text:
\section*{Introduction.}

The duality of finite-rank vector bundles $E$ is involutive, $(E^*)^*\cong E$, and so
may be said to have group $\mathbb{Z}_2$. In \cite{Mackenzie:1999} and \cite{Mackenzie:2005dts},
Mackenzie initiated the study of duality for multiple vector bundles, showing that the
group of dualization operations of a double vector bundle is the symmetric group of
order six, and stating that the corresponding group for triple vector bundles has
order 72.

Shortly before \cite{Mackenzie:2005dts} went to press, Gracia-Saz convinced Mackenzie
that the order of this group is 96. In the present paper we prove that this is indeed
so, by a method which we will subsequently extend to  $n$-fold vector
bundles \cite{Gracia-SazM:nfold}. The approach in \cite{Mackenzie:2005dts}
considered only the triple vector bundles themselves; in this paper, we show that the
correct question to pose is when two dualization operations are equal. The crux
of the method is to regard dualization operations as functors on appropriate categories,
and to regard two dualization operations as equal if they are naturally isomorphic as
functors.

One of the motivating examples at the inception of category theory \cite{MacLane:pres}
was the distinction between the isomorphisms which can be defined between a (finite-dimensional)
vector space $V$ and its dual by use of a basis, and the natural isomorphism which exists
between $V$ and $(V^*)^*.$ Our method extends this familiar idea to multiple vector
bundles. For ordinary vector bundles, local trivializations are transformed into each
other by maps into a general linear group, and dualizing the bundle corresponds to taking
the transpose in the group. Multiple vector bundles can always be globally decomposed --- that
is, they are always isomorphic to a combination of pullbacks of ordinary vector bundles ---  and
there is a corresponding group of what we call statomorphisms (see Definition \ref{df:stato3}).
In order to distinguish between two dualization operations for triple vector bundles
it is not sufficient to consider the action on the constituent bundles, as was done in
\cite{Mackenzie:2005dts} --- it is necessary to study their effect on the statomorphism group.

Double vector bundles have been used for many years in some accounts of connection theory
\cite{Besse} and in some approaches to theoretical mechanics \cite{Tulczyjew}. Their general
theory was initiated by Pradines \cite{Pradines:DVB}. In the 1990s double and triple vector
bundles began to arise in Poisson geometry, as a result of the relationships between Poisson
structures and Lie algebroids. At the simplest level, the dual of a Lie algebroid $A$ has a
Poisson structure, the linearity properties of which can be expressed as the condition
that the Poisson anchor $\pi^\#$ is a morphism of double vector bundles $T^*A^*\to TA^*$.
Secondly, given a Lie bialgebroid $(A,A^*)$, the cotangent double $T^*A^*\cong T^*A$ can play
a role corresponding to that of the classical Drinfel'd double of a Lie bialgebra. More
generally, the duality of double vector bundles, and the triple vector bundles associated with
them, are crucial to the compatibility conditions of \cite{Mackenzie:eddd} between the Lie
algebroid structures in a double Lie algebroid. The results of the present paper will be applied
elsewhere to the triple Lie algebroids which arise from double Lie algebroids, and in particular
to understanding the duals of thrice-iterated tangent bundles $T^3M$.
No familiarity with Lie algebroids or Poisson structures is needed in this paper.

The duality group for double vector bundles $(D;A,B;M)$ is the symmetric group $S_3$ and duality
operations in this case can be identified by their action on the three `building bundles': the
side bundles $A$ and $B$ and the core dual $C^*$. Indeed the duality group for double vector bundles
can be interpreted as those rotations of the cotangent triple $T^*D$ which preserve $T^*D$
and $M$ \cite{Mackenzie:2005dts}. In the triple case the corresponding result does not hold: there
are three nontrivial duality operations which preserve the building bundles but are not equivalent
to the identity. The significance of these operations deserves to be investigated further.

In \S\ref{sect:intro} we briefly review the basic notions for double vector bundles; some
readers may prefer to start with \S\ref{sect:df}. To introduce the methods of this paper and
of \cite{Gracia-SazM:nfold}, we begin by illustrating them on the known case of double vector
bundles. The duality theory in this case was set out in  \cite[\S1--\S3]{Mackenzie:2005dts}
but we reformulate it in \S\ref{sect:df} to clarify the problems which arise in the case of
triple vector bundles and to introduce the functorial point of view which we use in this
paper and in \cite{Gracia-SazM:nfold}. In Definition \ref{defin:dg2} we define the group $\dg_2$,
the group of duality functors modulo natural equivalence, which we usually refer to simply
as the \emph{duality group for double vector bundles}.

In \S\ref{sect:tvb} we set up a notation for triple vector bundles that will extend readily
to the $n$-fold case. The main work of the paper is in \S\ref{sect:dtvb}. This is the calculation
of the duality group $\dg_3$ for triple vector bundles, in terms of its action on the group of
statomorphisms (Definition \ref{df:stato3}). The final \S\ref{sect:group} provides some further
information on the structure of $\dg_3$. Throughout the paper we consider smooth vector bundles
of finite rank over the reals.

Multiple vector bundles in the setting of supergeometry are being developed by Voronov
\cite{Voronov:mtqm} and Grabowski \cite{GrabowskiR:mvbs}.

Our work on this paper began after Mackenzie had spoken on \cite{Mackenzie:2005dts} at
Poisson 2004 in Luxembourg. Gracia--Saz's research was partially supported by fellowships
from the \emph{Secretar\'ia de Estado de Universidades e Investigaci\'on del Ministerio
Espa\~{n}ol de Educaci\'on y Ciencia} and from the Japanese Society for the Promotion of
Science. Gracia--Saz would also like to thank the Pure Mathematics department of the
University of Sheffield for its hospitality during various visits, and both authors thank
the London Mathematical Society for funding one of these visits. The authors also very much
appreciate the comments of the referees.

\section{Review of double vector bundles}
\label{sect:intro}

A \emph{double vector bundle} $(D;A,B;M)$, as shown on the left of Figure~\ref{fig:dvbdef},
consists firstly of a manifold $D$ together with two vector bundle structures, on bases $A$ and $B$,
each of which is itself a vector bundle on base $M$, such that for each structure on $D$,
the structure maps (projection, addition, scalar multiplication) are vector bundle
morphisms with respect to the other structure. A \emph{morphism of double vector bundles}
from $(D;A,B;M)$ to $(D';A',B';M')$ is a system of maps
$\phi\colon\thinspace D\to D'$, $\phi_A\colon\thinspace A\to A'$, $\phi_B\colon\thinspace B\to B'$, $\phi_M\colon\thinspace M\to M'$
such that each of $(\phi,\phi_A)$, $(\phi,\phi_B)$, $(\phi_A,\phi_M)$ and $(\phi_B, \phi_M)$
is a morphism of vector bundles.

Two examples to keep in mind, also shown in Figure~\ref{fig:dvbdef}, are the tangent prolongation
of an ordinary vector bundle $A$, and the double vector bundle $A\bz B$ which is formed from two
vector bundles $A$ and $B$ on $M$ by giving the pullback manifold $A\times_M B$ the pullback vector
bundle structures $q_A^!B$ and $q_B^!A$. (We denote the pullback of $B$ over $q_A$ by $q_A^!B$
instead of $q_A^\star B$, so as not to confuse the symbol $\star$ with the many duals that appear
in this paper.) We use the notation $A \bz B$ to distinguish this double vector bundle from the
Whitney sum $A \oplus B$;  as manifolds they are the same, but we regard $A\oplus B$ as a vector
bundle over the base $M$, and $A\bz B$ as a vector bundle over the bases $A$ and $B$. The tangent
prolongation vector bundle $TA\to TM$ is formed by applying the tangent functor to the structure
maps of $A\to M$.

\begin{figure}[h]
\begin{picture}(340,80)(-20,-60)
\put(30,0){\dvb{D}{A}{B}{M}{q^D_A}{q^D_B}{q_A}{q_B}}
\put(130,0){\dvb{TA}{A}{TM}{M}{}{}{q_A}{}}
\put(230,0){\dvb{A \bz B}{A}{B}{M}{}{}{q_A}{q_B}}
\end{picture}
\caption{\label{fig:dvbdef}}
\end{figure}

We also need a further condition, which was not made explicit in \cite{Mackenzie:2005dts}
or \cite{Mackenzie:GT}.

\begin{defin}
A double vector bundle $(D;A,B;M)$ satisfies the \emph{splitting condition} if there is a
morphism of double vector bundles $\Sigma\colon\thinspace A\bz B\to D$ which preserves $A$ and $B$; such
a map is a \emph{splitting of} $D$.
\end{defin}

Equivalently, a splitting is a map $\Sigma\colon\thinspace A\times_M B\to D$ which is right-inverse
to the combination of the two projections $D\to A\times_M B$ and, when regarded as
$q_A^!B\to D$, is linear over $A$ and, when regarded as $q_B^!A\to D$, is linear over $B$.
We note below that the splitting condition is equivalent to the existence of a
decomposition, and as such it is the counterpart of requesting local triviality in the
definition of vector bundle.  We could request only a local splitting condition, but a standard
\v{Cech} cohomology argument shows that this implies the global splitting condition.
It can be shown that for double vector bundles the splitting condition is implied by the rest
of the definition \cite{GrabowskiR:mvbs}. In this paper, we include it as part of the definition.

It is standard that connections in a vector bundle $A\to M$ correspond to maps $A\bz TM\to TA$
which are linear in each variable and preserve $A$ and $TM$ (see Example \ref{ej:conns}). It is
easy to check (see below Definition \ref{defin:decom}) that the duals of a double vector
bundle which satisfies the splitting condition, also satisfy the condition.

We briefly recall the basic constructions with double vector bundles as used in
\cite[Chap.~9]{Mackenzie:GT}. In terms of $(D;A,B;M)$ as in Figure~\ref{fig:dvbdef},
we refer to $A$ and $B$ as the \emph{side bundles} of $D$, and to $M$ as the \emph{double
base}. In the two side bundles the addition, scalar multiplication and subtraction are
denoted by the usual symbols $+$, juxtaposition, and $-$. We distinguish the two
zero-sections, writing $0^A \colon\thinspace M\to A,\ m\mapsto 0^A_m$, and $0^B \colon\thinspace M\to B,\ m\mapsto
0^B_m$. We may denote an element $d\in D$ by $(d;a,b;m)$ to indicate that
$a = q^D_A(d),\ b = q^D_B(d),\ m = q_B(q^D_B(d)) = q_A(q^D_A(d)).$

In the \emph{vertical bundle structure} on $D$ with base $A$ the vector bundle operations
are denoted $+_A, {.}_A, {-}_A$, with $\tilo^A\colon\thinspace A \to D,\ a\mapsto\tilo^A_a$, for
the zero-section. In the \emph{horizontal bundle structure} on $D$ with base $B$ we
likewise write $+_B$ and so on. For $m\in M$ the \emph{double zero}
$\rule[-8pt]{0pt}{22pt}\tilo^A_{0^A_m} = \tilo^B_{0^B_m}$ is denoted
$\odot_m$ or $0^2_m$.

The map $D\to A*B$ which combines the two bundle projections is a morphism of double
vector bundles and the set of elements of $D$ which map to the double zeros of $A*B$
is denoted $C$ and called the \emph{core of} $D$. Thus
$$
C=\{c \in D ~|~ \exists m\in M \mbox{\ such that}\ q^D_B(c )= 0^B_m,\
q^D_A(c)=0^A_m\}.
$$
The core is an embedded submanifold of $D$ and has a well-defined vector bundle
structure on base $M$. The projection $q_C$ is the restriction of
$q_B\circ q^D_B = q_A\circ q^D_A$ and the addition and scalar multiplication
are the restrictions of either of the operations on $D$. This is the \emph{core
vector bundle of} $D$.

As an ordinary vector bundle, $D\to B$ has a dual which we denote $D^X$
or $D\duer B$. There is a second vector bundle structure on $D^X$, with base $C^*$,
and projection $\zeta$ given by
\begin{equation}                             \label{eq:unfprojH}
\langle \zeta(\Psi), c\rangle =
   \langle\Psi, \tilo^B_b +_B c\rangle
\end{equation}
where $c\in C_m,\ \Psi\colon\thinspace (q^D_B)^{-1}(b)\to\mathbb{R}$ and $b\in B_m.$ The
addition $+_{C^*}$ and scalar multiplication in
$D^X\to C^*$ are defined by
\begin{gather}
\langle\Psi +_{C^*}\Psi', d +_A d'\rangle =
   \langle\Psi,d\rangle + \langle\Psi',d'\rangle\\
\langle t ._{C^*}\Psi, t ._A d\rangle = t\langle\Psi,d\rangle,
\end{gather}
for suitable elements. The zero above $\kappa\in C^*_m$ is denoted $\tilo^{^{\star} B}_\kappa$
and is defined by
\begin{equation}
\langle\,\tilo^{^{\star} B}_\kappa, \tilo^A_a +_B c\rangle =
\langle\kappa, c\rangle
\end{equation}
where $a\in A_m,\ c\in C_m.$ The core of $D^X$ is $A^*$ with $\phi\in A_m^*$
defining $\Bar{\phi}\in D^X$ by
$$
\langle\Bar\phi, \tilo^A_a +_B c\rangle =
\langle\phi,a\rangle.
$$
With these structures $(D^X;C^*,B;M)$ is a double vector bundle with core
$A^*$. Likewise the dual of $D\to A$, denoted $D^Y$ or $D\duer A$, has a vector
bundle structure over $C^*$, making $(D^Y;A,C^*;M)$ a double vector bundle with
core $B^*$. These are displayed in Figure~\ref{fig:dvb}.

