text:
\section{Statement of result and background}\label{intro}

Our main result is the following immersion theorem for real projective
spaces.

\begin{thm}
If $e\ge7$, then $RP^{2^e-1}$ can be immersed in ${\mathbb R}^{2^{e+1}-e-7}$. \label{mainthm}\end{thm}

This improves, in these cases, by 1 dimension upon the result of Milgram
\cite{Mil}, who proved, by constructing bilinear maps, that if $n\equiv7$
mod 8,
then $RP^n$ can be immersed in ${\mathbb R}^{2n-\alpha(n)-4}$, where $\alpha(n)$ denotes the
number of 1s in the binary
expansion of $n$. In \cite[Theorem 1.2]{DM}, the first and fourth authors used
obstruction theory to prove that if $n\equiv7$ mod 8, then
$RP^n$ can be immersed in ${\mathbb R}^{2n-D}$, where $D=14,16,17,18$ if
$\alpha(n)=7,8,9,\ge10$. That result, with $n=2^e-1$, is
stronger than ours for $ e\le 12$. If $e\ge13$, then our result
improves on the result of \cite{DM}
by $e-13$ dimensions.
Thus \fullref{mainthm} improves on all known results by 1 dimension if
$e\ge14$.

In \cite{J}, James proved that $RP^{2^e-1}$ cannot be immersed in
${\mathbb R}^{2^{e+1}-2e-\delta}$ where $\delta=3,2,2,4$ for
$e\equiv 0,1,2,3$ mod 4. In \cite{GM}, Gitler and  Mahowald
announced an immersion result for $RP^{2^e-1}$
in dimension 1 greater than
that of James' nonimmersion, which would have been optimal. However, a
mistake in the argument of \cite{GM} was
pointed out by Crabb and Steer. The approach of our paper was initiated
by Mahowald around 1970 in an unpublished attempt to prove an optimal immersion of
$RP^{2^e-1}$. In order to improve our result to this extent, we would need
to show compatibility of our liftings with liftings given by the
Radon-Hurwitz theorem (\cite{Eck}).

\section{Outline of proof}\label{outlinesec}
In this section we outline the proof of \fullref{mainthm}. In subsequent
sections, we will fill in details.

If $\theta$ is a vector bundle over a compact connected space $X$, we
define the geometric dimension of $\theta$, denoted ${\on{gd}}(\theta)$, to be
the fiber dimension of $\theta$ minus the maximum number of linearly
independent sections of $\theta$. Equivalently, if $\dim(\theta)=n$, then
${\on{gd}}(\theta)$ equals the smallest integer $k$ such that the map
$X\mapright{\theta} BO(n)$ which classifies $\theta$ factors through
$BO(k)$.
The following lemma is standard (See eg Sanderson \cite[Theorem 4.2]{San}).
Here and throughout, $\xi_n$ denotes the Hopf line bundle over $RP^n$. We
will often write $P^n$ instead of $RP^n$, and will denote
the stunted space $P^n/P^{k-1}$ as $P^n_k$.
\begin{lem} Let $\phi(n)$ denote the number of positive integers $i$
satisfying $i\le n$ and $i\equiv0,1,2,4 \mod 8$.
Suppose $n>8$. Then $RP^n$ can be immersed in ${\mathbb R}^{n+k}$ if and only if
${\on{gd}}((2^{\phi(n)}-n-1)\xi_n)\le k$.\end{lem}
Thus \fullref{mainthm} will follow from the following result, to the proof
of which the remainder of this
paper will be devoted.

\begin{thm}\label{gdthm} If $e\ge7$, then
${\on{gd}}((2^{2^{e-1}-1}-2^e)\xi_{2^e-1})\le 2^e-e-6$.\end{thm}

The bulk of the work toward proving \fullref{gdthm} will be a determination
of upper bounds
for ${\on{gd}}(2^e\xi_n)$ for all $n\equiv7$ mod 8 by induction on $e$, starting with $e=7$. A similar
method could be employed for
all $n$, but we restrict to $n\equiv7$ mod 8 to simplify the already
formidable arithmetic. We let $A_k=RP^{8k+7}$,
and denote ${\on{gd}}(m\xi_{8k+7})$ by ${\on{gd}}(m,k)$.

The classifying map for $2^e\xi_{8k+7}$ will be
viewed as the following composite.
\begin{equation}\label{comp}A_k\mapright{d} (A_k\times
A_k)^{(8k+7)}\hookrightarrow\bigcup_j A_j\times A_{k-j}\mapright{f\times f}
BO_{2^{e-1}}\times BO_{2^{e-1}}\to BO_{2^e}.\end{equation}
Here $d$ is a cellular map homotopic to the diagonal map, $X^{(n)}$ denotes
the $n$--skeleton of $X$, and $f$ classifies
$2^{e-1}\xi$.
We write $BO_m$ for $BO(m)$ for later notational convenience.

As a first step, we would like to use \eqref{comp} to deduce that
$${\on{gd}}(2^e,k)\le\max\{{\on{gd}}(2^{e-1},j)+{\on{gd}}(2^{e-1},k-j):0\le j\le k\}.$$
In order to make this deduction, we need to know that the liftings of the
various $2^{e-1}\xi_{8j+7}$ to various
$BO_m$ have been made compatibly.

\begin{defin} If $\theta$ is a vector bundle over a filtered space
$X_0\subset\cdots\subset X_k$,
we say that $${\on{gd}}(\theta|X_{i})\le d_i\text{ compatibly for }i\le k$$ if
there is a commutative diagram
$$\begin{CD} X_{0}@>>> X_{1} @>>>\cdots@>>>X_{k}@.{}\\
@VVV @VVV @. @VVV@.\\
BO_{d_0}@>>> BO_{d_1} @>>> \cdots @>>> BO_{d_k}@>>>
BO_{\dim(\theta)}\label{compatdef}\end{CD}$$
where the map $X_k\to BO_{\dim(\theta)}$ classifies $\theta$, and the
horizontal maps are the usual inclusions.\end{defin}

\begin{rmk}
{\rm In our filtered spaces, we always assume that the inclusions are cofibrations.}
\end{rmk}

\begin{rmk}
{\rm Isomorphism classes of $n$--dimensional vector bundles over $X$ correspond to homotopy classes of maps of $X$ into $BO_n$. Thus one would
initially say that the diagram in \fullref{compatdef} commutes up to homotopy. However, by \fullref{xx}, we may interpret this diagram, and
other homotopy commutative diagrams that occur later, as being strictly commutative. To apply the lemma, we will often, at the outset, replace
maps $BO_n\to BO_{n+k}$ by homotopy equivalent fibrations.}
\end{rmk}

\begin{lem}\label{xx} If
$$\begin{CD} A@>f>> E\\ @ViVV @VVpV\\
X@>>g> B\end{CD}$$ commutes up to homotopy and $p$ is a fibration, then
$f$ is homotopic to a map $f'$ such that $p\circ f'=g\circ i$.\end{lem}

\begin{proof} Let $H \co A\times I\to B$ be a homotopy from $p\circ f$ to
$g\circ i$.
By the definition of fibration, there exists $\widetilde H \co A\times I\to E$
such that $p\circ\widetilde H=H$ and $\widetilde H|A\times 0=f$. Then
$\widetilde H|A\times 1$ is our desired $f'$.\end{proof}

If $X_0\subset\cdots\subset X_k$
and $Y_0\subset\cdots\subset Y_k$ are filtered spaces, we define, for $0\le
i\le k$,
$$(X\times Y)_i:=\bigcup_{j=0}^i X_j\times Y_{i-j}.$$
Then $(X\times Y)_0\subset\cdots\subset(X\times Y)_k$ is clearly a filtered space. We will prove the following general result in Section
\fullref{compatsec}.

\begin{prop}\label{compatprop} Suppose ${\on{gd}}(\theta|X_{i})\le d_i$
compatibly for $i\le k$ and
${\on{gd}}(\eta|Y_{i})\le d'_i$ compatibly for $i\le k$. For $0\le j\le k$, let
$e_j=\max(d_i+d'_{j-i}: 0\le i\le j)$. Then
${\on{gd}}(\theta\times\eta|(X\times Y)_{j})\le e_j$ compatibly for $j\le k$.
Moreover, if $X=Y$ and $\theta=\eta$, then the maps $(X\times
X)_{j}\mapright{f} BO_{e_j}$ can be chosen to satisfy $f\circ T=f$, where
$T \co X\times X\to X\times X$ interchanges factors.\end{prop}

We will begin an induction by deriving in \fullref{128prop} some compatible bounds for ${\on{gd}}(128,i)$. Proposition \fullref{compatprop} will,
after restriction under the diagonal map, allow us to prove ${\on{gd}}((\sum 2^{e_i})\xi_n)\le\max\{\sum{\on{gd}}(2^{e_i}\xi_{m_i}):\sum m_i=n\}$. These
bounds are not yet strong enough to yield new immersion results.
We must improve the bounds by taking advantage of paired obstructions. The following result will be proved in Section \fullref{compatsec}.

\begin{prop}\label{pairedprop} Let $BO_n[\rho]$ denote the pullback of
$BO_n$ and the $(\rho-1)$-connected cover $BO[\rho]$ over $BO$, and let
$s=\min(\rho+2m-1,4m-1)$.
\begin{enumerate}
\item There are equivalences $c'_1$ and $c'_2$ such that the following
diagram commutes.
$$\begin{CD} BO_{2m}[\rho]^{(s)}@>q_1>>
(BO_{2m}[\rho]/BO_{2m-1}[\rho])^{(s)}@>c'_1>>S^{2m}\\
@Vp_2VV @Vp_2'VV @ViVV\\
BO_{2m+1}[\rho]^{(s)}@>q_2>>
(BO_{2m+1}[\rho]/BO_{2m-1}[\rho])^{(s)}@>c'_2>>\Sigma P^{2m}_{2m-1}.
\end{CD}$$
Preparatory to the next two parts, we expand this diagram as follows, with
$c_i=c_i'\circ q_i$ and $(X,A)$ a finite
CW pair.
$$\begin{CD} A@>f_1>>BO_{2m-1}[\rho]^{(s)}@.{}\\
@VjVV @Vp_1VV @.\\
X@.BO_{2m}[\rho]^{(s)}@>c_1>> S^{2m}\\
@. @Vp_2VV @ViVV\\
{}@. BO_{2m+1}[\rho]^{(s)} @>c_2>>\Sigma P_{2m-1}^{2m}.
\end{CD}$$
\item Suppose $\dim(X)<s$, and we are given $X\mapright{f}
BO_{2m}[\rho]^{(s)}$ such that $f\circ j=p_1\circ f_1$ and $c_1\circ f$
factors as
$X\to X/A\mapright{g} S^{2m}$ with $[g]$ divisible by $2$ in
$[X/A,S^{2m}]$.\footnote{Note that $[X/A,S^{2m}]$ is in the stable range,
from
which it gets its group structure.} Then $p_2\circ f$ lifts to a map
$X\mapright{\ell}BO_{2m-1}[\rho]^{(s)}$ whose restriction to $A$ equals
$f_1$.
\item Suppose, on the other hand, $\dim(X)\le s$, and we are given
$X\mapright{f'} BO_{2m+1}[\rho]^{(s)}$ such that $f'\circ j=p_2\circ
p_1\circ f_1$ and $c_2\circ f'$ factors as $X\to X/A\mapright{g'}\Sigma
P_{2m-1}^{2m}$ with $[\Sigma g']$ divisible by $2$ in the stable group
$[\Sigma X/A,\Sigma^2 P_{2m-1}^{2m}]$. Then $f'$ is homotopic rel $A$ to a
map which lifts to $BO_{2m}[\rho]^{(s)}$.
\end{enumerate}
\end{prop}

In Section \fullref{indnsec}, we will implement \fullref{compatprop} and \fullref{pairedprop} to prove that the last part of
the following important result follows by induction on $e$ from the first five parts and its validity when $e=7$, while in Section
\fullref{numbersec}, we will establish the first five parts.

\begin{thm}\label{gthm} There is a function $g(e,k)$ defined for $e\ge7$
and $k\ge0$ satisfying the following.
\begin{enumerate}
\item If $k\ge2^{e-3}$, then $g(e,k)=2^e$.
\item For all $e$, $g(e,0)=g(e,1)=0$, and, if $2\le k\le 2^e$, then  $g(e,k)\ge 4k+4$.
\item If $0\le \ell\le k/2$, then $g(e+1,k)\ge g(e,\ell)+g(e,k-\ell)-1$.
\item If, for some $\ell$ with $0\le\ell\le k/2$, we have $g(e+1,k)=g(e,\ell)+g(e,k-\ell)-1$, then, for all $\ell$ with $0\le\ell\le (k-1)/2$, we have
$g(e,\ell)+g(e,k-1-\ell)<g(e+1,k)$ and, if also $k$ is even, then $g(e+1,k)\ge 2g(e,k/2)+1$.
\item For all $e$ and $k$, $g(e,k)\ge g(e,k-1)$. \item ${\on{gd}}(2^e,k)\le g(e,k)$ compatibly for all $k$.
\end{enumerate}\end{thm}

The function $g$ will be defined in  \eqref{complement} and \fullref{geq}. In \fullref{gtbl}, we list its values for
small values of the parameters. We prefer not to tabulate the values $g(e,k)=2^e$ when $k>2^{e-3}$.

