Maurizio Parton
Dipartimento di Matematica ``Leonida Tonelli''
via Filippo Buonarroti 2
I--56127 Pisa
math0009156.tex
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%\title{}
\author{ Maurizio Parton,\\
 Dipartimento di Matematica ``Leonida Tonelli'',\\
 via Filippo Buonarroti 2, I--56127 Pisa,\\
 Italy. E-mail:  {\tt parton@dm.unipi.it}%\\
}
\date{}


\begin{document}

\title{Explicit parallelizations on products of spheres}

%\author{Maurizio Parton}

\maketitle


%\tableofcontents

\section{Introduction}

It is a classical result in Algebraic Topology that spheres $S^n$
are
parallelizable if and only if their dimension is $n=1,3,7$.
%(see~\cite{AdaNEE}).
As for the products of two or more spheres
one has instead the following
result (\cite{KerCIG}):

\begin{teo}[Kervaire]\label{teokervaire}
The manifold $S^{n_1}\times\dots\times S^{n_r}$, $r\ge 2$, is
parallelizable if and only if at least one of the $n_i$ is odd.
\end{teo}

Kervaire's proof does not provide an explicit parallelization on
products of spheres.
%The article \cite{BruPEP} is
The only reference the
author knows to provide explicit parallelizations
is \cite{BruPEP}, that considers the cases when one of the spheres is of
dimension $1,3,5,7$,
and uses some specific arguments of these low dimensions. In \cite{BruPEP} the
general case is left as an open problem.

%
% on some products of
%spheres (namely, whenever one of the factors is $S^1,S^3,S^5,S^7$)
%is the paper \cite{BruPEP}.
%The general case is left as an open
%problem.

The aim of this paper is to write an explicit orthonormal
parallelization for all
parallelizable products of spheres, using an explicit
isomorphism with a trivial vector bundle obtained
following a hint of \cite{HirDTo}.

%
%In this paper the never-vanishing vector field on the
%odd-dimensional sphere is used
%to write an explicit
%isomorphism between $T(S^m\times S^n)$ and $S^m\times
%S^n\times\R^{m+n}$,
%following a hint of \cite{HirDTo}. An explicit orthonormal
%parallelization on $S^m\times
%S^n$ is then obtained
%pulling back the standard basis
%of $\R^{m+n}$.
%a series of hints contained in the book \cite{HirDTo}, one can
%use this vector field
%
%A suitable choice of  ingredients in the above construction
%gives an orthonormal frame on $S^m\times S^n$. If both $m$ and $n$ are
%odd, an almost-Hermitian
%structure on $S^m\times S^n$ is canonically associated to the
%frame by means of the standard
%unitary matrix
% whose only non-null entries are the blocks
%\[
%\left(\begin{array}{cc}
%    0&-1\\
%    1&0
%    \end{array}\right)
%\]
%on the main diagonal.
%In this paper it is shown, as an application of the explicit
%parallelization, that this almost-Hermitian
%structure on $S^m\times S^n$ is actually integrable, in fact
%that it coincides with a
%Calabi-Eckmann structure.

A description of some $G$-structures on $S^m\times S^n$
associated to this parallelization is given in \cite{ParPhT}.
%
%In section \ref{casogenerale} a relation between vector bundles
%(lemma \ref{lemmabundles}) is used to build up
%an isomorphism between
%$T(S^m\times S^n)$ and the trivial vector bundle
%$\varepsilon_{S^m\times S^n}^{m+n}$ of rank $m+n$
% on $S^m\times S^n$.
%The pull-back of suitably chosen sections of
%$\varepsilon_{S^m\times S^n}^{m+n}$ gives (theorem~\ref{teocomptes})
%an orthonormal
%parallelization $\mathcal{P}$ on $S^m\times
%S^{n}$, odd $n$,  which solves the
%problem left open in \cite{BruPEP}.
%
%In section \ref{casosemplice} the Calabi-Eckmann complex structure
%$I^{m,n}$ on $S^{m}\times S^{n}$, for odd $m,n$, is defined.
%Remark \ref{osscomptes}
%is the main block in the proof of theorem
%\ref{teo2comptes}: it is shown that when
%$n=1$, namely, in the Hopf case, the Calabi-Eckmann
% complex structure results to
%be the almost-Hermitian structure canonically
%associated to a natural
%parallelization, closely related to $\mathcal{P}$.
%Finally (theorem
%\ref{teo2comptes}) it is proven that
%$I^{m,n}$ is the structure canonically associated to $\mathcal{P}$.
%






\section{An explicit parallelization $\mathcal{B}$ on $S^m\times S^1$}\label{genm1}

Denote by $x=(x_i)$
the coordinates on $\R^{m+1}$, and let $S^m\subset\R^{m+1}$ be
given by
\[
\begin{split}
S^m&\ug\{x=(x_1,\dots,x_{m+1})\in\R^{m+1}\st
|x|^2=x_1^2+\dots+x_{m+1}^2=1\}.\\
%S^1&\ug\{y=(y_1,y_{2})\in\R^{2}\st
%|y|^2=y_1^2+y_{2}^2=1\}.
\end{split}
\]
The orthogonal projection of the standard coordinate
frame $\{\pa{x_i}\}_{i=1,\dots,m+1}$ to the sphere plays an
important role in the game, and deserves its own definition:
\begin{defi}
%\begin{definonumber}
%Let $S^m\subset\R^{m+1}$, and let
%$\{\pa{x_1},\dots,\pa{x_{m+1}}\}$ be the standard vector fields on
%$\R^{m+1}$.
The {\em $\imath^{\text{th}}$ meridian vector field} $M_i$
on $S^m$ is %the vector field given in any point $x\in S^m$ by the
%orthogonal
%projection of $\pa{x_i}$ to $T_x(S^m)\subset T_x(\R^{m+1})$:
\[
\begin{split}
M_i&\ug\text{orthogonal
projection of $\pa{x_i}$ on $S^m$}\qquad i=1,\dots,m+1.\\
%N_j&\ug\text{orthogonal
%projection of $\pa{y_j}$ on $S^{n}$}\in\Chi(S^m\times
%S^{n})\qquad j=1,\dots,n+1.
\end{split}
\]\EndDef
%\end{definonumber}
\end{defi}
%\begin{oss}
Let $M$ be
the normal versor field of $S^m\subset \R^{m+1}$, that is,
\[
M\ug\sum_{i=1}^{m+1}x_i\pa{x_i}.
\]
Since
\[
\langle \pa{x_i},M\rangle=x_i \qquad i=1,\dots,m+1,
\]
one obtains the following expression for $M_i$:
\begin{equation}\label{fundeq}
M_i=\pa{x_i}-x_i M\qquad i=1,\dots,m+1,
\end{equation}
and thus
\begin{equation}\label{fundeqscal}
\langle M_i,M_j\rangle=\delta_{ij}-x_i x_j\qquad i,j=1,\dots,m+1.
\end{equation}
%\end{oss}

%\section{Particular cases: $S^m\times S^1$, $S^m\times S^3$ and $S^m\times
%S^7$}\label{par137}



%\section{A parallelization on $S^m\times S^1$}\label{genm1}

Let $\Gamma$ be the cyclic infinite group of transformations
of $\R^{m+1}\setminus 0$ generated by the map $x\mapsto e^{2\pi}x$.
Denote by $H$ the
corresponding diagonal real Hopf manifold, that is,
the quotient manifold $(\R^{m+1}\setminus 0)/\Gamma$:
$H$ turns out to be diffeomorphic to $S^m\times
S^1$ by means of the map
induced by the projection $p$:
\[
\begin{array}{cll}\label{projm1}
\R^{m+1}\setminus 0 & \stackrel{p}{\longrightarrow} & S^m\times S^1\\
 x & \longmapsto &
 (x/|x|,\log|x| \mod{2\pi}).
\end{array}
\]