Observe that the notation $(D;A,B;M)$, and the diagram in the Figure, indicate
that a choice has been made between the two structures on $D$. We may say that such
a double vector bundle has been \emph{placed}, and that the \emph{flip} of $D$,
namely $(D^f;B,A;M)$ in which the two structures on $D$ have been interchanged,
has the opposite \emph{placement}.

\begin{figure}[h]
\begin{picture}(340,80)(-20,-60)
\put(30,0){\dvbc{D}{A}{B}{C}{M}}
\put(130,0){\dvbc{D^X}{C^*}{B}{A^*}{M}}
\put(230,0){\dvbc{D^Y}{A}{C^*}{B^*}{M}}
\end{picture}
\caption{\label{fig:dvb}}
\end{figure}

There is a non-degenerate pairing between $D^X$ and $D^Y$ given, with the conventions of
\cite{Mackenzie:GT}, by
\begin{equation}
\label{eq:3duals}
\,\rule[-0.1ex]{0.4ex}{0.7\baselineskip}\,\,\Phi, \Psi\,\,\rule[-0.1ex]{0.4ex}{0.7\baselineskip}\, =
\langle\Phi, d\rangle_A - \langle\Psi, d\rangle_B
\end{equation}
where $\Phi\in D^Y$ and  $\Psi\in D^X$ project to the same element of $C^*$ and $d$ is
any element of $D$ which can be paired with $\Phi$ over $A$ and with $\Psi$ over $B$.
This definition depends on the placement of $D$; the pairing for $D^f$ is the
negative of \eqref{eq:3duals}.

The pairing induces an isomorphism of double vector bundles
\begin{equation}
\label{eq:Q}
Q\colon\thinspace D^{XYX}\to D^f.
\end{equation}
This isomorphism is natural in the usual informal sense of the word and preserves
$A$ and $B$ but it induces minus the identity on the core. One can modify $Q$ so that
it will induce minus the identity on one of the side bundles instead, but not so that
it will induce the identity on all of $A$, $B$, and $C$. Accordingly,
\cite{Mackenzie:2005dts} does not identify $D^{XYX}$ with $D^f$. At the end of this
section we will show that although there does exist an isomorphism between $D^{XYX}$ and $D^f$
which induces the identity on $A$, $B$, and $C$ there is no canonical one.  This shows
that $XYX$ and $f$ are different as functors.

In \cite{Mackenzie:2005dts} the group of dualization operations, there denoted
$\EuScript{VB}_2$, is taken to be the group generated by $X$ and $Y$, subject to  $X^2 = I$
and $Y^2 = I$ and the identifications which follow from the existence of the non-degenerate
pairing between the two duals. Here we denote this group by $\dg_2$, for \emph{duality
functors}, and we provide a formal definition for it in \ref{defin:dg2}.

\section{Duality functors}
\label{sect:df}

In this section we will illustrate the methods which we will use in the triple and
$n$-fold cases by showing that $XYX$ and $YXY$ are equal as functors, but that $XYX$ and
$f$ are not. Indeed we will see that $f$ is not an element of the group of duality
functors.

\begin{defin}
\label{defin:stat}
Consider two double vector bundles\xspace $\ts{D}$ and $\ts{D'}$ which have the same side bundles $A$ and $B$,
and the same core bundle $C$. A \emph{statomorphism} $\phi\colon\thinspace D\to D'$ is a morphism
of double vector bundles\xspace which preserves $A$, $B$ and $C$.
\end{defin}

A statomorphism is necessarily an isomorphism $D\to D'$. All dualizations of a
statomorphism are also statomorphisms.

The role played for vector spaces by the choice of a basis, or equivalently the
choice of an isomorphism $V\to\mathbb{R}^n$, is played for multiple vector bundles by the
notion of decomposition. First recall the notion of a decomposed double vector bundle\xspace.

Let $A$, $B$ and $C$ be vector bundles over $M$. Consider their fibered product
\begin{equation*}
{A \bz B \bz C} = \{ (a,b,c)\in A \times B \times C \; \vert \;
q^A(a) = q^B(b) = q^C(c) \in M \}.
\end{equation*}
This has a pullback vector bundle structure over $A$:
\begin{equation*}
(a,b,c) \un{+}{A} (a,b',c') : = (a,b+b',c+c')
\end{equation*}
and a pullback vector bundle structure over $B$:
\begin{equation*}
(a,b,c) \un{+}{B} (a',b,c') : = (a+a',b,c+c')
\end{equation*}
With these structures ${A \bz B \bz C}$ is called the \emph{decomposed double vector bundle\xspace{}} with side bundles
$A$ and $B$ and core bundle $C$ and denoted $A\bz B\bz C$. The structures on
$A\bz B\bz C$ are $q_A^!(B\oplus C)$ and $q_B^!(A\oplus C)$. These need to be
distinguished from the Whitney sum bundle $A\oplus B\oplus C$ on base $M$.

In general, if $\ts{D}$ is a double vector bundle\xspace with side bundles $A$ and $B$ and core bundle $C$,
we say that $\ts{A\bz B\bz C}$ is the \emph{decomposed double vector bundle\xspace associated to $\ts{D}$} and
write $\ts{\decom{D}}: = \ts{A\bz B\bz C}.$

\begin{defin}  \label{defin:decom}
A \emph{decomposition} of $\ts{D}$ is a statomorphism $P\colon\thinspace\ts{D}\to\ts{\decom{D}}$.

The set of decompositions of the double vector bundle\xspace $\ts{D}$ will be denoted $\Dec(\ts{D})$.
\end{defin}

\begin{rem}
Let $U$ be an open subset of $M$ such that the restrictions of the side bundles
and the core over $U$ are trivializable.  If we compose a decomposition of $D$
restricted over $U$ with trivializations of $A$, $B$, and $C$, we obtain a statomorphism onto:
$$\dvbs{U \times V_A \times V_B \times V_C}{U \times V_A}{U \times V_B}{U}$$
where $V_A$, $V_B$, and $V_C$ are the fiber types of $A$, $B$, and $C$ respectively.
This produces convenient local coordinates for the double vector bundle $D$.  Thanks
to a \v{C}ech cohomology argument, the existence of local coordinates is equivalent
to the existence of a decomposition (which, as explained below, is equivalent to the
splitting condition).
\end{rem}

Given a decomposition $P$, there is a splitting $\Sigma(a,b) = P^{-1}(a,b,0)$ and conversely
given a splitting $\Sigma$, a decomposition is defined by $P^{-1}(a,b,c) =
\Sigma(a,b) +_A (c +_B\tilo_a).$

Denote by $\Stat(A\bz B \bz C)$ the group of statomorphisms from $A\bz B\bz C$ to itself;
that is, $\Stat(A\bz B \bz C) = \Dec(A \bz B \bz C)$.
Each element of $\Stat(A\bz B\bz C)$ is of the form
\begin{equation}  \label{eq:philambda}
\phi_\lambda(a,b,c) = (a,b,c+\lambda(a, b))
\end{equation}
where $\lambda\colon\thinspace A\otimes B\to C$ is a linear map. (Throughout the paper we will usually
write tuples in place of tensor elements, when they are the argument of a linear map.)
Accordingly $\Stat(A\bz B\bz C)$ is a group with composition
$\phi_\lambda\circ\phi_{\lambda'} = \phi_{\lambda+\lambda'}$
and can be identified with the additive group $\Gamma(A^*\otimes B^*\otimes C)$. (Here
and elsewhere in the paper, the tensor products are over $M$.) We do not
consider the $\cinf{M}$-module structure on $\Gamma(A^*\otimes B^*\otimes C)$.
\label{comment:affine}

The duals of $A\bz B\bz C$ are the decomposed double vector bundles\xspace $C^*\bz B\bz A^*$ and $A\bz C^*\bz B^*$.
Again, the statomorphisms from $C^*\bz B\bz A^*$ to itself form an abelian group
isomorphic to $\Gamma(C\otimes B^*\otimes A^*)$, and those from $A\bz C^*\bz B^*$ to itself
form an abelian group $\Gamma(A^*\otimes C\otimes B^*)$. In what follows we will identify
these three groups without comment and denote them by $G_2$.

Let $\ts{D}$ be a double vector bundle\xspace with side bundles $A$ and $B$ and core bundle $C$. Then
$G_2 = \Stat(A\bz B\bz C)$ acts on $\Dec(\ts{D})$ to the left, and the action is simple and
transitive. Thus $\Dec(\ts{D})$ is a (non--empty) $G_2$--torsor.

We now wish to consider the effect of the dualization operations on morphisms between
double vector bundles\xspace, and specifically on decompositions. We first define the relevant
category.

\begin{defin}
Let $\mathcal{C}$ be the category whose objects are double vector bundles, and whose
morphisms are statomorphisms of double vector bundles.
\end{defin}

For the rest of this section, fix three vector bundles $A$, $B$ and $C$ over $M$
and, for the moment, denote them by $E_1:=A$, $E_2 := B$ and $E_0 :=C^{\star}$.  (In the
triple case we will use an extension of this notation exclusively.)
Let $S_3$ be the group of permutations of the set $\{0,1,2\}$.
We will also write $\C^{op}$ or $\C^{-1}$ for the opposite category to $\C$.

We can now think of $X$, $Y$ and $f$ as functors:
\begin{equation}
\label{eq:XYf}
\vb{\mscr{C}}{\mscr{C}^{op}}{X},\qquad
\vb{\mscr{C}}{\mscr{C}^{op}}{Y},\qquad
\vb{\mscr{C}}{\mscr{C}}{f}
\end{equation}
The effect of $X$ on objects has already been defined
and the action on morphisms is as expected~:
Let $\phi\colon\thinspace\ts{D_1} \longrightarrow \ts{D_2}$ be a morphism in the category $\mscr{C}$.
Then $\phi$ is a statomorphism of double vector bundles\xspace. In particular
$\phi\colon\thinspace D_1\to D_2$ is a morphism of vector bundles over $B$ and so can be
dualized to $\phi^X\colon\thinspace D_2^X\to D_1^X$; this is also a statomorphism of double vector bundles\xspace
$\ts{D_2^X} \longrightarrow \ts{D_1^X}$ and is thus a morphism
$\ts{\phi^X}\colon\thinspace \ts{D_1^X} \longrightarrow \ts{D_2^X}$ in the category
$\mscr{C}^{op}$.

Note that, since every morphism in $\C$ is invertible, we can regard $X$ and $Y$
as functors $\C \to \C^{op}$ or as functors $\C^{op} \to \C$; hence, we can compose
them, and we can talk of the group of functors generated by $X$ and $Y$. Denote this
group by $\W_2$. Now for any $W \in \W_2$ there exists a unique permutation $\sigma \in S_3$
such that, if $D$ is a double vector bundle with side bundles $E_1$ and $E_2$, and with core
bundle $E_0^{\star}$, then $D^W$ is a double vector bundle with side bundles
$E_{\sigma (1)}$ and $E_{\sigma(2)}$, and with core bundle $E_{\sigma(0)}^{\star}$.
Define $\pi(W) := \sigma$,  and define $\varepsilon_W = \pm1 $ to be the signature
of $\pi(W)$. Now $\vb{\mscr{C}}{\mscr{C}^{ \varepsilon_W}}{W}$ is a functor.

The flip operation $f$ is most naturally considered as a covariant functor; it can be
defined as such for all morphisms of double vector bundles\xspace, not only for statomorphisms.

Finally, we recall the definition of natural isomorphism.

\begin{defin}
Let ${\tt Cat}_i$, $i=1,2$, be two categories, and let
$F, G \colon\thinspace {\tt Cat}_1 \longrightarrow {\tt Cat}_2$ be two functors.
A \emph{natural transformation}  $s\colon\thinspace F \longrightarrow G$ is a collection of morphisms
$s(O) \colon\thinspace F(O) \longrightarrow G(O)$ in ${\tt Cat}_2$ for every object $O$ in ${\tt Cat}_1$
such that, given any morphism $f\colon\thinspace O \longrightarrow O'$ in ${\tt Cat}_1$, the following diagram
is commutative:
\begin{equation*}
\dvb{F(O)}{G(O)}{F(O')}{G(O')}{s(O)}{F(f)}{G(f)}{s(O')}
\end{equation*}
If, in addition, $s(O)$ is an isomorphism in ${\tt Cat}_2$ for every object
$O$ in ${\tt Cat}_1$, the natural transformation $s$ is called a \emph{natural isomorphism}.
\end{defin}

Consider $W \in \W_2$ such that $\pi(W)= 1 \in S_3$. Thus $W$ is a word in $X$ and $Y$,
and as a functor, $W$ is $\mscr{C}\to\mscr{C}.$ We want criteria for a natural isomorphism
to exist between $W$ and the identity functor. We resume the notation of Figure~\ref{fig:dvb}
so as to facilitate relating this section to \cite{Mackenzie:GT}.

Let $\ts{D}$ be a double vector bundle\xspace with side bundles $A$ and $B$ and core $C$.
Since $\pi(W) = 1$, $\ts{D^W}$ is also a double vector bundle\xspace with side bundles $A$ and $B$ and
core $C$.  Choose a decomposition $P\colon\thinspace \ts{D} \to \ts{\decom{D}}$ and apply the functor
$W$ to it. We obtain a decomposition of $D^W$:
$$
P^W \colon\thinspace \ts{D^W} \longrightarrow \decom{\ts{D^W}} = \ts{{\decom{D}}^W} = \decom{D} ={A \bz B \bz C}.
$$
Hence we obtain a statomorphism $(P^W)^{-1} \circ P\colon\thinspace \ts{D}\to\ts{D^W}$.
This is our candidate for a natural isomorphism between the identity functor and the
functor $W$. In order for it to succeed, it should at least be independent of the choice
of decomposition. The next two theorems together show that this is also a sufficient
condition.

The following result is valid for any element of $\W_2$.
Part (i) has been stated already and is included here for reference.

\renewcommand{\theenumi}{{\rm (\roman{enumi})}}

\begin{thm} \label{thm:submain}
Let $W \in \W_2$ and let $\ts{D}$ be a double vector bundle\xspace with side bundles $A$ and $B$, and core $C$.
\begin{enumerate}
\item The spaces of decompositions $\Dec(\ts{D})$ and $\Dec(\ts{D^W})$ are $G_2$-torsors.

\item  Choose a decomposition $P_0 \in \Dec(\ts{D})$.  Then the bijection
$\vartheta_W \colon\thinspace \Dec(\ts{D}) \to \Dec(\ts{D^W})$ defined by
$$
\vartheta_W(P) = (P^W)^{\varepsilon_W},
$$
induces a group automorphism $\theta_W \colon\thinspace G_2 \to G_2$ such that
$$
\vartheta_W(\phi \circ P_0) = \theta_W(\phi) \circ \vartheta_W(P_0)
$$
for $\phi\in G_2$.
Moreover, $\theta_W$ does not depend on the choice of $P_0$ (that is, $\vartheta_W$ is a
morphism of $G_2$-torsors) nor on the choice of the double vector bundle $D$.