\begin{table} [ht!]
\begin{center}
\small{
\begin{tabular}{lr|rrrrrrrrrrrrrrrr}
&&&&&&$k$&&&&&&&&&&&\\
&&$1$&$2$&$3$&$4$&$5$&$6$&$7$&$8$&$9$&$10$&$11$&$12$&$13$&$14$&$15$&$16$\\
\hline
&$7$&$0$&$16$&$19$&$32$&$35$&$48$&$51$&$64$&$67$&$80$&$83$&$96$&$99$&$112$&$115$&$128$\\
&$8$&$0$&$15$&$18$&$32$&$34$&$47$&$50$&$64$&$66$&$79$&$82$&$96$&$98$&$111$&$114$&$128$\\
$e$&$9$&$0$&$14$&$17$&$31$&$33$&$46$&$49$&$64$&$66$&$78$&$81$&$95$&$97$&$110$&$113$&$128$\\
&$10$&$0$&$13$&$16$&$30$&$32$&$45$&$48$&$63$&$65$&$77$&$80$&$94$&$96$&$109$&$112$&$128$\\
&$11$&$0$&$12$&$16$&$29$&$31$&$44$&$47$&$62$&$64$&$76$&$79$&$93$&$95$&$108$&$111$&$127$\\
&$12$&$0$&$12$&$16$&$28$&$30$&$43$&$46$&$61$&$63$&$75$&$78$&$92$&$94$&$107$&$110$&$126$\\
&$13$&$0$&$12$&$16$&$27$&$29$&$42$&$45$&$60$&$62$&$74$&$77$&$91$&$93$&$106$&$109$&$125$\\
&$14$&$0$&$12$&$16$&$26$&$28$&$41$&$44$&$59$&$61$&$73$&$76$&$90$&$92$&$105$&$108$&$124$\\
\hline
&&&&&&$k$&&&&&&&&&&&\\
&&$17$&$18$&$19$&$20$&$21$&$22$&$23$&$24$&$25$&$26$&$27$&$28$&$29$&$30$&$31$&$32$\\
\hline
&$8$&$130$&$143$&$146$&$160$&$162$&$175$&$178$&$192$&$194$&$207$&$210$&$224$&$226$&$239$&$242$&$256$\\
&$9$&$130$&$142$&$145$&$159$&$161$&$174$&$177$&$192$&$194$&$206$&$209$&$223$&$225$&$238$&$241$&$256$\\
$e$&$10$&$130$&$141$&$144$&$158$&$160$&$173$&$176$&$191$&$193$&$205$&$208$&$222$&$224$&$237$&$240$&$256$\\
&$11$&$129$&$140$&$143$&$157$&$159$&$172$&$175$&$190$&$192$&$204$&$207$&$221$&$223$&$236$&$239$&$256$\\
&$12$&$128$&$139$&$142$&$156$&$158$&$171$&$174$&$189$&$191$&$203$&$206$&$220$&$222$&$235$&$238$&$255$\\
&$13$&$127$&$138$&$141$&$155$&$157$&$170$&$173$&$188$&$190$&$202$&$205$&$219$&$221$&$234$&$237$&$254$\\
&$14$&$126$&$137$&$140$&$154$&$156$&$169$&$172$&$187$&$189$&$201$&$204$&$218$&$220$&$233$&$236$&$253$
\end{tabular}
}
\caption{Values of $g(e,k)$ when $e\le14$ and $k\le32$.}
\label{gtbl}
\end{center}
\end{table}

In Section \fullref{normsec}, we apply the basic induction argument, \fullref{compatprop}, and the results for ${\on{gd}}(2^e\xi)$ in \fullref{gthm}
to prove the following result by induction on $t$. This clearly implies \fullref{gdthm} and hence \fullref{mainthm}.

\begin{prop}\label{1better} For $e\ge 7$ and $t\ge0$,
${\on{gd}}((2^e+2^{e+1}+\cdots+2^{e+t})\xi_{2^e-1})\le 2^e-e-6$.
\end{prop}

\section{Proof of general lifting results}\label{compatsec}
In this section, we prove \fullref{compatprop} and \fullref{pairedprop}. For the first one, we find it more convenient to
work with sections rather than geometric dimension.
\begin{thm} \label{sectthm}Let $X_0\subset\cdots\subset X_k$ and
$Y_0\subset\cdots\subset Y_k$
be filtered spaces, and let $\theta$ (resp. $\eta$) be a vector bundle over
$X_k$ (resp. $Y_k$). Suppose given $m_0$ (resp.~$n_0$) sections of
$\theta$ on $X_k$ (resp. $\eta$ on $Y_k$), of which the first $m_i$
(resp. $n_i$) are linearly independent (l.i.) on $X_{i}$ (resp. $Y_{i}$)
for
$0\le i\le k$. Let $$p_j=\min(m_{i}+n_{j-i}: 0\le i\le j).$$ Let
$$W_j=\bigcup_{i=0}^j X_i\times Y_{j-i}.$$ Then there are $p_0$ sections of
$\theta\times\eta$ on $W_k$ of which the first $p_j$ are linearly
independent on $W_{j}$ for $0\le j\le k$. Moreover, if $\ell+i\ge j$ and
$m_\ell+n_i\ge p_j$, then the first $p_j$ sections are l.i.~on
$X_\ell\times Y_i$.
\end{thm}

Note that we have $m_0\ge \cdots\ge m_k$, $n_0\ge \cdots\ge n_k$, and
$p_0\ge\cdots\ge p_k$.

The following result will be used in the final step of the proof of
\fullref{sectthm}.
\begin{lem} Suppose  $\theta$ is an $n$--dimensional
trivial vector bundle over a space $X$ with l.i.~sections $t_1,\ldots,t_n$.
Suppose $s_1,\ldots,s_r$ are l.i.~sections of $\theta$, each of
which is a linear combination with constant coefficients of the $t_i$. Then
there is a set $s_1,\ldots,s_r,s'_{r+1},\ldots,s'_n$ of linearly
independent sections of $\theta$, with all these sections being linear
combinations with constant coefficients of the $t_i$.\end{lem}

\begin{proof} Because of the constant-coefficient assumption, this is just
a consequence of
the result for vector spaces, that a basis for a subspace can be extended
to a basis for the
whole space.
\end{proof}

Note that the assumption about constant coefficients was required. For
example, the section $s(x)=(x,x)$
of $S\sp2\times {\mathbb R}\sp3$ cannot be extended to a set of three
l.i.~sections.

\begin{proof}[Proof of \fullref{sectthm}] Let $r_1,\ldots,r_{m_0}$ be the
given sections of $\theta$ on $X_k$,
and $s_1,\ldots,s_{n_0}$ the given sections of $\eta$ on $Y_k$. These are
considered as sections of $\theta\times\eta$ by using 0 on the other
component. Clearly $\{r_1,\ldots,r_{m_0},s_1,\ldots,s_{n_0}\}$ is a set of
$p_0$ sections on $W_k$ which is linearly independent on $W_0$. The
proof will proceed by finding $p_1$ linear combinations, always with
constant coefficients, of these sections which are l.i.~on $W_1$, then
$p_2$ linear combinations of these new sections which are l.i.~on $W_2$,
etc, until going into the last stage we have $p_{k-1}$ sections which
are l.i.~on $W_{k-1}$, and we find $p_k$ linear combinations of them which
are l.i.~on $W_k$. Now we apply the lemma repeatedly, starting with
the last $p_k$ sections. At the first step, we extend this set to a set of
$p_{k-1}$ sections l.i.~on $W_{k-1}$, and continue until going into
the last stage, where we have $p_1$ sections which are combinations of the
original $p_0$ sections and satisfy the conclusion of the theorem for
$1\le i\le k$. We apply the lemma one last time to extend the set of $p_1$
sections to the desired set of $p_0$ sections.

Here is an explicit algorithm for the sections described in the first half
of the preceding paragraph.
We may assume without loss of generality that $m_0\ge n_0$.

For $j$ from 0 to $k$,
\begin{itemize}
\item For $i$ from 1 to $p_j-n_0$ (resp. $p_j-m_0$), let $r_i^{(j)}=r_i$
(resp. $s_i^{(j)}=s_i$). (Note that if $n_0\ge p_j$,
then nothing happens at this step.)
\item For $i$ from $\max(1,p_j-n_0+1)$ to $\min(m_0,p_j)$, let both
$r_i^{(j)}$ and $s_{p_j+1-i}^{(j)}$
equal $r_i^{(j-1)}+s_{p_j+1-i}^{(j-1)}$.
\item Then the sections $r_i^{(j)}$ and $s^{(j)}_i$ constructed in the two
previous steps give the sections
which are l.i.~on $W_j$. (Each section constructed in the second step can
be counted as an $r$ or an $s$,
but is only counted once.)
\end{itemize}

We must show that these have the required linear independence. Before doing
so, we illustrate with
an example, computed by {\tt Maple}. Let $k=4$,
$[m_0,\ldots,m_4]=[11,6,4,1,0]$ and $[n_0,\ldots,n_4]=[10,8,3,2,0]$. Then
$[p_0,\ldots,p_4]=[21,16,14,9,7]$.
The 16 sections l.i.~on $W_1$ are
$$r_1,\ldots,r_6,\ r_7+s_{10},\ r_8+s_{9},\ r_9+s_8,\ r_{10}+s_7,\
r_{11}+s_6,\ s_5,\ldots,s_1.$$
The 14 sections l.i.~on $W_2$ are
\begin{eqnarray*}&&r_1,\ r_2,\ r_3,\ r_4,\ r_5+r_7+s_{10},\ r_6+r_8+s_{9},\
r_7+r_9+s_{10}+s_8,\ r_8+r_{10}+s_9+s_7,\\
&&r_9+r_{11}+s_8+s_6,\ r_{10}+s_7+s_5,\ r_{11}+s_6+s_4,\ s_3,\ s_2,\
s_1.\end{eqnarray*}
The 9 sections l.i.~on $W_3$ are
\begin{eqnarray*}&&r_1+r_6+r_8+s_{9},\  r_2+r_7+r_9+s_{10}+s_8,\
r_3+r_8+r_{10}+s_{9}+s_7,\\
&&r_4+r_9+r_{11}+s_8+s_6,\ r_5+r_7+r_{10}+s_{10}+s_7+s_5,\\
&& r_6+r_8+r_{11}+s_9+s_6+s_4,
r_7+r_9+s_{10}+s_8+s_3,\ r_8+r_{10}+s_9+s_7+s_2,\\\
&&r_9+r_{11}+s_8+s_6+s_1.\end{eqnarray*}
The 7 sections l.i.~on $W_4$ are
\begin{eqnarray*}&&r_1+r_3+r_6+2r_8+r_{10}+2s_9+s_7,\\
&&r_2+r_4+r_7+2r_9+r_{11}+s_{10}+2s_8+s_6,\\
&&r_3+r_5+r_7+r_8+2r_{10}+s_{10}+s_9+2s_7+s_5,\\
&&r_4+r_6+r_8+r_9+2r_{11}+s_{9}+s_8+2s_6+s_4,\\
&&r_5+2r_7+r_9+r_{10}+2s_{10}+s_8+s_7+s_5+s_3,\\
&&r_6+2r_8+r_{10}+r_{11}+2s_9+s_7+s_6+s_4+s_2,\\
&&r_7+2r_9+r_{11}+s_{10}+2s_8+s_6+s_3+s_1.\end{eqnarray*}

Now we continue with the proof. The property described in the first
paragraph of the proof, that
the sections claimed to be l.i.~on $W_j$ are linear combinations with constant coefficients of those
on $W_{j-1}$, is clear from
their inductive definition.

Next we easily show that if $i>p_j-n_0$, then
$$r_i^{(j)}=s_{p_j+1-i}^{(j)}=r_i+\sum_{\ell>i}c_\ell
r_\ell+s_{p_j+1-i}+\sum_{\ell>p_j+1-i}d_\ell s_\ell$$
with $c_\ell$ and $d_\ell$ integers. The point here is that the additional
terms have subscript
greater than $i$ or $p_j+1-i$. The proof is immediate from the inductive
formula
$$r_i^{(j)}=r_i^{(j-1)}+s_{p_j+1-i}^{(j-1)}$$
and the fact that $p_j\le p_{j-1}$. Indeed, from $r_i^{(j-1)}$ we obtain
terms $r_{\ge i}$ and
$s_{\ge p_{j-1}+1-i}$, and from $s_{p_j+1-i}^{(j-1)}$ we obtain terms
$s_{\ge p_j+1-i}$ and
$r_{\ge p_{j-1}-p_j+i}$.

Finally we show that the asserted sections are l.i.~on $W_j$. Let ${\mathbf x}\in
X_\ell\times Y_{j-\ell}$.
Note that $\{r_1({\mathbf x}),\ldots,r_{m_\ell}({\mathbf x})\}$ is l.i., as is
$\{s_1({\mathbf x}),\ldots,s_{n_{j-\ell}}({\mathbf x})\}$,
and that $p_j\le m_\ell+n_{j-\ell}$. If we form a matrix with columns
labeled
$$r_1,\ldots,r_{m_0},s_{n_0},\ldots,s_1,$$
and rows which express the sections, ordered as
\begin{equation}\label{secor}r_1^{(j)},
\ldots,r_{\min(m_0,p_j)}^{(j)},s^{(j)}_{p_j-m_0},\ldots,s_1^{(j)},\end{equation}
in terms
of the column labels, then, by the previous paragraph, the number of
columns is $\ge$ (usually strictly greater than) the number
of rows, the entry in position $(i,i)$ is 1 for $i\le\min(m_0,p_j)$, and
all entries to the left of these
1s are zero. If $i>\min(m_0,p_j)$, then all entries in the $r$-portion of row
$i$ are zero.
Moreover an analogous statement is true if the order of the rows and of the
columns are both reversed.
Thus there are 1s on the diagonal running
up from the lower right corner of the original matrix (for $\min(n_0,p_j)$
positions) and zeros to their right.

If a linear combination of our sections applied to ${\mathbf x}$ is 0, then the
triangular form of the matrix
implies that the first $m_\ell$ coefficients are 0, while the triangular
form looking up from the lower
right corner implies that the last $n_{j-\ell}$ coefficients are 0. Since
$p_j\le m_\ell+n_{j-\ell}$, this implies
that all coefficients are 0, hence the desired independence.

The same argument works for the last statement of the proposition. For $k$
satisfying $j\le k\le\ell+i$, replace
$W_k$ by $W_k\cup (X_{\ell}\times Y_i)$. Then everything goes through as
above.
\end{proof}

\begin{proof}[Proof of \fullref{compatprop}] Let $D=\dim(\theta)$ and
$D'=\dim(\eta)$. Then
$d_i$, $d'_i$, $e_i$, and $(X\times Y)_{i}$ of \fullref{compatprop}
correspond to  $D-m_i$, $D'-n_i$, $D+D'-p_i$, and $W_i$ of
\fullref{sectthm}, respectively. The compatible gd bounds may be
interpreted
as vector bundles $\theta_i$ over $X_i$ of dimension $d_i$ and isomorphisms
$\theta|X_i\approx \theta_i\oplus(D-d_i)$
and $\theta_{i}|X_{i-1}\approx\theta_{i-1}\oplus(d_{i}-d_{i-1})$. The
trivial subbundles yield, for all $i$, $D-d_i$
l.i.~sections of $\theta$ on $X_i$ such that the restrictions of the
sections on $X_i$ to $X_{i-1}$ are a subset of the sections on $X_{i-1}$.
Each of the sections on $X_0$ has a largest $X_i$ for which it is one of
the given l.i.~sections. By Atiyah
\cite[Section 1.4.1]{At}, this section on $X_i$ can be
extended over $X_k$ (although probably not as part of a linearly
independent set). Analogous statements are true for sections of $\eta|Y_i$.

By \fullref{sectthm}, there are $D+D'-e_0$ l.i.~sections of
$\theta\times\eta$ on $W_0$ of which the first $D+D'-e_i$ are l.i.~on
$W_i$. Taking orthogonal complements of the spans of the sections yields
the desired
compatible bundles on $W_i$ of dimension $e_i$, yielding the first part of
\fullref{compatprop}.

