\section{Stochastic flows and LeJan-Watanabe connections}
\label{section-1}
Consider a Stratonovich stochastic differential equation
\begin{equation}
\label{5}
dx_t=X(x_t)\circ dB_t +A(x_t)dt
\end{equation}
on $M$ driven by a $m$-dimensional Brownian motion $\{B_t, 0\le t<\infty\}$.
Here $X: M\times \R^m\to TM$ and $A: M\to TM$ are assumed to be smooth and
$X(x):= X(x,-):\R^m\to T_xM$ is assumed to be linear, surjective for each $x$
and to induce the inner product $\langle, \rangle_x$ on $T_xM$ given
by the Riemannian structure of $M$. This implies the Markov process of
solutions to (\ref{5}) has infinitesimal generator of the form
\begin{equation}
\label{6}
\A={1\over 2}\Delta +\LL_{Z}
\end{equation}
where $\LL_Z$ is Lie differentiation in the direction of some vector
field $Z$:
$$\LL_Z(f)(x)=df(Z(x)), \hskip 20pt x\in M.$$
The solution to (\ref{5}) from $x$ shall be denoted by
$\{\xi_t(x), 0\le t<\infty\}$. Write $x_t=\xi_t(x_0)$. We can, and will,
take versions which makes $\xi_t(x)$ continuous in $t$ and a $C^\infty$
diffeomorphism of $M$ in $x$.
\bigskip

Let $Y:TM\to \R^m$  be the adjoint of $X$, i.e. $Y_x=X(x)^*$, so that
$Y$ is the right inverse to $X$. As described in
 \cite{Elworthy-LeJan-Li-Tani}, or more generally in
 \cite{Elworthy-LeJan-Li-book}, $X$ induces a metric connection
 $\breve \nabla$ on $M$ defined by
\begin{equation}\label{7}
\breve \nabla_v U=X(x)d [x\mapsto Y_x(U(x))](v), \hskip 20pt v\in
T_xM\end{equation} for any smooth vector field $U$ on $M$. In
general this, which we call the {\it LeJan-Watanabe connection } of
our stochastic differential equation, has torsion. The torsion is given by
\begin{equation}
\label{8}\breve T(u,v)=X(x)dY(u,v),
\hskip 20 pt u.v\in T_xM
\end{equation}
where $dY$ refers to the exterior derivative of $Y$ considered as
an $\R^m$-valued differential 1-form on $M$, see Proposition 2.2.3
in \cite{Elworthy-LeJan-Li-book}. When the torsion vanishes $\breve \nabla$
 is  just the Levi-Civita connection, which we will always refer to
using $\nabla$, ${D\over \partial t}$ etc. There are two principal
classes of examples, Example 1.1 and Example 1.2 below, for which
 this holds.

\subsubsection*{Example 1.1. Gradient system.}
Here we consider an isometric immersion  $\alpha:M\to \R^m$ for
some $m$, e.g. by using Nash's theorem. Let $X(x): \R^m\to T_xM$
be the orthogonal projection identifying $T_xM$ with its image in
$\R^m$ under $d\alpha$. Then $Y_x: T_xM\to \R^m$ is just the
inclusion $(d\alpha)_x$ and (\ref{7}) is the classical formula for
the Levi-Civita connection of a sub-manifold of $\R^m$.


\subsubsection*{Example 1.2. Riemannian symmetric spaces.}
Let $M$ be a symmetric space $(K,H,\sigma)$. In particular $K$ is a Lie group
 acting transitively on $M$ and $H$ can be identified with the isotropy
 subgroup fixing the point $x_0$ of $M$, so that $k\mapsto kx_0$ gives a
diffeomorphism of $K/H$ with $M$. Assume that the Lie algebra
$\kk$ of $K$ has an inner product $\langle, \rangle_{\kk}$, say,
invariant under the adjoint action of $K$ on $\kk$. The action induces
a stochastic differential equation
$$X: M\times \kk \to TM$$ with
$X(x)e={d\over dt}[(\exp{(te)})\cdot x]_{t=0}$. It is surjective and so
induces a Riemannian structure $(\langle, \rangle_x, x\in M)$ on
$M$ which is $K$-invariant. The LeJan-Watanabe connection $\breve \nabla$
is the Levi-Civita connection for the Riemannian structure, as is proved in
 Corollary 1.4.9 of \cite{Elworthy-LeJan-Li-book}.
 \bigskip

An important class of examples which give connections other than the
Levi-Civita connections are the left invariant, and the right invariant
stochastic differential equations  on Lie groups. For $M$ a
Lie group with left invariant metric take the Lie algebra $\g$
as $\R^m$ and define $$X^L: G\times \g \to TG$$
to be left invariant with $X^L(e)\alpha=\alpha$, for $e$ the identity element
of $G$, i.e.
$$X^L(g)\alpha=TL_g(\alpha)$$
for $L_g: G\to G$ the left translation $x\mapsto g\cdot x$. Similarly if
 $R_g$ denotes right translation, $x\mapsto x\cdot g$, define
 $X^R:G\times \g \to TG$ by $X^R(g)\alpha=TR_g(\alpha)$.
 It is easy to see that the associated  connections in the sense of
 (\ref{7}) are just the canonical left and right  invariant connections,
  $\nabla^L$ and $\nabla^R$ say, of $G$, respectively. The flow $\xi_t^L$
 for the left invariant stochastic differential equation
 $dx_t=X^L(x_t)\circ dB_t$, where $B_t$ is a $\g$-valued Brownian motion,
is given by   $$\xi_t^L(x)=xg_t, \hskip 20 pt x \in G, t\ge 0.$$
 Here $g_t$ is the solution starting from the identity $e$ to
 $dg_t=X^L(g_t)\circ dB_t$. Similarly the solution $\xi_t^R(x)$,
  with $\xi_0^R(x)=x$  to the right invariant stochastic differential equation
$dx_t=X^R(x_t)\circ dB_t$ is given by  the right translation of $h_t$,
the solution starting from $e$. If the metric on $G$ is bi-invariant  both
stochastic differential equations, with $A\equiv 0$, have Brownian motions
as solutions. Also in that case the connections are torsion skew symmetric,
with $\nabla^L$ and $\nabla^R$ adjoint to each other. In particular the
torsions $T^L$ and $T^R$ are given by
\begin{equation}\label{torsion-of-Lie-groups}
T^L(u,v)\equiv -[U,V](g), \hskip 15pt  T^R(u,v)=[U,V](g),
\end{equation}  where $U$, $V$
are left invariant vector fields with $u=U(g)$ and $v=V(g)$.

In the bi-invariant case we can also treat $G$ as a symmetric space
by taking $K=G\times G$ with action $(g_1,g_2)x=g_1xg_2^{-1}$,
with $H=\Delta G\equiv\{ (g,g)\in K: g\in G\}$ the isotropy subgroup
fixing the identity $e$ of $G$. The symmetry map
 $\sigma:\g\times\g\to \g$ is given by
 $\sigma(\alpha, \beta)=(\beta, \alpha)$.
The stochastic differential equation induced by the action as described
in Example 1.2  above is just
\begin{equation}\label{sde-bi-invariant-Lie-group}
dx_t=TR_{x_t}\circ dB_t -TL_{x_t}\circ dB_t^\prime
\end{equation}
where $(B_t)$ and $(B_t^\prime)$ are two independent
Brownian motions on $\g$. The solution flow is then given by
$\xi_t(x)=g_tx(g_t^\prime)^{-1}$ where $g_t$ and $g_t^\prime$ are independent
 Brownian motions on $G$ and solutions to  the following right invariant
stochastic differential equations respectively:
$$dg_t=TR_{g_t}\circ dB_t, \hskip 18pt g_0=e,$$
$$dg_t^\prime=TR_{g_t^\prime}\circ dB_t^\prime, \hskip 18pt g_0=e,$$
The solutions $\xi_t(x)$, when $A\equiv 0$,  are also Brownian motions on $G$.

For more general homogeneous spaces $M=K/H$, if the
metric on $M$ is induced from an ad$_K$-invariant inner product
on $\kk$ and the orthogonal complement of the ker $X(x_0)$ is
ad$_H$ invariant then $\breve \nabla$, the connection associated to
the  stochastic differential equation defined as in Example 1.2 above,
is $K$-invariant. (see \cite{Elworthy-LeJan-Li-book}, Proposition 1.2.9).
Again in that case $$\xi_t(x)=k_t x, \hskip 18pt x\in M$$
for $k_t$ the solution to the equation $dk_t=TR_{k_t}\circ dB_t$ on $K$
 with initial point $k_0=e$.

\bigskip

\noindent{\bf Stochastic flows and differential forms.}
For $\{\xi_t: t\ge 0\}$ the flow of our stochastic differential equation
(\ref{5}) there are semigroups on differential forms defined by
$P_t\phi=\E\xi_t^*\phi$.
On $q$-forms the semigroup has infinitesimal generator $\A^{q}$
given on smooth forms by
$$\A^{q} \phi={1\over 2}\sum_{j=1}^m \LL_{X^j}\LL_{X^j}\phi+\LL_{A}(\phi)$$
for $\LL_{X^j}$ denoting the Lie differentiation in the direction $X^j$, etc
and with $X^j$ denoting the vector fields $X(\cdot)e^j$ for $e^1,\dots e^m$
the standard basis of $\R^m$. This comes from It\^o's formula, see
\cite{Elworthy-flow} or \cite{Elworthy-Stflour}. The generator can be written
\begin{equation}\label{11}
\A^q(\phi)=-(d\hat \delta+\hat \delta d)\phi+\LL_A(\phi)
\end{equation}
where, for $V\in \wedge^{q-1}TM$,
\begin{equation}
\label{hat-delta}
\begin{split}
\hat \delta \phi(V)=&-\sum_{j=1}^m(\iota_{X^j}\LL_{X^j}\phi)(V)\\
=&-trace \hat\nabla_-\phi(-,V),
\end{split}
\end{equation}
and $\hat \nabla$ is the adjoint connection to $\breve \nabla$,
in the sense of Driver \cite{Driver92} so the torsion of $\hat\nabla$
is $-\breve T$. See \cite{Elworthy-LeJan-Li-book}. This shows  $\A^q$
depends only on the connection $\breve \nabla$, the vector field $A$,  and  the Riemannian  metric on $TM$. In particular if $A\equiv 0$ in (\ref{5}) and  the associated
connection $\breve \nabla$ is the Levi-Civita connection we get  the usual
Hodge-Kodaira Laplacian, c.f. \cite{Kusuoka-degree}.



\section{Spaces of forms on the path space of $M$}

Fix $T>0$ and $x_0\in M$ and let $C_{x_0}M$ denote the Banach manifold of
continuous paths $\sigma:[0, T]\to M$ with $\sigma(0)=x_0$. Let $\mu_{x_0}$
be the measure induced on it by the  solution to (\ref{5}) starting
from $x_0$.