% and by
%$f$ the map from $H_{e^{2\pi}}$ to $S^m\times S^1$
%induced by the equivariant map
%\begin{equation}\label{eqequiv}
%\begin{array}{cll}
%\R^{m+1}\setminus 0 & \longrightarrow & S^m\times \R\\
% (x_1,\dots,x_{m+1})& \longmapsto &
% (x_1/|x|,
% \dots,x_{m+1}/|x|,\log|x|).
%\end{array}
%\end{equation}

%{\color{red}quelle reali non si chiamano Hopf... citare Vaisman-l.s.m?}

The standard coordinate frame $\{\pa{x_i}\}_{i=1,\dots,m+1}$ on $\R^{m+1}\setminus
0$ becomes $\Gamma$-equivariant when multiplied by the function $|x|$,
whence it defines a parallelization on $S^m\times S^1$.
This proves the following proposition\dots
\begin{pro}
$S^m\times S^1$ is parallelizable.
\end{pro}
\dots and enables us to give the following definition:
\begin{defi}\label{Bdef}
Define
${\cal B}=\{b_i\}_{i=1,\dots,m+1}$
as the frame  on $S^m\times S^1$ induced
by the $\Gamma$-equivariant frame $\{|x|\pa{x_i}\}_{i=1,\dots,m+1}$ on the
universal covering $\R^{m+1}$ of $S^m\times S^1$ by means of $p$:
%if $p$ denotes the canonical projection of $\R^{m+1}\setminus
%0$ on $H_{e^{2\pi}}$ and
%$f$ the map
%induced by \myref{eqequiv},
%Define the frame $\{b_1,\dots,b_{m+1}\}$ on $S^m\times S^1$ by
\[
b_i\ug
p_*(|x|\pa{x_i}(x))
\qquad{i=1,\dots,m+1}.
\]\EndDef
\end{defi}




%\begin{oss}\label{Bdef}
%Let $p$ be the canonical projection on $H_{e^{2\pi}}$.
%Define the frame $\{b_1,\dots,b_{m+1}\}$ on $S^m\times S^1$ by
%\[
%b_i\ug f_*(p_*(|x|\pa{x_i}))\qquad i=1,\dots,m+1.
%\]
%\end{oss}

%Notation $\{b_1,\dots,b_{m+1}\}$ is not by chance the same as in section
%\ref{secparexp}:

The following theorem explicitly describes the frame $\mathcal{B}$:
\begin{teo}\label{progeo}
Let $M_i$ be the $i^{\text{th}}$ meridian vector field
on $S^m\subset\R^{m+1}$.
%, and let $T$ be the
%vector field on $S^1$ given by multiplication by $i$,
%namely, $T=-y_2\pa{y_1}+y_1\pa{y_2}$.
Then
\begin{equation}\label{progeofor}
b_i=M_i+x_i\pa{\T} \qquad i=1,\dots,m+1.
\end{equation}
\end{teo}

\begin{D}
Look at $S^m\times S^1$ as a Riemannian submanifold
of $R^{m+1}\times S^1$, and in particular look at $T(S^m\times
S^1)=TS^m\times TS^1$ as a Riemannian subbundle of
$TR^{m+1}_{|_{S^m}}\times
TS^1$; this last is a trivial vector bundle and an orthonormal frame is
$\{\pa{x_1},\dots,\pa{x_{m+1}},\pa{\T}\}$. A computation then
shows  that
\begin{equation*}
\begin{split}
p_*=\di\frac{1}{|x|}\left((dx_1-x_1\omega)\otimes\pa{x_1}+\dots+
(dx_{m+1}-x_{m+1}\omega)\otimes\pa{x_{m+1}}+|x|\omega\otimes \pa{\T}\right),
\end{split}
\end{equation*}
where  $\omega$ is
the 1-form
given  by
\[
\omega\ug -d\left(\di\frac{1}{|x|}\right)=
\di\frac{1}{|x|^2}\left(x_1dx_1+\dots+x_{m+1}dx_{m+1}\right).
\]
Whence, the frame $\cal B$
in the point $p(x)=(x/|x|,\log|x| \mod{2\pi})$
is given by
\begin{equation*}
\begin{split}
\di\frac{1}{|x|^2}\left(-x_1x_i\pa{x_1}+\dots+(|x|^2-x_i^2)\pa{x_i}
+\dots-
x_{m+1}x_i\pa{x_{m+1}}+|x|x_i\pa{\T}\right)
\end{split}\qquad i=1,\dots,m+1,
\end{equation*}
that is,
the frame $\cal B$
in the point $(x,\T)\in S^m\times S^1$ is given by
\begin{equation}\label{bruni1}
\begin{split}
b_i&=\left(-x_1x_i\pa{x_1}+\dots+(1-x_i^2)\pa{x_i}+\dots-
x_{m+1}x_i\pa{x_{m+1}}+x_i\pa{\T}\right)\\
&=\pa{x_i}-x_i(x_1\pa{x_1}+\dots+
x_{m+1}\pa{x_{m+1}})+x_i\pa{\T}\stackrel{\myref{fundeq}}{=} M_i+x_i\pa{\T}.
\end{split}\qquad i=1,\dots,m+1.
\end{equation}
\end{D}
%{\color{red}1) troppi conti?}



%\begin{figure}
%\caption{$\pa{x_i}$ decompose in $M_i+x_i M$, where $M$ is the unitary
%normal vector field, and $f_*\circ p_*$ send $M$ to $\pa{\T}$.}
%\end{figure}

\begin{oss}
The notion of meridian vector field was given in
\cite{BruPEP}: it was used to describe a
parallelization on any product of  a
sphere by a parallelizable manifold. In this context, theorem
\ref{progeo} shows that the frame $\mathcal{B}$ given by definition \ref{Bdef}
coincide with that of \cite{BruPEP}.\EndRem
\end{oss}

\begin{oss}\label{remor}
The frame $\cal B$ is orthonormal with respect to the product
metric on $S^m\times S^1$ (use theorem \ref{progeo} and formula \myref{fundeqscal}).\EndRem
% But it is positively oriented
%with respect to the standard orientation on $S^m\times S^1$ if and
%only if $m$ is even.
\end{oss}




%{\color{red}2) formula troppo banale? pag. 6!!!}


The well-known bracket formula %(see for instance \cite[page 6]{KoNFD1})
\begin{equation}\label{kobano}
[fX,gY]=fg[X,Y]+f(Xg)Y-g(Yf)X
\end{equation}
gives the brackets of $\cal B$:
\begin{equation}\label{bracket1gen}
[b_i,b_j]=x_ib_j-x_jb_i\qquad i,j=1,\dots,m+1.
\end{equation}


%Let $\tau$ be the 1-form on $S^1$ dual to $T$, that is,
%\[
%\tau=-y_2dy_1+y_1dy_2.
%\]
By means of theorem \ref{progeo} and remark \ref{remor},
one obtains the coframe ${\cal B}^*\ug\{b^i\}_{i=1,\dots,m+1}$ dual to  $\cal B$ on
$S^m\times S^1$:
\begin{equation}\label{cobruni1}
\begin{split}
b^i&= \copa{i}+x_i d\T
\end{split}\qquad i=1,\dots,m+1.
\end{equation}

\begin{oss}
Since
\[
b_i=p_*(|x|\pa{x_i})\qquad i=1,\dots,m+1,
\]
the coframe ${\cal B}^*$ can be also described as the quotient of
the
$\Gamma$-invariant coframe on $\R^{m+1}\setminus 0$ given by
\[
\{|x|^{-1}\copa{i}\}_{i=1,\dots,m+1}.
\]\EndRem
\end{oss}