For the second part, first note that in the algorithm in the proof of
\fullref{sectthm}, if the $r$'s and $s$'s are equal, then the set of
sections constructed on each $W_i$ is invariant under the interchange map
$T$. Thus the same will be true of the orthogonal complement of their
span.\end{proof}

\begin{proof}[Proof of \fullref{pairedprop}] \qua
\begin{enumerate}
\item Let $F_1=S^{2m-1}$ denote the fiber of $BO_{2m-1}[\rho]\to
BO_{2m}[\rho]$. There is a relative
Serre spectral sequence for
\begin{equation}\label{SSS1}(CF_1,F_1)\to
(BO_{2m}[\rho],BO_{2m-1}[\rho])\to BO_{2m}[\rho].\end{equation}
The fibration $V_{2m}\to BO_{2m}[\rho]\to BO[\rho]$ shows that the bottom
class of $BO_{2m}[\rho]$ is in dimension $\ge\min(\rho,2m)$. The
spectral sequence of \eqref{SSS1} shows that $H_*(S^{2m})\to
H_*(BO_{2m}[\rho]/BO_{2m-1}[\rho])$ has cokernel beginning in dimension
$\ge s+1$,
and so the map is an $s$--equivalence. Thus the inclusion of the
$s$--skeleton of $BO_{2m}[\rho]/BO_{2m-1}[\rho]$ factors through $S^{2m}$ to
yield the map $c_1'$, which is an equivalence.

The second map is obtained similarly. A map $$\Sigma
P_{2m-1}^{2m}\mapright{\ell}BO_{2m+1}[\rho]/BO_{2m-1}[\rho]$$ is obtained as
the inclusion of
a skeleton of $CF_2/F_2$, where $F_2=V_{2m+1,2}$ is the fiber of
$BO_{2m-1}[\rho]\to BO_{2m+1}[\rho]$. The relative Serre spectral sequence
of
\begin{equation}(CF_2,F_2)\to (BO_{2m+1}[\rho],BO_{2m-1}[\rho]) \to
BO_{2m+1}[\rho]\label{SSS}\end{equation}
implies that $\on{coker}(\ell_*)$ begins in dimension $\ge s+1$, determined by
$$H_{2m}(CF_2,F_2)\otimes H_{\min(\rho,2m+1)}(BO_{2m+1}[\rho])$$  and
the first ``product" class in $H_{4m}(\Sigma V_{2m+1,2})$. The obtaining of
$c_2'$ now follows exactly as for $c_1'$.

\item Let $Q :=BO_{2m+1}[\rho]/BO_{2m-1}[\rho]$ and
$E:=\text{fiber}(BO_{2m+1}[\rho]\to Q)$.
The commutative diagram of fibrations
$$\begin{CD} V_{2m+1,2}@>>> BO_{2m-1}[\rho] @>>> BO_{2m+1}[\rho]\\
@VVV @VVV @VVV\\
\Omega Q@>>> E @>>> BO_{2m+1}[\rho]\end{CD}$$
implies the quotient $E/BO_{2m-1}[\rho]$ has the same connectivity as
$\Omega Q/V_{2m+1,2}$, which is 1 less than that
determined from \eqref{SSS}; that is, $E/BO_{2m-1}[\rho]$ is
$(s-1)$--connected. Thus, since $\dim(X)<s$, the vertical
maps in
$$\begin{CD}BO_{2m-1}[\rho]^{(s)}@>>>BO_{2m+1}[\rho]^{(s)}@>>>\Sigma
P_{2m-1}^{2m}\\
@VVV @VVV @VVV\\
E @>>> BO_{2m+1}[\rho] @>>> Q\end{CD}$$
are equivalences in the range relevant for maps from $X$, $A$, and $X/A$.
Since the bottom row is a fibration,
we may consider the top row to be one, too, as far as $X$ is concerned.

Since $g$ is divisible by 2, and $2\pi_{2m}(\Sigma P_{2m-1}^{2m})=0$, we
deduce that the composite
$$X/A\mapright{g} S^{2m} \mapright{i} \Sigma P_{2m-1}^{2m}$$
represents the 0 element of $[X/A,\Sigma P_{2m-1}^{2m}]$; ie the map is
null-homotopic rel $*$. There is a commutative diagram as in
\eqref{coffib} with the left sequence a cofiber sequence and the right
sequence a fiber sequence in the range of $\dim(X)$.
\begin{equation}\label{coffib}\begin{CD}
A @>f_1>> BO_{2m-1}[\rho]^{(s)}\\
@Vj_1VV @Vj_2VV\\
X @>p_2\circ f>> BO_{2m+1}[\rho]^{(s)}\\
@VqVV @VVV \\
X/A@>i\circ g>> \Sigma P_{2m-1}^{2m}\end{CD}\end{equation}
We have just seen that there is a basepoint-preserving homotopy
$$H \co X/A\times I\to \Sigma P_{2m-1}^{2m}$$
from $i\circ g$ to a constant map.
There is a commutative diagram
$$\begin{CD} X\times 0\cup A\times I@.\longrightarrow @.BO_{2m+1}[\rho]\\
@VVV @. @VVV\\
X\times I @>q\times I>> X/A\times I @>H>> \Sigma P_{2m-1}^{2m}\end{CD}$$ where the top map is $p_2\circ f$ on $X\times 0$ and $j_2\circ f_1$ on
each $A\times\{t\}$. By the Relative Homotopy Lifting Property of a fibration, there exists a map $\widetilde H \co X\times I\to
BO_{2m+1}[\rho]$ making both triangles commute. When $t=1$, it maps into $BO_{2m-1}[\rho]$, since it projects to the constant map at the
basepoint of $\Sigma P_{2m-1}^{2m}$.

\item
We use the fact that $2\cdot 1_{\Sigma P_{2m-1}^{2m}}$
factors as
$$\Sigma P_{2m-1}^{2m}\mapright{\text{col}}S^{2m+1}\mapright{\eta}
S^{2m}\hookrightarrow \Sigma P_{2m-1}^{2m}$$
to deduce that the composite
$$\Sigma X/A\mapright{\Sigma g'}\Sigma^2 P_{2m-1}^{2m}\mapright{\text{col}}
S^{2m+2}$$
is null-homotopic since $[\Sigma g']$ is divisible by 2. Note that we
needed to suspend once since if $\dim(X)=4m-1$, then $[X/A,\Sigma
P_{2m-1}^{2m}]$ might not have a group structure. Since
$$[X/A,S^{2m+1}]\mapright{\Sigma}[\Sigma (X/A),S^{2m+2}]$$
is bijective, we deduce that $X/A\mapright{{\text{col}\circ g'}} S^{2m+1}$
is null-homotopic.

An argument similar to the one in the beginning of the proof of (2) shows that $BO_{2m}[\rho]\to BO_{2m+1}[\rho]\to S^{2m+1}$ is a fibration
through dimension $\min(\rho+2m-1,4m)\ge s$. Since $\dim(X)\le s$, the lifting follows as in the proof of (2). \proved
\end{enumerate}
\end{proof}

\section[Inductive determination of a bound for gd(2^e,k)]{Inductive
determination of a bound for ${\on{gd}}(2^e,k)$}\label{indnsec}

In this section, we prove that part (6) of \fullref{gthm} follows from its first five parts, together with its validity for $e=7$.
We begin by proving the validity when $e=7$.
The following result is stronger than the required liftings for $e=7$; i.e., we have $m(k)\le g(7,k)$ and the inequality is strict
if $k$ is even with $4\le k\le 14$. The reason for beginning our induction with liftings weaker than the best results that we are able to prove
is to fit them into a simple formula that works for all values of $e$. Here and throughout we use the standard notation that $\nu(-)$ denotes
the exponent of 2 in an integer.

\begin{thm}\label{128prop} Let
$$m(k)=\begin{cases}0&k=0,1\\
16&k=2\\
8k-5&k\text{ odd, }\ 3\le k\le15\\
8k+\nu(k)-4&k\text{ even, }\ 4\le k\le16.\end{cases}$$
There are compatible liftings of $128\xi_{8k+7}$ to $BO_{m(k)}$ for $k\ge0$.\end{thm}
\begin{proof}
Let $H_k$ denote the Hopf bundle over quaternionic projective space $HP^k$.
Let $m'(k)=13$ if $k=2$, and otherwise $m'(k)=m(k)$.
We will use \cite[Theorem 1.1b]{HP} to prove
\begin{equation}\label{lifts}\text {there are compatible liftings of $32H_{2k+1}$ to $BO_{m'(k)}$ for $2\le k\le 16$.}\end{equation}

Three things are required to prove this. First we need that, for $k\le 15$ and  all $i\le 2k+1$
satisfying also $4i-1\ge m'(k)$,
$$\nu\tbinom{32}i\ge\nu(|\pi_{4i-1}(P_{m'(k)}\wedge bo)|).$$
This is easily verified using $\nu\tbinom{32}i=5-\nu(i)$ and, for $1\le\epsilon\le3$,
\begin{equation}\nu(|\pi_{4i-1}(P_{4a+\epsilon}\wedge bo)|)=\begin{cases}4-\epsilon&i=a+1\\
4&i=a+2\\ 8-\epsilon&i=a+3.\end{cases}\label{hts}\end{equation} For example, if $k$ is odd, we have $m'(k)=8k-5$. Then $a=2k-2$ and $\epsilon=3$ in
(\ref{hts}), and for $i=\langle 2k-1,2k,2k+1\rangle$, we have $\nu\tbinom{32}i=\langle 5,4,5\rangle$ and $\nu(|\pi_{4i-1}(P_{8k-5}\wedge
bo)|)=\langle 1,4,5\rangle$.

Secondly, we need that $\pi_{4i-1}(P_{m'(k)})\to\pi_{4i-1}(P_{m'(k)}\wedge bo)$ is injective for $i\le 2k+1$. This is obtained from Tables 8.4, 8.8,
8.14, 8.15, and 8.16 of \cite{Mem}. These show that for $m'(k)\equiv\langle3,7,13,14,15\rangle$ mod 16 and $4i-1\le m'(k)+\langle8,4,6,5,4\rangle$, the
asserted injectivity is true. Now the liftings follow from \cite[Theorem 1.1b]{HP}. If $k=16$, the lifting follows for dimensional reasons.

The third thing we need is compatibility. We must show that
$$\begin{CD}HP^{2k-1}@>>> BO_{m'(k-1)}\\
@VVV @VVV\\
HP^{2k+1}@>>> BO_{m'(k)}\end{CD}$$ commutes for $k\ge3$. The two composites agree stably, and so their obstructions to being homotopic lie in
$H^*(HP^{2k-1};\pi_*(V_{m'(k)}))$. If $k$ is even, then $8k-4<m'(k)$ so the groups are 0. If $k$ is odd, the result follows since
$\pi_{8k-4}(V_{8k-5})=0$.

We precede the compatible liftings of (\ref{lifts}) by the canonical maps $RP^{8k+7}\to HP^{2k+1}$, obtaining compatible liftings of
$128\xi_{8k+7}$ to $BO_{m(k)}$ for $k\ge2$. The bundle $128\xi_{15}$ is trivial. To insure compatibility of the liftings on $RP^{15}$ and $RP^{23}$,
we note that the obstructions to compatibility lie in $H^*(RP^{15};\pi_*(V_{16}))=0$.
This is why we use $m(k)=16$, rather than 13.
\end{proof}

Now we prove the induction step. Let
$$\rho(4a+b)=8a+2^b\text{ if }0\le b\le3.$$
It satisfies that $2^k\xi_n$ is nontrivial if and only if $n\ge\rho(k)$.
Let $\rho=\rho(e-1)$. Assume that we have obtained compatible liftings of $2^{e-1}\xi_{8k+7}$ to
$BO_{g(e-1,k)}[\rho]$ for all $k$. For $0\le k\le 2^{e-3}$, define $$g_1(e,k):=\max\{g(e-1,i)+g(e-1,k-i): \max(0,k-2^{e-4})\le i\le[k/2]\}.$$
Note that by \fullref{gthm}.(3),
\begin{equation}\label{gg1}g(e,k)\ge g_1(e,k)-1.\end{equation}

Recall $A_k=P^{8k+7}$, and let
$$(A\times A)_k=\bigcup_{i=0}^k A_i\times A_{k-i}.$$
Then by \fullref{compatprop} there are compatible symmetric liftings
$\ell_k$ of $2^{e-1}\xi\times 2^{e-1}\xi$ on $(A\times A)_{k}$ to
$BO_{g_1(e,k)}[\rho]$ for all $k$. We precede by compatible maps
$d_k:A_k\to (A\times A)_{k}$, cellular maps homotopic to the diagonal. The
composites
$A_k\stackrel{\ell_k\circ d_k}{\longrightarrow} BO_{g_1(e,k)}[\rho]$
are
compatible liftings of $2^e\xi_{8k+7}$ for all $k$.

By decreasing induction on $k$ starting with $k=2^{e-3}$, we will construct
compatible factorizations through
$BO_{g(e,k)}[\rho]$ of the maps $\ell_k\circ d_k$.
Assume inductively that, for all $j>k$, compatible factorizations, up to
homotopy rel $A_k$, of $\ell_j\circ d_j$
through $BO_{g(e,j)}[\rho]$ have been attained. If $g(e,k)\ge g_1(e,k)$,
then no factorization of $\ell_k\circ d_k$ is required, and so our
induction on $k$ is extended. So we may assume $g(e,k)=g_1(e,k)-1$.

Let $h=[k/2]$.  By \fullref{gthm}.(4),
\begin{equation}\label{2diff}g_1(e,k-1)\le g(e,k)-1.\end{equation}
By \eqref{2diff}, \fullref{gthm}.(4), and the last part of \fullref{sectthm}
(which is required for compatibility of the
lifts of $(A\times A)_{k-1}$ and $A_h\times A_h$ to $BO_{g(e,k)-1}$), we
have the commutative diagram below, similar to \eqref{coffib}.
$$\begin{CD}A_{k-1}@>d'>>(A\times A)_{k-1}\cup A_h\times
A_h@>>>BO_{g(e,k)-1}[\rho]^{(8k+7)}\\
@VVV @VVV @VVV\\
A_k @. (A\times A)_k @. BO_{g(e,k)}[\rho]^{(8k+7)}\\
@| @| @VVV\\
A_k@>d_k>>(A\times A)_{k}@>\ell_k>> BO_{g(e,k)+1}[\rho]^{(8k+7)}\\
@VVV @VVV @VcVV\\
A_k/A_{k-1}@>{\overline d}>>(A\times A)_{k}/((A\times A)_{k-1}\cup A_h\times A_h)@>{\overline{\ell}}>> C,\end{CD}$$ where $C=S^{g(e,k)+1}$ if $g(e,k)$ is odd, and
$C=\Sigma P^{g(e,k)}_{g(e,k)-1}$ if $g(e,k)$ is even. The maps labeled $d$ are cellular maps homotopic to the diagonal. The map $c$ is obtained
similarly to the first paragraph of the proof of \fullref{pairedprop}. Since $\dim(A_k)=8k+7$, the application of \fullref{pairedprop}
requires that
$$8k+7\le\min(\rho+g(e,k)-1,2g(e,k)-1).$$
The second follows from \fullref{gthm}.(2), while the first follows from $\rho\ge 2e-2$ and $g(e,k)\ge 8k-e+2$ since $e\ge8$.