We can identify the tangent space $T_\sigma C_{x_0}M$ to $C_{x_0}M$ at
$\sigma$ with the space of continuous $v:[0,T]\to TM$ with $v(0)=0$ such that
$v(t)\in T_{\sigma(t)}M$ for each $0\le t \le T$. For $q=1,2,\dots$, let
$\wedge^qT_\sigma C_{x_0}M$ be the space of anti-symmetric $q$-vectors at
$x_0$ completed using the largest reasonable cross norm, so that its dual
space can be identified with the Banach space of continuous alternating
$q$-linear maps
$\phi: T_\sigma C_{x_0}M\times \dots \times T_\sigma C_{x_0}M\to \R$.
Thus $q$-forms on $C_{x_0}M$ are sections of the dual bundle
$(\wedge^qT_\sigma C_{x_0}M)^*$, e.g. see \cite{Lang-book}.

 As an example of a $q$-vector on $C_{x_0}M$ consider $\wedge^q(T_{x_0}\xi)(U)$
evaluated on a fixed $\omega  \in \Omega$, where $U\in \wedge^q T_{x_0}M$. It
is then a $q$-vector at $\{x_t(\omega): 0\le t\le T\}$. For $q=1$ it is the
tangent vector $\{T_{x_0}\xi_t(\cdot, \omega)(U): 0\le t \le T\}$ to
$C_{x_0}M$ at $x_\cdot(\omega)$. For higher $q$ we identify $q$-vectors $V$
with certain continuous maps $V:[0,T]^q \to \otimes^q TM$ such that
$V_{t_1,t_2,\dots, t_q}\in T_{\sigma(t_1)}\otimes \dots \otimes T_{\sigma(t_q)}M$.
Then, generalizing to allow $U$ to be in $\wedge^{q+1}C([0,T];T_{x_0}M)$,
$$\wedge^{q+1}(T_{x_0}\xi)(U)_{t_1,\dots, t_{q+1}}
=\left(T_{x_0}\xi_{t_1}\otimes \dots\otimes T_{x_0}\xi_{t_{q+1}}\right)
\left(U_{t_1,\dots,t_{q+1}}\right).$$
As $\omega$ varies in $\Omega$ such $q+1$ vectors do not strictly speaking
form $q+1$ vector fields on $C_{x_0}M$. However we can obtain $q$-vector
fields from them by conditioning, `filtering out the redundant noise',
to give
$\sigma\mapsto \E\{\wedge^{q+1}(T_{x_0}\xi)(U)\left|
 \xi_\cdot(x_0, \omega)=\sigma \right.\}$.
As shorthand, we write:
\begin{equation}
\overline {\wedge^{q+1}(T_{x_0}\xi)}_\sigma (U)
 =\E\{\wedge^{q+1}(T_{x_0}\xi)(U)\left|
 \xi_\cdot(x_0, \omega)=\sigma \right.\}.
 \end{equation}
  See \cite{Elworthy-Yor},\cite{Elworthy-LeJan-Li-book}. Below we shall often
  use  $\overline f$ denote the conditional expectation of a random function
  $f$ with respect to the $\sigma$-algebra generated by $\{x_t: 0\le t<\infty\}$.

An important class of forms on $C_{x_0}M$ are the cylindrical forms. In
particular if $\phi$ is a $q$-form on $M$ and $t\in [0, T]$ we have
the cylindrical $q$-form $\phi^{t}$ on $C_{x_0}M$ given by
$\phi^t(V)=\phi(V_{t,\dots, t})$.

For $\psi$ a $(q-1)$-form on $M$ we see, with $P_t$ as in the last section,
\begin{equation}
P_t(d\psi)=\E(d\psi)\left(\wedge^q(T_x\xi_t)U\right)
=\int_{C_{x_0}M} (d\psi)^t(\overline{\wedge^q T_{x}\xi})(U))\,
 d\mu_{x_0}(\sigma).
\end{equation}
In general we say that an $L^2$ $q$-vector field $V$ {\it has a divergence}
if there is an $L^2$ $(q-1)$-vector field $div V$ on $C_{x_0}M$ such that
for all smooth cylindrical $(q-1)$-forms $\phi$ on $C_{x_0}M$ we
have
\begin{equation}
\label{14}
\int_{C_{x_0}M} d\phi(V)\, d\mu_{x_0}
=- \int_{C_{x_0}M} \phi(div V) \, d\mu_{x_0}.
\end{equation}

For $U=u^1\wedge\dots \wedge u^q$, a primitive vector,  the non-intrinsic Bismut
 type formula (\ref{4}) will follow from showing that
$\overline{\wedge^q(T_{x_0}\xi)}(U)$ has a divergence with
\begin{eqnarray*}
&&div\left(\overline{\wedge^qT_{x}\xi}_\sigma(U)\right)_{t,\dots, t}\\
&&={1\over t}\E\left\{
\sum_{j=1}^q (-1)^{j}\left(\int_0^t \langle T\xi_s (u^j), dx_s\rangle
\right) \wedge^{q-1}(T\xi_t)(u^1\wedge\dots \widehat{u^j}\dots \wedge u^q)
\right.\\
&& \hskip 30pt \left. \left\vert \; \xi_\cdot(x_\cdot)=\sigma \right. \right\}
\end{eqnarray*}
at least for a gradient stochastic differential equation with no drift.
Here we have adopted the convention that $\widehat{u^j}$ means the omission
of the vector $u^j$ in the tensor.
For a general $q$-vector $U$ in $\wedge^qT_{x_0}M$ this could be written
$$div\left(\overline{\wedge^qT_{x}\xi}_\sigma(U)\right)_{t,\dots, t}
=-{1\over t}  \overline{\wedge^{q-1}(T\xi_t)
\iota_{\int_0^t \langle T\xi_s -, dx_s\rangle}U}.$$
 Equation (\ref{1}) will follow, as in \cite{Elworthy-Li-forms-CR},
by calculating the conditional expectations. Note that we have used
another convention: if $\ell$ is a linear map on a vector space
 $$\iota_\ell(b^1\wedge\dots\wedge b^q)
 =\sum_{j=1}^q (-1)^{j+1}\ell(b^j)b^1
   \wedge\dots\wedge\widehat{b^j}\wedge\dots \wedge b^q.$$

In \cite{Elworthy-Li-Hodge-2}, see also \cite{Elworthy-Li-Hodge-1},
a general theory for forms on path spaces is given. There we define
Hilbert spaces $\HH^{(q)}$, densely included in
$\wedge^q T_\sigma C_{x_0}M$ and show that suitably regular sections of
$\HH^{(q)}$ do have a divergence and give evidence to suggest that
the divergence is a section of $\HH^{(q-1)}$, with proofs for $q=2,3$. This
is the major step in the construction of an $L^2$ deRham complex on
$C_{x_0}M$, with an $L^2$ Hodge decomposition and analogue of the
 Hodge-Kodaira Laplacian. For this construction we take a stochastic
differential equation (\ref{5}) on $M$ whose solutions are Brownian
motions and with its associated connection $\breve \nabla$ the Levi-Civita
connection, for example by taking an isometric embedding in $\R^m$
and  $A\equiv 0$. We take $(B_\cdot)$ to be the canonical Brownian
motion defined up to time $T$ so that  $C_0\R^m$ with Wiener measure $\PP$
is the underlying probability space. For fixed $x_0\in M$ there is the
It\^o map
\begin{eqnarray*}
\I: C_0\R^m &\to& C_{x_0}M\\
\I(\omega)_t&=&\xi_t(x_0,\omega).
\end{eqnarray*}
It is smooth in the sense of Malliavin Calculus with $H$-derivative
$$T_\omega\I: H\to TC_{x_0}M$$
defined for almost all $\omega\in C_0\R^m$.  Here $H$ is the Cameron-Martin
space; $H=L_0^{2,1}\left([0,T];\R^m\right)$. We have the formula
of Bismut
\begin{equation}
\label{formula-TI}
T\I(h)_t=T\xi_t\int_0^t \left(T\xi_s\right)^{-1}X(x_s)\dot h_s\; ds.
\end{equation}
By an $L^2$ $H$-$q$-vector field on $C_0\R^m$ we mean an element
$U\in L^2\left(C_0\R^m;\wedge^qH\right)$ where $\wedge^q H$ refers to
the completed tensor product using the usual Hilbert space cross norm.
Given such a vector field $U$ we can define
$$(\wedge^qT\I)(U): C_0\R^m \to \wedge^q TC_{x_0}M.$$
At $\omega\in C_0\R^m$ this is in $\wedge^q T_{x_\cdot(\omega)}C_{x_0}M$.

To obtain a $q$-vector field on $C_{x_0}M$ define, for $\sigma\in C_{x_0}M$
$$\overline{(\wedge^qT\I)(U)}_\sigma
=\E\left\{(\wedge^qT\I)(U)\left| x_\cdot=\sigma\right.\right\}.$$
Taking constant $U$, that is $U(\omega)\equiv h$, some $h\in H$ we obtain
continuous linear maps
$$\overline{(\wedge^qT\I)}_\sigma: \wedge^qH\to \wedge^qT_\sigma C_{x_0}M$$
defined by $$\overline{(\wedge^qT\I)}_\sigma(h)
 =\overline{(\wedge^qT\I)(h)}_\sigma.$$
We can then define $\HH^{q}_\sigma=Image(\overline{\wedge^q T\I}_\sigma)$
with its induced Hilbert space structure, inner product
 $\langle, \rangle_\sigma$, so that $(\overline{\wedge^q T\I})_\sigma$
becomes an isometry. These spaces are defined for almost all
 $\sigma\in C_{x_0}M$. It is shown in \cite{Elworthy-Li-Hodge-2} that
 they depend only on the Riemannian structure of $M$. The fact that
suitably regular $L^2$ sections of $\HH^{q}$ have a divergence can be
 deduced from the theory for Wiener spaces given by Shigekawa
\cite{Shigekawa-Hodge}.

