the invariant mass of the free particles is the length of the four vector .

In the nucleus the proton and neutron are not free, and quantum mechanics describes them as virtual particles which means that their four vectors are variable in length (see Feynman diagrams page), within the limits of the integration which  will give measurable quantities. This means that the virtual masses of protons and neutrons will be different at a given time, but this will depend on the model used to fit observed data, as the periodic table and the binding energy curve.


TITLE: Determine the limit of a series, involving trigonometric functions: $\sum \frac{\sin(nx)}{n^3}$ and $\frac{\cos(nx)}{n^2}$
QUESTION [5 upvotes]: I have $$\sum^\infty_{n=1} \frac{\sin(nx)}{n^3}.$$
I did prove convergence:
$0<\theta<1$
$$\left|\frac{\sin((n+1)x)n^3}{(n+1)^3\sin(nx)}\right|< \left|\frac{n^3}{(n+1)^3}\right|<\theta$$
Now I want to determine the limit. I found a similar proof but I need help understanding it; it goes like this. :
$$ F(x):=\sum^\infty_{n=1} \frac{\cos(nx)}{n^2}$$
As for this series we have uniform convergence. The series of derivatives: $$-\sum^\infty_{n=1} \frac{\sin(nx)}{n}$$ converges for every $\delta >0$ on the interval $[\delta, 2\pi-\delta]$ uniform against $\frac{x-\pi}{2}$
so, for every $x \in]0,2\pi[$ : $\displaystyle F'(x) = \frac{x-\pi}{2}$$\displaystyle F(x) = \left(\frac{x-\pi}{2}\right)^2+c,c\in \mathbb{R}$.
To determine the constant we calculate:
$$ \int^{2\pi}_0F(x)dx=\int^{2\pi}_0\left(\frac{x-\pi}{2}\right)^2dx+\int^{2\pi}_0cdx=\frac{\pi^3}{6}+2\pi c$$
(Question: Why can we do this do get the constant?)
Because $\int^{2\pi}_0cos(nx)dx= 0 \forall n≥1$ we have:
$$\int^{2\pi}_0F(x)dx = \sum^\infty_{n=1}\int^{2\pi}_0\frac{\cos(nx)}{n^2}=0,$$ so $c = -\frac{\pi^2}{12}$. (Question: How does he get to that term $\frac{\pi^2}{12}$?) With that we have proven, that
$$\sum^\infty_{n=1} \frac{\cos(nx)}{n^2}=\left(\frac{x-\pi}{2}\right)^2-\frac{\pi^2}{12}$$
If you can explain one of the questions about this proof, or if you know how to calculate the limit in my situation above, it would be cool if you leave a quick post here, thanks!

REPLY [3 votes]: $$\sum^\infty_{n=1}\frac{sin(nx)}{n^3}= ?$$
We know that $$\sum^\infty_{n=1}\frac{sin(nx)}{n}= \frac{x-\pi}{2}$$.
$$(\frac{sin(nx)}{n^3})''= (\frac{cos(nx)}{n^2})'=-\frac{sin(nx)}{n}$$
$$-\sum^\infty_{n=1}\frac{sin(nx)}{n}= -(\frac{x-\pi}{2})$$
$$ - \int \frac{x-\pi}{2}= -\int \frac{x}{2}-\frac{\pi}{2} = \frac{x^2}{4}+\frac{\pi x}{2}+c$$
$$\int\frac{x^2}{4}+\frac{\pi x}{2}+c=\frac{\pi x^2}{4}+\frac{x^3}{12}+xc$$
We can also determine the constant. We know that $-f'(0) = 0 + c$.
$$-\sum^\infty_{n=1} \frac{cos(n0)}{n^2}= - \sum^\infty_{n=1} \frac{1}{n^2} = -\frac{\pi^2} {6} = 0+c$$
$$-\int\frac{\pi^2} {6} = -\frac{x\pi^2} {6}$$ So
$$\sum^\infty_{n=1}\frac{sin(nx)}{n^3}= \frac{\pi x^2}{4}+\frac{x^3}{12}+(-\frac{x\pi^2}{6}) =  \frac{x\pi^2}{6}-\frac{\pi x^2}{4}+\frac{x^3}{12}$$


TITLE: The boundary of a manifold is a closed subset.
QUESTION [4 upvotes]: We want to show that the boundary $\partial M$ of an $n-$manifold M is a closed subset of the manifold. We show that its complement $M\setminus\partial M$ is open in $M$. Indeed,  each point $x \in M\setminus\partial M$ has an open neighborhood $V_x\subseteq M$ homeomorphic to $\mathbb R^n$. It remains only to show that $V_x$ lies entirely in $M\setminus \partial M$ which means that $V_x\cap \partial M=\emptyset$. Thank you for your help!
EDIT: In a manifold with boundary each point has an open neighborhood that is homeomorphic to $\mathbb R^n$ or to $\mathbb R^n_+=\{(x_1,\cdots,x_n)\in\mathbb R^n\;|\; x_n\ge 0\}$, the points who have open neighborhoods homeomorphic to $\mathbb R^n_+$ form the boundary of the manifold.

REPLY [6 votes]: This is obvious. Let $x \in M \setminus \partial M$. There exists an open neighborhood $V$ of $x$ in $M$ which is homeomorphic to $\mathbb R^n$. Then any $y \in V$ has the same property, hence $V \subset M \setminus \partial M$.
By the way, your definition of the boundary is not correct. Take $M = \mathbb R^n_+$. Then each $x \in \mathbb R^n_+$ has $\mathbb R^n_+$ as an open neighborhood which is homeomorphic $\mathbb R^n_+$. A correct definition is this: A point $p \in M$ is called a boundary point if there
exists a homeomorphism $\phi : U \to \mathbb R^n_+$ defined on an open neighborhood $U$ of $p$ in $M$ such that $\phi(p) \in \partial \mathbb R^n_+ =\mathbb R^{n-1} \times \{ 0\}$.


TITLE: if $Re(z\overline w)=|z||w|$ then $w=tz$ where $t\in \mathbb R$
QUESTION [3 upvotes]: I was given the following exercise in complex numbers:
Assume that for some $z,w \in \mathbb C$: $Re(z\overline  w)=|z||w|$
Show that this implies $w=tz$ where $t$ is some positive real.
What I did: we know that
$$Re(z\overline  w)=\frac{z\overline w+\overline z w}{2}=|z||w|$$
Then $$Re(z\overline  w)^2=(\frac{z\overline w+\overline z w}{2})^2=\frac{(z\overline w+\overline z w)(z\overline w+\overline z w)}{4}=\frac{z^2\overline w^2+2z\overline z w\overline w+\overline z^2w^2}{4}=|z|^2|w|^2=z \overline z w\overline w$$
Then overall we see that $z^2\overline w^2+\overline z^2w^2=2z\overline z w\overline w$
I don't see how this implies that $w=tz$. I don't see the connection.
What am I missing?

REPLY [3 votes]: You have
$$\frac{\overline zw+z\overline w}{2}=|z||w|$$
Dividing both sides by $z\overline z=|z|^2$, we get:
$$\frac12\left(\frac wz+\overline{\frac wz}\right)=\frac{|w|}{|z|}$$
or:
$$\operatorname{Re}\left(\frac wz\right)=\left|\frac wz\right|$$
If $\operatorname{Re}(x)=|x|$, then we know that $x$ is real. Thus, $\frac wz$ is real.

REPLY [0 votes]: Did you try writing $z$ and $w$ in the polar format? Writing
$$z=r_1e^{i\theta_1}\\
w=r_2e^{i\theta_2}$$
Makes it much easier to calculate what $z\overline{w}$ is.


text:
\begin{document}

\title[Universal enveloping algebras of nilpotent Lie algebras]{The isomorphism problem for universal enveloping algebras of
nilpotent Lie algebras}
\author{Csaba Schneider and Hamid Usefi}
\address[Schneider]{Centro de \'Algebra da Universidade de Lisboa\\
Av. Prof. Gama Pinto, 2\\ 1649-003 Lisboa\\ Portugal\hfill\break
Email: csaba.schneider@gmail.com\hfill\break
WWW: www.sztaki.hu/$\sim$schneider}
\address[Schneider]{Informatics Research Laboratory\\
Computer and Automation Research Institute\\
1518 Budapest Pf.\ 63\\
Hungary}
\address[Usefi]{Department of Mathematics\\
University of British Columbia\\
1984 Mathematics Road\\
Vancouver, BC, Canada, V6T 1Z2\hfill\break
Email: usefi@math.ubc.ca\hfill\break
WWW: www.math.ubc.ca/$\sim$usefi}

\thanks{The first author was supported by the FCT project
PTDC/MAT/101993/2008 and by the Hungarian
Scientific Research Fund (OTKA) grant~72845. The second author is supported by an NSERC PDF}

\begin{abstract}
In this paper we study the isomorphism problem for the
universal enveloping algebras of nilpotent Lie algebras. We prove that
if the characteristic of the underlying field is not~2 or~3, then
the isomorphism type of a nilpotent Lie algebra of dimension at most~6 is
determined by the isomorphism type of its universal enveloping algebra. Examples show that the
restriction on the characteristic is necessary.
\end{abstract}

\maketitle

\section{Introduction}\label{intro}
In this paper we examine the isomorphism problem for universal
enveloping algebras of Lie algebras. It is known that two non-isomorphic Lie
algebras may have isomorphic universal enveloping algebras; see for instance
\cite[Example~A]{ru}. However, all such known examples require that the
 characteristic
of the underlying field is a prime.  In this paper we focus on
nilpotent Lie algebras and prove the following main result.

\begin{theorem}\label{main}
Suppose that $\F$ is a field such that $\char\,\F\not\in\{2,3\}$. Then
the isomorphism type of a nilpotent Lie algebra of dimension at most~$6$
over~$\F$ is determined by the isomorphism type of its universal
enveloping algebra.
\end{theorem}

Theorem~\ref{main} is a consequence of Theorems~\ref{5main} and~\ref{6main}.
As shown by examples in Sections~\ref{dim5} and~\ref{dim6},
the requirement about the characteristic of the underlying field is necessary.
In addition to proving Theorem~\ref{main}, we classify the possible
isomorphisms between the universal enveloping algebras of nilpotent Lie
algebras of dimension at most~5 over an arbitrary field, and those of
dimension~6 over fields of characteristic different from~2 (see Theorems~\ref{5main} and~\ref{6main}).

