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\begin{document}
\title{Dorfman connections and Courant algebroids}

\author{M. Jotz}
\address{Georg-August-Universit\"at G\"ottingen\\
Mathematisches Institut\\
Bunsenstr. 3-5\\
37073 G\"ottingen\\
Germany}
\thanks{Supported  by the
\emph{Dorothea Schl\"ozer Programme} of the  University of G\"ottingen.}
\email{mjotz@uni-math.gwdg.de}

\subjclass[2010]{Primary ; Secondary }

\begin{abstract}
A linear connection $\nabla:\mathfrak{X}} \newcommand{\dr}{\mathbf{d}(M)\times\Gamma(E)\to\Gamma(E)$
on a vector bundle $E$ over a smooth manifold $M$
is tantamount to a splitting
$TE\xrightarrow{\sim}{} V\oplus H_\nabla$, where $V$ is the set of vectors tangent to the fibres of $E$.
Furthermore, the curvature of the connection measures the failure
of the horizontal space $H_\nabla$ to be involutive.
In this paper, we show that splittings $TE\oplus T^*E\xrightarrow{\sim}{}  (V\oplus V^\circ)\oplus L$
of the Pontryagin bundle over the vector bundle $E$
can be described in the same manner
via a certain class of maps
$\Delta:\Gamma(TM\oplus E^*)\times\Gamma(E\oplus T^*M)\to\Gamma(E\oplus T^*M)$.
Similarly to the tangent case, we find that, after the choice of a splitting,
 the Courant algebroid structure of $TE\oplus T^*E\to E$
can  be completely described by properties of the map $\Delta$.

The maps $\Delta$ in this correspondence theorem are particular examples of
connection-like maps that we define in this paper and name Dorfman connections.
Roughly said, these objects are to Courant algebroids what connections are to
Lie algebroids.

In a second part of this paper, we study splittings
of $TA\oplus T^*A$ over a Lie algebroid $A$, and we show
how $\mathcal LA$-Dirac structures on $A$ are in  bijection with a
 class of Manin triples over the base manifold $M$.
This has as special
cases the correspondences between Lie bialgebroids and
Poisson algebroids, and between IM-2-forms and presymplectic algebroids.
\end{abstract}
\maketitle

\tableofcontents

\section{Introduction}

Let us  start with a simple observation.
Take a subbundle $F\subseteq TM$ of the tangent space of a smooth
manifold $M$. Then the $\mathbb{R}} \newcommand{\bxi}{\boldsymbol{\xi}$-bilinear map
\[\tilde \nabla: \Gamma(F)\times \mathfrak{X}} \newcommand{\dr}{\mathbf{d}(M)\to \Gamma(TM/F),\]
\[\tilde \nabla_fX=\overline{[f, X}]\]
measures the failure of vector fields on $M$ to preserve $F$.
The subbundle $F$ is involutive if and only if
$\tilde\nabla_ff'=0$ for all $f,f'\in\Gamma(F)$,
and in this case, $\tilde \nabla$ induces
a flat connection
\[\nabla:\Gamma(F)\times\Gamma(TM/F)\to\Gamma(TM/F),\]
\[\tilde \nabla_f\bar X=\overline{[f, X}],\]
the \textbf{Bott connection} associated to $F$ \cite{Bott72}.

In the same manner, given a Courant algebroid $\mathsf E\to M$ with
bracket $\llbracket} \newcommand{\rb}{\rrbracket\cdot\,,\cdot\rb$, anchor $\rho$ and pairing
$\langle\cdot\,,\cdot\rangle$, and a subbundle $K\subseteq \mathsf E$,
one can define an $\mathbb{R}} \newcommand{\bxi}{\boldsymbol{\xi}$-bilinear map
\[\tilde\Delta:\Gamma(K)\times\Gamma(\mathsf E)\to \Gamma(\mathsf E/K),\]
\[\tilde\Delta_ke=\overline{\llbracket} \newcommand{\rb}{\rrbracket k,e\rb}.\] Again, we have $\tilde\Delta_kk'=0$ for all
$k,k'\in\Gamma(K)$ if and only if $K$ is a subalgebroid of $\mathsf
E$.  If $K$ is in addition isotropic, it is a Lie algebroid over $M$
and the pairing on $\mathsf E$ induces a pairing $K\times_M(\mathsf
E/K)\to \mathbb{R}} \newcommand{\bxi}{\boldsymbol{\xi}$.  The $\mathbb{R}} \newcommand{\bxi}{\boldsymbol{\xi}$-bilinear map
\[\Delta:\Gamma(K)\times\Gamma(\mathsf E/K)\to\Gamma(\mathsf
E/K),\]
\[\tilde\Delta_k\bar e=\overline{\llbracket} \newcommand{\rb}{\rrbracket k,e\rb}\]
that is induced by $\tilde \Delta$
is not a connection because
it is not $C^\infty(M)$-homogeneous in the first argument, but the obstruction to this
is, as we will see, measured by the pairing, the anchor of the Courant
algebroid and the de Rham derivative on $C^\infty(M)$.
This map  is an example of what is called in this paper a Dorfman
connection, namely the \textbf{Bott--Dorfman connection} associated to $K$.
In this paper,  Dorfman connections are defined, and some of their
properties and applications are studied.
Dorfman connections appear naturally in several situations related to Courant
algebroids  and
 play   a  role  similar to the one
that connections play for tangent bundles and Lie algebroids.

\medskip

It is for instance well-known that  a linear $TM$-connection $\nabla$
on a vector bundle $q_E\colon E\to M$
corresponds to

 a splitting $TE\xrightarrow{\sim}{} T^{q_E}E\oplus H_\nabla$. The failure of the horizontal space $H_\nabla$ to be involutive
is measured by the curvature of the connection, and the connection itself is, roughly said, nothing else than the Lie bracket
of horizontal and vertical vector fields, since it can be seen
as a projection
of the Bott connection
\[\nabla^{H_\nabla}: \Gamma(H_\nabla)\times\Gamma(TE/H_\nabla)\to \Gamma(TE/H_\nabla).
\]
in a sense that we will explain.

The first main result of this paper is the one-to-one correspondence in the same spirit
between splittings of the standard Courant algebroid over $E$ as
\[TE\oplus T^*E\xrightarrow{\sim}{} (T^{q_E}E\oplus(T^{q_E}E)^\circ)\oplus L_\Delta\]
and $TM\oplus E^*$-Dorfman connections $\Delta$ on $E\oplus T^*M$. The failure of $L$ to be isotropic (and so Lagrangian) relative
to the canonical pairing on $TE\oplus T^*E$ is equivalent
to the failure of the  dual  of the Dorfman connection (in the sense of connections) to be antisymmetric,
and the failure of $L$ to be closed under the Courant algebroid bracket is measured by the curvature of the Dorfman connection.
The Dorfman connection is the same as the Courant-Dorfman bracket on
horizontal and vertical sections.

We caracterize then double vector subbundles of $TE\oplus T^*E$ over
$E$ and a subbundle $U\subseteq TM\oplus E^*$.
Such double vector bundles can be described by triples
$U, K$and $\Delta$, where $\Delta$ is a Dorfman
connection and $K$ a subbundle of $E\oplus T^*M$.
For instance, we study Dirac structures on $E$ that define double
vector subbundles of the Pontryagin bundle.
We prove that maximal isotropy and integrability of the Dirac
structure depend only on (simple) properties
of the corresponding triple.

\bigskip

If the vector bundle $E=:A$ has a Lie algebroid structure $(q_A:A\to
M, \rho, [\cdot\,,\cdot])$,
then the Pontryagin bundle $TA\oplus T^*A$ has a naturally induced Lie
algebroid structure over $TM\oplus A^*$.
Given a $TM\oplus A^*$-Dorfman
connection $\Delta$ on $A\oplus T^*M$, we compute the representation up to homotopy
that corresponds to the splitting $TA\oplus T^*A\xrightarrow{\sim}{} (T^{q_A}A\oplus
(T^{q_A}A)^*)\oplus L_\Delta$
and describes the VB-algebroid $TA\oplus T^*A\to TM\oplus A^*$
\cite{GrMe10a}. This representation up to homotopy is in general
not the product of the two representations up to homotopy
describing $TA\to TM$ and $T^*A\to A^*$.

Knowing this, one can ask when a Dirac structure on
$A$, that is a double sub-vector bundle of $TA\oplus T^*A$,
is at the same time a  Lie subalgebroid of $TA\oplus T^*A\to TM\oplus A^*$
over its base $U\subseteq TM\oplus A^*$. In that case, the
Dirac structure has the induced structure of a double Lie algebroid,
and is called an \textbf{$\mathcal{LA}$-Dirac structure} on $A$.
A known example of an $\mathcal{LA}$-Dirac structure on a Lie algebroid $A$
is the graph of $\pi^\sharp:T^*A\to TA$, where
$\pi_A$ is the linear Poisson bivector field defined on $A$
by a Lie algebroid structure on $A^*$ such that $(A,A^*)$ is a Lie
bialgebroid \cite{MaXu00}. In that case, we know that the
$\mathcal{LA}$-Dirac
structure is completely encoded in the Lie bialgebroid, which
is itself equivalent to a  Courant algebroid with two transverse Dirac
structures \cite{LiWeXu97}.

The second standard example  is that of the graph of a linear
presymplectic form $\sigma^*\omega_{\rm can}\in \Omega^2(A)$, where
$\sigma:A\to T^*M$ is an IM-$2$-form \cite{BuCrWeZh04, BuCaOr09}.
Here also,  the $\mathcal{LA}$-Dirac structure is equivalent to the IM-$2$-form, and any
$\mathcal{LA}$-Dirac structure that is the graph of a presymplectic
form on $A$ arises in this manner.

The final, most easy, example is $F_A\oplus F_A^\circ$,
where $F_A\to A$ is an involutive subbundle that has at the same time
a Lie algebroid structure over some subbundle $F_M\subseteq TM$.
Here we know that the $\mathcal{LA}$-Dirac structure
corresponds to an infinitesimal ideal system, i.e.
a triple $(C,F_M,\nabla)$, where $C\subseteq A$ is a subalgebroid,
$F_M\subseteq TM$ and involutive subbundle and
$\nabla:\Gamma(F_M)\times\Gamma(A/C)\to\Gamma(A/C)$
a flat connection satisfying some properties that allow ones
to make sense of a quotient Lie algebroid $(A/C)/\nabla\to M/F_M$ \cite{JoOr12}.

\medskip
The second main result of this paper is a classification of
$\mathcal{LA}$-Dirac structures on Lie algebroids
via a certain class of  Manin pairs \cite{BuIgSe09} on the base space
of the Lie algebroid.
This result unifies the three correspondences summarized above.

\subsubsection*{Outline of the paper}
Some background on Courant algebroids and Dirac structures,
connections,
and double vector bundles is collected in the first section.
In the second section, Dorfman sections and dull algebroids are
defined,
and examples are listed.
In the third section, splittings of the standard Courant algebroid
$TE\oplus T^*E$ over a vector bundle $E$ are shown to be equivalent to
a certain class of $TM\oplus E^*$-Dorfman connections on $E\oplus
T^*M$.
Linear Dirac structures on the vector bundle $E\to M$ are studied via
Dorfman connections. In the fourth section, the geometric structures
on the two sides of the
standard $\mathcal{LA}$-Courant algebroid $TA\oplus T^*A$ over a Lie
algebroid
$A\to M$ are expressed via splittings of $TA\oplus T^*A$, and
$\mathcal{LA}$-Dirac structures on $A$ are classified via
Dorfman connections and some adequate vector bundles over the units
$M$.
Finally, it is shown that this data is equivalent to a Manin pair over
$M$, that is in a sense compatible with the Lie algebroid $A$.

\subsubsection*{Notation and conventions}
Let $M$ be a smooth manifold. We will denote by $\mathfrak{X}} \newcommand{\dr}{\mathbf{d}(M)$ and $\Omega^1(M)$ the
spaces of (local) smooth sections of the tangent and the cotangent bundle,
respectively. For an arbitrary vector bundle $E\to M$, the space of
(local) sections of $E$ will be written $\Gamma(E)$.
We write $p_M:TM\to M$, $c_M:T^*M \to M$ and $\pi_M:TM\oplus T^*M\to M$
for the canonical projections, and $q_E:E\to M$ for vector bundle maps.

The flow of a vector field $X\in\mathfrak{X}} \newcommand{\dr}{\mathbf{d}(M)$ will be written $\phi^X_\cdot$, unless specified otherwise.
Let $f:M\to N$ be a smooth map between two smooth manifolds $M$ and $N$.
Then two vector fields $X\in\mathfrak{X}} \newcommand{\dr}{\mathbf{d}(M)$ and $Y\in\mathfrak{X}} \newcommand{\dr}{\mathbf{d}(N)$ are said to be \textbf{$f$-related}
if $Tf\circ X=Y\circ f$ on $\dom(X)\cap f\inv(\dom(Y))$.
We write then $X\sim_f Y$.

Given a section $\xi$ of $E^*$, we will always write
$\ell_\xi:E\to \mathbb{R}} \newcommand{\bxi}{\boldsymbol{\xi}$ for the linear function associated to it,
i.e. the function defined by $e_m\mapsto \langle \xi(m), e_m\rangle$
for all $e_m\in E$. Given a section $e\in\Gamma(E)$,
we write $e^\uparrow\in\mathfrak{X}} \newcommand{\dr}{\mathbf{d}(E)$ for the derivation along $e$,
i.e. the vector field with complete flow $\mathbb{R}} \newcommand{\bxi}{\boldsymbol{\xi}\times E\to E$,
$(t,e_m')\mapsto e_m'+te(m)$.
Note that the vector fields $e^\uparrow$, for all $e\in\Gamma(E)$,
span the subbundle $V:=T^{q_E}E\subseteq TE$ and
$e^\uparrow$ is completely determined by $e^\uparrow(\ell_\xi)=q_E^*\langle \xi, e\rangle$
and $e^\uparrow(q_E^*\varphi)=0$ for all $\varphi\in C^\infty(M)$ and
$\xi\in\Gamma(E^*)$.
Note furthermore that $e^\uparrow(f_m)$ depends only on $e(m)$ and
$f_m$, so the expression $(e(m))^\uparrow(f_m)$ makes sense.
In the same manner, given a vector bundle morphism $\phi\in\Gamma(\Hom(E;E))$,
we can define $\phi^\uparrow\in\mathfrak{X}} \newcommand{\dr}{\mathbf{d}(E)$,
$\phi^\uparrow(e_m)=(\phi(e_m))^\uparrow(e_m)$ for all $e_m\in E$.

\subsubsection*{Acknowledgement}
The author wishes to thank Thiago Drummond and Cristian Ortiz for many
discussions that considerably influenced some parts of this paper,
which grew out of a project with them.  Many thanks also go out to
David Li-Bland, Rajan
Mehta and Marco Zambon for interesting discussions and remarks, and
especially to Kirill Mackenzie for interesting discussions, advice and
encouragements.

\section{Background definitions}
We recall first some necessary background notions on Courant algebroids,
of the double vector bundle structures on the tangent and cotangent spaces
 $TE$ and $T^*E$ of a vector bundle $E$, and on connections.

\subsection{Courant algebroids and Dirac structures}
A
Courant algebroid \cite{LiWeXu97,Roytenberg99} over a manifold $M$ is a vector bundle $\mathsf E\to
M$ equipped with a fibrewise nondegenerate symmetric bilinear form
$\langle\cdot\,,\cdot\rangle$, a bilinear bracket $[\cdot\,,\cdot]$ on
the smooth sections $\Gamma(\mathsf E)$, and a vector bundle map $\rho:
\mathsf E\to TM$ over the identity called the anchor, which satisfy the following
conditions
\begin{enumerate}
\item $[e_1, [e_2, e_3]] = [[e_1, e_2], e_3] + [e_2, [e_1, e_3]]$,
\item    $\rho(e_1 )\langle e_2, e_3\rangle = \langle[e_1, e_2], e_3\rangle
+ \langle e_2, [e_1 , e_3]\rangle$,
\item $[e_1, e_2] +[e_2, e_1] =\mathcal D\langle e_1 , e_2\rangle$
\end{enumerate}
for all $e_1, e_2, e_3\in\Gamma(\mathsf E)$ and $\varphi\in
C^\infty(M)$.
Here, we  use the notation $\mathcal D := \rho^*\circ\dr :
C^\infty(M)\to\Gamma(\mathsf E)$, using $\langle\cdot\,,\cdot\rangle$
to identify $\mathsf E$ with $\mathsf E^*$:
\[\langle \mathcal D\varphi, e\rangle=\rho(e)(\varphi)
\]
for all $\varphi\in
C^\infty(M)$ and $e\in\Gamma(E)$.
The following conditions
\begin{enumerate}
\setcounter{enumi}{3}
\item   $\rho([e_1, e_2]) = [\rho(e_1), \rho(e_2)]$,
\item    $[e_1, \varphi e_2] = \varphi [e_1 , e_2] + (\rho(e_1 )\varphi )e_2$
\end{enumerate}
are then also satisfied. They are often part of the definition in the
literature, but it was already observed in \cite{Uchino02}
that they follow from $(1)-(3)$.\footnote{Actually, they both follow
  immediately from $(2)$. To get $(4)$ replace $e_2$ by $\varphi e_2$
  in $(2)$, and to get $(5)$, replace $e_1$ by $[e_1,e_1']$:
an easy  computation yields then that $\rho[e_1,e_1']\langle e_2,
e_3\rangle=[\rho(e_1), \rho(e_1')]\langle e_2,
e_3\rangle$ for all $e_2,e_3\in\Gamma(E)$.}

\begin{example}\label{ex_pontryagin}
The direct sum $TM\oplus T^*M$
endowed with the projection on $TM$ as anchor map, $\rho=\pr_{TM}$,
the symmetric bracket
$\langle\cdot\,,\cdot\rangle$
given by
\begin{equation}
\langle(v_m,\alpha_m), (w_m,\beta_m)\rangle=\alpha_m(w_m)+\beta_m(v_m)
\label{sym_bracket}
\end{equation}
for all $m\in M$, $v_m,w_m\in T_mM$ and $\alpha_m,\beta_m\in T_m^*M$
and the \textbf{Courant-Dorfman bracket}
given by
\begin{align}
[(X,\alpha), (Y,\beta)]&=\left([X,Y], \ldr{X}\beta-\ip{Y}\dr\alpha\right)\label{wrong_bracket}\\
\end{align}
for all $(X,\alpha), (Y, \beta)\in\Gamma(TM\oplus T^*M)$,
yield the standard example of  a Courant algebroid (often called
the \emph{standard Courant algebroid over $M$}).
The map $\mathcal D: C^\infty(M)\to \Gamma(TM\oplus T^*M)$
is given by $\mathcal D f=(0, \dr f)$.
\end{example}

A \textbf{Dirac structure} $\mathsf D\subseteq \mathsf E$ is a subbundle
satisfying
\begin{enumerate}
\item $\mathsf D^\perp=\mathsf D$ relative to the pairing on $\mathsf E$,
\item $[\Gamma(\mathsf D), \Gamma(\mathsf D)]\subseteq \Gamma(\mathsf D)$.
\end{enumerate}
The rank of the Dirac bundle $\mathsf D$ is then  half the rank of $\mathsf E$,
and the triple \linebreak $(\mathsf D\to M, \rho\an{\mathsf D}, [\cdot\,,\cdot]\an{\Gamma(\mathsf D)\times\Gamma(\mathsf D)})$
is a Lie algebroid on $M$.
Dirac structures appear naturally in several contexts in geometry and geometric mechanics
(see for instance \cite{Bursztyn11} for an introduction to the geometry and applications of Dirac structures).

\subsection{The double vector bundles $TE$ and $T^*E$}
Consider a vector bundle $q_E:E\to M$. Then the tangent space $TE$ of $E$ has two vector bundle
structures. First, the usual vector bundle structure of the tangent space, $p_E:TE\to E$
and second the vector bundle structure $Tq_E:TE\to TM$,
with the addition defined as follows. If $x_{e_m}$ and $x_{e_m'}$
are such that $Tq_E(x_{e_m})=Tq_E(x_{e_m'})=:x_m\in TM$,
then there exist curves $c,c':(-\varepsilon, \varepsilon)\to E$
such that $\dot c(0)=x_{e_m}$, $\dot c'(0)=x_{e_m'}$ and $q_E\circ c=q_E\circ c'$.
The sum $x_{e_m}+_{Tq_E}x_{e_m'}$ is then defined as
\[x_{e_m}+_{Tq_E}x_{e_m'}=\left.\frac{d}{dt}\right\an{t=0}c(t)+_{q_E}c'(t)\in T_{e_m+e_m'}E.
\]
We get a double vector bundle
\begin{align*}
\begin{xy}
\xymatrix{
TE\ar[r]^{p_E}\ar[d]_{Tq_E} &E\ar[d]^{q_E}\\
TM\ar[r]_{p_M}&M
}\end{xy},
\end{align*}
that is, the structure maps of each vector bundle structure are vector bundle morphisms
relative to the other structure \cite{Mackenzie05}.

\bigskip

Dualizing $TE$ over $E$, we get
the double vector bundle
\begin{align*}
\begin{xy}
\xymatrix{
T^*E\ar[r]^{c_E}\ar[d]_{r_E} &E\ar[d]^{q_E}\\
E^*\ar[r]_{q_{E^*}}&M
}\end{xy}.
\end{align*}
The map $r_E$ is given as follows. For $\alpha_{e_m}$,
$r_E(\alpha_{e_m})\in E^*_m$,
\[\langle r_E(\alpha_{e_m}), f_m\rangle=\langle \alpha_{e_m}, \left.\frac{d}{dt}\right\an{t=0}e_m+tf_m\rangle
\]
for all $f_m\in E_m$.
The addition in $T^*E\to E^*$ is defined
as follows. If $\alpha_{e_m}$ and $\beta_{e_m'}$ are such that
$r_E(\alpha_{e_m})=r_E(\beta_{e_m'})=\xi_m\in E^*_m$, then the sum
$\alpha_{e_m}+_{r_E}\beta_{e_m'}\in T_{e_m+e_m'}^*E$
is given by
\[\langle  \alpha_{e_m}+_{r_E}\beta_{e_m'}, x_{e_m}+_{Tq_E}x_{e_m'}\rangle
=\langle  \alpha_{e_m}, x_{e_m}\rangle+\langle\beta_{e_m'}, x_{e_m'}\rangle.
\]
Note that $T^*_{e_m}E$ is generated by $\dr_{e_m}\ell_\xi$ and
$\dr_{e_m}(q_E^*\varphi)$ for $\xi\in\Gamma(E^*)$ and
$\varphi\in C^\infty(M)$.
We have $r_E(\dr_{e_m}\ell_\xi)=\xi(m)$ and
$r_E(\dr_{e_m}(q_E^*\varphi))=0^{E^*}_m$.
The sum $\dr_{e_m}\ell_\xi+_{r_E}\dr_{e_m'}\ell_\xi$
equals $\dr_{e_m+e_m'}\ell_\xi$.