The following is a specialization of his basic result to primitive
adapted $q$-vector fields.
\begin{theorem} [Shigekawa \cite{Shigekawa-Hodge}]
\label{theorem-Shigekawa}
For $q>1$, let $\phi:C_0\R^m\to (\wedge^q H)^*$ be an
H-differential $q$-form on Wiener space and $h^i: C_0\R^m \to H$,
$i=1$ to $q$ adapted vector fields such that
$h:=h^1(\cdot)\wedge\dots \wedge h^{q+1}(\cdot)$ is in
$\D^{1,2}(\wedge^{q+1}H)$. If also $\phi\in
\D^{1,2}\left((\wedge^qH)^*\right)$ then
$$\int_{C_0\R^m} d\phi(h)\; d\PP
=-\int_{C_0\R^m} \phi\left(div h\right)\, d\PP$$ where
\begin{eqnarray*}
div h&=&\sum_{j=1}^q (-1)^{j} \int_0^T \langle \dot h^j_s,
dB_s\rangle \left(h^1\wedge \dots \wedge \widehat{h^j}\wedge \dots
\wedge h^q \right)\\
&&-\sum_{1\le i<j\le q+1}(-1)^{i+j}[h^i,h^j]\wedge h^1 \wedge\dots
\wedge \widehat {h^i}\wedge\dots \wedge \widehat {h^j}\wedge\dots
\wedge  {h^{q+1}}.
\end{eqnarray*}
\end{theorem}
The assumptions of primitivity of $h$ or adaptness of the components could
have been omitted with Skorohod integrals replacing It\^o integrals,
 see \cite{Shigekawa-Hodge}.
\bigskip


The extension to forms on $C_{x_0}M$ in
\cite{Elworthy-Li-Hodge-1}, \cite{Elworthy-Li-Hodge-2} and the
proof we give below for formulae (1) and (2) is based on the
following observation:

\begin{theorem}\label{theorem-2.2}
Let $\psi$ be a smooth cylindrical $q$-form on $C_{x_0}M$. Then
for $h\in \D^{1,2}(\wedge^{q+1}H)$ of the form $h=h^1\wedge \dots
\wedge h^{q+1}$ with each $h^i$ adapted, the $q$+$1$-vector field
$\overline{\wedge^{q+1}T\I(h)}$ on $C_{x_0}M$ has a divergence:
\begin{equation}
\int_{C_{x_0}M} d\psi\left(\overline{\wedge^{q+1}T\I(h)}\right)
d\mu_{x_0} =-\int_{C_{x_0}M} \psi\left(div
\overline{\wedge^{q+1}T\I(h)}\right) d\mu_{x_0}
\end{equation}
and
\begin{equation}
\label{18}
div \left(\overline{\wedge^{q+1}T\I(h)}\right)
= \overline{\wedge^{q}T\I(div h)}.
\end{equation}
\end{theorem}

{\it Proof.}
  The pull back form $\I_*(\psi)$ is in
  $\D^{1,2}\left((\wedge^qH)^*\right)$. The integration by parts
  formula of Shigekawa, Theorem \ref{theorem-Shigekawa},  applies to give:
  \begin{eqnarray*}
&&\int_{C_{x_0}M} d\psi\left(\overline{\wedge^{q+1}T\I(h)}\right)
d\mu_{x_0}
=\int_{C_0\R^m} d(\I_*\psi)(h)\; d\PP\\
&=&-\int_{C_0\R^m} \I_*\psi\left(div h\right)\, d\PP
=-\int_{C_{x_0}M}\psi\left(\overline{\wedge^{q}T\I(div h)}\right)
d\mu_{x_0}. \\
\end{eqnarray*}
\hfill\rule{3mm}{3mm}


For other discussions of differential forms
 on path spaces see \cite{Jones-Leandre91}, \cite{Kusuoka-forms}
 and \cite{leandre-survey} together with the references there.


\section{Some families of $H$-valued vector fields}

{\bf 3.1 \hskip 5pt The Lie brackets.}
 Below we identify a family of vector fields whose Lie
 brackets behave particularly well.

\begin{proposition}\label{proposition-3.1}
Given tangent vectors $b_i \in T_{x_0}M, i=1,2$ and   $\rho: [0,T]\to \R$,
 a non-random $L^1$ function, set:
\begin{equation}\label{9}
h_{\cdot }^i=\int_0^{\cdot } \rho(r)Y_{x_r}\left( T\xi _r(b^i)\right) \;dr.
\end{equation}
Then the Lie bracket of the two vector fields $h^i$ on the Wiener space
 $C_0\left(\R^m\right)$ is given by:
\begin{equation}
\begin{split}
\frac d{dt}\left[ h^1,h^2\right]_t
 &=\rho(t) \left(\int_0^t \rho(r)\, dr \right)
         (dY)\left(T\xi _t(b^1), T\xi_t (b^2)\right) \\
 &+\rho(t)Y_{x_t}\left( T\xi_t \int_0^t  \rho(r) (T\xi_r)^{-1}  X(x_r)dY
    \left(T\xi_r(b^1),T\xi _r(b^2)\right)\, dr\right). \end{split}
\end{equation}
and so
\begin{equation}
\label{20}
T\I_t([h^1,h^2])
=\left(\int_0^t\rho(s)ds\right) T\xi_t\left(\int_0^t\rho(s) T\xi_s^{-1}
X(x_s)dY\left(T\xi_s(b^1), T\xi_s(b^2)\right)ds\right).
\end{equation}
\end{proposition}

{\it Proof.}  Denote by $D$ differentiation on Euclidean spaces, $D^H$ the
H-derivatives of Wiener functionals,  $\nabla$ covariant differentiation
using the Levi-Civita connection on $M$, and   $\nabla ^H$ the corresponding
covariant H-derivatives. Below we regularly abuse notation by omitting
$\omega$ in the argument, especially with $\xi_t$ and $T\xi_t$.
First observe that
\begin{equation}
\label{22a}
\begin{split}
T{\I}_t\left(h^i\right) (\omega) =&T\xi _t\int_0^t
(T\xi_r)^{-1} X(x_r(\omega))(\dot h_r^i)\,dr \\
 =&\left(\int_0^t \rho(r)dr\right)T\xi_t(b^i)
\end{split}
 \end{equation} and that
\begin{equation}
\label{derivative-of-Lie-bracket}
\begin{split} \frac d{dt}\left[ h^1,h^2\right] _t(\omega )
&=\frac d{dt}\left(
D^Hh_t^2(\omega )(h^1(\omega ))-D^Hh_t^1(\omega )(h^2(\omega ))\right) \\
&=D^H\dot{h}_t^2(\omega )(h^1(\omega ))-D^H\dot{h}_t^1(\omega )
(h^2(\omega)).
\end{split}
\end{equation}
Now
\begin{eqnarray*}
&&D^H\dot{h}_t^2(\omega )(h^1(\omega )) \\
&=&D^H\left( Y_{x_t}\left( T\xi _t(\rho(t)b^2)\right)
\right)(\omega) (h^1(\omega )) \\
&=&\rho(t) \left(\nabla_{h^1(\omega )}^H Y_{x_t}\right) \left(T\xi_t(b^2)\right)
  +   \rho(t) Y_{x_t(\omega )}\left(\nabla^H _{h^1(\omega )}
    \left(T\xi _t(b^2)\right) \right) \\
&=&\rho(t) \left(\nabla_{T_\omega{\I}_t(h^1)} Y\right)
  \left(T\xi _t(b^2)\right) +\rho(t)  Y_{x_t(\omega )}
\left( \nabla^H_{h^1(\omega )}\left(T\xi _t(b^2)\right)\right).
\end{eqnarray*}
By the expression (\ref{22a}) for $T\I_t(h^1)$ , we have
\begin{equation}
\label{derivative-of-h}
\begin{split}
D^H\dot{h}_t^2(\omega )(h^1(\omega ))
=&\rho(t)\left(\int_0^t\rho(r)dr\right)
   \nabla_{T\xi _t(b^1)} Y\left(T\xi_t(b^2)\right)\\
 & +\rho(t)  Y_{x_t(\omega )}\left(\nabla _{h^1(\omega )}^H\left(T\xi_t(b^2)\right)\right).
\end{split}
\end{equation}
To calculate the second term, consider the map
\begin{eqnarray*}
&\tilde {\I}_t:&M\times C_0\left( R^m\right) \rightarrow M\\
& \tilde \I_t(x,\omega)&=\hskip 8pt\xi_t(x)(\omega)
\end{eqnarray*}
with $T_{(x,\omega)}{\tilde\I_t}:T_xM\times L_0^{2,1}(\R^m)
          \to T_{\xi_t(x)(\omega)}M$
 its derivative in the sense of Malliavin calculus. Then
 $$T_{(x,\omega)} \tilde{\I}_t(u,h)
 =T_x\xi _t \int_0^t\left(T_x\xi_r\right)^{-1}
   X(\xi _r(x)) (\dot h_r)\,dr  +  T\xi_t(u),$$
  and $T_{( x_0,\omega)}{\tilde{\I}_t}(u,h(\omega))=T\I_t(h(\omega))+T\xi(u)$.
 Since $b^2$ is non-random,
\begin{eqnarray*}
&&Y_{x_t(\omega )}\left( \nabla _{h^1(\omega )}^H
   \left(T\xi _t(b^2)\right) \right)
=  Y_{x_t(\omega )}\left(\nabla^H_{h^1(\omega)} (T\xi_t)\right) (b^2)  \\
&=& Y_{x_t(\omega ))} \nabla _{b^2}
    \left[ T_{(-,\omega)}{\tilde{\I}}\left(0,h^1(\omega)\right) \right]_t.
\end{eqnarray*}
But
\begin{eqnarray*}
&& \nabla _{b^2} \left[ T_{(-,\omega)}{\tilde{\I}}
 \left(0,h^1(\omega)\right) \right]_t\\
&=&\nabla_{b^2}\left[ T_{-}\xi _t \int_0^t
   \left(T_{-}\xi_r\right) ^{-1}X(\xi _r(-,\omega ))
      \left( \dot{h}_r^1\left(\omega \right) \right) \,dr \right]  \\
&=&\nabla _{b^2}\left[ T_{-}\xi _t \int_0^t
   \left( T_{-}\xi_r\right) ^{-1} X(\xi _r(-))
   \rho(r)  Y_{x_r}\left(T\xi _r(b^1)\right)
     \,dr \right].
\end{eqnarray*}
Since $\left( T_{-}\xi _r\right) ^{-1}X(\xi _r(-,\omega ))
        Y_{\xi _r(-)}\left(T\_\xi _r\right)=Id$,