Little progress has ever been made on the isomorphism problem for
universal enveloping algebras. For a Lie algebra $L$, let $\uea L$ denote
its universal enveloping algebra (see Section~\ref{prelim} for the definitions).
Several invariants of $L$
are known to be determined by $\uea L$. For instance,
if $L$ is  finite-dimensional, then the (linear) dimension of $L$
coincides with the Gelfand-Kirillov dimension of $\uea L$ (see~\cite{KL}).
More recently~Riley and Usefi~\cite{ru} proved that the nilpotence of $L$ is
determined by $U(L)$ and, for a nilpotent $L$, the nilpotency
class of $L$ can be computed using $U(L)$.
Moreover the isomorphism type of $U(L)$ determines the isomorphism
type of the graded algebra $\gr L$ associated with the lower central series of
$L$ (see Section~\ref{prelim} for the definitions).
Malcolmson~\cite{mal}
showed that if $L$ is a 3-dimensional simple Lie algebra over
a field of characteristic not~2, then
$L$ is determined by $U(L)$ up to isomorphism.
Later Chun, Kajiwara, and Lee~\cite{ckl} generalized Malcolmson's result
to the class of all Lie algebras with dimension $3$
over fields of characteristic not~2.

By proving Theorem~\ref{main},
we verify  that the isomorphism
problem for universal enveloping algebras has a positive solution in the class
of nilpotent Lie algebras with dimension at most~6 over fields of
characteristic
different from 2~and~3.  The proof of this result relies on
the classification of nilpotent Lie
algebras with dimension at most 6. The classification of such Lie algebras
of dimension at most~5 has been known for a long time over
an arbitrary field. In dimension~6, several classifications have been published,
but they were often incorrect, and they usually only treated fields of
characteristic~0. Recently De Graaf~\cite{degraaf} published a
classification of 6-dimensional nilpotent Lie algebras over an arbitrary field
of characteristic not~2. As the classification by De Graaf has been obtained
making heavy use of computer calculations, and was checked
by computer for small
fields~\cite{sch}, we consider this classification as the most reliable in
the literature. The reason we do not treat 6-dimensional nilpotent Lie
algebras  over fields of characteristic~2 is that, in this case,
we do not know of a similarly reliable classification.

Our strategy in proving Theorem~\ref{main} is to determine all pairs
of nilpotent Lie algebras $L_1,\ L_2$ with dimension at most~6,
such that the graded
algebras $\gr{L_1}$ and $\gr{L_2}$ associated
with the lower central series
are isomorphic. We know from~\cite{ru}
that this is a necessary condition for the isomorphism $\uea{L_1}\cong
\uea{L_2}$. Such pairs can be read off
from the list of nilpotent Lie algebras with dimension at most~6
in~\cite{degraaf}. Next, for all such pairs,
we either argue that $U(L_1)$ cannot be isomorphic to $U(L_2)$, or
we exhibit an explicit isomorphism between $U(L_1)$ and $U(L_2)$.
Initially, computer experiments played a role in determining the isomorphisms
between universal enveloping algebras of nilpotent Lie algebras. Recent work
by Eick~\cite{eick} describes a practical algorithm to decide isomorphism between
finite dimensional nilpotent associative algebras.
Her algorithm was implemented in the {\sf ModIsom} package~\cite{modisom} of
the {\sf GAP} computational algebra system~\cite{gap} and the implementation
works for algebras with  dimensions more than 100 over small finite fields.
We used this implementation to decide isomorphisms between the
finite-dimensional, nilpotent quotients $\aug{L}/\augp Lk$ of the
augmentation ideals $\aug L$ of $U(L)$ (see Section~\ref{prelim} for notation).
We remark here an interesting observation. If $L$ is nilpotent of class $c$ then based on our calculations the isomorphism type of the quotient $\aug{L}/\augp L{c+1}$ determines the isomorphism type of $L$. So, the question remains whether $\aug{L}/\augp L{c+1}$
determines the isomorphism type of $L$ in all dimensions.

\section{Preliminaries}\label{prelim}

In this section we summarize some important facts about universal enveloping
algebras of Lie algebras; see~\cite{dix} for a more detailed background.
We assume from now on that Lie algebras are finite-dimensional, even though
most of the results referred to in this section hold for a larger class of
Lie algebras.

Let $L$ be a Lie algebra over a field $\F$.
The universal enveloping algebra $\uea L$ of $L$ is
defined as follows. For $i\geq 0$, let $L^{\otimes i}$ denote the $i$-fold tensor
power of $L$ and set
$$
\mathcal T=\bigoplus_{i=0}^\infty L^{\otimes i}.
$$
The space $L^{\otimes 0}$ is one-dimensional, generated by the unit of $\mathcal T$,
and is usually identified with $\F$.
The sum $\mathcal T$ can be considered as an algebra over $\F$ with respect to
the obvious multiplication defined on the generators  of $\mathcal T$ as
$$
(v_1\otimes \cdots\otimes v_k)\cdot (u_1\otimes \cdots\otimes u_l)=
v_1\otimes \cdots\otimes v_k\otimes u_1\otimes \cdots\otimes u_l,
$$
for all $v_1,\ldots,v_k,u_1,\ldots,u_l\in L$. The algebra
$\mathcal T$ is usually referred to as the {\em tensor algebra} of $L$.
Let $\mathcal I$ denote the
two-sided ideal of $\mathcal T$ generated by elements of the form
$[u,v]-u\otimes v+v\otimes u$.  Then the {\em universal enveloping algebra}
$\uea L$ of $L$ is defined as the quotient $\mathcal T/\mathcal I$.
We view $L=L^{\otimes 1}$ as a  Lie subalgebra of $\uea L$.
Universal enveloping algebras have the following universal property.

\begin{lemma}\label{Uni}
Suppose that $L$ is a Lie algebra, $A$ is an associative algebra, and let
$\varphi:L\rightarrow A$ be a Lie algebra homomorphism. Then there is a unique
associative algebra homomorphism $\overline \varphi:\uea L\rightarrow A$ such
that $\overline\varphi|_L=\varphi$.
\end{lemma}

The linear subspace spanned by a subset $X$ of a vector space
is denoted by $\la X\ra_{\F}$.
Recall that the {\em center} $Z(L)$ of a Lie algebra $L$ is the ideal
$$
\la x\in L\ |\ [x,a]=0, \mbox{ for all }a\in L\ra_{\F}.
$$
The center of the universal enveloping algebra plays an important role in
our arguments. The {\em center} $Z(A)$ of an associative
algebra $A$ is defined as
$$
Z(A)=\la x\in A\ |\ ax=xa, \mbox{ for all } a\in A \ra_{\F}.
$$
It is clear that the subalgebra of $U(L)$ generated by the center
$Z(L)$ of $L$ lies in $Z(U(L))$.

Note that $\mathcal T_+=L^{\otimes 1}\oplus L^{\otimes 2}\oplus\cdots$ is a two-sided ideal in
$\mathcal T$, and the image of $\mathcal T_+$ in $\uea L$ is referred to as
the {\em augmentation ideal} of $\uea L$ and is denoted by $\aug L$.
The following is proved in~\cite[Lemma~2.1]{ru}.

\begin{lemma}\label{augisom}
For Lie algebras $L$ and $K$, $\uea L\cong \uea K$ if and only if
$\aug L\cong\aug K$.
\end{lemma}

Investigating the  isomorphism
between $\uea L$ and $\uea K$ will often be carried out, using
Lemma~\ref{augisom}, through studying
the isomorphism between $\aug K$ and $\aug L$.
For $i\geq 1$, let $\augp Li$ denote the ideal of $\aug L$ generated by the
products
of elements of $\aug L$ with length at least $i$. This way we obtain
a descending series in $\aug L$:
$$
\aug L\geq \augp L2\geq\cdots.
$$
The sequence $\augp Li$ is a filtration on $\aug L$: for $x\in \W^i(L)$
and $y\in\W^j(L)$ we have that $xy\in\augp L{i+j}$.

A basis for $\augp Li$
can usually be constructed as follows. Let $\lcs Li$ denote the $i$-th term of
the {\em lower central series} of $L$; that is $L^1=L$, and, for $i\geq 1$,
$L^{i+1}=[L^i,L]$. For an element $v\in L$, we define
the {\em weight} $w(v)$ of $v$ as the largest integer $i$
such that $v\in\lcs Li$. A basis $\mathcal B=\{v_1,\ldots,v_d\}$ of a Lie algebra
$L$ is said to be {\em homogeneous} if the basis elements
with weight at least $i$ form a basis for $\lcs Li$.
An element of $\uea L$ of the form $m=v_{i_1}v_{i_2}\ldots v_{i_k}$
with $i_1\leq i_2\leq\cdots\leq i_k$ is said
to be a {\em Poincar\'e-Birkhoff-Witt monomial},
or more briefly, a
{\em PBW monomial}, in $\mathcal B$.
 The {\em weight} $w(m)$ of
such a monomial is defined as $w(v_{i_1})+w(v_{i_2})+\cdots+w(v_{i_k})$.

\begin{theorem}[Proposition~3.1(1) in \cite{R}]\label{basis-w}
Let $L$ be a  Lie algebra with a homogeneous
basis $\mathcal B$ and let $t\geq 1$.
Then  the set of all PBW monomials in $\mathcal B$
with weight at least $t$
forms an $\F$-basis for $\augp Lt$, for every $t\geq 1$.
\end{theorem}

In this paper we are interested to discover,
for a pair of nilpotent non-isomorphic
Lie algebras  $L_1$ and $L_2$, if $U(L_1)$ can be isomorphic to $U(L_2)$.
Often we use the graded algebras $\gr{L_i}$ associated with the lower
central series of the $L_i$ to rule out the isomorphism between
$\uea{L_1}$ and $\uea{L_2}$. For a Lie algebra $L$, $\gr L$ is defined
as the algebra on the linear space
$$
\gr L=L/\lcs L2\oplus \lcs L2/\lcs L3\oplus\cdots
$$
with respect to the multiplication given by the rule
$$
[x+\lcs L{i+1},y+\lcs L{j+1}]=[x,y]+\lcs L{i+j+1} \quad
\mbox{for all $x\in \lcs Li$ and $y\in \lcs Lj$}.
$$
The following is proved in~\cite[Proposition~4.1]{ru}.

\begin{theorem}\label{gr}
For any Lie algebra $L$, the isomorphism type of $\uea L$  determines the isomorphism type of $\gr L$. Consequently,
if $L$ is nilpotent of class $2$, then the isomorphism type of $\uea L$ determines
the isomorphism type of  $L$.
\end{theorem}

Suppose that $L_1$ and $L_2$ are nilpotent Lie algebras such that
$U(L_1)\cong U(L_2)$. Then Theorem~\ref{gr} implies, for all $i$, that
$\dim \lcs {L_1}i=\dim \lcs {L_2}i$, and that the nilpotency classes of $L_1$ and $L_2$
coincide. The second statement of Theorem~\ref{gr}
is an easy consequence of
the first statement if $L$ is finite-dimensional.
Indeed, if $L$ is a finite-dimensional Lie algebra
of nilpotency class $2$, then it is always isomorphic to $\gr L$.
The same assertion without a restriction on the dimension is proved
in~\cite[Section~5]{ru}.