\subsection{Basic facts about connections}
In this paper, connections will not
be defined on Lie algebroids, but more generally on \emph{dull algebroids}.
We make the following definition.
\begin{definition}
A \textbf{dull algebroid}
is a
 vector bundle  $Q\to M$ endowed with an  \emph{anchor},
i.e.  a  vector bundle morphism $\rho_Q:Q\to TM$
over the identity on $M$ and a bracket
$[\cdot\,,\cdot]_Q$ on $\Gamma(Q)$ with
\begin{equation}\label{anchor_preserves_bracket}
\rho_Q[q,q']_Q=[\rho_Q(q),\rho_Q(q')]
\end{equation}
for all $q, q'\in\Gamma(Q)$, and  satisfying the Leibniz identity
\begin{equation*}
[\varphi_1 q_1, \varphi_2 q_2]_Q=\varphi_1\varphi_2[q_1, q_2]_Q+\varphi_1\rho_Q(q_1)(\varphi_2)q_2-\varphi_2\rho_Q(q_2)(\varphi_1)q_1
\end{equation*}
for all $\varphi_1,\varphi_2\in C^\infty(M)$, $q_1, q_2\in\Gamma(Q)$.
\end{definition}
In other words, a dull algebroid is a \textbf{Lie algebroid}
if its bracket  is in addition skew-symmetric and satisfies the Jacobi-identity.

\medskip

Let $(Q\to M,\rho_Q, [\cdot,\cdot]_Q)$ be a dull algebroid
and $B\to M$ a vector bundle.
A $Q$-connection on $B$ is a map
\[\nabla:\Gamma(Q)\times\Gamma(B)\to\Gamma(B),
\]
with the usual properties.
By the properties of a dull algebroid, one can still make sense
of the curvature $R_\nabla$ of the connection, which is an element of
$\Gamma(Q^*\otimes Q^*\otimes B^*\otimes B)$.

The dual connection $\nabla^*:\Gamma(Q)\times\Gamma(B^*)\to\Gamma(B^*)$
is defined by
\[\langle \nabla^*_q\xi, b\rangle=\rho_Q(q)\langle \xi, b\rangle- \langle\xi, \nabla_qb\rangle
\]
for all $q\in\Gamma(Q)$, $b\in\Gamma(B)$ and $\xi\in\Gamma(B^*)$.

\subsubsection{The Bott connection associated to a subbundle $F\subseteq
  TM$}
Recall the definition of the Bott connection
associated to an involutive subbundle of $TM$:
Let  $F\subseteq TM$ be a subbundle, then the Lie bracket on vector fields on $M$
induces a map
\[\tilde\nabla^F:\Gamma(F)\times\Gamma(TM)\to\Gamma(TM/F),
\]
defined by
\[\tilde\nabla^F_XY=\overline{[X,Y]}.
\]
The subbundle $F$ is involutive if and only if
\[\tilde\nabla^F_XX'=0\quad \text{ for all }\quad X,X'\in\Gamma(F).
\]
In that case, the map $\tilde \nabla^F$
quotients to a flat connection
\[\nabla^F:\Gamma(F)\times\Gamma(TM/F)\to\Gamma(TM/F),\]
the Bott connection.
\subsubsection{The basic connections associated to a connection on a quasi Lie algebroid}\label{basic_connections}
Consider here a dull algebroid $(Q, \rho_Q, [\cdot\,,\cdot])$
together with a connection $\nabla:\mathfrak{X}} \newcommand{\dr}{\mathbf{d}(M)\times\Gamma(Q)\to\Gamma(Q)$.

Then there are $Q$-connections on $Q$ and $TM$, called the \textbf{basic connections}
and defined as follows \cite{CrFe05,ArCr11a}.
\[\nabla^{\rm bas}=\nabla^{{\rm bas},Q}:\Gamma(Q)\times\Gamma(Q)\to\Gamma(Q),\]
\[\nabla^{\rm bas}_qq'=[q,q']+\nabla_{\rho_Q(q')}q\]
and
\[\nabla^{\rm bas}=\nabla^{{\rm bas},TM}:\Gamma(Q)\times\mathfrak{X}} \newcommand{\dr}{\mathbf{d}(M)\to\mathfrak{X}} \newcommand{\dr}{\mathbf{d}(M),\]
\[\nabla^{\rm bas}_qX=[\rho_Q(q),X]+\rho_Q(\nabla_{X}q).\]
The \textbf{basic curvature} is the map
\[R_\nabla^{\rm bas}:\Gamma(Q)\times\Gamma(Q)\times\mathfrak{X}} \newcommand{\dr}{\mathbf{d}(M)\to\Gamma(Q),
\]
\[R_\nabla^{\rm
  bas}(q,q')(X)=-\nabla_X[q,q']+[\nabla_Xq,q']+[q,\nabla_Xq']+\nabla_{\nabla_{q'}^{\rm
    bas}X}q
-\nabla_{\nabla^{\rm bas}_qX}q'.
\]
The basic curvature is tensorial and we have the identities
\[ \nabla^{{\rm bas},TM}\circ \rho_Q=\rho_Q\circ \nabla^{{\rm bas},Q}, \qquad
\rho_Q\circ R_\nabla^{\rm bas}=R_{\nabla^{{\rm bas},TM}}\quad
\text{ and }\quad
R_\nabla^{\rm bas}\circ\rho_Q=R_{\nabla^{{\rm bas},Q}}.
\]

\subsubsection{Connections on a vector bundle $E$, splittings of $TE$ and the Lie bracket on $\mathfrak{X}} \newcommand{\dr}{\mathbf{d}(E)$.}\label{connections_and_splittings_of_TE}
We recall here the relation between a connection on a vector bundle $E$ and
the Lie bracket of vector fields on $E$.
\medskip

Let $q_E:E\to M$ be a vector bundle and
$\nabla:\mathfrak{X}} \newcommand{\dr}{\mathbf{d}(M)\times\Gamma(E)\to\Gamma(E)$ a connection.
Then for each $e_m\in E$ and $v_m\in TM$, we can define the vector
\[
\widetilde{v_m,e_m}=T_mev_m-\left.\frac{d}{dt}\right\an{t=0}e_m+t\nabla_{v_m}e\in T_{e_m}E
\]
for any section $e\in\Gamma(E)$ such that $e(m)=e_m$.
We have
\[\widetilde{v_m,e_m}(\ell_\xi)=v_m\langle \xi, e\rangle-\langle \xi(m),
\nabla_{v_m}e\rangle=\ell_{\nabla_{v_m}^*\xi}(e_m)\]
and
\[\widetilde{v_m,e_m}(q_E^*\varphi)=v_m(\varphi)
\]
for all $\varphi\in C^\infty(M)$ and $\xi\in\Gamma(E^*)$.
The set of all vectors in $TE$ defined in this manner is a subbundle $H_\nabla$
of $p_E:TE\to E$ that is in direct sum with the vertical space
$V:=T^{q_E}E=\{v_{e_m}\in TE\mid T_{e_m}q_Ev_{e_m}=0\}$:
\[TE\cong V\oplus H_\nabla\to E.\]

For each section $X\in\mathfrak{X}} \newcommand{\dr}{\mathbf{d}(M)$, the vector field $\tilde
X\in\Gamma(H_\nabla)\subseteq\mathfrak{X}} \newcommand{\dr}{\mathbf{d}(E)$ is defined by $\tilde X(e_m)=\widetilde{X(m), e_m}$.
For all functions $\varphi\in C^\infty(M)$ and sections $\xi\in\Gamma(E^*)$,
we have
\begin{equation}\label{ableitungen}
\tilde X(\ell_\xi)=\ell_{\nabla_{X}^*\xi}, \qquad \tilde
X(q_E^*\varphi)=q_E^*(X(\varphi)), \qquad f^\uparrow(\ell_\xi)=q_E^*\langle\xi,
e\rangle, \qquad f^\uparrow(q_E^*\varphi)=0.
\end{equation}
Conversely, consider a  splitting $TE\cong V\oplus H$
of $TE\to E$. Then, since $H\cong TE/V$
is isomorphic to the pullback
$q_E^!TM\to E$, we find
for each vector field  $X\in\mathfrak{X}} \newcommand{\dr}{\mathbf{d}(M)$ a unique section $\tilde X$ of $H$ such that
$\tilde X\sim_{q_E}X$.
Using this uniqueness, one can show
easily that the pair $(X,\tilde X)$ defines
a vector bundle morphism
\begin{align*}
\begin{xy}
\xymatrix{
E\ar[d]_{q_E}\ar[r]^{\tilde X}&TE\ar[d]^{Tq_E}\\
M\ar[r]_{X}&TM
}
\end{xy}
\end{align*}
and that we have the equality
\[\widetilde{\varphi\cdot X}=q_E^*\varphi\cdot \tilde X
\]
for all $\varphi\in C^\infty(M)$ and $X\in\mathfrak{X}} \newcommand{\dr}{\mathbf{d}(M)$.
Using this and
$\ell_{\varphi\cdot\xi}=q_E^*\varphi\cdot \ell_\xi$
for all $\varphi\in C^\infty(M)$ and $\xi\in\Gamma(E^*)$,
one can then \emph{define} a connection
$\nabla^H:\mathfrak{X}} \newcommand{\dr}{\mathbf{d}(M)\times\Gamma(E)\to \Gamma(E)$
by setting
\[\tilde X(\ell_\xi)=:\ell_{{\nabla^H}^*_X\xi}\]
for all $X\in\mathfrak{X}} \newcommand{\dr}{\mathbf{d}(M)$, $\xi\in\Gamma(E^*)$.

\medskip

This shows the correspondence of the two definitions of a connection;
the first as the map
\[\nabla:\mathfrak{X}} \newcommand{\dr}{\mathbf{d}(M)\times\Gamma(E)\to\Gamma(E),\]
the second as a splitting
\[TE\cong V\oplus H\to E.\]

Given $\nabla$ or $H_\nabla$, it is easy to see using the equalities
in \eqref{ableitungen} that
\begin{align*}
\left[\tilde X, \tilde
  Y\right]&=\widetilde{[X,Y|}-R_\nabla(X,Y)^\uparrow,\\
\left[\tilde X, e^\uparrow\right]&=(\nabla_Xe)^\uparrow,\\
\left[e^\uparrow,f^\uparrow\right]&=0
\end{align*}
for all $X,Y\in\mathfrak{X}} \newcommand{\dr}{\mathbf{d}(M)$ and $e,f\in\Gamma(E)$.
That is, the Lie bracket of vector fields on $E$ can be described
using the connection. The connection itself can also be seen as a
suitable quotient of the Bott connection $\nabla^{H_\nabla}$:
\[\nabla^{H_\nabla}_{\tilde X}\overline{e^\uparrow}=\overline{(\nabla_Xe)^\uparrow}
\]
for all $e\in\Gamma(E)$ and $X\in\mathfrak{X}} \newcommand{\dr}{\mathbf{d}(M)$, i.e.
the Bott connection associated to $H_\nabla$ restricts well to linear
(horizontal) and vertical sections.

\medskip

\begin{remark}
To any derivation $D:\Gamma(E)\to \Gamma(E)$ over a vector field $X\in\mathfrak{X}} \newcommand{\dr}{\mathbf{d}(M)$,
we can associate a vector field $\widehat{D}\in\mathfrak{X}} \newcommand{\dr}{\mathbf{d}(E)$ as follows:
\[\widehat{D}(q_E^*\varphi)=q_E^*(X(\varphi))\]
for all $\varphi\in C^\infty(M)$,
and
\[\widehat{D}(\ell_\xi)=\ell_{D^*\xi}
\]
for all $\xi\in\Gamma(E^*)$.
Here, $D^*:\Gamma(E^*)\to\Gamma(E^*)$ is the dual
of the derivation $D$, i.e.
\[\langle D^*(\xi), e\rangle=X\langle \xi,e\rangle-\langle \xi, D(e)\rangle
\]
for all $e\in\Gamma(E)$ and $\xi\in\Gamma(E^*)$.

We will use this notation in the paper. Given a connection $\nabla:\mathfrak{X}} \newcommand{\dr}{\mathbf{d}(M)\times\Gamma(E)\to\Gamma(E)$,
the vector field $\tilde X$ defined as above is just $\widehat{\nabla_X}$.
\end{remark}

\section{Dorfman connections: definition and examples}

\begin{definition}
Let $(Q\to M,\rho_Q,[\cdot\,,\cdot]_Q)$ be a dull algebroid.
 Let $B\to M$ be a vector bundle with a fiberwise
pairing $\langle\cdot\,,\cdot\rangle:Q\times_M B\to \mathbb{R}} \newcommand{\bxi}{\boldsymbol{\xi}$ and a map
$\dr_B: C^\infty(M)\to \Gamma(B)$ such that
\begin{equation}\label{compatibility_anchor_pairing}
\langle q, \dr_B\varphi\rangle=\rho_Q(q)(\varphi)
\end{equation}
for all $q\in\Gamma(Q)$ and $\varphi\in C^\infty(M)$.
Then $(B,\dr_B, \langle\cdot\,,\cdot\rangle)$  will be called a \textbf{pre-dual}
of $Q$ and
$Q$ and $B$
are said to be \textbf{paired by $\langle\cdot\,,\cdot\rangle$}.
\end{definition}

\begin{remark}
Note that if the pairing is nondegenerate, then $(B\to M,\dr_B, \langle\cdot\,,\cdot\rangle)$  is the \textbf{dual} of $(Q\to M,\rho_Q,[\cdot\,,\cdot]_Q)$
and $\dr_{Q^*}:C^\infty(M)\to\Gamma(Q^*)$ is \emph{defined}
by \eqref{compatibility_anchor_pairing}. We have then  $\dr_{Q^*}\varphi=\rho_Q^*\dr\varphi$,
i.e. $\langle \dr_{Q^*}\varphi, q\rangle =\rho_Q(q)(\varphi)$ for all $q\in\Gamma(Q)$ and $\varphi\in C^\infty(M)$.
\end{remark}

The main definition of this paper is the following.
\begin{definition}\label{the_def}
  Let $(Q\to M,\rho_Q,[\cdot\,,\cdot]_Q)$ be a dull algebroid
  and  $(B\to M, \dr_B, \langle\cdot\,,\cdot\rangle)$ a pre-dual of $Q$.
\begin{enumerate}
\item A \textbf{Dorfman ($Q$-)connection on $B$}
is an $\mathbb{R}} \newcommand{\bxi}{\boldsymbol{\xi}$-bilinear
  map
\begin{equation*}
  \Delta:\Gamma(Q)\times\Gamma(B)\to\Gamma(B)
\end{equation*}
such that
\begin{enumerate}
\item $\Delta_{\varphi q}b=\varphi\Delta_qb+\langle q, b\rangle \cdot \dr_B \varphi$,
\item $\Delta_q(\varphi b)=\varphi\Delta_qb+\rho_Q(q)(\varphi)b$,
\item $\rho(q)\langle q', b\rangle=\langle[q,q']_Q, b\rangle+\langle q', \Delta_qb\rangle$
\end{enumerate}
for all $\varphi\in C^\infty(M)$, $q,q'\in\Gamma(Q)$, $b\in\Gamma(B)$.
\item The curvature of $\Delta$ is the map
  \[R_\Delta:\Gamma(Q)\times\Gamma(Q)\to\Gamma(B^*\otimes B),\] defined
  on $q,q'\in\Gamma(Q)$ by
  $R_\Delta(q,q'):=\Delta_q\Delta_{q'}-\Delta_{q'}\Delta_q-\Delta_{[q,q']_Q}$.
\end{enumerate}
\end{definition}
The failure of a Dorfman connection to be a connection is hence measured by the map $\dr_B$ and the pairing
of $Q$ with $B$.
Before we go on with examples,we have to check that $R_\Delta(q,q')$ is an element of $\Gamma(B^*\otimes B)$
for all $q,q'\in\Gamma(Q)$. But this is a straightforward computation,
and we omit the proof of the following proposition.
\begin{proposition}\label{curvature_tensor}
Let $(Q\to M,\rho_Q,[\cdot\,,\cdot]_Q)$ be a dull algebroid and $(B,\dr_B,\langle\cdot\,,\cdot\rangle)$ a pre-dual of $Q$.
Let $\Delta$ be a Dorfman $Q$-connection on $B$.
Then:
\begin{enumerate}
\item For all $\varphi\in C^\infty(M)$ and $q,q'\in \Gamma(Q)$, $b\in \Gamma(B)$, we have
\[R_\Delta(q,q')(\varphi\cdot b)=\varphi\cdot R_\Delta(q,q').
\]
\item For all $q_1,q_2,q_3\in\Gamma(Q)$ and $b\in\Gamma(B)$, we have
\[\langle R_\Delta(q_1,q_2)(b),q_3\rangle=\langle[[q_1,q_2]_Q,q_3]_Q+[q_2,[q_1,q_3]_Q]_Q-[q_1,[q_2,q_3]_Q]_Q, b\rangle.
\]
In particular, if $Q$ is a Lie algebroid, the Dorfman connection is always flat in this sense.
\end{enumerate}
\end{proposition}

\begin{remark}
Note that this doesn't mean that the curvature of a Dorfman connection vanishes everywhere
if $Q$ is a Lie algebroid, since
the pairing of $Q$ and $B$ can be degenerate. We will see a trivial example for this phenomenon in Example \ref{trivial}.
\end{remark}

\begin{example}\label{trivial}
Let $(Q\to M, \rho_Q, [\cdot\,,\cdot]_Q)$ be a dull algebroid and
$B\to M$ a vector bundle. Take the pairing
$\langle\cdot\,,\cdot\rangle:Q\times_M B\to \mathbb{R}} \newcommand{\bxi}{\boldsymbol{\xi}$ and the map
$\dr_B:C^\infty(M)\to \Gamma(B)$ to be trivial. Then any $Q$-connection
 on $B$ is also a Dorfman connection.
\end{example}

\begin{example}
The  easiest non-trivial example of a Dorfman connection
is the map
\[\ldr{}:\Gamma(Q)\times\Gamma(Q^*)\to\Gamma(Q^*),
\]
\[\langle\ldr{q}\xi,q'\rangle=\rho_Q(q)\langle q, \xi\rangle-\langle\xi, [q,q']_Q\rangle,
\]
for a dull algebroid $(Q\to M, \rho_Q,[\cdot\,,\cdot]_Q)$
and its dual $(Q^*,\dr_{Q^*})$, i.e. with
the canonical pairing $Q\times_MQ^*\to\mathbb{R}} \newcommand{\bxi}{\boldsymbol{\xi}$ and
$\dr_{Q^*}=\rho_Q^*\dr:C^\infty(M)\to\Gamma(Q^*)$.

The third property of a Dorfman connection
is immediate by definition of $\ldr{}$ and the first two properties
are easily verified. The curvature vanishes if and only if $(Q, \rho_Q, [\cdot\,,\cdot]_Q)$ is a
Lie algebroid.
\end{example}

\bigskip

Let $(\mathsf E\to M, \rho:\mathsf E\to TM,
\langle\cdot\,,\cdot\rangle, [\cdot\,,\cdot])$ be  a Courant algebroid.  If $K$ is a subalgebroid
of $\mathsf E$, the (in general singular) distribution $S:=\rho(K)\subseteq TM$ is
algebraically involutive
and we can define the ``singular'' Bott connection
\[\nabla^S:\Gamma(S)\times\frac{\mathfrak{X}} \newcommand{\dr}{\mathbf{d}(M)}{\Gamma(S)}\to\frac{\mathfrak{X}} \newcommand{\dr}{\mathbf{d}(M)}{\Gamma(S)}
\]
by
\[\nabla^S_{s}\bar X=\overline{[s,X]}\]
for all $X\in\mathfrak{X}} \newcommand{\dr}{\mathbf{d}(M)$ and $s\in\Gamma(S)$.

The anchor $\rho:\mathsf E\to TM$ induces a map $\bar\rho:\Gamma(\mathsf E/K)\to
\mathfrak{X}} \newcommand{\dr}{\mathbf{d}(M)/\Gamma(S)$, $\bar\rho(\bar e)=\rho(e)+\Gamma(S)$.

\begin{proposition}\label{Bott_Dorfman}
  Let $\mathsf E\to M$ be a Courant algebroid and $K\subseteq \mathsf E$ an
  isotropic subalgebroid.  Then the map
\begin{align*}
  \Delta:\Gamma(K)\times\Gamma(\mathsf E/K)&\to\Gamma(\mathsf E/K)\\
  \Delta_k\bar e&=\overline{[k,e]}
\end{align*}
is a Dorfman connection. The dull algebroid structure on $K$ is its
induced Lie algebroid structure, the map $\dr_{\mathsf E/K}$ is just
$\mathcal D+\Gamma(K)$  and the pairing $\langle\cdot\,,\cdot\rangle:K\times_M (\mathsf E/K)\to \mathbb{R}} \newcommand{\bxi}{\boldsymbol{\xi}$
is the natural pairing induced by the pairing on $\mathsf E$.

We have
\[\bar\rho(\Delta_k\bar e)=\nabla^S_{\rho(k)}\bar\rho(\bar e)
\]
for all $k\in\Gamma(K)$ and $\bar e\in\Gamma(\mathsf E/K)$.
\end{proposition}

\begin{remark}
\begin{enumerate}
\item
Because of the analogy of the Dorfman connection in the last proposition
with the Bott connection defined by involutive subbundles of $TM$,
we name this  Dorfman connection the \emph{Bott--Dorfman connection
  associated to $K$}.
\item
Note that if $K$ is a Dirac structure $\mathsf D$ in $\mathsf E$, then  $\mathsf E/\mathsf D\simeq \mathsf D^*$ and the Dorfman connection is just the Lie algebroid derivative
of $\mathsf D$ on $\Gamma(\mathsf D^*)$.
\end{enumerate}
\end{remark}

\subsection{Example: reduction of Courant algebroids}
Let $\mathsf E\to M$ be a Courant algebroid and $K\subseteq E$ an
isotropic subalgebroid.  Choose $k,k'\in\Gamma(K)$ and
$e\in\Gamma(K^\perp)$.  Then the equality
\[\langle [k,e], k'\rangle=-\langle e, [k,k']\rangle+\rho(k)\langle e, k'\rangle=0
\]
shows that $[k,e]\in\Gamma(K^\perp)$.
Thus, the Dorfman connection in Proposition \ref{Bott_Dorfman}
restricts to a flat connection
\[\nabla:\Gamma(K)\times\Gamma(K^\perp/K)\to \Gamma(K^\perp/K).
\]
Assume that $\rho(K)\subseteq TM$ is simple, i.e. it has constant
rank, is hence Frobenius integrable and its space of leaves is a
smooth manifold such that the canonical projection is a smooth
surjective submersion.  The $\nabla$-parallel sections of $K^\perp/K$
are the sections that project to the quotient $(K^\perp/K)/\nabla\to
M/\rho(K)$ and the properties of $\nabla$:
 \begin{enumerate}
\item $\nabla$ is flat
\item $\rho$ intertwines $\nabla$ with the Bott connection
  $\nabla^{\rho(K)}$,
\end{enumerate}
can be used to show as in \cite{Zambon08} that, under necessary
regularity conditions, the quotient
has the structure of a Courant algebroid over $M/\rho(K)$.
(The $\nabla$-parallel sections of $K^\perp/K$ are exactly the sections that are
called basic in \cite{Zambon08}.)

\begin{example}
Consider the standard Courant algebroid $TM\oplus T^*M\to M$
over a smooth manifold $M$ and an involutive subbundle
$F\subseteq TM$.
The Dorfman connection associated to $F\oplus \{0\}\subseteq TM\oplus T^*M$
is given by
\[\Delta_X\overline{(Y,\beta)}=\overline{(\ldr{X}Y,\ldr{X}\beta)}
\]
for all $X\in\Gamma(F)$ and $(Y,\beta)\in\mathfrak{X}} \newcommand{\dr}{\mathbf{d}(M)\times\Omega^1(M)$.
Hence, from the preceding example, we recover the fact that
the sections of $TM\oplus T^*M$ that project
to the reduced Courant algebroid $T(M/F)\oplus T^*(M/F)\to (M/F)$
are the sections $(Y,\beta)$ of $TM\oplus F^\circ$ that are invariant under $F$,
i.e. such that $\ldr{X}(Y,\beta)\in\Gamma(F\oplus\{0\})$
for all $X\in\Gamma(F)$ (e.g. \cite{JoRa11b}).

