 how graphs of symplectic
automorphisms of $T^* \R^n$ can be turned into Lagrangian subspaces of
$T^* \R^{2n}$.

Let us identify the product $T^* \R^n \times T^* \R^n$ with $T^*
\R^{2n}$.  Then the graph of the linear involution $C$ is the conormal
space of the diagonal $\Delta_{\R^n}$ in $\R^n \times \R^n$,
\[
\graf C = N^* \Delta_{\R^n}.
\]
Moreover, the fact that $C$ is anti-symplectic easily implies that
a linear endomorphism $A : T^* \R^n \rightarrow T^* \R^n$ is
symplectic if  and only if the graph of $CA$ is a Lagrangian
subspace\footnote{Here $T^* \R^{2n}$ is endowed with its standard
  symplectic  structure. In symplectic geometry it is also customary
  to  endow the product of a symplectic vector space $(V,\omega)$ with
  itself by the symplectic structure $\omega \times (-\omega)$. With
  the  latter convention, the product of two Lagrangian subspaces is
  Lagrangian, and an endomorphism is symplectic if and only if its
  graph is  Lagrangian. When dealing with cotangent spaces and
  conormal  spaces it seems more convenient to adopt the former
  convention,  even if it involves the appearance of the involution
  $C$.}  of $T^* \R^{2n}$, which is true if and only if the graph of $AC$ is a
Lagrangian  subspace of $T^* \R^{2n}$.

Theorem 3.2 in \cite{rs93} implies that if $A$ is a path of symplectic
automorphisms of $T^* \R^n$ and $\lambda$, $\nu$ are paths of
Lagrangian  subspaces of $T^* \R^n$, then
\begin{equation}
\label{prd}
\mu( A \lambda, \nu) = \mu ( \graf AC, C\lambda \times \nu) = -
\mu(\graf  CA, \lambda \times C\nu).
\end{equation}

\section{Conormal bundles}

Let $M$ be a smooth manifold\footnote{Unless otherwise stated, manifolds and submanifolds are always assumed to have no boundary.} of dimension $n$, 
and let $T^*M$ be the
cotangent bundle of $M$ with projection $\tau^* : T^*M \to M$. The
cotangent bundle $T^*M$ carries the following canonical structures:
The {\em Liouville one-form\/} $\theta$ and the {\em Liouville
vector field\/} $\eta$ which can be defined by
\[
\theta(x) [\zeta] = x\big[ D\tau^*(x)[\zeta]\big]=
d\theta (x) [\eta, \zeta],  \qquad \ \forall\, x\in T^*M, \; \forall
\zeta  \in T_xT^*M,
\]
and the symplectic structure $\omega=d\theta$. Elements of $T^* M$ are
also denoted as pairs $(q,p)$, with $q\in M$, $p\in T_q^* M$. 

The {\em vertical subbundle} is the $n$-dimensional vector subbundle
of  $TT^*M$ whose fiber at $x\in T^* M$ is the linear subspace
\[
T_x^v T^*M := \ker D\tau^* (x)\subset T_x T^*M.
\]
Each vertical subspace $T_x^v T^*M$ is a Lagrangian subspace of the
symplectic vector space $(T_x T^*M,\omega_x)$.

If $Q$ is a smooth submanifold of $M$, the {\em conormal bundle} of
$Q$ is  defined by
\[
N^* Q := \set{x\in T^*M}{\tau^* (x)\in Q, \; x[\xi] = 0 \; \forall \xi
  \in  T_{\tau^*(x)} Q}.
\]
It inherits the structure of a vector bundle over $Q$ of dimension
$\codim  \, Q$. The conormal bundle of
the  whole $M$ is the zero-section, while the conormal bundle of a
point  $Q=\{q\}$ is $T^*_q M$. Moreover, the Liouville one-form $\theta$ vanishes identically on $N^* Q$, in particular $N^* Q$ is a Lagrangian 
submanifold of $T^*M$, i.e.\ its tangent space at every point $x$ is a
Lagrangian  subspace of $(T_x T^*M,\omega_x)$. Actually, the converse is also true:

\begin{prop}
\label{conorcar}
Let $M$ be a smooth $n$-dimensional manifold and let $L$ be an $n$-dimensional submanifold of $T^*M$ on which the Liouville one-form $\theta$ vanishes identically. Then the intersection of $L$ with the zero-section of $T^*M$ is a smooth submanifold $R$, and if $L$ is a closed subset of $T^*M$ then $L=N^* R$.
\end{prop}

\begin{proof}
{\em Claim 1. The intersection $R:= L\cap M$, where $M$ denotes the zero-section of $T^*M$, is a smooth submanifold.} The matter being local, up to choosing suitable local coordinates $q_1,\dots,q_n$ and conjugate coordinates $p_1,\dots,p_n$, we may assume that $M=\R^n$ and $L$ is a graph of the form
\begin{equation}
\label{locdes}
L = \set{(q,Q(q,p),P(q,p),p)}{q \in \R^k, \; p \in (\R^k)^{\perp} },
\end{equation}
where $\R^k$ denotes the subspace of $\R^n$ spanned by the first $k$ vectors of the standard basis, $0\leq k \leq n$, $Q$ is a smooth map into $\R^{n-k}$ -- the subspace of $\R^n$ spanned by the last $n-k$ vectors of the standard basis -- and $P$ is a smooth map into $(\R^{n-k})^{\perp}$. Here we are also using the fact that $L$ is Lagrangian, so that its tangent space at a given point can be represented as in Remark \ref{linlag}. 
Then the intersection of $L$ with the zero-section is the set
\begin{equation}
\label{inters}
R = L \cap ( \R^n \times (0) ) = \set{(q,Q(q,0),0,0)}{q\in \R^k \mbox{ such that } P(q,0)=0}.
\end{equation}
The fact that the Liouville one-form $\theta$ vanishes on $L$ is equivalent to the fact that the maps $Q=(Q_{k+1},\dots,Q_n)$ and $P=(P_1,\dots,P_k)$ satisfy the identity
\[
\sum_{j=1}^k P_j(q,p) \, dq_j + \sum_{j=k+1}^n p_j \, dQ_j (q,p) = 0, \quad \forall (q,p) \in \R^k \times (\R^k)^{\perp},
\]
from which we deduce the identities
\begin{equation}
\label{tv1}
P_j(q,p) + \sum_{h=k+1}^n p_h \frac{\partial Q_h}{\partial q_j} (q,p)  = 0,  \quad \forall (q,p) \in \R^k \times (\R^k)^{\perp}, \quad \forall j=1,\dots,k.
\end{equation}
In particular, (\ref{tv1}) implies that $P(q,0)=0$ for every $q\in \R^k$, so by (\ref{inters}) the intersection of $L$ with the zero-section is
\begin{equation}
\label{desR}
R = L \cap ( \R^n \times (0) ) = \set{(q,Q(q,0),0,0)}{q\in \R^k},
\end{equation}
which is a smooth submanifold.

\medskip

\noindent{\em Claim 2. The Liouville vector field $\eta$ is tangent to $L$.} In fact, for every $x\in L$ and every $\zeta\in T_x L$ we have
\[
\omega(x) [\eta(x),\zeta] = \theta(x)[\zeta] = 0,
\]
so $\eta(x