\item  The map $\W_2 \times \Wii\to G_2,\quad (W,\lambda) \mapsto \theta_W(\lambda),$
is a group action.
\end{enumerate}
\end{thm}

\noindent{\textsc{Proof.}\ }
(ii)  It is enough to prove the result for $W = X$, since the same proof will hold
for $Y$, and then for any word $W$ in $X$ and $Y$.

Consider $P \mapsto (P^X)^{-1},\ \Dec (\ts{D}) \to \Dec (\ts{D^X})$. For
$\phi\in\Stat(A\bz B\bz C)$ we have, by functoriality,
$$
\vartheta_X(\phi\circ P) = (\phi^X)^{-1}\circ (P^X)^{-1} = (\phi^X)^{-1}\circ \vartheta_X(P),
$$
so that $\theta_X(\phi) = (\phi^X)^{-1}$.

Let us write $\phi = \phi_{\lambda}$ for some $\lambda \colon\thinspace A \otimes B \to C$ as in \eqref{eq:philambda}.
Similarly, let us write $\theta_X(\phi) = \phi_{\lambda'}$ for some $\lambda' \colon\thinspace C^* \otimes B \to A^*$.
Notice that with our abuse of notation we can also write $\theta_W(\lambda) = \lambda'$.
Consider the following diagram:
\begin{equation*}
\xymatrix{D \ar[r]^{P} \ar[rd]_{\phi_{\lambda}\circ P} & \decom{D} \ar[d]^{\phi_{\lambda}} & \decom{D^X} \ar[r]^{P^X} \ar[d]_{\phi_{\lambda'}} & D^X  \\
& \decom{D} & \decom{D^X} \ar[ur]_{(\phi_{\lambda} \circ P)^X}   }
\end{equation*}

If $d \in D_b$ and $\delta \in D^X_b = (D \duer B)_b$, we can calculate the pairing of $d$ and $\delta$
in any decomposition.  That is:
\begin{equation*}
\langle \delta \vert d \rangle \; = \;
\langle  (P^X)^{-1}(\delta)       \vert  P(d)  \rangle \; = \;
\langle  \phi_{\lambda'} \circ (P^X)^{-1}(\delta) \vert  (\phi_{\lambda} \circ P)(d) \rangle
\end{equation*}
Let us write $P(d) = (a,b,c) \in {A \bz B \bz C}$ and $(P^X)^{-1}(\delta)= (\gamma,b,\alpha) \in C^* \bz B \bz A^*$.  Then:
$$
\langle (P^X)^{-1}(\delta) \vert P(d) \rangle =
\langle (\gamma,b,\alpha) \vert (a,b,c) \rangle =
\langle \alpha \vert a \rangle + \langle \gamma \vert c \rangle
$$
and
\begin{align*}
\langle \phi_{\lambda'} ((P^X)^{-1}(\delta)) \vert  \phi_{\lambda} (P(d)) \rangle  = &
\langle (\gamma',b,\alpha + \lambda(\gamma,b)) \vert (a,b,c+ \lambda(a,b))\rangle\\
 = & \langle \alpha \vert a \rangle + \langle \gamma \vert c \rangle +
\lambda(a,b,\gamma) + \lambda'(a,b,\gamma).
\end{align*}
In order for these to be equal we need $\lambda = - \lambda'$.  We have proved that $\theta_X$ is
minus the identity on $G_2$, and that $(\phi)^X = \phi$.
This completes the proof of (ii).

\medskip

(iii) is clear now that we know the map $\theta$ is well defined.
\pfend

The action of $\W_2$ on $G_2$ is not faithful.  However, if we restrict it to the kernel
of $\pi \colon\thinspace\W_2\to S_3$ then it becomes faithful, as the next theorem shows.

\begin{thm} \label{thm:main}
For $W \in \W_2$ with $\pi(W) = 1$, the following are equivalent:
\begin{enumerate}
\item $\theta_W$ is the identity on $\Wii$.
\item There exists a natural isomorphism between the identity functor and $W$.
\end{enumerate}
\end{thm}

\noindent{\textsc{Proof.}\ }
$\textrm{(i)} \implies \textrm{(ii)}$ For every double vector bundle\xspace $D$ in $\mscr{C}$ we want to
obtain a statomorphism $s(D) \colon\thinspace D \to D^W.$ Choose a decomposition $P \colon\thinspace D \to A\bz B\bz C$.
We also have $P^W \colon\thinspace D^W \to A\bz B\bz C$.  Compose them to define $s(D):=(P^W)^{-1} \circ P$.

We check first that $s(D)$ defined in this way does not depend on the choice of $P$.
If $P_1, P_2 \in \Dec(\ts{D})$, there is some $\lambda \in \Wii$ such that
$P_2 = \phi_{\lambda} \circ P_1$.  We then have
$(P_2)^W = \phi_{\theta_W(\lambda)} \circ (P_1)^W$.
Since $\theta_W = \operatorname{id}_{G_2}$, it follows that
$\phi_{\theta_W(\lambda)}=\phi_{\lambda}$, and the following diagram is commutative:
\begin{equation*}
\xymatrix{
D \ar[r]^(.3){P_1} \ar[rd]_{P_2} & {A \bz B \bz C} \ar[d]^{\phi_{\lambda}} \ar@{}[r]^(.7){(P_1)^W} &
D^W \ar[l] \ar[ld] \\
& {A \bz B \bz C} \ar@{}[ur]_{(P_2)^W}
}
\end{equation*}
which proves that $s(D)$ is well defined.

We need to check that $s$ is a natural transformation.  Let $f\colon\thinspace D_1 \to D_2$ be a
statomorphism of double vector bundles\xspace. Choose a decomposition $P_2\colon\thinspace D_2 \to {A \bz B \bz C}$ of $D_2$ and
consider the decomposition $P_1 :=P_2 \circ f$ of $D_1$.
Then the following diagram is commutative:
\begin{equation*}
\xymatrix{
\ts{D_1} \ar[r]_(.3){P_1}  \ar@/^1pc/[rr]^{s(D_1)} \ar[d]_{f} &
\ts{{A \bz B \bz C}} \ar@{}[r]_(.7){(P_1)^W} \ar[d]^{=} &
\ts{(D_1)^W} \ar[l] \ar[d]^{f^W} \\
\ts{D_2} \ar[r]^(.3){P_2} \ar@/_1pc/[rr]_{s(D_2)} &
\ts{{A \bz B \bz C}} \ar@{}[r]^(.7){(P_2)^W} &
\ts{(D_2)^W} \ar[l]
}
\end{equation*}

$\textrm{(ii)} \implies \textrm{(i)}$  Reciprocally, let $s$ be a natural
transformation from the identity functor to $W$.  Let $\ts{D}$ be a double vector bundle\xspace in $\mscr{C}$
and let $\decom{\ts{D}} = \decom{\ts{D^W}} = {\decom{\ts{D}}}^W$ be the
decomposed double vector bundle\xspace.  Let $P_1, P_2 \in \Dec(\ts{D})$ be two decompositions of $\ts{D}$
such that $P_2=\phi_{\lambda} \circ P_1$ for $\lambda \in \Wii$. Then the following
diagram is commutative:
\begin{equation*}
\xymatrix{
\ts{D} \ar[r]_{P_1} \ar[d]_{=} \ar@/^1pc/[rrr]^{s(D)} &
\ts{\Bar{D}} \ar[r]_{s(\Bar{D})} \ar[d]_{\phi_{\lambda}} &
\ts{\Bar{D}} \ar@{}[r]_{(P_1)^W} \ar[d]_{(\phi_{\lambda})^W} &
\ts{D} \ar[l] \ar[d]_{=} \\
\ts{D} \ar[r]_{P_2} &
\ts{\Bar{D}} \ar[r]_{s(\Bar{D})} &
\ts{\Bar{D}} \ar@{}[r]_{(P_2)^W} &
\ts{D} \ar[l]
}
\end{equation*}
Hence $\phi_{\theta_W(\lambda)} =(\phi_{\lambda})^W = \phi_{\lambda}$
and $\theta_W(\lambda) = \lambda$.
\pfend

We can now give a precise definition of $\dg_2$ and calculate it.

\begin{defin}
\label{defin:dg2}
The group $\dg_2$ is the group $\W_2$ of functors generated by $X$ and $Y$, quotiented over
natural isomorphism in $\mscr{C}$.
\end{defin}

From (\ref{eq:XYf}) we have $\sigma_1: = \pi(X) = (01)$ and $\sigma_2 : = \pi(Y) = (02)$
and it follows that $\pi \colon\thinspace \dg_2\to S_3$ is surjective. Define $K_3$ to be the
kernel of $\pi$. The following lemma is an easy exercise in algebra.

\begin{lem} \label{lem:gener}
Let $f\colon\thinspace G \to S$ be a surjective group homomorphism.  Let $g_1,\ldots,g_n$ be a set of
generators of $G$ and write $\sigma_i:=f(g_i)$.  Let
$\{ R_j(\sigma_1,\ldots,\sigma_n) \; \vert \; j=1,\ldots,m \}$ be a set of relations for
a presentation of $S$ with generators $\sigma_1,\ldots,\sigma_n$.
Then the kernel of $f$ is the normal subgroup of $G$ generated by
$\{ R_j(g_1,\ldots,g_n) \; \vert \; j=1,\ldots,m  \}$.
\end{lem}

A presentation of $S_3$ with generators $\sigma_1, \sigma_2$ is
\begin{equation*}
\langle  \sigma_1, \sigma_2 \; \vert \; \sigma_1^2,\, \sigma_2^2,\, (\sigma_1\sigma_2)^3 \rangle .
\end{equation*}
Hence, $K_3 = \ker \pi$ is the normal subgroup of $\dg_2$ generated by
\begin{equation*}
\{ X^2,\, Y^2,\, (XY)^3 \} .
\end{equation*}

In the proof of Theorem \ref{thm:submain} we showed that
$\theta_X = \theta_Y = - \operatorname{id}.$ From this we obtain
\begin{equation*}
\theta_{X^2} = \theta_{Y^2} = \theta_{(XY)^3} = \operatorname{id}.
\end{equation*}
By Theorem \ref{thm:main} it follows that $X^2 = Y^2 = (XY)^3 = 1$ in $\dg_2$.
Thus $K_3$ is the trivial group and $\dg_2 = S_3$, as in \cite{Mackenzie:2005dts}.
In the following sections we will extend this method to determine the group $\dg_3$.

We end the present section by demonstrating that, for any double vector bundle\xspace $D$, a choice of
statomorphism between $D^{XYX}$ and $D^f$ is equivalent to choosing a decomposition
of $D$.

First, let $P\colon\thinspace \ts{D} \longrightarrow \ts{{A \bz B \bz C}}$ be a decomposition. Applying the functors $XYX$
and $f$ we get statomorphisms
$$
P^{XYX}\colon\thinspace  B\bz A\bz C  \longrightarrow \ts{D^{XYX}}\quad\text{and}\quad
P^{f}\colon\thinspace \ts{D^f} \longrightarrow  B\bz A\bz C.
$$
Hence $\Phi(S) := P^{XYX} \circ P^f \colon\thinspace \ts{D^f} \longrightarrow \ts{D^{XYX}}$ is a statomorphism.
The interesting result is that those are all the statomorphisms.

\begin{prop}
Let $\mathcal{S}$ denote the set of all statomorphisms $\ts{D^f} \longrightarrow \ts{D^{XYX}}$. The map
$$
\Phi\colon\thinspace  \Dec(\ts{D}) \longrightarrow \mathcal{S}
$$
defined above is a bijection.
\end{prop}

\begin{proof}
Let $h\colon\thinspace \ts{D^f} \longrightarrow \ts{D^{XYX}}$ be a statomorphism.  We want to prove that there is
a unique $P \in \Dec(\ts{D})$ such that $\Phi(P) = h$.  First,
using notation similar to that in the proof of Theorem \ref{thm:submain}, we note that
the groups of statomorphisms of the decomposed double vector bundles\xspace $\ts{{A \bz B \bz C}}$ and
$B \bz A \bz C$ are canonically isomorphic to $G_2$.
\begin{equation*}
\begin{split}
\Tilde{\phi}: \; \Wii \longrightarrow & \Stat(B \bz A \bz C) \\
\lambda \; \; \mapsto & \; \Tilde{\phi}_{\lambda}: B \bz A \bz C \longrightarrow B \bz A \bz C \\
            & \quad \quad (b,a,c) \mapsto (b,a,c + \lambda(a,b))
\end{split}
\end{equation*}
We also know from the proof of Theorem \ref{thm:submain}(ii) that
$(\phi_{\lambda})^f = (\phi_{\lambda})^{XYX}=\Tilde{\phi}_{\lambda}$.

