text:
\begin{document}

\title[Action of a nilpotent group on its horofunction boundary]
{The action of a nilpotent group on its horofunction boundary
has finite orbits}
\date{\today}
\author{Cormac Walsh}
\address{
INRIA and CMAP, Ecole Polytechnique.
Postal address: CMAP, Ecole Polytechnique, 91128 Palaiseau C\'edex, France}
\email{cormac.walsh@inria.fr}

\subjclass[2000]{Primary 20F65; 20F18}

\keywords{group action, horoball, max-plus algebra, metric boundary,
Busemann function}

\begin{abstract}
We study the action of a nilpotent group $G$ with finite generating set $S$
on its horofunction boundary. We show that there is one finite orbit associated
to each facet of the polytope obtained by projecting $S$ into the infinite
component of the abelianisation of $G$.
We also prove that these are the only finite orbits of Busemann points.
To finish off, we examine in detail the Heisenberg group with its usual
generators.
\end{abstract}

\newcommand\isom{\operatorname{Isom}}

\maketitle

\section{Introduction}

Given a group acting by isometries on a space, one often finds it useful
to study the induced action on some boundary of the space,
for example the hyperbolic boundary of a Gromov hyperbolic space or the
ideal boundary of a CAT(0) space.
The hope is that the action on the boundary is simpler than the action on
the space itself.

Of course, any finitely generated group, when endowed with its word metric,
acts on itself, and then one can study the induced action on the horofunction
boundary, which exists for any metric space.

The case of $\Z^d$ with an arbitrary finite generating set was investigated by
Rieffel~\cite{rieffel_group}
in connection with his work on quantum metric spaces.
Essential to his work was that the action on the horoboundary is amenable
and that there are sufficiently many finite orbits.
In~\cite{develin_cayley}, Develin studied in greater detail the horoboundary
of $\Z^d$ and the action of $\Z^d$ on it.

In this paper, we study the case of a finitely generated nilpotent group.
We find that, just as in the case of $\Z^d$, there are always
finite orbits, whatever the generating set.

To state our results, we need the following construction.
Recall that the quotient $G/[G,G]$ of a group $G$ by its commutator subgroup
is an abelian group, called the \emph{abelianisation}.
If $G$ is finitely generated, then so is the abelianisation, which
can therefore be written as a direct product $Z_0 \times \Z^N$,
where $Z_0$ is a finite abelian group (the torsion subgroup) and $N\in\N$.
So we can define a projection map $\phi$ assigning to every element of $G$
its equivalence class in $(G/[G,G])/Z_0$.
We consider $\Z^N$ to be embedded in $\R^N$.
Let $P:=\conv(\phi(S))$ be the
smallest polytope containing the image of $S$ under the map $\phi$.

The Busemann points of the horofunction boundary are those points that are
the limits of geodesic paths.

\begin{theorem}
\label{thm:maintheorem}
Let $G$ be a nilpotent group with finite generating set $S$ and consider
the action of $G$ on its horofunction boundary coming from the word length
metric associated to $S$.
Then, there is a natural one-to-one correspondence between the finite orbits
of Busemann points and the facets of $P$.
\end{theorem}

When $G=\Z^d$, the map $\phi$ is just the identity map.
In this case, it was shown in~\cite{develin_cayley} that all boundary points
are Busemann and that there is a one-to-one
correspondence between the orbits and the proper faces of $P$.
The finite orbits correspond, of course, to the facets, that is,
the $d-1$-dimensional faces.
We will give an example later of a non-abelian nilpotent group
where this stronger correspondence breaks down.

We also have more precise results for the discrete Heisenberg group,
which has the following presentation:
\begin{align*}
H_3 := \langle a,b \mid [[a,b],a] = [[a,b],b] = e \rangle.
\end{align*}
Our results are for the word length metric coming from the usual set of
generators $S:=\{a,b,a^{-1},b^{-1}\}$.
We are helped by the fact that an exact formula for the
word length metric is known for this group and these generators.
We show directly that there are four finite orbits of Busemann points,
each consisting of a single horofunction. These horofunctions are the limits
of the four geodesic words
$a^{\epsilon_a} b^{\epsilon_b} a^{\epsilon_a} b^{\epsilon_b} \cdots$, with
$\epsilon_a =\pm 1$ and $\epsilon_b =\pm 1$.
We also show that every other Busemann point is the limit
of a geodesic path of the form $t\to c^m b^n a^{\pm t}$ or
$t\to c^{m\pm lt} b^{\pm t} a^l$.
Using the formula for the metric, all these Busemann points can be calculated
explicitly; see Theorems~\ref{thm:corners} and~\ref{thm:heisenberg_busemen}.

\section{General Nilpotent Groups}

The horofunction boundary of a metric space $(X,d)$ is defined as follows.
One assigns to each point $z\in X$ the function
$\psi_z:X\to \R$,
\begin{equation*}
\psi_z(x) := d(x,z)-d(b,z),
\end{equation*}
where $b$ is some base-point.
If $X$ is proper, then the map
$\psi:X\to C(X),\, z\mapsto \psi_z$ defines an embedding of $X$ into $C(X)$,
the space of continuous real-valued functions on $X$ endowed
with the topology of uniform convergence on compact sets.
The horofunction boundary is defined to be
\begin{align*}
X(\infty):=\closure\{\psi_z\mid z\in X\}\backslash\{\psi_z\mid z\in X\},
\end{align*}
and its elements are called horofunctions.
This definition seems to be due to Gromov~\cite{gromov:hyperbolicmanifolds}.
For more information, see ~\cite{bgs}, ~\cite{rieffel_group},
and~\cite{AGW-m}.

The action of the isometry group $\isom(X,d)$ of $X$ extends continuously
to an action by homeomorphisms on the horofunction compactification.
One can check that
\begin{align*}
g\cdot \xi(x) = \xi(g^{-1}\cdot x) - \xi(g^{-1}\cdot b),
\end{align*}
for any isometry $g$, point $x$, and horofunction $\xi$.

The metric spaces in which we will be interested in this paper are the Cayley
graphs of finitely generated nilpotent groups. Recall that the Cayley graph
of a group $G$ with generating set $S$ has a vertex for each element of $G$
and an edge between two vertices $u$ and $v$ if $u^{-1}v\in S$.
The word length distance $d(u,v)$ between $u$ and $v$ in $G$ is the length
of the shortest path between the corresponding vertices in the Cayley graph.
By construction, it is a left invariant metric.
We choose the identity element $e$ as the base-point.

A path is a sequence of group elements. We say that a path is
\emph{geodesic} if it is an isometry from $\Z$ to $G$.
It is known that geodesic paths converge to points in the horofunction
boundary.
In the present setting, since the metric is integer valued, we may define a
\emph{Busemann point} as the limit of a geodesic. See~\cite{rieffel_group}
and~\cite{AGW-m} for the definition for more general spaces.

For any finite word $w$ with letters in $S$, we denote by $\overline w$
the element of $G$ obtained by multiplying the letters of $w$,
and we say that $w$ \emph{represents} $\overline w$.
We denote by $|w|$ the length of $w$, that is, its number of letters.
We say that a finite word $w$ is a \emph{geodesic word}
if $|w|=d(e,\overline w)$, and that an infinite word is geodesic if each of
its finite prefixes is. Obviously, there is a one-to-one correspondence
between geodesic paths and pairs $(g,w)$, where $g$ is in $G$ (the starting
point of the path) and $w$ is an infinite geodesic word.
The \emph{steps} of a geodesic path are the letters of the corresponding
geodesic word. Continuing to use terminology of~\cite{develin_cayley},
we say that the \emph{direction} of a geodesic path is the set of letters
that occur infinitely often as steps.

We start off with an easy proposition saying when two geodesics converge to the
same Busemann point.
\begin{prop}
\label{prop:switch}
Let $\gamma_1$ and $\gamma_2$ be two geodesics in the Cayley graph of a
finitely generated group.
The following are equivalent:
\begin{itemize}
\renewcommand{\labelitemi}{(i)}
\item
$\gamma_1$ and $\gamma_2$ converge to the same Busemann point;
\renewcommand{\labelitemi}{(ii)}
\item
for all $N\in\N$, there is a geodesic agreeing with $\gamma_1$ up to time $N$
that has a point in common with $\gamma_2$ and eventually agrees with
$\gamma_1$ again;
\renewcommand{\labelitemi}{(iii)}
\item
there is a geodesic having infinitely many points in common with both
$\gamma_1$ and $\gamma_2$.
\end{itemize}
Moreover, the above statements hold if
\begin{itemize}
\renewcommand{\labelitemi}{(iv)}
\item
$\gamma_1$ and $\gamma_2$ have a point in common, and, for all $N\in\N$,
there is a geodesic agreeing with $\gamma_1$ up to time $N$ with the same limit
as $\gamma_2$, and a geodesic agreeing with $\gamma_2$ up to time $N$ with
the same limit as $\gamma_1$
\end{itemize}
\end{prop}
\begin{proof}
The implications (ii)$\Rightarrow$(iii) and (iii)$\Rightarrow$(i) are obvious.

Assume (i) holds, that is, $\gamma_1$ and $\gamma_2$ converge to some
horofunction $\xi$. Let $N\in\N$.
We can find $t_2$ large enough that
\begin{align*}
\xi(\gamma_2(0)) - \xi(\gamma_1(N))
   = d(\gamma_2(0),\gamma_2(t_2)) - d(\gamma_1(N), \gamma_2(t_2)).
\end{align*}
We can then find $t_1$ large enough that
\begin{align*}
\xi(\gamma_2(t_2)) - \xi(\gamma_1(N))
   = d(\gamma_2(t_2),\gamma_1(t_1)) - d(\gamma_1(N), \gamma_1(t_1)).
\end{align*}
Since $\gamma_2$ is a geodesic,
\begin{align*}
\xi(\gamma_2(0)) = d(\gamma_2(0),\gamma_2(t_2)) + \xi(\gamma_2(t_2)).
\end{align*}
It follows that
\begin{align*}
d(\gamma_1(N), \gamma_1(t_1))
   = d(\gamma_1(N), \gamma_2(t_2)) + d(\gamma_2(t_2),\gamma_1(t_1)).
\end{align*}
So there exists a geodesic agreeing with $\gamma_1$ up to time $N$,
passing through $\gamma_2(t_2)$, and agreeing with $\gamma_1$ after time $t_1$.
This establishes (ii).

Now assume that (iv) holds. So there is a geodesic $\gamma_3$ agreeing with
$\gamma_1$ up to any given time $N$ with the same limit as $\gamma_2$.
Applying the equivalence of (i) and (ii) just proved, we see there is a
geodesic $\gamma_4$ agreeing with $\gamma_1$ up to time $N$ and agreeing with
$\gamma_2$ after some time $M$.
A symmetrical argument shows that there is a geodesic $\gamma_5$
agreeing with $\gamma_2$ up to time $M$ and eventually agreeing with $\gamma_1$.
It is easy to prove that the path agreeing with $\gamma_4$ up to time $M$ and
thereafter agreeing with $\gamma_5$ is a geodesic satisfying the requirements
of (ii).
\end{proof}

We use $[a,b]$ to denote the commutator $a^{-1}b^{-1}ab$ of two elements
of a group. Also, for any two subgroups $A$ and $B$, we denote by
$[A,B]$ the subgroup generated by all $[a,b]$ with $a\in A$ and $b\in B$.
For any group $G$, the subgroup $[G,G]$ is called the
\emph{commutator subgroup}. It is always normal.

Recall that a group $G$ is \emph{nilpotent} if $\Gamma_n$ is trivial for
some $n\in\N$, where $\Gamma_i$; $i\in\N$ is the \emph{lower central series}
defined by $\Gamma_1:=G$ and $\Gamma_{i+1}:= [\Gamma_i,G]$ for $i\in\N$.

We will need the following two lemmas concerning finite-index subgroups
of nilpotent groups. They have similar proofs, which rely on the identities
\begin{align}
\label{eqn:commuteright}
[x,yz] &= [x,z] [x,y] [[x,y],z] \qquad\text{and} \\
\label{eqn:commuteleft}
[xy,z] &= [x,z] [[x,z],y] [y,z].
\end{align}

\begin{lemma}
\label{lem:comm_finite_index}
Let $H$ be a finite index subgroup of a finitely generated nilpotent group~$G$.
Then, $[H,H]$ has finite index in $[G,G]$.
\end{lemma}
\begin{proof}
We use an inductive argument.
The induction hypothesis is that there is a finite set $R_i\subset G$ such that
any $g\in[G,G]$ can be expressed as $g=h'r' \mod \Gamma_i$,
with $h'\in [H,H]$ and $r'\in R_i$.

The case when $i=2$ is trivial.

Now assume the hypothesis is true for some $i\ge 2$.
The assumption of the lemma implies
that, for each generator $s\in S$, there is some $n_s\ge 1$ and $h_s\in H$
such that $s^{n_s} = h_s$.
Let $C_i$ denote the set of simple commutators of weight $i$ of elements of the
generating set $S$.
In other words, $C_1 := S$ and
\begin{align*}
C_{j+1} := \big\{ [c,s] \mid \text{$c\in C_j$ and $s\in S$} \big\},
\qquad \text{for $j\ge 1$}.
\end{align*}
Let $c := [\dots[s_1, \dots] , s_i ]$ be an arbitrary element of $C_i$.
Using the identities~(\ref{eqn:commuteright}) and~(\ref{eqn:commuteleft}),
one can show that
\begin{align*}
c^{n_{s_1}\cdots n_{s_i}}
   &= [\dots[s_1^{n_{s_1}},\dots],s_i^{n_{s_i}}] \qquad\mod \Gamma_{i+1}  \\
   &= [\dots[h_{s_1} ,\dots], h_{s_i}] \qquad\mod \Gamma_{i+1}.
\end{align*}
So, letting $p_c:= n_{s_1}\cdots n_{s_i}$,
we see that $c^{p_c}$ is in $[H,H]$, modulo $\Gamma_{i+1}$.