The quotient $(A\times A)_{k}/(A_h\times A_h)$ equals $B\vee T(B)$,
where $T$ reverses the order of the factors, and
$B$ is the union of all cells $e^i\times e^j$ with $i<j$.
By the symmetry property of $\ell_k$, ${\overline{\ell}}|T(B)=({\overline{\ell}}|B)\circ T$. Since
$T\circ{\overline d} \simeq {\overline d}$, we conclude that
${\overline{\ell}}\circ{\overline d}$ is divisible by 2. Indeed, with $r_B$ denoting the
retraction onto $B$,
$$[{\overline{\ell}}\circ{\overline d}]=[({\overline{\ell}}|B)\circ r_B\circ{\overline d}]+[({\overline{\ell}}|T(B))\circ
r_{T(B)}\circ{\overline d}]$$
and we have $$[({\overline{\ell}}|T(B))\circ r_{T(B)}\circ{\overline d}]=[({\overline{\ell}}|T(B))\circ
T\circ r_B\circ {\overline d}]=[({\overline{\ell}}|B)\circ r_B\circ{\overline d}].$$

Thus, by \fullref{pairedprop}, $\ell_k\circ d_k$ is homotopic rel $A_{k-1}$
to a map which lifts to $BO_{g(e,k)}[\rho]$. Note that the
lifting into $BO_{g(e,k)-1}[\rho]$ was not needed if $g(e,k)$ is odd. We have extended our
inductive lifting hypothesis, and so  have proved that there are
compatible liftings of $A_k$ to $BO_{g(e,k)}[\rho]$ for all $k$. This
extends the induction on $e$ and proves \fullref{gthm}.(6), assuming the
first five parts of \fullref{gthm}.

\section{The function $g(e,k)$}\label{numbersec}

In this section, we define the function $g(e,k)$ which has been used in the
previous
sections, and prove the first five parts of \fullref{gthm}, its numerical
properties
which were already used to prove \fullref{gthm}.(6), its important geometrical
property.

We find it convenient to deal with the complementary function $G$ defined by
\begin{equation}\label{complement}G(e,k)=8k-g(e,k).\end{equation}
It has relatively small values, in which patterns are more readily apparent.
This function $G$ will be defined using several auxiliary functions.

We define a function $S$ for $k\ge2$ by
\begin{equation}\label{Sdef}S(k)=8k-13[\textstyle\frac{k+1}2]+2\alpha(k)+2\min(3,\nu(k-1))+\begin{cases} -1&k\equiv0\ (2)\\2&k\equiv1\ (8)\text{ and } \alpha(k)\ne2\\
4&\text{otherwise.}\end{cases}\end{equation}
Then $S(k)=8k-s(k)$, where $s(k)$ is the stable value of $g(e,k)$ when $e$ is sufficiently large.
The first values of $S$ are given by

\begin{center}
\begin{tabular}{c|cccccccccccccccccc|}
\hline
$k$&2&3&4&5&6&7&8&9&10&11&12&13&14&15&16&17&18&19\\
\hline
$S(k)$&4&8&7&13&12&16&13&21&18&22&21&27&26&30&25&33&30&34\\
\hline
\end{tabular}
\end{center}
It will occasionally be useful to set $S(1)=8$, consistent with $g(e,1)=0$.

Values of $k\equiv1$ mod 8 receive special treatment. They are excluded in the domain of some of
our functions. For example, for $k\not\equiv 1$ mod 8 with $k\ge2$, we define $V(k)$ by
$$V(k)=\alpha(k)-\begin{cases}2&k\equiv3\ (4)\\ 1&k\not\equiv3\ (4).\end{cases}$$
The reasons for defining some of these functions will be presented shortly.

We also define functions $\nu'$ and $R$ as follows.
$$\nu'(k)=\begin{cases}\nu(k)&k\text{ even}\\
-4&k\text{ odd,}\end{cases}$$
and, for $k\not\equiv1$ mod 8, \begin{equation}\label{Rdef}R(k)=S(k)+\nu'(k)-V(k).\end{equation}
The first few values of $R$ are given as follows.

\begin{center}
\begin{tabular}{c|cccccccccccccccc|}
\hline
$k$&2&3&4&5&6&7&8&10&11&12&13&14&15&16&18&19\\
\hline
$R(k)$&5&4&9&8&12&11&16&18&17&22&21&25&24&29&30&29\\
\hline
\end{tabular}
\end{center}

It will also be useful to introduce the notation $\langle n\rangle=\max(0,n)$.
We will frequently use the simple fact that for any number $X$,
\begin{equation}\label{X}X+\langle -X\rangle\ge0.\end{equation}

Now we can define our function $G$.
An integer $k$ is {\bf decomposable} if it can be written as $k=k_0+\cdots+k_r$ with $r\ge1$ and $\nu(k_i)>R(k_{i-1})$ for $1\le i\le r$.
Because each $k_i$  must be preceded in the binary expansion of $k$ by a long string of 0's,
it is clear that a decomposable integer has a unique maximal decomposition. The sum in (\ref{geq1}) is taken over all $k_i$, $i\ge1$,
in a maximal decomposition of $k$.
The first values of $k$ admitting a decomposition are 35, 66, and 67, with $k_0=3$, 2, 3,
respectively. A simple decomposition is a maximal decomposition with $r=1$.
 The first value of $k$ admitting a multiple decomposition is $2^{55}+35$ with $k_0=3$ and $k_1=32$.

\begin{defin}\label{geq}
If $2\le k\le 2^{e-3}$ and $(e,k)\ne(7,9)$, we define
$$G(e,k)=\begin{cases}\min(S(k),G'(e,k))&k\not\equiv1\ (8)\\
\min(S(k),6+G'(e,k-1))&k\equiv1\ (8),\end{cases}$$
where, for $k\not\equiv1$ mod 8,
\begin{equation}G'(e,k)=\langle e-6-\nu'(k)\rangle-\sum_{k_i}\langle\min(\nu(k_i),e-6)-R(k_{i-1})\rangle.\label{geq1}\end{equation}
The exceptional value is $G(7,9)=5$, not 6 as the formula would give.
\end{defin}

The terms in the sum in \eqref{geq1} will sometimes be called {\it deviations}.
We do not define $G(e,0)$, $G(e,1)$, or $G(e,k)$ for $k>2^{e-3}$; instead we just define the complementary function $g$ by $g(e,0)=g(e,1)=0$
and $g(e,k)=2^{e-3}$ for $k>2^{e-3}$, and observe that the crucial properties (3) and (4) in Theorem \ref{gthm} are easily seen to be
satisfied whenever these extreme values are involved.

Next we provide some general discussion of what led to the rather complicated formula for $g(e,k)$. First we describe what led to the
basic formula $g(e,k)\approx 8k-\langle e-6-\nu'(k)\rangle$, modified when $k\equiv1$ mod 8.
We began with the initial values $m(k)$ of \fullref{128prop} for $g(7,k)$ and used a computer
program implementing properties (3) and (4) of \fullref{gthm} to obtain bounds for $g(e,k)$ for larger $e$.
Except perhaps for the first few entries in a $k$-column, the values $8k-(e-6-\nu'(k))$ when $k\not\equiv1$ mod 8, and $g(e,8\ell+1)=g(e,8\ell)+2$,
were apparent until issues of stabilization, which we will discuss shortly, became involved. However, there was no apparent regular pattern
for the first few entries in each $k$-column. The formula $8k-\langle e-6-\nu'(k)\rangle$ was achieved after additional computer experimentation
as the simplest general formula satisfying $g(7,k)\ge m(k)$ and consistency with \fullref{gthm}.

Next we explain where $S(k)$ came from. It is related to the condition $g(e,k)\ge 4k+4$, which says that our lifting methods only work
in the stable range. In an earlier version of this paper, we used the triviality of $2^{\phi(n)}\xi_n$ to give 0 as the value of
$g(e,k)$ when $e>4k+3$, but we were unable to prove that this could be done compatibly with our other liftings; i.e. that the liftings
which we obtain inductively can be done so that their restrictions to appropriate skeleta are trivial. By forcing $g(7,2)=16$, we could,
as noted in the proof of \fullref{128prop}, guarantee that our liftings restrict to a trivial bundle on $P^{15}$, the case $k=1$.
For reasons of stability, we forced $g(e,2)\ge 12$ and $g(e,3)\ge 16$. Forcing $g(e,4)\ge20$ is not strong enough, since, with $g(15,2)=12$
and $g(15,4)=25$, we could not obtain $g(16,4)=24$ consistently with property (4) of \fullref{gthm}. Thus $g(e,4)=25$ for $e\ge15$; i.e. $s(4)=25$.
This translates to our value $S(4)=8\cdot 4-s(4)=7$.

To be consistent with \fullref{gthm}, our function $S$ must satisfy the inequalities of the following proposition, the proof of which
is straightforward, although somewhat tedious, and is omitted.
\begin{prop}\label{Sprop} The function $S$ defined in \eqref{Sdef} satisfies
$$S(i+j)\le S(i)+S(j)+1$$
and $$S(2i)\le 2S(i)-1,$$
with equality in the first if $i=2^t$ and $2\le j\le 2^t-1$ or $j=2^t+1$, and equality holds in the second if $i=2^t$.
Thus $S$ may be defined by $S(2)=4$, $S(3)=8$, and
$$S(k)=\min(S(i)+S(k-i)+1,\  2\le i<k/2, \ 2S(k/2)-1).$$
\end{prop}

To be consistent with property (3) of \fullref{gthm}, our function $G$ must satisfy the property stated in the next theorem,
the proof of which will occupy much of this section.

\begin{thm}\label{firstproperty} If $e\ge8$, and $2\le i,j\le2^{e-4}$, then
\begin{equation}G(e-1,i)+G(e-1,j)+1-G(e,i+j)\ge0.\label{Gineq}\end{equation}
\end{thm}

The stabilization given by $S(k)$ and the requirement \eqref{Gineq} are what lead to the complicated sum in \eqref{geq1}.
The first example of this is for $G(11,3)+G(11,32)+1-G(12,35)$. Since $G'(11,3)=9>S(3)=8$, we have $G(11,3)=8$. Also $G(11,32)=11-6-5=0$,
and $\langle 12-6-\nu'(35)\rangle=10$. Thus we must subtract 1 from $\langle 12-6-\nu'(35)\rangle$ in $G(12,35)$
in order that \eqref{Gineq} will hold. This is accounted for by
the decomposition of $35$ with $k_1=3$. The value $R(3)=4$ is the amount that $\nu(k-3)$ must exceed in order that the decomposition
affects the value of $G(e,k)$.

Note that 11 is the smallest value of $e$ for which $G(e,3)\ne\langle e-6-\nu'(3)\rangle$. This is obtained by solving
$$e-6-\nu'(3)=S(3)+1,$$
obtaining $e=11$.
We want $R(3)$ to be 1 less than the value of $t$ which satisfies
$$G(11,3)+G(11,2^t)+1-\langle 12-6-\nu'(2^t+3)\rangle=-1.$$
Here $G(11,3)-\langle 12-6-\nu'(2^t+3)\rangle$ necessarily equals $-2$:\ 1 from $12-11$, and 1 from $G(11,3)=G'(11,3)-1$. Thus we need $t$ to satisfy
$0=G(11,2^t)=11-6-t$, and so
$$R(3)=t-1=(S(3)+\nu'(3)+6)-6=S(3)+\nu'(3),$$
consistent with \eqref{Rdef}, since $V(3)=0$.

The way $V$ arises can be seen by comparing the requirements, for $t\ge5$,
$$G(e,2^t+5)\le G(e-1,2)+G(e-1,2^t+3)+1$$
and
$$G(e,2^t+5)\le G(e-1,5)+G(e-1,2^t)+1.$$
The first reduces to, for $e$ moderately large,
$$G(e,2^t+5)\le S(2)+e-6-t+S(3)=e+6-t,$$
while the second becomes
$$G(e,2^t+5)\le S(5)+e-6-t=e+7-t.$$
We must use the first condition because $S(2)+S(3)<S(5)$. The value $V(5)=1$ measures this. Our $V(k)$ satisfies that it is the largest
$r$ such that $k=i_0+\cdots+i_r$ with
$$S(i_0+\cdots+i_t)=S(i_0+\cdots+i_{t-1})+S(i_t)+1$$
for $1\le t\le r$.

This concludes our discussion of the rationale behind the definition of $G$ except for one more brief comment.
It was certainly to be expected that these modifications to the $G$-formula, given by the summands in \eqref{geq1},
would be cumulative. It was not {\it a priori} clear whether $R(k_{i-1})$ or $R(k_0+\cdots+k_{i-1})$ would be the appropriate part
of that formula. The answer will become apparent in Subcase 2d of the proof of \fullref{not8k+1}.

The following proposition will be needed shortly. The function $S'$ below will often be encountered in the
guise of $S'(k)=R(k)-\nu'(k)$.
\begin{prop} Let $S'(k)=S(k)-V(k)$. If $i,j,i+j\not\equiv1$ mod $8$, then
$$S'(i)+S'(j)\ge S'(i+j).$$\label{spprop}
Moreover, if $i<2^{\nu(j)}$, then equality is obtained.
\end{prop}
\begin{proof}
One easily verifies that
$$S'(k)=8k-13[\textstyle\frac{k+1}2]+\alpha(k)+\begin{cases}0&k\equiv0\ (2)\\
8&k\equiv3\ (4)\\
9&k\equiv5\ (8).\end{cases}$$
For $1\le m\le 4$, let $\phi_m$ denote the $m$th part of the above formula for $S'(k)$, and let $\psi_m(i,j)=\phi_m(i)+\phi_m(j)-\phi_m(i+j)$.
Then $$\psi_m(i,j)=\begin{cases}0&m=1\\
0&m=2,\ ij\text{ even}\\
-13&m=2,\ ij\text{ odd}\\
\nu\tbinom{i+j}i&m=3\\
\ge16&m=4,\ ij\text{ odd}\\
-1&m=4,\ i+j\equiv5 \ (8)\text{ and }i\text{ or }j\equiv 3\ (4)\\
\ge0&m=4, \text{otherwise.}\end{cases}$$
Since $\binom{i+j}{i}$ is even if $i+j\equiv5\ (8)$ and $i\equiv3\ (4)$, the inequality follows.