Note that $(TM\oplus F^\circ)/(F\oplus 0)\simeq (TM/F)\oplus F^\circ$
and $(TM/F)^*\simeq F^\circ$.
The connection
\[\nabla:\Gamma(F)\times \Gamma(TM/F\oplus F^\circ)\to \Gamma(TM/F\oplus F^\circ)
\]
is, modulo these identifications, just the product of the Bott
connection $\nabla^F$ and its dual
\[{\nabla^F}^*:\Gamma(F)\times \Gamma((TM/F)^*)\to \Gamma((TM/F)^*).
\]
\end{example}

\subsection{Generalized complex structures and Dorfman connections}
Let $V$ be a vector space. Consider a linear endomorphism $\mathcal J$
of $V\oplus V^*$ such that $\mathcal J^2=-\Id_V$ and $\mathcal J$ is
orthogonal with respect to the inner product
$$(X+\xi, Y+\eta)_V
=\xi(Y)+\eta(X), \quad \forall X,Y\in V, \; \xi, \eta \in V^* .$$ Such
a linear map is called a \emph{linear generalized complex structure}
by Hitchin \cite{Hitchin03}. The complexified vector
space $(V\oplus V^*)\otimes \mathbb{C}$ decomposes as the direct
sum $$(V\oplus V^* )\otimes \mathbb{C}=E_+\oplus E_-$$ of the
eigenbundles of $\mathcal J$ corresponding to the eigenvalues $\pm
\rm i$ respectively, i.e.,
$$E_{\pm}=\left\{(X+\xi)\mp \rm i \mathcal J(X+\xi) \mid X+\xi\in V\oplus V^* \right\}.$$
Both eigenspaces are maximal isotropic with respect to $\langle
\cdot\,,\cdot\rangle$ and they are complex conjugate to each other.

  The linear generalized complex structures are in 1-1 correspondence
  with the splittings $(V\oplus V^*)\otimes \mathbb{C}=E_+\oplus E_-$
  with $E_{\pm}$ maximal isotropic and $E_-=\overline{E_+}$.

\medskip

Now, let $M$ be a manifold and $\mathcal J$ a bundle endomorphism of
$TM\oplus T^*M$ such that $\mathcal J^2=-\Id_{TM\oplus TM}$, and $\mathcal J$ is
orthogonal with respect to $\langle \cdot\,,\cdot\rangle_M$.  Then
$\mathcal J$ is a \emph{generalized almost complex structure}.  In the
associated eigenbundle decomposition $$T_\mathbb{C} M\oplus
T_\mathbb{C}^*M =E_+\oplus E_- ,$$ if $\Gamma(E_+)$ is closed under
the (complexified) Courant bracket, then $E_+$ is a (complex) Dirac
structure on $M$ and one says that $\mathcal J$ is a \emph{generalized complex
  structure} \cite{Hitchin03, Gualtieri03}.  In this case, $E_-$ must
also be a Dirac structure since $E_-=\overline{E_+}$. Indeed $(E_+, E_-)$ is
a complex Lie bialgebroid in the sense of Mackenzie-Xu \cite{MaXu94},
in which $E_+$ and $E_-$ are complex conjugate to each other.

\medskip

Since $E_-=\overline{E_+}$ and $E_-\cap E_+=\{0\}$, we have a
vector bundle isomorphism
\[ \frac{T_{\mathbb C}M\oplus T_{\mathbb C}^*M}{E_-}\to E_+
\]
that
 is given by\footnote{To avoid confusions, we write
  in this subsection $\hat d$ for the class of
$d\in\Gamma(T_{\mathbb C}M\oplus T_{\mathbb C}^*M)$ in $(T_{\mathbb C}M\oplus T_{\mathbb C}^*M)/E_-$.}
\[ \hat d\mapsto \frac{1}{2} (d-\rm{i}\mathcal J(d)).
\]
The Dorfman connection
\[\Delta^{E_-}:\Gamma(E_-)\times\Gamma(E_+)\to\Gamma(E_+)
\]
is then simply given by
\[\Delta^{E_-}_{d-}d_+=\frac{1}{2}\left([d_-,d_+]-\rm{i}\mathcal J[d_-,d_+]\right).
\]
In the same manner,
\[\Delta^{E_+}:\Gamma(E_+)\times\Gamma(E_-)\to\Gamma(E_-)
\]
is given by
\[\Delta^{E_+}_{d+}d_-=\frac{1}{2}\left([d_+,d_-]+\rm{i}\mathcal J[d_+,d_-]\right)
\]
for all $d_-\in\Gamma(E_-)$ and $d_+\in\Gamma(E_+)$.

It would be interesting to study in more detail the properties of these two (flat!) Dorfman connections,
maybe in the spirit of the results in \cite{LaStXu08}.

\section{Splittings of $TE\oplus T^*E$}
Consider a vector bundle $q_E:E\to M$.
In this section, the vector bundle $TM\oplus E^*$ will always be anchored by the projection $\pr_{TM}:TM\oplus E^*\to TM$
and the dual $E\oplus T^*M$ will always be paired with $TM\oplus E^*$ by the canonical pairing.
The map $\dr_{E\oplus T^*M}:C^\infty(M)\to \Gamma(E\oplus T^*M)$ will consequently always be  given by
\[\dr_{E\oplus T^*M}=\pr_{TM}^*\circ \dr,
\]
i.e. \[\dr_{E\oplus T^*M}\varphi =(0,\dr \varphi )
\]
for all $\varphi \in C^\infty(M)$.
A Dorfman connection $\Delta$
will here always be a $TM\oplus E^*$-Dorfman connection on $E\oplus T^*M$,
with dual $\llbracket} \newcommand{\rb}{\rrbracket\cdot\,,\cdot\rb_\Delta$.
Note that since the pairing is nondegenerate, the  Dorfman connection is completely determined
by its dual structure, the associated bracket $\llbracket} \newcommand{\rb}{\rrbracket\cdot\,,\cdot\rb_\Delta$ and vice-versa.
Hence, we can say here  that a Dorfman connection
is equivalent to a dull algebroid $(TM\oplus E^*, \pr_{TM},\llbracket} \newcommand{\rb}{\rrbracket\cdot\,,\cdot\rb_\Delta)$.
It is easy to see, using  Proposition \ref{curvature_tensor}, that the curvature $R_\Delta$  will always vanish on $(TM\oplus E^*)\otimes(TM\oplus E^*)\otimes(0\oplus T^*M)$.

\bigskip
Recall from \S \ref{connections_and_splittings_of_TE} that an ordinary connection
\[\nabla:\mathfrak{X}} \newcommand{\dr}{\mathbf{d}(M)\times\Gamma(E)\to\Gamma(E)
\]
corresponds to
a splitting
\[TE=T^{q_E}E\oplus H_\nabla\to E.
\]
We show in this section that a Dorfman connection
\[\Delta:\Gamma(TM\oplus E^*)\times\Gamma(E\oplus T^*M)\to \Gamma(E\oplus T^*M)
\]
 is the same as a splitting
\[TE\oplus T^*E=(V\oplus V^\circ)\oplus L_\Delta
\]
where, as before, $V:=T^{q_E}E$.
The subbundle  $L_\Delta$ is an almost Dirac structure on $E$
if and only if the bracket $\llbracket} \newcommand{\rb}{\rrbracket\cdot\,,\cdot\rb_\Delta$ dual to the Dorfman connection is skew-symmetric
and the failure of $\Gamma(L_\Delta)$ to be closed under the Dorfman bracket is
 measured by the curvature $R_\Delta$.

\medskip
In the following, for any section $(f,\theta)$ of $E\oplus T^*M$, the
section $(f,\theta)^\uparrow$ of $V\oplus V^\circ$
is the pair defined by
\[(f,\theta)^\uparrow(e_m)=\left(\left.\frac{d}{dt}\right\an{t=0}e_m+tf(m), (T_{e_m}q_E)^*\theta(m)\right)
\]
for all $e_m\in E$.
Note that the pairs $(f,\theta)^\uparrow(e_m)$
for all $(f,\theta)\in\Gamma(E\oplus T^*M)$, span by construction the fiber $(V\oplus V^\circ)(e_m)$.

\subsection{The standard almost Dorfman connection associated to an usual connection on $E$.}\label{sec_standard_almost_dorfman}
We start in this subsection with a simple, motivating example.
\begin{definition}\label{standard_almost_dorfman}
  Let $E\to M$ be a vector bundle with a connection
  $\nabla:\mathfrak{X}} \newcommand{\dr}{\mathbf{d}(M)\times\Gamma(E)\to\Gamma(E)$.  Then the standard
  Dorfman connection associated to $\nabla$ is the map
  \[\Delta:\Gamma(TM\oplus E^*)\times\Gamma(E\oplus
  T^*M)\to\Gamma(E\oplus T^*M),\]
\begin{equation*}
  \Delta_{(X,\xi)}(e,\theta)=(\nabla_Xe,\ldr{X}\theta+\langle
  \nabla^*_\cdot \xi, e\rangle).
\end{equation*}
The dual bracket is in this case defined by
\[\llbracket} \newcommand{\rb}{\rrbracket(X,\xi), (Y,\eta)\rb_\Delta=([X,Y], \nabla_X^*\eta-\nabla_Y^*\xi)
\]
for all $(X,\xi), (Y,\eta)\in \Gamma(TM\oplus E^*)$.
\end{definition}

\begin{proposition}
  Let $E\to M$ be a vector bundle endowed with a connection $\nabla$.
\begin{enumerate}
\item The curvature of the standard Dorfman connection $\Delta$
  associated to $\nabla$
is given by
\[R_\Delta((X,\xi),(Y,\eta))=(R_\nabla(X,Y),
R_{\nabla^*}(\cdot,X)(\eta)-R_{\nabla^*}(\cdot,Y)(\xi)).\]
\item   $(TM\oplus E^*, \operatorname{pr}_{TM},
\llbracket} \newcommand{\rb}{\rrbracket\cdot\,,\cdot\rb_\Delta)$ is a Lie algebroid if  and only if $\nabla$ is flat.
\end{enumerate}
\end{proposition}

\begin{proof}
The first claim is proved by a straightforward computation. The second
statement follows then, using
Proposition \ref{curvature_tensor} and the fact that the pairing is
here nondegenerate.
\end{proof}

\begin{proposition}
Let $E\to M$ be a vector bundle endowed with a connection $\nabla$.
and let $\Delta$ be the
standard Dorfman connection
  associated to $\nabla$.
For any section $(X,\xi)\in\Gamma(TM\oplus E^*)$,
set $\widetilde{(X,\xi)}\in\mathfrak{X}} \newcommand{\dr}{\mathbf{d}(E)\times\Omega^1(E)$,
\begin{align*}
\widetilde{(X,\xi)}(e_m)&=\left(T_meX(m), \dr_{e_m}\ell_\xi\right)-\left(\Delta_{(X,\xi)}(e,0)\right)^\uparrow(e_m)\\
&=\left(T_meX(m), \dr_{e_m}\ell_\xi\right)-\left(\left.\frac{d}{dt}\right\an{t=0}e_m+t\nabla_Xe,(T_{e_m}q_E)^*\langle \nabla^*_\cdot \xi, e\rangle\right).
\end{align*}
The subbundle $L_\Delta$ spanned by these sections is equal to $H_\nabla\oplus H_\nabla^\circ$.
Hence, the standard Dorfman connection associated to a connection $\nabla$ is the same
as a splitting
\[TE\oplus T^*E\cong(V\oplus V^\circ)\oplus(H_\nabla\oplus H_\nabla^\circ),
\]
 the sum of a  Dirac structure and an almost Dirac structure.
\end{proposition}
\begin{proof}
We just have to check that the space spanned by the sections
$\widetilde{(X,\xi)}$  is equal to $H_\nabla\oplus H_\nabla^\circ$.
But since $\Delta_{(X,0)}(e,0)=(\nabla_Xe,0)$ and
$\Delta_{(0,\xi)}(e,0)=(0,\langle\nabla_\cdot^*\xi,e\rangle)$,
the subbundle that we are considering is spanned by the sections
$\widetilde{(X,0)}=(\widetilde X, 0)$ and
$\widetilde{(0,\xi)}=(0,\widetilde\xi)$, i.e. it is the direct sum of a subbundle
of $TE$ and a subbundle of $T^*E$. The tangent part is obviously equal
to $H_\nabla$ and the cotangent part is easily seen to be $H_\nabla^\circ$.
\end{proof}

\subsection{Almost Dorfman connection associated to a splitting $TE\oplus T^*E\cong(V\oplus V^\circ)\oplus L$}
Consider now  a splitting
\[TE\oplus T^*E\cong(V\oplus V^\circ)\oplus L\to E,
\]
where, as before, $V:=T^{q_E}E$.
Recall that the vector bundle morphism
\begin{equation*}
\begin{xy}
\xymatrix{
\Phi_E:=({q_E}_*, r_E):&TE\oplus T^*E\ar[r]\ar[d]& TM\oplus E^*\ar[d]\\
& E\ar[r]_{q_E}&M
}
\end{xy}
\end{equation*}
is a fibration of vector bundles over the projection $q_E:E\to M$.

That is,
\begin{equation*}
\begin{xy}
\xymatrix{
TE\oplus T^*E\ar[d]\ar[rr]^{q_E^!\Phi_E}& &q_E^!(TM\oplus E^*)\ar[d]\\%=E\oplus TM\oplus E^*\\
E\ar[rr]&&E
}
\end{xy}
\end{equation*}
is a surjective vector bundle morphism over the identity on $E$.
Since $V\oplus V^\circ$ is the kernel of $\Phi_E$,
the diagram above factors as
\begin{equation*}
\begin{xy}
\xymatrix{
L\simeq(TE\oplus T^*E)/(V\oplus V^\circ)
\ar[d]\ar[rr]^{\qquad q_E^!\Phi_E}& &q_E^!(TM\oplus E^*)\ar[d]\\%=E\oplus TM\oplus E^*\\
E\ar[rr]&&E
}
\end{xy}
\end{equation*}
and we find that for any section $(X,\xi)$ of $TM\oplus E^*$,
there exists
a unique section $\widetilde{(X,\xi)}$ of $L$
such that
\[\Phi_E\circ \widetilde{(X,\xi)}=(X,\xi)\circ q_E.
\]
Note that by the uniqueness of the section $\widetilde{(X,\xi)}$ of $L$ over $(X,\xi)$,
we have $\widetilde{\varphi \cdot(X,\xi)}=q_E^*\varphi \cdot \widetilde{(X,\xi)}$
for all $\varphi \in C^\infty(M)$.

We start by proving the following observation.
\begin{lemma}
Choose $(X,\xi), (Y,\eta)\in\Gamma(TM\oplus E^*)$
Then
\[e_m\mapsto \langle \widetilde{(X,\xi)}(e_m), (T_meY(m),\dr_{e_m}
\ell_\eta)\rangle-Y(m)\langle\xi,e\rangle,\]
where $e\in\Gamma(E)$ is such that $e(m)=e_m$,
defines a linear map on $E$.
\end{lemma}
\begin{proof}
Choose first $e,f\in \Gamma(E)$. Then we have
\[ \Phi_E\bigl(\widetilde{(X,\xi)}(e(m))+_{\Phi_E}\widetilde{(X,\xi)}(f(m))\bigr)=(X,\xi)(m)
\]
and
\[ \Phi_E\bigl(\widetilde{(X,\xi)}(e(m)+f(m))\bigr)=(X,\xi)(m).
\]
Since $\widetilde{(X,\xi)}(e(m))+_{\Phi_E}\widetilde{(X,\xi)}(f(m))$
and $\widetilde{(X,\xi)}(e(m)+f(m))$ are both elements of
$L_{e_m+f_m}$, we find by definition of $\widetilde{(X,\xi)}$
that
\[\widetilde{(X,\xi)}(e(m))+_{\Phi_E}\widetilde{(X,\xi)}(f(m))=\widetilde{(X,\xi)}(e(m)+f(m)).
\]
By definition of the addition in the $\Phi_E$-fibers,  we find also
\[(T_meY(m),\dr_{ e_m}
\ell_\eta)+_{\Phi_E}(T_mfY(m),\dr_{f_m}
\ell_\eta)=(T_m(e+f)Y(m),\dr_{e_m+f_m}
\ell_\eta)
\]
if $e,f\in\Gamma(E)$
are such that $e(m)=e_m$ and $f(m)=f_m$.
Again by definition of the addition
in the $\Phi_E$-fibers,
we get hence
\begin{align*}
&\langle \widetilde{(X,\xi)}(e(m)+f(m)), (T_m(e+f)Y(m),\dr_{e_m+f_m}
\ell_\eta)\rangle\\
=&\langle \widetilde{(X,\xi)}(e(m)), (T_meY(m),\dr_{ e_m}
\ell_\eta)\rangle
+
\langle \widetilde{(X,\xi)}(f(m)),
(T_mfY(m),\dr_{f_m}
\ell_\eta)\rangle.
\end{align*}
In particular, we find
\[\langle \widetilde{(X,\xi)}(r\cdot e(m)), (T_m(r\cdot
e)Y(m),\dr_{r\cdot e_m}
\ell_\eta)\rangle
=r\cdot\langle \widetilde{(X,\xi)}(e(m)), (T_meY(m),\dr_{ e_m}
\ell_\eta)\rangle
\]
for all $r\in\mathbb{N}} \newcommand{\smoo}{C^\infty(G)$. The same equality follows then for $r\in\mathbb{Q}$
and by continuity for all $r\in\mathbb{R}} \newcommand{\bxi}{\boldsymbol{\xi}$.

Choose  $\varphi \in C^\infty(M)$ and set $\varphi (m)=\alpha$. Then
\begin{align*}
&\langle \widetilde{(X,\xi)}(\alpha e_m), (T_m(\varphi \cdot e)Y(m),\dr_{\alpha e_m} \ell_\eta)\rangle-Y(m)\langle\xi,\varphi \cdot e\rangle\\
=&\langle \widetilde{(X,\xi) }(\alpha e_m), (T_m(\alpha
e)Y(m)+Y(m)(\varphi )e^\uparrow(\alpha e_m),\dr_{\alpha e_m}
\ell_\eta)\rangle -Y(m)\langle\xi,\varphi \cdot e\rangle\\
=&\langle \widetilde{(X,\xi) }(\alpha e_m), (T_m(\alpha
e)Y(m),\dr_{\alpha e_m}
\ell_\eta)
\rangle
-\alpha Y(m)\langle\xi, \cdot e\rangle\\
=&\alpha\cdot \bigl(
\langle \widetilde{(X,\xi) }(e_m), (T_m eY(m),\dr_{e_m}
\ell_\eta)
\rangle
-Y(m)\langle\xi, \cdot e\rangle
\bigr).
\end{align*}
Thus, we have shown that the function
\[e_m\mapsto \langle \widetilde{(X,\xi)}(e_m), (T_meY(m),\dr_{e_m}
\ell_\eta)\rangle-Y(m)\langle\xi,e\rangle\]
is well-defined, i.e. doesn't depend on the choice of the section $e$, and linear.
\end{proof}

Thus, we can consider the map
\[\llbracket} \newcommand{\rb}{\rrbracket\cdot\,,\cdot\rb_L:\Gamma(TM\oplus E^*)\times\Gamma(TM\oplus E^*)\to \Gamma(TM\oplus E^*)\]
defined by
\[\langle\llbracket} \newcommand{\rb}{\rrbracket(X,\xi), (Y,\eta)\rb_L,(e_m,0)\rangle=\langle \widetilde{(X,\xi)}(e_m), (T_meY(m),\dr_{e_m} \ell_\eta)\rangle-Y(m)\langle\xi,e\rangle,\]
for any section $e\in\Gamma(E)$ such that $e(m)=e_m$,
and \[ \pr_{TM}\llbracket} \newcommand{\rb}{\rrbracket(X,\xi), (Y,\eta)\rb_L=[X,Y].
\]

\begin{theorem}
Let $E\to M$ be a vector bundle and consider a splitting
\[TE\oplus T^*E\cong(V\oplus V^\circ)\oplus L\to E.
\]
The triple  $(TM\oplus E^*, \pr_{TM}, \llbracket} \newcommand{\rb}{\rrbracket\cdot\,,\cdot\rb_L)$, where
$\llbracket} \newcommand{\rb}{\rrbracket\cdot\,,\cdot\rb_L$ is defined as above, is a dull algebroid.
\end{theorem}

\begin{proof}
We compute for $e\in\Gamma(E)$, setting $\alpha:=\varphi (m)$:
\begin{align*}
\langle\llbracket} \newcommand{\rb}{\rrbracket(X,\xi), \varphi\cdot(Y,\eta)\rb_L, (e,0)\rangle=&\langle
\widetilde{(X,\xi)}(e_m), (T_me(\alpha Y(m)),\dr_{e_m}
(q_E^*\varphi \cdot \ell_\eta))\rangle-\alpha Y(m)\langle\xi,e\rangle\\
=&\alpha\langle
\widetilde{(X,\xi)}(e_m), (T_meY(m),\dr_{e_m}
\ell_\eta)\rangle-\alpha Y(m)\langle\xi,e\rangle
+X(\varphi )\langle \eta, e\rangle\\
=&\langle \varphi \llbracket} \newcommand{\rb}{\rrbracket(X,\xi), (Y,\eta)\rb_L+X(\varphi )(Y,\eta), (e,0)\rangle(m),\\
\langle\llbracket} \newcommand{\rb}{\rrbracket \varphi\cdot (X,\xi), (Y,\eta)\rb_L, (e,0)\rangle&=\langle
(q_E^*\varphi \widetilde{(X,\xi)})(e_m),
(T_meY(m),\dr_{e_m}\ell_\eta)\rangle\\
&\qquad\qquad-Y(m)(\varphi )\langle\xi,e\rangle(m)-\varphi (m)Y(m)\langle\xi,e\rangle\\
=&\langle \varphi \llbracket} \newcommand{\rb}{\rrbracket(X,\xi), (Y,\eta)\rb_L-Y(\varphi )(X,\xi), (e,0)\rangle(m).
\end{align*}
Since
\[\langle\llbracket} \newcommand{\rb}{\rrbracket(X,\xi), \varphi\cdot(Y,\eta)\rb_L, (0,\theta)\rangle=\langle \varphi \llbracket} \newcommand{\rb}{\rrbracket(X,\xi), (Y,\eta)\rb_L+X(\varphi )(Y,\eta), (0,\theta)\rangle\]
and
\[\langle\llbracket} \newcommand{\rb}{\rrbracket \varphi\cdot (X,\xi), (Y,\eta)\rb_L, (0,\theta)\rangle
=\langle \varphi \llbracket} \newcommand{\rb}{\rrbracket(X,\xi), (Y,\eta)\rb_L-Y(\varphi )(X,\xi), (0,\theta)\rangle
\]
hold by construction for all $\theta\in\Omega^1(M)$, we have shown the two Leibniz equalities.
Compatibility of $\pr_{TM}$ with $\llbracket} \newcommand{\rb}{\rrbracket\cdot\,,\cdot\rb_L$ is given by construction.
\end{proof}

\begin{corollary}\label{great}
Let $E\to M$ be a vector bundle and consider a splitting
\[TE\oplus T^*E\cong(V\oplus V^\circ)\oplus L\to E.
\]
Define the map
\[\Delta^L:\Gamma(TM\oplus E^*)\times\Gamma(E\oplus T^*M)\to\Gamma(E\oplus T^*M),
\]
\[\langle\Delta^L_{(X,\xi)}(e,\theta), (Y,\eta)\rangle=X\langle(e,\theta), (Y,\eta)\rangle -\langle \llbracket} \newcommand{\rb}{\rrbracket(X,\xi), (Y,\eta)\rb_L, (e,\theta)\rangle
\]
Then $\Delta^L$ is a Dorfman connection.
\end{corollary}

\begin{remark}\label{dorfman_boring}
Note that, by definition,
we have $\Delta^L_{(X,\xi)}(0,\theta)=(0,\ldr{X}\theta)$
for all $X\in \mathfrak{X}} \newcommand{\dr}{\mathbf{d}(M)$ and $\theta\in\Omega^¹(M)$.
To see this, choose $Y\in\mathfrak{X}} \newcommand{\dr}{\mathbf{d}(M)$ and compute
\[\langle \Delta^L_{(X,\xi)}(0,\theta), (Y,0)\rangle=X\langle\theta, Y\rangle-\langle \theta, [X,Y]\rangle.
\]
\end{remark}

\begin{proof}[Proof of Corollary \ref{great}]
By construction $\Delta^L$ is dual to the dull bracket $\llbracket} \newcommand{\rb}{\rrbracket\cdot\,,\cdot\rb_L$ on $\Gamma(TM\oplus E^*)$.
\end{proof}