As $\Gamma_i/\Gamma_{i+1}$ is abelian and generated by the elements of
$C_i \mod \Gamma_{i+1}$ we can write any $d\in \Gamma_i$ in the form
\begin{align*}
d = \prod_{c\in C_i} c^{m_c}    \qquad\mod\Gamma_{i+1},
\end{align*}
where the product is taken in any fixed order and $m_c\in \Z$ for all
$c\in C_i$. Therefore, from what we have just proved,
\begin{align*}
d = h \prod_{c\in C_i} c^{q_c}    \qquad\mod\Gamma_{i+1},
\end{align*}
for some $h\in[H,H]$, where $q_c$ is the remainder when $m_c$ is divided by
$p_c$. Let $R$ be the finite set
\begin{align*}
R:= \Big\{\prod_{c\in C_i} c^{q_c}
        \mid \text{$0\le q_c < p_c$ for all $c\in C_i$} \Big\}.
\end{align*}

By the induction hypothesis, there is a finite set $R_i\subset G$ such that
any $g\in[G,G]$ can be expressed as $g=h'r'd$, with $h'\in [H,H]$, $r'\in R_i$,
and $d\in \Gamma_i$.
So $g=h'dr' \mod \Gamma_{i+1}$. By the result of the previous paragraph,
$d =hr \mod \Gamma_{i+1}$ for some $h\in [H,H]$ and $r\in R$.
Therefore, $g=h'hrr' \mod \Gamma_{i+1}$, and so the induction hypothesis
is true for i+1.
\end{proof}

\begin{lemma}
\label{lem:group_finite_index}
Let $H$ be a subgroup of a finitely generated nilpotent group $G$,
and let $\Theta$ be the natural homomorphism from $G$ to its abelianisation
$G/[G,G]$. If $\Theta(H)$ has finite index in $\Theta(G) = G/[G,G]$, then
$H$ has finite index in $G$.
\end{lemma}
\begin{proof}
We use again an inductive argument.
The induction hypothesis here is that there is a finite set $R'\subset G$
such that any $g\in G$ can be expressed as $g=h'r' \mod \Gamma_i$,
with $h'\in H$ and $r'\in R'$.

By assumption, this is true for $i=2$.

Now assume it is true for some $i\ge 2$. The assumption of the lemma implies
that, for each generator $s\in S$, there is some $n_s\ge 1$ and $h_s\in H$
such that $\Theta(s^{n_s}) = \Theta(h_s)$.
Therefore, for all $s\in S$, there exists $g_s\in[G,G]$ such that
$s^{n_s} = h_s g_s$.

As in the proof of the previous lemma, let $C_i$ denote the set of simple
commutators of weight $i$ of elements of $S$, and consider an arbitrary
element $c := [\dots[s_1, \dots] , s_i ]$.
Using the identities~(\ref{eqn:commuteright}) and~(\ref{eqn:commuteleft}),
one can show as before that
\begin{align*}
c^{p_c}
   &= [\dots[s_1^{n_{s_1}} ,\dots],s_i^{n_{s_i}}] \qquad\mod \Gamma_{i+1}  \\
   &= [\dots[h_{s_1} ,\dots], h_{s_i}] \qquad\mod \Gamma_{i+1},
\end{align*}
where $p_c:= n_{s_1}\cdots n_{s_i}$.

It follows just as before that any $d\in \Gamma_i$ can be written as
$d =hr \mod \Gamma_{i+1}$ for some $h\in H$ and $r\in R$,
where $R$ is the finite set
\begin{align*}
R:= \Big\{\prod_{c\in C_i} c^{q_c}
        \mid \text{$0\le q_c < p_c$ for all $c\in C_i$} \Big\}.
\end{align*}

By the induction hypothesis there is a finite set $R'\subset G$ such that
any $g\in G$ can be expressed as $g=h'r'd$, with $h'\in H$, $r'\in R'$, and
$d\in \Gamma_i$.
So $g=h'dr' \mod \Gamma_{i+1}$.
But we have seen that $d =hr \mod \Gamma_{i+1}$ for some $h\in H$ and $r\in R$,
and therefore, $g=h'hrr' \mod \Gamma_{i+1}$.
This completes the induction step.
\end{proof}

\begin{lemma}
\label{lem:geo_facet}
Let $V$ be a subset of $S$ such that $\phi(V)$ is contained
in a facet of $P$. Then, any word with letters in $V$ is geodesic.
\end{lemma}
\begin{proof}
Let $F\subset \R^n$ be the facet into which $V$ is mapped,
and let $f:\R^n \to \R$ be the linear functional defining $F$, that is, such
that $f(x)=1$ for $x\in F$.
So, if $z_0 z_1\cdots$ is a word with letters in $V$,
then $f\after\phi(\overline z_0\cdots \overline z_{n-1})=n$ for all $n\in\Z$.
Let $y_0 y_1\cdots$ be a word with letters in $S$ such that
$\overline y_0\cdots \overline y_{m-1} = \overline z_0\cdots \overline z_{n-1}$
for some $n$ and $m$ in $\Z$. Since $f(x)<1$ for $x\in P\backslash F$,
we have $m\ge f\after\phi(y_0\cdots y_{m-1}) = n$. It follows
that $z_0 z_1\cdots$ is a geodesic word.
\end{proof}

\begin{lemma}
\label{lem:rearrange}
Let $H$ be a nilpotent group generated by a finite set $V$.
Then, for any $g\in H$, we can write $g\overline x = \overline y$,
where $x$ and $y$ are words with letters in $V$.
\end{lemma}
\begin{proof}
We will show the result is true for each of the groups $H/\Gamma_i$; $i\ge 2$,
using induction.

It is clearly true for the group $H/\Gamma_2= H/[H,H]$, since this group is
abelian.

Now assume the result is true for $H/\Gamma_i$ for some $i\ge 2$.

Let $C_i$ denote the set of simple commutators of weight $i$ of elements of $V$.
In other words, $C_1 := V$ and
\begin{align*}
C_{i+1} := \big\{ [c,s] \mid \text{$c\in C_i$ and $s\in V$} \big\},
\qquad \text{for $i\ge 1$}.
\end{align*}
From the induction hypothesis,
for every $c\in C_{i-1}$, there exists words $x_c$ and $y_c$ with letters in
$V$ such that $c\overline x_c  = \overline y_c $ mod $\Gamma_i$.
For each $c\in C_{i-1}$, let $g_c \in \Gamma_i$ be such that
\begin{align}
\label{eqn:interchange}
c\overline x_c  = \overline y_c  g_c.
\end{align}

Consider the word
\begin{align*}
w:= \prod_{\substack{c\in C_{i-1} \\ s\in V}}
                 (y_c s^{n(c,s)}x_c)(x_c s^{m(c,s)} y_c),
\end{align*}
where the product is taken in any fixed order, and $n(c,s)$ and $m(c,s)$
are non-negative integers for all $c\in C_{i-1}$ and $s\in V$.
Clearly, the letters of $w$ are in $V$.

Using~(\ref{eqn:interchange}) and the fact that $[c,h]$ and $g_c$ are
central in $H/\Gamma_{i+1}$ for every $c\in C_{i-1}$ and $h\in H$, we get
\begin{align*}
\overline y_c  s^{n(c,s)}\overline x_c
   &= c\overline x_c  g_c^{-1} s^{n(c,s)} c^{-1} \overline y_c  g_c,
          \qquad \mod \Gamma_{i+1} \\
   &= \overline x_c  [c,\overline x_c ] s^{n(c,s)} [c,s]^{n(c,s)}\overline y_c ,
          \qquad \mod \Gamma_{i+1}.
\end{align*}
Similarly,
\begin{align*}
\overline x_c  s^{m(c,s)}\overline y_c
   &= \overline y_c  [\overline x_c ,c] s^{m(c,s)} [s,c]^{m(c,s)}\overline x_c ,
          \qquad \mod \Gamma_{i+1}.
\end{align*}
So
\begin{align*}
\overline w:= \prod_{\substack{c\in C_{i-1} \\ s\in V}}
                [c,s]^{n(c,s)-m(c,s)}
            \prod_{\substack{c\in C_{i-1} \\ s\in V}}
                (\overline x_c  s^{n(c,s)} \overline y_c)
                (\overline y_c  s^{m(c,s)} \overline x_c),
   \qquad \mod \Gamma_{i+1}.
\end{align*}
Since the group $\Gamma_i/\Gamma_{i+1}$ is generated by $C_i\mod\Gamma_{i+1}$,
and is in the center of $H/\Gamma_{i+1}$, any $h\in\Gamma_i$ can be
written ($\!\!\!\!\mod \Gamma_{i+1}$) as the first factor in the
right-hand-side of the above equation,
with an appropriate choice of $n$ and $m$.
The second factor can be represented by a word $z$ with letters in $V$.
So $h\overline z = \overline w \mod \Gamma_{i+1}$.

Now let $g\in H$. By the induction hypothesis, $g\overline x = \overline y
\mod \Gamma_i$ for some words $x$ and $y$ with letters in $V$.
In other words, $g\overline x h=\overline y$ for some $h\in\Gamma_i$.
As we have just seen, $h\overline z = \overline w \mod \Gamma_{i+1}$
for some words $z$ and $w$ with letters in $V$.
So $g\overline x\overline w = \overline y \overline z \mod \Gamma_{i+1}$.
This completes the induction step.
\end{proof}

We are now ready to construct geodesic paths leading to the Busemann points
lying in finite orbits.

Let $F$ be a facet of $P$, and let $V$ be the set of generators mapped into
$F$ by $\phi$. We denote by $\langle V\rangle$ the subgroup of $G$ generated
by the elements of $V$.
Of course, $\langle V\rangle$ is also nilpotent.
So, by Lemma~\ref{lem:rearrange}, for each $g\in\langle V\rangle$, there exist
words $x_g$ and $y_g$ with letters in $V$ such that
$g\overline x_g = \overline y_g$.
Let $C_i$ be the set of simple commutators of weight $i$ of elements of $V$.
Take a finite word $w_F$ with letters in $V$ such that
$x_c$ and $y_c$ appear as subwords, for every $c\in\bigcup_i C_i$.
The infinite word $w_F w_F \cdots$ is geodesic, by Lemma~\ref{lem:geo_facet}.
We define $\Lambda_F:\N\to G$ to be the geodesic path starting at the identity,
having steps given by this word. Since $\Lambda_F$ is geodesic, it converges
to a Busemann point of the horoboundary, which we denote by $\xi_F$.

\begin{lemma}
\label{lem:premult}
Let $F$ be a facet of $P$, and let $V$ be the set of generators mapped into
$F$ by $\phi$. Then, for any $g\in\langle V \rangle$, there is an $n\in\N$
such that $g\evalword^n$ can be written as a product of elements of $V$.
\end{lemma}
\begin{proof}
Let $\Gamma_i(\langle V\rangle)$ denote the decending central series of
$\langle V\rangle$, and let $C_i$ denote the set of simple commutators
of weight $i$ of elements of $V$.

We consider the groups $\langle V\rangle/\Gamma_i(\langle V\rangle)$; $i\ge 2$
and use induction.

The case when $i=1$ is trivial.

Now assume that $g\evalword^n$ can be written (mod $\Gamma_i(\langle V\rangle)$)
in the desired form.
That is to say, $g\evalword^n = \overline v c$ for some word $v$
with letters in $V$ and some $c\in\Gamma_i(\langle V\rangle)$.
We know that $c$ can be written as a product of simple commutators
and their inverses:
\begin{align*}
c = c_1^{\epsilon_1} \cdots c_m^{\epsilon_m},
\qquad\text{mod $\Gamma_{i+1}(\langle V\rangle)$},
\end{align*}
where $m\in\N$, and $c_j\in C_i$ and $\epsilon_j=\pm 1$
for all $j\in\{1,\dots,m\}$.
Recall that $\Gamma_i(\langle V\rangle)/\Gamma_{i+1}(\langle V\rangle)$
is in the center of $\langle V\rangle/\Gamma_{i+1}(\langle V\rangle)$.
Suppose that $\epsilon_j=1$ for some $j$.
Recall that $\word$ has subwords $x_{c_j}$ and $y_{c_j}$ satisfying
$c_j \overline x_{c_j} = \overline y_{c_j}$.
So by replacing the subword $x_{c_j}$ in
$\word$ by $y_{c_j}$, we obtain a word with letters in $V$ that represents
$c_j \evalword$ mod $\Gamma_{i+1}(\langle V\rangle)$.
Similarly, for any $j$ for which $\epsilon_j = -1$, we replace $y_{c_j}$
by $x_{c_j}$ to obtain a word with letters in $V$ representing
$c_j^{-1} \evalword$ mod $\Gamma_{i+1}(\langle V\rangle)$.
We concatenate all these words to get a word $u$ with letters in $V$
representing $c \evalword^m$ mod $\Gamma_{i+1}(\langle V\rangle)$.
So $vu$ represents $g\evalword^{n+m}$ mod $\Gamma_{i+1}(\langle V\rangle)$.
This concludes the induction step.
\end{proof}

\begin{lemma}
\label{lem:stabilises}
Let $F$ be a facet of $P$, and let $V$ be the set of generators mapped into
$F$ by $\phi$.
Then, the horofunction $\xi_F$ is invariant under the subgroup
$\langle V \rangle$.
\end{lemma}
\begin{proof}
Let $g\in\langle V \rangle$ and $n\in\N$.
Since $g[g,\evalword^n] \in \langle V \rangle$, we can find,
by Lemma~\ref{lem:premult},
$m\in\N$ such that $g[g,\evalword^n]\evalword^m$ is represented
by some word $v$ with letters in $V$.

Similarly,
we can find $l \in \N$ such that $[\evalword^{n+m},g]g^{-1}\evalword^{l}$
is represented by a word $v'$ with letters in $V$.