For the second part, one easily sees that, if $i<2^{\nu(j)}$, then $\psi_m(i,j)=0$ for $1\le m\le4$. When $m=4$, it is true
because $i\equiv i+j$ mod 8 (or $\nu(j)=2$ and $i=2$ or 3).
\end{proof}

We now begin the lengthy proof of \fullref{firstproperty}. In order to keep the number of cases and subcases within
reason, we split the theorem into two parts. Most of the work will go into proving the following result.
\begin{thm}\label{not8k+1} If $e\ge8$,  $2\le i,j\le2^{e-4}$, and $i,j,i+j\not\equiv1$ mod 8, then
\eqref{Gineq} holds.
\end{thm}
\begin{proof} We divide into cases depending upon whether $S(i)$ and/or decompositions are involved.
\smallskip

\noindent {\bf Case 1}: Neither $i$ nor $j$ decomposes,  $G(e-1,i)\ne S(i)$, and $G(e-1,j)\ne S(j)$.
In this case, the LHS of \eqref{Gineq} becomes
\begin{equation}\label{lara}\ge\langle e-7-\nu'(i)\rangle+\langle e-7-\nu'(j)\rangle+1-\langle e-6-\nu'(i+j)\rangle.\end{equation}
By considering separately the four subcases (a) $i$ and $j$ odd, (b) $i$ odd, $j$ even, (c) $\nu(j)>\nu(i)>0$, and
(d) $\nu(i)=\nu(j)>0$, one easily shows that \eqref{lara} is $\ge0$ in each subcase. Note that if $i+j$ decomposes,
then the LHS of \eqref{Gineq} is greater than \eqref{lara}, and so we need not worry about this possibility here.

\smallskip
\noindent {\bf Case 2}: $G(e-1,i)=S(i)$ and $i$ does not decompose.

{\bf Subcase 2a}: Also, $G(e-1,j)=S(j)$. Then the LHS of \eqref{Gineq} is $\ge S(i)+S(j)+1-S(i+j)\ge0$, by \fullref{Sprop}. The remaining subcases of Case 2 now assume that $G(e-1,j)<S(j)$.

\smallskip
{\bf Subcase 2b}: $j$ does not decompose, and $\nu(j)\le\nu(i)$. Then $\nu'(i+j)\ge\nu'(j)$, and so the LHS of \eqref{Gineq} is
$$\ge S(i)+\langle e-7-\nu'(j)\rangle+1-\langle e-6-\nu'(i+j)\rangle\ge S(i)>0.$$

\smallskip
{\bf Subcase 2c}: $j$ does not decompose, and $\nu(j)>\nu(i)$. We allow for the possibility that $i$ might
serve as the bottom part of a decomposition of $i+j$. This will be true if $\nu(j)$ is sufficiently large.
Because of our $\langle-\rangle$-notation, our analysis is valid regardless.
This time $\nu'(i)=\nu'(i+j)$, and so the LHS of \eqref{Gineq} is
$$\ge S(i)+\langle e-7-\nu(j)\rangle+1-\langle e-6-\nu'(i)\rangle+\langle\min(\nu(j),e-6)-S(i)-\nu'(i)+V(i)\rangle.$$
If $\nu(j)\le e-7$, this is $\ge V(i)\ge0$. If $\nu(j)\ge e-6$, it simplifies to
\begin{equation}\label{2c}\ge V(i)+1+e-6-\nu'(i)-\langle e-6-\nu'(i)\rangle.\end{equation}
Since $j\le 2^{e-4}$ and $\nu(i)<\nu(j)$, we have $\nu'(i)\le e-5$, and so \eqref{2c} is $\ge V(i)\ge0$.

\smallskip
{\bf Subcase 2d}: $j$ admits a decomposition. We  consider a 2-stage decomposition $j=j_0+j_1+2^tA$ with
$A$ odd and $\nu(j_1)>R(j_0)$.
It will be clear that the argument here can be adapted to a longer decomposition. Letting $D\ge0$ denote
any amount added for a decomposition of $i+j$,
the LHS of \eqref{Gineq} becomes, using \fullref{spprop},
\begin{eqnarray}&&S(i)+( e-7-\nu'(j))-(\nu(j_1)-R(j_0))\nonumber\\
&&\quad-(\min(t,e-7)-R(j_1))+1-\langle e-6-\nu'(i+j)\rangle+D\label{minD}\\
&=&S'(i)+V(i)+S'(j_0)+S'(j_1)+\nu'(i+j)-\min(t,e-7)+D\nonumber\\
&\ge&V(i)+S'(i+j_0+j_1)+\nu'(i+j)-\min(t,e-7)+D\nonumber\\
&=&V(i)+R(i+j_0+j_1)-\nu'(i+j_0+j_1)+\nu'(i+j)-\min(t,e-7)+D\label{2d}.\end{eqnarray}
We will discuss later the removal of the $\langle-\rangle$ at the first step.

We will show below that \begin{equation}\label{V}V(i)-\nu'(i+j_0+j_1)+\nu'(i+j)>0.\end{equation}
Assuming this, the only way that \eqref{2d} could be negative is if $\min(t,e-7)>R(i+j_0+j_1)$.
But if this is the case, then $(i+j_0+j_1)+2^tA$ is a decomposition of $i+j$, which makes
 $D\ge\min(t,e-6)-R(i+j_0+j_1)$. If $i+j_0+j_1$ decomposes
further, that only adds more to $D$. Thus, assuming \eqref{V}, we obtain that \eqref{2d} is $\ge0$.

We now prove \eqref{V}. The only way it could possibly be negative is if $i=2^tB-j_0-j_1$ with
$B$ even. Then the LHS of \eqref{V} becomes
\begin{eqnarray*}&\ge& \alpha(2^tB-j_0-j_1)-2-(t+\nu(B))+t\\
&=&\alpha(B-1)+t-\alpha(j_0+j_1-1)-2-\nu(B)\\
&>&0\end{eqnarray*}
since $\alpha(B-1)\ge\nu(B)$ and $t\ge R(j_0+j_1)>>\alpha(j_0+j_1-1)$.

Regarding the removal of $\langle-\rangle$ above: if $\nu'(i+j)>e-6$, then  \eqref{minD} becomes
\begin{eqnarray*}&\ge&S(i)+e-7-\nu'(j)-\nu(j_1)+R(j_0)-\min(t,e-7)+R(j_1)+1\\
&=&S(i)+(e-7-\min(t,e-7))+(R(j_1)-\nu(j_1))+(R(j_0)-\nu'(j_0))+1\\
&>&0\end{eqnarray*}
because each of its terms is nonnegative.

\smallskip
\noindent{\bf Case 3}: $G(e-1,i)=S(i)$ and $i$ decomposes. Although the decomposition
of $i$ does not affect the value of $S(i)$, it could affect the value of $G(e,i+j)$
by affecting the decomposition of $i+j$. In the analogues of Subcases 2a and 2b, the
decomposition of $i+j$ was not needed, and so a decomposition of $i$ cannot affect the validity.

\smallskip
{\bf Subcase 3a}: $j$ does not decompose and $\nu(j)>\nu(i)$.

\smallskip
{\bf Subsubcase 3ai}: $i$ admits a simple decomposition. Let $i=i_0+2^t\alpha$ with $\alpha$ odd
and $t>R(i_0)$.
If $\nu(j)\ge t$, then, considering $i_0+(2^t\alpha)+j$ as a possible decomposition of $i+j$, the LHS of \eqref{Gineq} becomes
\begin{eqnarray*}&\ge&S(i)+\langle e-7-\nu(j)\rangle+1-\langle e-6-\nu'(i)\rangle\\
&&\quad +\langle\min(t,e-6)-R(i_0)\rangle+\langle\min(\nu(j),e-6)-R(2^t\alpha)\rangle.\end{eqnarray*}
This exceeds the amount analyzed in Subcase 2c by
\begin{equation}t-R(i_0)-R(2^t\alpha)+R(i_0+2^t\alpha).\label{tR}\end{equation}
Since, in the notation of \fullref{spprop}, $S'=R-\nu'$, and $\nu'(i_0)=\nu'(i_0+2^t\alpha)$, then
\eqref{tR} equals $S'(i_0+2^t\alpha)-S'(i_0)-S'(2^t\alpha)=0$ by \fullref{spprop}.

If, on the other hand, $\nu(j)<t$, then we don't need $i+j$ to be decomposable, since the LHS of
\eqref{Gineq} $$\ge S(i)+\langle e-7-\nu(j)\rangle+1-\langle e-6-\nu'(i)\rangle=S(i)+\nu'(i)-\nu(j)>0,$$
since $S(i)>t+4>\nu(j)+4$. (The $+4$ is included because of the possibility that $\nu'(i)=-4$.)

\smallskip
{\bf Subsubcase 3aii}: $i$ admits a multiple decomposition. If $\nu(j)\le S(i)+\nu'(i)$, then, as in the preceding paragraph,
we do not need a decomposition of $i+j$ in order to satisfy \eqref{Gineq}. If, on the other hand, $\nu(j)>S(i)+\nu'(i)$, then
the result follows as in the first paragraph of Subcase 3ai, using additivity of $S'$ on disjoint decompositions.

\smallskip
{\bf Subcase 3b}: $i$ and $j$ both decompose exactly once. Let $i=i_0+2^m\beta$ with $\beta$ odd and $m> R(i_0)$, and $j=j_0+2^t\alpha$ with $\alpha$ odd and
$t> R(j_0)$.

If $m>t$, then we can consider $i+j$ as
$(i_0+j_0)+2^t\alpha+2^m\beta$. It is possible that $\langle m-R(2^t\alpha)\rangle$ might contribute to $G(e,i+j)$, but even if it does, we do not need it.
The situation is similar to Subcase 2d. Using the $\langle t-R(j_0)\rangle$ and $\langle t-R(i_0+j_0)\rangle$ parts of $G(e-1,j)$ and $G(e,i+j)$,
respectively, the LHS of \eqref{Gineq} simplifies to
$$\ge S(i)-\nu'(j_0)+\nu'(i_0+j_0)+R(j_0)-R(i_0+j_0),$$
which is very positive. (It would be $\ge V(i_0)$ by \fullref{spprop} if $S(i)$ were replaced by the much smaller number $S(i_0)$.)
Keeping in mind that $2^{e-3}\ge i+j$, we will usually omit, from now on, explicit consideration of the possibility that $e-6<\nu(k_{i-1})$ in \eqref{geq1}.
In Subcase 4d, there is a detailed discussion of a delicate case in which we consider carefully what happens when $e-6$ is larger than
the relevant 2-exponent .

If $m=t$, then a very similar argument works. Because the decomposition of $i+j$ now is $(i_0+j_0)+2^p\gamma$ with $p>t$,
and this exponent appears with a $+$ sign in $-G(e,i+j)$, the LHS of \eqref{Gineq} is even larger than it was when $m>t$.

Now suppose $m<t$. We use $(i_0+j_0+2^m\beta)+2^t\alpha$ as our trial decomposition of $i+j$. If it is not a true
decomposition, then the $\langle-\rangle$ will take care of it.

The LHS of \eqref{Gineq} becomes
\begin{eqnarray*} &\ge&S(i)+(e-7-\nu'(j_0)) -(t-R(j_0))+1\\
&&\ -(e-6-\nu'(i_0+j_0))+\langle t-R(i_0+j_0+2^m\beta)\rangle\\
&\ge&S(i)-\nu'(j_0)+R(j_0)+\nu'(i_0+j_0)-R(i+j_0)\\
&=&V(i)+S'(i)+S'(j_0)-S'(i+j_0)\\
&\ge&V(i).\end{eqnarray*}

{\bf Subcase 3c}: At least one of $i$ and $j$ decomposes more than once. The argument is very similar to that of Subcase 3b.
The only reason for separating them is to use 3b as a warmup for 3c.
Let $i=i_0+\cdots+i_r$ and $j=j_0+\cdots+j_s$ be maximal decompositions.

If $\nu(j_s)\le \nu(i_r)$, then the LHS of \eqref{Gineq} is, without using any decomposition of $i+j$,
\begin{eqnarray*} &\ge&S(i)-\nu'(j)-\sum_{k=1}^s(\nu(j_k)-R(j_{k-1}))+\nu'(i+j)\\
&\ge&S(i)+\sum_{k=0}^{s-1}(R(j_k)-\nu'(j_k))-\nu(j_s)\\
&\ge& S(i)-\nu(i_r)\\
&>\!>&0.\end{eqnarray*}

If $\nu(i_r)<\nu(j_s)$, first suppose the only decomposition of $i+j$ is the simple decomposition $K+j_s$ with $K=i+j_0+\cdots+j_{s-1}$.
Then the LHS of \eqref{Gineq} is
\begin{eqnarray*}&\ge&S(i)-\nu'(j)-\sum_{k=1}^s(\nu(j_k)-R(j_{k-1}))+\nu'(i+j)+\nu(j_s)-R(K)\\
&=&R(i)+V(i)-\nu'(i)+\sum_{k=0}^{s-1}(R(j_k)-\nu'(j_k))+\nu'(K)-R(K)\\
&\ge&V(i)\end{eqnarray*}
by \fullref{spprop}.

If $i+j$ decomposes more finely, say as $A+B+j_s$, then $-R(K)$ is replaced by $-R(B)+\nu(B)-R(A)$. But these are equal by the
second part of \fullref{spprop}, noting that $\nu'(A+B)=\nu'(A)$.

\medskip
{\bf Case 4}: $S(-)$ not involved, $i$ decomposes, $j$ doesn't. Recall $i,j\le 2^{e-4}$. We assume that $i$ admits a decomposition
as $i_0+i_1+i_2$. The nature of our argument will show that the conclusion will also be true for longer
decompositions. The LHS of \eqref{Gineq} becomes
\begin{equation} \label{case4} e-6-\nu'(i_0)-\nu(i_1)+R(i_0)-\nu(i_2)+R(i_1)+\langle e-7-\nu'(j)\rangle+Y,\end{equation}
where $Y=-G(e,i+j)$. We use \eqref{X} often in what follows.

\smallskip
{\bf Subcase 4a}: $\nu(j)<\nu(i)$. Then, using a decomposition $i+j=(i_0+j+i_1)+(i_2)$, we obtain
\begin{equation}\label{-1}Y\ge-(e-6-\nu'(j))+\langle\nu(i_2)-R(i_0+j+i_1)\rangle.\end{equation}
If there is an additional decomposition of $i+j$ as $(i_0+j)+(i_1)+(i_2)$, then by the second part of
\fullref{spprop}, $R(i_0+j+i_1)=R(i_0+j)+R(i_1)-\nu(i_1)$, and so the same expression is obtained.
Then \eqref{case4} is
\begin{equation}\label{-2}\ge(e-7-R(i_0+j+i_1))+(R(i_0)-\nu'(i_0))+(R(i_1)-\nu(i_1))>0,\end{equation}
since if the $\langle-\rangle$ in \eqref{-1} is $>0$, then $$e-7\ge\nu(i_2)-2\ge R(i_0+j+i_1)-2,$$
but the $(R-\nu)$-expressions are $>2$. If the $\langle-\rangle$ in \eqref{-1} is 0, then the first
part of \eqref{-2} is replaced by $(e-7-\nu(i_2))\ge-2$.