A straightforward computation gives the structure equations for $\cal
B$:
\begin{equation}
db^i=\copa{i}\wedge d\T\stackrel{\myref{cobruni1}}{=}b^i\wedge{d\T}\qquad i=1,\dots,m+1,
\end{equation}
where the 1-form $d\T$ is related to ${\cal B}^*$ by
\[
d\T=\sum_{i=1}^{m+1}x_ib^i.
\]



The following lemma is trivial to prove, but will be useful:
\begin{lem}\label{lemgen}
For each permutation $\pi$ of $\{1,\dots,m+1\}$, the automorphism
of $\R^{m+1}\setminus 0$ given by $(x_1,\dots,x_{m+1})\mapsto
(x_{\pi(1)},\dots,x_{\pi({m+1})})$ is $\Gamma$-equivariant. The induced
diffeomorphism is
\dismap{f_\pi}{S^m\times S^1}{S^m\times
S^1}{(x_1,\dots,x_{m+1},\T)}{(x_{\pi(1)},\dots,x_{\pi({m+1})},\T),}
and $df_\pi(b_{\pi(i)})=b_i$.
\end{lem}



\section{The Hopf fibration $S^3\rightarrow S^2$ extends $\mathcal{B}$ to
$S^m\times S^3$}\label{genm3}

Denote by $y=(y_j)$
the coordinates on $\R^{4}$, and let $S^3\subset\R^{4}$ be
given by
\[
\begin{split}
S^3&\ug\{y=(y_1,\dots,y_{4})\in\R^{4}\st
|y|^2=y_1^2+\dots+y_{4}^2=1\}.\\
%S^1&\ug\{y=(y_1,y_{2})\in\R^{2}\st
%|y|^2=y_1^2+y_{2}^2=1\}.
\end{split}
\]

Let $T=T_1,T_2,T_3$ be the vector fields on $S^3$ given
by multiplication by $i,j,k\in\Ha=\R^4$ respectively, that is,% (see formulas \myref{base}),
\begin{equation}
\begin{split}
T=T_1&=-y_2\pa{y_1}+y_1\pa{y_2}-y_4\pa{y_3}+y_3\pa{y_4},\\
T_2&=-y_3\pa{y_1}+y_4\pa{y_2}+y_1\pa{y_3}-y_2\pa{y_4},\\
T_3&=-y_4\pa{y_1}-y_3\pa{y_2}+y_2\pa{y_3}+y_1\pa{y_4}.
\end{split}
\end{equation}

It is well-known that $S^3$ can be foliated in $S^1$'s, by means
of the Hopf fibration $S^3{\rightarrow}S^2$. Whence,
one has a foliation of $S^m\times S^3$ in $S^m\times S^1$'s, and
section \ref{genm1} gives $m+1$ vector fields tangent to the
leaves: they can be completed to a parallelization of $S^m\times S^3$
by means of a suitable parallelization of $S^3$, as it is now going
to be shown in the following proposition:
%Everything glue right,
%as we are now going to show.



\begin{pro}\label{parm3}
$S^m\times S^3$ is parallelizable.
\end{pro}

\begin{D}
One would like to use definition \ref{Bdef}
to define $m+1$ vector fields on $S^m\times S^3$. The problem
is that there is not a canonical identification of the fiber of
the Hopf fibration $S^3{\rightarrow}S^2$ with $S^1$,
whence one has not a canonical angular coordinate on fibers.
But formula \myref{progeofor}
 of theorem \ref{progeo}
only needs a unitary and tangent to fibers vector field on $S^3$
to be used: this is just what $T$ is.
Whence, define
${\cal B}\ug\{b_i\}_{i=1,\dots,m+3}$ by
\begin{equation}\label{parsms3}
\begin{split}
b_i&\ug M_i+x_i T\qquad i=1,\dots,m+1,\\
b_{m+j}&\ug T_j\qquad j=2,3,
\end{split}
\end{equation}
where $M_i$ is the $\imath^{\text{th}}$ meridian vector field on
$S^m$,
to obtain the wished frame on $S^m\times S^3$.
\end{D}

\begin{oss}\label{remor3}
The frame $\cal B$ is orthonormal with respect to the product
metric on $S^m\times S^3$ (use formulas \myref{parsms3} and formula \myref{fundeqscal}).\EndRem
%But it is positively oriented
%with respect to the standard orientation on $S^m\times S^3$ if and
%only if $m$ is even.
\end{oss}

The same argument used in section \ref{genm1} gives the brackets of $\cal
B$:
%(see also formulas \myref{Ebrackets}):
\begin{equation}\label{bracket3gen}
\begin{split}
[b_i,b_j]&=x_ib_j-x_jb_i\qquad i,j=1,\dots,m+1,\\
[b_i,b_{m+2}]&=-2x_ib_{m+3}\qquad i=1,\dots,m+1,\\
[b_i,b_{m+3}]&=2x_ib_{m+2}\qquad i=1,\dots,m+1,\\
[b_{m+2},b_{m+3}]&=-2T=-2\sum_{i=1}^{m+1}x_ib_i.
\end{split}
\end{equation}

Let $\tau=\tau_1,\tau_2,\tau_3$ be the 1-forms on $S^m\times S^3$ dual to
$T=T_1,T_2,T_3$ respectively.
The coframe ${\cal B}^*\ug\{b^i\}_{i=1,\dots,m+3}$
is given by
\begin{equation}\label{cobruni3}
\begin{split}
b^i&= x_i \tau+\copa{i}\qquad i=1,\dots,m+1,\\
b^{m+j}&=\tau_j\qquad j=2,3.
\end{split}
\end{equation}


Differently from $S^m\times S^1$, the 1-form $\tau$ is
not closed, so structure equations are a bit more complicated:
\begin{equation}\label{struc3}
\begin{split}
db^i&=b^i\wedge{\tau}+2x_ib^{m+2}\wedge b^{m+3}\qquad
i=1,\dots,m+1,\\
db^{m+2}&=2b^{m+3}\wedge\tau,\\
db^{m+3}&=-2b^{m+2}\wedge\tau,
\end{split}
\end{equation}
where the 1-form $\tau$ is related to ${\cal B}^*$ by
\[
\tau=\sum_{i=1}^{m+1}x_ib^i.
\]

\begin{oss}
The same argument used above for $S^m\times S^3$ can be applied to the
Hopf fibration $S^7\rightarrow \Cp{3}$ to obtain a
frame on $S^m\times S^7$. Nevertheless, formulas in this case are much more
complicated.\EndRem
\end{oss}


Proposition \ref{parm3} and the previous remark can be easily generalized:
\begin{teo}[\cite{BruPEP}]
Let $N^n$ be any parallelizable $n$-dimensional manifold. Then $S^m\times N$ is
parallelizable.
\end{teo}

\begin{D}
Let $T=T_1,T_2,\dots,T_n$ be a frame on $N$. The required
parallelization is thus given by
\begin{equation*}
\begin{split}
b_i&\ug M_i+x_i T\qquad i=1,\dots,m+1,\\
b_{m+j}&\ug T_j\qquad j=2,\dots,n,
\end{split}
\end{equation*}
where $M_i$ is the $\imath^{\text{th}}$ meridian vector field on
$S^m$.
\end{D}