The following lemma can be found in~\cite[Lemma~5.1]{ru}.

\begin{lemma}\label{L^i}
Let $L$ and $K$ be Lie algebras and let $\varphi: \aug L\to \aug K$ be
an isomorphism.
Then $\varphi(L^i+\augp L{i+1})=K^i+\augp K{i+1}$, for every positive integer $i$.
\end{lemma}

\section{Nilpotent Lie algebras with dimension at most 5}\label{dim5}

In this section we determine all isomorphisms between the universal
enveloping algebras of nilpotent Lie algebras
with dimension at most 5.
A classification of such Lie algebras is well known and
can be found, for instance, in~\cite{degraaf}. The main result of this section
is the following.

\begin{theorem}\label{5main}
Let $L$ and $K$ be nilpotent Lie algebras of dimension at most~$5$ over a field $\F$ such that $\uea{L}\cong\uea{K}$.
Then one of the following must hold:
\begin{itemize}
\item[(i)] $L\cong K$;
\item[(ii)] $\char\,\F=2$; further $L$ and $K$ are isomorphic to the
Lie algebras $L_{5,3}$ and $L_{5,5}$ or they are isomorphic to the Lie algebras
$L_{5,6}$ and $L_{5,7}$ in Section~$5$ of~\cite{degraaf}.
\end{itemize}
\end{theorem}

Since there is a unique isomorphism class of nilpotent Lie  algebras
with dimension 1, and there is a unique such class with dimension 2,
the isomorphism problem of universal enveloping algebras is trivial in these
cases. Up to isomorphism,
there are two nilpotent Lie algebras with dimension $3$, an abelian,
and a non-abelian. By Theorem~\ref{gr} their universal enveloping algebras
must be non-isomorphic. The number of isomorphism classes of
4-dimensional nilpotent Lie algebras
is 3. One of these algebras is abelian, the second has nilpotency class 2, and
the third has nilpotency class 3. Again, by Theorem~\ref{gr},
their universal enveloping algebras are pairwise non-isomorphic.
This proves Theorem~\ref{5main} for Lie algebras of dimension at most~4.

There are 9 isomorphism classes of nilpotent Lie algebras with dimension 5 and
they are listed at the beginning of Section~4 in~\cite{degraaf}.
To simplify notation, we denote the Lie algebra $L_{5,i}$ in De Graaf's list by
$L_i$.
Inspecting this list, we find that the sequence  $(\dim\lcs L1,\dim\lcs L2,\ldots)$
for these Lie algebras, after omitting the tailing zeros,
are  as follows: $(5)$, $(5,1)$, $(5,2,1)$, $(5,1)$, $(5,2,1)$,
$(5,3,2,1)$, $(5,3,2,1)$, $(5,2)$, $(5,3,2)$. Therefore Theorem~\ref{gr}
implies that if $U(L_i)\cong U(L_j)$ then either $i=j$, or $\{i,j\}=\{3,5\}$,
or $\{i,j\}=\{6,7\}$. The Lie algebras that are involved in these
possible isomorphisms are as follows:
\begin{eqnarray*}
L_3&=&\left<x_1,x_2,x_3;x_4;x_5\ |\ [x_1,x_2]=x_4,\ [x_1,x_4]=x_5\right>;\\
L_5&=&\left< x_1,x_2,x_3;x_4;x_5\ |\ [x_1,x_2]=x_4,\ [x_1,x_4]=x_5,\
[x_2,x_3]=x_5\right>;\\
L_6&=&\left<x_1,x_2;x_3;x_4;x_5\ |\ [x_1,x_2]=x_3,\ [x_1,x_3]=x_4,\ [x_1,x_4]=x_5,\ [x_2,x_3]=x_5\right>;\\
L_7&=&\left<x_1,x_2;x_3;x_4;x_5\ |\ [x_1,x_2]=x_3,\ [x_1,x_3]=x_4,\ [x_1,x_4]=x_5\right>.
\end{eqnarray*}
Products that are zero are omitted from the multiplication tables above.
For instance $[x_1,x_3]=0$ in $L_3$ and in $L_5$.
Note that the bases of the Lie algebras given above
are homogeneous.  Further, basis elements of different weights are separated by a semicolon. The
multiplication tables given in~\cite{degraaf} are somewhat different from the
ones above, as we changed the orders of certain basis elements in order to
work with homogeneous bases.
A possible source of confusion is that
the symbols $x_i$ are used to denote elements of different Lie algebras,
but we believe that using different letters or introducing a subscript or
superscript would unnecessarily complicate the notation.

\begin{lemma}\label{5-dim}
If $\F$ is a filed of characteristic not 2, then  $\Omega(L_3)/\Omega^4(L_3)\not\cong
\Omega(L_5)/\Omega^4(L_5)$ and $\Omega(L_6)/\Omega^5(L_6)\not\cong
\Omega(L_7)/\Omega^5(L_7)$; consequently $\W(L_3)\not\cong\W(L_5)$
and $\W(L_6)\not\cong\W(L_7)$.
Otherwise, $\W(L_3)\cong \W(L_5)$ and $\W(L_6)\cong \W(L_7)$.
\end{lemma}
\begin{proof}
Suppose that $\char\,\F\neq 2$ and  set $B_i=\Omega(L_i)/\Omega^4(L_i)$,
for $i=3, 5$. First we show that $B_3\not\cong B_5$.
We claim that
$Z(B_5)\leq (B_5)^2$ while $Z(B_3)\not\leq (B_3)^2$, which
will imply that $B_3\not\cong B_5$.
As $x_3\in Z(B_3)\setminus(B_3)^2$, the second assertion of the claim
is valid.
In order to prove the first assertion, let $w\in Z(B_5)$.
We write $w$ as a linear combination of PBW monomials:
$$
w=\sum_{i_1\leq\cdots\leq i_n}\alpha_{i_1,\ldots,i_n}x_{i_1}\cdots x_{i_n}.
$$
Then $w\equiv\alpha_1 x_1+\alpha_2x_2+\alpha_3x_3\pmod{(B_5)^2}$.
Since $w$ is a central element in $B_5$, we have $[x_1, w]=[x_2, w]=0$,
and so
\begin{align*}
0=[x_1, w]&\equiv\alpha_2[x_1, x_2]= \alpha_2 x_4\pmod{(B_5)^3},\\
0=[x_2, w]&\equiv\alpha_1[x_2, x_1]=- \alpha_1 x_4\pmod{(B_5)^3}.
\end{align*}
Hence, $\alpha_1=\alpha_2=0$.
Since $x_5\in \lcs{L_5}3$, we have
\begin{align*}
0=[x_2, w]&=\alpha_3[x_2, x_3]+\alpha_{1,1} [x_2, x_1^2]+\alpha_{1,2}[x_2, x_1x_2]+\alpha_{1,3}[x_2, x_1x_3] \\
&=\alpha_3x_5+\alpha_{1,1}(-2x_1x_4+x_5)-\alpha_{1,2}x_2x_4-\alpha_{1,3}x_3x_4.
\end{align*}
Since $\char  \F\neq 2$, we deduce that $\alpha_3=0$. Thus, $w\in(B_5)^2$ as claimed.

Now set $B_i=\Omega(L_i)/\Omega^5(L_i)$, for $i=6, 7$.  Suppose, to the contrary, that
 $f:B_7 \rightarrow B_6$ is an isomorphism. Since $B_7$
is generated by $x_1$ and $x_2$, the map $f$ is determined by
the images $f(x_1)$ and $f(x_2)$.
As above, let us write $f(x_1)$ and $f(x_2)$ as linear combinations of
PBW monomials:
\begin{align*}
f(x_1)&=\sum_{i_1\leq\cdots\leq i_n}\alpha_{i_1,\ldots,i_n}x_{i_1}\cdots x_{i_n},\\
f(x_2)&=\sum_{i_1\leq\cdots\leq i_n}\beta_{i_1,\ldots,i_n}x_{i_1}\cdots x_{i_n}.
\end{align*}
As
\begin{align}\label{f[x1,x2]}
f([x_1,x_2])\equiv(\alpha_1\beta_2-\alpha_2\beta_1)[x_1,x_2]= (\alpha_1\beta_2-\alpha_2\beta_1)x_3\pmod{(B_6)^3},
\end{align}
we obtain that $\alpha_1\beta_2-\alpha_2\beta_1\neq 0$. Equation (\ref{f[x1,x2]}) also gives
$$
0=f([x_1,x_2,x_2])\equiv \beta_1(\alpha_1\beta_2-\alpha_2\beta_1)[x_1,x_2,x_1]=
-\beta_1(\alpha_1\beta_2-\alpha_2\beta_1)x_4 \pmod{(B_6)^4}.
$$
Since $\alpha_1\beta_2-\alpha_2\beta_1\neq 0$, we must have $\beta_1=0$.
This implies that $\alpha_1\neq 0$ and $\beta_2\neq 0$.

Furthermore, modulo $(B_6)^4$, we have
\begin{multline*}
f([x_1,x_2])\equiv
\alpha_1\beta_2x_3+(\alpha_1\beta_3+\alpha_2\beta_{1,1}-\alpha_{1,1}\beta_2)x_4+\\
(\alpha_1\beta_{1,2}-2\alpha_2\beta_{1,1}+2\alpha_{1,1}\beta_2)x_1x_3+
(2\alpha_1\beta_{2,2}-\alpha_2\beta_{1,2}+\alpha_{1,2}\beta_2)x_2x_3.
\end{multline*}

Hence $f([x_1,x_2,x_2])$ is equal to the following:
\begin{multline*}
(-\alpha_1\beta_2^2+\alpha_1\beta_2\beta_{1,1})x_5
-2\alpha_1\beta_2\beta_{1,1}x_1x_4-\alpha_1\beta_2\beta_{1,2} x_2x_4+\beta_2(\alpha_1\beta_{1,2}-2\alpha_2\beta_{1,1}+2\alpha_{1,1}\beta_2)x_3x_3.
\end{multline*}
As $[x_1,x_2,x_2]=0$ in $\W(L_7)$, we must have $f([x_1,x_2,x_2])=0$.
Since  $\char(\F)\neq 2$ and $\alpha_1\beta_2\neq 0$, we get $\beta_{1,1}=0$.
Thus the coefficient
of $x_5$ in $f([x_1,x_2,x_2])$ is $-\alpha_1\beta_2^2$. This, implies that $
f([x_1,x_2,x_2])\neq 0$, which is a contradiction. Thus $B_6\not\cong B_7$, as claimed.

