Proposition 57

Let $G$ be a Lie group of finite dimension $n$. Choose a basis for $\bigwedge^n T_e(G)^*$; by means of the right (resp. left) trivialization of $\bigwedge^n T(G)^*$, we can identify this vector bundle with the trivial vector bundle $G \times K$, so that the transpose of a scalar differential operator is identified with a scalar differential operator.

Then, if $u \in U(G)$, the transpose of $L_u$ (resp. $R_u$) is $L_u^\nu$ (resp. $R_u^\nu$).

We shall consider the case where $\Lambda^n T(G)^*$ has been trivialized using a right invariant form $\omega$.

Suppose that the proposition has been proved for elements $u_1, u_2$ of $U(G)$. Then,

and hence the proposition is true for $u_1 * u_2$. It therefore suffices to prove the proposition when $u \in T_e(G)$. Now $L_u$ is defined by $G$ operating on $G$ on the right (no. 6) and hence $\theta_{L_u} \omega = 0$ since $\omega$ is right invariant (\emph{Differentiable and Analytic Manifolds}, R, 8.4.5); therefore, if $f$ is an analytic function in an open neighbourhood of $e$ with values in $K$, then $\theta_{L_u}(f \omega) = (\theta_{L_u} f) \omega$ (\emph{Differentiable and Analytic Manifolds}, R, 8.4.8). Using the identifications made and \emph{Differentiable and Analytic Manifolds}, R, 14.4.1, the transpose of $L_u$ is $-L_u$, that is $L_u^\nu$.