\begin{eqnarray*}
&&\nabla _{b^2}\left[T_{\left(-,\omega \right) }{\tilde{\I}}
\left(0, h^1(\omega )\right) \right] _t\\
&=&\left(\int_0^t\rho(r)dr \right) \nabla _{b^2}\left( T\xi _t\right)(b^1)\\
&&-T_{x_0}\xi_t\int_0^t\left(T_{x_0}\xi _r\right)^{-1}X(x_r)
    \nabla _{b^2}\left(\rho(r) Y_{\xi _r(- )}
     \left( T_{-}\xi _r(b^1)\right) \right)  \\
&=& \left(\int_0^t\rho(r)dr \right)
     \nabla _{b^2}\left( T\xi _t\right)(b^1)
     -T_{x_0}\xi _t\int_0^t\rho(r)
   \left(T_{x_0}\xi _r\right)^{-1}
  \nabla_{b^2} \left(T_{-}\xi _r\right)(b^1)\\
&& -T_{x_0}\xi _t\int_0^t \rho(r)\left( T_{x_0}\xi _r\right) ^{-1}
     X(x_r)\nabla _{b^2} \left( Y_{\xi _r(- )}\right)
  \left(T_{x_0}\xi _r(b^1)\right)   \\
  &=&  \left(\int_0^t\rho(r)dr \right)
  \, \nabla _{b^2}\left( T\xi _t\right) \left( b^1\right)
   - T\xi _t\int_0^t \rho(r) \left(T\xi _r\right)^{-1}
  \nabla_{b^2}\left(T\xi _r\right)(b^1) \;dr\\
&& -  T\xi_t \int_0^t \rho(r) \left( T\xi_r\right)^{-1}
     X(x_r)\left(\nabla_{T\xi _r(b^2)} Y\right)
     \left( T\xi _r(b^1)\right) \;dr,
\end{eqnarray*}
 and so
\begin{eqnarray*}
&&\nabla _{b^2}\left[ T_{(-,\omega)}\tilde\I
   \left(0, h^1(\omega )\right) \right] _t-\nabla _{b^1}
  \left[ T_{\left(-,\omega \right) }\tilde{\I}
         \left(0, h^2(\omega )\right)\right] _t \\
&=&\left(\int_0^t \rho(r)dr\right)
 \left( \nabla _{b^2}\left( T\xi _t\right) \left( b^1\right)
-\nabla_{b^1}\left(T\xi _t\right) \left( b^2\right) \right)\\
&& - T\xi _t\int_0^t \rho(r) \left( T\xi _r\right)^{-1}
      \left( \nabla_{b^2}\left( T\xi _r\right)(b^1)
       -\nabla _{b^1}\left( T\xi_r\right)(b^2) \right) \;dr \\
&& -T\xi _t\int_0^t  \rho(r)\left( T\xi _r\right) ^{-1}
   X(x_r) dY\left( T\xi _r(b^2), T\xi _r(b^1)\right)\;dr \\
&=& -T\xi _t  \int_0^t\rho(r) \left(T\xi _r\right)^{-1}
   X(x_r) dY\left( T\xi _r(b^2),T\xi _r(b^1)\right)\;dr.
\end{eqnarray*}
Finally combining (\ref{derivative-of-Lie-bracket}), (\ref{derivative-of-h})
and the last identity, we have

\begin{eqnarray*}
&&\frac d{dt}\left[ h^1,h^2\right]_t(\omega ) \\
&=& \rho(t)\left(\int_0^t\rho(r)dr\right)
    \left(\nabla_{T\xi _t(b^1)} Y\right)
      \left( T\xi _t(b^2)\right)\\
&&-\rho(t)\left(\int_0^t\rho(r)dr\right)
    \left(\nabla_{T\xi _t(b^2)} Y \right)
      \left( T\xi _t(b^1)\right)\\
 &&  -  \rho(t)Y_{x_t}\left( T\xi _t  \int_0^t\rho(r) \left(T\xi _r\right)^{-1}
   X(x_r) dY\left( T\xi _r(b^2),T\xi _r(b^1)\right)\;dr\right)\\
&=&\rho(t)\left(\int_0^t\rho(r)dr\right)
    dY\left(T\xi _t(b^1),T\xi _t(b^2)\right)\\
 &&  + \rho(t)Y_{x_t}\left( T\xi _t  \int_0^t\rho(r) \left(T\xi _r\right)^{-1}
   X(x_r) dY\left( T\xi _r(b^1),T\xi _r(b^2)\right)\;dr\right).
\end{eqnarray*}

To prove (\ref{20}), simply do an integration by parts:

\begin{eqnarray*}
T\I_t([h^1,h^2])&=&T\xi_t\int_0^t\rho(s)\left(\int_0^s\rho(r)dr\right)
(T\xi_s)^{-1}XdY\left(T\xi _s(b^1), T\xi_s (b^2)\right) ds\\
&&+T\xi_t\int_0^t \rho(s)\int_0^s  \rho(r) (T\xi_r)^{-1} XdY
    \left(T\xi_r(b^1),T\xi _r(b^2)\right)\, dr ds\\
    &=&T\xi_t\int_0^t\rho(s)\left(\int_0^s\rho(r)dr\right)
(T\xi_s)^{-1}XdY\left(T\xi _s(b^1), T\xi_s (b^2)\right) ds\\
&&+T\xi_t\left(\int_0^t \rho(s)ds\right) \int_0^t  \rho(r) (T\xi_r)^{-1} XdY
    \left(T\xi_r(b^1),T\xi _r(b^2)\right)\, dr ds\\
    &&-T\xi_t\int_0^t\left( \int_0^s \rho(r)dr\right)  \rho(s) (T\xi_s)^{-1} XdY
    \left(T\xi_s(b^1),T\xi _s(b^2)\right)\,ds\\
&=&\left(\int_0^t\rho(s)ds\right) T\xi_t\left(\int_0^t\rho(s)(T\xi_s)^{-1}XdY
\left(T\xi_s(b^1), T\xi_s(b^2)\right)ds\right).
\end{eqnarray*}


\hfill\rule{2mm}{2mm}

Note that if $\breve T$ is the torsion of the connection $\breve \nabla$,
then  $X(dY)={\breve T}$ by  (\ref{8}). Consequently,
\begin{corollary} For $h^i$ defined by (\ref{8}) in Proposition
  \ref{proposition-3.1},
 \begin{equation}
 \begin{split}X(x_t){d\over dt}[h^1,h^2]_t
 =&\rho(t) \left(\int_0^t \rho(r)\, dr \right)
        \breve T\left(T\xi _t(b^1), T\xi_t (b^2)\right) \\
 &+\rho(t)  T\xi_t \int_0^t  \rho(r) (T\xi_r)^{-1} \breve T
    \left(T\xi_r(b^1),T\xi _r(b^2)\right)\, dr \end{split}
\end{equation}
and
\begin{equation}
\label{24}
T\I_t([h^1,h^2])
=\left(\int_0^t\rho(s)ds\right) T\xi_t\left(\int_0^t\rho(s) T\xi_s^{-1}
\breve T\left(T\xi_s(b^1), T\xi_s(b^2)\right)ds\right).
\end{equation}
\end{corollary}


We shall now return to our examples.

\subsubsection*{Example 3.1. Gradient system.}

In Example 1.1, where the stochastic differential equation is a gradient
system, we have $dY\equiv 0$. Consequently
 $\frac d{dt}\left[ h^1,h^2\right]_t(\omega )$ vanishes and so does
$\left[ h^1,h^2\right]_t(\omega )$. Indeed in this case if we set
$$h^v=\int_0^\cdot Y_{x_r}\left(T\xi_r(v)\right)\;dr$$
for $v\in T_{x_0}M$, we have a  commuting family
$\{h^v: v\in T_{x_0}M\}$  of $\HH$-valued vector fields on $C_0\R^m$.
Note that if $\alpha: M\to \R^m$ is the immersion defining our
stochastic differential equation, then $\alpha\circ \xi_t$ is a random
map of $M$ into $\R^m$ and
$\dot h_t^v=d(\alpha\circ \xi_t)(v)=(\xi_t)^*(d\alpha)(v)$.

\subsubsection*{Example 3.2. The Lie group case.}

For $M$ a Lie group with left invariant stochastic differential equation
corresponding to a bi-invariant metric as  discussed in \S\ref{section-1},
if  $v\in \g$ and $x_0=e$ the identity element, we have
\begin{eqnarray*}
h_\cdot^i&:=&\int_0^\cdot \rho(r) Y^L_{x_r}\left(T\xi_r^L(b^i)\right)\, dr\\
&=&\int_0^\cdot \rho(r) TL^{-1}_{x_r}TR_{x_r}(b^i)\, dr\\
&=&\int_0^\cdot \rho(r) ad(x_r^{-1})(b^i)\,dr
\end{eqnarray*}
In this case
$$T\I_t(h^i)=TR_{x_t}\int_0^t ad(x_r)\dot h_r^i \,dr
=TR_{x_t}\left(\int_0^t\rho(r)dr b^i\right).$$

 Let  $T^R$ and $T^L$ be the torsions for the left and right invariant
connections respectively. From above, (\ref{torsion-of-Lie-groups}),
the fact that $T\xi_t^L(u)=TR_{x_t}(u)$, and by the right invariance of $TR$,
 \begin{eqnarray*}
{d\over dt}[h^1,h^2]_t
 &=&-\rho(t) \left(\int_0^t \rho(r)\, dr \right)
       TL^{-1}_{x_t} TR_{x_t} [b^1,b^2] \\
&&- \rho(t) TL_{x_t}^{-1}TR_{x_t}\int_0^t  \rho(r) [b^1,b^2]\, dr\\
&=&-2\rho(t) \left(\int_0^t \rho(r)\, dr \right) ad(x_t^{-1})[b^1,b^2]\\
&=&-{d \over dt}\left(\int_0^t\rho(r)dr\right)^2 ad(x_t^{-1})[b^1,b^2]\\
&=&-ad(x_t^{-1}){d \over dt}
\left[\int_0^t\rho(r)dr b^1, \int_0^t\rho(r)dr b^2\right].
\end{eqnarray*}
Note that
\begin{eqnarray*}
T\I_t([h^1,h^2])
&=&-TR_{x_t}\int_0^t  ad(x_r)({d \over dr}[h^1,h^2]_r)dr\\
&=&-TR_{x_t}\int_0^t 2\rho(r) \left(\int_0^r \rho(s) ds\right) dr[b^1, b^2] \\
&=&-TR_{x_t} [\int_0^t \rho(r)dr b^1, \int_0^t \rho(r)drb^2].
\end{eqnarray*}


\subsubsection*{Example 3.3.  Lie group as a symmetric space.}
We now consider the stochastic differential equation
(\ref{sde-bi-invariant-Lie-group}):
$$dx_t=TR_{x_t}\circ dB_t -TL_{x_t}\circ dB_t^\prime$$
 on the Lie group $G$ with bi-invariant metric. The derivative flow
of the solution $\xi_t$ is given by $T\xi_t=TR_{(g_t^\prime)^{-1}}TL_{g_t}$.
Note that in our notation the map, giving the equation
(\ref{sde-bi-invariant-Lie-group}),
 $X(g):\g\times \g\to T_xM$ is given by
 $X(g)(\alpha_1,\alpha_2)=TR_g(\alpha_1)-TL_g(\alpha_2)$ with inverse
 $Y:TG\to \g\times \g$ given, for $u\in T_gG$, by
 \begin{eqnarray*}
 Y(u)&=&(TR_g^{-1}(u), -TL_g^{-1}(u))\\
 &=&(Y^R u, -Y^Lu).
\end{eqnarray*}
 Thus if $u, v\in T_gG$, for $T^R$ and $T^L$ the torsion tensors as in
 (\ref{torsion-of-Lie-groups}),
 \begin{eqnarray*}
 dY(u,v)&=&\left( TR_g^{-1}T^R(u,v), TL_g^{-1}T^L(u,v) \right)\\
 &=&\left( [Y^R(u), Y^R(v)], \; [Y^L(u), Y^L(v)] \right)
 \end{eqnarray*}
so
\begin{eqnarray*}
{d\over dt}[h^1,h^2]_t
&=&\rho(t)\left(\int_0^t\rho(r)dr\right)
 \left(\;[Y^R \left(TL_{g_t}TR_{g_t^\prime}^{-1}b^1\right),
                Y^R (TL_{g_t}TR_{g_t^\prime}^{-1}b^2)],\right.\\
&&\hskip 100pt \left.[Y^L (TL_{g_t}TR_{g_t^\prime}^{-1}b^2),
    Y^L (TL_{g_t}TR_{g_t^\prime}^{-1}b^1])\;\right).
    \end{eqnarray*}
Now \begin{eqnarray*}
Y^R \left(TL_{g_t}TR_{(g_t^\prime)}^{-1}b^i\right)
&=&\left(TR_{g_t}(g_t^\prime)^{-1}\right)^{-1}TL_{g_t} TR_{g_t^\prime}^{-1}b^i\\
&=&\left( TR_{g_t^\prime}^{-1}\circ TR_{g_t}\right)^{-1}
TL_{g_t} TR_{g_t^\prime}^{-1}b^i\\
&=&ad(g_t)b^i.
\end{eqnarray*}
while
\begin{eqnarray*}
Y^L \left(TL_{g_t}TR_{(g_t^\prime)}^{-1}b^i\right)
&=&\left(TL_{g_t}(g_t^\prime)^{-1}\right)^{-1}
   TL_{g_t} TR_{g_t^\prime}^{-1}b^i\\
 &=&TL_{g_t^\prime}TR_{g_t^\prime}^{-1} b^i=ad(g_t^\prime)b^i,
 \end{eqnarray*}
so
\begin{eqnarray*}
{d\over dt}[h^1,h^2]_t
&=&\rho(t)\left(\int_0^t\rho(r)dr\right)
\left(ad(g_t)[b^1,b^2], ad(g_t^\prime)[b^1,b^2]\right)\\
&\in& \g\times \g.
\end{eqnarray*}
In this case $h_t^i=\left(\int_0^t\rho(r)ad(g_r)b^i\, dr,
-\int_0^t\rho(r)ad(g_r^\prime)b^i\,dr\right)$.
\bigskip