Let us now assume that $\char\,\F=2$, and prove $\W(L_3)\cong\W(L_5)$ and $\W(L_6)\cong \W(L_7)$.
Note that the map from $L_3$ to $\W(L_5)$ induced by  $x_1\mapsto x_1$, $x_2\mapsto x_2$,
$x_3\mapsto x_3+x_1^2$ is a Lie homomorphism. Hence, by Lemma~\ref{Uni},
it can be extended to
a homomorphism $\varphi$ from $\W(L_3)$ to $\W(L_5)$.
Since $x_1$, $x_2$, and $x_3+x_1^2$ form
a generating set for $\W(L_5)$, $\varphi$ is onto. We need to show that
$\varphi$ is injective. Since
$\dim \W(L_3)/\W^i(L_3) = \dim \W(L_5)/\W^i(L_5)$ (see Theorem~\ref{basis-w}),
the homomorphism $\varphi$ induces an isomorphism
$$
\varphi_i:\W(L_3)/\W^i(L_3) \to \W(L_5)/\W^i(L_5),
$$
 for every $i\geq 1$.
Let  $x$ be a non-zero element in $\W(L_3)$ such that $\varphi(x)=0$. By Theorem \ref{basis-w}, there exists a positive integer $i$  such that $x\in \W^{i-1}(L_3)\setminus  \W^i(L_3)$.
As $\varphi_i$ is an isomorphism, we get $\varphi_i(x)\neq 0$,
which is a contradiction. Hence $\varphi$ is injective, and so $\W(L_3)\cong \W(L_5)$.

One can show precisely the same way that  $x_1\mapsto x_1$, $x_2\mapsto x_2+x_1^2$ extends to an isomorphism from
$\W(L_7)$ to $\W(L_6)$.
\end{proof}

Lemma~\ref{5-dim} in combination with Lemma~\ref{augisom},
and the argument preceding it proves Theorem~\ref{5main}.

\section{6-dimensional nilpotent Lie algebras}\label{dim6}

The isomorphisms between universal enveloping algebras of nilpotent Lie
algebras of dimension~6 in characteristic different from~2 are described by the following
theorem.

\begin{theorem}\label{6main}
Let  $L$ and $K$ be nilpotent Lie algebras of dimension~$6$ over a field $\F$ of characteristic not 2. If  $\uea{L}\cong\uea{K}$, then one of the following must hold.
\begin{itemize}
\item[(i)] $L\cong K$.
\item[(ii)] $\char\,\F=3$; further $L$ and $K$ are isomorphic to one
of the following pairs of Lie algebras in~\cite[Section~5]{degraaf}:
$L_{6,6}$ and $L_{6,11}$;
$L_{6,7}$ and $L_{6,12}$; $L_{6,17}$ and $L_{6,18}$; $L_{6,23}$ and $L_{6,25}$.
\end{itemize}
\end{theorem}

Throughout this section $\F$ denotes a field with characteristic different from 2.
The proof of Theorem~\ref{6main} is presented in this section.
De Graaf~\cite{degraaf} lists the isomorphism types of 6-dimensional
nilpotent Lie algebras over an arbitrary field $\F$ of characteristic different
from~2. De Graaf denotes these Lie algebras by $L_{6,i}$ or
$L_{6,i}(\varepsilon)$ where $1\leq i\leq 26$ and $\varepsilon$ is a field
element. To simplify notation and to distinguish
between the Lie algebras of Sections~\ref{dim5} and~\ref{dim6}
we will denote $L_{6,i}$ with $K_i$ and $L_{6,i}(\varepsilon)$ with
$K_i(\varepsilon)$.
Inspecting the list of Lie algebras, we obtain that the isomorphisms
among the graded Lie algebras associated with the lower central series of
these Lie algebras are as follows:

\begin{enumerate}
\item $\gr{K_{3}}\cong\gr{K_5}\cong \gr{K_{10}}$;\label{iso1}
\item $\gr{K_{6}}\cong\gr{K_7}\cong \gr{K_{11}}\cong\gr{K_{12}}\cong\gr{K_{13}}$; \label{iso2}
\item $\gr{K_{14}}\cong\gr{K_{16}}$;\label{iso3}
\item $\gr{K_{15}}\cong\gr{K_{17}}\cong \gr{K_{18}}$;\label{iso4}
\item $\gr{K_{23}}\cong\gr{K_{25}}$;\label{iso5}
\item $\gr{K_{24}(\varepsilon)}\cong\gr{K_{9}}, \mbox{ for all } \varepsilon\in\F$.\label{iso6}
\end{enumerate}

Let $L$ and $K$ be 6-dimensional  non-isomorphic nilpotent
Lie algebras over $\F$. If $\uea L\cong\uea K$ then, by Theorem~\ref{gr}, $L$ and $K$ must both occur in one of the lines~\eqref{iso1}--\eqref{iso6} above.
In Lemmas \ref{dim51}-\ref{lem-iso6} we examine the possible isomorphisms
between the universal enveloping algebras of the Lie algebras that
occur in one of the families above. Using Lemma \ref{augisom} we only examine
the possible isomorphisms between the augmentation ideals.

 As with the 5-dimensional Lie algebras, we change the multiplication tables of the
algebras presented in~\cite{degraaf} in order to work with homogeneous bases.
The bases elements of different weights are separated by  a semicolon. Further,
as in Section~\ref{dim5}, we omit products of the form
$[x_i,x_j]=0$  from the multiplication tables of the Lie algebras.
For $i=3,\ 5,\ 6,\ 7,\ 9,\ 10,\ 11,\ 12,\ 13,\ 14,\ 15,\ 16,\ 17,\
18,\ 23,\ 25$ we set $U_i=\uea{K_i}$ and $\Omega_i=\aug{K_i}$.
Further, let $U_{24}(\varepsilon)$
and $\Omega_{24}(\varepsilon)$ denote $\uea{K_{24}(\varepsilon)}$ and
$\aug{K_{24}(\varepsilon)}$, respectively.

\subsection{Family~\eqref{iso1}}

First we deal with the isomorphism $\gr{K_{3}}\cong\gr{K_5}\cong \gr{K_{10}}$
where
\begin{eqnarray*}
K_3&=&\left<x_1,x_2,x_3,x_4;x_5;x_6\ |\ [x_1,x_2]=x_5,\ [x_1,x_5]=x_6\right>;\\
K_5&=&\left<x_1,x_2,x_3,x_4;x_5;x_6\ |\ [x_1,x_2]=x_5,\ [x_1,x_5]=x_6,\
[x_2,x_3]=x_6\right>;\\
K_{10}&=&\left<x_1,x_2,x_3,x_4;x_5;x_6\ |\ [x_1,x_2]=x_5,\ [x_1,x_5]=x_6,\
[x_3,x_4]=x_6\right>.
\end{eqnarray*}

\begin{lemma}\label{dim51}
The algebras $\Omega_3$, $\Omega_5$, and $\Omega_{10}$ are pairwise non-isomorphic.
\end{lemma}
\begin{proof}
For $i=3,\ 5,\ 10$, let $B_i=\W_i/(\W_i)^4$.
It is enough to prove that the centers $Z_i=Z(B_i)$ have different dimensions.
The quotient $B_i$ is spanned by the images of the
PBW monomials with weight at most 3 and
we will identify such a monomial with
its image.

We claim that $Z_3=\left<x_3,x_4,(B_3)^3\right>_{\F}$. Set
$C=\left<x_3,x_4,(B_3)^3\right>_{\F}$.
Clearly,  $C\leq Z_3$. Let $z\in Z_3$. Then $z$ is a linear
combination of PBW monomials with weight at most~3.
As usual, $\alpha_{i_1,\ldots,i_n}$ is  the coefficient of $x_{i_1}\cdots x_{i_n}$ in the PBW representation of $z$. We may assume without
loss of generality that all monomials in $C$ occur with coefficient zero.
First we compute that
$$
[x_1,z]=\alpha_2 x_5+\alpha_{1,2} x_1x_5+2\alpha_{2,2}x_2x_5+\alpha_{2,3} x_3x_5+\alpha_{2,4} x_4x_5+\alpha_{5}x_6.
$$
We deduce that $\alpha_2=\alpha_{1,2}=\alpha_{2,2}=\alpha_{2,3}=\alpha_{2,4}=\alpha_{5}=0$.
Now
$$
[z,x_2]=\alpha_1x_5+2\alpha_{1,1}x_1x_5-\alpha_{1,1} x_6+\alpha_{1,3}x_3x_5+\alpha_{1,4}x_4x_5.
$$
This implies that  $\alpha_1=\alpha_{1,1}=\alpha_{1,3}=\alpha_{1.4}=0$.
Therefore $Z_3=C$ as claimed. Since $x_3$, $x_4$, $x_3x_3$, $x_3x_4$, $x_4x_4$
together with the PBW monomials with weight~3 form a basis for $C$, we
obtain that $\dim C=\dim Z_3=30$.
Similar argument shows that $\dim Z_5=29$, and $\dim Z_{10}=28$.
Thus the $Z_i$ have different dimensions, as required.
\end{proof}

\subsection{Family \eqref{iso2}}
We examine the following family of Lie algebras.

\begin{eqnarray*}
K_6&=&\left<x_1,x_2,x_3;x_4;x_5;x_6\ |\ [x_1,x_2]=x_4,\ [x_1,x_4]=x_5,\ [x_1,x_5]=x_6,\ [x_2,x_4]=x_6\right>;\\
K_7&=&\left<x_1,x_2,x_3;x_4;x_5;x_6\ |\  [x_1,x_2]=x_4,\ [x_1,x_4]=x_5,\ [x_1,x_5]=x_6\right>;\\
K_{11}&=&\left<x_1,x_2,x_3;x_4;x_5;x_6\ |\ [x_1,x_2]=x_4,\ [x_1,x_4]=x_5,\ [x_1,x_5]=x_6,\ [x_2,x_4]=x_6,\right.\\
&&\left.[x_2,x_3]=x_6\right>;\\
K_{12}&=&\left<x_1,x_2,x_3;x_4;x_5;x_6\ |\ [x_1,x_2]=x_4,\ [x_1,x_4]=x_5,\ [x_1,x_5]=x_6,\ [x_2,x_3]=x_6\ \right>;\\
K_{13}&=&\left<x_1,x_2,x_3;x_4;x_5;x_6\ |\ [x_1,x_2]=x_4,\ [x_1,x_4]=x_5,\ [x_1,x_5]=x_6,\ [x_2,x_3]=x_5,\right.\\
&&\left. [x_4,x_3]=x_6\right>.
\end{eqnarray*}