Consider the path $\gamma$ starting at the identity with steps given by the
word $\word^n v v' \word\word\cdots$.
By Lemma~\ref{lem:geo_facet}, it is a geodesic.
It agrees with the geodesic path $\facetgeo_F$ up to time $n|\word|$.
But $\evalword^n\overline v = g\evalword^{n+m}$, and
so $\gamma$ has a point in common with the path $g\facetgeo_F$ at
time $|\word^n v|$. We also have that
$\evalword^n \overline v \overline v' = \evalword^{n+m+l}$.
Therefore, $\gamma$ agrees with $\facetgeo_F$ after time $|\word^n v v'|$.

We now use Proposition~\ref{prop:switch} to deduce that $\facetgeo_F$ and
$g \facetgeo_F$ converge to the same limit $\xi_F$.
\end{proof}

\begin{theorem}
\label{thm:stabiliser}
Let $F$ be a facet of $P$, and let $V$ be the set of generators mapped into
$F$ by $\phi$. Then, the stabiliser of $\xi_F$ is $\langle V \rangle$.
\end{theorem}
\begin{proof}
We have already seen in Lemma~\ref{lem:stabilises} that $\langle V \rangle$
stabilises $\xi_F$.

Let $g\in G$ be such that $g\xi_F=\xi_F$. Consider the geodesics $\facetgeo_F$
and $g\facetgeo_F$. Since they have the same limit $\xi_F$, there exists
a geodesic $\gamma$ having infinitely many points in common with each, by
Proposition~\ref{prop:switch}.
So there exist $t_0$, $t_1$, and $t_2$ in $\N$ with $t_0\le t_1\le t_2$,
such that $\gamma(t_0)=\facetgeo_F(s_0)$, $\gamma(t_1)=g\facetgeo_F(s_1)$,
and $\gamma(t_2)=\facetgeo_F(s_2)$ for some $s_0$, $s_1$, and $s_2$ in $\Z$.
Let $f:\R^n \to \R$ be the linear functional defining the facet $F$,
so that $f(x)=1$ for $x\in F$ and $f(x)<1$ for $x\in P\backslash F$.
Since the steps of $\facetgeo_F$ lie in $V$, we have
\begin{align*}
f\after\phi(\facetgeo_F(s_2)) - f\after\phi(\facetgeo_F(s_0)) = s_2 - s_0.
\end{align*}
But since $\gamma$ is a geodesic, $s_2 - s_0 = t_2 - t_0$.
We conclude that
\begin{align*}
f\after\phi(\gamma(t_2)) - f\after\phi(\gamma(t_0)) = t_2 - t_0.
\end{align*}
Using the fact that $f\after\phi$ is a homomorphism and that
$f\after\phi(x)\le 1$ for $x\in S$, we get that every step $x$ of $\gamma$
between $t_0$ and $t_2$ satisfies $f\after\phi(x)=1$, that is, they all
lie in~$V$.

Consider the finite path that starts at the identity, is identical to
$\facetgeo_F$ up to time $s_0$, follows $\gamma$ from
$\gamma(t_0)=\facetgeo_F(s_0)$ to $\gamma(t_1)=g\facetgeo_F(s_1)$, and
then follows $g\facetgeo_F$ backwards along to $g$. The steps of this path
lie entirely within $V\union V^{-1}$. We deduce that $g\in\langle V \rangle$.
\end{proof}

\begin{corollary}
For every facet $F$ of $P$, the Busemann point $\xi_F$ lies in a finite orbit.
\end{corollary}
\begin{proof}
Recall that $Z_0$ is the torsion subgroup of the abelianisation of $G$.
Choose a subset $R$ of $G$ containing one representative of each
equivalence class in $Z_0$.

Since $F$ is a facet of $P$, the set of vectors $\phi(V)$ generates a finite
index subgroup of $\Z^N$. So there is a finite subset $R'$ of $G$ such that,
for any $g\in G$, there exists $r'\in R'$ and $v\in\langle V\rangle$ satisfying
$\phi(g)=\phi(v)\phi(r)$.

So $\Theta(g) = \Theta(vr) \mod Z_0$, where $\Theta$ is the natural
homomorphism from $G$ to its abelianisation $G/[G,G]$.
This is equivalent to $\Theta(g) = \Theta(v)\Theta(rr')$ for some $r\in R$.
We conclude from this that $\Theta(\langle V\rangle)$ has finite index in
$\Theta(G)$. So, by Lemma~\ref{lem:group_finite_index}, $\langle V\rangle$
has finite index in $G$. But, by the theorem, $\langle V\rangle$ is exactly
the stabiliser of $\xi_F$, and so this Busemann point has a finite orbit.
\end{proof}

We have just shown that to every facet of $P$ there corresponds a finite
orbit of Busemann points. To prove the correspondence in the opposite
direction, we will need the following lemma. Note that its hypothesis that
the orbit is finite is not needed in the abelian setting~\cite{develin_cayley}.

\begin{lemma}
\label{lem:projects_to_facet}
Suppose the limit of a geodesic path $\gamma$ lies in a finite orbit.
Then, the projection map $\phi$ maps the direction of $\gamma$ into some
facet of $P$.
\end{lemma}
\begin{proof}
Let $u_0 u_1 \cdots$ be the geodesic word giving the steps of $\gamma$.
For each $n\in\N$, let $\gamma^n$ be the geodesic starting at the identity
and having steps given by the word $u_n u_{n+1} \cdots$.
The limit of each $\gamma^n$ is in the same orbit as that of $\gamma$,
which we have assumed finite. Therefore, there is some subsequence
of geodesics $\gamma^{n_i}$ all of whose limits are equal.
Call the common limit $\xi$.

For each $i\in\N$, define the finite word $w_i:=u_{n_i} \cdots u_{n_{i+1}-1}$,
and let $W:=\{w_i \mid i\in\N \}$ be the set of all these words.
Fix $i\in\N$. Since $\gamma^{n_i}$ is eventually just a copy of
$\overline w_i \gamma^{n_{i+1}}$ shifted in time, these two geodesics have
the same limit. But both $\gamma^{n_i}$ and $\gamma^{n_{i+1}}$ converge
to $\xi$. Therefore, $w_i$ represents a group element in the stabiliser
of $\xi$.

Now consider an infinite concatenation $z := z_0 z_1 \cdots z_m \cdots$ of
words $z_m$; $m\in\N$, each in $W$. Given any $m\in\N$, the word $z$ can be
written
\begin{align*}
z = z_0 \cdots z_{m-1} u_{n_i} \cdots u_{n_{i+1}-1} z_{m+1} \cdots
\end{align*}
because $z_m = w_i$ for some $i\in\N$.
Since $p:= \overline z_0 \cdots \overline z_{m-1}$
is in the stabiliser of $\xi$,
we have $\xi(pg) = \xi(p) + \xi(g)$ for all $g\in G$.
So
\begin{align*}
\xi(\overline z_0  \cdots \overline z_{m-1} \overline u_{n_i}
          \cdots \overline u_{n_i +l-1})
   = \xi(\overline z_0 \cdots \overline z_{m-1})
        + \xi(\overline u_{n_i} \cdots \overline u_{n_i +l-1}),
\end{align*}
for $l\in\{0,\ldots,n_{i+1}-n_i\}$.
But since $\gamma^{n_i}$ is a geodesic converging to $\xi$, we have
$\xi(\overline u_{n_i} \cdots \overline u_{n_i +l-1})= -l$.
Since this reasoning is valid for any $m\in\N$, we conclude that $\xi$,
evaluated along the path $\lambda$, decreases at every step, where
$\lambda$ is the path starting at the identity with steps given by the word
$z$. It follows from this that $\lambda$ is a geodesic.
We have thus shown that every concatenation of words from $W$ is a geodesic
word.

Take now sufficiently many words $z_0, \ldots, z_{n-1}$ from $W$ that every
generator in the direction of $\gamma$ appears in at least one.
Let $a:=\phi(\overline z_0 \cdots \overline z_{n-1} )$,
and let $b$ be the intersection
of the line $0a$ in $\R^N$ with the boundary of $P$. Of course, $b$ has
rational coordinates and lies in some facet $F$ of $P$.
So, by Lemma~7 of~\cite{develin_cayley}, $b$ may be expressed as an affine
combination of the vertices of $F$, with non-negative rational coefficients.
Since $a$ is a rational multiple of $b$, we may express $a$ in the same way.
Multiplying through by the common denominator, we can find a positive integer
$l$ as large as we wish, such that $la$ is a linear combination of the vertices
of $F$ with non-negative integer coefficients. So there exists a word $x$ with
letters in $\phi^{-1}(F) \intersection S$ such that $\phi(\overline x)=la$.

Recall that $Z_0$ is the torsion subgroup of the abelianisation $G/[G,G]$.
Let $T\subset G$ be such there is exactly one element of $T$ in each
of the equivalence classes in $Z_0$. Of course, $|T| = |Z_0| < \infty$.

We can now write $(\overline z_0 \cdots \overline z_{n-1})^l = \overline xtc$,
with $t\in T$ and $c\in[G,G]$.

We have assumed that the stabiliser of $\xi$, which
we now call $U$, has finite index in $G$.
Therefore, by Lemma~\ref{lem:comm_finite_index}, $[U,U]$ has finite index in
$[G,G]$. So we can write $c=ru$, with $r$ in some finite subset $R$ of $G$,
and $u$ in $[U,U]$. Because $U$ stabilises $\xi$, the restriction of $\xi$
to $U$ is a homomorphism into $\Z$. In particular, $\xi(u^{-1})=0$.
So, for $i\in\N$ large enough,
$d(u^{-1},\gamma^{n_0}(i))= d(e,\gamma^{n_0}(i)) = i$.

Let
\begin{align*}
D:= \max\{ d(e,t'r') \mid \mbox{$t'\in T$ and $r' \in R$} \}.
\end{align*}
Of course, $D$ is finite.
We have
\begin{align*}
d(e,\overline x tru \gamma^{n_0}(i))
         & \le d(e,\overline x) + d(e,tr) + d(e,u\gamma^{n_0}(i)) \\
         & \le |x| + D + i.
\end{align*}
But $|x| = l f(a)$, where $f:\R^N\to \R$ is the linear functional such that
$f(g)=1$ for all $g\in F$.
Also, since $(z_0 \cdots z_{n-1})^l w_0 w_1 w_2 \cdots$ is a geodesic word,
\begin{align*}
d(e,(\overline z_0 \cdots \overline z_{n-1})^l\gamma^{n_0}(i))
                   = l|z_0 \cdots z_{n-1}| + i.
\end{align*}
Therefore  $l f(a) + D \ge l |z_0 \cdots z_{n-1}|$.
Since $l$ can be made arbitrarily large, $f(a) \ge |z_0 \cdots z_{n-1}|$.
We may now apply the fact that $f\after \phi(s) \le 1$ for every $s\in S$
to deduce that $f\after \phi(g)= 1$ for every letter $g$ in
$z_0 \cdots z_{n-1}$, which includes every letter in the direction of $\gamma$.
Therefore, all these letters are mapped by $\phi$ to $F$.
\end{proof}

\begin{theorem}
\label{all_finite_orbits}
Every finite orbit of Busemann points can be written
$\{g\xi_F \mid g\in G\}$ for some facet $F$ of $P$.
\end{theorem}
\begin{proof}
Let $\gamma_1$ be a geodesic converging to a point in a finite orbit.
By Lemma~\ref{lem:projects_to_facet}, the image of the direction of $\gamma_1$
under $\phi$ lies in some facet $F$ of $P$. By, if necessary, premultiplying
by an appropriate factor and removing an initial section of the path, we can
find a geodesic path $\gamma$ converging to a point $\xi$ in the same orbit
as the limit of $\gamma_1$ that starts at the identity and for which the image
under $\phi$ of every step lies in $F$.
Let the word $\word$ be defined as above. Since $\xi$ is in a finite orbit,
there exists some $m\in\N$ with $m>0$ such that $\evalword^m\xi$ is equal
to $\xi$. So $\evalword^{mn}\xi=\xi$ for all $n\in\N$.
Let $z_0 z_1 \cdots$ be the geodesic word giving the steps of $\gamma$,
and let $\gamma_2$ be the geodesic path starting at the identity and having
steps given by $\word^{mn}z_0 z_1 \cdots$ for some large $n\in\N$.
By Lemma~\ref{lem:geo_facet}, $\gamma_2$ is a geodesic. Also, $\gamma_2$ agrees
with $\facetgeo_F$ up to time $nm|\word|$, which can be made as large as one
likes. Finally, $\gamma_2$ converges to $\xi$, and so has the same limit as
$\gamma$.

We have proved one half of condition~(iv) of Proposition~\ref{prop:switch}.
To prove the other, take $n$ as large as you want and let
$g:= \overline z_0 \cdots \overline z_n$.
Clearly, $g^{-1} \in \langle V \rangle$,
where $V$ is the set of generators mapped into $F$ by $\phi$.
Therefore, by Lemma~\ref{lem:premult}, there is some $m\in\N$ such that
$g^{-1} \evalword^m$ can be represented by some word $v$ with letters in $V$.
By Lemma~\ref{lem:geo_facet}, the path starting at the identity and having
steps given by $z_0 \cdots z_n v \word \word \cdots$ is a geodesic.
This path agrees with $\gamma$ up to time $n$, and it eventually agrees with
$\facetgeo_F$ because $\overline z_0 \cdots \overline z_n v = \evalword^m$.

It now follows from Proposition~\ref{prop:switch} that $\gamma$ and
$\facetgeo_F$ have the same limit, in other words, that $\xi=\xi_F$.
\end{proof}

We finish this section by showing how the correspondence in the abelian setting
between proper faces of $P$ and orbits of Busemann points breaks down for
general finitely-generated nilpotent groups.

\textbf{Example 1.}
Consider the following group of nilpotency class 3:
\begin{align*}
G:= \langle a,b \mid [a,g] = [b,g] = [a,h] = [b,h] = e \rangle,
\end{align*}
where $c=[a,b]$, $g:=[a,c]$, and $h:=[b,c]$.