\noindent
{\bf 3.2 \hskip5pt A family of $\HH^2$-valued vector fields. }
In general when the associated connection $\breve \nabla$ for (\ref{5})
is the Levi-Civita connection $\nabla$ and $A\equiv 0$ we expect that
$\overline{\wedge^qT\I(h^1\wedge \dots \wedge h^q)}_\sigma \in \HH^q_\sigma$
for almost all $\sigma \in C_{x_0}M$. It follows from \cite{Elworthy-Li-Hodge-1}
that this holds for $q=1$ and from \cite{Elworthy-Li-Hodge-2} for $q=2$ with sufficiently regular $h$, and a
proof for $q=2$ for our special $h^i$'s  is given below in Proposition
\ref{proposition-3.3} as an illustrative
example. For general $q$ we have,
$$\overline{\wedge^q(T\I)(h^1\wedge \dots \wedge h^q)}
=\E\left\{\wedge^q(T\xi_\cdot)\left(b^{\rho,1}_\cdot\wedge \dots \wedge
b_\cdot^{\rho,q}\right) \left| x_s: 0\le s \le T \right.\right\}$$
where $b_t^{\rho,j}=\left(\int_0^t \rho(r)dr\right) b^j$ and
$(T\xi_\cdot)(b_\cdot^{\rho,j})$ is just the vector field \\
$\{(T\xi_t)(b_t^{\rho,j}): 0\le t \le T\}$
along $\xi_\cdot(x_0)$. From \cite{Elworthy-LeJan-Li-book} or
\cite{Elworthy-Yor}  for gradient systems, we can deduce that for
 $0<t_1<\dots <t_q\le T$
\begin{equation}
\label{25}
\begin{split}
&\overline{\wedge^qT\I(h^1\wedge \dots \wedge h^q)}_{t_1,\dots,t_q}\\
=&\int_0^{t_1}\rho(r)dr\int_0^{t_2}\rho(r)dr\dots \int_0^{t_q}\rho(r)dr
 \cdot \left(\1\otimes \dots \otimes W_{t_q}^{t_{q-1}} \right)\\
&\left(\1\otimes \dots \otimes {W^{(2)}}^{t_{q-2}}_{t_{q-1}}\right)
\dots W_{t_1}^{(q)}(b^1\wedge \dots \wedge b^q).
\end{split}
\end{equation}
Here ${W^{(q)}}^s_t=W_t^{(q)}\circ (W_s^{(q)})^{-1}:
 \wedge^qT_{x_s}M \to \wedge^qT_{x_t}M$
 with $W^s_t={W^{(1)}}^s_t$, for  $W^{(q)}$ the damped parallel transport
 defined by (\ref{3}).

 Let us recall the characterization of $\HH^1$ and $\HH^2$. For $q=2$ it is shown
  in \cite{Elworthy-Li-Hodge-2}, that
 $$\HH^{2}_\sigma=\left\{U+Q_\sigma(U) \quad |\quad
 U\in \wedge^2 \HH^{1}_\sigma\right\}\subset \wedge^2T_\sigma C_{x_0}M $$
 where $Q_\sigma: \wedge^2\HH^1_\sigma \to \wedge^2 T_\sigma C_{x_0}M$
 is defined by
\begin{equation}\label{Q}
Q(U)_{s,t}=(\1\otimes W^s_t) W_s^{(2)} \int_0^s (W_r^{(2)})^{-1}
 \RR(U_{r,r})\, dr, \hskip 15pt 0\le s \le t \le T
\end{equation}
 for $\RR: \wedge^2TM \to \wedge^2 TM$ the curvature operator.


 The space $\HH^1$ is  the `Bismut tangent space', \cite{Jones-Leandre91},
but with a different Hilbert space structure. In fact elements of
$\HH^1$ are exactly vector fields $v$ in $M$ along $\sigma$
 such that $\|v\|_\sigma:=\int_0^T\left \vert {\D\over dt}v_t\right
   \vert^2_{\sigma(t)} \; dt<\infty$ where
\begin{equation}
{\D\over d t}v_t:={D\over d t}v_t+{1\over 2}Ric^{\#}(v_t)
=W_t^{-1}{d\over dt}W_t^{-1}V_t.
\end{equation}


From this it follows that $\wedge^2 \HH^1_\sigma$ consists of those $U$
in $\wedge^2T_\sigma C_{x_0}M$ for which
$$\left(\wedge^2(W_\cdot)^{-1}(U)\right)_{s,t}
=\int_0^s\int_0^t H(r_1,r_2) \, dr_1 dr_2$$
for some $H$ in $L^2\left([0,T]\times [0,T]; T_{x_0}M\otimes T_{x_0}M\right)$.

\begin{proposition}
\label{proposition-3.3}
Let $\lambda\in L_0^{2,1}\left([0,T];\R\right)$.  Take $V$ in
$\wedge^2T_{x_0}M$ and $V^\lambda \in \wedge^2 L_0^{2,1}T_{x_0}M$ defined by
$V^\lambda_{s,t}=\lambda(s)\lambda(t)V$.
If $A\equiv 0$ in (\ref{5}) and the associated connection is the Levi-Civita
connection, the vector field $Z$, where
$$Z=\E\left\{\wedge^2(T\xi_\cdot)(V^\lambda) \vert x_s, 0\le s \le T\right\}$$
forms a section of $\HH^2$. It is equal to $U+Q(U)$ for $Q$ as above
and $U$ given by
\begin{equation}
\label{28}
\begin{split}
U_{s,t}&=(\1\otimes W^s_t)\lambda(s)\lambda(t) W_s^{(2)}(V)\\
&-(W_s\otimes W_t)      \int_0^s    \lambda^2(r) \wedge^2(W_r)^{-1}
\RR\left( W_r^{(2)}(V)\right)\, dr,
\end{split}
\end{equation}
for all $0\le s\le t\le T$. It has divergence
\begin{equation}
\label{divergence}
\begin{split}
(div Z)_t=&-\int_0^T \left(\langle \frac{\D} {\partial r}-, dx_r\rangle
\otimes \1\right) Z_{r,t}\\
&+W_t\int_0^t\left(\langle -, dx_r \rangle
\otimes (W_r)^{-1})\RR(\1\otimes W^r_t)\right)^{-1}Z_{r,t}.
\end{split}
\end{equation}
(For an interpretation of the, apparently adapted, integrals, see the proof below.)
\end{proposition}

{\it Proof.}
From (\ref{25}), for $0\le s\le t\le T$,
\begin{equation}
\label{32a}
Z_{s,t}=\lambda(s)\lambda(t)(\1\otimes W^s_t)W_s^{(2)}(V).
\end{equation}
So
\begin{equation}\label{29}
\left[(\wedge^2 W_\cdot)^{-1} Z
            \right]_{s,t}
=\lambda(s)\lambda(t)(\wedge^2 W_s)^{-1}W_s^{(2)}(V).
\end{equation}
Setting
\begin{equation}\label{30}
k(r)=\wedge^2(W_r^{-1})W_r^{(2)}(V)\in \wedge^2T_{x_0}M
\end{equation}
we see that
$$\left[(\wedge^2 W_\cdot)^{-1} Z
            \right]_{s,t} =\lambda(s)\lambda(t)k(s\wedge t),
\hskip 15pt \hbox{ for all $s,t\in [0,T]$}.$$

Set $$\tilde U=\wedge^2(W^{-1}_\cdot)(U)$$
We only need to show that
\begin{equation}\label{31}
\tilde U+(\wedge^2 W_\cdot^{-1})Q(U) =\lambda(s)\lambda(t)k(s\wedge t)
\end{equation}
 and that $U$ belongs to $\wedge^2 \HH^1$.