\begin{lemma}\label{fam2lem}
The enveloping algebras $\W_6$, $\W_7$, $\W_{11}$, $\W_{12}$, and
$\W_{13}$ are pairwise non-isomorphic provided that $\char\,\F\neq 3$.
If $\char\,\F=3$ then $\W_6\cong \W_{11}$ and $\W_7\cong \W_{12}$ and there
is no more isomorphism among the algebras in this family.
\end{lemma}
\begin{proof}
We claim, for $i\in\{6,\ 7,\ 11,\ 12\}$,  that $\W_{13}\not\cong \W_i$.
Note that $L_3\cong K_i/K_i^4$, while $L_5\cong K_{13}/K_{13}^4$,
where $L_3$ and $L_5$ are 5-dimensional Lie algebras defined in
Section~\ref{dim5}. Thus, $\W(L_3)/\W^5(L_3)\cong\Omega_i/(K_i^4+\Omega_i^5)$ and
$\W(L_5)/\W^5(L_5)\cong\Omega_{13}/(K_{13}^4+\Omega^5_{13})$.
Suppose that $f: \Omega_i\rightarrow \Omega_{13}$ is an isomorphism.
By Lemma~\ref{L^i}, $f(K_i^4+\Omega_i^5)=K_{13}^4+\Omega_{13}^5$. Hence, $f$
induces an isomorphism between $\Omega_{i}/(K_i^4+\Omega_i^5)$ and
 $\Omega_{13}/(K_{13}^4+\Omega_{13}^5)$.
Thus,
$$
\W(L_3)/\W^5(L_3)\cong \W(L_5)/\W^5(L_5).
$$
 However Lemma~\ref{5-dim} shows that $\W(L_3)/\W^4(L_3)\ncong \W(L_5)/\W^4(L_5)$. This contradiction implies that  $\Omega_{13}\not\cong\Omega_i$, as claimed.

Next we show that $\Omega_6\not\cong \Omega_7$ and that $\Omega_{11}\not\cong \Omega_{12}$.
First we argue that $\Omega_6\not\cong \Omega_7$. Suppose on the contrary that
$f:\Omega_7\rightarrow \Omega_6$ is an isomorphism. Then $f$ is
determined by the images $f(x_1)$, $f(x_2)$, $f(x_3)$.
Write $f(x_i)$ as a linear combination of PBW monomials and
let $\alpha_{i_1,\ldots,i_n}$, $\beta_{i_1,\ldots,i_n}$ and
$\gamma_{i_1,\ldots,i_n}$ denote the coefficients of $x_{i_1}\cdots x_{i_n}$
in $f(x_1)$, $f(x_2)$, $f(x_3)$, respectively.
Since $x_3\in Z(\Omega_7)$, we have
$$
0=[x_1,f(x_3)]\equiv\gamma_2x_4\bmod(\Omega_6)^3\quad\mbox{and}\quad
0=[x_2,f(x_3)]\equiv-\gamma_1x_4\bmod(\Omega_6)^3.
$$
Therefore $\gamma_1=\gamma_2=0$.
Let us compute modulo $(\Omega_6)^4$ that
$$
0=f([x_2,x_1,x_2])\equiv
\beta_1(\alpha_1\beta_2-\beta_1\alpha_2)x_5.
$$
If $\alpha_1\beta_2-\beta_1\alpha_2= 0$, then,
as $\gamma_1=\gamma_2=0$, it follows that $f(x_i)$ are linearly dependent modulo $(\Omega_6)^2$, which is
impossible. Thus $\beta_1=0$ and $\alpha_1\beta_2\neq 0$.
Now computation shows, modulo $(\Omega_6)^5$, that
\begin{multline*}
0=f([x_1,x_2,x_2])\equiv\\(-\alpha_1\beta_2^2+\alpha_1\beta_2\beta_{11})x_6-
2\alpha_1\beta_2\beta_{1,1}x_1x_5-\alpha_1\beta_2\beta_{1,2}x_2x_5-
\alpha_1\beta_2\beta_{1,3}x_3x_5+\delta\beta_2 x_4x_4
\end{multline*}
where $\delta=\alpha_1\beta_{1,2}-2\alpha_{2}\beta_{1,1}+2\alpha_{1,1}\beta_2$.
Considering the coefficients of $x_6$ and $x_1x_5$, we deduce that either $\alpha_1=0$ or $\beta_2=0$,
which is a contradiction.

Let us now show that $\Omega_{11}\not\cong \Omega_{12}$.
Assume by contradiction that $f:\Omega_{12}\rightarrow \Omega_{11}$ is
an isomorphism. Then $f$ is determined by the images
$f(x_1)$, $f(x_2)$, and $f(x_3)$.
As above, we write $f(x_i)$ as a linear combination of PBW monomials and
we let $\alpha_{i_1,\ldots,i_n}$, $\beta_{i_1,\ldots,i_n}$ and
$\gamma_{i_1,\ldots,i_n}$ denote the coefficients of $x_{i_1}\cdots x_{i_n}$
in $f(x_1)$, $f(x_2)$, $f(x_3)$, respectively.
Let us compute modulo
$(\Omega_{11})^3$ that
$$
0=f([x_1,x_3])\equiv(\alpha_1\gamma_2-\gamma_1\alpha_2)x_4
$$
and
$$
0\equiv f([x_2,x_3])\equiv(\beta_1\gamma_2-\beta_2\gamma_1)x_4.
$$
If the vector $(\gamma_1,\gamma_2)$ is not zero,
then the vectors $(\alpha_1,\alpha_2)$ and $(\beta_1,\beta_2)$
must be its scalar multiples and so the $f(x_i)$
are linearly dependent modulo $(\Omega_{11})^2$. Hence $\gamma_1=\gamma_2=0$.
Now we get a contradiction using the same argument as in the previous paragraph.

Now assume that $\char\,\F \neq 3$.
Set  $B_i=\Omega_i/(\Omega_i)^5$, for $i=6,\ 7,\ 11,\ 12$.
Let $Z_i$ denote the center of the $B_i$.
We claim that $\dim Z_6=\dim Z_7=29$ and $\dim Z_{11}=\dim Z_{12}=28$.
We only compute $\dim Z_7$, as the computation
of the other $Z_i$ is very similar.
We claim that
$$
Z_7=\left<x_3,x_3x_3,x_3x_3x_3,(B_7)^4\right>_{\F}.
$$
Set $C=\left<x_3,x_3x_3,x_3x_3x_3,(B_7)^4\right>_{\F}$. Clearly, $C\leq Z_7$.
Let $z\in Z_7$ and write $z$ as a linear combination of PBW monomials with
weight at most 4.
We may assume that all PBW monomials in $C$ occur with coefficient
$0$ in $z$. Let $\beta_{i_1,\ldots,i_k}$ be the coefficient of $x_{i_1}\cdots
x_{i_k}$. Then $\beta_3=\beta_{3,3}=\beta_{3,3,3}=0$.
First we obtain that
$$
0=[z,x_5]=\beta_1[x_1,x_5]=\beta_1x_6,
$$
and so $\beta_1=0$. Also, $[z,x_1^3]=-3\beta_2x_1^2x_4+3\beta_2x_1x_5-\beta_2x_6$. So, $\beta_2=0$.
Next compute that
\begin{multline*}
[z,x_1^2]=-2\beta_{1,2}x_1^2x_4-4\beta_{2,2}x_1x_2x_4-2\beta_{2,3}x_1x_3x_4
\\+
(\beta_{1,2}-2\beta_4)x_1x_5
+2\beta_{2,2}x_2x_5+\beta_{2,3}x_3x_5+2\beta_{2,2}x_4x_4+\beta_4x_6.
\end{multline*}
This
implies that $\beta_{1,2}=\beta_{2,2}=\beta_{2,3}=\beta_4=0$.
Further,
$$
[z,x_2^2]=4\beta_{1,1}x_1x_2x_4-2\beta_{1,1}x_2x_5-2\beta_{1,1}x_4x_4+
2\beta_{1,3}x_2x_3x_4.
$$
Thus $\beta_{1,1}=\beta_{1,3}=0$.
Then
\begin{multline*}
0=[z,x_1]=-\beta_{1,1,2}x_1x_1x_4-2\beta_{1,2,2}x_1x_2x_4-\beta_{1,2,3}
x_1x_3x_4-3\beta_{2,2,2}x_2x_2x_4-2\beta_{2,2,3}x_2x_3x_4\\
-\beta_{2,3,3}x_3x_3x_4
-\beta_{1,4}x_1x_5-\beta_{2,4}x_{4}^2-\beta_{2,4}x_2x_5-\beta_{3,4}x_3x_5-\beta_5x_6,
\end{multline*}
and so $\beta_{1,1,2}=\beta_{1,2,2}=\beta_{1,2,3}=\beta_{2,2,2}=
\beta_{2,2,3}=\beta_{2,3,3}=\beta_{1,4}=\beta_{2,4}=\beta_{3,4}=
\beta_5=0$.
Further
\begin{multline*}
0=[z,x_2]=3\beta_{1,1,1}x_1x_1x_4-3\beta_{1,1,1}x_1x_5+\beta_{1,1,1}x_6\\+2\beta_{1,1,3}x_1x_3x_4
-\beta_{1,1,3}x_3x_5+\beta_{1,3,3}x_3x_3x_4.
\end{multline*}
Hence $\beta_{1,1,1}=\beta_{1,1,3}=\beta_{1,3,3}=0$.
Thus $z\in C$, and so $C=Z_7$ as required. Similar
calculations show that
\begin{eqnarray*}
Z_6&=&\left<x_3,x_3x_3,x_3x_3x_3,(B_6)^4\right>_{\F};\\
Z_{11}&=&\left<x_3x_3,x_3x_3x_3,(B_6)^4\right>_{\F};\\
Z_{12}&=&\left<x_3x_3,x_3x_3x_3,(B_6)^4\right>_{\F}.
\end{eqnarray*}
Thus the dimensions of $Z_i$ are as claimed. Hence, if $\char\,\F \neq 3$ then,
 $\W_{6}\not\cong \W_{11},\ \W_{12}$ and  $\W_{7}\not\cong \W_{11},\ \W_{12}$.

Combining the results of the last two paragraph, we obtain that the algebras
$\W_6$, $\W_7$, $\W_{11}$, $\W_{12}$, $\W_{13}$
are pairwise non-isomorphic if $\char\,\F\neq 3$.