Any element of $G$ can be written in the form $g^i h^j c^k b^l a^m$,
with $i$, $j$, $k$, $l$, and $m$ in $\Z$. We may take the projection map $\phi$
to be $\phi(g^i h^j c^k b^l a^m) := (l,m)$. The image of the generating set
$\{a,b,a^{-1},b^{-1}\}$ is a square. Let $F$ be the facet defined by the
corners $\{(1,0), (0,1)\}$.

Define the word $x:= ab$. We will show that, although $xxx\cdots$ and
$\facetgeo_F$ have the same directions, their limits differ.
\begin{lemma}
\label{lem:example_perm}
Let $w$ be any word with letters in $\{a,b\}$ such that
$\overline w = g^m h^n {\overline x}^l$ for some $m,n \in \Z$ and $l\in\N$.
Then, $m-n$ is positive unless $w=x^l$.
\end{lemma}
\begin{proof}
Represent $w$ as a piecewise affine curve that starts at the origin and takes
a unit step in the $x$-direction for each letter $b$, and unit step in the
$y$-direction for each letter $a$.
The curve will obviously finish at $(l,l)$.
Write $\overline w$ in the form $g^i h^j c^k b^l a^l$.
It is not hard to see that $k$ is the number of unit squares under the curve.
After further thought, one also sees that $j$ is the sum of the $x$-coordinates
of the upper right corners of these squares, and that $i$ is the sum of the
$y$-coordinates.
So the center of gravity, taking the mass of each unit square to be
concentrated on its upper right corner, is $(j/k, i/k)$.
We now redraw the figure with the axes rotated through a $45^\circ$ angle,
as in Figure~\ref{fig:water}. We see that choosing $w$ to minimise $i-j$
is the same as finding the curve between $(0,0)$ and $(l,l)$ having $(l+1)l/2$
unit squares underneath it that minimises the height of the center of mass
of these squares. But it is obvious that there is a unique minimising
curve, namely that obtained from $x^l$. The conclusion follows.
\end{proof}

\begin{figure}
\input{water.pstex_t}
\caption{Diagram for the proof of Lemma~\ref{lem:example_perm}.}
\label{fig:water}
\end{figure}

\begin{prop}
\label{prop:example}
The limit $\eta$ of $xxx\cdots$ is different from $\xi_F$.
\end{prop}
\begin{proof}
Let $l\in\N$ and let $w$ be a geodesic word with letters in
$\{a,b,a^{-1},b^{-1}\}$ representing $g^{-1} \overline x^l$.
By Lemma~\ref{lem:example_perm}, either $a^{-1}$ or $b^{-1}$ occurs as a
letter of $w$. Also, the excess of $a$s over $a^{-1}$s in $w$ equals $l$,
as does the excess of $b$s over $b^{-1}$s.
Therefore, $d(e, g^{-1}\overline x^l) = |w| > 2l$.
We conclude that
\begin{align*}
\eta(g) = \lim_{l\to\infty} \Big( d(g,\overline x^l) - d(e,\overline x^l) \Big)
        >0.
\end{align*}

However, $\{a,b\}$ generates $G$, and so $\xi_F$ is a fixed point
by Theorem~\ref{thm:stabiliser}. So $\xi_F$ is a homomorphism from $G$ to $\Z$.
In particular, $\xi_F(g) = 0$. Therefore, $\eta$ and $\xi_F$ differ.
\end{proof}

\section{The boundary of the discrete Heisenberg group}

The discrete Heisenberg group $H_3$ is the group of $3\times 3$ upper triangular
matrices of the form
\begin{align*}
\begin{pmatrix} 1 & x & z\\ 0 & 1 & y\\ 0 & 0 & 1\\ \end{pmatrix},
\qquad\text{with $x$, $y$, and $z$ in $\Z$}.
\end{align*}
It is the simplest non-abelian nilpotent group. Let
\begin{align*}
a:=\begin{pmatrix} 1 & 1 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ \end{pmatrix},
\qquad
b:=\begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 1\\ 0 & 0 & 1\\ \end{pmatrix},
\qquad\text{and}\quad
c:=\begin{pmatrix} 1 & 0 & 1\\ 0 & 1 & 0\\ 0 & 0 & 1\\ \end{pmatrix}.
\end{align*}
Observe that $ab=bac$, that is, $c$ is the commutator of $a$ and $b$,
and that the center of $H_3$ is the cyclic group generated by $c$.
The following is a presentation of $H_3$:
\begin{align*}
H_3 := \langle a,b \mid [[a,b],a] = [[a,b],b] = e \rangle.
\end{align*}
All elements of $H_3$ can be written in the form $c^z b^y a^x$,
with $x,y,z\in\Z$.

We will calculate all the Busemann points of this group with the word length
metric coming from the standard generating set $S:=\{a,b,a^{-1},b^{-1}\}$.
Our work relies on the following formula for this metric,
found by Blach{\`e}re~\cite{blachere}.
\begin{theorem}[\cite{blachere}]
\label{thm:formula}
Let $g:= c^z b^y a^x$ be an element of $H_3$.
Its distance to the identity with respect to the generating set
$S:=\{a,b,a^{-1},b^{-1}\}$ is
\begin{itemize}
\renewcommand{\labelitemi}{I.}
\item
if $zyx\ge 0$, and
   \begin{itemize}
   \renewcommand{\labelitemii}{I.1.}
   \item
   if $\max(x^2,y^2) \le |z|$, then
      \begin{align*}
      d(e,g) = 2\big\lceil 2\sqrt{|z|}\big\rceil - |x| - |y|;
      \end{align*}
   \renewcommand{\labelitemii}{I.2.}
   \item
   if $\max(x^2,y^2) \ge |z|$, and
       \begin{itemize}
          \renewcommand{\labelitemiii}{I.2.1.}
          \item
          $|xy| \ge |z|$, then
            \begin{align*}
            d(e,g) = |x| + |y|;
            \end{align*}
          \renewcommand{\labelitemiii}{I.2.2.}
          \item
          $|xy| \le |z|$, then
            \begin{align*}
            d(e,g) = 2\lceil \min(|z/x|,|z/y|) \rceil + \big| |x| - |y| \big|;
            \end{align*}
       \end{itemize}
   \end{itemize}
\renewcommand{\labelitemi}{II.}
\item
if $zyx\le 0$, and
   \begin{itemize}
   \renewcommand{\labelitemii}{II.1.}
   \item
   if $\max(x^2,y^2) \le |z| + |xy|$, then
      \begin{align*}
      d(e,g) = 2\big\lceil 2\sqrt{|z|+|xy|}\big\rceil - |x| - |y|;
      \end{align*}
   \renewcommand{\labelitemii}{II.2.}
   \item
   if $\max(x^2,y^2) \ge |z| + |xy|$, then
         \begin{align*}
         d(e,g) = 2\lceil \min(|z/x|,|z/y|) \rceil + |x| + |y|.
         \end{align*}
   \end{itemize}
\end{itemize}
\end{theorem}

Given a word $y$, we  call the pairs of consecutive letters of $y$
the \emph{transitions} of $y$. The transitions $ab$, $ba^{-1}$, $a^{-1}b^{-1}$,
and $b^{-1}a$ are said to be positive. The transitions obtained by reversing
the letters of these are called negative.
Observe that the group element corresponding to a positive transition is equal
to $c$ times the element corresponding to the negative transition obtained
by reversing the order of the letters. So, if one takes a word and transforms
it by reversing an equal number of positive and negative transitions,
the resulting word is a geodesic if and only if the original was.

\begin{prop}
\label{prop:h3geodesics}
Every infinite geodesic word in $H_3$ either has two elements of $S$
as letters or consists of a finite prefix followed by a single letter repeated.
\end{prop}
\begin{proof}
Let $y$ be an infinite word that has at least three elements of $S$ as letters,
at least two of which appear infinitely often.

We wish to show that $y$ is not a geodesic.
Obviously, this is the case if it is not freely reduced, so we assume
the contrary. Our above assumptions on $y$ then imply that there are either
infinitely many positive transitions in $y$ or there are
infinitely many negative transitions.

Since $y$ has at least three elements of $S$ as letters, it must have
as letters both some generator and its inverse.
By repeatedly swapping this generator with one of its neighbours,
we can bring it to a position adjacent to its inverse.
If $y$ has both infinitely many positive and infinitely many negative
transitions,
then we can compensate for these swaps by reversing transitions elsewhere
so that the total number of positive transitions reversed equals the
total number of negative ones reversed.
Since the resulting word is clearly not a geodesic,
it follows that $y$ is not either.

So we will assume without loss of generality that there are infinitely
many positive transitions and finitely many negative ones.
Therefore $y$ can be written in the form
\begin{align*}
y = R a^{i_0} b^{j_0} a^{-i_1} b^{-j_1} a^{i_2}\cdots,
\end{align*}
where $(i_n)$ and $(j_n)$ are sequences of positive integers, and $R$ is some
finite initial word.

Consider what happens if $(i_n)$ and $(j_n)$ eventually become, respectively,
constants $i$ and $j$. Define the word $w:=a^i b^j a^{-i} b^{-j}$.
Observe that $\overline w = c^{ij}$. So, the word $www\cdots$ is not geodesic
since, by the formula of Theorem~\ref{thm:formula}, $d(e,\overline w^n)$
grows like the square root of $n$, rather than linearly.
Therefore, in this case, $y$ is not geodesic.

If $(i_n)$ and $(j_n)$ are not both eventually constant, then
we can find $n\ge 1$ such that $i_n < i_{n+1}$ or $j_n < j_{n+1}$.
Assume that the latter holds; the former case can be handled similarly.
Also assume without loss of generality that $n$ is even.
So $y$ can be written
\begin{align*}
y = R \cdots b^{-j_{n-1}} (a^{i_n}) b^{j_n} a^{-i_{n+1}} b^{-j_{n+1}} \cdots.
\end{align*}
By moving the bracketed clump of $a$'s forward, we get a word
\begin{align*}
z = R \cdots b^{-j_{n-1}} b^{j_n} a^{-i_{n+1}}
                             b^{-j_n} (a^{i_n}) b^{-j_{n+1}+j_n} \cdots,
\end{align*}
which is clearly not a geodesic.
However, $z$ is obtained from $y$ by reversing the letters of $i_n j_n$
positive transitions and then reversing the letters of the same number of
negative transitions. It follows that $y$ is not geodesic either.
\end{proof}

\newcommand\epsilona{{\epsilon_a}}
\newcommand\epsilonb{{\epsilon_b}}
\newcommand\sign{\operatorname{sign}}

\newcommand\corner{\xi}
\newcommand\ahoro{\eta}
\newcommand\bhoro{\zeta}

\begin{theorem}
\label{thm:corners}
Let $\epsilona$ and $\epsilonb$ be in $\{-1,+1\}$, and let $\Omega$ be the set
of all words having letters in $\{a^\epsilona,b^\epsilonb\}$ with both letters
occurring infinitely often.
Then, all words in $\Omega$ are geodesic, and the corresponding paths
all converge to the same Busemann point, given by
\begin{align*}
\corner^{\epsilona \epsilonb}(c^z b^y a^x) := -\epsilona x - \epsilonb y.
\end{align*}
\end{theorem}
\begin{proof}
Let $w$ be a finite word having set of letters $\{a^\epsilona,b^\epsilonb\}$,
and let $\overline w=c^z b^y a^x$ be the associated element of the group.
If $z\neq 0$, then $\sign z = \sign y. \sign x = \epsilonb \epsilona$.
Therefore, $zyx\ge0$.
Also, $|z|\le |xy|$. We deduce from the word length formula of
Theorem~\ref{thm:formula} that $d(e,\overline w) = |x| + |y|$, which is exactly
the number of letters in $w$. So $w$ is geodesic. It follows that every
word in $\Omega$ is geodesic.

Let $w_1$ and $w_2$ be words in $\Omega$, and let $\gamma_1$ and $\gamma_2$
be the corresponding geodesic paths starting at the identity.
Choose $N\in\N$, and let $u$ be the prefix of $w_1$ of length $N$.
Since $w_2$ has each of the letters $a^\epsilona$ and
$b^\epsilonb$ infinitely often, we can find a prefix $v$ of $w_2$ that
has more $a^\epsilona$'s and $b^\epsilonb$'s than $u$.
So there exists a word $x$ with letters in $\{a^\epsilona,b^\epsilonb\}$
of length $|v|-|u|$ such that $\overline u \overline x = c^m \overline v$
for some $m\in\Z$. Since $w_2$ has infinitely many positive transitions
and infinitely many negative transitions, we can write it $w_2=vy\omega$,
where $\omega$ is an infinite word (in $\Omega$) and $y$ is a finite word
satisfying $\overline y = c^m \overline z$, for some $z$ with letters
in $\{a^\epsilona,b^\epsilonb\}$.
By the first part of the proposition, $uxz\omega$ is a geodesic word.
The corresponding geodesic path agrees with $\gamma_1$ up to
time $N$ and agrees with $\gamma_2$ after time $|uxz|$. A similar argument
shows that there is also a geodesic path agreeing with $\gamma_2$ up to time
$N$ and eventually coinciding with $\gamma_1$.
We now apply Proposition~\ref{prop:switch} to deduce that $\gamma_1$
and $\gamma_2$ converge to the same limit.