  Now for any  $K \colon [0,T]\times \Omega\to \wedge^2TM$ over
 $\{x_t: 0\le t\le T\}$
 \begin{equation}\label{32}
 \wedge^2(W_r){d \over dr}\left(\wedge^2(W_r)^{-1}K_{r}\right)
 =\RR (K_{r})+  W^{(2)}_r{d \over dr}[(W_r^{(2)})^{-1}K_r]
 \end{equation}
when the derivative exists, so
$k^\prime(s)=\wedge^2(W_s^{-1})\RR \left(W_s^{(2)}(V)\right)$
and consequently by (\ref{28})
\begin{equation}\label{33}
\tilde U_{s,t}=\lambda(s)\lambda(t)k(s\wedge t)
-\int_0^{s\wedge t} \lambda^2(r)k^\prime(r)\, dr,
\end{equation}
 for $s,t\in [0,T]$. Integrating by parts,
\begin{eqnarray}
\nonumber
\tilde U_{s,t}&=&\lambda(s)\lambda(t)k(s\wedge t)
  -\lambda^2(s\wedge t)k(s\wedge t)
 + 2 \int_0^{s\wedge t}\lambda^\prime(r)\lambda(r)k(r)\,dr\\
&=&2 \int_0^{s\wedge t}\lambda^\prime(r)\lambda(r)k(r)\,dr
  +(\lambda(s)\lambda(t)-\lambda^2(s\wedge t))k(s\wedge t).
  \label{36}
\end{eqnarray}
From this it is  easy to see that
 \begin{equation}
\label{34}
\tilde U_{\bar s \bar t}=\int_0^{\bar s}ds \int_0^{\bar t}dt
\left(\lambda^\prime(s)\lambda^\prime(t)k(s\wedge t)
+\lambda(s\wedge t)\lambda^\prime(s\vee t) k^\prime(s\wedge t)\right)
\end{equation}
since for  $\bar s \le \bar t$ one has
\begin{equation}
\begin{split}
&\int_0^{\bar s}ds \int_0^{\bar t}dt
\left(\lambda^\prime(s)\lambda^\prime(t)k(s\wedge t)
+\lambda(s\wedge t)\lambda^\prime(s\vee t) k^\prime(s\wedge t)\right)\\
=&\int_0^{\bar s}\left\{ \int_0^{s}
\left(\lambda^\prime(s)\lambda^\prime(t)k(t)
+\lambda(t)\lambda^\prime(s) k^\prime(t)\right)dt\right.\\
& \left.+\int_s^{\bar t}  \left(\lambda^\prime(s)\lambda^\prime(t)k(s)
+\lambda(s)\lambda^\prime(t) k^\prime(s)\right)dt\right\}\\
=&   \int_0^{\bar s}\lambda^\prime(s)\lambda(s)k(s)\,ds
+\int_0^{\bar s}(\lambda(\bar t)-\lambda(s)){d \over ds}(\lambda(s)k(s))\,ds\\
=& 2 \int_0^{\bar s}\lambda^\prime(s)\lambda(s)k(s)\,ds
+(\lambda(\bar t)-\lambda(\bar s))\lambda(\bar s)k(\bar s).
\end{split}
\end{equation}
This shows that $U\in \wedge^2\HH^1$. Next note that
$$\tilde U_{s,s}=2\int_0^s \lambda^\prime(r)\lambda(r)k(r)\, dr$$
by (\ref{36}). By the definition,  (\ref{Q}),  of $Q$, (\ref{30}),
and using (\ref{32}) again  we see that  for $0\le s\le t\le T$,

\begin{eqnarray*}
&&\left[\wedge^2(W_\cdot^{-1})Q(U)\right]_{s,t}
:=\wedge^2(W_s^{-1})W_s^{(2)}\int_0^s (W_r^{(2)})^{-1}
\RR\left(U_{rr}\right) dr\\
&=&\wedge^2(W_s^{-1})W_s^{(2)}\int_0^s (W_r^{(2)})^{-1}\wedge^2(W_r)
 {d\over dr}\tilde U_{r,r}\, dr\\
 &&-\wedge^2(W_s^{-1})W_s^{(2)}\int_0^s{d \over dr}
 \left[(W_r^{(2)})^{-1} U_{rr}\right]\,dr\\
 &=&\wedge^2(W_s^{-1})W_s^{(2)}\int_0^s 2 (W_r^{(2)})^{-1}\wedge^2(W_r)
  \lambda^\prime(r)\lambda(r)k(r)\, dr-\tilde U_{s,s}\\
&=&\wedge^2(W_s^{-1})W_s^{(2)}\int_0^s 2
  \lambda^\prime(r)\lambda(r)V\, dr-\tilde U_{s,s}\\
  &=&\lambda^2(s)k(s)-2\int_0^s \lambda^\prime(r)\lambda(r)k(r)\, dr \\
&=&\int_0^s  [\lambda(r)]^2 k^\prime(r) \, dr.
\end{eqnarray*}
Finally add (\ref{33}) to the above to obtain the required identity
(\ref{31}).

To obtain (\ref{divergence}) we can assume $V$ to be primitive,
$V=b^1\wedge b^2$ say. By (\ref{18}) the divergence of $Z$ is just the
conditional expectation $\bar J$ of $J$ for $J=T\I(div(h^1\wedge h^2))$
where $\dot h_t^j=\dot \lambda(t)Y_{x_t}T\xi_t(b^j)$.
By Theorem \ref{theorem-Shigekawa} and Proposition \ref{proposition-3.1}
\begin{eqnarray*}
J_t&=&T\I_t\left(
-\int_0^T\langle \dot h_r^1, dB_r\rangle h_\cdot^2
+\int_0^T\langle \dot h_r^2, dB_r \rangle h_\cdot^1\right)\\
&=&J_t^0+J_t^1
\end{eqnarray*}
where
\begin{eqnarray*}
J_t^0&=&T\I_t\left(-\int_t^T\langle \dot h_r^1, dB_r\rangle h^2_\cdot
+\int_t^T\langle \dot h_r^2, dB_r\rangle h_\cdot^1\right)\\
&=&-\int_t^T\left\langle \dot \lambda(r)T\xi_r^t(T\xi_t(b^1)), dx_r
\right\rangle  \lambda(t) T\xi_t(b^2)\\
&&+\int_t^T \left\langle \dot \lambda(r) T\xi_r^t (T\xi_t(b^2)), dx_r
 \right\rangle \lambda(t)T\xi_t(b^1).
\end{eqnarray*}
Therefore
$$\E\left\{ J_t^0\vert \; \F^{x_0}\vee \F_t\right\}
=-\lambda(t)\int_t^T \dot\lambda(r)\iota_{\langle W_r^t-,dx_r\rangle}
\wedge^2(T\xi_t)(b^1\wedge b^2)$$
giving
$$\bar J_t^0=\int_t^T\left(\1\otimes \left\langle \frac{\D}{\partial r}-,
dx_r\right\rangle \right) \, Z_{t,r}$$
by (\ref{32a}).

The second term $J_t^1$ could be treated as in Corollary \ref{cor-invariant}
below, but to get the form (\ref{divergence}) observe
$$J_t^1=- \int_0^t \lambda(t)\dot \lambda(r)
\left(\langle-, dx_r\rangle \otimes \1\right)
\left(\wedge^2(T\xi)(b^1\wedge b^2)\right)_{rt}$$
and so by (\ref{25})
$$\bar J_t^1=-\int_0^t \lambda(t)\dot \lambda(r)
\left(\langle-, dx_r\rangle \otimes \1\right)
\left(\1\otimes W_t^r)W_r^{(2)}(b^1\wedge b^2)\right)_{rt}.$$
Here we can interpret the, apparently adapted, integral by using
$W^r_t=W_t(W_r)^{-1}$ and taking the term $(\1\otimes W_t)$
outside of the integral.
Now
\begin{eqnarray*}
(\frac{\D}{\partial r}\otimes \1)Z_{rt}
&=&\frac{D}{\partial r}\left[ \lambda(r)\lambda(t)
(\1\otimes W^r_t)W^{(2)}_r(b^1\wedge b^2)\right]\\
&&+\frac{1}{2}(Ric^{\#}\otimes \1)Z_{rt}
\end{eqnarray*}
giving
\begin{eqnarray*}
&&\dot\lambda(r)\lambda(t)(\1\otimes W^r_t)W^{(2)}_r(b^1\wedge b^2)\\
&=&(\frac{\D}{\partial r}\otimes \1)Z_{rt}-\frac{1}{2}(Ric^{\#}\otimes \1)Z_{rt}\\
&&-\lambda(r)\lambda(t)(1\otimes W_t)\frac{D}{\partial r}
(\1\otimes (W_r)^{-1})W_r^{(2)}(b^1\wedge b^2),
\end{eqnarray*}
but
\begin{eqnarray*}
&&\frac{D}{\partial r}(\1\otimes (W_r)^{-1})W_r^{(2)}(b^1\wedge b^2)\\
&&=(W_r\otimes \1)\frac{D}{\partial r}(\wedge^2 W_r^{-1})
W_r^{(2)}(b^1\wedge b^2)
-\frac{1}{2}(Ric^{\#}\otimes \1)(\1\otimes W_r^{-1})
W_r^{(2)}(b^1\wedge b^2)\\
&&=(\1\otimes W_r^{-1})\RR W_r^{(2)}(b^1\wedge b^2)
-\frac{1}{2}(Ric^{\#}\otimes \1)(\1\otimes W_r^{-1})W_r^{(2)}(b^1\wedge b^2)
\end{eqnarray*}
by (\ref{32}), and so
$$\dot\lambda(r)\lambda(t)(\1\otimes W^r_t)W^{(2)}_r(b^1\wedge b^2)
=(\frac{\D}{\partial r}\otimes \1)Z_{rt}
-(\1\otimes W_t^r)\RR (\1\otimes W_t^r)^{-1}Z_{rt}.$$
The result follows noting that $(\1\otimes W^r_t)^{-1}Z_{rt}$
is $\F_r^{x_0}$-measurable and the factor $(\1\otimes W_t)$
in $(\1\otimes W^r_t)=(\1\otimes W_t)(\1\otimes (W_r)^{-1})$
can be taken out of the stochastic integrals to make them It\^o
integrals.

\hfill \rule{3mm}{3mm}



\bigskip

\noindent{\bf Remark 3.3.}
\vskip 2pt

(1).   The formula (\ref{28}) in Proposition \ref{proposition-3.3} demonstrates the
alternative characterisation of $\HH^2$ from \cite{Elworthy-Li-Hodge-2}
 which states that $V$ is in $\HH^2$ if and only if $V-\pathR(V)$ is in
 $\wedge^2 \HH^1$  where $\pathR$ is the curvature operator of the damped
 Markovian connection on $\HH^1$.
\bigskip

(2). Note that for the proof of (\ref{28}) we could allow
$V: C_{x_0}M\to \wedge^2T_{x_0}M$ to be nonconstant provided it is in
$L^{1+\epsilon}$ for some $\epsilon>0$. Our filtration in calculating
(\ref{25}) can be chosen to be $\{\F_t \vee \F^{x_0}: 0\le t\le T\}$ as
 in \cite{Elworthy-Li-Ito}.
Now let $V: [0,T]\times [0,T] \to \wedge^2 T_{x_0}M$ be such that
$V_{s,t}=\int_0^s\int_0^t Z_{a,b}\, da\, db$ some
$Z\in L^2\left([0,T]\times [0,T]\times C_{x_0}M\to \wedge^2T_{x_0}M\right)$
symmetric in $(s,t)\in [0,T]\times [0,T]$. As a corollary we see,
 for $0\le s\le t \le T$,
$$\E\{\wedge^2(T\xi_\cdot)(V)\vert \,x_s: 0\le s\le t\le T\}=U+Q(U)$$
for $$U_{s,t}=(\1\otimes W^s_t)W^{(2)}_s(V_{s,t})
-(W_s \otimes W_t)
\int_0^s \wedge^2(W_r)^{-1}\RR \left(W^{(2)}_r(V_{rr})\right)\, dr.$$
Moreover $U$ is a section of $\wedge^2\HH^1$ and so
$U+Q(U)$ is a section of $\HH^2$. This follows from the proposition
by polarization for
$V_{s,t}=\left(\lambda^1(s)\lambda^2(t)+\lambda^2(s)\lambda^1(t)\right)b$,
 some $b\in L^2(C_{x_0}M;\wedge^2T_{x_0}M)$ and then for general $V$ by
 continuity since we can consider such $V$ as elements of the completed
tensor product
$[\odot^2L_0^{2,1}([0,T];\R)]\otimes L^2(C_{x_0}M;\wedge^2T_{x_0}M)$,
where $\odot$ refers to the symmetric tensor product. Note the mapping
\begin{eqnarray*}
&&\odot^2L_0^{2,1}\left([0,T];\R\right)
 \otimes L^2(C_{x_0}M;\wedge^2T_{x_0}M)\\
&&\stackrel{\Theta}{\longrightarrow}
L^2\left(C_{x_0}M; \wedge^2(L_0^{2,1}([0,T];\R)\otimes T_{x_0}M)\right)\\
&&\simeq L^2\left(C_{x_0}M; \wedge^2(L_0^{2,1}(T_{x_0}M)\right)
\end{eqnarray*}
given by
\begin{eqnarray*}
&&\Theta\left((u\odot v)\otimes f(\alpha\wedge\beta)\right)(\sigma)\\
&&={1\over 2} f(\sigma)\left[(u\otimes \alpha)\wedge (v\otimes \beta)
+(v\otimes \alpha)\wedge (u\otimes \beta)\right],\sigma\in C_{x_0}M,
\end{eqnarray*}
 where $f(\cdot)(\alpha\wedge \beta)
\in L^2\left(C_{x_0}M;\wedge^2T_{x_0}M\right)
\simeq L^2(C_{x_0}M;\R)\otimes \wedge^2 T_{x_0}M$,\\
$\alpha, \beta\in T_{x_0}M$, $f\in L^2(C_{x_0}M;\R)$.