\smallskip
{\bf Subcase 4b}: $\nu(i)\le \nu(j)<R(i_0)$. In this case, which is very similar to 4a,
$$Y\ge-(e-6-\nu'(i))+\langle\nu(i_2)-R(i_0+j+i_1)\rangle,$$
because if there is an additional decomposition of $i+j$ as $(i_0+j)+(i_1)+(i_2)$, then
$R(i_0+j+i_1)=R(i_0+j)+R(i_1)-\nu(i_1)$, and so the expression for $Y$ is unchanged.
Then \eqref{case4} is
$$\ge(S'(i_0)+S'(i_1)+S'(j)-S'(i_0+i_1+j))+(e-7-R(j))>0.$$
In the remaining subcases, we deal with a maximum possible decomposition of $i+j$, realizing, as in 4a and 4b,
that if the decomposition must be amalgamated, the expression is not changed.

\smallskip
{\bf Subcase 4c}: $R(i_0)\le \nu(j)<\nu(i_1)$. Then
$$Y\ge-(e-6-\nu'(i))+\langle\nu(j)-R(i_0)\rangle+\langle\nu(i_1)-R(j)\rangle+\langle\nu(i_2)-R(i_1)\rangle,$$
and so \eqref{case4} is
$$\ge(e-7-\nu(i_1))+\langle\nu(i_1)-R(j)\rangle>0.$$

\smallskip
{\bf Subcase 4d}: $\nu(i_1)\le \nu(j)<\nu(i_2)$. Then
$$Y\ge-(e-6-\nu'(i))+\langle\nu(i_1)-R(i_0)\rangle+\langle\nu(j)-R(i_1)\rangle+\langle\nu(i_2)-R(j)\rangle,$$
and so \eqref{case4} is
\begin{equation}\label{4d}\ge(e-7-\nu(i_2))+\langle\nu(i_2)-R(j)\rangle>0.\end{equation}

As noted in Subcase 3b, we are usually not paying explicit attention to the possibility
that $e-6\le\nu(i_2)$ (in the situation in this subcase, 4d). Here it does warrant our
attention. We might have $i_2=2^{e-5}$, $2^{e-6}$, or $3\cdot2^{e-6}$, and then it would
seem that \eqref{4d} might not be valid.

If $i_2=2^{e-5}$, then $\langle \nu(i_2)-R(i_1)\rangle$ in the above analysis is replaced by
$\langle e-7-R(i_1)\rangle$. This decrease of 2 compensates for the fact that $e-7-\nu(i_2)=-2$
in \eqref{4d}. Similarly, if $\nu(i_2)=e-6$, then $\langle\nu(i_2)-R(i_1)\rangle$ is replaced by
$\langle e-7-R(i_1)\rangle$, compensating for $e-7-\nu(i_2)=-1$.

\smallskip
{\bf Subcase 4e}: $ \nu(i_2)<\nu(j)$. Then
$$Y\ge-(e-6-\nu'(i))+\langle\nu(i_1)-R(i_0)\rangle+\langle\nu(i_2)-R(i_1)\rangle+\langle\nu(j)-R(i_2)\rangle,$$
and so \eqref{case4} is
$$\ge\langle e-7-\nu(j)\rangle+\langle\nu(j)-R(i_2)\rangle>0.$$

\smallskip
{\bf Case 5}: $S(-)$ not involved, both $i$ and $j$ decompose.  We consider here a typical
example in which both $i$ and $j$ decompose twice. It should be clear that the general case
will work out in the same way. We assume that $i=i_0+i_1+i_2$ and $j=j_0+j_1+j_2$
are decompositions. Then
\begin{eqnarray*}G(e-1,i)+G(e-1,j)+1&=&
e-6-\nu'(i_0)-\nu(i_1)+R(i_0)-\nu(i_2)+R(i_1)\\&&+e-7-\nu'(j_0)-\nu(j_1)
+R(j_0)-\nu(j_2)+R(j_1)
\end{eqnarray*}
We assume without much loss of generality that $\nu(j_2)>\nu(i_2)$ and $\nu(i_0)<\nu(j_0)$.

\smallskip
{\bf Subcase 5a}: $\nu(j_2)<R(i_0+i_1+i_2+j_0+j_1)$. We use no decomposition of $i+j$.
We obtain that
\begin{eqnarray*}&&G(e-1,i)+G(e-1,j)+1-G(e,i+j)\\
&\ge&R(i_0)+S'(i_1)-\nu(i_2)+e-7+S'(j_0)+S'(j_1)-\nu(j_2)\\
&=&S'(i_0)+S'(i_1)+S'(i_2)+S'(j_0)+S'(j_1)+\nu'(i_0)-R(i_2)+e-7-\nu(j_2)\\
&\ge&R(i_0+i_1+i_2+j_0+j_1)-R(i_2)+e-7-\nu(j_2)\\
&>\!>&0,\end{eqnarray*}
since $e-7-\nu(j_2)\ge-2$ while $R(i_0+i_1+i_2+j_0+j_1)-R(i_2)>\!>0$.

\smallskip
{\bf Subcase 5b}: $\nu(j_2)>R(i_0+i_1+i_2+j_0+j_1)$. We use a decomposition of $i+j$
as $(i_0+i_1+i_2+j_0+j_1)+(j_2)$. We discuss afterward the usual argument regarding what
happens if it decomposes more finely. Similarly to Subcase 5a, we obtain
\begin{eqnarray*}&&G(e-1,i)+G(e-1,j)+1-G(e,i+j)\\
&\ge&S'(i_0)+S'(i_1)+S'(i_2)+S'(j_0)+S'(j_1)-S'(i_0+i_1+i_2+j_0+j_1)\\
&&\ +e-7-R(i_2)\\
&\ge&0\end{eqnarray*}
using \fullref{spprop} and $$e-7\ge\nu(j_2)-2\ge R(i_0+i_1+i_2+j_0+j_1)-2>\!>R(i_2).$$

Further decomposition of $i_0+i_1+i_2+j_0+j_1$ into 2-adically disjoint parts does not
change the expression, using the second part of \fullref{spprop}, similarly to the argument
in Subcases 4a and 4b.
\end{proof}

The following result will be useful in some subsequent proofs. In particular,
\fullref{gthm}.(5) is an immediate consequence.
\begin{prop}\label{minus1} For $e\ge7$ and $2\le k<2^{e-3}$,
$$G(e,k+1)-G(e,k)\begin{cases}=8&k\equiv0\ (8),\ \alpha(k)=1,\ e\ge S(k)+\nu(k)+8\\
=7&k\equiv0\ (8),\ \alpha(k)=1,\ e= S(k)+\nu(k)+7\\
=6&k\equiv0\ (8),\text{ otherwise}\\
\le-1&k\equiv1\ (8)\\
\le6&\text{otherwise.}\end{cases}$$
\end{prop}
\begin{proof} We begin by noting that the result is true for the limiting values, $S(k)$,
since they are easily shown to satisfy
\begin{equation}S(k+1)-S(k)\begin{cases}=8&k=2^e,\ e\ge3\\
=6&k\equiv0\ (8),\ \alpha(k)>1\\
=6&k\equiv 4\ (8)\\
=4&k\equiv2\ (4)\\
\le-1&k\equiv1\ (8)\\
=-1&k\equiv3,5\ (8)\\
\le-3&k\equiv7\ (8).\end{cases}\label{Scases}\end{equation}
The case $k\equiv0$ mod 8 of the proposition follows easily from \eqref{Scases}
and the definitions.

We next handle the case $k=8\ell+1$. If $\nu(\ell)\ge3$, then $8\ell+2$ admits a
decomposition with $k_0=2$. Any additional portions of a decomposition of $8\ell+2$
will occur identically in $8\ell$. Thus, in this case, with $\nu=\nu(8\ell)\ge6$,
$$G(e,8\ell+2)-G(e,8\ell+1)=e-7-\langle\min(\nu,e-6)-5\rangle-(6+\langle e-6-\nu\rangle).$$
This is $\le-2$, regardless of the sign of $e-6-\nu$.

Now assume $\nu(\ell)<3$. If $8\ell$ admits a decomposition as $k_0+2^t\alpha$ with $\alpha$ odd,
then we consider $(k_0+2)+2^t\alpha$ as a possible decomposition of $8\ell+2$. Any additional
portions of a decomposition of $8\ell+2$ occur identically in $8\ell$. For $v=\nu(\ell)=0$, 1, or 2,
we obtain
\begin{equation}G(e,8\ell+2)-G(e,8\ell+1)=e-13-\langle e-9-v\rangle-\langle D-2+v\rangle+\langle D\rangle,\label{DE}\end{equation}
where $D=\min(t,e-6)-R(k_0)$. Here we have used the easily-verified fact that
if $k_0\equiv0$ mod 8, then $R(k_0+2)-R(k_0)=5-\nu(k_0)$. One easily checks that
\eqref{DE} is $\le -2$ for any $e$ and $D$, since $0\le v\le 2$.

For $\tau=[2,3,4,5,6,7]$ and $k=8\ell+\tau$, we have, for $e>7$,
$$\langle e-6-\nu'(k+1)\rangle-\langle e-6-\nu'(k)\rangle=[5,-6,6,-5,5,\le-5],$$
and, if $k$ admits a simple decomposition $k_0+2^t\alpha$ with $\alpha$ odd,
$$\langle m-R(k_0)\rangle -\langle m-R(k_0+1)\rangle\le[0,5,0,4,0,5].$$
Here $m=\min(e-6,t)$. As before, higher deviations will cancel in the difference.
Thus $G(e,k+1)-G(e,k)$, which is the sum of the two displays of this paragraph, is $\le6$,
as claimed.
\end{proof}

Now we can complete the proof of \fullref{firstproperty} by proving.
\begin{thm}\label{1mod8} Theorem \ref{firstproperty} is true when
$i$ or $j$ or $i+j$ is $\equiv1$ mod $8$.\end{thm}
\begin{proof} Again we divide into cases.

\smallskip
{\bf Case 1}: Only $i\equiv1$ mod 8. We have
\begin{eqnarray}&&\nonumber G(e-1,i)+G(e-1,j)+1-G(e,i+j)\\
&=&\bigl(G(e-1,i)-G(e-1,i-1)\bigr)-\bigl(G(e,i+j)-G(e,i-1+j)\bigr)\nonumber\\
&&+\bigl(G(e-1,i-1)+G(e-1,j)+1-G(e,i-1+j)\bigr)\label{diff}\\
&\ge&0,\nonumber\end{eqnarray}
since the first $(-)$ in \eqref{diff} is $\ge6$ by \fullref{minus1}, the second is
$\le6$ by \fullref{minus1}, and the third is $\ge0$ by \fullref{not8k+1}.

\smallskip
{\bf Case 2}: both $i$ and $j \equiv1$ mod 8. This follows by an argument similar to
that of Case 1.

\smallskip
{\bf Case 3}: $i$ and $i+j\equiv1$ mod 8. This follows from the validity for $(i-1,j)$
similarly to Case 1. Usually $G(e-1,i)-G(e-1,i-1)=6$ and $G(e,i+j)-G(e,i-1+j)=6$, and so the inequality follows as in \eqref{diff}. If $G(e,i+j)-G(e,i-1+j)>6$, then $G(e-1,i)=S(i)$ and $G(e-1,j)=S(j)$, and so
$$G(e-1,i)+G(e-1,j)+1-G(e,i+j)\ge S(i)+S(j)+1-S(i+j)\ge0$$
by \fullref{Sprop}.

\smallskip
{\bf Case 4}: $i+j\equiv1$ mod 8, while $i,j\not\equiv1$ mod 8. If $G(e,i+j)-G(e,i+j-1)>6$, then
$G(e-1,i)=S(i)$, $G(e-1,j)=S(j)$, and $G(e,i+j)\le S(i+j)$, and so the result follows from \fullref{Sprop}.
So we may now assume $G(e,i+j)-G(e,i+j-1)=6$.
Without loss of generality,
assume $i$ is odd and $j$ is even.

First, we assume $i\equiv3$ mod 4. By the proof of \fullref{minus1},
$G(e,i)-G(e,i-1)=4$ or 5, and if $i$ is indecomposable, then $G(e,i)-G(e,i-1)=
4$ if and only if $G(e,i-1)=S(i-1)$.
Thus the result will follow as in \eqref{diff} once we show that if $i,j\equiv2$ mod 4 and $i+j\equiv0$ mod 8, then \eqref{Gineq} is satisfied with
1 to spare, and with 2 to spare if $G(e,i)-G(e,i-1)=4$.

The basic value of the LHS of \eqref{Gineq} in this case is
\begin{equation}\label{e-8}\langle e-8\rangle+\langle e-8\rangle+1-\langle e-v\rangle\end{equation}
with $v\ge9$. This equals 1 if $e=7$ or 8, while for $e\ge9$, it is $\ge e-6$.
The smallest $e$ for which the LHS of \eqref{Gineq} does not equal \eqref{e-8}
is $e=12$, when $i=2$.

Neglecting temporarily the effect of deviations, the desired conclusion is obtained
since it is true at the onset of $S(i)$ and will continue to be true as $e$ increases,
since now $G(e-1,j)$ and $G(e,i+j)$ will both increase by 1 each time. When $G(e-1,j)$ achieves
a value of $S(j)$, then the LHS of \eqref{Gineq} is
$$\ge S(i)+S(j)+1-S(i+j)>2$$
for the congruences being considered here.

When deviations are taken into account, the fact that makes it work is the easily-verified fact that
\begin{equation}\label{26}R(8\ell+2)+R(8\ell'+6)-R(8\ell+8\ell'+8)=1+\nu\bigl(\tbinom{\ell+\ell'}{\ell}\bigr).
\end{equation}
Suppose, for example, that $i=i_0+2^t\alpha$ and $j=j_0+2^u\beta$ are decompositions with $\alpha$ and $\beta$ odd, and $t<u\le e-7$.
The LHS of \eqref{Gineq} becomes
$$\ge e-8+e-7-(t-R(i_0))-(u-R(j_0))-(e-v)+\langle t-R(i_0+j_0)\rangle+\langle  u-R(2^t\alpha)\rangle$$
with $v\ge9$. Using \eqref{26}, this is
\begin{eqnarray*}&\ge&e+v-14+R(i_0+j_0)-t+\nu\bigl(\tbinom{i_0+j_0}{i_0}\bigr)+\langle t-R(i_0+j_0)\rangle-u\\
&\ge& v-7\\
&\ge&2,\end{eqnarray*}
since $e-7\ge u$ and using \eqref{X}.
Other situations involving decompositions work out similarly.

