===== recycled on Sun Oct 27 09:02:00 EDT 2013 by drorbn on Debian-1207 ======


More accurately, let $G$ be
a finite dimensional Lie group and let $\frakg$ be its Lie algebra, let
$j:\frakg\to\bbR$ be the Jacobian of the exponential map $\exp:\frakg\to
G$, and let $\Phi:\Fun(G)\to\Fun(\frakg)$ be given by $\Phi(f)(x):=
j^{1/2}(x)f(\exp x)$. Then if $f,g\in\Fun(G)$ are Ad-invariant and
supported near the identity, then
\[ \Phi(f)\star\Phi(g) = \Phi(f\star g). \]