Let us pick a decomposition $P_1\in \Dec(\ts{D})$. Then
$((P_1)^f)^{-1} \circ h \circ ((P_1)^{XYX})^{-1}$ is an automorphism of
$B \bz A \bz C$, and hence equals $\Tilde{\phi}_{\mu}$ for some $\mu \in \Wii$.
\begin{equation*}
\xymatrix{
\ts{D}^f \ar[r]^{h} \ar[d]_{(P_1)^f} & \ts{D}^{XYX} \ar@{}[d]^{(P_1)^{XYX}} \\
\ts{B \bz A \bz C} \ar[r]_{\Tilde{\phi}_{\mu}} & \ts{B \bz A \bz C} \ar[u]
}
\end{equation*}
We know that every decomposition of $\ts{D}$ is of the form $\phi_{\lambda} \circ P_1$ for a
unique $\lambda \in \Wii$.  So we want to prove that there is a unique $\lambda \in \Wii$
such that $h = \Phi(\phi_{\lambda} \circ P_1)$.  Noting that
$$
h  = (P_1)^{XYX} \circ \Tilde{\phi}_{\mu} \circ (P_1)^f
$$
we have that
\begin{multline*}
\Phi(\phi_{\lambda} \circ P_1) =
(\phi_{\lambda} \circ S_1)^{XYX} \circ (\phi_{\lambda} \circ P_1)^f =
(P_1)^{XYX} \circ (\phi_{\lambda})^{XYX} \circ (\phi_{\lambda})^f \circ (P_1)^f  \\
= (P_1)^{XYX} \circ \Tilde{\phi}_{\lambda} \circ \Tilde{\phi}_{\lambda} \circ (P_1)^f =
(P_1)^{XYX} \circ \Tilde{\phi}_{2 \lambda} \circ (P_1)^f
\end{multline*}
and so $h = \Phi(\phi_{\lambda} \circ P_1)$ if and only if $2 \lambda = \mu.$
This completes the proof.
\end{proof}

To summarize: there is a canonical isomorphism between $D^{XYX}$ and $D^f$, but it is not
a statomorphism;  there are statomorphisms between $D^{XYX}$ and $D^f$, but not canonical
ones, as choosing one such statomorphism is equivalent to choosing a decomposition; the
functors $XYX$ and $f$ are not naturally isomorphic through statomorphisms.  Because of
this, we regard $XYX$ and $f$ as distinct  functors.

\begin{ej}
\label{ej:conns}
It is worthwhile to consider the significance of decompositions for the tangent
prolongation of an ordinary vector bundle.

Let $\vb{A}{M}{q}$ be a vector bundle. Applying the tangent functor, we obtain a double vector bundle\xspace:
\begin{equation*}
\dvb{TA}{A}{TM}{M}{\pi_A}{Tq}{q}{\pi_M}
\end{equation*}
with side bundles  $A$, and $TM$. The core consists of those $X\in TA$
which are annulled by both $T(q)$ and $\pi_A$; that is, $C$ is the set of vertical
vectors along the zero section, and may be identified with $A$.

A splitting of $TA$ is a connection in $A$ in the usual sense; see, for example,
\cite{Besse}. Given a splitting
\begin{equation*}
\Sigma\colon\thinspace TM\times_M A\to TA,
\end{equation*}
and a vector field $x$ on $M$, define a vector field $\Tilde{x}$ on $A$ by
$$
\Tilde{x}(e) = \Sigma(x(q(e)),e).
$$
It follows from the properties of $\Sigma$  that $x\mapsto\Tilde{x}$
is a connection in $A$. Conversely a connection in $A$ induces a decomposition,
and $\Dec(TA)$ can be identified with the space of connections in $A$.

As is well known, connections in $A$ form an affine space with model space
$\Gamma(T^* M \otimes\operatorname{End}(A))$. This includes the $G_2$-torsor structure.

The dual over $TM$, namely $(TA)^X$, can be identified with the tangent
double vector bundle $T(A^*)$ \cite[9.3.2]{Mackenzie:GT} and so
Theorem \ref{thm:submain}(ii) includes the correspondence between connections
in $A$ and connections in $A^*$.
\end{ej}

\section{Triple vector bundles}
\label{sect:tvb}

We begin by recalling the definition of a triple vector bundle. See
\cite{Mackenzie:2005dts} for fuller details. We use a notation that extends readily
to the $n$-fold case.

\begin{figure}[h]
\setlength{\unitlength}{1cm}
\begin{picture}(15,5)(-0.5,-0.5)
\put(0,4)
{\smallcube{E_{1,2,3}}{E_{2,3}}{E_{1,3}}{E_3}{E_{1,2}}{E_2}{E_1}{M}}
\put(7,3.6){\xymatrix@=2mm{
&& E_{3,12}\ar[dddd] &&\\
& E_{2,31}\ar[rrdd] &&&\\
E_{1,23}\ar[rrrr] &&&& E_{23}\\
&&& E_{13} &\\
&& E_{12} &&\\
}}
\end{picture}
\caption{\ \label{fig:tvb}}
\end{figure}

\begin{defin}
A \emph{triple vector bundle} $E$ is a system as on the left of Figure~\ref{fig:tvb}
where each arrow represents a vector bundle structure, such that each face
is a double vector bundle, and which satisfies the splitting condition stated
below.

A \emph{morphism of triple vector bundles} $\phi\colon\thinspace E\to F$ is a system of maps
$\phi_I\colon\thinspace E_I\to F_I$, for all subsets $I$ of $\{1,2,3\}$, such that for all
nonempty subsets $I$ and each $k\in I$, $\phi_I$ is a morphism of vector
bundles over $\phi_{I\backslash\{k\}}$.
\end{defin}

We always read figures as in Figure~\ref{fig:tvb} with
$(E_{1,2,3};E_{1,2}, E_{2,3};E_2)$ at the rear and $(E_{1,3};E_1,E_3;M)$ coming
out of the page toward the reader. Notice that when a nonempty subscript has $k$
commas, $0\leq k\leq 2$, the space which it denotes has $k+1$ vector bundle
structures. We sometimes denote $M$ by $E_\emptyset$ for uniformity.

The three structures of double vector bundle on $E_{1,2,3}$ are the \emph{upper}
double vector bundles, and $E_{1,2}, E_{2,3}, E_{3,1}$ are the \emph{lower} double
vector bundles. We refer to $(E_{1,2};E_1, E_2;M)$ as the \emph{floor} of
$E$ and to $(E_{1,2,3};E_{1,3}, E_{2,3};E_3)$ as the \emph{roof} of $E$,
with \emph{left, right, front} and \emph{back} having their usual meanings.

The cores of the lower double vector bundles are denoted
$E_{12},\, E_{23},\, E_{31},$ and the cores of the upper double
vector bundles are denoted $E_{3,12},\,E_{1,23},\,E_{2,31}.$
The latter are vector bundles over the former, as indicated on the
right of Figure~\ref{fig:tvb}, and  form three \emph{core double vector
bundles}, $(E_{1,23}; E_1,E_{23};M)$, $(E_{2,31}; E_2,E_{31};M)$, and
$(E_{3,12}; E_3,E_{12};M)$, Each of the three core double vector bundles
has the same core, denoted $E_{123}$ and called the \emph{ultracore}.
The seven vector bundles $E_1,\,E_2,\,E_3,\,E_{12},\,E_{23}$, $E_{31},\,E_{123}$
are called the \emph{building bundles of} $E.$ We denote them collectively
by $E_\bullet$.

\begin{ej}
The tangent prolongation of a double vector bundle $(D;A,B;M)$ is the triple
vector bundle $TD$ shown in Figure~\ref{fig:tvbexs}(a). The three core double
vector bundles are $(D;A,B;M)$, $(D;A,B;M)$ and $(TC;C.TM;M)$, where $C$ is the
core of $D$. The ultracore is $C$.

\begin{figure}[h]
\begin{center}
\subfloat[]
{\smallcube{TD}{TB}{TA}{TM}{D}{B}{A}{M}}
\qquad
\subfloat[]
{\xymatrix@=5mm{
                             & & E_{2,3}\ar'[d][dd] \ar[dr] &\\
& E_{1,3}   \ar[rr] \ar[dd]    & & E_3 \ar[dd]  \\
E_{1,2}\ar'[r][rr] \ar[dr]  &       & E_2 \ar[dr] &\\
& E_1  \ar[rr] & & M\\
}}
\caption{\ \label{fig:tvbexs}}
\end{center}
\end{figure}

Suppose given three double vector bundles arranged as in Figure~\ref{fig:tvbexs}(b)
and a vector bundle $E_{123}$ on $M$.
Define the manifold $E'_{1,2,3}$ to be the pullback of the diagram; that is, $E'_{1,2,3}$ consists of
triples $(e_{1,2},\,e_{1,3},\,e_{2,3})\in E_{1,2}\times E_{1,3}\times E_{2,3}$ such that
$$
q^{1,2}_1(e_{1,2}) = q_1^{1,3}(e_{1,3}),\quad
q^{1,2}_2(e_{1,2}) = q_2^{2,3}(e_{2,3}),\quad
q^{1,3}_1(e_{1,3}) = q_3^{1,3}(e_{1,3}),
$$
using an obvious notation for the projections $q_k^{i,j}$. Finally, define
$E_{1,2,3}$ to be the manifold of tuples
$(e_{1,2},\,e_{1,3},\,e_{2,3},\,e_{123})\in E'_{1,2,3}\times E_{123}$ such that
$(e_{1,2},\,e_{1,3},\,e_{2,3})$ projects to the same point of $M$ as does $e_{123}$.
Define a vector bundle structure on $E_{1,2,3}$ with base $E_{1,2}$ by
\begin{multline*}
(e_{1,2},\,e_{1,3},\,e_{2,3},\,e_{123}) \add{1,2} (e_{1,2},\,e'_{1,3},\,e'_{2,3},\,e'_{123})\\
= (e_{1,2},\,e_{1,3}\add{1}e'_{1,3},\,e_{2,3}\add{2}e'_{2,3},\,e_{123}+e'_{123}),
\end{multline*}
and scalar multiplication analogously. With the similar structures over $E_{1,3}$ and
$E_{2,3}$, this makes $E_{1,2,3}$ a triple vector bundle with ultracore $E_{123}.$
Thus any diagram of the form Figure~\ref{fig:tvbexs}(b) can be completed to a triple
vector bundle with an arbitrary ultracore.

Given an indexed set of vector bundles
$E_\bullet = \{E_1,\,E_2,\,E_3,\,E_{12},\,E_{23},\,E_{31},\,E_{123}\}$ over base $M$, performing
this construction with the decomposed double vector bundles $E_{i,j} = E_i\bz E_j\bz E_{ij}$
yields the \emph{decomposed triple vector bundle} $\Bar{\BE}$ with the $E_\bullet$ as building
bundles.  (Note that the bar here means something slightly different from the notation
$\decom{D}$ for a double vector bundle $D$.  In both cases the bar means that we build a
decomposed $n$-fold vector bundle, but in one case we start with a general $n$-fold vector
bundle, while in the other we start with a set of building bundles.)
\end{ej}

\begin{defin}
\label{df:stato3}
Let $E$ and $F$ be triple vector bundles with the same building bundles
$E_\bullet$. A \emph{statomorphism} from $E$ to $F$ is an isomorphism
$\phi\colon\thinspace E\to F$ which induces the identity on each of the building bundles.

A triple vector bundle $E$ satisfies the \emph{splitting condition} if there is a
statomorphism of triple vector bundles $\phi\colon\thinspace E\to \Bar{\BE}$;
such a map is a \emph{splitting of} $E$.
\end{defin}

\begin{prop}
If a double vector bundle $(D;A,B;M)$ satisfies the splitting condition, then its
tangent prolongation $TD$ does also.
\end{prop}

\noindent{\textsc{Proof.}\ }
Taking the tangent of a decomposition of $D$ we have a diffeomorphism
$TD\to TA\times_{TM} TB\times_{TM} TC$. Using decompositions of $TA, TB$ and $TC$, this
gives a diffeomorphism
$$
TD\cong (A\bz A\bz TM)\times_{TM} (B\bz B\bz TM) \times_{TM} (C\bz C\bz TM)
\cong A\bz A\bz B\bz B\bz C\bz C\bz TM,
$$
where $\bz$ denotes pullback over $M$. Setting $E_1 = A,\ E_2 = B,\ E_3 = TM,\ E_{12} = C,$
$E_{13} = A,\,E_{23} = B,\, E_{123} = C$ we have a decomposition of $TD$.
\pfend

The duals of a triple vector bundle are set out in \cite[\S7]{Mackenzie:2005dts}; see
Figure~\ref{fig:tvbXYZ}. Here we need a different notation to use for the duals of decomposed
triple vector bundles. For any subset $I\subseteq\{0,1,2,3\}$, for the duals of building
bundles, write
$$
E_I^* = E_{I^c},
$$
where $I^c$ us the complement of $I$.
In particular $E_0 = E_{123}^*$ is the dual of the ultracore.
The effect of dualization on the building bundles is shown in Table~\ref{table:tvbXYZ}. In
this table the middle column shows the duals of the ultracores, rather than the ultracores themselves.

\begin{figure}[h]
\setlength{\unitlength}{10mm}
\begin{picture}(15,6)(1,0)
\put(0,5){\xymatrix@!=1pc{
\save[]+<-1.5cm,0.1cm>*\txt<8pc>{$E^X = {}$} \restore
{E_{1,2,3}\duer E_{2,3}} \ar[rr] \ar[dd] \ar[dr] & & E_{2,3} \ar'[d][dd] \ar[dr] &\\
& E_{3,12}\duer E_3 \ar[rr] \ar[dd]         & & E_3 \ar[dd]  \\
E_{2,31}\duer E_2  \ar'[r][rr] \ar[dr]         & & E_2 \ar[dr] &\\
& E_0  \ar[rr] & & M\\
}}
\put(8.5,5){\xymatrix@!=0.4pc{
&& E_{1,3}\duer E_3\ar[dddd] &&\\
& E_{1,2}\duer E_2\ar[rrdd] &&&\\
E_{1,23}\duer E_{23}\ar[rrrr] &&&& E_{23}\\
&&& E_{03} &\\
&& E_{02} &&\\
}}
\end{picture}
\caption{The $X$ dual of a general triple vector bundle.\label{fig:tvbXYZ}}
\end{figure}

Define $\dg_3$ to be the group of functors generated by $X, Y, Z$ up to natural
isomorphism, that is, two functors are considered the same if there is
a natural isomorphism between them through statomorphisms. (We will normally
denote elements of $\dg_3$ by representatives.) As in the double case,
there is a group homomorphism $\pi \colon\thinspace\dg_3\to S_4$, defined by the action of
$X$, $Y$ and $Z$ on $E_1$, $E_2$, $E_3$ and $E_0$. We have
$$
\pi(X) = (01),\quad
\pi(Y) = (02),\quad
\pi(Z) = (03).
$$

\newcolumntype{Z}{>{$}c<{$}}
\newcolumntype{L}{>{$}l<{$}}
\newcolumntype{R}{>{$}r<{$}}

\begin{table}[h]
\begin{center}
\begin{tabular}{||L||R|R|R|R|R|R|R||}
\hline\hline
&E_1&E_2&E_3&E_0&E_{12}&E_{23}&E_{31}\\
\hline\hline
X &E_0&E_2&E_3&E_1&E_{02}&E_{23}&E_{03}\\\hline
Y &E_1&E_0&E_3&E_2&E_{01}&E_{03}&E_{31}\\\hline
Z &E_1&E_2&E_0&E_3&E_{12}&E_{02}&E_{01}\\\hline
\hline
\end{tabular}
\end{center}
\caption{The action on the building bundles, with the ultracores
replaced by their duals. \label{table:tvbXYZ}}
\end{table}

Write $\sigma_1, \sigma_2, \sigma_3$ for the images of $X$, $Y$ and $Z$. These generate
$S_4$ and so $\pi$ is surjective. Write $K_4$ for the kernel of $\pi$.
A presentation for $S_4$ in terms of these generators is
\cite{CoxeterM:1965}:
$$
\langle \sigma_1,\,\sigma_2,\,\sigma_3 \ |\ \sigma_i^2,\ (\sigma_i\sigma_j)^3,\
(\sigma_i\sigma_{j}\sigma_i\sigma_k)^2\rangle,
$$
where $i$, $j$, and $k$ are distinct.
Hence, by Lemma \ref{lem:gener}, $K_4$ is the normal subgroup generated by
the
elements in the list
\begin{equation}
\label{eq:k4gens}
X^2,\,Y^2,\,Z^2,\,(XY)^3,\,(YZ)^3,\,(ZX)^3,\, (XYXZ)^2,\,(YZYX)^2,\,(ZXZY)^2.
\end{equation}

A priori, and according to our presentation, the list should contain further elements, but
they are not needed since $XYX = YXY$ implies $(XYXZ)^2 = (YXYZ)^2$ and similar conditions.