The case $i\equiv5$ is handled similarly.
\end{proof}

Next we verify the first part of \fullref{gthm}.(4). In fact the conclusion of that theorem is true without
regard for the hypothesis.
\begin{thm}\label{nohyp} If $i,j\le 2^{e-3}$ and $i+j+1\le2^{e-2}$, then
$$g(e,i)+g(e,j)<g(e+1,i+j+1).$$
\end{thm}
\begin{proof} We prove the equivalent statement, with $i$, $j$, and $e$ as in the hypothesis,
\begin{equation}\label{equiv}G(e,i)+G(e,j)+8>G(e+1,i+j+1).\end{equation}
By \fullref{firstproperty} and \fullref{minus1}, we have
$$G(e,i)+G(e,j)+8\ge G(e+1,i+j)+7>G(e+1,i+j+1)$$
unless $i+j+1=2^t+1$ with $t\ge3$ and $G(e+2,i+j+1)=S(i+j+1)$. In this case, it will
also be true that $G(e,i)=S(i)$ and $G(e,j)=S(j)$. Thus it suffices to show
$$S(i)+S(2^t-i)+8>S(2^t+1).$$
This follows readily from the definition of $S$. The smallest value of $S(i)+S(2^t-i)$ occurs
when $i=2^{t-1}$ and is $3\cdot2^{t-1}+2$, while $S(2^t+1)=3\cdot2^{t-1}+9$.
\end{proof}

The second part of \fullref{gthm}.(4) follows from the following result.
\begin{thm}\label{2k} For $k\le 2^{e-3}$, $G(e+1,2k)\le 2G(e,k)$ with equality if and only if
$G(e+1,2k)=G(e,k)=0$, which occurs if and only if $$k\in\{2^{e-3},\ 2^{e-4}, \ 2^{e-5},\ 3\cdot2^{e-5},\ 2^{e-6}\alpha\}\text{ with }\alpha\in\{1,3,5,7\}.$$
If equality occurs, then
$$G(e+1,2k)<G(e,\ell)+G(e,2k-\ell)+1$$
for all $\ell$.
\end{thm}
\begin{proof} The second sentence follows immediately from the first, since
$$0<G(e,\ell)+G(e,2k-\ell)+1.$$

For basic values, we have
$$2G(e,k)-G(e+1,2k)=\begin{cases}2(e-2)-(e-6)&k\text{ odd}\\
\langle e-6-\nu(k)\rangle&k\text{ even.}\end{cases}$$
This is clearly $\ge0$, and $=0$ in exactly the cases claimed.

If $G(e,k)=S(k)$, then
$$2G(e,k)-G(e+1,2k)\ge 2S(k)-S(2k)=\begin{cases}2\alpha(k)-1&k\text{ even}\\
12&k=2^t+1,\ t\ge3\\
2\alpha(k)+4&k\equiv1\ (8),\ \alpha(k)\ne2\\
4\nu(k-1)+2\alpha(k)-4&k\equiv3,5,7\ (8).\end{cases}$$
This is $>0$.

Suppose $k=k_0+2^t\alpha$ is a simple decomposition, with $\alpha$ odd and $e-6\ge t$.
If $k$ is even, then $2R(k)=R(2k)+\alpha(k-1)$, and so
\begin{eqnarray*}&&2G(e,k)-G(e+1,2k)\\
&=&e-6-\nu(k)-2(t-R(k_0))+\langle t+1-R(2k_0)\rangle\\
&=&e-5-t+\alpha(k_0)-1+R(2k_0)-t-1+\langle t+1-R(2k_0)\rangle\\
&\ge&1,\end{eqnarray*}
using \eqref{X}.
If $k\equiv3,5,7$ mod 8, then
$$2R(k)=R(2k)+4\nu(k-1)+\alpha(k)-\begin{cases}10&k\equiv3\ (4)\\ 12&k\equiv1\ (4).\end{cases}$$
Then
\begin{eqnarray*}&&2G(e,k)-G(e+1,2k)\\
&=&e+2-2(t-R(k_0))+\langle t+1-R(2k_0)\rangle\\
&\ge& e+3-t+R(2k_0)-t-1+\langle t+1-R(2k_0)\rangle+4\nu(k_0-1)+\alpha(k_0)-12\\
&>&0.\end{eqnarray*}
The situation when $t>e-6$ and the case of higher deviations are handled similarly.

Finally, we have
\begin{eqnarray*}&&2G(e,8\ell+1)-G(e+1,16\ell+2)\\
&\ge&2(G(e,8\ell)+6)-G(e+1,16\ell)-(G(e+1,16\ell+1)-G(e+1,16\ell))\\
&&\ -(G(e+1,16\ell+2)-G(e+1,16\ell+1))\\
&\ge&12+0-8-(-1)\\
&>&0.\end{eqnarray*}
\end{proof}

Finally, we verify part (2) of \fullref{gthm}. We have
$$g(e,k)=8k-G(e,k)\ge 8k-S(k)\ge13[\textstyle\frac{k+1}2]-2\alpha(k)-10.$$
This is $\ge 4k+4$ for $k\ge7$, while for $k<7$ we verify directly that $8k-S(k)\ge 4k+4$.

\section{A bound for geometric dimension of normal bundle}\label{normsec}
In this section, we prove the following key result, a main ingredient in the proof of our geometric dimension result, \fullref{1better}, which has
already been seen to imply our immersion theorem.
\begin{thm} If $e\ge7$ and $t\ge 1$ and $k_0+\cdots+k_{t-1}=2^{e-3}-1$, then \begin{equation}\label{immsum}\sum_{i=0}^{t-1} G(e+i,k_i)\ge e-2.
\end{equation}\label{gdsum}\end{thm}

\begin{rmk} {\rm The integers $k_i$ in this theorem are nonnegative, but possibly zero. Some examples in which equality is obtained are}
\begin{itemize}
\item $G(e,2^{e-3}-1)$;
\item $G(e,2^{e-4}-1)+G(e+1,2^{e-4})$;
\item $G(e,2^{e-5}-1)+G(e+1,3\cdot2^{e-5})$;
\item $G(e,3\cdot2^{e-5}-1)+G(e+1,2^{e-5})$;
\item $G(e,2^{e-5}-1)+G(e+1,2^{e-5})+G(e+2,2^{e-4})$;
\item $G(e,2^{e-4}-1)+G(e+1,0)+G(e+2,2^{e-4})$.
\end{itemize}
\end{rmk}

Before proving the theorem, we provide the easy deduction of \fullref{1better}.
\begin{proof}[Proof of \fullref{1better}] From \fullref{gdsum} and \eqref{complement},  we obtain
\begin{equation}\sum_{i=0}^{t-1} g(e+i,k_i)\le (2^e-8)-(e-2)=2^e-e-6.\label{ge}\end{equation}
Let $e$ be fixed, and for $t\ge1$ and $0\le\ell\le 2^{e-3}-1$, let
$$M(t,\ell)=\max\biggl(\sum_{i=0}^{t-1} g(e+i,k_i):\ k_0+\cdots+k_{t-1}=\ell\biggr).$$
Then $M(t,\ell)=\max(M(t-1,i)+g(e+t-1,\ell-i):\ 0\le i\le\ell)$. Using \fullref{compatprop}, induction on $t$, and \fullref{gthm}.(6),
we obtain that for all $t$ and $\ell\le 2^{e-3}-1$
$${\on{gd}}((2^e+\cdots+2^{e+t-1}),\ell)\le M(t,\ell)$$
compatibly for all $\ell$. By \eqref{ge}, $M(t,2^{e-3}-1)\le 2^e-e-6$. Since ${\on{gd}}(n,k)={\on{gd}}(n\xi_{8k+7})$, we obtain the conclusion
of \fullref{1better}.
\end{proof}

The proof of \fullref{gdsum} is expedited by the following lemma.
\begin{lem} Let $d>0$. If $G(e,i)<S(i)$ and $G(e+d,j)<S(j)$, then
$$G(e,i)+G(e+d,j)\ge G(e,i+j).$$\label{d>0}
\end{lem}
\begin{proof} This follows exactly as in the proofs of Cases 1, 4, and 5 of
\fullref{not8k+1} and the proof of \fullref{1mod8}. In those results, there was an
extra 1 on the LHS, but the larger $e$-components here more than compensate for that.
\end{proof}
\begin{rmk}{\rm Lemma \ref{d>0} is not always true when $S(-)$ is involved.
For example, if $e\ge15$, then $G(e,2)+G(e+1,3)=12<13=G(e,5)$.}
\end{rmk}
\begin{proof}[Proof of \fullref{gdsum}] Let $\cal S$ denote the set of those $k_i$
for which $G(e+i,k_i)=S(k_i)$. This includes cases in which $k_i=0$ or $k_i=1$.
If $\cal S$ is empty, then the result follows by induction from
\fullref{d>0}, since $G(e,2^{e-3}-1)=e-2$. Let $K=\sum_{k_i\in\cal S}k_i$. We split the LHS of \eqref{immsum} as
\begin{equation}\sum_{k_i\in\cal S}G(e+i,k_i)+\sum_{k_i\not\in\cal S}G(e+i,k_i).\label{splitsum}\end{equation}
Since, as is easily proved, $S(k)\ge\frac32k$, the first half of \eqref{splitsum} is $\ge\frac32K$, while \fullref{d>0}
implies that the second half of \eqref{splitsum} is $$\ge G(e,2^{e-3}-1-K)\ge e-6-\nu'(K+1)-D(e,2^{e-3}-1-K),$$
where $D(-,-)$ denotes the deviation, i.e., the sum in \eqref{geq1}. Now the desired inequality reduces to
\begin{equation}\label{32}\textstyle\frac 32K\ge \nu'(K+1)+4+D(e,2^{e-3}-1-K).\end{equation}
If $K\ne1,3$, this inequality is true, usually with much to spare. Indeed, $K\ge\nu'(K+1)+4$ if $K\ne1,3$,
and \begin{equation}\label{34}\tfrac12K\ge D(e,2^{e-3}-1-K).\end{equation}
To see \eqref{34}, note that for $D(e,k)$ to be positive due to a single deviation, then $k=2^t\alpha+k_0$ with
$t> R(k_0)>k_0$, $\alpha$ odd, and $D(e,k)=t-R(k_0)$. For such $k$, if $k=2^{e-3}-1-K$, then $K\ge 2^t-1-k_0$, and so the difference
in \eqref{34} is $$\ge\tfrac12(2^t-1-k_0)-(t-R(k_0))=(\tfrac12(2^t-1)-t)+(R(k_0)-\tfrac12k_0)>0,$$
and a similar analysis applies when multiple deviations are involved. When $K=1,3$, \eqref{32} is true if the LHS is replaced by $S(K)=8$.
\end{proof}

\def\rule{.6in}{.6pt}{\rule{.6in}{.6pt}}

redpajama_set_name: RedPajamaArXiv


Illustration of an ice cream cone with yellow ice cream and yellow text outlined in black over the top all on a white background.
In the year from 1985 to 1986, Donvier ice cream makers made up 30% of the ice cream makers sold. Manufactured by Nikkal Industries in Virginia Beach, Virginia, the compact, non-electric machines were advertised on their ease of use. They required no electricity, no salt, no ice, and just a few turns of a handle. Donvier machines worked by the use of "Chillfast", a non-toxic refrigerant that was sealed inside an aluminum tub. Once the tub was frozen, the ingredients would freeze and stay cold upon contact. In 1986, a half-pint Donvier with the design featured on this button was released for customers who wanted to make smaller batches of ice cream. Donvier also made small machines designed for children, which included instructions and recipes that young readers could follow.


text:
Mark Iossifowitsch Grajew (; Moscou,  – ) foi um matemático russo.

Foi palestrante convidado do Congresso Internacional de Matemáticos em Moscou (1966: Theory of Representation of Groups com Alexandre Kirillov.

Ligações externas

Graev bei mathnet.ru

Matemáticos da Rússia
Matemáticos da Rússia do século XX
Naturais de Moscou

redpajama_set_name: RedPajamaWikipedia


0426 267 300 - We come to you, from Sydney Harbour to Central Coast!
Bird Electrical is proud to provide sales, installation and service for Yacht Controller wireless yacht manoeuvering systems. Bird Electrical is the premium installation agent for NSW based.
Why use an Authorised Garmin Supplier?
Why use an authorised Garmin dealer ? Well, we were recently approached by a client who was intent on removing his Garmin equipment from his relatively.
Peter Bird will be on the Garmin Stand at the Sydney International Boat Show 2018. The show starts on Thursday the 2nd of August and Peter will be.
Bird Electrical has recently opened a second workshop at Fenwicks Marina providing coverage for the Hawkesbury region. Bird Electrical can provide licensed electrical installation and maintenance, electronics.
New Garmin GPSMAP models now available Further expanding Garmin's popular GPSMAP multifunction display series, the new 752xs/952xs touchscreen and 1022xsv/1222xsv rotary control/keyed units are now available.