%\section{$S^m\times S^7$}\label{genm7}


\section{The general problem:
when is a product of spheres parallelizable?}\label{bruintro}

%It is a classical result in Algebraic Topology that spheres $S^n$
%are
%parallelizable if and only if their dimension is $n=1,3,7$.
%%(see~\cite{AdaNEE}).
%When one looks at the product of two or more spheres $S^{n_1}\times\dots\times
%S^{n_r}$, $r\ge 2$, one gets instead a parallelizable manifold
%only under the condition that at least one of the $n_i$ be odd.
%This was proved by Kervaire in the fifties (see\cite{KerCIG}):

The proof of the theorem of Kervaire cited in the introduction is
here sketched:
%\begin{teo}\label{teokervaire}
%The manifold $S^{n_1}\times\dots\times S^{n_r}$, $r\ge 2$, is
%parallelizable if and only if at least one of the $n_i$ is odd.
%\end{teo}

\begin{Dsketch}[ of theorem \ref{teokervaire} (Kervaire)]
\begin{enumerate}
\item Show by induction there exists an embedding of $S^{n_1}\times\dots\times
S^{n_r}$ in $\R^{n_1+\dots+n_r+1}$. This is true for $r=1$. Let
\[
\map{f=(f_1,\dots,f_{n_1+\dots+n_{r-1}+1})}{S^{n_1}\times\dots\times
S^{n_{r-1}}}{\R^{n_1+\dots+n_{r-1}+1}}
\]
be the embedding given by
the inductive hypothesis, where $f$ is chosen in such a way that $f_1\ge
0$. Let $u\in S^{n_1}\times\dots\times
S^{n_{r-1}}$, and let $(\xi_1,\dots,\xi_{{n_r}+1})\in S^{n_r}$: the
embedding $f$ is thus given by
\[
\begin{array}{cll}
S^{n_1}\times\dots\times S^{n_r} &
\stackrel{f}{\longrightarrow} & \R^{n_1+\dots+n_r+1}\\
(u,(\xi_1,\dots,\xi_{{n_r}+1})) & \longmapsto & (f_2(u),\dots,f_{n_1+\dots+n_{{r-1}+1}}(u),
\xi_1\sqrt{f_1(u)},\dots,\xi_{{n_r}+1}\sqrt{f_1(u)});
\end{array}
\]
\item
suppose without any loss of generality
that the odd dimension is not $n_1$, and observe that
the degree of the Gauss map of the embedding $f$ built in 1.\ is
given by %({\color{red}see ???})
\[
\chi(D^{n_1+1}\times S^{n_2}\times\dots\times S^{n_r})=
\chi(D^{n_1+1})\chi(S^{n_2})\dots\chi(S^{n_r})=0,
\]
where $D^{n_1+1}$ denotes a topological disk of dimension $n_1+1$;
\item
denote by $G_{k,n}$ and $V_{k,n}$ the Grassmannian
and the Stiefel-Whitney manifold of oriented $k$-planes  and
oriented orthonormal frames in $\R^{k+n}$,
respectively. The tangential map
\[
S^{n_1}\times\dots\times S^{n_r}\longrightarrow G_{n_1+\dots+n_r,1}
\]
is null-homotopic, since by 2.\ the Gauss map is;
\item
last, denote by $P(S^{n_1}\times\dots\times S^{n_r})$ the
principal bundle of $S^{n_1}\times\dots\times S^{n_r}$, and look
at the following diagram to end the proof:
\[
\xymatrix{
P(S^{n_1}\times\dots\times S^{n_r})\ar@{.>}[r] \ar@{.>}[d] &
V_{n_1+\dots+n_r,1}\ar[d] \\
S^{n_1}\times\dots\times S^{n_r}\ar[r] &
G_{n_1+\dots+n_r,1}
}
\]
\end{enumerate}
\end{Dsketch}

%{\color{red}inserire argomentazione di Kervaire}

%\begin{oss}
%The proof of theorem \ref{teokervaire} gives no hint on how to
%explicitly build the parallelization.
%\end{oss}


Note that, due to the
homotopy theory considerations, the above proof is not very suitable
to write down explicit parallelizations on products of spheres.

%The only reference the
%author knows where explicit parallelization on some products of
%spheres are given, is the article \cite{BruPEP}, where the frame is
%given whenever one of the factor sphere is $S^1,S^3,S^5,S^7$, and
%the writing of frames in the general case is left as an open
%problem.

Another proof of Kervaire's theorem can be developed using
a series of hints contained in the book \cite[exercises 3,4,5 and 6 of section
4.2]{HirDTo}. Details of such a proof, as developed by the
author, are given in the following.

%Using this proof, an explicit
%parallelization on products of two spheres is obtained in the next section.

%We now prove a more general statement,
%following a hint given in \cite[exercises 3,4,5 and 6 of section
%4.2]{HirDTo}. This proof
%
%
%The problem solved by Kervaire in theorem \ref{teokervaire} is
%given as a sequence of exercises in \cite[exercises 3,4,5 and 6 of section
%4.2]{HirDTo}. These exercises
%
%is the tool we use in
%next section to explicitly give the parallelization for $S^m\times
%S^n$, $n$ odd, which we shall mainly be concerned with
%for the rest of the chapter.
%We thus give another proof of theorem \ref{teokervaire}
%following the Hirsch approach.

In what follows, $\varepsilon_B^k$ denotes
the trivial vector bundle of rank $k$ with base
space $B$; moreover, whenever $\al$ is a vector bundle,
$E(\al),p_{\al},B(\al)$ denote the total space, the projection and
the base space of \al\ respectively.



%Whether a product of more than one sphere is parallelizable, is a
%problem given in \cite[exercises 3,4,5 and 6 of section
%4.2]{HirDTo}. Being the main topics of the rest of the chapter, we
%here shall give a solution following the Hirsch approach.
%In what follows,
%the trivial vector bundle of rank $n$ with base
%space $B$ will be denoted by $\varepsilon_B^n$.

\begin{lem}\label{lembun}
Let $\al$ be a vector bundle.
%Let $\varepsilon$ and $\varepsilon'$ trivial
%vector bundles of rank 1 respectively over $B(\al)$ and $B(\be)$.
The Whitney sum $\al\oplus\varepsilon_{B(\al)}^k$ is described by
\[
\begin{split}
E(\al\oplus\varepsilon_{B(\al)}^k)&\iso E(\al)\times\R^k,\\
p_{\al\oplus\varepsilon_{B(\al)}^k}(e,v)&=p_{\al}(e),\\
B(\al\oplus\varepsilon_{B(\al)}^k)&=B(\al).
\end{split}
\]
\end{lem}

\begin{D}
The Whitney sum $\al\oplus\varepsilon_{B(\al)}^k$ is given by the pull-back
of $\al\times\varepsilon_{B(\al)}^k$ by means of the diagonal map
$B(\al)\rightarrow B(\al)\times
B(\al)$ (see for instance \cite[page
27]{MiSCCl}). Then
\[
E(\al\oplus\varepsilon_{B(\al)}^k)
=\{(e,b,v,b)\in E(\al)\times B(\al)\times
\R^k\times B(\al)\st
p_{\al}(e)=b\}
\]
and the thesis follows.
\end{D}