Now suppose that  $\char\,\F=3$.
We are required to show that $\W_{6}\not\cong \W_{12}$,
$\W_{7}\not\cong \W_{11}$, $\W_6\cong \W_{11}$ and that
$\W_7\cong \W_{12}$.
Suppose that $f:\Omega_{12}\rightarrow\Omega_6$ is an isomorphism. Write $f(x_2)$
and $f(x_4)$ as linear combinations of PBW monomials and
let
$\beta_{i_1,\ldots,i_n}$ and $\gamma_{i_1,\ldots,i_n}$ denote the coefficient
of $x_{i_1}\cdots x_{i_n}$ in $f(x_2)$ and $f(x_4)$ respectively.
By Lemma~\ref{L^i}, $f(x_4)\equiv\gamma_4 x_4\pmod{(\Omega_6)^3}$.
Then, modulo $(\Omega_6)^4$,
$0=f([x_2,x_4])\equiv\beta_1\gamma_4x_5$, which gives that $\beta_1=0$.
As $x_4=[x_1,x_2]$, we find that
$$
\gamma_{1,1,1}=\gamma_{1,1,2}=\gamma_{1,1,3}=
\gamma_{1,2,2}=\gamma_{1,2,3}=\gamma_{1,3,3}=\gamma_{2,2,2}=\gamma_{2,2,3}=
\gamma_{2,3,3}=\gamma_{3,3,3}=0.
$$
Then
$$
0=f([x_2,x_4])=(\beta_2\gamma_4-\beta_{1,1}\gamma_4)x_6-\beta_2\gamma_{1,4}x_4x_4+\beta_{1,2}\gamma_4x_2x_5+\beta_{1,3}\gamma_4x_3x_5-\beta_{1,1}\gamma_4 x_1x_5.
$$
This implies that $\beta_2\gamma_4=0$, and in turn that $\beta_2=0$.
However, this gives that $f(x_1),\ f(x_2),\ f(x_3)$ are linearly
dependent modulo $(\Omega_6)^2$, which is impossible. Hence
$\W_{6}\not\cong \W_{12}$, and very similar
argument shows that $\W_{7}\not\cong \W_{11}$.

Finally we need to show that if $\char\,\F=3$, then $\Omega_6\cong \Omega_{11}$ and $\Omega_7\cong \Omega_{12}$. The argument
presented in the proof of Lemma~\ref{5-dim}
shows that the map $x_1\mapsto x_1$,
$x_2\mapsto x_2$, and $x_3\mapsto x_3+x_1^3$ can be extended to
isomorphisms between $\Omega_6$ and $\Omega_{11}$ and between
$\Omega_7$ and $\Omega_{12}$.
\end{proof}

\subsection{Family~\eqref{iso3}}

\begin{eqnarray*}
K_{14}&=&\left<x_1,x_2;x_3;x_4;x_5;x_6\ |\ [x_1,x_2]=x_3,\ [x_1,x_3]=x_4,\
[x_1,x_4]=x_5, [x_2,x_5]=x_6\right.\\
&&\left.\ [x_3,x_4]=-x_6,\ [x_2,x_3]=x_5\right>;\\
K_{16}&=&\left<x_1,x_2;x_3;x_4;x_5;x_6\ |\ [x_1,x_2]=x_3,\ [x_1,x_3]=x_4,\
[x_1,x_4]=x_5,\ [x_2,x_5]=x_6,\right.\\
&&\left.[x_3,x_4]=-x_6\right>.
\end{eqnarray*}

\begin{lemma}\label{lem-iso3}
The algebras $\W_{14}$ and $\W_{16}$ are not isomorphic.
\end{lemma}
\begin{proof}
Note that $K_{14}$ and $K_{16}$ are algebras with maximal class. Further,
$K_{14}/(K_{14})^5 \cong L_6$ and $K_{16}/(K_{16})^5\cong L_7$.
Now the argument presented in the first paragraph of the proof
of Lemma~\ref{fam2lem} shows that $\W_{14}\not\cong \W_{16}$.
\end{proof}

\subsection{Family~\eqref{iso4}}
\begin{eqnarray*}
K_{15}&=&\left<x_1,x_2;x_3;x_4;x_5;x_6\ |\ [x_1,x_2]=x_3,\ [x_1,x_3]=x_4,\
[x_1,x_4]=x_5,\ [x_2,x_3]=x_5,\right.\\
&&\left.[x_1,x_5]=x_6,\ [x_2,x_4]=x_6\right>;\\
K_{17}&=&\left<x_1,x_2;x_3;x_4;x_5;x_6\ |\ [x_1,x_2]=x_3,\ [x_1,x_3]=x_4,\
[x_1,x_4]=x_5,\ [x_1,x_5]=x_6,\right.\\
&&\left.[x_2,x_3]=x_6\right>;\\
K_{18}&=&\left<x_1,x_2;x_3;x_4;x_5;x_6\ |\ [x_1,x_2]=x_3,\ [x_1,x_3]=x_4,\
[x_1,x_4]=x_5,\ [x_1,x_5]=x_6\right>.
\end{eqnarray*}

\begin{lemma}
If $\char\,\F\neq 3$ then $\W_{15}$, $\W_{17}$ and $\W_{18}$ are pairwise
non-isomorphic; otherwise the only isomorphism among
these algebras is $\W_{17}\cong \W_{18}$.
\end{lemma}
\begin{proof} Note that
$K_{15}/(K_{15})^5\cong L_6$ and $K_{17}/(K_{17})^5\cong L_7 \cong
K_{18}/(K_{18})^5$.
Applying the same method as in the proof of Lemma~\ref{fam2lem} yields that
$\W_{15}\not\cong \W_{17}$ and $\W_{15}\not\cong \W_{18}$.

Suppose that $\char\,\F\neq 3$, and let us show that $\W_{17}\not\cong \W_{18}$.
Assume, by contradiction, that $f: \W_{17}\to \W_{18}$ is an isomorphism.
Then $f$ is determined by the images $f(x_1)$ and $f(x_2)$.
Let $f(x_1)\equiv\a_{1}x_1 +\a_{2}x_2$ and $f(x_2)\equiv\b_{1}x_1 +\b _{2}x_2$  modulo $\W_{18}^2$.
Let $\delta=\a_1\b_2-\a_2\b_1$. Clearly, $\delta \neq 0$. Note that $f(x_3)\equiv\delta x_3 \pmod{\W_{18}^3}$.
Thus,  $f(x_4)\equiv\a_1\delta x_4 \pmod{\W_{18}^4}$ and  $f(x_6)\equiv\a_1^3 \delta x_6 \pmod{\W_{18}^6}$.
Also, $f([x_2, x_3])\equiv\b_1\delta x_4 \pmod{\W_{18}^4}$. We deduce that $\b_1=0$.
Now we write
\begin{align*}
f(x_2)& \equiv \b_2x_2+ u_0+x_1u_1+x_1^2u_2+x_1^3u_3 \pmod{\W_{18}^4},\\
f(x_3)&\equiv \delta x_3+ v_0+x_1v_1+x_1^2v_2+x_1^3v_3 \pmod{\W_{18}^4},
\end{align*}
where each $u_i$ and $v_i$ is a linear combination of (possibly trivial) PBW  monomials that do not involve $x_1$.
Since $f(x_3)\equiv \delta x_3 \pmod{\W_{18}^3}$, we deduce that weight of $v_0+x_1v_1+x_1^2v_2$ is at least 3. Similarly, weight of
$u_0+x_1u_1$ is at least 2. Note that $[u_0, v_i]=[u_i, v_0]=0$. So,
\begin{align}\label{brack-3-2}
0 \equiv f([x_3, x_2])&\equiv \b_2\sum_{i=1}^3 [x_1^i, x_2]v_i - \delta \sum_{j=1}^2[x_1^j, x_3]u_j \pmod{\W_{18}^5}.
\end{align}
Expanding out the commutators in Equation (\ref{brack-3-2}), we observe that $x_1^2x_3v_3$ is the unique term in Equation (\ref{brack-3-2}) that has  the highest exponent of $x_1$. Since $\char(\F)\neq 3$, this can happen only if $v_3\in \W_{18}$. Thus,
$x_1^2x_3v_3\in \W_{18}^5$.  The highest exponent of $x_1$ in Equation (\ref{brack-3-2}) then  appears in
$x_1x_3v_2$ and $x_1x_4u_2$. So, these terms have to cancel out with each other. We deduce that
$u_2\in \W_{18}$ and $v_2\in \W_{18}^2$.  Now Equation (\ref{brack-3-2}) reduces to
$\b_2 x_3v_1 \equiv\delta  x_4u_1 \pmod{\W_{18}^5}.$ This implies that $u_1\in \W_{18}^2$ and $v_1\in \W_{18}^3$.
So, if we  write
\begin{align*}
f(x_2)& \equiv\b_2x_2+ u_0+x_1u_1+x_1^2u_2+x_1^3u_3+x_1^4u_4 \pmod{\W_{18}^5},\\
f(x_3)& \equiv\delta x_3+ v_0+x_1v_1+x_1^2v_2+x_1^3v_3+x_1^4v_4 \pmod{\W_{18}^5},
\end{align*}
then each $u_i$ and $v_i$ is a linear combination of (possibly trivial) PBW  monomials that do not involve $x_1$,
$u_0$ has weight at least 2, $v_0$ has weight at least 3,  weight of $x_1v_1+x_1^2v_2+x_1^3v_3+x_1^4v_4$ is at least 4, and  weight of
$x_1u_1+x_1^2u_2$ is at least 3. Thus,
\begin{align}\label{brack-3-3}
-\a_1^3 \delta x_6=f([x_3, x_2])& \equiv\b_2\sum_{i=1}^4 [x_1^i, x_2]v_i - \delta \sum_{j=1}^3[x_1^j, x_3]u_j \pmod{\W_{18}^6}.
\end{align}

Arguing as in Equation (\ref{brack-3-2}), we deduce that Equation (\ref{brack-3-3}) reduces to the following:
\begin{align*}
-\a_1^3 \delta x_6& \equiv\b_2 x_3 v_1 - \delta x_4u_j  \pmod{\W_{18}^6}.
\end{align*}
The later is possible only if $\a_1=0$ or $\delta=0$. Since we have already proved that $\b_1=0$ it follows that
$\delta=0$ which is a contradiction.