Consider the sequence $h_n := (a^\epsilona b^\epsilonb)^n$.
By the previous part of the proposition, this sequence converges to the common
limit of the geodesics in $\Omega$. We have
\begin{align*}
h_n = c^{\epsilona\epsilonb n(n+1)/2} b^{\epsilonb n} a^{\epsilona n}.
\end{align*}
Let $g:= c^z b^y a^x$ be an arbitrary point in $H_3$. We have
\begin{align*}
d(g,h_n) = d(e,g^{-1} h_n)
      = d(e,c^{yx-z-\epsilonb xn + \epsilona\epsilonb n(n+1)/2}
            b^{\epsilonb n - y}
            a^{\epsilona n - x}).
\end{align*}
Consider what happens when $n$ is large. The exponents of $c$, $b$, and $a$
will have the signs $\epsilona\epsilonb$, $\epsilonb$, and $\epsilona$,
respectively; therefore, their product will be positive.
The square of the exponents of both $a$ and $b$ will be approximately
twice the absolute value of that of $c$, as will the absolute value of
the product of the exponents of $a$ and $b$.
Looking at the formula of Theorem~\ref{thm:formula},
we see that, for large $n$, the relevant case is I.2.1,
and so $d(g,h_n)=|\epsilonb n - y| + |\epsilona n - x|$.
In particular $d(e,h_n)=2n$. So the limiting horofunction is
$\corner^{\epsilona \epsilonb}(c^z b^y a^x) := -\epsilona x - \epsilonb y$.
\end{proof}

\begin{theorem}
\label{thm:heisenberg_busemen}
The following functions are Busemann points of $H_3$:
\begin{align*}
\ahoro^{\epsilon}_{m,n}(c^k b^j a^i)
   &:= -\epsilon i + |j-n| - |n|
        + 2 J \big(\epsilon (j - n), (j-n)i - (k-m)\big) - 2J(-\epsilon n,m), \\
\bhoro^{\epsilon}_{m,l}(c^k b^j a^i)
   &:= -\epsilon j + |i-l| - |l|
        + 2 J \big(-\epsilon (i - l), jl - (k-m)\big) - 2J(\epsilon l,m),
\end{align*}
where $\epsilon\in\{-1,+1\}$, $m, n, l\in\Z$, and
\begin{align*}
J(u,v) := \begin{cases}
   1, & \text{if $v\neq 0$ and $uv\ge 0$,} \\
   0, & \text{otherwise}, \\
   \end{cases}
\qquad\text{for $u,v\in\Z$}.
\end{align*}
Together with $\corner^{++}$, $\corner^{-+}$, $\corner^{++}$,
and $\corner^{--}$, these are the only Busemann points.
\end{theorem}
\begin{proof}
Choose $\epsilon\in\{-1,+1\}$ and $m, n\in\Z$, and define the path
$\gamma^\epsilon_{m,n}(t) := c^m b^n a^{\epsilon t}$, for $t\in\Z$.
By the formula of Theorem~\ref{thm:formula},
\begin{align*}
d(\gamma^\epsilon_{m,n}(0),\gamma^\epsilon_{m,n}(t)) = d(e,a^{\epsilon t}) = t.
\end{align*}
Therefore $\gamma^\epsilon_{m,n}$ is a geodesic starting at $c^m b^n$.

Let $g:=c^k b^j a^i$ be an element of $H_3$. We have
\begin{align*}
d(g,\gamma^\epsilon_{m,n}(t)) = d(e,c^{m-k+(j-n)i} b^{n-j} a^{\epsilon t-i}).
\end{align*}
Write
\begin{align*}
x_t:=\epsilon t-i, \qquad
y:= n-j, \qquad
\text{and $z:= m-k + (j-n) i$}.
\end{align*}
So $|x_t|$ tends to infinity as $t$ tends to infinity, whereas $y$ and $z$
remain constant.

Therefore the only relevant cases in the formula of Theorem~\ref{thm:formula}
when $t$ is large are I.2.1.~and II.2.
If $z=0$, then both cases give the same result:
$|x_t|+|y|$. If $z\neq 0$, then we have case I.2.1~if $\epsilon yz > 0$,
in which case $d(g,\gamma^\epsilon_{m,n}(t))=|x_t|+|y|$,
or case II.2.~if $\epsilon yz\le 0$, which
gives $d(g,\gamma^\epsilon_{m,n}(t))=|x_t|+|y|+2$, for $x_t$ large enough.
To sum up:
\begin{align*}
d(g,\gamma^\epsilon_{m,n}(t)) = |x_t|+|y|+2 J(-\epsilon y,z).
\end{align*}
Subtracting from this the value obtained when $i=j=k=0$, and taking the limit
as $t$ tends to infinity, we see that the path $\gamma^\epsilon_{m,n}$ converges
to the horofunction $\ahoro^{\epsilon}_{m,n}$ defined above.
This horofunction is a Busemann point since $\gamma^\epsilon_{m,n}$ is a
geodesic.

Similar reasoning shows that the path
$\lambda^\epsilon_{m,l}(t):= c^{m+\epsilon tl} b^{\epsilon t} a^l$ converges to
$\bhoro^\epsilon_{m,l}$, for all $\epsilon\in\{-1,+1\}$ and  $m, l\in\Z$, and
that the limit is a Busemann point.

By Proposition~\ref{prop:h3geodesics}, every infinite geodesic in $H_3$ either
uses exactly two generators or uses all but one only a finite number
of times. In the former case, the geodesic converges to one of
$\corner^{++}$, $\corner^{-+}$, $\corner^{+-}$, and $\corner^{--}$,
by Theorem~\ref{thm:corners}.
In the latter, the geodesic must eventually coincide with either
$\gamma^\epsilon_{m,n}$ or $\lambda^\epsilon_{m,l}$ for some value of
$\epsilon$ and of $m$ and $n$, or $m$ and $l$. Therefore, in this case,
the geodesic converges to the corresponding $\ahoro^{\epsilon}_{m,n}$
or $\bhoro^\epsilon_{l,m}$.
\end{proof}

The action of $H_3$ on each of the Busemann points can easily be calculated:
\begin{align*}
(c^z b^y a^x) \ahoro^\epsilon_{m,n} &= \ahoro^\epsilon_{m+z+nx,n+y}, \\
(c^z b^y a^x) \bhoro^\epsilon_{m,l} &= \bhoro^\epsilon_{m+z-y(l+x),l+x}, \\
(c^z b^y a^x) \corner^{\epsilona,\epsilonb} &= \corner^{\epsilona,\epsilonb}.
\end{align*}
So, in particular, each of the Busemann points $\corner^{++}$, $\corner^{+-}$,
$\corner^{-+}$, and $\corner^{--}$ is fixed by the action of $H_3$.
We have here an illustration of Theorem~\ref{thm:maintheorem}.
The abelianisation of $H_3$ is isomorphic to $\Z^2$. One can take the map
$\phi$ to be $\phi(c^z b^y a^x) := (x,y)$. The image of $S$ under this map is
the set $\{(1,0),(0,1),(-1,0),(0,-1)\}$, and so $P$ is the square with these
points as corners. Of course, $P$ has four facets, corresponding to the four
fixed Busemann points.

In \cite{winweb_busemann}, Webster and Winchester conjectured that
there is a boundary point or points of the form
$\lim_{i\to\infty}\ahoro^+_{m_i,n_i}$,
where $m_i$ and $n_i$ grow without bound as $i$ tends to infinity,
with $m_i\ge \alpha n_i$ eventually for any $\alpha$, and that this point
or these points are fixed under the action of $H_3$.
Using the formula for $\ahoro^+_{m,n}$ from Theorem~\ref{thm:heisenberg_busemen}
one can calculate that there is exactly one point of this form,
namely $\corner^{++}$, which is indeed a fixed point.

\bibliographystyle{plain}
\bibliography{nilpotent}

\end{document}


TITLE: Balls are randomly taken out of a box of 4 red and 6 green balls, without replacement, until we have a ball of each colour.
QUESTION [1 upvotes]: What is the probability of

Needing exactly two draws
Needing two or three draws
Needing to make exactly three draws if the first two balls drawn are of the same colour

For part 1. I computed that the number of ways we can choose a red followed by a green or vice versa is $|A|=6\cdot4$. The sample space itself has $10\choose2$ items so $$P(A)=\frac{6\cdot4}{10\choose2}=\frac{8}{15}$$ Part 2. and 3. posed the same issue for me. I approached part 2. with two events $A$: It takes two throws, $B$: It takes three throws. Then those two events are mutually exclusive so $P(A\cup B)=P(A)+P(B)$ which we are looking for. Here is the issue for 2. and 3.. I don't know how to calculate the probability that the first two are of the same colour. I thought that it may be $$\frac{{6\choose4}+{4\choose2}}{10\choose2}$$ But even if this were right I don't know how to mathematically express the fact that the 3rd one must be different. My theory is that (assuming the above is correct) it may be $$\frac{{6\choose4}\cdot{4\choose1}+{4\choose2}\cdot{6\choose 1}}{10\choose3}$$
EDIT 1
I got solutions for part 1. and 3. but am still wrapping my head around 2. For 1. you have the event $A$: The balls are of different colours. For $A$ we have $|A|=6\cdot4+4\cdot6$ and a sample space $|\Omega|=10\cdot9$ hence $P(A)=\frac{8}{15}$ similar to the above. For 3. we need to find $P(B|C)$. We know that $$P(B|C)=\frac{P(B\cap C)}{P(C)}$$. Furthermore $P(B\cap C)=P(B)$ because $B\cap C$ is the event that three balls are chosen. If three balls are chosen then the first two must have been the same colour. Hence $$P(B|C)=\frac{P(B)}{P(C)}=\frac47$$
EDIT 2
Question 2. has been resolved as well. Given that all of the events involved are pairwise disjoint the answer is just the sum of the prob that two are drawn and three are drawn. $$P(A)+P(B)=\frac{12}{15}$$

REPLY [1 votes]: Alternative approach:
First see my comments, following the original question.  This response is an extension of those comments.  Normally, probability problems such as this are best attacked with Combinatorics, which is represented by Graham Kemp's answer.
However, the order that the questions (1,2,and 3) are posed is suspicious.  The problem composer may have intended that the problem solver try to see the relationships among the questions, and use that to guide not only the order that the questions are attacked, but whether Combinatorics might reasonably be bypassed.
The analysis below, will exclusively stay with a probability of events approach, rather than Combinatorics.  Also, the analysis below will give the details indicated in my comments.

Balls are randomly taken out of a box of 4 red and 6 green balls, without replacement, until we have a ball of each colour.

What is the probability of

Needing exactly two draws
Needing two or three draws
Needing to make exactly three draws if the first two balls drawn are of the same colour

I will attack question 1 first.  Then question 3.  Then question 2.
$\underline{\text{Question 1}}$
The only way of needing exactly two draws is if the 1st and 2nd draw are not the same color.  Letting $R$ denote the event of selecting a red ball, and $G$ denote the event of selecting a green ball, the probability of $RG$ occuring + probability of $GR$ occurring is
$$\left[\frac{4}{10} \times \frac{6}{9}\right]
+ \left[\frac{6}{10} \times \frac{4}{9}\right] = \frac{48}{90} = \frac{8}{15}. \tag1 $$
In the first LHS term, after the red ball is selected, there are $9$ balls left, $6$ of which are green.  So (1) above represents conditional probabilities.  That is, after the red ball was selected, the probability that the 2nd draw would be green had changed from $\displaystyle ~\frac{6}{10}~$ to $\displaystyle ~\frac{6}{9}$.

$\underline{\text{Question 3}}$
I advise attacking this question next, since it will be used to answer Question 2.
Let $E_x$ denote the event that the first two draws were the same color.  Let $E_y$ denote the event that the third draw is of a different color from the 1st $2$ draws (i.e. that the 3rd draw is the other color).  Then Question 3 is specifically asking for $p(E_y|E_x)$.
This means that you are being asked about two mutually exclusive events:

3rd draw green, assuming 1st two draws red.
Probability is $~\displaystyle \frac{6}{8}$, since at this point, it is assumed that there are $8$ balls left, $6$ of which are green.
3rd draw red, assuming 1st two draws green.
Probability is $~\displaystyle \frac{4}{8}.$

However, here you have a problem.  Although you can compute $p(E_y|E_{RR})$ and $p(E_y|E_{GG})$ the problem requires (in effect) a weighted average.  So, Bayes Theorem must be employed.
Using the syntax of $(R,S)$ to denote that events $R$ and $S$ both occur, formally, you must compute :
$$\frac{p\left(E_y,E_{RR}\right) + p\left(E_y,E_{GG}\right)}{p\left(E_{RR}\right) + p\left(E_{GG}\right)}. \tag2 $$
Letting $E_{3R}$ and $E_{3G}$ denote the events of 3rd draw red and 3rd draw green, respectively, (2) above equals
$$\frac{p\left(E_{3G},E_{RR}\right) + p\left(E_{3R},E_{GG}\right)}{p\left(E_{RR}\right) + p\left(E_{GG}\right)}. \tag3 $$
In (3) above, the numerator is
$$\left[\frac{4}{10} \times \frac{3}{9} \times \frac{6}{8}\right] + \left[\frac{6}{10} \times \frac{5}{9} \times \frac{4}{8}\right] = \frac{72 + 120}{720} = \frac{4}{15}.$$
In (3) above, the denominator does not need to be computed.  This is because the denominator, the probability of the first two draws being the same color, may be inferred from Question 1.  That is, Question 1 asked for the probability of the complementary event, and was computed to be $~\displaystyle \frac{8}{15}.$
Therefore, in (2) above, you immediately know that the denominator equals $~\displaystyle 1 - \frac{8}{15} = \frac{7}{15}.$
Therefore (3) above evaluates to
$$\frac{\frac{4}{15}}{\frac{7}{15}} = \frac{4}{7}. \tag4$$

$\underline{\text{Question 2}}$
This question is attacked last, because virtually no further Math computations are required here.  This question is asking for the combined probability of two mutually exclusive events:

Exactly $2$ draws needed.
From Question 1, probability equals $~\displaystyle \frac{8}{15}.$

Exactly $3$ draws needed.
This probability may be inferred, using the computations/analysis already done in Questions 1 and 3.

Let $E_1$ denote the event that the first two draws are the same color.  Let $E_2$ denote the event that the 3rd draw is a different color from the first $2$ draws.
Then, the probability of exactly $3$ draws being needed equals
$$p(E_1) \times p(E_2 | E_1).$$
As discussed in Question 3, $~\displaystyle p(E_1) = 1 -  \frac{8}{15} = \frac{7}{15}.$
As answered directly, in Question 3, $~\displaystyle p(E_2 | E_1) = \frac{4}{7}.$
Therefore, $~\displaystyle p(E_1) \times p(E_2 | E_1) =
\frac{7}{15} \times \frac{4}{7} = \frac{4}{15}.$
Therefore, the overall answer to question 2 is
$$\frac{8}{15} + \frac{4}{15} = \frac{12}{15} = \frac{4}{5}.$$

I do not normally recommend a probability of events approach to a problem of this sort.  Normally, a Combinatorics approach is best.  However, in my opinion, Probability Theory requires that the student's intuition be stretched.  In this unusual situation, because of the interplay between the questions, the probability of events approach is better suited to stretching the student's intuition.