%The following corollary is the heart in the
%construction of the frame:
\begin{cor}\label{lemmabundles}
Let $\al,\be$ be vector bundles. Then, for any $k\ge 0$,
\[
\al\times(\be\oplus\varepsilon_{B(\be)}^k)\iso
(\al\oplus\varepsilon_{B(\al)}^k)\times\be.
\]
\end{cor}

\begin{D}
Observe that
\[
\begin{split}
E(\al\times(\be\oplus\varepsilon_{B(\be)}^k))&\iso E(\al)\times
E(\be\oplus\varepsilon_{B(\be)}^k)\stackrel{\ref{lembun}}{\iso} E(\al)\times
E(\be)\times\R^k,\\
E(\al\oplus\varepsilon_{B(\al)}^k)\times\be&\iso E(\al\oplus\varepsilon_{B(\al)}^k) \times E(\be)
\stackrel{\ref{lembun}}{\iso} E(\al)\times\R^k\times
E(\be),
\end{split}
\]
and use the obvious isomorphism.
\end{D}


\begin{teo}\label{probundles}
%Let $\varepsilon$ and $\varepsilon'$ be trivial vector bundles of rank
%1 over respectively $M$ and $N$, and
Suppose $M^m$ and $N^n$ satisfy the following
properties:
\begin{enumerate}
\item $T(M)\oplus\varepsilon_M^1$ is trivial;
\item $T(N)\oplus\varepsilon_N^1$ is trivial;
\item there is a non-vanishing vector field on $N$.
\end{enumerate}
Then $M\times N$ is parallelizable.
\end{teo}

\begin{D}
Let $\nu$ be a complement in $T(N)$ of the non-vanishing vector field on
$N$, that is,
\begin{equation}\label{c}
T(N)\iso\nu\oplus\varepsilon_N^1.
\end{equation}
Then
\begin{equation}\label{chain}
\begin{split}
T(M\times N)&\iso T(M)\times T(N)\stackrel{\myref{c}}{\iso}
T(M)\times(\nu\oplus\varepsilon_N^1)\\
&\stackrel{\ref{lemmabundles}}{\iso}
(T(M)\oplus\varepsilon_M^1)\times\nu
\stackrel{\text{1.}}{\iso}
\varepsilon_M^{m+1}\times\nu\\
&\stackrel{\ref{lemmabundles}}{\iso}\varepsilon_M^{m-1}\times(\nu\oplus\varepsilon_N^2)
\stackrel{2.}{\iso}
\varepsilon_M^{m-1}\times\varepsilon_N^{n+1}%\iso\varepsilon_{M\times N}^{m+n}
\end{split}
\end{equation}
\end{D}

\begin{oss}\label{ossshort}
Whenever $N$ is itself parallelizable, formula \myref{chain} can
be shortened:
\begin{equation}\label{chainshort}
\begin{split}
T(M\times N)&\iso T(M)\times T(N)\stackrel{}{\iso}
T(M)\times\varepsilon_N^{n}\\
&\stackrel{}{\iso}
(T(M)\oplus\varepsilon_M^1)\times\varepsilon_N^{n-1}
\stackrel{}{\iso}
\varepsilon_M^{m+1}\times\varepsilon_N^{n-1}.
\end{split}
\end{equation}\EndRem
\end{oss}



The embedding $S^n\subset\R^{n+1}$ gives the triviality of
$T(S^n)\oplus\varepsilon_{S^n}^1$; whenever $n$ is odd, a non-vanishing vector
field on $S^n\subset\C^{(n+1)/2}$
is given by the complex multiplication. Thus, the following:
\begin{cor}\label{corbundles}
Let $n$ be any positive odd integer. Then the manifold $S^m\times S^n$
is parallelizable.
\end{cor}
%\end{enumerate}
%\end{D}
%Last, the main theorem:

And finally:

{\noindent\nopagebreak\em Second proof of theorem \ref{teokervaire}: }
Apply $r-1$ times the corollary \ref{lemmabundles} to show
that $T(S^{n_2}\times\dots\times S^{n_r})\oplus\varepsilon_{S^{n_2}\times\dots\times
S^{n_r}}^1$ is a trivial vector bundle, and use theorem
\ref{probundles}.
\EndDim



\section{An explicit parallelization $\mathcal{P}$ for products of 2
spheres}\label{secparexp}

An explicit parallelization $\mathcal{B}$ has already been found on $S^m\times S^n$,
for $n=1,3,7$, in the previous sections. Can one use theorem
\ref{probundles} to explicitly find a parallelization on any
parallelizable $S^m\times
S^n$? Answer is positive.



The trick in theorem \ref{probundles} is simple: split $TN$ by
means of the never-vanishing vector field, then use the trivial
summand to parallelize $TM$, and last detach a rank 2 trivial
summand to parallelize the remaining part of $TN$. Remark
\ref{ossshort} simply says that if $N$ is itself parallelizable,
one can avoid to detach the rank 2 trivial summand from $M$, using
the parallelization of $N$ instead.

{\em Here and henceforth, $n$ is supposed to be the odd dimension in $S^m\times S^n$.}

%The same pattern can be followed
%to explicitly give a parallelization of $S^m\times S^n$, for
%any odd $n$.
%This is just what we are going to do in this section.

%\subsection{Some notations}

Denote by $y=(y_j)$
the coordinates on $\R^{n+1}$, and let $S^n\subset\R^{n+1}$ be
given by
\[
\begin{split}
%S^m&\ug\{x=(x_1,\dots,x_{m+1})\in\R^{m+1}\st
%x_1^2+\dots+x_{m+1}^2=1\},\\
S^n&\ug\{y=(y_1,\dots,y_{n+1})\in\R^{n+1}\st
y_1^2+\dots+y_{n+1}^2=1\}.
\end{split}
\]


Being $n$ odd, a never-vanishing vector field, and hence a versor
field,
is defined on $S^n$: {\em
here and henceforth, $T$ denotes the versor field  on $S^n$
given by
multiplication by $i$ in $\C^{(n+1)/2}$, namely,
\begin{equation}\label{eqtjexp}
T\ug
-y_2\pa{y_1}+y_1\pa{y_2}+\dots-y_{n+1}\pa{y_n}+y_n\pa{y_{n+1}}.
%\sum_{j=1}^{n+1}t_j\pa{y_j}
\end{equation}
}
When a shorter form of $T$ is needed, $t_j$ denotes the
coordinates of $T$, that is,
\begin{equation}
T=
\sum_{j=1}^{n+1}t_j\pa{y_j}
\end{equation}
where $t_j$ is given by
%multiplication by $i$ in $\C^{(n+1)/2}$, namely,
\begin{equation}\label{eqtj}
t_j=\left\{\begin{split}
-&y_{j+1}\qquad \text{if }j\text{ is
odd,}\\
&y_{j-1}\qquad \text{if }j\text{ is
even.}
    \end{split}\right.
\end{equation}

Moreover, denote by $N$ the normal versor field of
$S^n\subset\R^{n+1}$ (recall that $M$ denotes the normal versor field of
$S^m\subset\R^{m+1}$):
\begin{equation}
\begin{split}
%M\ug\sum_{i=1}^{m+1}x_i\pa{x_i}\qquad\text{and}\qquad
N\ug\sum_{j=1}^{n+1}y_j\pa{y_j}.
\end{split}
\end{equation}

It is convenient to think of $T(S^m\times
S^n)=TS^m\times TS^n$ as a Riemannian subbundle of
$T\R^{m+1}_{|_{S^m}}\times T\R^{n+1}_{|_{S^n}}$;
this last is trivial, and an orthonormal frame is
$\{\partial_{x_1},\dots,\partial_{x_{m+1}},\partial_{y_1},\dots,\partial_{y_{n+1}}\}$.