From Table~\ref{table:tvbXYZ} it is clear that the action of $X,Y,Z$ on the set of
building bundles is determined by the action on the set $\{0,1,2,3\}$ of indices, and so acts
as $S_4$.   In particular, $W$ acts trivially on the set of building bundles if and only if
$\pi(W)=1$.   As a consequence, since two triple vector bundles are statomorphic if and only
if they have the same building bundles, we conclude that $W \in K_4$ if and only if $E$ and
$E^W$ are statomorphic for every triple vector bundle $E$ (albeit not canonically, unless
$W=1$).

\begin{ej}
\label{ej:its}
For the tangent prolongation of a double vector bundle $(D;A,B;M)$ with core $C$, as shown in
Figure~\ref{fig:tvbexs}(a), two duals are shown in Figure~\ref{fig:dtp}. The first is the cotangent
double of $D$ and is described in detail in \cite{Mackenzie:2005dts}. In (b), $T_A^\bullet D\to TA$
denotes the dual of the tangent prolongation bundle $TD\to TA$ and likewise $T^\bullet C\to TM$ is the
dual of $TC\to TM$.

The duality between $C$ and $C^*$ induces a duality between $TC\to TM$ and $T(C^*)\to TM$ and
induces an isomorphism $T(C^*)\to T^\bullet C$ \cite{MackenzieX:1994,Mackenzie:GT}. Similarly there
is an isomorphism $T(D\duer A)\to T_A^\bullet D$. These combine into an isomorphism from the triple
vector bundle in (b) to the tangent prolongation of $(D\duer A;A,C^*;M)$.

In the case where $D$ is the double tangent bundle $T^2M$, the $X, Y$ and $Z$ duals of $T^3M$
may be canonically identified with $T^*(T^2M)$, $T(T^*TM)$ and $T^2(T^*M)$. There are also
canonical isomorphisms from the first of these to the second and third, and to the three
triple vector bundles $T^*(T^*TM)$, $T^*(TT^*M)$ and $T(T^*T^*M)$. These isomorphisms are of a
different character to those treated in this paper and will be studied elsewhere.

\begin{figure}[h]
\begin{center}
\subfloat[]
{\smallcube{T^*D}{D\duer B}{D\duer A}{C^*}{D}{B}{A}{M}}
\qquad
\subfloat[]
{\smallcube{T_A^\bullet D}{T^\bullet C}{TA}{TM}{D\duer A}{C^*}{A}{M}}
\caption{See Example \ref{ej:its}. \label{fig:dtp}}
\end{center}
\end{figure}

\end{ej}

\section{Decompositions of triple vector bundles}
\label{sect:dtvb}

We know from \S\ref{sect:intro} that the first six elements listed in
\eqref{eq:k4gens} are equal to the identity.
In addition, the group generated by the last three elements is normal in $\dg_3$.  This
can be proven by a direct calculation: using the fact that the first six elements are equal
to the identity, it follows that:
\begin{equation}  \label{eq:actiononK4}
\begin{split}
X(XYXZ)^{2}X^{-1} = (ZXZY)^{-2}, \\
Y(XYXZ)^2Y^{-1}=(YZYX)^{-2},  \\
 Z(XYXZ)^2Z^{-1}=(XYXZ)^{-2}.
\end{split}
\end{equation}
Hence, $K_4$ is the subgroup generated by the last three elements in \eqref{eq:k4gens}.
To calculate it we use a triple analogue
of \ref{thm:main}.

We start by determining the statomorphisms from a decomposed triple vector bundle to
itself. Denote the set of decompositions of the triple vector bundle $E$ by $\Dec(E)$.

\begin{prop}
\label{prop:statos}
Each statomorphism $\phi$ from $\Bar{\BE}$ to itself is of the form
\begin{multline}  \label{eq:e123}
\phi(e_1,e_2,e_3,e_{12},e_{13},e_{23},e_{123})
= (e_1,\,e_2,\,e_3,\\
e_{12}+\gamma(e_1,\, e_2),\,e_{13}+\beta(e_1,\, e_3),\,
e_{23}+\alpha(e_2,\, e_3),\\
e_{123}+\nu(e_3,\, e_{12})+\lambda(e_1,\, e_{23})+\mu(e_2,\, e_{13})+\rho(e_1,\, e_2,\, e_3))
\end{multline}
where
\begin{gather*}
\gamma\colon\thinspace E_1\otimes E_2\to E_{12},\quad
\beta\colon\thinspace E_1\otimes E_3\to E_{13}, \quad
\alpha\colon\thinspace E_2\otimes E_3\to E_{23}, \\
\lambda\colon\thinspace E_1\otimes E_{23}\to E_{123},\quad
\mu\colon\thinspace E_2\otimes E_{13}\to E_{123},\quad
\nu\colon\thinspace E_3\otimes E_{12}\to E_{123},
\end{gather*}
and $\rho\colon\thinspace E_1\otimes E_2\otimes E_3\to E_{123}$ are linear maps.
\end{prop}

Let $G_3$ denote the set of all such $(\gamma,\beta,\alpha,\lambda,\mu,\nu,\rho)$.
This is a group under the composition
\begin{equation}
(\gamma',\beta',\alpha',\lambda',\mu',\nu',\rho')\,(\gamma,\beta,\alpha,\lambda,\mu,\nu,\rho)
= (\gamma'',\beta'',\alpha'',\lambda'',\mu'',\nu'',\rho'')
\end{equation}
where $\gamma'' = \gamma'+\gamma,\,\dots,\,\nu'' = \nu'+\nu$ and
\begin{multline*}
\rho''(e_1,e_2,e_3) =
\rho(e_1,\, e_2,\, e_3) + \rho'(e_1,\, e_2,\, e_3)\\
+\lambda'(e_1,\,\alpha(e_2,\, e_3)) +
\mu'(e_2,\,\beta(e_1,\, e_3)) +
\nu'(e_3,\,\gamma(e_1,\, e_2)).
\end{multline*}

Given two decompositions $P$ and $P_0$ of a triple vector bundle $E$, there is a
unique statomorphism $\phi\in G_3$ such that $P = \phi\circ P_0$.
Thus $\Dec(E)$ is a simply transitive $G_3$--space, or torsor.

As in the double case, we define $\C$ to be the category whose objects are triple
vector bundles, and whose morphisms are statomorphisms. We denote by $\C^{op}$ or by
$\C^{-1}$ its opposite category. The dualization operations $X$, $Y$ and $Z$ can now
be regarded as functors $\mscr{C} \to \mscr{C}^{op}$ or as functors
$\mscr{C}^{op} \to \mscr{C}$.

Let $W \in \dg_3$. There is a unique permutation $\sigma \in S_4$ such that, if $E$ is
a triple vector bundle with building bundles $\BE$, then $E^W$ is a triple vector bundle
with building bundles
$$
E_{\sigma(1)},\,
E_{\sigma(2)},\,
E_{\sigma(3)},\,
E_{\sigma(1)\sigma(2)},\,
E_{\sigma(1)\sigma(3)},\,
E_{\sigma(2)\sigma(3)},\,
E_{\sigma(1)\sigma(2)\sigma(3)}.
$$
Define $\pi (W) := \sigma$ and define $\varepsilon_W = \pm 1$ to be the signature of $\pi(W)$.
We can now look at $W$ as a functor $\mscr{C} \to \mscr{C}^{\varepsilon_W}$.
We now give the triple analogue of Theorem \ref{thm:submain}.

We know that $\Dec(E)$ is a torsor modelled on $G_3$, which is (as a set) the
sum of the seven $\cinf{M}$-modules
\begin{gather*}
\Gamma(E_1^*\otimes E_2^*\otimes E_{12}),\quad
\Gamma(E_1^*\otimes E_3^*\otimes E_{13}), \quad
\Gamma(E_2^*\otimes E_3^*\otimes E_{23}),\\
\Gamma(E_1^*\otimes E_{23}^*\otimes E_{123}),\quad
\Gamma(E_2^*\otimes E_{13}^*\otimes E_{123}),\quad
\Gamma(E_3^*\otimes E_{12}^*\otimes E_{123}),\\
\Gamma(E_1^*\otimes E_2^*\otimes E_3^*\otimes E_{123}).
\end{gather*}
In the same way it is easy to establish that $\Dec(E^X)$ is a torsor modelled on a group
which is the sum of the modules
\begin{gather*}
\Gamma(E_0^*\otimes E_2^*\otimes E_{02}),\quad
\Gamma(E_0^*\otimes E_3^*\otimes E_{03}), \quad
\Gamma(E_2^*\otimes E_3^*\otimes E_{23}),\\
\Gamma(E_0^*\otimes E_{23}^*\otimes E_{023}),\quad
\Gamma(E_2^*\otimes E_{03}^*\otimes E_{023}),\quad
\Gamma(E_3^*\otimes E_{02}^*\otimes E_{023}),\\
\Gamma(E_0^*\otimes E_2^*\otimes E_3^*\otimes E_{023}),
\end{gather*}
(see Table \ref{table:tvbXYZ}). Up to rearrangements of
the tensor products, this is a permutation of the first list: for example,
$E_0^*\otimes E_2^*\otimes E_{02}$, the first space in the second list, is
$E_{123}\otimes E_2^*\otimes E_{13}^*$, a rearrangement of
the fifth space in the first list.
A similar result holds for $E^Y$ and $E^Z$ and thus for $E^W$ for any word $W\in\dg_3$.

In this way one proves the following result.

\begin{prop}
\label{thm:submain3-i}
Let $W \in \dg_3$ and let $E$ be a triple vector bundle
with building bundles $\BE$.
Then there is a canonical representation of the space of decompositions
$\Dec(E^W)$ as a torsor modelled on $G_3$.
\end{prop}

We now calculate $\theta_X \colon\thinspace G_3 \to G_3$.
First, we explain some notation.  For any  $I \subseteq \{0,1,2,3\}$, we write $e_I$ for a generic element
of $E_I$.  Given a linear map $\gamma \colon\thinspace E_1 \otimes E_2 \to E_{12}$, we can also think of it, for instance,
as a map $\gamma \colon\thinspace E_{30} \otimes E_2 \to E_{023}$; we can also think of it as simply
$\gamma \in \Gamma(E_1^* \otimes E_2^* \otimes E_{03}^*)$.   With this abuse of notation, we can write:
$$
\pair{e_{03}}{\gamma(e_1, e_2)}= \pair{e_1}{\gamma(e_2, e_{03})} = \gamma(e_1, e_2, e_{03}).
$$
If, in addition, $\nu \in \Gamma(E_{0}^* \otimes E_{3}^* \otimes E_{12}^*)$, there is a linear map
$\gamma \nu \in \Gamma(E_0^* \otimes E_1^* \otimes E_2^* \otimes E_3^*)$ defined by:
$$
\gamma \nu (e_0, e_1, e_2, e_3) := \pair{\nu(e_0, e_3)}{\gamma(e_1, e_2)} =
\pair{e_3 }{  \nu \left( e_0, \gamma(e_1, e_2) \right) } = \ldots
$$

Let $g = (\gamma,\beta,\alpha,\lambda,\mu,\nu,\rho)$ be an element of $G_3$ and let
$\phi = \phi_g \colon\thinspace \Bar{\BE} \to \Bar{\BE}$ be the corresponding statomorphism.
Then $\theta_X(\phi) \colon\thinspace \Bar{\BE}^X \to \Bar{\BE}^X$ is another statomorphism.
We can write $\theta_X(\phi) = \phi_{\tilde{g}}$  for some element
$\Tilde{g} = (\Tilde{\mu}, \Tilde{\nu}, \Tilde{\alpha}, \Tilde{\lambda}, \Tilde{\gamma}, \Tilde{\beta}, \Tilde{\rho})$
in $G_3$.  In order to describe $\theta_X$, we want to write $\Tilde{g}$ in terms of $g$.