Edit
Although I do not recommend that the following approach be used to (indirectly) answer Question 2, this approach will expand the student's intuition.  Since the answer to Question 2 is $~\displaystyle \frac{4}{5}~$, you can infer that the probability of the complementary event, needing more than $3$ draws, is $~\displaystyle \frac{1}{5}.~$
This implies that there is an alternate method of computing the answer to Question 2.  You could compute $1 - p(\text{needing more than 3 draws}).$
The probability of needing more than $3$ draws can be computed using Questions 1 and $3$ as follows:
First, you calculate the probability that $2$ draws are insufficient.  This is $~\frac{7}{15}.~$
Then, you calculate (in effect) the probability that $3$ draws are insufficient, given that $2$ draws were insufficient.
Question 3 had you compute the complementary probability, which was $~\displaystyle \frac{4}{7}$.
Therefore, the probability that $3$ draws are insufficient, given that $2$ draws are insufficient is $~\displaystyle \frac{3}{7}.$
Therefore, the overall probability of needing more than $3$ draws is $~\displaystyle \frac{7}{15} \times \frac{3}{7} = \frac{3}{15} = \frac{1}{5}.$
Therefore, Question 2 may be answered by reasoning that the probability of needing at most $3$ draws is $~\displaystyle 1 - \frac{1}{5} = \frac{4}{5}.$


TITLE: Alternative formula for parametric integration
QUESTION [0 upvotes]: Given a circle $x^2+y^2=R^2$ for which we would like to calculate its area
One way is using the function: $y=\sqrt{R^2-x^2}$ and integrating as follows:
$2\int_{-R}^R\sqrt{R^2-x^2}\,dx=\pi R^2$
The second way is to use parametric equation $x=R\cos t$ and $y=R\sin t$ with $0\le t \le 2\pi$ and integrate as follows:
$2\int_{\pi}^0y\frac{dx}{dt}\,dt=-2R^2\int_{\pi}^0\sin^2 t\,dt=\pi R^2$
Note that you need to integrate from $t=\pi$ to $t=0$ which corresponds to the integration from $x=-R$ to $x=R$.
However, I found another formula for parametric integration in one of my notebooks, and I don't know where does it come from, but it works like magic:
$\frac{1}{2}\int_{t_{1}}^{t_{2}} (x\frac{dy}{dt}-y\frac{dx}{dt})\,dt$
Not only that it works, with this formula you need to integrate from $t=0$ to $t=2\pi$ to get the right answer:
$\frac{1}{2}\int_{0}^{2\pi} R^2(\cos^2t+\sin^2t)\,dt=\pi R^2$.
Can someone prove this formula? And why do the integration can be done over the whole circle and not half of it?
Thank you

REPLY [0 votes]: The following formula:
$$\frac{1}{2}\int_{0}^{2\pi} R^2(\cos^2t+\sin^2t)\,dt=\pi R^2$$

is simply obtained by integration from $0 \to 2\pi$ of the infinitesimal slice:
$$\frac12 \rho \cdot \rho \cdot d\theta=\frac12 (R^2 \cos^2 \theta+R^2 \sin^2 \theta)\cdot d\theta=\frac12 R^2 (\cos^2 \theta+\sin^2 \theta) \cdot d\theta=\frac12 R^2 \cdot d\theta$$


TITLE: A cochain-level Adem relation?
QUESTION [18 upvotes]: The original paper on Steenrod squares, Steenrod's "Products of cocycles and extensions of mappings", 1947, uses an explicit combinatorial formula for the squares in terms of simplicial cochains: given a simplicial cochain $\alpha$ on some simplicial set, Steenrod defines a cochain $\mathrm{Sq}^i\alpha$; if $\alpha$ is a cocycle, so is $\mathrm{Sq}^i\alpha$, and the cohomology class of $\mathrm{Sq}^i\alpha$ only depends on that of $\alpha$, and so $\mathrm{Sq}^i$ descends to a (linear!) map on cohomology. (In general, $\mathrm{Sq}^i$ is not linear on cochains and does not commute with $\mathrm d$, so this is a nontrivial statement. But it is not too hard to find an operation that agrees with $\mathrm{Sq}^i$ on cocycles and that does commute with $\mathrm d$, and its failure to be linear, it is not hard to show, is a total derivative.) (Steenrod makes his definition with $\mathbb Z$-coefficients, but I will only care about $\mathbb Z/2$ coefficients, and so assume that throughout.)
The well-known Adem relations, due originally to Adem,  "The iteration of the Steenrod squares in algebraic topology", 1952 and beautifully explained by Bullett and Macdonald, "On the Adem relations", 1982, are relations of the form $0 = \sum_{\text{certain pairs }(i,j)} \mathrm{Sq}^i \mathrm{Sq}^j$ that hold in cohomology. The first few are
 $\mathrm{Sq}^1 \mathrm{Sq}^1 = 0$, $\mathrm{Sq}^1\mathrm{Sq}^2 = \mathrm{Sq}^3 \mathrm{Sq}^0$, $\mathrm{Sq}^1 \mathrm{Sq}^3 = 0$, $\mathrm{Sq}^3 \mathrm{Sq}^1 = \mathrm{Sq}^2 \mathrm{Sq}^2$, ...
At the cochain level, these relations do not hold on the nose. The fact that they hold on cohomology just says that, if $\alpha$ is a cocycle, then $\sum_{\text{certain pairs }(i,j)} \mathrm{Sq}^i (\mathrm{Sq}^j(\alpha))$ is a coboundary, which of course depends on $\alpha$ (and on the particular relation in question).
Is there a combinatorial formula (like the one Steenrod gives for $\mathrm{Sq}^i$) that takes in $\alpha$ a cocycle and produces a primitive for $\sum_{\text{certain pairs }(i,j)} \mathrm{Sq}^i (\mathrm{Sq}^j(\alpha))$? Where can I find it? I care only about the early Adem relations up to $\mathrm{Sq}^3\mathrm{Sq}^1 = \mathrm{Sq}^2\mathrm{Sq}^2$.
A related question was asked by Kapustin. Note that I am not after more categorical descriptions of the Adem relations and the Steenrod squares like those available in the many excellent answers to this old MO question — I really do want a combinatorial description that fits into the framework of Steenrod's original paper.

REPLY [4 votes]: Explicit homotopies for a cochain-level Adem relation were first worked out in:
Greg Brumfiel, Anibal M. Medina-Mardones, John Morgan. A Cochain Level Proof of Adem Relations in the Mod 2 Steenrod Algebra. 2020
The odd-prime case is still open.


TITLE: principal ideal and ideal generated by subset S
QUESTION [0 upvotes]: Let R be a commutative ring with unity and a in R. Then $<a>$=aR=Ra
We can easily show that aR=Ra, aR ⊆ $<a>$ and Ra ⊆ $<a>$
Next I am trying to show that $<a>$ ⊆ aR
We khow that If S={a} then
$<a>$={na+ra+as+xay|n∈Z and r,s,x,y∈R}
let x ∈ $<a>$ then
x=na+ra+as+xay for some n∈Z and r,s,x,y∈R.
next I want to show that x is also belongs to aR.
But I got stuck.I can workout the part $  ra+as+xay  ∈ aR $ but but I got stuck because of $na$ .I hope someone help me with problem.

REPLY [0 votes]: Since $R$ is a ring with unity, $n$ can be replaced by a sum of $1$s and $-1$s (eg $2 a = a + a = (1 + 1) \cdot a$).
Now say $x=na + ra + as + xay$, by commutativity we get
$$x = na + ra + sa +  xy a = \underbrace{(n + r + s + xy)}_{\in R} a$$


TITLE: Problem about Hahn-Banach Theorem
QUESTION [2 upvotes]: Could someone help me with the following problem?
Let be $X$ a real normed space , $f,g\in{X^*}$ ($\|f\|=\|g\|=1$) and  $0<r\leq1$  such that $|f(x)|\leq{r}$ $\forall{x\in{\ker(g)}},\|x\|\leq{1}$.
Prove that either $\|f-g\|\leq{2r}$ or $\|f+g\|\leq{2r}$.
I am trying to use Hahn-Banach Theorem to $f_{\ker(g)}$, but I don't get nothing .
Thanks.

REPLY [2 votes]: Well, as long as we're willing to change the question, it's not hard to show that $\|f+g\|\le 1+r$ or $\|f-g\|\le 1+r$.
Let $Z$ be the kernel of $g$. We have $\|f|_Z\|\le r$,  so Hahn-Banach shows that there exists $F$ with $F|_Z=f|_Z$ and $\|F\|\le r$.
Now $F-f$ vanishes on the kernel of $g$, so  there exists $c$ with $$F-f=cg.$$Note that $|c|=\|F-f\|\le 1+r$.
If $0\le c\le 1+r$ then $$\|f+g\|\le\|f+cg\|+|c-1|\le r+1.$$
Similarly if $-1-r\le c\le 0$ then $\|f-g\|\le 1+r$.


TITLE: Required bits to communicate a partial order?
QUESTION [1 upvotes]: Suppose that you have a ranking (i.e. a strict complete partial order) over $n$ different objects, so that the objects can be ordered as $a>b>\cdots>n$. You want to communicate the exact order you have to a outside server. The question is how many bites do you need to do it? We are interested in the minimal number of bits that the task requires.
My initial guess is that you need $\frac{n(n-1)}{2}$ bites to transfer the information, because the partial order is equivalent to pairwise comparison among all the elements of ${1,...,n}$. One can verify that there are exactly $\frac{n(n-1)}{2}$ such binary relations.
Yet a recent paper claims that the partial order can be transmited in $\log_2(n!)$ bits. I understand that this follows from Shannon's entropy, but I just can't see the precise way to allocate the binary digits to express the ranking. Any ideas or explanations?
Intuitive answers are very welcome. Thanks,

REPLY [2 votes]: Assuming that the ranking is strict and linear, transmitting it amounts to transmitting a permutation $\sigma$ of $[n]=\{1,2,\ldots,n\}$. For $k,\ell\in[n]$ write $k\prec\ell$ if and only if $k$ precedes $\ell$ in $\sigma$. For each $k\in[n]$ let
$$p(k)=|\{\ell\in[n]:\ell<k\text{ and }\ell\prec k\}|\;;$$
$0\le p(k)\le k-1$. The number $p(k)$ specifies where $k$ must be inserted amongst the numbers $\{1,\ldots,k-1\}$, so $\sigma$ can be reconstructed from the sequence $\langle p(2),p(3),\ldots,p(n)\rangle$, which can be expressed in
$$\sum_{k=2}^n\lceil\lg k\rceil\approx\lg(n!)$$
bits.


TITLE: How a set whose existence is guarantied by one formula can be substituted into another formula?
QUESTION [1 upvotes]: I try to combine the Power Set Axiom and the Axiom Schema of Specification to prove that for any given set there exist another set that contains the given one as its only element. Moreover, I would like to do it as formal / strict as possible.
So, I start from the following two statements:
$\forall X : \exists P : \forall p : [(p \in P) \Leftrightarrow (p \subseteq X)]$
$\forall X : \forall B : \exists Y : \forall y : [(y \in Y) \Leftrightarrow ((y \in B) \wedge (y = X))]$
One can think of these two statements as "functions". In the first case we take a set X and for this set we can "generate" set P. In the second case we take sets X and B and "generate" set Y. Now I want to combine these "functions". In more details, in the second formula for all X I would like to use its P (as defined by the first formula) instead of B. Or, in other words, in the second formula for any X, I can use any B and I decided to use P as B.
So, the first part of my question is: How can I substitute B by P formally? In other words, how can I write P instead of B in the second formula such that P comes in there together with its definition provided by the first formula?
The second part of my question (assuming that we can write a formula described by the first part of the question) is if we can make such step-by-step transformations of this formula to get the following formula:
$\forall X : \exists Y : \forall y : [(y \in Y) \Leftrightarrow (y = X)]$

REPLY [2 votes]: This is accomplished via existential instantiation, with universal instantiation also playing a key role in the proof.
Specifically, we reason as follows:

Introduce a symbol $u$ standing for an arbitrary set; we want to show, using the $\mathsf{ZFC}$ axioms, that $\exists Z\forall z(z\in Z\leftrightarrow z=u)$. If we can do this we'll apply universal generalization to conclude $\forall X\exists Z\forall z(z\in Z\leftrightarrow z=X)$ as desired.

By powerset and universal instantiation, we get $\exists P\forall p(p\in P\leftrightarrow p\subseteq u)$.

Now we apply existential instantiation: this lets us introduce a new symbol $v$ and the formula $\forall p(p\in v\leftrightarrow p\subseteq u)$.

And this $v$ we can plug directly into an instance of the separation scheme as desired, via - again - universal instantiation.

Of course there are many different proof systems out there (e.g. compare natural deduction and Hilbert systems), but the above sketch will go through with only minor elaboration in any of them that I'm aware of.


TITLE: Bragg-Williams theory of phase transition
QUESTION [2 upvotes]: Hi I am currently studying Bragg wiliam theory and I dont understand how they derive the equations f(T,m). Actually i dont understand how they got S/N given by equation 8. Can someone explain?