The argument in the proof of Lemma~\ref{5-dim} shows, for $\char(\F)=3$, that
the map  $x_1\mapsto x_1$, $x_2\mapsto x_2+x_1^3$ can be extended to
 an isomorphism between $\W_{17}$ and $\W_{18}$.
\end{proof}

\subsection{Family~\eqref{iso5}}
\begin{eqnarray*}
K_{23}&=&\left<x_1,x_2,x_3;x_4,x_5;x_6\ |\ [x_1,x_2]=x_4,\ [x_1,x_4]=x_6,\
[x_1,x_3]=x_5,\ [x_2,x_3]=x_6\right>;\\
K_{25}&=&\left<x_1,x_2,x_3;x_4,x_5;x_6\ |\ [x_1,x_2]=x_4,\ [x_1,x_4]=x_6,\
[x_1, x_3]=x_5\right>.
\end{eqnarray*}

\begin{lemma}
The algebras  $\W_{23}$ and $\W_{25}$ are not isomorphic.
\end{lemma}
\begin{proof}
Suppose that $\char\,\F\neq 3$ and assume, by contradiction, that  $f:\W_{25} \to \W_{23}$ is an isomorphism.
For $i=1,\ 2,\ 3$, write $f(x_i)$ as a linear combination of PBW
monomials and assume that $\alpha_{i_1,\ldots,i_n}$, $\beta_{i_1,\ldots,i_n}$,
and $\gamma_{i_1,\ldots,i_n}$ are the coefficients of $x_{i_1}\cdots x_{i_n}$
in $f(x_1)$, $f(x_2)$, and $f(x_3)$, respectively.
Then
$$
0=f([x_2, x_3])\equiv(\beta_1\gamma_2-\beta_2\gamma_1)x_4 +
(\beta_1\gamma_3-\beta_3\gamma_1)x_5 \pmod{\W_{23}^3},
$$
and hence $\beta_1\gamma_2-\beta_2\gamma_3=\beta_1\gamma_3-\beta_3\gamma_1=0$.
Since $f$ induces an isomorphism between $\Omega_{25}/(\Omega_{25})^2$
and
$\Omega_{23}/(\Omega_{23})^2$, we obtain that $\beta_1=\gamma_1=0$.

Note that
$$
f(x_5)=f([x_1, x_3])\equiv\alpha_1\gamma_2x_4+\a_1\gamma_3x_5 \pmod{\W_{23}^3}.
$$
Thus,
$0=f([x_1, x_5])\equiv\a_1^2\gamma_2x_6\pmod{(\W_{23})^4}$, which gives that
$\gamma_2=0$.
Now we calculate $f([x_2, x_3])$ modulo $(\W_{23})^4$
to show that $\beta_2\gamma_3=0$.
As $[x_4, \W_{23}]\leq (\W_{23})^4$, we have, modulo $(\W_{23})^4$,  that
\begin{multline*}
f([x_2, x_3])\equiv
\beta_2\gamma_3x_6+ \beta_2\gamma_{1,1}(x_6-2x_1x_4) -\beta_2\gamma_{1,2}
x_2x_4  -\beta_2\gamma_{1,3} x_3x_4\\
+2(\beta_{1,1}\gamma_3 -\beta_3\gamma_{1,1}) x_1x_5 + (\beta_{1,2}\gamma_3-\beta_3\gamma_{1,2})x_2x_5 +(\beta_{1,3}\gamma_3 -\beta_3\gamma_{1,3}) x_3 x_5.
\end{multline*}
This gives that $\beta_2\gamma_3=0$ which implies that the images
$f(x_2)$, $f(x_2)$, $f(x_2)$
are linearly dependent modulo $(\Omega_{23})^2$, which is a contradiction.
\end{proof}

\subsection{Family~\eqref{iso6}}
We consider the following Lie algebras:
\begin{eqnarray*}
K_{9}&=&\left<x_1,x_2,x_3;x_4;x_5,x_6\ |\ [x_1,x_2]=x_4,\ [x_1,x_4]=x_5,\ [x_2,x_4]=x_6\right>;\\K_{24}(\varepsilon)
&=&\left<x_1,x_2,x_3;x_4;x_5,x_6\ |\ [x_1,x_2]=x_4,\ [x_1,x_4]=x_5,\
[x_1,x_3]=\varepsilon x_6,\ [x_2,x_4]=x_6,\right.\\
&&\left.[x_2,x_3]=x_5\right>.\\
\end{eqnarray*}
The family $K_{24}(\varepsilon)$ is a parametric family of Lie
algebras such that $K_{24}(\varepsilon_1)\cong K_{24}(\varepsilon_2)$ if
and only if there is a $\nu\in\F$ such that
$\varepsilon_1\nu^2=\varepsilon_2$ (see~\cite{degraaf}).

\begin{lemma}\label{lem-iso6}
The algebras   $\W_9$ and $\W_{24}(\varepsilon)$ are not isomorphic, for all  $\varepsilon\in\F$.  Further,  $\W_{24}(\varepsilon_1)\cong \W_{24}(\varepsilon_2)$ if and only if $K_{24}(\varepsilon_1)\cong K_{24}(\varepsilon_2)$,
for every $\varepsilon_1,\ \varepsilon_2
\in\F$.
\end{lemma}
\begin{proof}
Since $x_3\in Z(\Omega_9)\setminus(\Omega_9)^2$, we have that
$Z(\Omega_9)\not\leq (\Omega_9)^2$. Let $\varepsilon\in\F$.
We claim that $Z(\Omega_{24}(\varepsilon))\leq (\Omega_{24}(\varepsilon))^2$.
Let $z\in Z(\Omega_{24}(\varepsilon))$ and write $z$ as a linear combination
of PBW monomials in which $\alpha_{i_1,\ldots,i_n}$ denotes the coefficient
of $x_1\cdots x_n$.
First we compute, modulo $(\Omega_{24}(\varepsilon))^3$, that
$0=[z, x_3]\equiv\alpha_1\varepsilon x_6+ \alpha_2 x_5$, which gives that
$\alpha_2=0$.
Then,
$$
[z,x_2]\equiv \alpha_1 x_4+(-\alpha_3-\alpha_{1,1})x_5+2\alpha_{1,1}x_1x_4+\alpha_{1,2}x_2x_4+\alpha_{1,3}x_3x_4 \pmod{\Omega^4_{24}(\varepsilon)},
$$
which shows that $\alpha_1=\alpha_3=0$. Hence $z\in (\Omega_{24}(\varepsilon))^2$,
and so $Z(\Omega_{24}(\varepsilon))\leq (\Omega_{24}(\varepsilon))^2$,
as claimed. This implies that $\Omega_9\not\cong \Omega_{24}(\varepsilon)$.

Let us now prove the second assertion of the lemma.
Without loss of generality, we assume that $\e_2\neq 0$.
Let $K=K_{24}(\e_2)$ and $\W=\W(K)$.
Suppose that $f: \Omega_{24}(\varepsilon_1)\to \W$ is an algebra isomorphism.
As usual, for $i=1,\ 2,\ 3$, we write the images $f(x_i)$ as linear
combinations of PBW monomials, and let $\alpha_{i_1,\ldots,i_n}$,
$\beta_{i_1,\ldots,i_n}$ and $\gamma_{i_1,\ldots,i_n}$ denote the
coefficients of $x_{i_1}\cdots x_{i_n}$ in $f(x_1)$, $f(x_2)$, and
$f(x_3)$, respectively.
Since $f([x_1, x_3]),\ f([x_2,x_3])\in \W^3$, we deduce that $\a_1\gamma_2-\a_2\gamma_1=0$ and that
$\beta_1\gamma_2-\beta_2\gamma_1=0$. Since the images
$f(x_1)$, $f(x_2)$, $f(x_2)$ are linearly independent
modulo $\Omega^2$, this gives that $\gamma_1=\gamma_2=0$.

Set $\delta=\a_1\beta_2-\a_2\beta_1$. Since $f(x_1)$, $f(x_2)$,
and $f(x_3)$ are linearly independent modulo $\Omega^2$,
we have $\delta\neq 0$. Further,
$$
f(x_4)=[f(x_1), f(x_2)]\equiv\delta[x_1, x_2]=\delta x_4 \pmod{\W^3}.
$$
Thus,  $f([x_1,x_4])\equiv\alpha_1\delta x_5+\alpha_2\delta x_6 \pmod{\W^4}$.
So, modulo $\W^4$, we have,
\begin{multline*}
f([x_2,x_3])\equiv(\beta_1\gamma_4+\beta_2\gamma_3+\beta_2\gamma_{1,1})x_5+
(\varepsilon_2\beta_1\gamma_3-\beta_1\gamma_{2,2}+\beta_2\gamma_{1,2}+\beta_2\gamma_4)x_6+\\
(\beta_1\gamma_{1,2}-2\beta_2\gamma_{1,1})x_1x_4+
(2\beta_1\gamma_{2,2}-\beta_2\gamma_{1,2})x_2x_4+
(\beta_1\gamma_{2,3}-\beta_2\gamma_{1,3})x_3x_4.
\end{multline*}
As $f([x_1,x_4])=f([x_2,x_3])$ we obtain that following equations:
\begin{eqnarray}
\label{11}\beta_1\gamma_4+\beta_2\gamma_3+\beta_2\gamma_{1,1}&=&\alpha_1\delta;\\
\label{12}\varepsilon_2\beta_1\gamma_3-\beta_1\gamma_{2,2}+\beta_2\gamma_{1,2}+\beta_2\gamma_4 &=& \alpha_2\delta;\\
\label{13}\beta_1\gamma_{1,2}-2\beta_2\gamma_{1,1}&=&0;\\
\label{14}2\beta_1\gamma_{2,2}-\beta_2\gamma_{1,2}&=&0;\\
\label{15}\beta_1\gamma_{2,3}-\beta_2\gamma_{1,3}&=&0.
\end{eqnarray}
Now we use the relation $\varepsilon_1[x_2,x_4]=[x_1,x_3]$. So, modulo $\W^4$, we have
\begin{align*}
f([x_2,x_4])\equiv& \beta_1\delta x_5+\beta_2\delta x_6,\\
f([x_1,x_3])\equiv& (\alpha_1\gamma_4+\alpha_2\gamma_3+\alpha_2\gamma_{1,1})x_5+
(\varepsilon_2\alpha_1\gamma_3-\alpha_1\gamma_{2,2}+\alpha_2\gamma_{1,2}+\alpha_2\gamma_4)x_6+\\
&(\alpha_1\gamma_{1,2}-2\alpha_2\gamma_{1,1})x_1x_4+
(2\alpha_1\gamma_{2,2}-\alpha_2\gamma_{1,2})x_2x_4+
(\alpha_1\gamma_{2,3}-\alpha_2\gamma_{1,3})x_3x_4.
\end{align*}
We get the following equations:
\begin{eqnarray}
\label{21}\alpha_1\gamma_4+\alpha_2\gamma_3+\alpha_2\gamma_{1,1} &=&
\varepsilon_1\beta_1\delta;\\
\label{22}\varepsilon_2\alpha_1\gamma_3-\alpha_1\gamma_{2,2}+\alpha_2\gamma_{1,2}+\alpha_2\gamma_4 &=& \varepsilon_1\beta_2\delta;\\
\label{23}\alpha_1\gamma_{1,2}-2\alpha_2\gamma_{1,1}&=&0;\\
\label{24}2\alpha_1\gamma_{2,2}-\alpha_2\gamma_{1,2}  & =& 0;\\
\label{25}\alpha_1\gamma_{2,3}-\alpha_2\gamma_{1,3}&=&0.
\end{eqnarray}
Equations~\eqref{13} and~\eqref{23} imply that
$\gamma_{1,1}=\gamma_{1,2}=0$. Similarly $\gamma_{2,2}=
\gamma_{2,3}=\gamma_{1,3}=0$. Thus the system of
equations above are reduced to the
following:
\begin{eqnarray}
\label{31}\beta_1\gamma_4+\beta_2\gamma_3&=&\alpha_1\delta;\\
\label{32}\varepsilon_2\beta_1\gamma_3+\beta_2\gamma_4 &=& \alpha_2\delta;\\
\label{33}\alpha_1\gamma_4+\alpha_2\gamma_3 &=&
\varepsilon_1\beta_1\delta;\\
\label{34}\varepsilon_2\alpha_1\gamma_3+\alpha_2\gamma_4 &=& \varepsilon_1\beta_2\delta.
\end{eqnarray}
Set
\begin{eqnarray*}
p_1&=&(-1/2)\varepsilon_1\beta_2\delta^{-2}-
(1/2)\varepsilon_2\alpha_1\gamma_3\delta^{-3};\\
p_2&=&(1/2)\alpha_1\gamma_4\delta^{-3} + \alpha_2\gamma_3\delta^{-3};\\
p_3&=&\alpha_2\delta^{-2} - (1/2)\beta_2\gamma_4\delta^{-3}\\
p_4&=&(-1/2)\alpha_1\delta^{-2} - (1/2)\beta_2\gamma_3\delta^{-3}.
\end{eqnarray*}
We can check that
\begin{multline*}
p_1(\beta_1\gamma_4+\beta_2\gamma_3-\alpha_1\delta)+p_2(\varepsilon_2\beta_1\gamma_3+\beta_2\gamma_4-\alpha_2\delta)+\\
p_3(\alpha_1\gamma_4+\alpha_2\gamma_3 -\varepsilon_1\beta_1\delta)+p_4(\varepsilon_2\alpha_1\gamma_3+\alpha_2\gamma_4 - \varepsilon_1\beta_2\delta)=\varepsilon_1 - \varepsilon_2\gamma_3^2\delta^{-2}.
\end{multline*}
Thus, considering that $\delta\neq 0$,
the equation $\varepsilon_1 = \varepsilon_2\gamma_3^2\delta^{-2}$ follows
from the equations~\eqref{31}--\eqref{34}. However this implies that
$K_{24}(\varepsilon_1)\cong K_{24}(\varepsilon_2)$ as claimed.
\end{proof}