As we explained with double vector bundles, if $d \in \Bar{\BE}$ and $\delta \in \Bar{\BE}^X = \Bar{\BE} \duer E_{2,3}$
are two elements that project to the same element in $E_{2,3}$ (so that they can be paired over $E_{2,3}$), then we
shall have $\pair{\delta}{d} = \pair{\theta_{X}(\phi)(\delta)}{\phi(d)}$.  Let us write
$$
d=(e_1, e_2, e_3, e_{12}, e_{13}, e_{23}, e_{123}),\qquad
\delta=(e_0, e_2, e_3, e_{02}, e_{03}, e_{23}, e_{023}).
$$
Then $\phi(d)$ is given by \eqref{eq:e123}, whereas
\begin{multline*}
\theta_X(\phi) (\delta)
= (e_0,\,e_2,\,e_3,
e_{02}+\Tilde{\mu}(e_0,\, e_2),\,e_{03}+\Tilde{\nu}(e_0,\, e_3),\,
e_{23}+\Tilde{\alpha}(e_2,\, e_3),\\
e_{023}+\Tilde{\beta}(e_3,\, e_{02})+\Tilde{\lambda}(e_0,\, e_{23})+\Tilde{\gamma}(e_2,\, e_{03})+\Tilde{\rho}(e_0,\, e_2,\, e_3))
\end{multline*}
First we notice that, in order to be able to pair $\theta_{X}(\phi)(\delta)$
and ${\phi(d)}$ over $E_{2,3}$, we need to have $\Tilde{\alpha} = \alpha$.
Now:
$$
\pair{\delta}{d} =
\pair{e_{023}}{e_1} + \pair{e_{03}}{e_{12}} + \pair{e_{02}}{e_{13}} + \pair{e_0}{e_{123}},
$$
and $\pair{\theta_{X}(\phi)(\delta)}{\phi(d)}$ is equal to
\begin{multline*}
\pair{e_{023}}{e_1} + \Tilde{\lambda}(e_0,  e_{23}, e_1) + \Tilde{\gamma}(e_{03},,e_2, e_1) +
\Tilde{\beta}(e_{02}, e_3, e_1)  + \Tilde{\rho}(e_0, e_2, e_3, e_1)\\
+ \pair{e_{03}}{e_{12}} + \Tilde{\nu}(e_{03}, e_2, e_1) + \gamma(e_0, e_3, e_{12})
+ \gamma \Tilde{\nu} (e_0, e_2, e_3, e_1) + \pair{e_{02}}{e_{13}}\\
+ \Tilde{\mu}(e_{02}, e_3, e_1) + \beta(e_0, e_2, e_{13}) + \beta \Tilde{\mu}(e_0, e_2, e_3, e_1) + \pair{e_0}{e_{123}}\\
+ \lambda(e_0, e_{23}, e_1) + \mu(e_0, e_2, e_{13}) + \nu(e_0, e_3, e_{12}) + \rho(e_0, e_2, e_3, e_1).
\end{multline*}
For these two expressions to be equal we must have:
\begin{equation*}
\begin{split}
& \lambda + \Tilde{\lambda} = 0, \quad \quad \mu + \Tilde{\mu} = 0, \quad \quad \nu + \Tilde{\nu} = 0, \\
& \beta + \Tilde{\beta} = 0 , \quad \quad \gamma + \Tilde{\gamma} = 0, \quad \quad \rho + \Tilde{\rho} + \gamma \Tilde{\nu} + \beta \Tilde{\mu} = 0.
\end{split}
\end{equation*}

To conclude we solve these equations and we obtain the action of $\theta_X$ on $G_3$, which is summarized in the
first row of Table \ref{table:XYZ}.
The actions of $Y$ and $Z$ are obtained in the same way.
Recall that the notation $\gamma\nu + \beta\mu - \rho$ indicates the map $\Tilde{\rho}$ given by
$$
\Tilde{\rho}(v_0,e_2,e_3) = \gamma(\nu(v_0,e_3),e_2) + \beta(\mu(v_0,e_2),e_3) - \rho(v_0,e_2,e_3).
$$

\begin{table}[h]
\begin{center}
\begin{tabular}{|R||R|R|R|R|R|R|Z||R||}
\hline
g & \gamma & \beta &\alpha &\lambda & \mu & \nu &\rho  & 1,2,3\\
\hline
\theta_X(g) & -\mu & -\nu & \alpha & -\lambda & -\gamma & -\beta & \gamma\nu + \beta\mu - \rho & 0,2,3\\
\hline
\theta_Y(g) & -\lambda & \beta & -\nu & -\gamma & -\mu & -\alpha & \alpha\lambda +\gamma\nu - \rho & 1,0,3\\
\hline
\theta_Z(g) & \gamma & -\lambda & -\mu & -\beta & -\alpha & -\nu & \alpha\lambda + \beta\mu - \rho & 1,2,0\\
\hline
\end{tabular}
\end{center}
\caption{The action of $X,Y,Z$ on $G_3$. In the final column, $i,j,k$ indicates that the domain of $\rho$
is $E_i\otimes E_j\otimes E_k$; this is needed in the proof of Proposition \ref{prop:xyxz}. \label{table:XYZ}}
\end{table}

The next two results can now be proved in the same way as in \S\ref{sect:intro}.

\begin{thm} \label{thm:submain3}
Let $W \in \dg_3$ and let $E$ be a triple vector bundle in $\mscr{C}$.
\begin{enumerate}
\item  The bijection $\vartheta_W \colon\thinspace \Dec(E) \to \Dec(E^W)$ defined by
$$
\vartheta_W(P) = (P^W)^{\varepsilon_W},
$$
is a map of torsors and the associated group automorphism $\theta_W \colon\thinspace G_3\to G_3$ does not depend on the
choice of triple vector bundle $E$.

\item  The map $\dg_3 \times G_3\to G_3,\quad (W,g) \mapsto \theta_W(g),$
is a group action.
\end{enumerate}
\end{thm}

\begin{thm} \label{thm:main3}
For $W \in \dg_3$ with $\pi(W) = 1$, the following are equivalent:
\begin{enumerate}
\item $\theta_W$ is the identity on $G_3$.
\item There exists a natural isomorphism between the identity functor and $W$.
\end{enumerate}
\end{thm}

Using Table~\ref{table:XYZ} we can determine which words act as the identity.

\begin{prop}
\label{prop:xyxz}
The words $(XYXZ)^2,\ (YZYX)^2,\ (ZXZY)^2$ have order $2$ in $\dg_3$, and
\begin{equation}
\label{eq:K4}
(YZYX)^2 (XYXZ)^2= (ZXZY)^2.
\end{equation}
\end{prop}

\noindent{\textsc{Proof.}\ }
Consider $(XYXZ)^2$. From Table~\ref{table:XYZ} the first six columns of Table~\ref{table:xyxz2}
are easily established.

\begin{table}[H]
\begin{center}
\begin{tabular}{|R||R|R|R|R|R|R|Z||}
\hline
 & \gamma & \beta &\alpha &\lambda & \mu & \nu &\rho  \\
\hline
X & -\mu & -\nu & \alpha & -\lambda & -\gamma & -\beta & \gamma\nu + \beta\mu - \rho\\
\hline
YX & \mu & \alpha & -\nu & \gamma & \lambda & -\beta & \rho - \beta\mu -\gamma\nu\\
\hline
XYX & -\gamma & \alpha & \beta & -\mu & -\lambda & \nu & -\rho\\
\hline
XYXZ & -\gamma & \mu & \lambda & -\alpha & -\beta & -\nu & \rho - \alpha\lambda - \beta\mu\\
\hline
(XYXZ)^2 & \gamma & -\beta & -\alpha & -\lambda & -\mu & \nu & \rho\\
\hline
\end{tabular}
\end{center}
\caption{Calculation of the action of $(XYXZ)^2$. We omit $\theta$ from the notation.\label{table:xyxz2}}
\end{table}

Now consider the final column. The operation of $X$ sends $\rho$ to $\gamma\nu + \beta\mu - \rho$,
considered as a map $E_0\otimes E_2\otimes E_3\to E_{023}$. Applying $Y$ to this we get
\begin{equation}
\label{eq:rhoXY}
(-\lambda)(-\alpha) + \beta(-\mu) - (\alpha\lambda + \gamma\nu - \rho),
\end{equation}
which should be considered as a map $E_0\otimes E_1\otimes E_3\to E_{013}.$

The term $\lambda\alpha$ in (\ref{eq:rhoXY}), when considered as a map $E_0\otimes E_1\otimes E_3\to E_{013}$,
is
$$
\langle (\lambda\alpha)(e_0,e_1,e_3)\,\vert\,  e_2\rangle = \langle \lambda(e_1,\alpha(e_2,e_3))\,\vert\,  e_0\rangle.
$$
On the other hand, the term $\alpha\lambda$ in (\ref{eq:rhoXY}), when considered as a map $E_0\otimes E_1\otimes E_3\to E_{013}$,
is
$$
\langle (\alpha\lambda)(e_0,e_1,e_3)\,\vert\,  e_2\rangle = \langle \alpha(\lambda(e_0,e_1),e_3))\,\vert\,  e_2\rangle.
$$
These are equal and so we may briefly write $\alpha\lambda = \lambda\alpha$. Thus (\ref{eq:rhoXY})
simplifies to $\rho - \beta\mu -\gamma\nu.$

The remaining three entries in the $\rho$ column are obtained in the same way, and it is now
clear that $(XYXZ)^2$ has order 2.

For reference, we list the actions of all three elements in Table~\ref{table:K4}.

\begin{table}[h]
\begin{center}
\begin{tabular}{|R||R|R|R|R|R|R|Z||}
\hline
         & \gamma & \beta &\alpha &\lambda & \mu & \nu &\rho  \\
\hline
(XYXZ)^2 & \gamma & -\beta & -\alpha & -\lambda & -\mu & \nu & \rho\\
\hline
(YZYX)^2 & -\gamma & -\beta & \alpha & \lambda & -\mu & -\nu & \rho\\
\hline
(ZXZY)^2 & -\gamma & \beta & -\alpha & -\lambda & \mu & -\nu & \rho\\
\hline
\end{tabular}
\end{center}
\caption{Actions on $G_3$ of the nonidentity elements of $K_4$. \label{table:K4}}
\end{table}

We prove (\ref{eq:K4}) by applying $XYX = YXY$ and its conjugates repeatedly to the LHS.
Eventually we arrive at $(YZXZ)^2$. Since this has order 2 and each of $X,Y,Z$ have order 2,
this is equal to $(ZXZY)^2$.
\pfend

We remark that, given the action of $X$ and the action of $Y$ on $G_3$, there are two ways
to understand how the action of the composition $YX$ should be calculated.  They are duals
of each other, and they correspond to thinking in terms of ``frames'' or thinking in
terms of ``coordinates''.  If the reader is obtaining different calculations, this may be
the reason.  They are equivalent, however, and they both should agree on $K_4$.

From $(XYXZ)^2\neq I$ it follows that $XYX$ and $Z$ do not commute. Notice however that
the actions of $XYX$ and $Z$ on the set of building bundles do commute. The action of
$XYX$ on the decomposed triple vector bundle, as distinguished from the action on the statomorphisms,
takes place entirely in the roof double vector bundle (in terms of Figure~\ref{fig:tvb}) and
preserves the floor double vector bundle. The action on $G_3$ however shows that the operations
of $XYX$ and $Z$ are `entangled'.

From \ref{prop:xyxz} it follows that $K_4$ is the Klein 4-group. As an immediate
consequence we have:

\begin{thm}
The order of $\dg_3$ is $96.$
\end{thm}

In \cite{Mackenzie:2005dts} the order of $\dg_3$ (there denoted $\EuScript{VB}_3$) was given
as 72. Further, it was stated that $(XYZ)^4 = I$. Applying Table~\ref{table:XYZ}, the action
of $(XYZ)^4$ on $G_3$ is as shown in Table~\ref{table:xyz4}, so $(XYZ)^4$ has order 2.
More specifically, $(XYZ)^4=(ZXZY)^2$ is a non-trivial element of $K_4$.

\begin{table}[h]
\begin{center}
\begin{tabular}{|R||R|R|R|R|R|R|Z||}
\hline
         & \gamma & \beta &\alpha &\lambda & \mu & \nu &\rho  \\
\hline
(XYZ)^4 & -\gamma & \beta & -\alpha & -\lambda & \mu & -\nu & \rho \\
\hline
\end{tabular}
\end{center}
\caption{Action of $(XYZ)^4$ on $G_3$. \label{table:xyz4}}
\end{table}

\begin{thm} \  \label{thm:descriptionDG3}
\begin{enumerate}
\item The group $\dg_3$ is an
extension of $S_4$ by $K_4$; that is, there is a short exact sequence
$\ses{K_4}{\dg_3}{S_4}$.
\item  As an $S_4$--module, $K_4$ is isomorphic to the normal subgroup of $S_4$
$$\{  1, (12)(30), (23)(10), (13)(20) \}$$ with action by conjugation.
\item  The extension $\ses{K_4}{\dg_3}{S_4}$ is not split.
\end{enumerate}
\end{thm}

\noindent{\textsc{Proof.}\ }
(i) has already been established, and (ii) follows from \eqref{eq:actiononK4}.
We now prove (iii).

Assume the extension is split.  Let $f \colon\thinspace \dg_3 \to K_4 \rtimes S_4$ be an isomorphism,
where $\rtimes$ represents semidirect product.   Let $\sigma=(12)(30)$.
Then $f(XYXZ) = (u, \sigma)$ for some element $u \in K_4$.  Using the semidirect product rule:
$$f((XYXZ)^2)=\left(u,\sigma\right)^2=(u (\sigma \cdot u), \sigma^2) = (u^2, 1) = (1, 1),$$
where we have used the fact that $\sigma$ acts trivially on $K_4$.  We have reached
a contradiction.
\pfend

The three nontrival elements of $K_4$ are in some ways comparable to the dualization of
an ordinary vector bundle $A$: isomorphisms between $A$ and $A^*$ exist but are not natural,
and likewise there are statomorphisms between a triple vector bundle $E$ and $E^W$ where
$W\in K_4,\, W\neq 1,$ but there is no canonical statomorphism. However, whereas ordinary
dualization is contravariant, the elements of $K_4$ are covariant functors.

For arbitrary $n$--fold vector bundles, (i) of Theorem \ref{thm:descriptionDG3} will still
be true \cite{Gracia-SazM:nfold}. Instead of (ii) we will give a combinatorial description
of $K_{n+1}$.  The corresponding extension will be split if and only if $n$ is even.

\section{The group $\dg_3$}
\label{sect:group}

In this section we identify $\dg_3$ as a semi-direct product and give its normal and
conjugacy class structure.