REPLY [2 votes]: Well, as usual the entropy is the logarithm of the number of microstates. If the total number of spins is $N$ and there are $N_{\rm up}$ "$+$ spins", then this corresponds to a total number of possible configurations given by $\frac{N!}{N_{\rm up}!(N-N_{\rm up})!} = {}^NC_{N_{\rm up}}$ (the number of ways of choosing which of the $N$ spins are "$+$ spins"). This is exactly the first identity in (8).
The second identity in (8) follows from the fact that the magnetization density is
$$
m= \frac{N_{\rm up} - N_{\rm down}}{N} = \frac{2 N_{\rm up} - N}{N} = 2 \frac{N_{\rm up}}{N} - 1,
$$
so that
$$
N_{\rm up} = \frac{(1+m)N}{2}.
$$
Of course, this way of computing the entropy totally ignores the fact that the different configurations do not have the same energy, which is why the Bragg-Williams theory only provides an approximation.
Then, they go on getting another approximation, this time for the energy, writing
$$
E = -J\sum_{\langle i,j \rangle} \sigma_i\sigma_j \approx -J\sum_{\langle i,j \rangle} m^2 = -\frac12 JNz m^2,
$$
since there are $\frac12 N z$ pairs of nearest neighbors.
Once they have the (approximate) entropy and energy, they compute the free energy by the usual thermodynamic relation.


text:
\begin{document}

\maketitle

\vspace{0.1cm}

{\bf Abstract:} We reduce the problem of the projective normality of polarized abelian varieties to check the rank of very explicit matrices. This allow us to prove some results on normal generation of primitive line bundles on abelian threefolds and fourfolds. We also give two situations where the projective normality always fails. Finally we make some conjecture. \\
{\bf Mathematics Subject Classifications (2000):} Primary, 14K05;
secondary, 14C20, 14K25, 14N05.\\ {\bf Key Words:} Abelian varieties,
 projective normality, theta functions.

\vspace{0.1cm}

\section{Introduction}\label{intro}

Let $(A,L)$ be a polarized abelian variety of dimension $g$. We are interested in studying the projective normality of the embedding given by the linear system $|L|$. The problem is completely solved when $g=1$ or $g=2$ (see \cite{bila}, \cite {la}, \cite{iy1}, \cite{fu}). For higher dimension, most of the existent results are refered to powers of ample line bundle:

\begin{enumerate}

\item If $L=M^n$ with $M$ ample and $n\geq 3$ then $L$ is normally generated (see \cite{bila}).

\item If $L=M^2$ with $M$ of characteristic $c$, then $L$ is normally generated if and only if no point of $t^*_{\bar{c}}K(L)$ is a base point of $M$ (see \cite{bila}).

\item If $L$ is of type $(2,d_2,\ldots,d_g)$ and $d_g\leq 4$  then $L$ is never normally generated; if $(A,L)$ is generic of type $(2,d_2,\ldots,d_g)$ and  $d_g>4$ then $L$ is normally generated (see \cite{ru}).

\end{enumerate}

On the other hand, the new  techniques of  Mukai-Fourier transform  and M-regularity have been applied by G.Pareschi an M.Popa  to attack these problems (see  \cite{papo1}, \cite{papo2}). However,  it  seems that they only  work well with powers of ample line bundles.

Thus, if  $L$ is primitive, very little is known. In \cite{iy}, J.Iyer proves that if $A$ is simple and $h^0(L)>2^g g!$ then $L$ is normally generated. First she prove that it suffices to study the $2$-normality. Then, she takes a suitable isogeny $A\lrw B$ such that $L$ descends to a principal polarization $M$ on $B$. In this way, the problem is reduced to check the surjectivity of some multiplication maps between translations of the line bundle $M$.

Here, by using the canonical theta functions, we develop this idea. The main theorem reduces the question of $2$-normality to check the rank of some matrices. Their elements are canonical theta functions evaluated in $0$. The most important fact is that we can check the projective normality of any line bundle making a explicit computation. We use this Theorem in two ways.

Firstly, we use the main Theorem to show two situations where the projective normality always fails. If $(A,L)$ is a $g$-dimensional polarized abelian variety of type $(d_1,\ldots,d_g)$ with some $d_i=2$, we prove:

\begin{enumerate}

\item If $d_j\leq 4$, $1\leq j\leq g$  then $L$ is not normally generated. This extends the result of E.Rubei in \cite{ru}.

\item If $h^0(L)=2^{g+1}$, then $L$ is not normally generated. \qed

\end{enumerate}

Secondly, by using a computer we exhibit explicit examples of projectively normal abelian varieties.   Note, that a necessary condition for the projective normality is $h^0(L)\geq 2^{g+1}-1$. Moreover, by the result of J.Iyer we only have to check the cases where $h^0(L)\leq 2^g g!$. Because the projective normality is an open condition, these examples allow us to prove the normal generation in the generic case for some fixed types. For abelian threefolds we prove the following:

\begin{enumerate}

\item If $(A,L)$ is generic of type $(d_1,d_2,d_3)$ with $h^0(L)\geq 15$ and
$
(d_1,d_2,d_3)\not\in \{(1,2,8),(1,3,6),(2,2,4),(2,4,4)\}
$
then $L$ is  normally generated.

\item If $L$ is of type $(1,2,8)$, $(2,2,4)$ or $(2,4,4)$ then it is never normally generated.

\end{enumerate}

Note, that the open case is when $L$ is of type $(1,3,6)$.

For abelian fourfolds we prove:

\begin{enumerate}

\item If $(A,L)$ is generic of type $(d_1,d_2,d_3,d_4)$ with $h^0(L)\geq 31$ and $$
\begin{array}{rl}
{(d_1,d_2,d_3,d_4)\not\in }&{\{(1,1,2,16),(1,2,2,8), (1,2,4,4),}\\
{}&{\,(1,3,3,6),(2,2,2,4), (2,2,4,4), (2,4,4,4)\}}\\
\end{array}
$$ then $L$ is normally generated.

\item If $L$ is of type $(1,1,2,16)$, $(1,2,2,8)$, $(1,2,4,4)$, $(2,2,2,4)$, $(2,2,4,4)$ or $(2,4,4,4)$ then it is never normally generated.

\end{enumerate}

Now the open case is when $L$ is of type $(1,3,3,6)$.

{\bf Acknowledgements: } I thank K. Hulek for introducing me in these topics.

{\bf Notations:} We will work over the field of the complex numbers. We will follow the definitions and notation of the book \cite{bila}.

\section{The main Theorem.}

Let $(B,M_0)$ be a principal polarized variety with $B=\CC^g/\Lambda$ and $M_0=L(H,\chi_0)$. Suppose that $M_0$ is of characteristic $0$. Let $\lambda_1,\ldots,\lambda_g,\mu_1,\ldots,\mu_g$ be the corresponding  symplectic base.

We will denote by $V_1$ and $V_2$ the real spaces generated respectively by $\lambda_1,\ldots,\lambda_g$ and $\mu_1,\ldots,\mu_g$. Any $x\in \CC^g$ decomposes in a unique way  $x=x_1+x_2$ with $x_i\in V_i$.  Let $\Lambda_i=V_i\cap \Lambda$.

Let $D=(d_1,\ldots,d_g)$ with $d_i|d_{i+1}$ be a type. Let $$\Lambda'=\langle \lambda_1,\ldots,\lambda_g,d_1\mu_1,\ldots,d_g\mu_g \rangle.$$ We consider the abelian variety $A=\CC^g/\Lambda'$. There is a natural projection $p:A\lrw B$. It is clear that the line bundle
$L=p^*M_0$ on $Y$ has type $D$ and  characteristic $0$.

Let $K_1=\{\frac{n_1}{d_1}\lambda_1+\ldots+\frac{n_g}{d_g}\lambda_g|\,n_i\in \ZZ\, ,1\leq i\leq g\}/\Lambda_1$. We have the following isomorphisms (see \cite{iy}):
$$
H^0(L)=\sum_{w\in K_1}H^0(t^*_w M_0); \qquad
H^0(L^2)=\sum_{w\in K_1}H^0(t^*_{\frac{w}{2}} M_0^2).
$$

\begin{teo}\label{iyer1}

The map $H^0(L)\otimes H^0(L)\lrw H^0(L^2)$ factorizes through the maps:
$$
\sum_{u\in K_1}H^0(t_{u}^*M_0)\otimes H^0(t_{-u+w}^*M_0)\lrw H^0(t_{\frac{w}{2}}^*M_0^2)
$$
where $w$ runs over $K_1$.

\end{teo}
{\bf Proof:} See \cite{iy}. \qed

Let us study explicitly the surjectivity of these factorizations. We will denote by $\xi^c_w$ the canonical theta functions for the line bundle $M_0$ and by $\vartheta^c_w$ the canonical theta functions of $M_0^2$. Let $u,w\in K_1$. We know that  $H^0(t_{u}^*M_0)$ and $H^0(t_{-u+w}^*M_0)$ are generated respectively by the functions $\xi_u^0$ and $\xi_{-u+w}^0$. Thus, from the Multiplication Formula (\cite{bila},7.1.3):
$$
\xi_u^0 \xi_{-u+w}^0=\sum_{z\in Z_2} \vartheta_{u+z-\frac{w}{2}}^0(0) \vartheta^{\frac{w}{2}}_z
$$
where $Z_2=B_2\cap V_1$. Now, we can obtain the following result:

\begin{teo}\label{principal}

The line bundle $L$ is $2$-normal if and only if for each $w\in K_1/2K_1$ the matrix:
$$
\left(\vartheta_{u+z-\frac{w}{2}}^0(0)\right)_{u\in K_1,z\in Z_2}
$$
has rank $2^g$.

\end{teo}
{\bf Proof:} From the previous arguments and Theorem \ref{iyer1} it follows that $L$ is $2$-normal exactly when the matrices
$$
\left(\vartheta_{u+z-\frac{w}{2}}^0(0)\right)_{u\in K_1,z\in Z_2}
$$
have rank $2^g$ for any $w\in K_1$. But, it is clear that if we take $w,w'\in K_1$ such that $w-w'\in 2K_1$ then the two corresponding matrices have the same rank. \qed

\section{Cases where the projective normality fails.}

In this section we will use the main Theorem to give two situations where the projective normality fails. We will suppose that there is a $d_j$ equal to $2$. In particular, let $d_{i_0}=2$ where $i_0$ is the minimum index verifying this condition. Take $w=\lambda_{i_0}/2$. We decompose $K_1$ in two subgroups:
$$
\begin{array}{l}
{K_{11}=\{\frac{n}{2}\lambda_{i_0}|\, n\in \ZZ\}/\Lambda_1}\\
{K_{12}=\{\frac{n_{i_0+1}}{d_{i_0+1}}\lambda_1+\ldots+\frac{n_g}{d_g}\lambda_g|\,n_i\in \ZZ\, ,i_0+1\leq i\leq g\}/\Lambda_1}\\
\end{array}
$$
We have the following lemma:

\begin{lemma}
With the previous notation, the  matrices:
$$
\left(\vartheta_{u+z-\frac{w}{2}}^0(0)\right)_{u\in K_1,z\in Z_2}\hbox { and }
\left(\vartheta_{u+z-\frac{w}{2}}^0(0)\right)_{u\in K_{12},z\in Z_2}
$$
have the same rank.
\end{lemma}
{\bf Proof:} Let $u=u_1+u_2$, with $u_1\in K_{11}$ and $u_2\in K_{12}$. Suppose that $u_1=\frac{\lambda_{i_0}}{2}$. Then, since $M_0$ is of characteristic $0$ and by using the properties of the canonical theta functions:
$$
\vartheta_{u+z-\frac{w}{2}}^0(0)=\vartheta_{\frac{\lambda_{i_0}}{2}+u_2+z-\frac{\lambda_{i_0}}{4}}^0(0)=\vartheta_ {-\frac{\lambda_{i_0}}{2}-u_2-z+\frac{\lambda_{i_0}}{4}}^0(0)=
\vartheta_{-u_2+z-\frac{w}{2}}^0(0)
$$ \qed

By this result we will center our attention in the matrix:
$$
Q=\left(\vartheta_{u+z-\frac{w}{2}}^0(0)\right)_{u\in K_{12},z\in Z_2}
$$
with $w=\frac{\lambda_{i_0}}{2}$.

We decompose the group $Z_2$ in :
$$
\begin{array}{l}
{Z_{21}=Z_2\cap \langle \frac{\lambda_{i_0}}{2} \rangle}\\
{Z_{22}=Z_2\cap \langle \frac{\lambda_{1}}{2},\ldots,\frac{\lambda_{i_0-1}}{2},\frac{\lambda_{i_0+1}}{2},\ldots,\frac{\lambda_{g}}{2} \rangle}\\
\end{array}
$$
this induces a decomposition for the matrix $Q$:
$$
Q=\left(\matrix{Q_1 & Q_2}\right)
$$
with
$$
Q_1=\left(\vartheta_{u+z-\frac{w}{2}}^0(0)\right)_{u\in K_{12},z\in Z_{22}},\qquad Q_2=\left(\vartheta_{u+z+\frac{\lambda_{i_0}}{2}-\frac{w}{2}}^0(0)\right)_{u\in K_{12},z\in Z_{22}}
$$

Let $u\in K_{12}, z\in Z_{22}$. Then:
$$
\begin{array}{rl}
{\vartheta_{u+z-\frac{w}{2}}^0(0)}&{=\vartheta_{u+z-\frac{\lambda_{i_0}}{4}}^0(0)=\vartheta_{-u-z+\frac{\lambda_{i_0}}{4}}^0(0)=}\\
{}&{=\vartheta_ {u+(-2u-z)+\frac{\lambda_{i_0}}{2}-\frac{\lambda_{i_0}}{4}}^0(0)}\\
\end{array}
$$
If $u$ is a $4$-torsion point of $K_{12}$ then $-2u-z$ is an element of $Z_{22}$. Therefore, in this case, the sum of the columns of $Q_1$ is the same that the sum of the columns of $Q_2$. We deduce that the matrix $Q$ has not maximal rank $2^g$. Thus, we have prove the following Theorem:

\begin{teo}\label{fail1}

Let $(A,L)$ be a $g$-dimensional polarized abelian variety of type $(d_1,\ldots,d_g)$, with $d_j\leq 4$, $1\leq j\leq g$ and some $d_i=2$. Then $L$ is not normally generated. \qed