We end this subsection with a proposition relating directly the Dorfman connection $\Delta^L$ with the subbundle
$L\subseteq TE\oplus T^*E$.
\begin{proposition}\label{relation_Delta_L}
Let $E\to M$ be a vector bundle and consider a splitting
\[TE\oplus T^*E\cong(V\oplus V^\circ)\oplus L\to E.
\]
Choose $(X,\xi)\in\Gamma(TM\oplus E^*)$.
Then the corresponding section of $L$ is given by
\[\widetilde{(X,\xi)}(e_m)=\left( T_meX(m), \dr_{e_m}\ell_\xi\right)-\Delta^L_{(X,\xi)}(e,0)^\uparrow(e_m)
\]
for all $e_m\in E$ and $e\in\Gamma(E)$ such that $e(m)=e_m$.
\end{proposition}

\begin{proof}
Since $\Phi_E\widetilde{(X,\xi)}(e_m)=(X,\xi)(m)=\Phi_E(T_meX(m), \dr_{e_m}\ell_\xi)$
for $e_m\in E$, there exists
a pair $(f,\theta)\in\Gamma(E\oplus T^*M)$ such that
\[\widetilde{(X,\xi)}(e_m)=(T_meX(m), \dr_{e_m}\ell_\xi)+(f,\theta)^\uparrow(e_m).
\]
We compute for $(Y,\eta)\in\Gamma(TM\oplus E^*)$:
\begin{align*}
\langle (f,\theta), (Y,\eta)\rangle(m)&=\langle
(f,\theta)^\uparrow(e_m), (T_meY(m), \dr_{e_m}\ell_\eta)\rangle\\
&=\langle\widetilde{(X,\xi)}(e_m)-(T_meX(m), \dr_{e_m}\ell_\xi) ,
(T_meY(m), \dr_{e_m}\ell_\eta)\rangle\\
&=\langle \llbracket} \newcommand{\rb}{\rrbracket(X,\xi), (Y,\eta)\rb_L,(e_m,0)\rangle +Y(m)\langle \xi, e\rangle-X(m)\langle
\eta,e\rangle -Y(m)\langle \xi, e\rangle\\
&=\langle \llbracket} \newcommand{\rb}{\rrbracket(X,\xi), (Y,\eta)\rb_L,(e_m,0)\rangle-X(m)\langle
\eta,e\rangle \\
&=-\langle \Delta^L_{(X,\xi)}(e,0), (Y,\eta)\rangle(m).
\end{align*}
\end{proof}

\subsection{The converse construction}

Let $E\to M$ be a vector bundle and consider a Dorfman connection
\[\Delta:\Gamma(TM\oplus E^*)\times\Gamma(E\oplus T^*M)\to
\Gamma(E\oplus T^*M).\]

We define the subset
\[L_\Delta\subseteq TE\oplus T^*E\]
by
\[L_\Delta(e_m)=\left\{\left.\left(T_me X(m), \dr
    \ell_\xi(e_m)\right)-\Delta_{(X,\xi)}(e,0)^\uparrow(e_m)\right| (X,\xi)\in\Gamma(TM\oplus E^*)\right\}
\]
for any section $e\in\Gamma(E)$ such that $e(m)=e_m$.

For any pair $(X,\xi)\in\Gamma(TM\oplus E^*)$,
we will write $\widetilde{(X,\xi)}\in\Gamma_E(TE\oplus T^*E)=\mathfrak{X}} \newcommand{\dr}{\mathbf{d}(E)\times\Omega^1(E)$ for the section defined by
\[\widetilde{(X,\xi)}(e_m)=\left(T_me X(m), \dr
    \ell_\xi(e_m)\right)-\Delta_{(X,\xi)}(e,0)^\uparrow(e_m)
\]
for all $e_m\in E$.
Note that $\Phi_E\circ \widetilde{(X,\xi)}=(X,\xi)\circ q_E$.

\begin{proposition}
$L_\Delta$ is a well-defined subbundle of $TE\oplus T^*E\to E$
such that $(V\oplus V^\circ)\oplus L_\Delta\xrightarrow{\sim}{} TE\oplus T^*E$.
\end{proposition}

\begin{proof}
We show that the fiber over $e_m$ doesn't depend on the choice of the
section $e\in\Gamma(E)$ such that $e(m)=e_m$.
Note first that for any pair $(f,\theta)\in\Gamma(E\oplus T^*M)$,
we have
\begin{align*}
\left\langle \left(T_me X(m), \dr
    \ell_\xi(e_m)\right)-\Delta_{(X,\xi)}(e,0)^\uparrow(e_m),
  (f,\theta)^\uparrow(e_m)\right\rangle
&= \langle \theta(m), X(m)\rangle +\langle \xi(m), f(m)\rangle.
\end{align*}
This pairing doesn't depend on $e$.

Choose then any connection $\nabla$ on $E$
and a pair $(Y,\eta)\in\Gamma(TM\oplus E^*)$.
Consider the pair
\[(Y_\nabla,\eta_\nabla)(e_m):=\left(T_meY(m)-\left.\frac{d}{dt}\right\an{t=0}e_m+t\nabla_{Y(m)}e,
\dr \ell_\eta(e_m)-(T_{e_m}q_E)^*\langle \nabla^*_\cdot\eta,
e\rangle\right).\]
Then
\begin{align*}
&\left\langle \left(T_me X(m), \dr
    \ell_\xi(e_m)\right)-\Delta_{(X,\xi)}(e,0)^\uparrow(e_m),
  (Y_\nabla,\eta_\nabla)(e_m)\right\rangle\\
=\,& X(m)\langle \eta, e\rangle -\langle  \nabla^*_{X(m)}\eta,
e\rangle -\langle \eta(m), \pr_{E}\Delta_{(X,\xi)}(e,0)\rangle\\
&+Y(m)\langle \xi, e\rangle -\langle  \pr_{T^*M}\Delta_{(X,\xi)}(e,0),
Y(m)\rangle +\langle \xi(m), \nabla_{Y(m)}e\rangle\\
=\,& X(m)\langle \eta, e\rangle -\langle  \nabla^*_{X(m)}\eta,
e(m)\rangle -\langle (Y,\eta)(m), \Delta_{(X,\xi)}(e,0)\rangle+\langle
\nabla_{Y(m)}^*\xi, e(m)\rangle\\
=\,& \langle \llbracket} \newcommand{\rb}{\rrbracket(X,\xi), (Y,\eta)\rb_\Delta, (e(m), 0)\rangle-\langle  \nabla^*_{X(m)}\eta,
e(m)\rangle +\langle \nabla_{Y(m)}^*\xi, e(m)\rangle.
\end{align*}
Again, this does only depend on the value $e_m$ of $e$ at $m$.

Since pairs $(Y_\nabla,\eta_\nabla)$ and $(f,\theta)^\uparrow$ span
the whole of $TE\oplus T^*E$ and the pairing is nondegenerate, we have shown that $L_\Delta$ is well-defined.
The second claim is immediate, using the fact that $\Phi_E\circ \widetilde{(X,\xi)}=(X,\xi)$
for all $(X,\xi)\in\Gamma(TM\oplus E^*)$ and that $V\oplus V^\circ$
is spanned by the sections $(e,\theta)^\uparrow$ for $(e,\theta)\in\Gamma(E\oplus T^*M)$.
\end{proof}

The results in the last two subsections are summarized in the following theorem.
\begin{theorem}
Let $q_E:E\to M$ be a Lie algebroid.
The maps
\begin{align*}
\Delta&\mapsto L_\Delta,\\
\Delta^L&\mapsfrom L
\end{align*}
define a bijection
\begin{equation*}
\left\{\begin{array}{c}
(TM\oplus E^*)\text{-Dorfman connections }\\
\Delta \text{  on } E\oplus T^*M
\end{array} \right\}
\leftrightarrow
 \left\{\begin{array}{c}
\text{ Splittings }\\ TE\oplus T^*E\cong(V\oplus V^\circ)\oplus L
\end{array}\right\}.
\end{equation*}
\end{theorem}

Since a $(TM\oplus E^*)$-Dorfman connection $\Delta$ on $E\oplus T^*M$
is the same as a dull algebroid structure $(\pr_{TM}, \llbracket} \newcommand{\rb}{\rrbracket\cdot\,,\cdot\rb_\Delta)$
on $TM\oplus E^*$, we can reformulate this bijection as follows:
\begin{equation*}
\left\{\begin{array}{c}
\text{Dull algebroids }\\
(TM\oplus E^*, \pr_{TM}, \llbracket} \newcommand{\rb}{\rrbracket\cdot\,,\cdot\rb)
\end{array} \right\}
\leftrightarrow
 \left\{\begin{array}{c}
\text{ Splittings }\\ TE\oplus T^*E\cong(V\oplus V^\circ)\oplus L
\end{array}\right\}.
\end{equation*}

\subsection{The canonical pairing and the Courant-Dorfman bracket on $TE\oplus T^*E$}
We show in this section that the failure of a splitting $L$
of $TE\oplus T^*E$ to be Lagrangian
is equivalent to the failure of $\llbracket} \newcommand{\rb}{\rrbracket\cdot\,,\cdot\rb_L$
to be skew-symmetric, and the failure of its set of sections to be closed under the Courant-Dorfman
bracket is measured by the curvature of $\Delta$.

Here and later, we will need
the following notation.
Let $E\to M$ be a vector bundle and consider a Dorfman connection
$\Delta:\Gamma(TM\oplus E^*)\times\Gamma(E\oplus T^*M)\to\Gamma(E\oplus T^*M)$.
We call $\Skew_\Delta\in\Gamma((TM\oplus
E^*)\otimes(TM\oplus E^*)\otimes E^*)$
the tensor
defined by
\[\Skew_\Delta(v_1,v_2)=\pr_{E^*}(\llbracket} \newcommand{\rb}{\rrbracket v_1, v_2\rb_\Delta+\llbracket} \newcommand{\rb}{\rrbracket v_2, v_1\rb_\Delta)
\]
for all $v_1,v_2\in\Gamma(TM\oplus E^*)$. By the Leibniz identity,
this is indeed $C^\infty(M)$-linear in both arguments.
Note that the $TM$-part of $\llbracket} \newcommand{\rb}{\rrbracket v_1, v_2\rb_\Delta+\llbracket} \newcommand{\rb}{\rrbracket v_2, v_1\rb_\Delta$
always vanishes since the Lie bracket of vector fields is skew-symmetric.

\begin{theorem}\label{Lagrangian}
  Let $\Delta:\Gamma(TM\oplus E^*)\times\Gamma(E\oplus T^*M)\to\Gamma(E\oplus T^*M)$ be a Dorfman connection
and choose $v, v_1,v_2\in\Gamma(TM\oplus E^*)$ and $\sigma,\sigma_1, \sigma_2\in\Gamma(E\oplus T^*M)$.
Then
 \begin{enumerate}
\item $\left\langle \tilde v_1, \tilde v_2\right\rangle=\ell_{
  \Skew_\Delta(v_1, v_2)}$,
\item $\left\langle \tilde v,
    \sigma^\uparrow\right\rangle=q_E^*\langle v, \sigma\rangle$,
\item $\left\langle \sigma_1^\uparrow, \sigma_2^\uparrow\right\rangle=0$.
\end{enumerate}
\end{theorem}

\begin{proof}
Since the second and third equalities are immediate, we prove only the first one.
We write $v_1=(X,\xi)$, $v_2=(Y,\eta)$ and compute for
any section $e\in\Gamma(E)$:
\begin{align*}
&\left\langle \left(T_me X(m), \dr
    \ell_\xi(e_m)\right)-\Delta_{(X,\xi)}(e,0)^\uparrow(e_m),
\left(T_me Y(m), \dr
    \ell_\eta(e_m)\right)-\Delta_{(Y,\eta)}(e,0)^\uparrow(e_m)\right\rangle\\
=\,&X(m)\langle \eta, e\rangle
-\langle\pr_{T^*M}\Delta_{(Y,\eta)}(e,0), X(m)\rangle- \langle
\eta(m), \pr_E\Delta_{(X,\xi)}(e,0)\rangle\\
&+ Y(m)\langle \xi, e\rangle
-\langle\pr_{T^*M}\Delta_{(X,\xi)}(e,0), Y(m)\rangle- \langle
\xi(m), \pr_E\Delta_{(Y,\eta)}(e,0)\rangle\\
=\,&\left(X\langle \eta, e\rangle
-\langle\Delta_{(Y,\eta)}(e,0), (X,\xi)\rangle+ Y\langle \xi, e\rangle
-\langle\Delta_{(X,\xi)}(e,0), (Y,\eta)\rangle\right)(m)\\
=\,&\langle(e,0), \llbracket} \newcommand{\rb}{\rrbracket v_2, v_1\rb_\Delta+\llbracket} \newcommand{\rb}{\rrbracket v_1, v_2\rb_\Delta\rangle.
\end{align*}
\end{proof}

\begin{corollary}
The bracket $\llbracket} \newcommand{\rb}{\rrbracket\cdot\,,\cdot\rb_\Delta$ associated to a Dorfman connection $\Delta$
is skew-symmetric, if and only if $L_\Delta$ is Lagrangian. The corresponding splitting
\[ TE\oplus T^*E\cong(V\oplus V^\circ)\oplus L_\Delta\]
is then the direct sum of the Dirac structure $V\oplus V^\circ$ and the almost Dirac
structure $L_\Delta$.
\end{corollary}

\begin{proof}
Since the rank of $L_\Delta$ is equal to the dimension of $E$ as a manifold, we have only
to show that $L_\Delta$ is isotropic if and only if $\llbracket} \newcommand{\rb}{\rrbracket\cdot\,,\cdot\rb_\Delta$ is skew-symmetric.
But this is immediate by the preceding theorem.
\end{proof}

\bigskip

Next, we will see how the Dorfman connection
encodes the Courant-Dorfman bracket on linear and core sections.
The next theorem shows how
 integrability of $L_\Delta$ is related to the curvature $R_\Delta$ of the  Dorfman connection.
\begin{theorem}\label{super}
Choose $v, v_1, v_2$ in $\Gamma(TM\oplus E^*)$ and $\sigma, \sigma_1, \sigma_2\in\Gamma(E\oplus T^*M)$.
Then
\begin{enumerate}
\item  $\left[\sigma_1^\uparrow, \sigma_2^\uparrow\right]=0$,
 \item  $\left[\widetilde{v}, \sigma^\uparrow\right]=\left(\Delta_{v}\sigma\right)^\uparrow$,
\item $
\left[\widetilde{v_1},\widetilde{v_2}\right]
=\widetilde{\llbracket} \newcommand{\rb}{\rrbracket v_1, v_2\rb_\Delta}-R_\Delta(v_1, v_2)(\cdot,0)^\uparrow$.
\end{enumerate}
\end{theorem}
The proof of this theorem is quite long and technical, it can be found in
Appendix \ref{proof_of_super}.
\begin{remark}\label{twisted_shit}
\begin{enumerate}
\item If the Courant-Dorfman bracket is twisted by a linear closed $3$-form $H$ over
a map $\bar H:TM\wedge TM\to E^*$ \cite{BuCa12}, then the
bracket $[\tilde v_1,\tilde v_2]$ will be linear over $\llbracket} \newcommand{\rb}{\rrbracket v_1, v_2\rb_{\bar H}=\llbracket} \newcommand{\rb}{\rrbracket v_1, v_2\rb+(0,\bar H(X_1,X_2))$.
Note that the Dorfman connection dual to this bracket
is $\Delta^{\bar H}_v\sigma=\Delta_v\sigma+(0,\langle \bar H(X,\cdot), e\rangle)$.
A more careful study of general exact Courant algebroids \cite{Roytenberg99} over vector bundles and
the corresponding twistings of the Dorfman connections and dull algebroids corresponding
to splittings of $TE\oplus T^*E$ will be done later.
\item The \emph{Courant bracket}, i.e. the anti-symmetric counterpart of
  the Courant-Dorfman bracket,
is given by
\begin{enumerate}
\item  $\left[\sigma_1^\uparrow, \sigma_2^\uparrow\right]_C=0$,
 \item  $\left[\widetilde{v},
     \sigma^\uparrow\right]_C=\left[\widetilde{v},
     \sigma^\uparrow\right]-(0, \frac{1}{2}q_E^*\dr\langle
   v,\sigma\rangle)=\left(\Delta_{v}\sigma-(0, \frac{1}{2}\dr\langle
   v,\sigma\rangle)\right)^\uparrow$,
\item $
\left[\widetilde{v_1},\widetilde{v_2}\right]_C
=\widetilde{\llbracket} \newcommand{\rb}{\rrbracket v_1, v_2\rb_\Delta}-R_\Delta(v_1,
v_2)(\cdot,0)^\uparrow-(0, \frac{1}{2}\dr\ell_{\Skew_\Delta(v_1,v_2)})$.
\end{enumerate}
Our choice of working with non anti-symmetric Courant algebroids
is because of the fact that Dorfman connections describe naturally the
Courant-Dorfman bracket: recall Remark \ref{dorfman_boring}
and Proposition \ref{relation_Delta_L}. Here, we see from (b) that
working with a ``Courant connection'' starting from a splitting
like as we did would yield much more complicated formulas
and axioms for the connection.
This is also why we chose to call the Dorfman
connections after I. Dorfman.
\end{enumerate}
\end{remark}

The following corollary of Theorem \ref{super}
is immediate.

\begin{corollary}
Let $E\to M$ be a vector bundle
and consider a splitting $TE\oplus T^*E=(V\oplus V^\circ)\oplus L$.
Then the horizontal space $L$ is a Dirac structure if and only if the
corresponding dull algebroid $(TM\oplus E^*,\pr_{TM},\llbracket} \newcommand{\rb}{\rrbracket\cdot\,,\cdot\rb_L)$
is a Lie algebroid.
\end{corollary}
We will study more general (non-horizontal) Dirac structures on $E$ in the next section.
Before that, we end this subsection with some examples.

\begin{example}\label{example_easy}
Recall that an ordinary  connection
on a vector bundle $E\to M$ is a splitting $TE\cong V\oplus H_\nabla$.
We have seen in Section \ref{sec_standard_almost_dorfman} that the Dorfman connection
\[\Delta:\Gamma(TM\oplus E^*)\times\Gamma(E\oplus
T^*M)\to\Gamma(E\oplus T^*M),
\]
\[\Delta_{(X,\xi)}(e,\theta)=(\nabla_Xe,\ldr{X}\theta+\langle\nabla_\cdot^*\xi,e\rangle)\]
corresponds to the splitting
\[TE\oplus T^*E\cong (V\oplus V^\circ)\oplus(H_\nabla\oplus H_\nabla)^\circ.
\]
\end{example}

\begin{example}\label{lie_algebroid_dual}
Consider a dull algebroid $(A,\rho, [\cdot\,,\cdot])$ with
\emph{skew-symmetric bracket}. We construct a $TM\oplus A$-connection $\Delta$ on $A^*\oplus T^*M$,
hence corresponding to a splitting $TA^*\oplus T^*A^*=(V\oplus V^\circ)\oplus L_\Delta$ of the Pontryagin bundle of $A^*$.
Take any connection $\nabla:\mathfrak{X}} \newcommand{\dr}{\mathbf{d}(M)\times \Gamma(A)\to\Gamma(A)$
and recall the definition of the basic connection $\nabla^{\rm bas}:\Gamma(A)\times\Gamma(A)\to\Gamma(A)$
associated to $\nabla$ and the  dull algebroid structure on $A$:
\[\nabla_a^{\rm bas}b=[a,b]+\nabla_{\rho(b)}a
\]
for all $a,b\in\Gamma(A)$.
The Dorfman connection
\[\Delta: \Gamma(TM\oplus A)\times\Gamma(A^*\oplus T^*M)\to
\Gamma(A^*\oplus T^*M)\]
is defined by
\[\Delta_{(X,a)}(\xi,\theta)=\left(\langle \xi, \nabla_\cdot^{\rm
    bas}a\rangle+\nabla_X^*\xi-\rho^*\langle \nabla_\cdot a,
  \xi\rangle,
\ldr{X}\theta+ \langle \nabla_\cdot a, \xi\rangle
\right).
\]
The bracket $\llbracket} \newcommand{\rb}{\rrbracket\cdot\,,\cdot\rb_\Delta$ on sections of $TM\oplus A$ is then
given by
\[\llbracket} \newcommand{\rb}{\rrbracket(X,a), (Y,b)\rb_\Delta=\left([X,Y],
  \nabla_Xb-\nabla_Ya+\nabla_{\rho(b)}a-\nabla_{\rho(a)}b+[a,b]
\right).
\]
Since it is skew-symmetric, the horizontal space $L_\Delta$ is in this
case Lagrangian.
The projection $\pr_{TM}$ intertwines obviously this bracket with the
Lie bracket of vector fields.
The curvature of this Dorfman connection is given by
\begin{align}
\langle R_\Delta((X,a),(Y,b))(\xi,\theta), (Z,c)\rangle
=&- \langle \llbracket} \newcommand{\rb}{\rrbracket(X,a),\llbracket} \newcommand{\rb}{\rrbracket(Y,b), (Z,c)\rb_\Delta\rb_\Delta+{\rm c.p.}, (\xi,\theta)\rangle\nonumber\\
=&-\langle \bigl(R_\nabla(X,Y)c- R_\nabla(\rho(a),Y)c\bigr) +{\rm c.p.}, \xi\rangle\label{curvature_poisson}\\
&-\langle \bigl(R_\nabla(\rho(a),\rho(b))c-R_\nabla(X,\rho(b))c\bigr) +{\rm c.p.}, \xi\rangle\nonumber\\
&-\langle \bigl(R^{\rm bas}_\nabla(a,b)Z- R^{\rm bas}_\nabla(a,b)\rho(c)\bigr) +{\rm c.p.}, \xi\rangle\nonumber\\
&-\langle [a,[b,c]]+[b,[c,a]]+[c,[a,b]], \xi\rangle.\nonumber
\end{align}
The proof of this formula is a rather long, but straightforward computation that is ommitted here.