Denote by $N_j$ the $\jmath^{\text{th}}$ meridian vector field on
$S^n$ (recall that $M_i$ denotes the $\imath^{\text{th}}$ meridian vector field on
$S^m$):
\[
\begin{split}
%M_i&\ug\text{orthogonal
%projection of $\pa{x_i}$ on $S^m$}\qquad i=1,\dots,m+1.\\
N_j&\ug\text{orthogonal
projection of $\pa{y_j}$ on $S^{n}$}\qquad j=1,\dots,n+1.
\end{split}
\]

The tangent space in a point  $(x,y)\in S^m\times S^n$ is thus
given by an euclidean  vector subspace
\[
T_xS^m\oplus T_yS^n\subset
\R^{m+1}\oplus\R^{n+1},
\]
which is generated by the
$m+n+2$ vectors $\{M_1(x),\dots,M_{m+1}(x),N_1(y),\dots,N_{n+1}(y)\}$.

One also has
\begin{equation}\label{uga}
T_xS^m\oplus\langle M(x)\rangle_\R
=\R^{m+1}\qquad\text{and}\qquad T_yS^n\oplus\langle N(y)\rangle_\R
=\R^{n+1}.
\end{equation}

As in formula \myref{fundeq}, one obtains
\begin{equation}\label{fundeqold}
\begin{split}
\pa{x_i}&=M_i+x_iM\qquad i=1,\dots,m+1,\\
\pa{y_j}&=N_j+y_jN\qquad j=1,\dots,n+1.
\end{split}
\end{equation}

%We shall need a trivialization for
%$T(S^m)\oplus\varepsilon_{S^m}^1$, which is given by the embedding
%$S^n\subset\R^{n+1}$ gives the triviality of
%$T(S^n)\oplus\varepsilon_{S^n}^1$
Moreover, denote by $T(y)^\perp$ the vector subspace of $T_y(S^n)$
which is orthogonal to $T(y)$:
\begin{equation}\label{ugb}
\langle T(y)\rangle_\R\oplus T(y)^\perp=T_yS^n.
\end{equation}

%\subsection{The frame
%$\cal P$ on $S^m\times S^n$, for $n$ odd}

{\em In what follows, some
computation on the vector space $T_x(S^m)\oplus
T_y(S^n)$ is done. For the sake of simplicity, the argument of
vector fields is omitted, that is, $T$ stands for $T(y)$, $M$ stands for
$M(x)$ etc\dots}

Formula \myref{chain} in theorem \ref{probundles} gives the following chain
of pointwise isomorphisms:
\newlength{\temp}
\settowidth{\temp}{$\scriptstyle\myref{ugb}$}
\begin{equation}\label{chainlinear}
\begin{split}
T_x(S^m)\oplus T_y(S^n)&\stackrel{\myref{ugb}}{=}
T_x(S^m)\oplus\langle T\rangle_\R\oplus T^\perp\\
&\stackrel{\makebox[\temp]{$\scriptstyle\al$}}{\iso}
T_x(S^m)\oplus\langle M\rangle_\R\oplus T^\perp\\
&\stackrel{\myref{uga}}{=}
\R^{m+1}
\oplus T^\perp\\
&\stackrel{\makebox[\temp]{$\scriptstyle\be$}}{\iso}\R^{m-1}\oplus \langle N\rangle_\R\oplus\langle T\rangle_\R
\oplus
T^\perp\\
&\stackrel{\myref{ugb}}{=}\R^{m-1}\oplus \langle N\rangle_\R\oplus T_yS^n\\
&\stackrel{\myref{uga}}{=}
\R^{m-1}\oplus\R^{n+1},
\end{split}
\end{equation}
where $\al$
is defined by
\[
\al(T)\ug M,
\]
and $\be$ is defined by
\[
\be(\pa{x_m})\ug N,\qquad\be(\pa{x_{m+1}})\ug T.
\]


Pulling back to $T_x(S^m)\oplus T_y(S^n)$ the $m-1$ generators
$\{\partial_{x_1},\dots,\partial_{x_{m-1}}\}$
of $\R^{m-1}$ one obtains
\begin{equation}\label{framea}
\begin{split}
\pa{x_i}&\stackrel{\myref{fundeq}}{=}
M_i+x_iM\\
&\stackrel{\al^{-1}}{\longmapsto}M_i+x_iT
\end{split}\qquad i=1,\dots,m-1,
\end{equation}
whereas pulling back to $T_x(S^m)\oplus T_y(S^n)$ the $n+1$ generators
$\{\partial_{y_1},\dots,\partial_{y_{n+1}}\}$
of $\R^{n+1}$ one obtains the more complicated formulas
\begin{equation}\label{frameb}
\begin{split}
\pa{y_j}&\stackrel{\myref{fundeq}}{=}
N_j+y_jN\\
&\stackrel{\phantom{\myref{fundeq}}}{=}\langle N_j,T\rangle T+(N_j-\langle N_j,T\rangle T)+y_jN\\
&\stackrel{\be^{-1}}{\longmapsto}
\langle N_j,T\rangle \pa{x_{m+1}}+(N_j-\langle N_j,T\rangle T)+y_j\pa{x_{m}}\\
&\stackrel{\myref{fundeq}}{=}
\langle N_j,T\rangle (M_{m+1}+x_{m+1}M)+(N_j-\langle N_j,T\rangle T)+y_j(M_{m}+x_{m}M)\\
&\stackrel{\al^{-1}}{\longmapsto}
\langle N_j,T\rangle (M_{m+1}+x_{m+1}T)+(N_j-\langle N_j,T\rangle T)+y_j(M_{m}+x_{m}T)
\end{split}j=1,\dots,n+1.
\end{equation}

The following theorem applies the above argument to $S^m\times
S^n$, odd $n$, in order to obtain an explicit frame on it:
\begin{teo}
Let $n$ be odd, and let $T=\sum_{j=1}^{n+1}t_j\pa{y_j}$ be the
tangent versor field on $S^n$
given by formula \myref{eqtj}.
Also, let $\{M_i\}_{i=1,\dots,m+1}$ and $\{N_j\}_{j=1,\dots,n+1}$
be the meridian vector fields on $S^m$
and $S^n$ respectively. Last, let $M$ and $N$ be the normal versor
fields of $S^m\subset\R^{m+1}$
and $S^n\subset\R^{n+1}$ respectively.
The product $S^m\times S^n$ is parallelized by the
 frame ${\cal P}\ug\{p_1,\dots,p_{m+n}\}$ given by
\begin{equation}\label{frame}
\begin{split}
p_i&\ug M_i+x_iT\qquad i=1,\dots,m-1,\\
p_{m-1+j}&\ug y_j M_{m}+t_j M_{m+1}+(t_j x_{m+1}+y_j
x_{m}-t_j)T+N_j
\qquad j=1,\dots,n+1.
\end{split}
\end{equation}
Moreover, $\cal P$ is orthonormal with respect to the standard metric on $S^m\times S^n$.
\end{teo}

\begin{D}
Observe that
\[
\langle N_j,T\rangle\stackrel{\myref{fundeq}}{=}\langle \pa{y_j}-y_jN,T\rangle=\langle \pa{y_j},T\rangle=t_j\qquad j=1,\dots,n+1
\]
and use
formulas \myref{framea} and
\myref{frameb}
to obtain \myref{frame}.
The orthonormality can be proved by observing that both \al\ and
\be\ in \myref{chainlinear} are isometries. But one can also
directly
check the $p_i$'s,
taking into
account formula \myref{fundeqscal}.
%\[
%\begin{split}
%\langle M_i,M_{j}\rangle&=
%\langle \pa{x_i}-x_iM,\pa{x_j}-x_jM\rangle=\delta_{ij}-x_ix_j\\
%\langle N_i,N_j\rangle&=
%\langle \pa{y_i}-y_iN,\pa{y_j}-y_jN\rangle=\delta_{ij}-y_iy_j.
%\end{split}
%\]
\end{D}

\section{The frames $\cal P$ and $\cal B$ on $S^m\times S^1$ and  $S^m\times S^3$}

If $n=1,3$ or $7$, remark \ref{ossshort} can be used to obtain a
parallelization simpler than $\cal P$
on $S^m\times S^n$. If $n=1,3$ this parallelization is just the one given in
sections $\ref{genm1}, \ref{genm3}$ respectively, which was called $\cal B$.
In this section
relations between $\cal B$ and $\cal P$ are exploited.