\end{teo}

On the other hand, we consider the matrix:
$$
Q_1-Q_2=\left(\vartheta_{u+z-\frac{w}{2}}^0(0)-\vartheta_ {u+z+\frac{\lambda_{i_0}}{2}-\frac{w}{2}}^0(0)\right)_{u\in K_{12},z\in Z_{22}}
$$
We decompose $K_{12}$ in the following way:
$$
K_{12}=K_{12}^0\cup K_{12}^1
$$
where $K_{12}^0$ are the $2$-torsion points of $K_{12}$ and $K_{12}^1$ the rest of them. Each  $u\in K_{12}^1$ has a unique inverse $-u$ on $K_{12}^1$.  Furthermore:

- If $u\in K_{12}^0$ then $(Q_1-Q_2)_{u,z}=0$ for any $z\in Z_{22}$, because:
$$
\vartheta_{u+z+\frac{\lambda_{i_0}}{2}-\frac{\lambda_{i_0}}{4}}^0(0)=\vartheta_{-u-z-\frac{\lambda_{i_0}}{2}+\frac{\lambda_{i_0}}{4}}^0(0)=\vartheta_{u+z-\frac{\lambda_{i_0}}{4}}^0(0)
$$

- If $u\in K_{12}^1$ then $(Q_1-Q_2)_{u,z}=-(Q_1-Q_2)_{-u,z}$ for any $z\in Z_{22}$, because:
$$
\begin{array}{rl}
{\vartheta_{u+z-\frac{w}{2}}^0(0)-\vartheta_{u+z+\frac{\lambda_{i_0}}{2}-\frac{w}{2}}^0(0)}&{=\vartheta_{-u-z+\frac{w}{2}}^0(0)-\vartheta_{-u-z-\frac{\lambda_{i_0}}{2}+\frac{w}{2}}^0(0)=}\\
{}&{=\vartheta_{-u+z+\frac{\lambda_{i_0}}{2}-\frac{w}{2}}^0(0)-\vartheta_{-u+z-\frac{w}{2}}^0(0)}\\
\end{array}
$$

We see that   $Q_1-Q_2$ has at least $\#K_{12}^0$ null rows and $\frac{1}{2}\#K_{12}^1$ pairs of rows with opposite sign. Let $h^0(L)=2^{g+1-{i_0}} n$, then $\#K_{12}=2^{g-{i_0}} n$ and $\#K_{12}^0=2^{g-{i_0}}$. From this:
$$
rank(Q_1-Q_2)\leq \frac{1}{2}\#K_{12}^1=\frac{1}{2}(\#K_{12}-\#K_{12}^0)=2^{g-{i_0}-1} n-2^{g-{i_0}-1}
$$
When this rank is less than $2^{g-1}$ then $Q$ has not maximal rank, so $L$ is not normally generated:
$$
rank(Q_1-Q_2)<2^{g-1}\iff n<2^{i_0}+1 \iff n\leq 2^{i_0}\iff h^0(L)\leq 2^{g+1}
$$
Since a necessary condition for the projective normality is $h^0(L)\geq 2^{g+1}-1$, we have proof the following heorem:

\begin{teo}\label{fail2}

Let $(A,L)$ be a $g$-dimensional polarized abelian variety of type $(d_1,\ldots,d_g)$, with some $d_i=2$. If $h^0(L)=2^{g+1}$, then $L$ is not normally generated. \qed

\end{teo}

\section{Projective normality for $g=3$ and $g=4$.}

In this section we will use the main Theorem \ref{principal} to study the projective normality of abelian threefolds and fourfolds making explicit computations. Since this is an open condition we will give explicit examples where the projective normality holds. This will prove the projective normality in the generic case.

To apply the Theorem \ref{principal}, we have to compute $\vartheta_{c}^0(0)$ with $c\in V_1$. We will write this functions by using the classical theta functions. We will follow the Chapter 8 of \cite{bila}. Let $Z\in M_g(\CC)$ be a symmetric matrix with $Im(Z)>0$ and let $c\in \CC^g$ such that $c=Zc^1$, with $c^1\in \RR^g$. By the Lemma (8.5.2., \cite{bila}) and the properties of the theta functions we have:
$$
\begin{array}{rl}
{\vartheta_{c}^0(0)}&{=\vartheta_{0}^c(0)=\vartheta\left[\matrix{c^1 \cr 0}\right](0,Z)=}\\
{}&{=\e{\pi \imath ^tc^1Zc^1}\vartheta\left[\matrix{0 \cr 0}\right](Zc^1,Z)=\e{\pi \imath ^tc^1Zc^1}\theta(Z,Zc^1)}\\
\end{array}
$$
where $\theta$ is the Siegel theta function:
$$
\theta(Z,v)=\sum_t \e{\pi \imath ^tt Z t +2 \pi\imath ^t t v}
$$
Now, we can rewrite the Theorem \ref{principal}.
Let $I=\{(\frac{n_1}{d_1},\ldots,\frac{n_g}{d_g})|\,0\leq n_i< d_i\, , 1\leq i\leq g\}$, $I'=\{(\frac{n_1}{d_1},\ldots,\frac{n_g}{d_g})|\,0\leq n_i< m.c.d.(d_i,2)\, , 1\leq i\leq g\}$ and $J=\{(\frac{a_1}{2},\ldots,\frac{a_g}{2})|,0\leq a_1,\ldots,a_g\leq 1\}$. Then:

\begin{teo}

The line bundle $L$ is $2$-normal if and only if for any $w\in I'$ the matrices:
$$
\left(\e{\pi\imath^t (i+j-\frac{w}{2})Z(i+j-\frac{w}{2})}\theta(Z,Z(i+j-\frac{w}{2}))\right)_{i\in I,j\in J}
$$
have rank $2^g$. \qed

\end{teo}

The classical Theta functions can be evaluated with the help of a computer. In particular Mathematica 5.0 (\cite{wo}) includes an implementation of the Siegel theta function. Furthermore, in \cite{deheboho} algorithms for their computation  are given. However, one can simplify the situation working on a particular case. We can suppose that $Z=X+Y$ where $X$ is a symmetric integer matrix and $Y=k Id$, with $k\in \CC$ and $Im(k)>0$. Now, by the Theta Transformation Formula (8.6.1, \cite{bila}):
$$
\e{\pi \imath ^tc^1Zc^1}\vartheta\left[\matrix{0 \cr 0}\right](Zc^1,Z)=\e{\pi \imath ^tc^1Zc^1}\vartheta\left[\matrix{0 \cr 0}\right](Zc^1,Y)
$$
Because $Y$ is a diagonal matrix, the functions $\vartheta\left[\matrix{0 \cr 0}\right](Zc^1,Z)$ can be decomposed in a product of elliptic functions.

We have applied this method for $g=3$, and the decomposition:
$$
X=\left(\matrix{0 & 0 & 1 \cr 0 & 0 & 2 \cr 1 & 2 & 0}\right);\qquad
Y=(1+\sqrt{1/3} \imath) Id $$
We have checked all the possibilities for the type $(d_1,d_2,d_3)$ when $7\leq d_1d_2d_3\leq 2^3\cdot 3!$. In other case, the results of Iyer \cite{iy} provide the projective normality. We have obtained the following result:

\begin{teo}

Let  $(A,L)$ be a generic abelian threefold of type $(d_1,d_2,d_3)$ with $h^0(L)\geq 15$. If  $
(d_1,d_2,d_3)\not\in \{(1,2,8),(1,3,6),(2,2,4),(2,4,4)\}$ then $L$ is normally generated.

\end{teo}

Moreover, applying the Theorems \ref{fail1} and \ref{fail2} for $g=3$ we obtain the following corollary:

\begin{cor}

An abelian threefold $(A,L)$ of type $(1,2,8),(2,2,4)$ or $(2,4,4)$ is never projectively normal. \qed

\end{cor}

For $g=4$ we have used the  decomposition:
$$
X=\left(\matrix{0 & 0 & 0 & 1 \cr 0 & 0 & 0 & 2 \cr 0 & 0 & 0 & 3 \cr 1 & 2 &3 & 0}\right);\qquad
Y=(1+\sqrt{1/3} \imath) Id
$$
Now, we have checked all the possibilities for the type $(d_1,d_2,d_3,d_4)$ when $15\leq d_1d_2d_3d_4\leq 2^4\cdot 4!$. We have obtained the following results:

\begin{teo}
Let  $(A,L)$ be a generic abelian fourfold of type $(d_1,d_2,d_3,d_4)$ with $h^0(L)\geq 31$. If  $$
\begin{array}{rl}
{(d_1,d_2,d_3,d_4)\not\in }&{\{(1,1,2,16),(1,2,2,8), (1,2,4,4),}\\
{}&{\,(1,3,3,6),(2,2,2,4), (2,2,4,4), (2,4,4,4)\}}\\
\end{array}
$$ then $L$ is normally generated. \qed
\end{teo}

In this case, from  Theorems \ref{fail1} and \ref{fail2} it follows:

\begin{cor}

If $(A,L)$ is an abelian fourfold of type $(1,1,2,16)$, $(1,2,2,8)$, $(2,2,2,4)$, $(2,2,4,4)$ or $(2,4,4,4)$ then $L$ is not normally generated. \qed

\end{cor}

Note, that the open cases for $g=3$ and $g=4$ are respectively the types $(1,3,6)$ and $(1,3,3,6)$. We have not found examples of projectively normal polarized abelian varieties of theses types. On the other hand, we have explored some other cases for $g>4$. Thus, we make the following conjectures:

\begin{conj}

Let $(A,L)$ be a polarized abelian variety of dimension $g$. If $L$ is of type $(1,3,\cdots,3,6)$ then $L$ is not normally generated. \qed

\end{conj}

\begin{conj}

Let $(A,L)$ be a polarized abelian variety of dimension $g$. If $L$ is of type $(1,\cdots,1,d)$ with $d\geq 2^{g+1}-1$ then $L$ is  normally generated. \qed

\end{conj}

config: arxiv


TITLE: Question o Trigonometrical equation
QUESTION [0 upvotes]: I'm know confused by an equation involving $tan$
Given: $tan\bigl(x - \dfrac{\pi}{4}) \le 1$ for $-\pi < x < \pi$.
Adjusting the bound for $x - \dfrac{\pi}{4}$ it becomes $\dfrac{-5\pi}{4} < x < \dfrac{3\pi}{4}$.
Finding $x = 1$ boundaries, $x = \dfrac{-3\pi}{4}, \dfrac{\pi}{4}$.
Resulting in intervals with $tan\bigl(x - \dfrac{\pi}{4}) <= 1$ are:
$-\pi < x < -\dfrac{\pi}{2}, -\dfrac{\pi}{4} < x < \dfrac{\pi}{2}, \dfrac{3\pi}{4} < x < \pi$
But wolfram gives different results for where all my $\dfrac{\pi}{2}$ bound should be $\dfrac{\pi}{4}$.
Can someone show me what's wrong ?

REPLY [1 votes]: At the beginning, you may be wrong.
Let $x-\frac{\pi}{4}=X$, then we have
$$-\frac{5}{4}\pi\lt X\lt \frac{3}{4}\pi.$$
Don't confuse with $x$. You then need to solve for $X$ such that
$$-\frac{5}{4}\pi\lt X\lt \frac{3}{4}\pi\ \ \text{and}\ \ \tan X\le 1.$$


TITLE: Limits for sequences and the $e$ method
QUESTION [2 upvotes]: After doing a lot of homework, I've realized that I don't really get how to use the $\left(1+\frac{1}{n}\right)^n\to e$ method, which shows in how I can't solve indeterminate limits of the type $1^\infty$. For example:
$$\lim_{n\to\infty}{\left(\frac{1}{2}+\frac{2}{n}\right)^n}$$
Then I would do as follows:
$$\lim_{n\to\infty}{\left(\frac{1}{2}+\frac{1}{2}+\frac{2}{n}-\frac{1}{2}\right)^n}$$
$$\lim_{n\to\infty}{\left(1+\frac{4-n}{2n}\right)^n}$$
$$\lim_{n\to\infty}{\left(1+\frac{1}{\frac{2n}{4-n}}\right)^n}$$
And, from here on, I would not know what to do. Thus, the question is:
How am I supposed to work with the exponents?

REPLY [1 votes]: Another approach:
If $\lim_{x\to{+\infty}} f(x)^{g(x)}$ is as $costant^{+\infty}$ which is indeterminate form then: $$\lim_{x\to{+\infty}} f(x)^{g(x)}=e^{\lim_{x\to +\infty}\big(f(x)-1\big)g(x)}$$


TITLE: Isometries are affine transformations
QUESTION [1 upvotes]: I want to show that, if $(M,\mathrm{g})$ is a Riemannian manifold, $\nabla$ is the covariant derivative from the Levi-Civita connection, and $f:M\to M$ is an isometry, then $f_*(\nabla_XY)=\nabla_{f_*X}f_*Y$. I can demonstrate that this works if I resort to coordinate expressions, but I don't want to do that.
At first, I thought to use the fact that $\nabla$ is metric, so that $$\nabla_Zf^*\mathrm{g}(X,Y)=\mathrm{g}(f_*(\nabla_ZX),f_*Y)+\mathrm{g}(f_*X,f_*(\nabla_ZY))=\mathrm{g}(\nabla_ZX,Y)+\mathrm{g}(X,\nabla_ZY),$$ but I can't seem to get anywhere with it.
Any hints as to how I might approach this?

REPLY [3 votes]: Assume that $f\colon M\to M$ is a diffeomorphism satisfying $f^*g  = g$.
It's not enough just to use the fact that $\nabla$ is a metric connection; you also have to use the fact that it's torsion-free. The idea of the proof is to define $\widetilde \nabla\colon \mathfrak X(M)\times \mathfrak X(M) \to \mathfrak X(M)$ by
$$
\widetilde \nabla_XY = (f^{-1})_*(\nabla_{f_*X}{f_*Y}).
$$
Then  you can show that $\widetilde\nabla$ is a connection on $M$ that is compatible with $g$ and torsion-free. By the uniqueness of the Riemannian connection, $\widetilde \nabla$ must be equal to $\nabla$.