We will see in the next subsection the signification of this example
in terms of the linear almost Poisson structure  defined on $A^*$ by the skew-symmetric dull algebroid structure.
\end{example}

\begin{example}\label{IM_2_form}
Consider a vector bundle $E\to M$ endowed with a
vector bundle morphism $\sigma:E\to T^*M$ over the identity and
a connection $\nabla:\mathfrak{X}} \newcommand{\dr}{\mathbf{d}(M)\times\Gamma(E)\to\Gamma(E)$.
Define the Dorfman connection
\[\Delta: \Gamma(TM\oplus E^*)\times\Gamma(E\oplus T^*M)\to
\Gamma(E\oplus T^*M)
\]
by
\[\Delta_{(X,\xi)}(e,\theta)=(\nabla_Xe,
\ldr{X}(\theta-\sigma(e))+\langle \nabla_\cdot^*(\sigma^*X+\xi), e\rangle
+\sigma(\nabla_Xe)).
\]
The bracket $\llbracket} \newcommand{\rb}{\rrbracket\cdot\,,\cdot\rb_\Delta$ on sections
of $TM\oplus E^*$ is here given by
\[\llbracket} \newcommand{\rb}{\rrbracket (X,\xi), (Y,\eta)\rb_\Delta=([X,Y], \nabla_X^*(\eta+\sigma^*Y)-\nabla_Y^*(\xi+\sigma^*X)-\sigma^*[X,Y]).
\]
In this case also, $L_\Delta$ is  Lagrangian.

Here also, we give the curvature of the Dorfman connection in terms of the
Jacobiator of the associated bracket:
\begin{equation}\label{curvature_sigma}
\llbracket} \newcommand{\rb}{\rrbracket(X,\xi), \llbracket} \newcommand{\rb}{\rrbracket(Y,\eta), (Z,\gamma)\rb_\Delta\rb_\Delta+{\rm c.p.}= \Bigl(0, R_{\nabla^*}(X,Y)(\gamma+\sigma^*Z)+{\rm c.p.}\Bigl).
\end{equation}
We will see in the next section how this Dorfman connection is related
to the $2$-form $\sigma^*\omega_{\rm can}\in\Omega^2(E)$, where
$\omega_{\rm can}$ is the canonical symplectic form on $T^*M$.
\end{example}

\subsection{Dirac structures and Dorfman connections}
In  this subsection, we will consider  sub- double vector bundles
\begin{equation*}
\begin{xy}
\xymatrix{
D\ar[d]\ar[r]&U\ar[d]\\
E\ar[r]&M
}\end{xy}
\qquad \text{ of } \qquad
\begin{xy}
\xymatrix{
TE\oplus T^*E\ar[d]\ar[r]&TM\oplus E^*\ar[d]\\
E\ar[r]&M
}\end{xy}
\end{equation*}
The intersection of such a sub- double vector bundle $D$ with the vertical
space $V\oplus V^\circ$ always has constant rank on $E$
and there is a subbundle $K\subseteq E\oplus T^*M$
such that $D\cap (V\oplus V^\circ)$
is spanned over $E$ by the sections $k^\uparrow$
for all $k\in \Gamma(K)$. To see  that, use for instance
\cite{Mackenzie05}.
We will call $K$ the \textbf{core} of $D$.
The following proposition
follows from this observation.
\begin{proposition}
Let $E$ be a vector bundle endowed with a sub-double vector bundle
$D\subseteq TE\oplus T^*E$ over $U\subseteq TM\oplus E^*$ and with
core $K\subseteq E\oplus T^*M$.
Then there exists a   Dorfman connection $\Delta$
such that $D$ is spanned by the sections $k^\uparrow$,
for all $k\in\Gamma(K)$ and $\widetilde{u}$ for all $u\in\Gamma(U)$.
\end{proposition}
The   Dorfman connection $\Delta$ is then said to be
\emph{adapted} to $D$. Conversely,
given a Dorfman connection and two subbundles
$U\subseteq TM\oplus E^*$ and $K\subseteq E\oplus T^*M$,
we call $D_{U,K,\Delta}$ the sub-double vector bundle
that is spanned by $\sigma^\uparrow$,
for all $\sigma\in\Gamma(K)$ and $\widetilde{u}$ for all $u\in\Gamma(U)$.

\begin{definition}
Two Dorfman connections $\Delta,\Delta'$ are said to be $(U,K)$-\emph{equivalent}
if $(\Delta-\Delta')(\Gamma(U)\times\Gamma(E\oplus 0)\subseteq\Gamma(K)$.
\end{definition}

The following proposition shows
that this defines an $(U,K)$-equivalence relation on the set of   Dorfman connections.
We will write $[\Delta]_{U,K}$, or simply $[\Delta]$ since there will
never be a risk of confusion, for the $(U,K)$-class of the Dorfman connection $\Delta$.
The triple $(U,K,[\Delta])$ will be called a VB-triple in the following.
By the next proposition, VB-triples
are in one-one correspondence
with sub-double vector bundles of $TE\oplus T^*E\to E$.
\begin{proposition}
Choose two   Dorfman connections
$\Delta, \Delta'$ and assume that $\Delta$ is adapted to $D$.
Then $\Delta'$ is adapted to $D$ if and only if
$\Delta$ and $\Delta'$ are $(U,K)$-equivalent.
\end{proposition}

\begin{proof}
Assume that $\Delta$ is adapted to $D$. Then $D$ is spanned by the sections $\widetilde{u}^\Delta$ and
$\sigma^\uparrow$ for all $\sigma\in\Gamma(K)$ and $u\in\Gamma(U)$.
If $\Delta$ and $\Delta'$ are $(U,K)$-equivalent,
we have $\widetilde u^\Delta-\widetilde u^{\Delta'}=k^\uparrow$ for some $k\in\Gamma(K)$.
This implies immediately that $\Delta'$ is adapted to $D$.
The converse implication can be proved in a similar manner.
\end{proof}

The following theorem follows immediately from the results in the
preceding subsection.
\begin{theorem}
Let $D$ be a sub- double vector bundle of $TE\oplus T^*E$ over
$U\subseteq TM\oplus E^*$ and $K\subseteq E\oplus T^*M$ and choose
a
Dorfman connection $\Delta$ that is adapted to $D$.
Then
\begin{enumerate}
\item $D$ is isotropic if and only if $\Skew_\Delta\an{U\otimes U}=0$ and $K\subseteq U^\circ$.
\item $D$ is Lagrangian if and only if  $\Skew_\Delta\an{U\otimes U}=0$
 and $K=U^\circ$.
\item $\Gamma(D)$ is closed under the Courant-Dorfman bracket
if and only if
\begin{enumerate}
\item $\Delta_uk\in\Gamma(K)$ for all $u\in \Gamma(U)$,
$k\in\Gamma(K)$,
\item $\llbracket} \newcommand{\rb}{\rrbracket\Gamma(U), \Gamma(U)\rb_\Delta\subseteq \Gamma(U)$,
\item
$R_\Delta\Bigl(U\otimes U\otimes(E\oplus T^*M)\Bigr)\subseteq K$.
\end{enumerate}
\end{enumerate}
\end{theorem}

\begin{proof}
This is an immediate corollary of the results in the preceding subsection,
using the fact that $R_\Delta\Bigl((TM\oplus E^*)\otimes (TM\oplus E^*)\otimes(0\oplus T^*M)\Bigr)=0$.
(To see this, use Proposition \ref{curvature_tensor} and the fact that the anchor is $\pr_{TM}$.)
\end{proof}

\begin{corollary}\label{Dirac_triples}
Let $D$ be a sub- double vector bundle of $TE\oplus T^*E$ over
$U\subseteq TM\oplus E^*$ and $K\subseteq E\oplus T^*M$ and choose
a Dorfman connection $\Delta$ that is adapted to $D$.
Then
\begin{enumerate}
\item $D$ is an isotropic subalgebroid of $TE\oplus T^*E\to E$ if and only if
\begin{enumerate}
\item $U\subseteq K^\circ$,
\item $\Delta_uk\in\Gamma(K)$ for all $u\in \Gamma(U)$,
$k\in\Gamma(K)$,
\item $(U, \pr_{TM}\an{U}, \llbracket} \newcommand{\rb}{\rrbracket\cdot\,, \cdot\rb_\Delta\an{\Gamma(U)\times\Gamma(U)})$ is a skew-symmetric dull algebroid.
\item the induced  Dorfman connection
\[\bar\Delta:\Gamma(U)\times\Gamma((E\oplus T^*M)/K)\to \Gamma((E\oplus T^*M)/K)
\]
is flat.
\end{enumerate}
 \item $D$ is a Dirac structure if and only if
\begin{enumerate}
\item $U=K^\circ$,
\item $\Delta_uk\in\Gamma(K)$ for all $u\in \Gamma(U)$,
$k\in\Gamma(K)$,
\item $(U,\pr_{TM}\an{U},
  \llbracket} \newcommand{\rb}{\rrbracket\cdot\,,\cdot\rb_\Delta\an{\Gamma(U)\times\Gamma(U)})$ is a Lie algebroid.
\end{enumerate}
\end{enumerate}
\end{corollary}

Note that in the second situation, the induced Dorfman connection
$\bar\Delta$
 is just  the Lie derivative
\[\ldr{}=\bar\Delta:\Gamma(U)\times\Gamma(U^*)\to \Gamma(U^*),
\]
which flatness is equivalent to the fact that the restriction of
$\llbracket} \newcommand{\rb}{\rrbracket\cdot\,,\cdot\rb_\Delta$ to $\Gamma(U)$ satisfies the Jacobi-identity.

\begin{remark}
\begin{enumerate}
\item Note that $K=U^\circ$ and $\Delta_uk\in\Gamma(K)$ for all $u\in\Gamma(U)$,
$k\in\Gamma(K)$ imply together that the dull bracket
restricts to a bracket on $\Gamma(U)$.
\item Using the following
Proposition \ref{brackets_equal_on_U} and Remark \ref{dorfman_boring}, it can be checked directly that if the conditions in (2) of Corollary \ref{Dirac_triples}
are satisfied for $\Delta$, then they are also satisfied for any $\Delta'$ that is $(U,K)$-equivalent
to $\Delta$.
\item We will say that  $(U,K,[\Delta])$ is a Dirac triple
if the corresponding sub-double vector bundle $D_{(U,K,[\Delta])}$ is a Dirac structure on $E$.
By the considerations above, we find that Dirac sub- double vector bundles
of $TE\oplus T^*E\to E$ are in one-one correspondence with Dirac triples.
\end{enumerate}
\end{remark}

\begin{proposition}\label{brackets_equal_on_U}
Let $E\to M$ be a vector bundle and choose a VB-triple $(U,K,[\Delta]_{U,K})$
such that $U=K^\circ$.
Then for any two representatives $\Delta,\Delta'\in [\Delta]_{U,K}$, we have
\[\llbracket} \newcommand{\rb}{\rrbracket u_1, u_2\rb_\Delta=\llbracket} \newcommand{\rb}{\rrbracket u_1, u_2\rb_{\Delta'}
\]
for all $u_1,u_2\in\Gamma(U)$.
\end{proposition}

\begin{proof}
Since $\pr_{TM}\llbracket} \newcommand{\rb}{\rrbracket u_1, u_2\rb_\Delta=[\pr_{TM}u_1, \pr_{TM}u_2]=\pr_{TM}\llbracket} \newcommand{\rb}{\rrbracket u_1, u_2\rb_{\Delta'}$,
we need only to check that
\[\langle \llbracket} \newcommand{\rb}{\rrbracket u_1, u_2\rb_\Delta, (e,0)\rangle=\langle\llbracket} \newcommand{\rb}{\rrbracket u_1, u_2\rb_{\Delta'}, (e,0)\rangle
\]
for all $e\in\Gamma(E)$.
But this is immediate by the hypothesis, the duality of $\Delta$ and $\llbracket} \newcommand{\rb}{\rrbracket\cdot\,,\cdot\rb_\Delta$ and the definition of $(U,K)$-equivalence.
\end{proof}

\bigskip

We conclude this section with a study of our recurrent examples.

\begin{example}\label{foliation_example}
In the situation of Example \ref{example_easy}, choose
two subbundles $F_M\subseteq TM$ and $C\subseteq E$.
Set $U:=F_M\oplus C^*$ and $K:=C\oplus F_M^\circ=U^\circ$.
The sub-double vector bundle $D_{U,K,\Delta}$
corresponding to $U$, $K$ and the standard Dorfman connection
associated to $\nabla$ is then the direct sum of a subbundle $F_E\subseteq TM$,
with  $C_E\subseteq T^*E$.
Since $U=K^\circ$, we get immediately that $C_E=F_E^\circ$ and
$D_{(U,K,[\Delta])}$ is the trivial almost Dirac structure $F_E\oplus F_E^\circ$.
An application of Corollary \ref{Dirac_triples} to this situation yields that
 $F_E\oplus F_E^\circ$ is Dirac if and only if
\begin{enumerate}
\item $F_M$ is involutive,
\item $\nabla_Xc\in\Gamma(C)$ for all $X\in\Gamma(F_M)$ and $c\in\Gamma(C)$
and
\item the induced connection $\tilde \nabla:\Gamma(F_M)\times \Gamma(E/C)\to\Gamma(E/C)$
is flat.
\end{enumerate}
 Since $F_E\oplus F_E^\circ$ is Dirac if and only if $F_E\subseteq TE$ is involutive,
we recover one of the results in \cite{JoOr12}.
\end{example}

\begin{example}\label{lie_algebroid_dual_2}
In the situation of  Example \ref{lie_algebroid_dual}, consider
$U=\graphe(\rho:A\to TM)$ and $K=\graphe(-\rho^*:T^*M\to
A^*)=U^\circ$.
A straightforward computation shows that
$\Delta_{(\rho(a), a)}(-\rho^*(\omega),\omega)=(-\rho^*({\nabla^{\rm
    bas}_a}^*\omega), {\nabla^{\rm bas}_a}^*\omega)\in\Gamma(K)$ for all $
a\in\Gamma(A)$ and $\omega\in\Omega^1(M)$.
Furthermore, we have
\[ \llbracket} \newcommand{\rb}{\rrbracket(\rho(a),a), (\rho(b),b)\rb_\Delta=(\rho([a,b]), [a,b])
\]
for all $a,b\in\Gamma(A)$, which shows that $(U,\pr_{TM},
\llbracket} \newcommand{\rb}{\rrbracket\cdot\,,\cdot\rb_\Delta)$
is a Lie algebroid if and only if $A$ is a Lie algebroid.
We have:
\begin{align*}
\bar\Delta_{(\rho(a),a)}\overline{(\xi,0)}&=\overline{\left(\langle \xi, \nabla_\cdot^{\rm
    bas}a\rangle+\nabla_{\rho(a)}^*\xi-\rho^*\langle \nabla_\cdot a,
  \xi\rangle, \langle \nabla_\cdot a, \xi\rangle
\right)}\\
&=\overline{\left(\langle \xi, \nabla_\cdot^{\rm
    bas}a\rangle+\nabla_{\rho(a)}^*\xi, 0
\right)}=\overline{\left(\ldr{a}\xi, 0
\right)}.
\end{align*}
Finally, the right-hand side of
\eqref{curvature_poisson}
vanishes for $(\rho(a), a), (\rho(b), b), (\rho(c),c)\in\Gamma(U)$ and
arbitrary $(\xi,\theta)$  if and only if $A$ is a Lie algebroid.

Hence, we find that the sub-double vector bundle $D$ of $TA^*\oplus T^*A^*$
associated to $U, K$ and $\Delta$ is an almost Dirac structure on
$A^*$ and a Dirac structure if and only if $A$ is a Lie algebroid.
The vector bundle $D\to A^*$ is in fact the graph of the vector bundle morphism
\[\pi_{A}^\sharp:T^*A^*\to TA^*
\]
associated to the linear amost Poisson structure defined on $A^*$ by the skew-symmetric dull
algebroid structure on $A$.
Indeed, $D$ is spanned by the sections $k^\uparrow$ for $k\in
\Gamma(K)$ and $\widetilde{u}$ for $u\in\Gamma(U)$,
or, equivalently, by the sections
\[ (-\rho^*\theta^\uparrow, q_{A^*}^*\theta)
\]
for $\theta\in\Omega^1(M)$
and
\[ (\widetilde{\rho(a)}, \tilde a)
\]
for $a\in \Gamma(A)$, where
\begin{align*}
\widetilde{\rho(a)}(\xi_m)&=T_m\xi(\rho(a)(m))-\left.\frac{d}{dt}\right\an{t=0}\xi_m+t\cdot\ldr{a}\xi(m)\\
\tilde a(\xi_m)&=\dr_{\xi_m} l_a.
\end{align*}
But by Appendix \ref{appendix_linear_Poisson}, these are exactly the
sections $(\pi_A^\sharp(q_{A^*}^*\theta), q_{A^*}^*\theta)$ and
$(\pi_A^\sharp(\dr l_a), \dr l_a)$.
\end{example}

\begin{example}\label{IM_2_form_2}
Consider, in the situation of Example \ref{IM_2_form},
$U=:\graphe(-\sigma^*:TM\to E^*)$ and $K=\graphe(\sigma:E\to T^*M)$.
Then $U=K^\circ$ by definition and
since
\[\Delta_{(X,-\sigma^*X)}(e,\sigma(e))=(\nabla_Xe,\sigma(\nabla_Xe))
\]
by definition, we find that $\Delta_uk\in\Gamma(K)$ for all
$u\in\Gamma(U)$ and $k\in\Gamma(K)$.
Furthermore, we have $\llbracket} \newcommand{\rb}{\rrbracket (X,-\sigma^*X),
(Y,-\sigma^*Y)\rb_\Delta=([X,Y], -\sigma^*[X,Y])$ for all
$X,Y\in\mathfrak{X}} \newcommand{\dr}{\mathbf{d}(M)$ and $U$ is a Lie algebroid (isomorphic to $TM$ with the
Lie bracket of vector fields). Alternatively, the  Jacobiator in \eqref{curvature_sigma} is easily seen to
vanish on this type of sections.
This shows that the double vector subbundle $D\subseteq TE\oplus T^*E$
defined
by $U,K$ and $\Delta$ is a Dirac structure.

By the considerations in Appendix \ref{pullback_canonical_symplectic},
$D$ is the graph of the vector bundle morphism $TE\to T^*E$
defined by the closed $2$-form $\sigma^*\omega_{\rm can}$.
\end{example}

\begin{example}
We now combine Examples   \ref{foliation_example} and
\ref{IM_2_form_2} to recover an example in \cite{JoRa12a}.

We consider the vector bundle $T^*M\to M$ endowed
with a $TM$-connection $\nabla$
and the Dorfman connection
\[\Delta:\Gamma(TM\oplus TM)\times \Gamma(T^*M\oplus T^*M)\to
\Gamma(T^*M\oplus T^*M),\]
\[\Delta_{(X,Y)}(\theta,\omega)=(\nabla_X\theta,
\ldr{X}(\omega-\theta)+\langle \nabla^*_\cdot(X+Y),\omega\rangle+\nabla_X\theta).
\]
Consider a subbundle $F\subseteq TM$ and $U:=\{(x,-x)\mid x\in
F\}\subseteq TM\oplus TM$. The annihilator $K=U^\circ$
is then given by $K=\{(\alpha,\beta)\in T^*\oplus T^*M\mid
\alpha-\beta\in F^\circ\}$.

Note that by Example \ref{IM_2_form}, the dull bracket
on $TM\oplus TM$ is skew-symmetric. In fact, it is easy to see that
its restriction to $U$  is just the Lie bracket of vector fields
\[ \llbracket} \newcommand{\rb}{\rrbracket (X,-X), (Y,-Y)\rb_\Delta=([X,Y],-[X,Y])
\]
for all $X,Y\in\Gamma(F)$.
Hence, we know already that the sub-double vector bundle
$D_{(U,K,[\Delta])}$ is an almost Dirac structure on $T^*M$.
An easy computation using Appendix \ref{pullback_canonical_symplectic}
yields that
\[D_{(U,K,[\Delta])}
(\alpha)
=\{(x_\alpha, \omega_{\rm can}^\flat(x_\alpha)+\theta_\alpha)\mid
x_\alpha\in\mathcal F(\alpha), \theta_\alpha\in \mathcal F^\circ(\alpha)
\}
\]
for all $\alpha\in T^*M$,
where $\mathcal F=(Tc_M)\inv(F)$.
Assume that  $M$ is the configuration space of a nonholonomic mechanical
system and $F$ the constraints distribution. If $L$ is the Lagrangian
of the system, then the pullback of the Dirac structure that we find
to the contraints submanifold $\mathbb FL(F)\subseteq T^*M$
is one of the frameworks proposed in \cite{JoRa12a} for the study of
the nonholonomic system.
\end{example}

\section{  Dorfman connections and Lie algebroids}
We consider in this section a Lie algebroid $(A\to M, \rho, [\cdot\,,\cdot])$ and a   Dorfman connection
\[\Delta: \Gamma(TM\oplus A^*)\times\Gamma(A\oplus T^*M)\to \Gamma(A\oplus T^*M)
\]
with corresponding dull bracket $\llbracket} \newcommand{\rb}{\rrbracket\cdot\,,\cdot\rb_{\Delta}$ and
anchor $\pr_{TM}$ on $TM\oplus A^*$.

We will work here with the map
\[\Omega:\Gamma(TM\oplus A^*)\times\Gamma(A)\to\Gamma(A\oplus T^*M)
\]
\[\Omega_{(X,\xi)}a=\Delta_{(X,\xi)}(a,0)-(0,\dr\langle\xi, a\rangle).
\]
 $\Omega$ has the following properties
\begin{enumerate}
\item $\Omega_{\varphi (X,\xi)}a=\varphi \Omega_{(X,\xi)}a$,
\item $\Omega_{(X,\xi)}(\varphi a)=\varphi \Omega_{(X,\xi)}a+X(\varphi )(a,0)-\langle \xi,
  a\rangle (0,\dr \varphi )$
\end{enumerate}
for all $\varphi\in C^\infty(M)$, $a\in\Gamma(A)$ and $(X,\xi)\in\Gamma(TM\oplus A^*)$.

For each $a\in\Gamma(A)$, we have two derivations over $\rho(a)$:
\[\ldr{a}:\Gamma(A\oplus T^*M)\to\Gamma(A\oplus T^*M),\]
\[\ldr{a}(b,\theta)=([a,b], \ldr{\rho(a)}\theta)\]
and
\[\ldr{a}:\Gamma(TM\oplus A^*)\to\Gamma(TM\oplus A^*)\]
\[\ldr{a}(X,\xi)=([\rho(a),X], \ldr{a}\xi).\]
Note that
\begin{align*}
\ldr{\varphi a}(b,\theta)=\varphi \ldr{a}(b,\theta)
+(-\rho(b)(\varphi)a,\langle \theta, \rho(a)\rangle \dr\varphi).
\end{align*}

Finally, note that there is a ``Dorfman-like'' bracket
$[\cdot\,,\cdot]_D$ on sections of $A\oplus T^*M$:
\[  [(a,\theta), (b,\omega)]_D=([a,b], \ldr{\rho(a)}\omega-\ip{\rho(b)}\dr\theta)
\]
for $(a,\theta), (b,\omega)\in\Gamma(A\oplus T^*M)$.
We have
\begin{equation}\label{Dorfman_skew-sym}
 [\sigma_1, \sigma_2]_D+[\sigma_2, \sigma_1]_D=(0, \dr\langle \sigma_1, (\rho, \rho^*)\sigma_2\rangle)\footnote{We will write
$\langle \cdot\,,\cdot\rangle_D$ for this pairing of $A\oplus T^*M$ with itself, i.e. $\langle\sigma_1,\sigma_2\rangle_D=\langle \sigma_1, (\rho, \rho^*)\sigma_2\rangle$ for all
$\sigma_1,\sigma_2\in\Gamma(A\oplus T^*M)$.}
\end{equation}
and the Jacobi identity in Leibniz form
\begin{equation}\label{Dorfman_Jacobi}
[\sigma_1, [\sigma_2, \sigma_3]] = [[\sigma_1, \sigma_2], \sigma_3] + [\sigma_2, [\sigma_1, \sigma_3]]
\end{equation}
for all $\sigma_1,\sigma_2,\sigma_3\in\Gamma(A\oplus T^*M)$.
Note that $(A\oplus T^*M, \rho\circ\pr_{A}, [\cdot\,,\cdot]_D,\langle \cdot\,,\cdot\rangle_D)$
is not a Courant algebroid because $\langle \cdot\,,\cdot\rangle_D$ is in general degenerate.