%\subsection{The frame $\cal P$ and $\cal B$ on $S^m\times S^1$}
Let $n=1$.
Formula \myref{frame} gives the frame ${\cal P}=\{p_1,\dots,p_{m+1}\}$
on $S^m\times S^1$, whereas the frame $\cal B$ is given by formula \myref{progeofor}.
Clearly,
\[
p_i=b_i\qquad i=1,\dots,m-1.
\]
Since $\pa{\T}=-y_2\pa{y_1}+y_1\pa{y_2}=T$, one obtains
\[
\begin{split}
\langle N_1,\pa{\T}\rangle&=\langle \pa{y_1}-y_1N,-y_2\pa{y_1}+y_1\pa{y_2}\rangle=-y_2,\\
\langle N_2,\pa{\T}\rangle&=\langle \pa{y_2}-y_2N,-y_2\pa{y_1}+y_1\pa{y_2}\rangle=y_1,
\end{split}
\]
and thus
\[
\begin{split}
N_1&=-y_2T,\\
N_2&=y_1T.
\end{split}
\]
Whence
\[
\begin{split}
p_{m}&=y_1(M_{m}+x_{m}T)-y_2(M_{m+1}+x_{m+1}T)+y_2T-y_2T=y_1b_m-y_2b_{m+1},\\
p_{m+1}&=y_2(M_{m}+x_{m}T)+y_1(M_{m+1}+x_{m+1}T)-y_1T+y_1T=y_2b_m+y_1b_{m+1},
\end{split}
\]
and one gets
\begin{equation}\label{changebtop}
{\cal P}={\cal B}\left(\begin{array}{ccc|cc}
 & & & 0 & 0\\
 &I_{m-1}& &\vdots &\vdots \\
 & &  & 0& 0\\ \hline
 0&\cdots &0 &  y_1 & y_2 \\
 0&\cdots &0  & -y_2 & y_1
                \end{array}\right)
\end{equation}


%{\color{red}2) formula troppo banale? pag. 6!!!}


Brackets of $\cal P$ are thus easily obtained by means of formulas
\myref{changebtop}, \myref{bracket1gen} and \myref{kobano}:
\begin{equation}\label{bracket1genp}
\begin{split}
[p_i,p_j]&=x_ip_j-x_jp_i\qquad i,j=1,\dots,m-1\\
[p_i,p_m]&=(-x_my_1+x_{m+1}y_2)p_i+x_ip_m-x_ip_{m+1}\qquad
i=1,\dots,m-1\\
[p_i,p_{m+1}]&=(-x_my_2-x_{m+1}y_1)p_i+x_ip_m+x_ip_{m+1}\qquad
i=1,\dots,m-1\\
[p_m,p_{m+1}]&=(x_m(y_1-y_2)-x_{m+1}(y_1+y_2))p_m +(x_m(y_1+y_2)+x_{m+1}(y_1-y_2))p_{m+1}
\end{split}
\end{equation}


Formula \myref{changebtop} gives the frame ${\cal P}^*$
dual to $\cal P$:
\[
\begin{split}
p^i&=b^i\qquad i=1,\dots,m-1,\\
p^{m}&=y_1b^m-y_2b^{m+1},\\
p^{m+1}&=y_2b^m+y_1b^{m+1}.
\end{split}
\]

The structure equations for $\cal P$ are thus obtained
by a straightforward computation:
\begin{equation}\label{diff1genp}
\begin{split}
dp^i&=\copa{i}\wedge\tau=p^i\wedge{\tau}\qquad i=1,\dots,m-1,\\
dp^m&=p^m\wedge\tau+p^{m+1}\wedge\tau,\\
dp^{m+1}&=p^{m+1}\wedge\tau-p^m\wedge\tau,
\end{split}
\end{equation}
where
$\tau$ is given by
\[
\tau=\sum_{i=1}^{m+1}x_ib^i=\sum_{i=1}^{m-1}x_ip^i+(x_my_1-x_{m+1}y_2)p^m+
(x_my_2+x_{m+1}y_1)p^{m+1}.
\]

%\subsection{The frame $\cal P$ and $\cal B$ on $S^m\times S^3$}

Let $n=3$.
Formula \myref{frame} gives the frame ${\cal P}=\{p_1,\dots,p_{m+3}\}$
on $S^m\times S^3$, whereas the frame $\cal B$ is given by formula \myref{parsms3}.
Clearly,
\[
p_i=b_i\qquad i=1,\dots,m-1.
\]
Denote by ``$(*)_{\text{$\jmath^{\text{th}}$}}$'' the
$\jmath^{\text{th}}$ coordinate of $*$.
Since
\[
\begin{split}
\langle N_j-t_jT,T\rangle&=0\\
\langle N_j-t_jT,b_{m+2}\rangle&=(b_{m+2})_{\text{$\jmath^{\text{th}}$}},\\
\langle N_j-t_jT,b_{m+3}\rangle&=(b_{m+3})_{\text{$\jmath^{\text{th}}$}},\\
\end{split}\qquad j=1,\dots,4,
\]
one gets
\[
p_{m-1+j}=y_jb_m+t_jb_{m+1}+(b_{m+2})_{\text{$\jmath^{\text{th}}$}}b_{m+2}
+(b_{m+3})_{\text{$\jmath^{\text{th}}$}}b_{m+3}\qquad j=1,\dots,4.
\]
Whence
\begin{equation}\label{changebtop3}
{\cal P}={\cal B}\left(\begin{array}{ccc|cccc}
 & & & 0 & 0 & 0 & 0\\
 &I_{m-1}& &\vdots &\vdots &\vdots &\vdots \\
 & &  & 0& 0 & 0 & 0\\ \hline
 0&\cdots &0 &  y_1 & y_2 & y_3 & y_4 \\
 0&\cdots &0  & -y_2 & y_1 & -y_4 & y_3 \\
  0&\cdots &0  & -y_3 & y_4 & y_1 & -y_2 \\
 0&\cdots &0  & -y_4 & -y_3 & y_2 & y_1 \\
                \end{array}\right)
\end{equation}


%{\color{red}2) formula troppo banale? pag. 6!!!}


Brackets of $\cal P$ can be obtained by means of a not straightforward computation
using formulas
\myref{changebtop3}, \myref{bracket3gen} and \myref{kobano}. One can
also refer to the next section, where general formulas for
$\mathcal{P}$ are given.
% We
%give them in appendix \ref{appform3}.