Reflection is the examination of our own actions and experiences in order to learn from them. Writing is a vehicle for reflection, and helps to give permanence to our thoughts and ideas at a particular point in time.
By recording thoughts, ideas and emotions you will become more aware of the changes over time in your ways of thinking and doing. This is why you may be asked to do reflective writing as part of your course: it helps others to clearly understand what you have learned, and how you plan to use this in developing future work.


text:
Ballet training in the UK
By balletla,
February 8, 2014 in Doing Dance
balletla
I have just been watching a clip of a 10 year old Korean girl performing in a ballet competition on pointe, at a level that is way beyond anything I would expect to see in the UK. I know everyone will go on about the dangers of going on pointe at a young age but I think it is more than that. These kids are obviously being pushed in all areas.
With a significant proportion of RBS Upper School and both the Royal Ballet and Birmingham Royal Ballet made up of dancers from China, Japan and other Asian countries, should we be asking whether our training institutions are still fit for purpose. I know we provide 'safe' training in this country but our students are obviously significantly behind when selections for ballet companies are being made.
Anjuli_Bai 16 posts
aileen 19 posts
drdance 24 posts
mimi66 18 posts
I've seen this clip before, usually I find many of the youtube 'baby ballerina' clips slightly gymnasticy and vulgar at the very best and downright scary at the worst. This little one seems to have sound technique and a lovely quality. Yes, we all freak out when we hear of girls going en pointe before the age of 11 etc, but maybe exceptions should be made in certain cases.....Interesting point Ribbons, the safety of a young dancers body is paramount, which hopefully leads to longevity in their career.....(but if they're not getting the contracts there's no career full stop......?)
drdance
This thread raises a many issues for me (apologies for the spilling out of many of them now!)
Personally I'd have enjoyed that little girls' performance just as much if she'd been on demi pointe; there's no reason for her to be on pointe but she's clearly pretty strong so only time will tell if she's going to suffer for being on pointe at such a young age
Watching with my *very* critical eye, I don't think her turnout was particularly well maintained, and a short tutu would reveal this more
Her performance was quite nicely 'immature' ie it didn't show a mature style of dancing but then she's a little girl and actually it was nice that it wasn't overly mature for her age
I agree that in the UK the majority of dance training does not really push children to advanced levels when they are young and I think we can learn an awful lot from training of children in countries like USA, Australia and in North Asia especially in terms of the physical training but it does need to be done with care.
To be able to dance to this level, you need a lot of physical strength, flexibility and technical skill and control. To get this anyone needs to be training daily and training needs to be very focussed (and as I said before, with immense care)
I think that in the UK there is a general culture of frowning upon pushing young children, celebrating success, promoting viruosity. We're very 'British' in not wanting to be big headed or bratty so we inherently pass this message onto our children. Consequently our young dancers might not actually strive to be the best in the same way or at least part of them doesn't want to be the best perhaps.
In the countries where we see more successful children, there is more competition as there is a larger population. We think the competition for places at vocational schools etc is steep but imagine being in country with a population 10 times larger than our own!
Is it actually necessary for the all-round welfare of our children, for them to be performing full variations, en pointe, age 10?!
I agree with everything you said drdance, but thinking competitively, if this little lady is pulling off doubles en pointe at 10, what on earth is she going to be capable and quite comfortable doing at the age of 14/16/18...which could possibly clinch her that all important contract over someone who had been introduced to fouettes en pointe (for example) at the age of 15/16.....I'm the very first person to shout out that dance is an art and not a sport etc etc and I'm no fan of over exaggerated extensions and a million pirouettes 'just for the sake of it', but it's such a horribly competitive market and if the companies want higher, faster, stronger, then that's what you have to be able to do....x
She is beautiful. I would suggest more gymnastics for stretching in our country. I know the Russian schools do gymnastics for ballet.
Anjuli_Bai
Location: San Diego, California, USA
Li Cunxin's book is well worth reading.
I would much rather the slow nurture of talent.
How many of these pre-pubescent "ballerinas" end up with careers of any length and fame for artistry?
I wonder what the real cost is? Thinking in statistical terms, what is the "number needed to treat" here? How many damaged children to produce each successful dancer? What is happening to their general education whilst this training is going on, and what happens to those who "fail"? One wonders if children like this have actually chosen their path, or have they been selected at a very early age on the basis of physique/genetics etc, bearing in mind that the political and cultural situations are very different in different countries. It puts me in mind of the stories one hears about the Eastern Bloc gymnasts of the 70s and 80s.
I'm sure there are things that the UK schools can learn from those in other countries, but I would hope they would be very selective abut what aspects they would adopt. Even outstanding results don't always justify the means.
I totally agree with you - I think you summed it up beautifully here.
The only thing I will add is that the Japanese student coming over are outstanding, is it because they are the elite or the way they are trained???
I agree with you too Anjuli regarding both the nurturing of talent, longevity of career and artistic quality (and will google the book you mention) However, we have to face the cold, hard reality that after many years of training and in most cases great financial expense, many students are simply not being offered contracts. (Of course, there have never been enough places available for the graduates...)
Pups mum. I understand, but it doesn't have to be taken to the extremes. For example I used to observe and take class with a fantastic Cuban teacher. He would get his 12 year old (vocational students) to turn...(and boy do the Cubans know how to teach turns !) Obviously they had already mastered the basics of a pirouette, but at the end of class he'd encourage them to do doubles, triples, the next year fouettes....Now these girls were by no means executing perfect turns first of all, but they had no fear....and a couple of years down the line, wow, they were incredible.
I think it's the training matched with the sheer hard work and determination. DS has some Japanese students at his school and you never, ever see them give up.
Great discussion ! Thanks Ribbons !
Thecatsmother
My take is always never to judge a cake until it is fully baked so to speak. There are so many factors here and some kids who are amazing at 11 pale into insignificance as other dancers strengthen over the passage of time.
This is an issue I have always had with dance festivals as it is often the shorter and easier to strengthen type bodies who will win when in the long term there are many with more difficult to train physiques who actually have the most potential.
I have noticed Japanese dancers seem to a lot less fearful in terms of turning because they tend to start turning earlier. I think there is some merit in starting to get the sense of turning early when the fear factor is less present. By this I mean not pirouettes on pointe but just the general feel of turns so they become fun.
Like a good wine, some dancers take longer to reach their peak than others but that does not make the ones who peak earlier better dancers. There are also dangers of picking up bad technical habits by starting on pointe too young as compensation tends to occur to make up for the lack of strength and experience.
Lisa O`Brien
Location: Newry,County Down Northern Ireland
I lived and worked in Japan for 3 years as a dancer [not ballet] in the early 1990`s. There was very little interest in Classical Ballet in the country then.If in one of our dance routines,there was to be a classical pas de deux,the management would tell us to change it dramatically as people were not interested in ballet. I think the interest therefore must only have come about in the last few decades. When the Japanese decide to do something they do it to excel at it. Their school hours and work load are absolutely gruelling,and it has always been traditional that the children would partake in extra curricular activities as well such as piano lessons or violin.The ballet lessons,therefore are probably just a natural extension of their education for some children nowadays.
s.dancer
I think the children in these countries such as Japan and USA do have more intense training than in the UK probably giving them a better chance of making it. Yes they are young to be on pointe and also the dancers in their countries also do a lot more pointe! I have been on pointe since I was 11, and now I have just turned 14 have only just started more intense pointe such as more turns an soutenus even though I am at a vocational school! Sometimes I wish I was training in America
taxi4ballet
Is it becoming more difficult for UK students to compete with overseas students when auditioning for upper schools, and if so - why?
IInteresting point Ribbons, the safety of a young dancers body is paramount, which hopefully leads to longevity in their career.....(but if they're not getting the contracts there's no career full stop......?)
Agree with all of the points above. The thing is, there are very set points in time when UK students are competing against these more advanced students i.e. Upper School auditions at 16 and Company auditions 3 years later. They are being judged at that moment in time against their peers. I don't think the selectors are thinking much about 'maturing fine wine' at that point sadly
I haven't had first hand experience of the Upper School audition process for quite a few years (I was referring more to company auditions), but I would hazard a guess that the wealth of information that the internet brings, schools going overseas to audition and ease of travel makes the world a much smaller place. The information about schools and auditions that these 'youngsters' have at their fingertips is incredible. In the '80's and early '90's if you wanted to find a job in say, Germany for example, you'd get a railcard, book a tour, stay in hostels and knock on doors to see if they'd let you do class ! Now you can go to one of a number of websites and find out whats on and when within a matter of minutes ! x
Katymac
It's interesting what Balleteacher is saying about 'turning', as you know DD started Ballet late - but her teacher has always been impressed with her turns (both Ballet & modern). I've put this down to the fact that as a Ballroom/Latin dancer she has turned, at speed, for years and spotted easily - for Viennese Waltz it's essential. So maybe other genre have useful skills for Ballet.
I'll be interested to see if, when she starts working in a couple whether this is also a bit she is good at because of past experience
I'm afraid I don't know enough about Ballet or even dance training to comment on ages or speed of training
Location: South West
I have been surprised that it does not appear to be the norm for vocational school (from 11) dancers to be given a personal training/stretching/ conditioning programme. I know my dd would benefit from a certain focus on specific aspects of her training and I am sure that this is the same for most children. Classes can cover so much, but personalisation would be so helpful. I know it might be difficult to manage, but not impossible with some good organisation. Any thoughts why schools don't seem to see the benefit of this (or do they and it has passed me by?)
I think they do if it becomes apparent a child has a particular need
My dd isn't injured but was advised by her ballet teacher to see the physio which she has been doing weekly (covered by insurance) & it's having a real effect.
They also seem to be doing a lot of Pilates in class.
Karen - perhaps it's for the same reason schools don't have a dedicated specialist in this area on staff full time (ie cost). I know of a schools that do not even have physiotherapists on staff, but require students to have medical insurance. Once injured, students see a private physiotherapist (who may or may not visit the school) and the medical insurance company covers the cost of treatment. In my opinion prevention is better than cure but I don't think it's fair that a student had to get injured before the gain access to a personalized injury prevention/ rehab programme.
Dds physio is covered by the insurance even though its injury prevention related.
We just had to see a gp who wrote a referral.
I think that vocational schools (UK) have come a long, long way over the last decade or so in terms of the importance of nutrition advice, injury prevention and cross training...I think that it is so important for ballet students to do cross training (pilates, gyrrotonics, weight training etc) to enhance the ballet technique and to strengthen the 'under used' muscles, tendons etc which may not be essential in a ballet class, but when weak, lead to over compensation on other muscles etc and thus injury. Dance science and medicine may be still in relative 'infancy' but it is invaluable. Finance, however is obviously a big issue...
The issue with finance isn't really finance, it's all to do with how the directors/governors of the school choose to allocate the money that they do have ie priorities. Until cross-training, injury prevention, body conditioning/fitness training that is NOT ballet class etc IS a priority, it won't be funded.
LinMM
Totally agree Ellie.
I have lived in Japan for a while - from what my dancing friends there told me, things are rather different from what we in the West sometimes assume.
Their interests for ballet goes back longer than we might think. Ballet as a performing art was introduced around 1880's and after the russian revolution in 1917 russian dancers who fled from the communists government settled in Japan via Siberia and strated to teach ballet there. Ballet is not exactly a popularlist pass time there, but that is the same in the UK.
It has been a popular thing for the girls to take ballet lessons for about 50 years now. Nowadays you see young dancers who are taught by their grandmothers.
With regards to foreign students in vocational school here, one must keep in mind that those students were elite students in their own country already.
In addition, one tends to forget that Japan is twice as populous as the UK.
Japanese students here generally try harder, because they feel they are physically disadvantaged in comparison to European students so they know they have to be better technically and artistically.
By the way, at least nowadays, no serious ballet studios in Japan put little girls on pointe until they are over 11. At least that's what I heard.
I don't know what it is like in South Korea... I do hear that kids there do get "pushed" by parents a lot.
Edited February 8, 2014 by mimi66
With regards to the clip. She does seem to have a very nice presentation apart from other skills and doesn't look as if she is struggling on pointe which you do see on some awful YouTube videos. However she could just be exceptionally talented.....even by Korea's standards!!
Wasn't Olga Korbet crippled by arthritis by the time she was 30? Admittedly her "career" was over by then and she had all the benefits which came with it from the then Soviet system but it is a considerable price to pay I think and one which should be considered when training young girls/boys. What happens down the line a bit?
Maybe there are some dancers who just don't care in the end though. A dancing career is usually shortish for the majority of dancers anyway and they would be happy to have this chance of an early career and trade possible future mobility for it.....in the worse case scenario that is.
My feeling is the problem of burn out could come if there is too much advanced dancing at an earlier age......more emotional than physical I mean. But perhaps some dont mind a shorter career in ballet and don't want to be having a long career for 20 years or more who knows? But if a longer career is envisaged then its probably best not to push too much at an earlier age......the body can only take so much punishment over time......but for exceptional cases.
I think lovely Marienuela Nunez started at a very young age and is still going from strength to strength but she may be one of these exceptions.
Marienela Nunez was dancing professionally in Argentina at 14 and only went to the RBS (for about a year) because she was too young to join the RB (ie she was not yet 16). I read that Laura Morera was able to do many fouettés when she arrived at the RBS at 11. Maina Gielgud has said that she was asked to get pointe shoes after her first couple of lessons at 8 (or possibly 9) and she did not have any problems with her feet and danced professionally at an early age. Children do seem to be more advanced at an earlier age abroad.
Maina Gielgud has said that she was asked to get pointe shoes after her first couple of lessons at 8 (or possibly 9) and she did not have any problems with her feet and danced professionally at an early age. Children do seem to be more advanced at an earlier age abroad.
This did make me chuckle slightly - I don't know how old Maina Gielgud is now but having seen her in and around ENB one might assume that dancing has taken a considerable toll on her joints! Walking seemed tough at times! But her arms & head when rehearsing were still beautiful
One would hope that injury prevention is already a high priority!!!

redpajama_set_name: RedPajamaCommonCrawl


Notice: HRI follows the standard ACM Policy on Prior Publication and Simultaneous Submissions.
Send feedback and questions to pc@hri2017.org or to the chairs of your preferred theme.
Best wishes, and we look forward to seeing your successful submissions at HRI 2017.


The best way to get where you want to go is to know where you current are. And we've made it simple.
Give us some quick info about your website in the form.
We'll provide you with an in-depth analysis of your current web presence WITHIN A FEW HOURS of your request (unless it's after hours, then give us until the next business day).
After we've sent over your report, we'll schedule a time to meet with you to explain what it all means.
Best Performers for the Keywords you provided.
Backlink Profile including Toxic Backlinks Point to your Website.
SEO Factors That Influence Your Website's Visibility On Search Engines.
Where your Website Currently Ranks for the Keywords You Chose.
Your Website's Compliance with Google Requirements.
Usability Issues Affecting Your Website like Mobile Friendly and Page Speed Score.


text:
Q: With this rule , how can I say that the same field is greater than or less than 0? Code php:
<?php echo $prod['gross_total'] >= 0 ?>
<?php echo $prod['gross_total'] <= 0 ?>

In the same field have the values , $300, $350, $-100, want to get these values and put on the table as higher values for the left and lower values for the right side of the table.
entries | outputs
$300    | $-100
$350

A: With the ternary operator you could use:
$prod['gross_total']=-1;
echo $prod['gross_total']>0 ? 'more than 0' : 'less than or equal to 0';

the 3 parts are as follows:

*

*Check for true/false

*Do if True

*Do if false

With a switch you can do this:
$prod['gross_total']=-1;
switch (TRUE) {
    case ($prod['gross_total']>0):
        //do this code
        echo "positive";
        break;
    case ($prod['gross_total']<0):
        //do this code
        echo "negative";
        break;
    default:
    //do nothing
        echo "neutral";
}

With array sorting:
$prod['gross_total']=array("1" => "3", "2" => "-1", "3" => "-4", "4" => "4");
asort($prod['gross_total']);
foreach ($prod['gross_total'] as $key => $val) {
    echo "$key = $val
";
}

A: You could use PHP's spaceship operator to do multiple comparisons in one statement.
(expr) <=> (expr)

Returns 0 if both expressions are equal, 1 if the left is greater, and -1 if the right is greater.  So you could use it like this:
echo $prod['gross_total'] <=> 0;

//echoes 0 if $prod['gross_total'] = 0,
//echoes 1 if $prod['gross_total'] > 0
//echoes -1 if $prod['gross_total'] < 0

redpajama_set_name: RedPajamaStackExchange