\subsection{The basic connections associated to $\Delta$}
\begin{proposition}\label{prop_def_basic_connections}
The two maps
\begin{align*}
\nabla^{\rm bas}: \Gamma(A)\times\Gamma(TM\oplus A^*)&\to \Gamma(TM\oplus A^*)\\
\nabla^{\rm bas}_a(X,\xi
&=(\rho,\rho^*)(\Omega_{(X,\xi)}a)+\ldr{a}(X,\xi)
\end{align*}
and
\begin{align*}
\nabla^{\rm bas}: \Gamma(A)\times\Gamma(A\oplus T^*M)&\to \Gamma(A\oplus T^*M)\\
\nabla^{\rm
  bas}_a(b,\theta
&=\Omega_{(\rho,\rho^*)(b,\theta)}a+\ldr{a}(b,\theta)
\end{align*}
are connections in the usual sense.
\end{proposition}

\begin{proof}
Recall the notation $\dr_{A^*} \varphi :=\rho^*\dr \varphi $ for all $\varphi \in C^\infty(M)$.
We compute for $\varphi\in C^\infty(M)$, $a, b\in\Gamma(A)$, $\xi\in\Gamma(A^*)$,
$\theta\in\Omega^1(M)$ and $X\in\mathfrak{X}} \newcommand{\dr}{\mathbf{d}(M)$:
\begin{align*}
\nabla^{\rm bas}_{\varphi a}(X,\xi)&=(\rho,\rho^*)(\Omega_{(X,\xi)}(\varphi a))+([\varphi \rho(a),X], \ldr{\varphi a}\xi)\\
&=\varphi \cdot \nabla^{\rm bas}_a(X,\xi)+(\rho,\rho^*)(X(\varphi ) a,- \langle\xi, a\rangle\dr \varphi  )+ (-X(\varphi )\rho(a), \langle \xi, a\rangle \dr_{A}^* \varphi )\\
&=\varphi \cdot \nabla^{\rm bas}_a(X,\xi),\\
\nabla^{\rm bas}_{\varphi a}(b,\theta)&=\Omega_{(\rho,\rho^*)(b,\theta)}(\varphi a)
+\ldr{\varphi a}(b,\theta)\\
&=\varphi \cdot\nabla^{\rm bas}_a(b,\theta)+ (\rho(b)(\varphi )\cdot a, 0)-\langle
\theta, \rho(a)\rangle(0,\dr \varphi )- (\rho(b)(\varphi )a,0)+\langle\theta, \rho(a)\rangle(0,\dr \varphi )\\
&=\varphi \cdot\nabla^{\rm bas}_a(b,\theta),\\
\nabla^{\rm bas}_{a}\left(\varphi (X,\xi)\right)&=(\rho,\rho^*)(\Omega_{\varphi (X,\xi)}a)+\ldr{a}(\varphi (X, \xi))=\varphi \nabla^{\rm bas}_{a}(X,\xi)
+\rho(a)(\varphi )(X, \xi)\\
\nabla^{\rm bas}_{a}\left(\varphi (b,\theta)\right)&=\Omega_{(\rho,\rho^*)\varphi (b,\theta)}a+\ldr{a}(\varphi (b,\theta))
=\varphi \nabla^{\rm bas}_{a}(b,\theta)+\rho(a)(\varphi )(b, \theta).
\end{align*}
\end{proof}

The following proposition is easily checked, and shows that the connections are in general not dual to each other.
\begin{proposition}\label{prop_basic_connections}
We have
\begin{equation}\label{not_dual}
\langle \nabla_a^{\rm bas}v, \sigma\rangle
+\langle v, \nabla_a^{\rm bas}\sigma\rangle
=\rho(a)\langle v, \sigma\rangle-\langle \Skew_\Delta(v,
(\rho,\rho^*)\sigma), a\rangle
\end{equation}
and
\begin{equation}\label{intertwined}
\nabla_a^{\rm bas}(\rho,\rho^*)\sigma=(\rho,\rho^*)\nabla_a^{\rm
    bas}\sigma
\end{equation}
for all $a\in \Gamma(A)$, $v\in\Gamma(TM\oplus A^*)$ and $\sigma\in\Gamma(A\oplus T^*M)$.
\end{proposition}

\begin{definition}
The connections in Proposition \ref{prop_def_basic_connections} will be called
the \textbf{basic connections} associated to $\Delta$.
We will sometimes also write $\nabla_\sigma^{\rm bas}:=\nabla_{\pr_A\sigma}^{\rm bas}$
for a section $\sigma\in\Gamma(A\oplus T^*M)$.
\end{definition}

\begin{proposition}\label{prop_tensoriality}
The  map
\[R_\Delta^{\rm bas}:\Gamma(A)\times\Gamma(A)\times\Gamma(TM\oplus A^*)\to\Gamma(A\oplus T^*M)
\]
given by
\begin{align*}
R_\Delta^{\rm bas}(a,b)(X,\xi)=-\Omega_{(X,\xi)}[a,b] +\ldr{a}\left(\Omega_{(X,\xi)}b\right)-\ldr{b}\left(\Omega_{(X,\xi)}a\right)
                                                    + \Omega_{\nabla^{\rm bas}_b(X,\xi)}a-\Omega_{\nabla^{\rm bas}_a(X,\xi)}b.
\end{align*}
is  tensorial, i.e. a section of \[A^*\otimes
  A^*\otimes (A\oplus T^*M)\otimes (A\oplus T^*M).\]
\end{proposition}

\begin{definition}
The tensor $R_\Delta^{\rm bas}$ will be called the \textbf{basic curvature} associated to $\Delta$.
\end{definition}

\begin{proof}[Proof of Proposition \ref{prop_tensoriality}]
We compute for $v=(X,\xi)\in\Gamma(TM\oplus A^*)$ and $a,b\in\Gamma(A)$:
\begin{align*}
  R_\Delta(\varphi a,
  b)v=&-\Omega_{v}\left(\varphi[a,b]-\rho(b)(\varphi)a\right)
  +\ldr{\varphi a}\Omega_{v}b\\
&-\ldr{b}\left(\varphi\Omega_{v}a+X(\varphi)a-\langle
    \xi, a\rangle(0,\dr \varphi)\right)+ \Omega_{\nabla^{\rm bas}_bv}(\varphi
  a)-\Omega_{\varphi\nabla^{\rm bas}_{a}v}b\\
=&\varphi R_\Delta(a, b)v-X(\varphi)([a,b],0)+\langle \xi,
[a,b]\rangle
(0,\dr\varphi)+\rho(b)(\varphi)\Omega_{v}a\\
&+X(\rho(b)(\varphi))(a,0)-\langle
\xi, a\rangle (0, \dr (\rho(b)(\varphi)))\\
&  -\left((\rho\circ\pr_A)(\Omega_vb)(\varphi)a,-\langle
  \Omega_vb, (a,0)\rangle\dr\varphi\right)\\
&-\rho(b)(\varphi)\Omega_{v}a-\rho(b)(X(\varphi))(a,0)-X(\varphi)([b,a],0)\\
&+\rho(b)\langle
    \xi, a\rangle(0,\dr \varphi)+\langle
    \xi, a\rangle(0,\ldr{\rho(b)}\dr \varphi)\\
  & +((\rho\circ\pr_A)(\Omega_vb)+[\rho(b),X])(\varphi)
  (a,0)-\langle \Omega_vb, (a,0)\rangle(0, \dr\varphi)\\
&-\langle
  \ldr{b}\xi,a\rangle(0, \dr\varphi)\\
=&\varphi R_\Delta(a, b)v,\\
 R_\Delta(a, b)(\varphi v)=&\varphi R_\Delta(a, b)v
-\rho(a)(\varphi)\Omega_{v}b+\rho(b)(\varphi)\Omega_{v}a-\rho(b)(\varphi)\Omega_{v}a+\rho(a)(\varphi)\Omega_{v}b\\
=&\varphi R_\Delta(a, b)v.
\end{align*}
\end{proof}

\begin{proposition}\label{basic_curvatures}
The basic curvature has the following properties:
\begin{enumerate}
\item $R_{\nabla^{\rm bas}}=R^{\rm bas}_\Delta\circ
  (\rho,\rho^*)$,
\item $R_{\nabla^{\rm bas}}=(\rho,\rho^*)\circ R^{\rm bas}_\Delta$.
\end{enumerate}
\end{proposition}

\begin{proof}
\begin{enumerate}
\item For $\sigma\in\Gamma(A\oplus T^*M)$ and $a,b\in\Gamma(A)$,
  we have
\begin{align*}
R_\Delta(a,b)((\rho,\rho^*)\sigma)=&-\Omega_{(\rho,\rho^*)\sigma}[a,b] +\ldr{a}\left(\Omega_{(\rho,\rho^*)\sigma}b\right)-\ldr{b}\left(\Omega_{(\rho,\rho^*)\sigma}a\right)\\
                                               &     + \Omega_{\nabla^{\rm bas}_b(\rho,\rho^*)\sigma}a-\Omega_{\nabla^{\rm bas}_a(\rho,\rho^*)\sigma}b\\
=&-\Omega_{(\rho,\rho^*)\sigma}[a,b] -\ldr{[a,b]}\sigma+\ldr{a}\left(\Omega_{(\rho,\rho^*)\sigma}b+\ldr{b}\sigma\right)-\ldr{b}\left(\Omega_{(\rho,\rho^*)\sigma}a+\ldr{a}\sigma\right)\\
                                               &     +                                             \Omega_{\nabla^{\rm
                                                   bas}_b(\rho,\rho^*)\sigma}a-\Omega_{\nabla^{\rm
                                                   bas}_a(\rho,\rho^*)\sigma}b\\
=&-\nabla_{[a,b]}^{\rm bas}\sigma+\nabla_a^{\rm bas}\nabla_b^{\rm
  bas}\sigma-\nabla_b^{\rm bas}\nabla_a^{\rm bas}\sigma=R_{\nabla^{\rm bas}}(a,b)\sigma.
\end{align*}
\item The second equality is shown in the same manner.
\end{enumerate}
\end{proof}

\subsection{The Lie algebroid structure on  $TA\oplus T^*A\to TM\oplus A^*$}
Consider a Lie algebroid $A$ and a    Dorfman connection
\[\Delta:\Gamma(TM\oplus A^*)\times\Gamma(A\oplus T^*M)\to \Gamma(A\oplus T^*M).
\]
Then, for any section $a\in\Gamma(A)$, we define
\[\Sigma_a\in\Gamma_{TM\oplus A^*}(TA\oplus T^*A)
\]
 by
\begin{align*}
\Sigma_a(v_m,\xi_m)&=(T_mav_m,\dr_{a_m}
\ell_\xi)-\Delta_{(X,\xi)}(a,0)^\uparrow(a_m)
\end{align*}
for any choice of section $(X,\xi)\in\Gamma(TM\oplus A^*)$ such that
$(X,\xi)(m)=(v_m,\xi_m)$.
That is, we have
\[\Sigma_a=(Ta, R(\dr \ell_a))-\Omega_\cdot a^\dagger=\tilde a-\Omega_\cdot a^\dagger
\]
for all $a\in \Gamma(A)$ (see the description of the Lie algebroid structure on $T^*A\to A^*$ in Appendix \ref{big_lie_algebroids}).

\begin{theorem}\label{rep_up_to_hom}
The Lie algebroid structure on $TA\oplus T^*A\to TM\oplus A^*$ can be caracterized as follows, where we write
$\Theta: TA\oplus T^*A\to T(TM\oplus A^*)$ for its anchor:
\begin{enumerate}
\item $[\Sigma_a,\Sigma_b]=\Sigma_{[a,b]}-R_\Delta^{\rm bas}(a,b)^\uparrow$,
\item $[\Sigma_a,\sigma^\dagger]=(\nabla_a^{\rm bas}\sigma)^\dagger$,
\item $[\sigma_1^\dagger, \sigma_2^\dagger]=0$,
\item $\Theta(\Sigma_a)=\widehat{\nabla_a^{\rm bas}}\in\mathfrak{X}} \newcommand{\dr}{\mathbf{d}(TM\oplus A^*)$,
\item $\Theta(\sigma^\dagger)=((\rho,\rho^*)\sigma)^\uparrow \in\mathfrak{X}} \newcommand{\dr}{\mathbf{d}(TM\oplus A^*)$.
\end{enumerate}
\end{theorem}

\begin{remark}
In other words, $(\rho,\rho^*):A\oplus T^*M\to TM\oplus A^*$, the basic connections $\nabla^{\rm bas}$ and
the basic curvature $R_\Delta^{\rm bas}$ define the representation up to homotopy describing the VB-Lie algebroid
structure on $TA\oplus T^*A\to TM\oplus A^*$  in terms of  the splitting
given by $\Delta$ (see \cite{GrMe10a}).
\end{remark}

\begin{proof}[Proof of Theorem \ref{rep_up_to_hom}]
The proof is this theorem is just checking of the formulas, using the
description of the Lie algebroid structure on $TA\oplus T^*A\to
TM\oplus A^*$ that can be found in Appendix \ref{big_lie_algebroids}.

We start with the Lie algebroid brackets. Choose $a,b\in\Gamma(A)$ and
$\sigma\in\Gamma(A\oplus T^*M)$.
We have, using Proposition \ref{structure_of_TAT*A}:
\begin{enumerate}
\item \begin{align*}
[\Sigma_a, \Sigma_b]&=\left[\tilde a-\Omega_\cdot a^\dagger,
\tilde b-\Omega_\cdot b^\dagger\right]\\
&=\widetilde{[a,b]}-(\ldr{a}\Omega_\cdot
b)^\dagger+(\ldr{b}\Omega_\cdot a)^\dagger+(\Omega_\cdot
b\circ(\rho,\rho^*)\circ \Omega_\cdot a- \Omega_\cdot
a\circ(\rho,\rho^*)\circ \Omega_\cdot b)^\dagger\\
&=\Sigma_{[a,b]}-R_\Delta(a,b)^\dagger,
\end{align*}
since we have, for all $v\in \Gamma(TM\oplus A^*)$:
\begin{align*}
&-(\ldr{a}\Omega_\cdot
b)(v)+(\ldr{b}\Omega_\cdot a)(v)+(\Omega_\cdot
b\circ(\rho,\rho^*)\circ \Omega_\cdot a)(v)- (\Omega_\cdot
a\circ(\rho,\rho^*)\circ \Omega_\cdot b)(v)\\
=&-\ldr{a} \Omega_vb+\Omega_{\ldr{a}v}b+\ldr{b}
\Omega_va-\Omega_{\ldr{b}v}a
+\Omega_{(\rho,\rho^*)\Omega_v a}b- \Omega_{(\rho,\rho^*)\Omega_v
  b}a\\
=&-\ldr{a} \Omega_vb+\ldr{b}
\Omega_va
+\Omega_{\nabla_a^{\rm bas}v}b- \Omega_{\nabla_b^{\rm bas}v}a
=-R_\Delta(a,b)v-\Omega_v[a,b].
\end{align*}
\item
  $[\Sigma_a,\sigma^\dagger]=(\ldr{a}\sigma)^\dagger+\Omega_{(\rho,\rho^*)\sigma}a^\dagger=(\nabla_a^{\rm
    bas}\sigma)^\dagger$.
\end{enumerate}
For the anchor map, we compute:
\begin{enumerate}
\setcounter{enumi}{3}
\item $\Theta(\Sigma_a)(\ell_\sigma)=\ell_{\ldr{a}\sigma-(\Omega_\cdot a)^*((\rho,\rho^*)\sigma)}$, which yields
the desired equality
since \begin{align*}
\langle(\Omega_\cdot a)^*((\rho,\rho^*)\sigma),v\rangle
&=\langle\Omega_v a, (\rho,\rho^*)\sigma\rangle=\langle(\rho,\rho^*)\Omega_v a, \sigma\rangle=\langle\nabla_a^{\rm bas}v-\ldr{a}v, \sigma\rangle\\
&=\rho(a)\langle v, \sigma\rangle-\langle\Skew_\Delta(v,(\rho,\rho^*)\sigma),a\rangle-\langle v, \nabla_a^{\rm bas}\sigma\rangle-\langle\ldr{a}v,\sigma\rangle\\
&=-\langle\Skew_\Delta(v,(\rho,\rho^*)\sigma),a\rangle-\langle v, \nabla_a^{\rm bas}\sigma\rangle+\langle v, \ldr{a}\sigma\rangle
\end{align*}
and consequently
\begin{align*}
\langle v, \ldr{a}\sigma-(\Omega_\cdot a)^*((\rho,\rho^*)\sigma)\rangle=
\langle v, {\nabla_a^{\rm bas}}^*\sigma\rangle
\end{align*}
by (1) of Proposition \ref{prop_basic_connections}.
\end{enumerate}
The remaining equalities follow from Proposition \ref{structure_of_TAT*A}.
\end{proof}

\begin{theorem}\label{subalgebroids}
Consider a Lie algebroid $A$ and a Dorfman connection
\[\Delta:\Gamma(TM\oplus A^*)\times\Gamma(A\oplus T^*M)\to \Gamma(A\oplus T^*M).
\]
Let $U\subseteq TM\oplus A^*$ and $K\subseteq A\oplus T^*M$ be subbundles. Then
the sub-double vector bundle $D_{(U,K,[\Delta])}$ is a subalgebroid of $TA\oplus T^*A\to TM\oplus A^*$
over $U$ if and only if:
\begin{enumerate}
\item $(\rho,\rho^*)(K)\subseteq U$,
\item $\nabla_a^{\rm bas}k\in\Gamma(K)$ for all $a\in\Gamma(A)$ and $k\in\Gamma(K)$,
\item  $\nabla_a^{\rm bas}u\in\Gamma(U)$ for all  $a\in\Gamma(A)$ and $u\in\Gamma(U)$,
\item $R_\Delta^{\rm bas}(a,b)u\in\Gamma(K)$ for all $u\in\Gamma(U)$, $a,b\in\Gamma(A)$.
\end{enumerate}
\end{theorem}

\begin{proof}
Assume that  $D_{(U,K,[\Delta])}\to U$ is a subalgebroid of $TA\oplus T^*A\to TM\oplus A^*$. Then we have
$((\rho,\rho^*)k)^\uparrow\an{U}=\Theta(k^\dagger\an{U})\in\mathfrak{X}} \newcommand{\dr}{\mathbf{d}(U)$ and $\widehat{\nabla_a^{\rm bas}}\an{U}=\Theta(\tilde a\an{U})\in\mathfrak{X}} \newcommand{\dr}{\mathbf{d}(U)$
for all $a\in\Gamma(A)$ and $k\in\Gamma(K)$.
This is the case if and only if $((\rho,\rho^*)k)^\uparrow(\ell_{l})\an{U}=0$
and $\widehat{\nabla_a^{\rm bas}}(\ell_{l})\an{U}=0$ for all $l\in\Gamma(U^\circ)$.
Since $((\rho,\rho^*)k)^\uparrow(\ell_{l})=\pi^*\langle (\rho,\rho^*)k, l\rangle$ and
$\widehat{\nabla_a^{\rm bas}}(\ell_{l})=\ell_{{\nabla_a^{\rm bas}}^*l}$,
we find that $(\rho,\rho^*)k$ must be a section of $U$ and ${\nabla_a^{\rm bas}}^*l\in\Gamma(U^\circ)$ for all $l\in\Gamma(U^\circ)$.
But the latter is equivalent to $\nabla_a^{\rm bas}u\in\Gamma(U)$ for all $u\in\Gamma(U)$.
We  have in the same manner
$(\nabla_a^{\rm bas}k)^\dagger\an{U}=[\tilde a, k^\dagger]\an{U}\in\Gamma(D_{(U,K,[\Delta])})$
and
$\bigl(\widetilde{[a,b]}-R_\Delta^{\rm bas}(a,b)^\dagger\bigr)\an{U}=[\tilde a, \tilde b]\an{U}\in\Gamma(D_{(U,K,[\Delta])})$
for all $a,b\in\Gamma(A)$ and $k\in\Gamma(K)$. But this is only the case if
$\nabla_a^{\rm bas}k\in\Gamma(K)$ and, since $\widetilde{[a,b]}\an{U}\in\Gamma(D_{(U,K,[\Delta])})$,
$R_\Delta^{\rm bas}(a,b)^\dagger\an{U}\in\Gamma(D_{(U,K,[\Delta])})$.
The lattest holds only if  $R_\Delta^{\rm bas}(a,b)u\in\Gamma(K)$ for all $u\in\Gamma(U)$.

\medskip

The converse implication is shown in the same manner.
\end{proof}

\subsection{$\mathcal {LA}$-Dirac structures in $TA\oplus T^*A$}
In this subsection and the next, we will study in more details the triples $(U,K,[\Delta]_{U,K})$ associated
to Dirac structures on $A$ that are at the same time Lie subalgebroids of $TA\oplus T^*A\to TM\oplus A^*$.
We call such a Dirac structure an $\mathcal {LA}$-Dirac structure on $A$.

\begin{theorem}\label{morphic_Dirac_triples}
Consider a Lie algebroid $A$ and a    Dorfman connection
\[\Delta:\Gamma(TM\oplus A^*)\times\Gamma(A\oplus T^*M)\to \Gamma(A\oplus T^*M).
\]
Let $U\subseteq TM\oplus A^*$ and $K\subseteq A\oplus T^*M$ be subbundles. Then
$D_{(U,K,[\Delta])}$ is a Dirac structure in $TA\oplus T^*A\to A$ and a
subalgebroid of $TA\oplus T^*A\to TM\oplus A^*$
over $U$ if and only if $(U,K,[\Delta])$ is a $\mathcal {LA}$-Dirac triple, i.e. if and only if:
\begin{enumerate}
\item $K=U^\circ$
\item $(\rho,\rho^*)(K)\subseteq U$,
\item $(U,\pr_{TM}, \llbracket} \newcommand{\rb}{\rrbracket\cdot\,,\cdot\rb_\Delta)$ is a Lie algebroid,
\item $\nabla_a^{\rm bas}k\in\Gamma(K)$ for all $a\in\Gamma(A)$ and $k\in\Gamma(K)$,
\item $R_\Delta^{\rm bas}(a,b)u\in\Gamma(K)$ for all $u\in\Gamma(U)$, $a,b\in\Gamma(A)$.
\end{enumerate}
\end{theorem}

\begin{proof}
This theorem follows from (2) in Corollary  \ref{Dirac_triples} and Theorem \ref{subalgebroids}.
Note that if $U=K^\circ$, $(\rho,\rho^*)K\subseteq U$ and $(U,\pr_{TM}, \llbracket} \newcommand{\rb}{\rrbracket\cdot\,,\cdot\rb_\Delta)$ is a Lie algebroid, then $\nabla^{\rm bas}_a$ preserves
$\Gamma(U)$ if and only if $\nabla^{\rm bas}_a$ preserves $\Gamma(K)$. So (2) and (3) in Theorem \ref{subalgebroids} become the same condition.
\end{proof}

\begin{remark}
A triple $(U,K,[\Delta]_{U,K})$ satisfying (1)--(5) in Theorem \ref{subalgebroids} will
be called an $\mathcal {LA}$-Dirac triple on $A$. We have found a one-one correspondence
between $\mathcal {LA}$-Dirac triples on $A$ and $\mathcal {LA}$-Dirac structures on the Lie algebroid $A$.
\end{remark}

\begin{example}\label{lie_algebroid_dual_3}
Consider again Examples \ref{lie_algebroid_dual} and \ref{lie_algebroid_dual_2}.
Assume that $A^*$ has itself also a Lie algebroid structure with anchor $\rho_*$ and
bracket $[\cdot\,,\cdot]_*$.
For simplicity, we  switch the roles of $A$ and $A^*$ in  Examples \ref{lie_algebroid_dual} and \ref{lie_algebroid_dual_2}.
Here, the second condition
is equivalent to
\begin{equation}\label{antisym}
\rho_*\circ\rho^*=-\rho\circ\rho_*^*
\end{equation}
We assume in the following that this condition is satisfied.