\section*{Acknowledgments}
The second author would like to thank the  Centro de \'Algebra da Universidade de Lisboa for hosting his visit in April 2010 while this work was completed.

config: arxiv


TITLE: Meeting management
QUESTION [6 upvotes]: A friend wants to have ten meetings of six people every day for five days with no pair of people meeting twice. Is this possible? It appears to be a question about maximal decomposition of a complete graph on 60 vertices into sets of 10 disjoint complete graphs on 6 vertices such that no edge is used twice. Does anyone know a useful theorem or, better, a way of constructing an example?

REPLY [4 votes]: Look at the edit history for an old idea (if you wish).
Here is a schedule which runs for 8 days. Each row is an agenda for a day. It consists of 60 digits in 6 groups of 10 (for readability). Note that I will usually count days, people and meetings each day starting from 0 ( except that the schedule below starts with day 1.) Day 2 begins 0123456777.8777 which tells us that people 7,8,9,11,12, and 13 make up  meeting 7 that day. You may confirm that no two of those six people is ever in a meeting together on another day.
0001000234.5167513847.9497368284.1239662875.3198562431.7649587529
0123456777.8777465293.8508631463.0984282401.9598350211.5946032961
0123456465.2738999699.9630784282.4017578350.2115746032.7615803418
0123456724.0185873502.1999499928.6157034172.6846528375.0763146308
0123456157.4603276158.0341826799.9299950863.1463078428.2401757835
0123456630.7842824017.5783502115.7460399919.9934182674.6527385086
0123456276.1580341826.7465273850.8631463078.4299909997.8350211574
0123456783.5021157460.3276158034.1826746527.3850863149.9979998240
The method of generation is systematic (I thought it might work at least for 5 days and was pleased to get 8). I only sketch it here, still it took less time to do (with Maple) than it does to describe.
Begin with the field of $9$ elements which could be thought of as $\mathbb{Z}_3[i]$, the Gaussian integers $\mod{3}$. Code it somehow, I used $0 \rightarrow 0$ and $(1+i)^{j-1} \rightarrow j$ for $1 \le j \le 8$.
Using the method of Kevin Costello, one can construct a 9 day schedule of meetings for 63 people which has 9 meetings of 7 people each day (and no pair is ever in the same meeting twice.) What we will do is take 80 of those 81 meetings, reduce each one to six people and arrange the 80 meetings into eight days each with ten meetings. Call a meeting big if it has seven people and good once it is reduced to six people.
On day $0$, meeting $j$ has people $7j,7j+1, \cdots,7j+6$ so the ninth meeting (meeting $8$)  is $56,57,58,59,60,61,62$ Go through days 1 through 8 and erase people 60,61,62 (note that no two of them are ever in the same meeting after day 0). At this stage each of those eight days has nine meetings, three are good and six are big. We wish to remove one person from each big meeting to make them good and use the removed people for a tenth good meeting. To have a chance of doing this appropriately, we look for a meeting from day 0 which has only one member already in a good meeting. If we manage to do this then the other six people are sure to be distributed one in each big meeting. Use them to make a tenth good meeting.
An example may help: Here is the schedule for day 4 after people $60,61,62$ are removed (note that I have sorted the people in each meeting and sorted the list of meetings according to least person in the meeting)
$\small [0, 10, 18, 26, 34, 50, 58], [1, 11, 20, 21, 31, 37, 54], [2, 8, 19, 28, 39, 45],
[3, 16, 27, 35, 47, 53, 57], [4, 9, 24, 36, 42, 55],$ $\small [5, 13, 17, 22, 32, 44, 49],[6, 25, 30, 40, 43, 52, 56], [7, 15, 33, 38, 48, 51], [12, 14, 23, 29, 41, 46, 59]$
The eighteen people who are already in good meetings are:
2  4 * 7  8  9 * 15  19 *24 *28  33  *36  38  39  *42  45  48  51  55 so the only day 0 meeting which only includes one of them is [21,22,23,24,25,26,27]. Removing person 24 from that group and the other six people from their groups yelds
$\small [0, 10, 18, 34, 50, 58], [7, 15, 33, 38, 48, 51], [12, 14, 29, 41, 46, 59],
 [1, 11, 20, 31, 37, 54], [2, 8, 19, 28, 39, 45],$
$\small [3, 16, 35, 47, 53, 57], [4, 9, 24, 36, 42, 55], [5, 13, 17, 32, 44, 49], [6, 30, 40, 43, 52, 56], [21, 22, 23, 25, 26, 27]$
0123456724.0185873502.1999499928.6157034172.6846528375.0763146308 is the resulting code appearing as line 4 above. The first meeting explains the digit 0 in positions 0,10,18,34,50,58.
As it turns out, each day from 1 to 8 is compatible with exactly one meeting from day 0 and vice versa, perhaps there is an explanation for this, but I am satisfied that it happened this once.
Reviewing this I see that I could have pushed the construction by Kevin to get $10$ days with $90$ meetings  (I think.) It does not seem that this would allow 10 days of meetings with $60$ people but I have not carefully checked (nor do I plan to.)


TITLE: Probability - Airplane overselling tickets
QUESTION [0 upvotes]: Few days ago, I came across a question for probability in one of the interview.
Question :

The same small commuter plane has 30 seats. The probability that any
  particular passenger will not show up for a flight is 0.1, independent
  of other passengers. The average fare paid by a passenger who succeeds
  in boarding is $100. A passenger who shows up but cannot board is given 200 dollars and a free flight later on.
Given that the airline can sell 30 + m tickets, how should the airline
  set m to maximize expected revenue?

I did not get how to solve this probability question. Can somebody please suggest a approach to solving this kind of questions ?

REPLY [2 votes]: Assuming that the free flight is equivalent to selling a ticket for a future flight which is then not sold to somebody else and there is no refund, then it is worth $\$100$ and the compensation has a total cost to the airline of $\$300$.
The total gross revenue for a flight is $\$100\times(30+m)$ and the expected compensation is $\displaystyle \sum_{i = 1}^m \$300\times i \times {30+m \choose 30+i} 0.9^{30+i} 0.1^{m-i}$ , with the expect net revenue being the difference between these.
There are various ways of finding the maximum expected net revenue.  One would be to work out the values for sensible values of $m$: since $30/0.9 \approx 33.3$, it might worth looking at $m=3$ as this will have the plane often full or almost full.  Then look at $m$ marginally more or less, i.e. $m=2$ or $4$.


TITLE: The eigenspinors for the spin operator in the $x$-direction?
QUESTION [1 upvotes]: $$S_x= \frac{\hbar}{2}\quad\begin{pmatrix}0&1\\1&0\end{pmatrix}\quad$$
$$S_x{X_+}^{(x)}=\frac{\hbar}{2}{X_+}^{(x)}$$
How can I find the eigenvalue of $S_x$?
My book says
$$
  \left| \begin{array}{cc}
-\lambda &  \frac{\hbar}{2} \\
\frac{\hbar}{2} & -\lambda\\\end{array} \right|=0 $$
So $\lambda=\frac{\hbar}{2} or \frac{-\hbar}{2}$ and therefore
$$S_x{X_+}^{(x)}=\frac{\hbar}{2}{X_+}^{(x)}$$ and $$S_x{X_-}^{(x)}=\frac{-\hbar}{2}{X_-}^{(x)}$$
My question is why do we need determinant=0? And what does the ${-\lambda}$ in the diagonal stand for?

REPLY [1 votes]: $\lambda$ stands for the eigenvalue.
Eigenvalue equation is:
$S_xX=\lambda X$
$S_xX-\lambda X=0$
$(S_x-\lambda I)X=0$
Since X is eigenfunction, we seek solutions for $det(S_x-\lambda I)=0$
\begin{align}
(S_x-\lambda I)=
\begin{bmatrix}
 0 & \frac{\hbar}{2} \\
 \frac{\hbar}{2} &0
\end{bmatrix}
-
\begin{bmatrix}
 \lambda & 0 \\
 0 & \lambda
\end{bmatrix}
=
\begin{bmatrix}
 - \lambda & \frac{\hbar}{2} \\
 \frac{\hbar}{2} & - \lambda
\end{bmatrix}
\end{align}
So to find eigenvalues you should just solve this:
\begin{align}
\begin{vmatrix}
 - \lambda & \frac{\hbar}{2} \\
 \frac{\hbar}{2} & - \lambda
\end{vmatrix}=0
\end{align}