%[p_i,p_{m-1+j}]=(-x_my_j-x_{m+1}t_j)p_i\mp x_ip_{m-1+j\pm
%1}+x_i(y_1y_j-y_2t_j)p_{m}+x_i(y_2y_j+y_1t_j)p_{m+1}+
%x_i(y_3y_j-y_4t_j)p_{m+2}+x_i(y_4y_j+y_3t_j)p_{m+3}

Formula \myref{changebtop3} gives the frame ${\cal P}^*$
dual to $\cal P$:
\begin{equation*}
{{\cal P}^*}={{\cal B}^*}\left(\begin{array}{ccc|cccc}
 & & & 0 & 0 & 0 & 0\\
 &I_{m-1}& &\vdots &\vdots &\vdots &\vdots \\
 & &  & 0& 0 & 0 & 0\\ \hline
 0&\cdots &0 &  y_1 & y_2 & y_3 & y_4 \\
 0&\cdots &0  & -y_2 & y_1 & -y_4 & y_3 \\
  0&\cdots &0  & -y_3 & y_4 & y_1 & -y_2 \\
 0&\cdots &0  & -y_4 & -y_3 & y_2 & y_1 \\
                \end{array}\right)
\end{equation*}


\section{General formulas for $\mathcal{P}$}

Recall that $T=\sum_{j=1}^{n+1}t_j\pa{y_j}$. Set
\[
\begin{split}
X_m&\ug M_m+x_mT,\\
X_{m+1}&\ug M_{m+1}+x_{m+1}T,\\
C_{j,k}&\ug y_jt_k-y_kt_j\qquad j,k=1,\dots,n+1,\\
D_{j,k}&\ug 2C_{j,k}\underbrace{\mp\delta_{k,j\pm 1}}_{j\text{ }\frac{\text{odd}}{\text{even}}}
\underbrace{\pm\delta_{j,k\pm 1}}_{k\text{ }\frac{\text{odd}}{\text{even}}}
\qquad j,k=1,\dots,n+1.
\end{split}
\]

Formulas \myref{frame} easily give
\[
\begin{split}
\sum_{j=1}^{n+1}y_jp_{m-1+j}&=M_m+x_mT=X_m,\\
\sum_{j=1}^{n+1}t_jp_{m-1+j}&=M_{m+1}+x_{m+1}T=X_{m+1}.
\end{split}
\]
A hard calculation then gives
\begin{equation}\label{genbracket}
\begin{split}
[p_i,p_{j}]&=x_ip_j-x_jp_i\qquad i,j=1,\dots,m-1,\\
[p_i,p_{m-1+j}]&=-(y_jx_m+t_jx_{m+1})p_i\\
&\underbrace{\mp
x_i p_{m-1+j\pm 1}}_{j\text{ }\frac{\text{odd}}{\text{even}}}
+x_iy_jX_m+x_it_jX_{m+1}\qquad i=1,\dots,m-1,j=1,\dots,n+1,\\
%&=(-x_my_j-x_{m+1}t_j)p_i\mp x_ip_{m-1+j\pm
%1}+\sum_{k=1}^{n+1}x_i(y_ky_j+t_kt_j)p_{m-1+k}\qquad\text{for }j
%\stackrel{\scriptstyle\text{odd}}{\scriptstyle\text{even}}\\
%\frac{\text{odd}}{\text{even}}\\
[p_{m-1+j},p_{m-1+k}]&=D_{j,k}\sum_{i=1}^{m-1}x_ip_i+y_jp_{m-1+k}-y_kp_{m-1+j}\\
&+(x_mD_{j,k}-x_{m+1}C_{j,k})X_m+((x_{m+1}-1)D_{j,k}+x_mC_{j,k})X_{m+1}\\
&+\underbrace{(\mp y_jx_m\mp t_jx_{m+1}\pm
t_j)p_{m-1+k\pm 1}}_{k\text{ }\frac{\text{odd}}{\text{even}}}\\
&+\underbrace{(\pm y_kx_m\pm t_kx_{m+1}\mp
t_k)p_{m-1+j\pm 1}}_{j\text{ }\frac{\text{odd}}{\text{even}}}
\qquad j,k=1,\dots,n+1.\\
\end{split}
\end{equation}












% Call $\{T=T_1,T_2,\dots,T_{n}\}$ an orthonormal
%frame
%on $S^n$: using formula
%\myref{chainshort} instead of \myref{chain} leads to the
%orthonormal frame ${\cal B}\ug\{b_1,\dots,b_{m+n}\}$ given by
%\begin{equation}\label{frameshort}
%\begin{split}
%b_i&\ug M_i+x_iT\qquad i=1,\dots,m+1,\\
%b_{m+1+j}&\ug T_{j+1}\qquad j=1,\dots,n-1.
%\end{split}
%\end{equation}
%Remark the simplest case $n=1$:
%\begin{equation}\label{frameshort1}
%\begin{split}
%b_i&\ug M_i+x_iT\qquad i=1,\dots,m+1.
%\end{split}
%\end{equation}
%\end{oss}


\begin{acknowledgements}
This paper is part of a Ph.~D.\
thesis, and the author wishes in a special way
to thank Paolo Piccinni for the
motivation and the constant help.
\end{acknowledgements}


%\begin{address}
%Maurizio Parton, {\tt parton@dm.unipi.it},\\
%Dipartimento di Matematica ``Leonida Tonelli'',\\
%via Filippo Buonarroti 2, I--56127 Pisa,\\
%Italy.%\\
%%e-mail address: {\tt parton@dm.unipi.it}
%\end{address}


%\nocite{BrGRMH,PedHMa,KoSHPE,BruPEP,CMSCGS}
\begin{thebibliography}{Bru92}

\bibitem[Bru92]{BruPEP}
M.~Bruni.
\newblock Sulla parallelizzazione esplicita dei prodotti di sfere.
\newblock {\em Rend. di Mat., serie {VII}}, 12:405--423, 1992.

\bibitem[Hir88]{HirDTo}
M.~W. Hirsch.
\newblock {\em Differential Topology}, volume~33 of {\em GTM}.
\newblock Springer-Verlag, third edition, 1988.

\bibitem[Ker56]{KerCIG}
M.~Kervaire.
\newblock Courbure int\'egrale g\'en\'eralis\'ee et homotopie.
\newblock {\em Math. Ann.}, 131:219--252, 1956.

\bibitem[MS74]{MiSCCl}
J.~W. Milnor and J.~D. Stasheff.
\newblock {\em Characteristic Classes}.
\newblock University Press, Princeton, 1974.

\bibitem[Par00]{ParPhT}
M.~Parton.
\newblock {Hermitian and special structures on products of spheres}.
\newblock 2000.
\newblock tesi di dottorato, preliminary version.

\end{thebibliography}

\end{document}

math0009156.extracted_bib
\begin{thebibliography}{Bru92}

\bibitem[Bru92]{BruPEP}
M.~Bruni.
\newblock Sulla parallelizzazione esplicita dei prodotti di sfere.
\newblock {\em Rend. di Mat., serie {VII}}, 12:405--423, 1992.

\bibitem[Hir88]{HirDTo}
M.~W. Hirsch.
\newblock {\em Differential Topology}, volume~33 of {\em GTM}.
\newblock Springer-Verlag, third edition, 1988.

\bibitem[Ker56]{KerCIG}
M.~Kervaire.
\newblock Courbure int\'egrale g\'en\'eralis\'ee et homotopie.
\newblock {\em Math. Ann.}, 131:219--252, 1956.

\bibitem[MS74]{MiSCCl}
J.~W. Milnor and J.~D. Stasheff.
\newblock {\em Characteristic Classes}.
\newblock University Press, Princeton, 1974.

\bibitem[Par00]{ParPhT}
M.~Parton.
\newblock {Hermitian and special structures on products of spheres}.
\newblock 2000.
\newblock tesi di dottorato, preliminary version.

\end{thebibliography}