We have also:
\[\Omega_{(\rho_*(\xi),\xi)}a=\left(\ldr{\xi}a-\rho_*^*\langle \nabla^*_\cdot \xi,
  a\rangle,\langle \nabla^*_\cdot \xi, a\rangle
\right)-(0,\dr\langle\xi,a \rangle)
=\left(\ip{\xi}\dr_Aa+\rho_*^*\langle  \xi,
  \nabla_\cdot a\rangle,-\langle  \xi, \nabla_\cdot  a\rangle
\right),
\]
for all $\xi\in\Gamma(A^*)$ and $a\in\Gamma(A)$
and so
\begin{align*}
\Omega_{(\rho,\rho^*)(-\rho_*^*\theta,\theta)}a&=\Omega_{(\rho_*(\rho^*\theta),\rho^*\theta)}a=\left(\ip{\rho^*\theta}\dr_Aa+\rho_*^*\langle \rho^*\theta,
  \nabla_\cdot a\rangle,-\langle  \rho^*\theta, \nabla_\cdot  a\rangle
\right).
\end{align*}
for all $\theta\in\Omega^1(M)$.
In particular, if $\theta=\dr\varphi$ for some $\varphi\in C^\infty(M)$, we get:
\begin{align*}
\nabla^{\rm bas}_a(-\rho_*^*\dr\varphi, \dr\varphi)
&=\Omega_{(\rho,\rho^*)(-\rho_*^*\dr\varphi,\dr\varphi)}a+\ldr{a}(-\rho_*^*\dr\varphi,\dr\varphi)\\
&=\left(\ip{\rho^*\dr\varphi}\dr_Aa+\rho_*^*\langle \rho^*\dr\varphi,
  \nabla_\cdot a\rangle  -[a,\rho_*^*\dr\varphi], -\langle  \rho^*\dr\varphi, \nabla_\cdot  a\rangle
+\ldr{\rho(a)}(\dr\varphi)
\right)\\
&=\left(\ip{\dr_{A^*}\varphi}\dr_Aa+\rho_*^*\langle \dr_{A^*}\varphi,
  \nabla_\cdot a\rangle  -[a,\dr_A\varphi], -\langle  \dr_{A^*}\varphi, \nabla_\cdot  a\rangle
+\dr(\rho(a)(\varphi))
\right).
\end{align*}
Thus, using the first condition, $\nabla^{\rm bas}_a(-\rho_*^*\dr\varphi, \dr\varphi)\in\Gamma(K)$ if and only if
\[\langle \left(\ip{\dr_{A^*}\varphi}\dr_Aa+\rho_*^*\langle \dr_{A^*}\varphi,
  \nabla_\cdot a\rangle  -[a,\dr_A\varphi], -\langle  \dr_{A^*}\varphi, \nabla_\cdot  a\rangle
+\dr(\rho(a)(\varphi))
\right), (\rho_*\xi,\xi)\rangle=0
\]
for all $\xi\in\Gamma(A^*)$.
But this pairing equals
\begin{align*}
&\langle \left(\ip{\dr_{A^*}\varphi}\dr_Aa+\rho_*^*\langle \dr_{A^*}\varphi,
  \nabla_\cdot a\rangle  -[a,\dr_A\varphi], -\langle  \dr_{A^*}\varphi, \nabla_\cdot  a\rangle
+\dr(\rho(a)(\varphi))
\right), (\rho_*\xi,\xi)\rangle\\
=\,&-\langle  \dr_{A^*}\varphi, \nabla_{\rho_*(\xi)}  a\rangle+\rho_*(\xi)\rho(a)(\varphi)
+(\dr_Aa)(\dr_{A^*}\varphi,\xi)
+\langle \dr_{A^*}\varphi,
  \nabla_{\rho_*(\xi)} a\rangle  -\langle\xi, [a,\dr_A\varphi]\rangle\\
=\,&\rho_*(\xi)\rho(a)(\varphi)+\rho_*(\dr_{A^*}\varphi)\langle \xi,a\rangle
-\rho_*(\xi)\langle\dr_{A^*}\varphi, a\rangle-\langle a, [\dr_{A^*}\varphi,\xi]_*\rangle-\langle\xi, [a,\dr_A\varphi]\rangle\\
=\,&\langle(\rho_*\circ\rho^*)(\dr\varphi),\dr(\langle \xi,a\rangle)\rangle
+\rho_*(\xi)\langle
a,\dr_{A^*}\varphi\rangle-\langle\ldr{\xi}a,\dr_{A^*}\varphi\rangle-\rho(a)\langle\xi, \dr_A\varphi\rangle
+\langle \ldr{a}\xi,\dr_A\varphi\rangle \\
=\,&\langle\dr\varphi,(\rho\circ\rho^*_*)\dr(\langle \xi,a\rangle)\rangle
+\rho_*(\xi)\rho(a)(\varphi)-\rho(\ldr{\xi}a)(\varphi)-\rho(a)\rho_*(\xi)(\varphi)
+\rho_*(\ldr{a}\xi)(\varphi)\\
=\,&\bigl([\rho_*(\xi),\rho(a)]+\rho_*(\ldr{a}\xi)-\rho(\ldr{\xi}a)+\rho(\dr_A\langle \xi,a\rangle)\bigr)(\varphi).
\end{align*}
Since $\varphi$ was arbitrary, we have shown
that the fourth condition is satisfied
if and only if
\begin{equation}\label{compatibility_of_derivations}
[\rho(a),\rho_*(\xi)]-\rho_*(\ldr{a}\xi)+\rho(\ldr{\xi}a)=\rho(\dr_A\langle \xi,a\rangle)
\end{equation}
for all $a\in\Gamma(A)$ and $\xi\in\Gamma(A^*)$.
Thus, we have found until here
\eqref{antisym} and \eqref{compatibility_of_derivations},
which are shown in \cite[Theorem 12.1.9]{Mackenzie05}
to imply the fact that $(A,A^*)$ is a Lie bialgebroid.

We conclude by showing that the last condition, on the basic curvature, follows as well
from \eqref{antisym} and \eqref{compatibility_of_derivations}.
Since $\Omega_{(\rho_*(\xi),\xi)}a=(\ip{\xi}\dr_Aa, 0)-(-\rho_*^*\langle  \xi,
  \nabla_\cdot a\rangle,\langle  \xi, \nabla_\cdot  a\rangle)$ and $K=U^\circ$, we find
\[\langle \Omega_{(\rho_*(\xi),\xi)}a, (\rho_*\eta,\eta)\rangle=(\dr_Aa)(\xi,\eta)
\]
for all $a\in\Gamma(A)$, $\xi,\eta\in\Gamma(A^*)$.
The fourth condition together with the first identity in Proposition \ref{prop_basic_connections}
and the first and third conditions imply that
$\nabla_a^{\rm bas}u\in\Gamma(A)$ for all $u\in\Gamma(U)$.
Hence,
\begin{align*}
\nabla_a^{\rm bas}(\rho_*(\xi),\xi)&=(\rho,\rho^*)\left(\ip{\xi}\dr_Aa+\rho_*^*\langle  \xi,
  \nabla_\cdot a\rangle,-\langle  \xi, \nabla_\cdot  a\rangle\right)+\ldr{a}(\rho_*(\xi),\xi)\\
&=\left(\rho_*(-\rho^*\langle  \xi, \nabla_\cdot  a\rangle+\ldr{a}\xi),-\rho^*\langle  \xi, \nabla_\cdot  a\rangle+\ldr{a}\xi\right)
\end{align*}
and
\[\langle \Omega_{\nabla^{\rm bas}_a(\rho_*\xi,\xi)}b, (\rho_*\eta,\eta)\rangle=
(\dr_Ab)(-\rho^*\langle  \xi, \nabla_\cdot  a\rangle+\ldr{a}\xi,\eta)
\]
for all $a,b\in\Gamma(A)$, $\xi,\eta\in\Gamma(A^*)$.

We get hence that
\begin{align*}
\langle R_\Delta^{\rm bas}(a,b)(\rho_*\xi,\xi), (\rho_*\eta,\eta)\rangle
=\,&(\dr_A[a,b])(\xi,\eta) -\langle\ldr{a}(\ip{\xi}\dr_Ab+\rho_*^*\langle  \xi,
  \nabla_\cdot b\rangle,-\langle  \xi, \nabla_\cdot  b\rangle), (\rho_*\eta,\eta)\rangle\\
&\,+\langle\ldr{b}(\ip{\xi}\dr_Aa+\rho_*^*\langle  \xi,
  \nabla_\cdot a\rangle,-\langle  \xi, \nabla_\cdot  a\rangle), (\rho_*\eta,\eta)\rangle\\
&\,-(\dr_Aa)(-\rho^*\langle  \xi, \nabla_\cdot  b\rangle+\ldr{b}\xi,\eta)
+(\dr_Ab)(-\rho^*\langle  \xi, \nabla_\cdot  a\rangle+\ldr{a}\xi,\eta)\\
=\,&(\dr_A[a,b])(\xi,\eta) -\rho(a)(\dr_Ab(\xi,\eta))+\dr_Ab(\xi,\ldr{a}\eta)\\
&+\langle(\rho_*^*\langle  \xi,
  \nabla_\cdot b\rangle,-\langle  \xi, \nabla_\cdot  b\rangle), \ldr{a}(\rho_*\eta,\eta)\rangle\\
&+\rho(b)(\dr_Aa(\xi,\eta))-\dr_Aa(\xi,\ldr{b}\eta)
-\langle(\rho_*^*\langle  \xi,
  \nabla_\cdot a\rangle,-\langle  \xi, \nabla_\cdot  a\rangle), \ldr{b}(\rho_*\eta,\eta)\rangle\\
&\,-(\dr_Aa)(-\rho^*\langle  \xi, \nabla_\cdot  b\rangle+\ldr{b}\xi,\eta)
+(\dr_Ab)(-\rho^*\langle  \xi, \nabla_\cdot  a\rangle+\ldr{a}\xi,\eta)\\
=\,&(\dr_A[a,b]-[a,\dr_Ab]+[b,\dr_Aa])(\xi,\eta)\\
&+\langle  \xi,\nabla_{\rho_*(\ldr{a}\eta)} b\rangle-\langle  \xi, \nabla_{[\rho(a),\rho_*\eta]}  b\rangle\\
&-\langle  \xi,\nabla_{\rho_*(\ldr{b}\eta)} a\rangle+\langle  \xi, \nabla_{[\rho(b),\rho_*\eta]}  a\rangle\\
&\,+\rho_*\rho^*\langle  \xi, \nabla_\cdot  b\rangle(\langle a,\eta\rangle)
-\rho_*\eta\langle  \xi, \nabla_{\rho(a)}  b\rangle+\langle a,[\eta,\rho^*\langle  \xi, \nabla_\cdot  b\rangle]_*\rangle\\
&\,-\rho_*\rho^*\langle  \xi, \nabla_\cdot  a\rangle(\langle b,\eta\rangle)
+\rho_*\eta\langle  \xi, \nabla_{\rho(b)}  a\rangle-\langle b,[\eta, \rho^*\langle  \xi, \nabla_\cdot  a\rangle]_*\rangle\\
=\,&(\dr_A[a,b]-[a,\dr_Ab]+[b,\dr_Aa])(\xi,\eta)\\
&+\langle  \xi,\nabla_{\rho_*(\ldr{a}\eta)} b\rangle-\langle  \xi, \nabla_{[\rho(a),\rho_*\eta]}  b\rangle\\
&-\langle  \xi,\nabla_{\rho_*(\ldr{b}\eta)} a\rangle+\langle  \xi, \nabla_{[\rho(b),\rho_*\eta]}  a\rangle\\
&\,+\langle  \xi, \nabla_{\rho(\dr_A\langle a,\eta\rangle)} b\rangle
-\langle  \xi, \nabla_{\rho(\ldr{\eta}a)}  b\rangle\\
&\,-\langle  \xi, \nabla_{\rho(\dr_A\langle b,\eta\rangle)} a\rangle
+\langle  \xi, \nabla_{\rho(\ldr{\eta}b)}  a\rangle.
\end{align*}
By \eqref{compatibility_of_derivations}, the second and the fourth
lines, and the third and the fifth lines cancel each other.
We find hence that the last condition is satisfied if and only if
$(A,A^*)$ is a Lie bialgebroid. Hence,
$(U,K,[\Delta])$
is an $\mathcal{LA}$-Dirac triple if and only if
$(A,A^*)$
is a Lie bialgebroid, and so the graph of $\pi_A$
is mrophic and Dirac if and only if $(A,A^*)$
is a Lie bialgebroid. This recovers a result in \cite{MaXu00}.
\end{example}

\begin{example}\label{IM_2_form_3}
In the situation of Examples \ref{IM_2_form} and \ref{IM_2_form_2},
assume furthermore that $E=:A$ is a Lie algebroid.
To avoid confusions,
we write $\nabla^{A}$
for the $A$-basic connections
induced on $A$ and $TM$ by the Lie algebroid structure on $A$ and the
connection $\nabla$,
 and $R_{\nabla}^A$ for the basic curvature associated to it.
\medskip

The second condition of the last theorem reads here
\[(\rho,\rho^*)(a,\sigma(a))=(\rho(a),-\sigma^*\rho(a))
\]
for all $a\in\Gamma(A)$, i.e.
\[ \rho^*\circ\sigma=-\sigma^*\circ\rho.
\]
This is exactly the first axiom defining an IM-$2$-form $\sigma:A\to
T^*M$ \cite{BuCrWeZh04,BuCaOr09}:
\[\langle \sigma(a),\rho(b)\rangle=-\langle\rho(a), \sigma(b)\rangle
\]
for all $a,b\in\Gamma(A)$.
We next compute $\nabla^{\rm bas}_a(b,\sigma(b))$.
We have
\[\Omega_{(X,-\sigma^*X)}a=(\nabla_Xa,-\ldr{X}\sigma(a)+\sigma(\nabla_Xa))+(0,\dr\langle\sigma(a),X\rangle)
=(\nabla_Xa,-\ip{X}\dr\sigma(a)+\sigma(\nabla_Xa))
\]
and as a consequence
\[\nabla_a^{\rm
  bas}(b,\sigma(b))=\Omega_{(\rho,\rho^*)(b,\sigma(b))}a+\ldr{a}(b,\sigma(b))
=(\nabla_{\rho(b)}a+[a,b], \ldr{\rho(a)}\sigma(b)-\ip{\rho(b)}\dr\sigma(a)+\sigma(\nabla_{\rho(b)}a)).
\]
We find hence that $\nabla_a^{\rm
  bas}(b,\sigma(b))\in\Gamma(K)$ if and only
if $([a,b],
\ldr{\rho(a)}\sigma(b)-\ip{\rho(b)}\dr\sigma(a))\in\Gamma(K)$,
i.e. if and only if
\[ \sigma([a,b])=\ldr{\rho(a)}\sigma(b)-\ip{\rho(b)}\dr\sigma(a).
\]
Since this is the second axiom in the definition of an IM-$2$-form, we
recover the fact that the graph of
$(\sigma^*\omega_{\rm can})^\flat:TA\to T^*A$
is a subalgebroid of $TA\oplus T^*A\to TM\oplus A^*$ over $U=\graphe(-\sigma^*)$
only if $\sigma:A\to T^*M$ is an IM-$2$-form.
To show the equivalence, we show that the last condition in the last
theorem follows here again from the four other.
We have, for $a,b\in\Gamma(A)$ and $X,Y\in\mathfrak{X}} \newcommand{\dr}{\mathbf{d}(M)$:
\begin{align*}
\Omega_{(X,-\sigma^*X)}a
&=-(0,\ip{X}\dr\sigma(a))+(\nabla_Xa,\sigma(\nabla_Xa)),
\end{align*}
hence
\begin{align*}
\ldr{b}\Omega_{(X,-\sigma^*X)}a&=-(0,\ldr{\rho(b)}\ip{X}\dr\sigma(a))+([b,\nabla_Xa],\ldr{\rho(b)}\sigma(\nabla_Xa))\\
&=-(0,\ip{[\rho(b),X]}\dr\sigma(a)+\ip{X}\ldr{\rho(b)}\dr\sigma(a))+([b,\nabla_Xa],\sigma([b,\nabla_Xa])+\ip{\rho(\nabla_Xa)}\dr\sigma(b))
\end{align*}
and
\begin{align*}
\nabla_a^{\rm
  bas}(X,-\sigma^*X)=-(\rho,\rho^*)(0,\ip{X}\dr\sigma(a))+(\rho,\rho^*)(\nabla_Xa,\sigma(\nabla_Xa))
+\ldr{a}(X,-\sigma^*X)
\end{align*}
which equals $(\nabla_a^AX,-\sigma^*(\nabla_a^AX))$
since $\nabla_a^{\rm bas}u\in\Gamma(U)$ for all $u\in\Gamma(U)$.

We get hence
\begin{align*}
 R_\Delta^{\rm bas}(a,b)(X,-\sigma^*X)
=&\bigl( R_\nabla^A(a,b)(X), \sigma(R_\nabla^A(a,b)(X))\bigr)\\
&+(0,-\ip{X}\dr\sigma([a,b])
+\ip{[\rho(a),X]}\dr\sigma(b)+\ip{X}\ldr{\rho(a)}\dr\sigma(b)-\ip{\rho(\nabla_Xb)}\dr\sigma(a)\\
&\quad -\ip{[\rho(b),X]}\dr\sigma(a)-\ip{X}\ldr{\rho(b)}\dr\sigma(a)+\ip{\rho(\nabla_Xa)}\dr\sigma(b)
+\ip{\nabla^A_bX}\dr\sigma(a)-\ip{\nabla^A_aX}\dr\sigma(b)
)\\
=&\bigl( R_\nabla^A(a,b)(X), \sigma(R_\nabla^A(a,b)(X))\bigr)-(0,\ip{X}\dr(\sigma([a,b])-\ldr{\rho(a)}\sigma(b)
+\ip{\rho(b)}\dr\sigma(a) ))\\
=&\bigl( R_\nabla^A(a,b)(X), \sigma(R_\nabla^A(a,b)(X))\bigr)\in\Gamma(K).
\end{align*}
\end{example}

\bigskip

We continue with the study of $\mathcal{LA}$-Dirac triples.
We first observe that if  $(U,K,\Delta)$ is a $\mathcal {LA}$-Dirac triple, then $K$ inherits a Lie algebroid structure.
\begin{theorem}\label{K_algebroid}
Consider an $\mathcal{LA}$-Dirac triple $(U,K,[\Delta]_{U,K})$.
Then \[(K, \rho_K:=\rho\circ\pr_A, [\cdot\,,\cdot]_D\an{\Gamma(K)\times\Gamma(K)})\]
is a Lie algebroid
and the map
$(\rho, \rho^*):K\to U$ is a Lie algebroid morphism.
\end{theorem}

We need the following two lemmas, which will also be useful later.
\begin{lemma}\label{basic_like_eq}
The equality
\begin{equation}\label{basicLike_eqq}
\nabla_{\sigma_1}^{\rm bas}\sigma_2=-[\sigma_2,\sigma_1]_D+\Delta_{(\rho,\rho^*)(\sigma_2)}\sigma_
\end{equation}
holds for all $\sigma_1,\sigma_2\in\Gamma(A\oplus T^*M)$.
\end{lemma}

\begin{proof}Write $\sigma_1=(a_1,\theta_2)$ and $\sigma_2=(a_2,\theta_2)\in\Gamma(A\oplus T^*M)$.
Then:
\begin{align*}
\nabla_{\sigma_1}^{\rm bas}\sigma_2&=\nabla_{a_1}^{\rm bas}\sigma_2=\Omega_{(\rho,\rho^*)\sigma_2}a_1+\ldr{a_1}\sigma_2\\
&=\Delta_{(\rho,\rho^*)\sigma_2}\sigma_1-(0,\dr\langle\theta_2,\rho(a_1)\rangle)-(0,\ldr{\rho(a_2)}\theta_1)+([a_1,a_2], \ldr{\rho(a_1)}\theta_2)\\
&=\Delta_{(\rho,\rho^*)\sigma_2}\sigma_1-[\sigma_2, \sigma_1]_D.
\end{align*}
\end{proof}

\begin{lemma}\label{complicated_eq}
Let $(U,K,[\Delta]_{U,K})$ be an $\mathcal{LA}$-Dirac triple. Then, for all $v\in\Gamma(TM\oplus A^*)$
and $\tau,\sigma\in\Gamma(A\oplus T^*M)$:
\begin{equation}\label{complicated_eqq}
\langle (\rho,\rho^*)\Delta_{v}\tau-\llbracket} \newcommand{\rb}{\rrbracket v,(\rho,\rho^*)\tau\rb_\Delta-\nabla_{\tau}^{\rm bas}v,\sigma\rangle=\langle\nabla_{\sigma}^{\rm bas}v, \tau\rangle.
\end{equation}
\end{lemma}
This yields the following corollary.
\begin{corollary}\label{eq_for_morphism}
Let $(U,K,[\Delta]_{U,K})$ be an $\mathcal{LA}$-Dirac triple. Then, for all $u\in\Gamma(U)$
and $k\in\Gamma(K)$:
\begin{equation*}
(\rho,\rho^*)\Delta_{u}k=\llbracket} \newcommand{\rb}{\rrbracket u,(\rho,\rho^*)k\rb_\Delta+\nabla_{k}^{\rm bas}u.
\end{equation*}
\end{corollary}
\begin{proof}
By Lemma \ref{complicated_eq}, we have
\[\langle (\rho,\rho^*)\Delta_{u}k-\llbracket} \newcommand{\rb}{\rrbracket u,(\rho,\rho^*)k\rb_\Delta-\nabla_{k}^{\rm bas}u,\sigma\rangle=\langle\nabla_{\sigma}^{\rm bas}u, k\rangle\]
for all $\sigma\in\Gamma(A\oplus T^*M)$.
Since  $\nabla_\sigma^{\rm bas}$ preserves $\Gamma(U)$ by  Theorem \ref{morphic_Dirac_triples}, this vanishes.
\end{proof}