In this generality the stochastic integrals in (\ref{divergence})
will need to be treated more carefully and if $V$ is non-random an
additional term involving its H-derivative will be involved in the divergence.
\bigskip

\noindent{\bf 3.3 \hskip 5pt
From Integration by parts to  Bismut type formulae.}
First we shall give an extension of the non-intrinsic formula (\ref{4}).
For this consider a general non-degenerate stochastic differential equation
(\ref{5}) with smooth coefficients. For $\{\xi_t: 0\le t <\infty\}$, its
flow of diffeomorphisms of the manifold, consider the semigroup
described in \S\ref{section-1}, given by
$$P_t\phi=\E \xi_t^*(\phi).$$
Since $M$ is compact $d(P_t\phi)=P_t(d\phi)$ when $\phi$ is $C^1$.
The generator $\A$ of $\{P_t\}_{t\ge 0}$ is given on smooth forms by
(\ref{11}) above.

\begin{proposition}
For any bounded measurable form $\phi$ on $M$ and $L^1$ function
 $\rho:[0,t]\to \R$ with $\int_0^t \rho(r)\,dr\not =0$,
\begin{equation}
\label{Bismut-formula}
\begin{split}
&d(P_t\phi)(b^1\wedge \dots \wedge b^{q+1})\\
&=\left(\int_0^t\rho(r)dr\right)^{-1}
  \E  \sum_{j=1}^{q+1}(-1)^{j+1}\int_0^t \rho(s)
\langle T\xi_s(b^j), X(x_s)dB_s\rangle_{x_s} \cdot\\
&   \hskip 130pt \xi_t^*(\phi)(b^1\wedge\dots
 \wedge\widehat{b^j}\wedge \dots \wedge b^{q+1})\\
 &-\left(\int_0^t\rho(r)dr\right)^{-1}
 \sum_{1\le i<j\le q+1}(-1)^{i+j+1}\; \E \xi_t^*(\phi)\\
&  \left(\int_0^t\rho(s)T\xi_s^{-1}
 \left(\breve T\left(T\xi_s(b^i),T\xi_s(b^{j})\right)\right) ds
 \wedge b^1\wedge \dots \wedge\widehat{b^i} \dots
 \wedge\widehat{b^j} \dots \wedge b^{q+1}\right),
\end{split}
\end{equation}
where $\breve T:TM \oplus TM \to TM$ is the torsion of the  connection
$\breve \nabla$ given by (\ref{8}).
\end{proposition}
\noindent
{\it Proof.}
Define $h^i, i=1$ to $q+1$, by (\ref{9}). Set
 $h=h^1\wedge \dots \wedge h^{q+1}$ and $b=b^1\wedge\dots\wedge b^{q+1}$.
Arguing as in Theorem \ref{theorem-2.2}, but without taking conditional
 expectations we see, for $\phi$ a $C^1$ $q$-form,
\begin{eqnarray*}
d(P_t\phi)(b)&=&P_t(d\phi)(b)\\
&=& \E \xi_t^*(d\phi)(b)
=\left(\int_0^T \rho(r)dr\right)^{-q-1} \E
   d\phi\left(\wedge^{q+1}(T\I)(h)_{t,\dots, t}\right)\\
&=& -    \left(\int_0^T \rho(r)dr\right)^{-q-1} \E
  \phi\left(\wedge^{q}(T\I)(div h)_{t,\dots, t}\right).
\end{eqnarray*}
By Shigekawa's result, Theorem \ref{theorem-Shigekawa},
 formula (\ref{Bismut-formula})
follows using Proposition \ref{proposition-3.1},
 equation (\ref{formula-TI}), (\ref{24}), and the fact
that $\int_0^t\langle \dot h^i_s, dB_s\rangle
=\int_0^t\rho(s)\langle T\xi_s(b^i), X(x_s)dB_s\rangle_{x_s}$.
By continuity it also holds for bounded measurable $\phi$.
\hfill\rule{3mm}{3mm}

\bigskip

To obtain an intrinsic formula from (\ref{Bismut-formula}), we shall
take conditional expectations, which can easily be done if
the torsion $\breve T=X(dY)$ is invariant under the flow (e.g.
 for certain homogeneous spaces). In this case
formula (\ref{Bismut-formula}) becomes:

\begin{equation}
\begin{split}
&d(P_t\phi)(b^1\wedge \dots \wedge b^{q+1})\\
&=\left(\int_0^t\rho(r)dr\right)^{-1}
  \E  \sum_{j=1}^{q+1}(-1)^{j+1}\int_0^t \rho(s)
\langle T\xi_s(b^j), X(x_s)dB_s\rangle _{x_s}\cdot\\
&   \hskip 130pt \xi_t^*(\phi)(b^1\wedge\dots
 \wedge\widehat{b^j}\wedge \dots \wedge b^{q+1})\\
 &-\sum_{1\le i<j\le q+1}(-1)^{i+j+1}
  \E\left( \xi_t^*(\phi)\left( \breve T(b^i,b^j)
 \wedge b^1\wedge \dots \wedge\widehat{b^i} \dots
 \wedge\widehat{b^j} \dots \wedge b^{q+1}\right)\right).
\end{split}
\label{40}
\end{equation}
Let $\breve W_t^{A,q}:\wedge^q T_{x_0}M \to \wedge^q T_{x_t}M$,  the damped
 parallel translation of $q$ vectors, be defined by:
\begin{equation}\left\{
\begin{split}
{\hat D\over \partial s}(\breve W_s^{A,q}(V_0))&
 =-{1\over 2}{\breve \RR}_{x_0}^q\left(\breve W_s^{A,q}(V_0)\right)
+d\wedge^q \left(\breve \nabla A\right)\left(\breve W_s^{A,q}(V_0) \right)\\
\breve W_0^{A,q}(V_0)&=V_0.
\end{split}\right.
\end{equation}
Here ${\hat D\over \partial s}$ refers to covariant differentiation
using the connection $\hat \nabla$, the adjoint connection of
$\breve \nabla$. Since the conditional expectation of $\wedge^qT\xi_t$
is given by
$$\overline{\wedge^q T\xi_t}(-)=\breve W_t^{A,q}(-),$$
 by Theorem 3.3.7 in \cite{Elworthy-LeJan-Li-book}, or Theorem A in
 \cite{Elworthy-Yor} for the Levi-Civita connection, we only
 need to worry about the first term on the right hand side of
 equation (\ref{40}).
 Set
 $$U_t=\int_0^t \rho(s) \sum_{j=1}^{q+1}(-1)^{j+1}
 \langle T\xi_s(b^j), X(x_s)dB_s\rangle \cdot \wedge^q T\xi_t
 (b^1\wedge\dots \wedge\widehat{b^j}\wedge \dots \wedge b^{q+1})$$
 and $$V_t^j= \wedge^q T\xi_t
 (b^1\wedge\dots \wedge\widehat{b^j}\wedge \dots \wedge b^{q+1}).$$
 Then,  e.g.  by (3.3.10) in \cite{Elworthy-LeJan-Li-book},

 \begin{eqnarray*}
 {\hat D}U_t
 &=&\sum_{j=1}^{q+1}(-1)^{j+1}\rho(t)\langle T\xi_t(b^j), X(x_t)dB_t\rangle \cdot
 V_t^j\\
&&+\sum_{j=1}^{q+1}(-1)^{j+1}
\int_0^t \rho(s) \langle T\xi_s(b^j), X(x_s)dB_s\rangle \cdot
d\wedge^q\left(\breve \nabla X(-)dB_t)(V_t^j)\right)\\
&&+\sum_{j=1}^{q+1}(-1)^{j+1}
\int_0^t \rho(s) \langle T\xi_s(b^j), X(x_s)dB_s\rangle \cdot\\
&&\hskip 50pt
  \left(-{1\over 2}(\breve R^q)^*(V_t^j)dt
+d\wedge^q\breve \nabla A(V_t^j)dt\right)\\
&=&\sum_{j=1}^{q+1}(-1)^{j+1}\rho(t)\langle T\xi_t(b^j), X(x_t)dB_t\rangle \cdot
V_t^j\\
&&+ d\wedge^q\left(\breve \nabla X(-)dB_t)(U_t)\right)
-{1\over 2}(\breve R^q)^*(U_t)dt
+d\wedge^q\breve \nabla A(U_t)dt,
 \end{eqnarray*}
Taking the conditional expectation of the above equation,
as in the proof of Proposition 3.3.7 in \cite{Elworthy-LeJan-Li-book}
or that of Theorem A in \cite{Elworthy-Yor}, we have