First, consider the words $a=(ZXY)^2$, $b=(XYZ)^2$, and $c=(YZX)^2$.  Let $H$ be the
subgroup of $\dg_3$ generated by $a$, $b$, and $c$.  These three elements have order
4, commute with each other, and satisfy $abc=1$.   Thus $H \cong \mathbb{Z}_4 \times \mathbb{Z}_4$.
A direct calculation shows that $H$ is a normal subgroup of $\dg_3$.   Second,
consider $\dg_2 \cong S_3$ as a subgroup of $\dg_3$ generated by $X$ and $Y$.
Then $S_3$ acts on $H$ by permuting $a$, $b$, and $c$.
In addition, $H \cap \dg_2 = \{ 1 \}$, producing:

\begin{thm}
$\dg_3$ is isomorphic to the semidirect product $(\mathbb{Z}_4 \times \mathbb{Z}_4) \rtimes S_3$,
as described above.
\end{thm}

All normal subgroups of $\dg_3$ can be obtained with help of the homomorphism
$\pi \colon\thinspace \dg_3 \to S_4$.

\begin{thm}
The normal subgroups of $\dg_3$, apart from the trivial ones and $K_4 = \ker \pi$, are
\begin{itemize}
\item $\pi^{-1} (A_4)$, which has index 2.  This consists of the words in $X$, $Y$,
and $Z$ that have an even number of letters.  Alternatively, this consists of the
duality functors which are covariant.
\item The subgroup $H$ described above, which has index 6, and which is also
$\pi^{-1}(V)$, where $V$ is the normal subgroup of $S_4$ of order 4.
\end{itemize}
\end{thm}
\noindent{\textsc{Proof.}\ }
It is clear that these are normal subgroups.  We obtained that there are
no others by use of  GAP \cite{GAP449}.
\pfend

Also from GAP, or equally by hand calculation,
we find that there are nine nontrivial
conjugacy classes. In Table \ref{table:cc} we have listed their size, representative
elements, their orders, and the action of these elements on $G_3$.
We can read all the normal subgroups from this table.  Apart from the identity, $K_4$ consists
of the first conjugacy class, $H$ consists of the first four conjugacy classes,
and $\pi^{-1}(A_4)$ consists of the first five conjugacy classes.

\begin{table}[h]
\begin{center}
\begin{tabular}{||Z|R|R|R|R|R|R|R|R|Z||}
\hline
& \textrm{s}  & \textrm{o} & \alpha & \beta &\gamma &\lambda & \mu & \nu &\rho  \\
\hline\hline
  (XYXZ)^2 & 3 & 2 & -\alpha & -\beta & \gamma & -\lambda & -\mu & \nu & \rho \\
\hline
 (XYZ)^2 & 3 & 4 & -\lambda & \beta & \nu & \alpha & \mu & -\gamma & \rho - \alpha \lambda - \gamma \nu \\
\hline
 (ZYX)^2 & 3 & 4 & \lambda & \beta & -\nu & -\alpha & \mu & \gamma & \rho - \alpha \lambda - \gamma \nu \\
\hline
 XZXY & 6 & 4 &  \lambda & -\mu & \nu & -\alpha & -\beta & -\gamma & \rho - \alpha \lambda - \gamma \nu \\
\hline
 XY & 32 & 3 & \beta & -\nu & \lambda & \mu & \gamma & -\alpha & \rho - \alpha \lambda - \gamma \nu \\
\hline
 Z & 12 & 2 &  -\mu & -\lambda & \gamma & -\beta & -\alpha & -\nu & -\rho + \alpha \lambda + \beta \mu \\
\hline
 XYZYXZY & 12 & 4 & \mu & -\lambda & -\gamma & \beta & -\alpha & \nu & -\rho + \alpha \lambda + \beta \mu \\
\hline
 XYZ & 12 & 8 & -\gamma & -\mu & -\lambda & \nu & -\beta & \alpha & -\rho + \alpha \lambda + \beta \mu \\
\hline
 ZYX & 12 & 8 & \nu & -\mu & -\alpha &  \gamma & -\beta & \lambda & -\rho + \beta \mu + \gamma \nu \\
\hline
\end{tabular}
\end{center}
\caption{Conjugacy class structure of $\dg_3$. The leftmost column gives a representative
element of each class, followed by the size of the class (s), the order of any element of
the class (o), and the action of the representative on $G_3$.\label{table:cc}}
\end{table}

To conclude, there is a faithful linear representation of $\dg_3$ on $\mathbb{R}^6$.
Neglecting the $\rho$ column of Table~\ref{table:XYZ}, the action of $X,Y,Z\in \dg_3$ on $G_3$ defines
a linear action on $\mathbb{R}^6$ by matrices with entries $0, +1$ or $-1$.
From Table~\ref{table:cc}, the only element of $\dg_3$ to fix all of
$\gamma,\dots,\nu$ is the identity.

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  \def$'${$'$}
\providecommand{\bysame}{\leavevmode\hbox to3em{\hrulefill}\thinspace}
\providecommand{\MR}{\relax\ifhmode\unskip\space\fi MR }
\providecommand{\MRhref}[2]{%
  \href{http://www.ams.org/mathscinet-getitem?mr=#1}{#2}
}
\providecommand{\href}[2]{#2}

redpajama_set_name: RedPajamaArXiv


this bunny likes bananas more.
a pretty girl French speaking.
Do you have problems by decorating your Christmas tree? Take a look to those professionals' tips for have the best Christmas tree.
See a summary of the Red Bull competition of break dance development in Tokyo.
Great magic trick performed by Paul Annett., enjoy it.
Really amazing trick, you'll stun your friends with this.
The BEST way to fix every damaged CDs.
Now we are going to show you some infarcting compilation of ghost, poltergeist and diabolic videos, do not show it if you are sensitive.


Feel free to come by for the whole thing or drop in when you have time.
Need a ride to the event or willing to drive others? Please share in the comments!
Please review Girl Develop It's Code of Conduct prior to the event.
Note: This venue is handicap accessible, as well as 100% ADA compliant.


text:
pardés in Hebrew:
orchard, grove, fruit garden
paradesi in Tamil:
Foreigner, stranger … 2. Sojourner, pilgrim, traveller;
What the latter definition leaves out is that paradesi is almost always used as a derogatory word. Are the two words cognates? The Hebrew word's roots are Proto-Iranian as the etymology describes, while the Tamil word is probably derived from Sanskrit. Proto-Iranian and Sanskrit are both theorized to have developed from Proto-Indo-European. They are probably false friends — true cognates with different meanings due to semantic change.
My interest in these two words began when I learned about the Paradesi Jews of India — is it an accident that a word with negative connotations ties back to people who faced persecution and antisemitism? Maybe, but the sad truth today is that few people in Chennai are aware of its Jewish past. History is the story of people - many stories are tragic but deserve to be remembered.
Don't Make Me Talk
John Siracusa in episode 19 of his pre-eminent podcast Hypercritical:
What were the earliest mass-market PC interfaces like?…they were like conversations…you'd tell it what to do, it gives information back to you…That was the basic paradigm until the Macintosh…What it gave you was not a conversation, but a thing…you could poke the thing and see how it reacts…it worked like a physical object…
I really liked the comparison of early command-prompt user interfaces to conversations. It struck me that today's AI assistants (Alexa, Google Assistant, Siri) are all based around having conversations. If these systems ever approached anything close to human intelligence and common-sense, perhaps having a conversation is the best way to interact with AI. But I wonder if there is a better interface to interact with AI? What's the next leap from conversational AI? Perhaps Augmented Reality is the answer — artificial intelligence dispersed in our lived reality, giving us glancable information and the illusion of a physical object we can interact with. I wonder if this is why Apple is so bullish on AR as well.
Or maybe the best way to interact with artificial intelligence is the same way we interact with other people — using conversations.
Stop Using MTurk
Why now? My thought process was simple: My research output thus far has been based on underpaid labour. Do I want that to continue?
For me, the answer is no, hence why I'm going to stop using MTurk. Perhaps there are alternatives that are fair to workers, that lead to good wages and job security. Linguistic data is important, and more so when it comes from a diverse population. But I don't think paying miniscule amounts of money to obtain it is something to be proud of. We can do better.
I've been thinking more about this topic after a research seminar discussion – would ensuring that workers get paid atleast $10 per hour fix the problem? It certainly ensures that they get paid a decent wage by you. A few requesters making the decision to pay more(and advertising their decision) isn't going to make the median wage go up, nor does it do much to fix the numerous other issues with how the platform treats workers. It does help in a small way, and its the least evil thing you can do on the platform.
Ongoing Genocide
Howard Zinn in A People's History of the United States (emphasis mine):
One can lie outright about the past. Or one can omit facts which might lead to unacceptable conclusions. Morison does neither. He refuses to lie about Columbus. He does not omit the story of mass murder; indeed he describes it with the harshest word one can use: genocide.
This passage, and especially the last line, made an unusual impression on me when I first read it, and still does. Perhaps it's the fact that one word needs to do so much work. How can one word possibly convey the magnitude of death, destruction, and suffering? On the other hand, genocide, along with words like racist, do seem to work. Those in power (and the privileged) seem more upset over being labelled racist, or being accused of genocide, than the crimes themselves.
I was reminded of the passage when reading a report describing the forced labour of Uyghurs in China, which used the term — cultural genocide (or ethnocide) to describe the Chinese government's practises. Words matter, and the crimes against Kashmiris and Dalits need to be called for what they are, regularly and loudly.
Can the decades of military occupation of Kashmir, and the torture and killing of Kashmiris be captured by these words? Or the centuries of discrimination and persecution directed at the Dalit community? Calling these crimes for what they truly are — ongoing genocide, may not be adequate to describe the extent of the horrors that the people have experienced, nor the generational trauma, but it is a good start.
older>

redpajama_set_name: RedPajamaCommonCrawl


Soybean Oil, Coconut Oil, Fragrance, Tocopherol, Pentaclethra Macroloba Seed Oil, Sunflower Seed Oil, Rosemary Extract, Coumarin, Citric Acid.
It's really light and gets absorbed quickly.
Soothes my itchy scalp and I love it for this one.
It's easy to wash off.
It's paraben and silicone free.
The previous bottle of mine is almost over, so I can say that it does make my hair soft.
I love leaving the oil overnight. FYI, I always oil (be it any) my hair at night and wash them in the morning.
The smell is a little overwhelming.
It does not reduce hair fall so I think, it does not make my hair strong.
Though the packaging looks quite nice with a flip open cap and all, but the bottle gets all messy and oily after frequent use. Not so travel friendly I guess.
Being a coconut oil, the first ingredient should had been Coconut Oil not Soybean Oil.
Its not pure coconut oil, contains other oils as well, which might be good for your hair though. But incase you were looking for pure coconut oil.
My first bottle of The Body Shop Rainforest Coconut Hair Oil was almost over when I was sent a sample for its review. Now that I have used enough of the product, I can say that the oil won't help you with hair fall (didn't help me), but if the horrible climate is giving you an itchy scalp like me, it's worth trying. Also it does make my hair soft and smooth.
If you looking for pure organic coconut oil without any artificial fragrance, I think you should give this one a skip.
p.s. ~ product sent by pr for consideration. However my views are not affected and are honest as always.


"America the Beautiful" was the theme for the 2018 contest for the 4th Congressional District. Click for information on 4th Congressional District contest information.
A reception was held on May 3, 2018 at BI Design Studios in El Dorado Hills to honor the 2018 winners. Students, parents, friends and classmates were welcome to attend and Congressman McClintock was on hand to congratulate the winners and present Congressional Certificates of Recognition.
First Place Winner: Cassidy Harootunian, "Beauty of the American People"
Second Place: Hannah Lucas, "Coming Together"
Third Place: Mariya Grabovska, "America's Heroes"
Honorable Mention: Sarah Edwards, "Opportunities"
Honorable Mention: Candice Leung, "Paradise"
Honorable Mention: Samantha Khieu, "We are the People"
Honorable Mention: Korryn Taylor, "United"


Learn from international lighting wizard Dave Montizambert, how to create incredible portrait and fashion lighting on models, clients, friends or family, in your living-room using nothing more than an inexpensive work-light and readily available repurposed house-hold items in place of lighting modifiers, stands and rigging. This video's six lighting setups are designed to expose you to the creative possibilities of lighting and to inspire you into painlessly learning the necessary lighting principles that can be applied to all lighting situations. During the six shoots, you will have an over-the-shoulder view, from many angles, of the homemade lighting setups as they unfold – you will see exactly where each piece of rigging is placed and why, plus you will see it all again in detail using Dave's patented intricate lighting diagrams and image capture progressions. All this plus a bonus lesson on main-light placement for best facial lighting.
This fast loading program contains 2.7 hours of live action video.
Mac/Windows (videos in clips folder: for transfer to your iPad/iPhone etc).
An overview of DIY Studio Lighting and some important thoughts on approaching your homemade photography techniques.
Dave demonstrates some fundamental lighting principals on the iconic face of Mozart.
Imagine creating a studio quality portrait complete with backlight, fill light and key light- using only a living room window in the daytime and a few household items.
Learn how to achieve this epic, high contrast top lit shot by utilizing a ladder, a shelf, a broom, some sheets and an inexpensive hardware store work light.
If you have kids, or know anyone who might want high quality photographs of their children without spending thousands of dollars, don't miss this lesson!
An in depth look at creating a studio portrait in your living room using things that exist in every home.
High contrast dramatic photographs are always in high demand. Learn an easy, inexpensive way to do this using only one light.
Taking a high quality self portrait by yourself can be a very difficult thing to achieve. Watch the lighting wizard himself, Dave Montizambert, design a self portrait using a number of techniques and tricks that will de-mystify the process.
Dave shares some final insights and gives resources for further learning.