\begin{proof}[Proof of Lemma \ref{complicated_eq}]
We write $\tau=(b,\theta)$ and $v=(X,\xi)$. Then we have for any $\sigma=(a,\omega)\in\Gamma(A\oplus T^*M)$:
\begin{align*}
&\langle (\rho,\rho^*)\Delta_{v}\tau-\llbracket} \newcommand{\rb}{\rrbracket v,(\rho,\rho^*)\tau\rb_\Delta-\nabla_{b}^{\rm bas}v, \sigma\rangle\nonumber\\
=&\langle \Delta_v\tau-\Omega_vb, (\rho,\rho^*)\sigma\rangle
-\langle \ldr{b}v,\sigma\rangle
+\langle \llbracket} \newcommand{\rb}{\rrbracket(\rho,\rho^*)\tau,  v\rb_\Delta, \sigma\rangle-\langle \Skew_\Delta((\rho,\rho^*)\tau,  v), a\rangle\nonumber\\
=&\langle (0, \ldr{X}\theta+\dr\langle\xi, b\rangle), (\rho,\rho^*)\sigma\rangle
-\langle \ldr{b}v,\sigma\rangle
+\rho(b)\langle  v, \sigma\rangle-\langle v, \Delta_{(\rho,\rho^*)\tau}\sigma\rangle-\langle \Skew_\Delta((\rho,\rho^*)\tau,  v), a\rangle\nonumber \\
=&\langle \ldr{X}\theta, \rho(a)\rangle+\rho(a)\langle\xi, b\rangle
+\langle  v, \ldr{b}\sigma\rangle-\langle v,
\nabla_{\sigma}^{\rm bas}\tau-[\sigma,\tau]_D+(0,\dr\langle\sigma,(\rho^*,\rho)\tau\rangle)\rangle\nonumber\\
&-\langle \Skew_\Delta((\rho,\rho^*)\tau,  v), a\rangle\hspace*{7cm}  \text{by Lemma \ref{basic_like_eq}}\nonumber\\
=&-\langle v, \nabla_{\sigma}^{\rm bas}\tau\rangle+\langle \ldr{X}\theta, \rho(a)\rangle+\rho(a)\langle\xi, b\rangle
+\langle  v, \ldr{b}\sigma\rangle\nonumber\\
&+\langle\xi,[a,b]\rangle
+\langle \ldr{\rho(a)}\theta-\ldr{\rho(b)}\omega, X\rangle+X\langle\omega, \rho(b)\rangle
-X\langle\sigma,(\rho^*,\rho)\tau\rangle-\langle \Skew_\Delta((\rho,\rho^*)\tau,  v), a\rangle\nonumber\\
=&-\langle v, \nabla_{\sigma}^{\rm bas}\tau\rangle-\langle \theta, [X,\rho(a)]\rangle+\rho(a)\langle v, \tau\rangle
+\langle  v, \ldr{b}\sigma\rangle\nonumber\\
&+\langle\xi,[a,b]\rangle
-\langle\theta, [\rho(a), X]\rangle-\langle \ldr{\rho(b)}\omega, X\rangle-\langle \Skew_\Delta((\rho,\rho^*)\tau,  v), a\rangle\nonumber\\
=&-\langle v, \nabla_{\sigma}^{\rm bas}\tau\rangle+\rho(a)\langle v, \tau\rangle
-\langle \Skew_\Delta((\rho,\rho^*)\tau,  v), a\rangle=\langle\nabla_{\sigma}^{\rm bas}v, \tau\rangle.
\end{align*}
\end{proof}

\begin{proof}[Proof of Theorem \ref{K_algebroid}]
By \eqref{Dorfman_skew-sym}, the equality $U=K^\circ$ and the
inclusion $(\rho,\rho^*)(K)\subseteq U$,
the bracket $[\cdot,\cdot]_D$ is skew-symmetric on sections of $K$. Choose $k_1=(a_1,\theta_1),k_2\in\Gamma(K)$.
Then, by Lemma \ref{basic_like_eq}, we have
\[[k_1, k_2]_D=\nabla_{a_1}^{\rm bas}k_2-\Delta_{(\rho,\rho^*)k_2}k_1
\]
Since by Theorem \ref{morphic_Dirac_triples}, the sections $\nabla_{a_1}^{\rm bas}k_2$ and $\Delta_{(\rho,\rho^*)k_2}k_1$
are elements of $\Gamma(K)$, we find that $[k_1, k_2]_D\in\Gamma(K)$.
The Jacobi-identity follows directly from \eqref{Dorfman_Jacobi}.

\bigskip

We show next that $(\rho,\rho^*):K\to U$
is a Lie algebroid morphism.
We have
\[\rho_U\circ(\rho,\rho^*)=\pr_{TM}\circ(\rho,\rho^*)=\rho\circ\pr_A=\rho_K\]
and, for all $k_1,k_2\in\Gamma(K)$, using Lemma \ref{eq_for_morphism}, Lemma \ref{basic_like_eq} and Proposition \ref{prop_basic_connections}:
\begin{align*}
\llbracket} \newcommand{\rb}{\rrbracket(\rho,\rho^*)k_1, (\rho,\rho^*)k_2\rb_\Delta&=(\rho,\rho^*)\Delta_{(\rho,\rho^*)k_1}k_2-\nabla_{k_2}^{\rm bas}(\rho,\rho^*)k_1\\
&=(\rho,\rho^*)\left(\Delta_{(\rho,\rho^*)k_1}k_2-\nabla_{k_2}^{\rm bas}k_1\right)\\
&=(\rho,\rho^*)\left([k_1,k_2]_D\right).
\end{align*}
\end{proof}

 \begin{example}
\begin{enumerate}
\item In the case of a $\mathcal {LA}$-Dirac triple as in Example \ref{lie_algebroid_dual_3},
we have $K=\{(-\rho_*^*\theta,\theta)\mid \theta\in T^*M\}$
and $U=\{\rho_*(\xi),\xi)\mid \xi\in A^*\}$.
The Lie algebroid structure on $K$ is just the Lie algebroid defined by the graph of the anchor
of the Lie algebroid $(T^*M)_\pi$ with $\pi^\sharp=-\rho\circ  \rho_*^*=\rho_*\circ\rho^*$
and the fact that
\[ (\rho,\rho^*):K\to U
\]
recovers the fact that, for a Lie bialgebroid, the map
$\rho^*:T^*M\to A^*$ is a Lie algebroid morphism (see for instance \cite{Mackenzie05}, Proposition 12.1.13).
\item In the case of a $\mathcal {LA}$-Dirac triple as in Example
  \ref{IM_2_form_3}, we have $[(a,\sigma(a)), (b,\sigma(b))]_D=([a,b],
  \ldr{\rho(a)}\sigma(b)-\ip{\rho(b)}\dr\sigma(a))
=([a,b],\sigma([a,b]))\in\Gamma(K)$ by the results in Example
\ref{IM_2_form_2}. Hence, $K$ is a Lie algebroid. The map
$(\rho,\rho^*)$
sends $(a,\sigma(a))$
to $(\rho(a), -\sigma^*\rho(a))\in\Gamma(U)$
since $\rho^*\circ\sigma=-\sigma^*\circ\rho$.
The fact that $(\rho,\rho^*)$ is a Lie algebroid morphism also follows
from this equality and
the fact that $\rho$ is is a Lie algebroid morphism.
\end{enumerate}
\end{example}

\section{The Manin pair associated to an $\mathcal{LA}$-Dirac structure}
We conclude this paper by showing that the infinitesimal description
of an $\mathcal{LA}$-Dirac structures on Lie algebroids is a
Manin pair \cite{BuIgSe09}
$(C,U)$, where $C\to M$ is a Courant  algebroid that is in a
particular sense compatible with $A$ and  such that the Dirac
structure $U$ in $C$
can be seen as a subbundle of $TM\oplus A^*$ with anchor $\pr_{TM}$.

We start by describing the representations up to homotopy describing
the two sides of a Dirac algebroid.

\subsection{The two VB-Lie
  algebroid structures on an $\mathcal{LA}$-Dirac structure}\label{rep_up_to}
Let $A\to M$ be a Lie algebroid and $D_A$ a $\mathcal {LA}$-Dirac structure on
$A$,
hence corresponding to a $\mathcal{LA}$-Dirac triple $(U,K,[\Delta]_{U,K})$.

The Lie algebroid structure $D_{(U,K,[\Delta])}\to A$ is described by the
representation up to homotopy \cite{GrMe10a}
defined by
the vector bundle morphism
$\pr_A:K\to A$,  the connections
\[(\pr_A\circ\,\Omega):\Gamma(U)\times\Gamma(A)\to\Gamma(A)
\]
and
\[\Delta:\Gamma(U)\times\Gamma(K)\to\Gamma(K),\]
and the curvature $R_\Delta:\Gamma(U^*\otimes U^*\otimes A^*\times
K)$.

The Lie algebroid structure of the other side $D_{(U,K,[\Delta])}\to U$
is described by the complex $(\rho,\rho^*):K\to U$ with the
basic connections $\nabla^{\rm bas}:\Gamma(A)\times\Gamma(U)\to
\Gamma(U)$ and $\nabla^{\rm bas}:\Gamma(A)\times\Gamma(K)\to
\Gamma(K)$
and the basic curvature $R_\Delta^{\rm bas}\in\Gamma(A^*\otimes
A^*\otimes U^*\otimes K)$.

\begin{theorem}\label{difficult_computation}
Let $A\to M$ be a Lie algebroid and $(K,U,\Delta)$ a $\mathcal {LA}$-Dirac
triple. Then:
\begin{enumerate}
\item Condition (5) of Theorem
\ref{morphic_Dirac_triples} is equivalent to:
 \begin{equation*}\label{bialgebroid1}
\nabla^{\rm bas}_\sigma\llbracket} \newcommand{\rb}{\rrbracket u, v\rb_\Delta-\llbracket} \newcommand{\rb}{\rrbracket \nabla^{\rm bas}_\sigma u,
v\rb_\Delta-\llbracket} \newcommand{\rb}{\rrbracket u, \nabla^{\rm bas}_\sigma v\rb_\Delta+\nabla^{\rm
  bas}_{\Delta_u\sigma}-\nabla^{\rm bas}_{\Delta_v\sigma}u=-(\rho,\rho^*)R_\Delta(u,v)\sigma
\end{equation*}
for all $u,v\in\Gamma(U)$ and $\sigma\in\Gamma(A\oplus T^*M)$.
\item The equality
\begin{align*}\label{bialgebroid2}
&\Delta_u[\sigma_1,\sigma_2]_D-[\Delta_u\sigma_1,\sigma_2]_D-[\sigma_1,\Delta_u\sigma_2]_D
+\Delta_{\nabla_{a_1}^{\rm bas}u}\sigma_2-\Delta_{\nabla_{a_2}^{\rm
    bas}u}\sigma_1+(0,\dr\langle\sigma_1,\nabla^{\rm bas}_{a_2}u\rangle)\\
=\,&-R_\Delta^{\rm bas}(a_1,a_2)u
\end{align*}
holds for all $\sigma_1,\sigma_2\in\Gamma(A\oplus T^*M)$ with
$\pr_A(\sigma_i)=:a_i$ and all $u\in\Gamma(U)$.
\end{enumerate}
\end{theorem}

The proof of these formulas is quite long, but straightforward. It can
be found in Appendix \ref{proof_courant}.

A double Lie algebroid is not only a double vector bundle
with VB-algebroid structures on both sides such that the VB-structure
maps on each side are Lie algebroid morphisms
of the other side. There is an additionnal
more complicated criterion that one needs to check,
namely the existence of a Lie bialgebroid over the dual of the core of
the double vector bundle \cite{Mackenzie11}.
The two formulas in the last theorem seem to be part of  the compatibility conditions on the two
representations up to homotopy, that are necessary for
this Lie bialgebroid condition to be satisfied.

\subsection{The Courant algebroid associated to an $\mathcal {LA}$-Dirac triple}\label{the_courant_alg}
Assume that the triple $(U,K,\Delta)$ is an  $\mathcal {LA}$-Dirac triple and
consider the vector bundle
\begin{equation}\label{def_of_C}
C:=\frac{U\oplus(A\oplus T^*M)}{\operatorname{graph}(-(\rho,\rho^*)\an{K})}\to M.
\end{equation}

We will write $u\oplus \sigma$ for the class in $C$ of a pair $(u,\sigma)\in\Gamma(U\oplus(A\oplus T^*M))$.
It is easy to check that
\begin{equation}\label{pairing}
\langle\langle u_1\oplus \sigma_1, u_2\oplus\sigma_2\rangle\rangle_C:=\langle u_1,\sigma_2\rangle+\langle u_2,\sigma_1\rangle+\langle\sigma_1,(\rho,\rho^*)\sigma_2\rangle
\end{equation}
defines a symmetric fiberwise pairing
 $\langle\langle\cdot\,,\cdot\rangle\rangle_C$
on $C$.

\begin{proposition}
The pairing $\langle\langle\cdot\,,\cdot\rangle\rangle_C$ is
nondegenerate and
the vector bundle $C$ is isomorphic to $U\oplus U^*$.
\end{proposition}

\begin{proof}
The nondegeneracy of $\langle\langle\cdot\,,\cdot\rangle\rangle_C$
is easy to check.

Define the map $\iota: U\to C$,
$u\mapsto u\oplus 0$ and $\pi: C\to U^*$,
$\pi(u\oplus \sigma)(v)=\langle\langle u\oplus \sigma, v\oplus
0\rangle\rangle_C$
for all $u\oplus\sigma\in C$, $v\in U$.
The map $\iota$ is obviously injective.
Assume that $\psi(u\oplus\sigma)=0$. Then
$\langle \sigma, v\rangle =0$ for all $v\in U$.
Hence, $\sigma\in K$ and
$u\oplus\sigma=(u+(\rho,\rho^*)\sigma)\oplus
0\in\operatorname{im}(\pi)$.
A dimension count yields that $\pi$ is surjective,
and since the sequence
\[ 0\rightarrow U\rightarrow C\rightarrow U^*\rightarrow 0\]
is exact, we are done.
\end{proof}

Set \[c:C\to TM,\]
\[c(u\oplus\sigma)=\pr_{TM}(u)+\rho\circ\pr_{A}(\sigma).\]
\begin{theorem}\label{Courant_algebroid}
Assume that $(U,K,[\Delta])$ is an $\mathcal {LA}$-Dirac triple.
Then $C$ is a Courant algebroid with anchor
$c$, pairing $\langle\langle\cdot\,,\cdot\rangle\rangle_C$
and bracket
\[\llbracket} \newcommand{\rb}{\rrbracket\cdot\,,\cdot\rb:\Gamma(C)\times\Gamma(C)\to\Gamma(C),
\]
\begin{align}\label{bracket_on_C}
&\llbracket} \newcommand{\rb}{\rrbracket u_1\oplus\sigma_1, u_2\oplus\sigma_2\rb\\
=\,&\left(\llbracket} \newcommand{\rb}{\rrbracket u_1,u_2\rb_\Delta+\nabla_{\sigma_1}^{\rm bas}u_2-\nabla_{\sigma_2}^{\rm bas}u_1\right)\oplus\left( [\sigma_1,\sigma_2]_D+\Delta_{u_1}\sigma_2-\Delta_{u_2}\sigma_1
+(0,\dr\langle \sigma_1,u_2\rangle)\right).\nonumber
\end{align}
The map \[\mathcal D=\,c^*\circ \dr:C^\infty(M)\to \Gamma(C)\]
is given by
\[f\mapsto 0\oplus(0,\dr f).
\]
\end{theorem}

The proof of this theorem can be found in the appendix.

\begin{remark}
\begin{enumerate}
\item It is easy to check that the Courant algebroid structure only depends on the
$(U,K)$-equivalence class of $\Delta$.
\item This construction has some similarities with the one of matched
  pairs
of Courant algebroids in \cite{GrSt12}. It would be interesting to
understand the relation between the two constructions.
\end{enumerate}
\end{remark}

\begin{definition}
\begin{enumerate}
\item A Manin pair over a manifold $M$ is a pair $(E,D)$
of vector bundles over $M$, such that $E$ is a Courant algebroid and
$D$ a Dirac structure in $E$ \cite{BuIgSe09}.
\item Let $(A\to M, \rho, [\cdot\,,\cdot])$ be a Lie algebroid. An
 $A$-Manin pair  is a Manin pair $(C,U)$ over $M$, where
\begin{enumerate}
\item $U\subseteq TM\oplus A^*$ is a subbundle such that
  $(\rho,\rho^*)(U^\circ)\subseteq U$,
\item $C$ is the vector bundle
\[\frac{U\oplus(A\oplus T^*M)}{\operatorname{graph}(-(\rho,\rho^*)\an{U^\circ})}\to M
\]
endowed with the anchor $c=\pr_{TM}\oplus\rho\circ\pr_A$
and the pairing $\langle\langle\cdot\,,\cdot\rangle\rangle_C$.
\item The bracket of $C$ satisfies $\llbracket} \newcommand{\rb}{\rrbracket 0\oplus \sigma_1,
0\oplus\sigma_2\rb=0\oplus[\sigma_1,\sigma_2]_D$
for all $\sigma_1,\sigma_2\in\Gamma(A\oplus T^*M)$.
\end{enumerate}
\end{enumerate}
\end{definition}

The final theorem and second main result of this paper is the following:
\begin{theorem}
  Let $A$ be a Lie algebroid over a manifold $M$. Then there is a one-one correspondence
between $\mathcal{LA}$-Dirac structures on $A$ and $A$-Manin pairs.
\end{theorem}

\begin{proof}
We have already seen that $\mathcal{LA}$-Dirac structures on a Lie
algebroid $A$ are in
bijection with $\mathcal{LA}$-Dirac triples on $A$. We show here that there
is a one-one correspondence between $\mathcal{LA}$-Dirac triples on
$A$
and $A$-Manin pairs.
We have seen in Theorem \ref{Courant_algebroid} how to associate an
$A$-Manin pair to an $\mathcal{LA}$-Dirac triple on $A$.

Conversely, choose an $A$-Manin pair $(C,U)$ and set
$K:=U^\circ\subseteq A\oplus T^*M$.
Then $U\simeq U\oplus 0$ is a Dirac structure in $C$
and there is hence an induced Dorfman connection
\[\Delta^U:\Gamma(U)\times\Gamma(C/U)\to \Gamma(C/U).
\]
It is easy to verify that the map $C/U\to (A\oplus T^*M)/U^\circ$
sending $\overline{u\oplus\sigma}\in C/U$ to $\bar\sigma\in (A\oplus
T^*M)/U^\circ$
is an isomorphism of vector bundles. Using the Leibniz equality in
both arguments, extend the Lie algebroid bracket
of $U$ to a dull algebroid structure on $TM\oplus A^*$ with anchor
$\pr_{TM}$.
It is easy to see that the corresponding $TM\oplus A^*$-Dorfman
connection $\Delta$
on $A\oplus T^*M$ satisfies $\Delta_uk\in\Gamma(K)$ for all
$k\in\Gamma(K)$ and $u\in\Gamma(U)$, and that the induced $U$-Dorfman
connection on the quotient $(A\oplus T^*M)/K$ is equal to $\Delta^U$.
Furthermore, for two dull extentions of $[\cdot\,,\cdot]_U$,
we find that the corresponding Dorfman connections are
$(U,K)$-equivalent. Hence, we can write $\Delta^U=[\Delta]$.
We check that $(U,K,\Delta^U)$ is an $\mathcal{LA}$-Dirac triple.

For this, we only check that the Courant bracket of $C$ is defined as
in \eqref{bracket_on_C}. The proof of Theorem
\ref{Courant_algebroid}
shows that all the conditions in Theorem \ref{morphic_Dirac_triples}
are then satisfied.

Choose $\tau=(a,\theta)\in\Gamma(A\oplus T^*M)$, $\sigma=(b,\omega)\in\Gamma(A\oplus T^*M)$ and
$u=(X,\xi)\in\Gamma(U)$.
We want to compute $v=v(\tau,u)\in\Gamma(U)$
such that $\llbracket} \newcommand{\rb}{\rrbracket u\oplus 0, 0\oplus \tau\rb=v\oplus\Delta_u\tau$.
Note first that
\[\llbracket} \newcommand{\rb}{\rrbracket u\oplus 0, 0\oplus \tau\rb+\llbracket} \newcommand{\rb}{\rrbracket 0\oplus \tau, u\oplus 0\rb
=\mathcal D\langle\langle u\oplus 0, 0\oplus
\tau\rangle\rangle_C=\mathcal D\langle \tau,u\rangle.
\]
The map $\mathcal D:C^\infty(M)\to\Gamma(C)$
is given by
\[\langle\langle u\oplus\sigma, \mathcal D \varphi\rangle\rangle_C=
(\pr_{TM}(u)+\rho\circ\pr_A(\sigma))\varphi\]
for all $u\oplus \sigma\in\Gamma(C)$,
i.e. $\mathcal D\varphi=0\oplus(0,\dr\varphi)$.
Then, by the Leibniz property of the Courant algebroid bracket on $C$,
we find
\begin{align*}
\rho(a)\langle u, \sigma\rangle&=c(0\oplus\tau)\langle\langle u\oplus
0, 0\oplus\sigma\rangle\rangle_C\\
&=\langle\langle\llbracket} \newcommand{\rb}{\rrbracket 0\oplus\tau, u\oplus
0\rb, 0\oplus\sigma\rangle\rangle_C
+\langle\langle u\oplus
0, \llbracket} \newcommand{\rb}{\rrbracket 0\oplus\tau, 0\oplus\sigma\rb\rangle\rangle_C\\
&=\langle\langle (-v)\oplus(-\Delta_u\tau+(0,\dr\langle \tau, u\rangle)) , 0\oplus\sigma\rangle\rangle_C
+\langle\langle u\oplus
0, 0\oplus(\ldr{a}\sigma+(0,-\ip{\rho(b)}\dr\theta))\rangle\rangle_C\\
&=-\langle v,\sigma\rangle-\langle
(\rho,\rho^*)\Omega_ua,\sigma\rangle-\langle\ldr{X}\theta,\rho(b)\rangle+\rho(b)\langle\theta,X\rangle+\langle
u, \ldr{a}\sigma\rangle
-\dr\theta(\rho(b),X)
\end{align*}
This leads to
\[ -\langle v,\sigma\rangle=\langle
(\rho,\rho^*)\Omega_ua,\sigma\rangle+\langle \ldr{a}u, \sigma\rangle
\]
and, since $\sigma$ was arbitrary, we have shown that
\[\llbracket} \newcommand{\rb}{\rrbracket u\oplus 0, 0\oplus \tau\rb=(-\nabla^{\rm
  bas}_\tau u)\oplus\Delta_u\tau.\]
\end{proof}

\begin{example}
Consider an $\mathcal{LA}$-Dirac triple as in Example
\ref{lie_algebroid_dual_3}.
The vector bundle morphisms
\[\Psi: C\to A\oplus A^*, \qquad
\Psi((\rho_*\xi,\xi)\oplus(a,\theta))=( a+\rho_*^*\theta, \xi+\rho^*\theta)
\]
and
\[\Phi:A\oplus A^*\to C, \qquad \Phi(a,\xi)=(\rho_*\xi,\xi)\oplus(a,0)
\]
are well-defined and inverse to each other.
A straightforward computation using the
considerations in Examples \ref{lie_algebroid_dual} and \ref{lie_algebroid_dual_3} shows that
$C$ is isomorphic to the Courant algebroid structure on $A\oplus A^*$
induced by the Lie bialgebroid $(A,A^*)$ \cite{LiWeXu97}\footnote{In
  \cite{LiWeXu97}, the authors work with the definition of Courant
  algebroids with antisymmetric brackets. Here, we get the
  corresponding Courant algebroid as we chose to define them.}
\end{example}

\begin{example}
Consider now an $\mathcal{LA}$-Dirac triple as in Example
\ref{IM_2_form_3}.
Again, we find that the vector bundle morphisms
\[\Pi: C\to TM\oplus T^*M, \qquad
\Pi((X,-\sigma^*X)\oplus(a,\theta))=(X+\rho(a), \theta+\sigma(a))
\]
and
\[\Theta:TM\oplus T^*M\to C, \qquad
\Theta(X,\theta)=(X,-\sigma^*X)\oplus (0,\theta)
\]
are well-defined and inverse to each other.
Here, one gets immediately  that $C$ and the standard Courant algebroid $TM\oplus
T^*M$ are isomorphic via these maps.
\end{example}

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