\begin{eqnarray*}
 {\hat D}\overline{U_t}
&=&\overline{\sum_{j=1}^{q+1}(-1)^{j+1} \rho(t)
\langle T\xi_t(b^j), X(x_t)dB_t\rangle \cdot \wedge^q T\xi_t
 (b^1\wedge\dots \wedge\widehat{b^j}\wedge \dots \wedge b^{q+1})}\\
&&-{1\over 2}(\breve R^q)^*(\overline{U_t})dt
+d\wedge^q\breve \nabla A(\overline{U_t})dt\\
&=&\rho(t)\iota_{\langle-, X(x_t)dB_t\rangle}
   \overline{\wedge^{q+1} T\xi_t(b^1\wedge\dots \wedge b^{q+1})}
   -{1\over 2}(\breve R^q)^*(\overline{U_t})dt
+d\wedge^q\breve \nabla A(\overline{U_t})dt\\
&=&\rho(t)\iota_{\langle-,X(x_t)dB_t\rangle}
   \breve W_t^{A,q+1}(b^1\wedge\dots \wedge b^{q+1})
   -{1\over 2}(\breve R^q)^*(\overline{U_t})dt
+d\wedge^q\breve \nabla A(\overline{U_t})dt\\
&=&\rho(t)\iota_{\langle-,\breve{\paral_t} d\breve B_t\rangle}
   \breve W_t^{A,q+1}(b^1\wedge\dots \wedge b^{q+1})
   -{1\over 2}(\breve R^q)^*(\overline{U_t})dt
+d\wedge^q\breve \nabla A(\overline{U_t})dt
 \end{eqnarray*}
 Here $\breve{\paral_t}$ denotes parallel translation corresponding
 to the connection $\breve \nabla$ and $(\breve B_s)$ is the stochastic
 anti-development Brownian motion on $T_{x_0}M$, i.e. the martingale
 part of $\int_0^\cdot \breve{\paral_s}^{-1}\circ dx_s$.
Solve the equation to obtain
$$\overline{U_t}
  =\breve W_t^{A,q}\int_0^t \left( \breve W_s^{A,q}\right)^{-1}
   \rho(s) \iota_{\langle -,\breve {\parals_s} d\breve B_s\rangle}
   \breve W_s^{A,q+1}(b^1\wedge\dots \wedge b^{q+1}) \,ds.$$
Finally we arrive at:
\begin{corollary}
\label{cor-invariant}
Suppose that the torsion $\breve T\equiv XdY$ is invariant under the flow
 $\xi_t$. Then for $b^i\in T_{x_0}M, i=1,\dots, q+1$,
\begin{equation}
\begin{split}
&d(P_t\phi)(b^1\wedge\dots \wedge b^{q+1})\\
&=\left(\int_0^t\rho(r)dr\right)^{-1}
  \E\phi\left(\breve W_t^{A,q} \int_0^t \rho(s)(\breve W_s^{A,q})^{-1}
  \iota_{\langle -,\breve {\parals_s} d\breve B_s\rangle}
  \breve W_s^{A,q+1}(b^1\wedge\dots \wedge b^{q+1})\right)\\
 & -\phi \left( \breve W_t^{A,q}
     \left(\sum_{1\le i<j\le q+1}(-1)^{i+j+1} \breve T(b^i,b^j)\wedge
     b^1\wedge \dots \wedge\widehat{b^i} \dots
 \wedge\widehat{b^j} \dots \wedge b^{q+1}\right)\right).
\end{split}
\end{equation}
\end{corollary}

Note that in the non-invariant case, an analogous proof to that of Corollary
\ref{cor-invariant} leads to the intrinsic formula below.
 If $V$ is a $q+1$ vector, we define $\iota_{\breve T}V$
to be the operator from $\wedge^{q+1}TM \to \wedge^qTM$  which restricted
 to primitive vectors is given by:
$$\iota_{\breve T}(b^1\wedge \dots \wedge b^{q+1})=
\sum_{1\le i<j\le q+1}(-1)^{i+j+1}
     \breve T(b^i,b^j)\wedge
     b^1\wedge \dots \wedge\widehat{b^i} \dots
 \wedge\widehat{b^j} \dots \wedge b^{q+1}.$$

\begin{corollary}
Let $b$ be a $q+1$ vector in $\wedge^{q+1}T_{x_0}M$, then
\begin{equation}
\begin{split}
&\left(\int_0^t\rho(r)dr\right)\, d(P_t\phi)(b)\\
=&
  \E\phi\left(\breve W_t^{A,q} \int_0^t \rho(s)(\breve W_s^{A,q})^{-1}
 \left( \iota_{\langle -, \breve {\parals_s} d\breve B_s\rangle}
  \breve W_s^{A,q+1}(b)+\iota_{\breve T}   \breve W_s ^{A,q+1}(b)\,ds\right)\right).
\end{split}
\end{equation}
\end{corollary}

\noindent{\it Proof.}
We only need to worry about the last term of (\ref{Bismut-formula}), since
 the previous term is as in  Corollary \ref{cor-invariant}. For this
\begin{eqnarray*}
&&\sum_{1\le i< j\le q+1}(-1)^{i+j+1}\\
&&\E(\xi_t^*)(\phi)\left(\int_0^t\rho(s)T\xi_s^{-1}
 \left(\breve T\left(T\xi_s(b^i),T\xi_s(b^{j})\right)\right) ds
 \wedge b^1\wedge \dots \wedge\widehat{b^i} \dots
 \wedge\widehat{b^j} \dots \wedge b^{q+1}\right)\\
 &&=\E\phi\left(\sum_{1\le i<j\le q+1}(-1)^{i+j+1}
 T\xi_t\int_0^t\rho(s)T\xi_s^{-1}
 \left(\breve T\left(T\xi_s(b^i),T\xi_s(b^{j})\right)\right)ds \;
 \wedge \right.\\
&&\hskip 60pt \left. T\xi_t(b^1)\wedge \dots \wedge\widehat{T\xi_t(b^i)} \dots
 \wedge\widehat{T\xi_t(b^j)} \dots \wedge T\xi_t(b^{q+1})\right).
 \end{eqnarray*}
Set
\begin{eqnarray*}
U_t&=&\sum_{1\le i< j\le q+1}(-1)^{i+j+1}
 T\xi_t\int_0^t\rho(s)T\xi_s^{-1}
 \left(\breve T\left(T\xi_s(b^i),T\xi_s(b^{j})\right)\right)ds \;
 \wedge \\
&& T\xi_t(b^1)\wedge \dots \wedge\widehat{T\xi_t(b^i)} \dots
 \wedge\widehat{T\xi_t(b^j)} \dots \wedge T\xi_t(b^{q+1}).
 \end{eqnarray*}
Then after covariant differentiation and filtering we have
\begin{eqnarray*}
 {\hat D}\overline{U_t}
&=&\sum_{1\le i<j\le q+1}(-1)^{i+j+1}\rho(t)\cdot\\
&&\overline{ \breve T\left(T\xi_t(b^i),T\xi_t(b^j)\right)\wedge
 T\xi_t(b^1)\wedge \dots \wedge\widehat{T\xi_t(b^i)} \dots
 \wedge\widehat{T\xi_t(b^j)} \dots \wedge T\xi_t(b^{q+1})}\\
&&-{1\over 2}(\breve R^q)^*(\overline{U_t})dt
+d\wedge^q\breve \nabla A(\overline{U_t})dt\\
&=&\rho(t)\overline{\iota_{\breve T}
   \wedge^{q+1} T\xi_t(b^1\wedge\dots \wedge b^{q+1})}
   -{1\over 2}(\breve R^q)^*(\overline{U_t})dt
+d\wedge^q\breve \nabla A(\overline{U_t})dt\\
&=&\rho(t)\iota_{\breve T}
   \breve W_t^{A,q+1}(b)
   -{1\over 2}(\breve R^q)^*(\overline{U_t})dt
+d\wedge^q\breve \nabla A(\overline{U_t})dt,
 \end{eqnarray*}
 giving
 $$\bar U_t=\breve W_t^{A,q}\int_0^t \rho(s)\left(\breve W_s^{A,q}\right)^{-1}
   \iota_{\breve T}   \breve W_s ^{A,q+1}(b)\,ds.$$
The required equation now follows.   \hfill\rule{3mm}{3mm}

\subsubsection*{Special cases}
\begin{enumerate}
\item[(1)]
 When the connection $\breve \nabla$ defined by (\ref{7})  is the
Levi-Civita connection and $\rho(t)\equiv 1$, formula (\ref{Bismut-formula})
essentially reduces to (\ref{4}), but in this case $P_t$ has generator
 given by ${1\over 2}\Delta+\LL_A$ on smooth forms.

\item[(2)]
For a left invariant stochastic differential equation on a Lie group
$G$ with bi-invariant metric and $A\equiv 0$,
 formula  (\ref{Bismut-formula}) reduces to
\begin{equation}\label{formula-Li-group}
\begin{split}
&d(P_t\phi)(b^1\wedge\dots \wedge b^{q+1})\\
&=\left(\int_0^t\rho(r)dr\right)^{-1}
\E\left(\left\langle \int_0^t ad(x_s)dB_s, -\right\rangle \wedge
R_{x_t}^*(\phi)\right)(b^1\wedge\dots\wedge b^{q+1})\\
&-\sum_{1\le i<j\le q+1} (-1)^{i+j} \E\, R^*_{x_t}(\phi)
\left([b^i, b^j]\wedge b^1\wedge\dots \widehat{b^i}\wedge \dots
 \widehat{b^j}\wedge \dots b^{q+1}\right),
\end{split}
\end{equation}
for $x_0=e$ (so $b^j\in \g$, each $j$).
In this case the generator of $P_t$ is ${1\over 2}trace \nabla^R\nabla^R$
by (2.4.3) of \cite{Elworthy-LeJan-Li-book}.

\bigskip

 Formula  (\ref{formula-Li-group}) is intrinsic. It could have been deduced
from the path space integration by parts formula of \cite{Fang-Franchi97}.
The $q=0$ case was given in \cite{Elworthy-Li-JFA}.

\item[(3)] Another computable example
comes when $A\equiv 0$ and $X$ is chosen so that $\breve \nabla$ has
torsion $\breve T(u,v)={2\over n-1}(v\wedge u)Z(x)$ for
$u, v\in T_xM$ and $Z$ a fixed vector field on $M$. Here $v\wedge u$
is the operator such that
$(v\wedge u)Z(x)=\langle v,Z(x) \rangle_x u- \langle u, Z(x)\rangle_x v$.
This connection was used by \cite{Ikeda-Watanabe}. In
\cite{Elworthy-LeJan-Li-book}, example 2.3.5 (though there is a minor
misprint in the formula written there), the generator on $q$-forms
is shown to be
${1\over 2}\Delta+\LL_{[{2(q-1)\over n-1}-1]Z}-{2\over n-1}\iota_Zd$.
Set $Z^{\#}=\langle Z(x), -\rangle_x$. The term involving the torsion in
 (\ref{Bismut-formula}) reduces to
\begin{eqnarray*}
&&\left(\int_0^t\rho(r)dr\right)^{-1}{4\over n-1}
\E\,\xi_t^*(Z^{\#}\wedge \phi)(b^1\wedge\dots \wedge b^{q+1})\\
&&={4\over (n-1)\int_0^t\rho(r)\,dr}P_t(Z^{\#}\wedge\phi).
\end{eqnarray*}
In this case we have:
\begin{equation}
\label{41}
\begin{split}
&d(P_t\phi)(b^1\wedge \dots \wedge b^{q+1})\\
&=\left(\int_0^t\rho(r)dr\right)^{-1}
  \E\,\phi\left( \int_0^t \rho(s)  \iota_{\langle -,
  { \breve {\parals_s}} d \breve{B}_s\rangle}
  W_s^{q+1}(b)\right)\\
 & +  {4\over (n-1)\int_0^t\rho(r)\,dr}\left(Z^{\#}\wedge \phi\right)
  \left(b^1\wedge\dots\wedge b^{q+1}\right).
\end{split}
\end{equation}
\end{enumerate}