Helicopter Mosquito XE. Specifications. A photo.
Home » Blog » other » Helicopter Mosquito XE. Specifications. A photo.
Mosquito XE - Canadian single ultra-light helicopter designed in 2004, the company «Mosquito Helicopters».
Draft aircraft Mosquito XE become established in Canadian aviarazrabotchikami 2002 year. As a starting base, aircraft manufacturers of the company «Mosquito Helicopters» used its previous model helicopter - Mosquito AirWhich, despite its limited functionality, it has become very popular both within Canada itself, and beyond.
His first test test aircraft model Mosquito XE held in 2004 year, and in view of the fact that all tests have been completed very successfully, the project was almost immediately sent into production, which continues to be to this day.
Thanks to its compact dimensions, the helicopter Mosquito XE is a fairly high-speed ultra-light aircraft, which has a high maneuverability. Due to the fact that the functionality of the helicopter is very limited, the aircraft is positioned exclusively as a private aircraft, however, in contrast to the model Mosquito Air, the helicopter has become more efficient in terms of its flight performance, and almost completely replaced with the market model Mosquito Air.
Today helicopters Mosquito XE edition is carried out exclusively as sets and kits for self-assembly prospective buyers, however, initially, «Mosquito Helicopters» company has built about 40 aircraft, and only after that the project was transferred to the production as sets.
A distinctive feature of the helicopter Mosquito XE, compared with the model Mosquito Air, it is enclosed cab, which eliminates the discomfort for the pilot, which is especially characteristic in the performance of flight at high speeds. Due to the accessories of the aircraft to the class of single aircraft, the pilot only sleeps on a board, with, transport any cargo in this model is also available.
The ultra-lightweight aircraft model Mosquito XE is equipped with a twin-cylinder piston aircraft engine of the Compact Radial Engines MZ 202 brand, capable of developing thrust in 64 hp, which in turn allows the aircraft to accelerate to maximum flight speed in 130 km / h time as the maximum distance a helicopter can fly over is 145 kilometers.
Power: 64 hp (Depending on version).
Existe a possibilidade do fabricante aceitar sócio investidor?
"Arista" AOP 6 / 9 / 11. A photo. Characteristics.


text:
Q: Having trouble using tkinter in python to create a simple gui for my simple chatbot So... i am wanting to turn my console chatbot into a GUI chatbot and whenever i do it seems to never be working... ill paste an original code(which runs in console) then i will paste the same code but added some tkinter codes to turn it into gui instead of console app(just clarifying again, my gui code really doesnt work). what was my mistake in the gui code and how to make it work.
import random
#THIS IS THE CONSOLE APP CODE
greetings = ['hola', 'hello', 'hi', 'Hi', 'hey!','hey']
random_greeting = random.choice(greetings)

question = ['How are you?','How are you doing?']
responses = ['Okay',"I'm fine"]
random_response = random.choice(responses)

while True:
    userInput = input(">>> ")
    if userInput in greetings:
        print(random_greeting)
    elif userInput in question:
        print(random_response)
    else:
        print("I did not understand what you said")

alright here is the potential gui code that i tried
import random
import tkinter
#GUI CODE
t = tkinter.Tk()
userInput = tkinter.Entry(t)
userInput.pack()
def cb():
    greetings = ['hola', 'hello', 'hi', 'Hi', 'hey!','hey']
    random_greeting = random.choice(greetings)

    question = ['How are you?','How are you doing?']
    responses = ['Okay',"I'm fine"]
    random_response = random.choice(responses)

    while True:
        if 'hello' in userInput.get():
            lab = tkinter.Label(t,text =random_greeting)
            lab.pack()
        elif 'how are you' in userInput.get():
            labb = tkinter.Label(t,text =random_response)
            labb.pack()
        else:
            labc = tkinter.Label(t,text ="I did not understand what you said")
            labc.pack()
button = tkinter.Button(t, text="Enter",command=  cb)
button.pack()

tkinter.mainloop()

( the purpose of this chatbot code is really so i could understand GUI more)
Thanks!

A: There's a lot of room for improvement in this code, but this should get you heading in the right direction.
It's a common convention in well-written Tkinter code to do import tkinter as tk. That lets us write tk instead of tkinter, which saves typing, and makes it easy to change the code to run on Python 2 if we need to do that: just change the import statement to import Tkinter as tk. And it's much better than doing from tkinter import *, because it lets us see what names we've imported from Tkinter, and it doesn't dump over a hundred names into our namespace.
As PRMoureu mentioned in the comments, we don't want a while loop in the callback function: we just need to respond to the string that's in the Entry widget when the user presses the button.
We create a single Label widget to hold the bot's response text, and we can update it with the Labels.config` method.
import random
import tkinter as tk

root = tk.Tk()
user_input = tk.Entry(root)
user_input.pack()

greetings = ['hola', 'hello', 'hi', 'Hi', 'hey!', 'hey']
question = ['How are you?', 'How are you doing?']
responses = ['Okay', "I'm fine"]
huh = "I did not understand what you said"

def cb():
    user_text = user_input.get()
    if user_text in greetings:
        bot_text = random.choice(greetings)
    elif user_text in question:
        bot_text = random.choice(responses)
    else:
        bot_text = huh
    output.config(text=bot_text)

button = tk.Button(root, text="Enter", command=cb)
button.pack()

output = tk.Label(root, text='')
output.pack()

tk.mainloop()

It takes a little while to get used to GUI programming, because the control flow is not what you're used to from console programming. With code that runs in the console, the code does what it wants to do, when it wants to do it, and the user responds. In GUI code, we set everything up and then wait for events generated by the user's actions, and then our code responds to those actions. This is called event-driven programming. Although it can be a bit disorienting at first, with practice you'll soon get the hang of it. ;)

We don't really need that button. Instead, we can bind our callback to the Entry widget so that it's called when the Enter key is pressed inside the Entry widget. We have to modify the signature of the callback because it will now receive an Event object when it's called. But we don't need to change anything inside the callback function body because we aren't actually using the data in the Event object.
Here's the new version:
import random
import tkinter as tk

root = tk.Tk()
user_input = tk.Entry(root)
user_input.pack()

greetings = ['hola', 'hello', 'hi', 'Hi', 'hey!', 'hey']
question = ['How are you?', 'How are you doing?']
responses = ['Okay', "I'm fine"]
huh = "I did not understand what you said"

def cb(event):
    user_text = user_input.get()
    if user_text in greetings:
        bot_text = random.choice(greetings)
    elif user_text in question:
        bot_text = random.choice(responses)
    else:
        bot_text = huh
    output.config(text=bot_text)

user_input.bind("<Return>", cb)
output = tk.Label(root, text='')
output.pack()

tk.mainloop()

redpajama_set_name: RedPajamaStackExchange


text:
Coachbuilt Cars
Exceptional Cars
Great Cars
Porter Profiles
Martin Brundle
Murray Walker
Philip Porter
Stephen Archer
Cristián Bertschi
Phillip Bingham
Keith Bluemel
Gordon Bruce
Mark Cole
Sir John Egan
John Elmgreen
Bob Exelby
Alan Fearnley
Mike Gould
Simon Ham
Richard Heseltine
J. Mac Hulbert
Nick Hull
Ray Hutton
Michael Kliebenstein
Julius Kruta
Rinsey Mills
Clive Neville
Doug Nye
Chas Parker
Martin Port
Chris Rooke
Steve Saxty
Serge Vanbockryck
Ian Wagstaff
Mick Walsh
April 17, 2020 Porter Press
Stirling Moss - Some Personal Memories
By Mark Cole - Motorsport Commentator and Author
I know that you will be as sad as I in losing our hero. I was fortunate to grow up in Tring, Hertfordshire, where Stirling lived with his family at White Cloud Farm, and on Monday mornings during school holidays, my 10-year old pals and I would hang on the gate waiting for him to return from his latest race; he would always stop and talk to us.
My passion for racing grew from that, and Stirling Moss's Book of Motorsport Sport (1955) was an early Christmas present. I first met him through our neighbour Norman Ward, who raced Ford Zodiac saloon cars, and it was he, Stirling and Mike Keele (who went on to build Keele Karts) who started the karting at Long Marston airfield. The USAAF was still in residence there in the 1950s, and American airmen brought over these new-fangled go-karts from California, and it became an international craze. The circuit had two corners I remember well, Wardy's Wiggle and Stirling's Bump.
Later Norman took me to Silverstone regularly, where we would meet Stirling (who always had time to introduce me, still in short trousers, to his friends including Mike Hawthorn). Stirling was always very amenable, and in the early 1980s I ghosted his motoring column in Harpers for a few months, visiting him at his hi-tech Shepherd's Market house.
I loved your Stirling Moss scrapbooks and those, along with many other Moss books, autographs and models, I will treasure for the rest of my life.
Michael Kliebenstein - Author and Automobilist
When I asked him in which car he would like to be driven at London Classic Car Show in London, he immediately pointed at my 1922 Rolls-Royce Silver Ghost.
He said that he had lots of respect for the immense quality of the early Silver Ghosts and that they played numerous and diverse roles in the history of Great Britain. But above all that they were masterpieces of automotive design. And he added the old Rolls-Royce mantra that "... obviously the quality remains long after the price has been forgotten"!
He enjoyed the drive and I was very taken by the fact that he really appreciated the old Ghost. Other cars we gave him to choose from were a Blower Bentley, a Ferrari 275 GTS and a Maserati Tipo 26 B. A truly great man who appreciated history and quality so much.
Philip Porter - Author, Publisher and Close Friend
For me, it all starts in 1956. I was just five and my parents took me to Silverstone for the International Trophy Meeting. Stirling - my hero was driving a DB3S and for Vanwall. I was no doubt a real pain because I excitedly pointed him out to my parents on every lap he passed our grandstand.
To every schoolboy in the '50s and '60s, he was the perfect role model: daring, courageous, patriotic, successful, dated beautiful women and often won against the odds.
Fast-forwarding to the late '90s, he accepted my invitation to become Patron of our Jaguar XK Club - 'On a strictly non-participatory basis, Old Boy'. He then came to every one of our Annual Dinners, at the House of Commons, for at least the next 10 years. I would interview him after dinner and the capacity audience of 170 would love it. We had so much fun and our friendship rapidly grew.
I have so many memories of our time together, sometimes just us, sometimes with our wives. We had so much fun and laughed so much. As I got to know him better and better, our friendship grew to the stage where I could tease him. Of course, he would give as good as he got.
One of the things I treasured most was the absolute trust. We did four books together - the four Scrapbooks we published. We never even had a contract.
As to money, Stirling could be very careful but also very generous. On one occasion about 14 years ago I, together with colleagues Abi (who designs these Newsletters) and Clare, were at the house for two days - I was marking in Stirling's scrapbooks what was to be copied and the girls were scanning like mad. At one point Clare was making tea in the kitchen and Stirling 'told her off' for not using a tea bag more than once!
However, on a number of occasions we went out for dinner, sometimes the three of us, or four of us, or in a larger party. Almost every time, Stirling would insist on paying for everyone. In fact, he had usually had a discreet word with the Maitre d' on the way in and made his wishes clear. Very generous and you couldn't argue because it was already sorted.
I used to take a pile of books to Stirling and Susie's house in Mayfair for Stirling to sign. They were in slipcases in individual book boxes in cartons of six. I would park in the garage, carry the boxes through Stirling's office, take them up in the infamous lift to the first floor and into the main sitting room. There I would take the individual book boxes out of the cartons, the book and slipcase out of their boxes and the books out of the slipcases. It was chaos with piles of cartons, book boxes, slipcases and books everywhere. I had wrecked their tidy room.
Stirling would then come and sign them in batches and I would repeat the process in reverse: books into slipcases, into boxes, into cartons, into lift, into car. On occasions the lift, which would become famous for all the wrong reasons, would refuse to work for me. Susie would come and shout at it, kick it and thump it. The lift then obeyed instructions but it was a cussed creature.
After Stirling famously fell down the lift shaft and, remarkably, survived and made a pretty good recovery, I suggested he put a trampoline at the bottom in case it happened again. 'Can't do that, Old Boy - I'd hit my head on the roof!'
All the publicity at that time helped our book sales. I said to him, 'When sales tail off, I'm going to push you down again!'
At some stage, we had changed our Range Rover for a Discovery - the original one with a raised roof at the back. We carefully checked the height of the garage and decided it would just fit. I carefully reversed in and began the process described in tedious detail above. Not being very bright, I had not thought what might happen to said Discovery when it became rather lighter. It wedged itself against the ceiling of the garage.
Rather sheepishly, I told Stirling, who was in the room directly above, what the current situation was. His splendid response was: 'I thought I felt the floor rising!'
For at least 10 years, in spite of his packed schedule, Stirling and Susie would visit our stand at the Goodwood events, at both Festival and Revival, twice a day for 20 minutes or often more on the Friday, Saturday and Sunday to sign our Stirling Moss Scrapbooks. For at least six years, Murray Walker would do the same, signing the Scrapbook we did together.
For obvious reasons, they would be scheduled to avoid clashing. On occasions, though, the track timings, which impacted on Stirling's schedule, would have been affected by accidents or whatever. Thus several times they would be on the stand together, which was great fun and lovely for them as they were great friends.
While we had two queues and were trying to cope with all the sales, our wonderful Rob Schulp would be outfront, drumming up sales. 'Come and meet the legend that is Stirling Moss and get your book signed. Come and meet the man with the most famous voice in motor racing...' When Murray first saw Rob in action, he said to me afterwards, 'That chap is worth his weight in gold.' True.
Rob had some great lines. Memorably, when they were both on the stand, Rob referred to them to his audience, as 'The Ant and Dec of motorsport!'
One of Susie Moss's favourite stories she told me involved both Stirling and Murray. One evening, they gave a talk together on a cruise ship. Next day, a lady went up to Murray and said, 'You were awfully good last night. And that commentator chappie wasn't bad either'!
I recall being at Goodwood Revival in, I guess, 1999. Stirling's 70th birthday was being celebrated and he was driven round the track by Damon Hill dressed as a chauffeur. The car was an enormous Cadillac and Stirling was in the back with a Marilyn Monroe lookalike. It had been raining earlier but luckily had now stopped.
After completing their lap, Damon drove into the paddock and happened to stop right by where I was standing. The rear door was opened and, as there was a puddle adjacent, Stirling gallantly lifted the young lady and placed her on the ground beyond the puddle.
I heard him mutter, very quietly so no-one else would hear, 'Just like the old days: pick 'em up, drop 'em immediately'!
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items are available in all shapes sizes and uses, all are highly collectable around the World.
pottery was chiefly known for producing high-quality porcelain figurines and have become highly sought after in the collectables market.
More will be added as they come into my possession.