How to show that $\pi^*(g)=\chi(\det g)^{-1}$? I am reading the lecture notes. On page 14, how could we show that $\pi^*(g)=\chi(\det g)^{-1}$? I think that $\langle \pi^*(g)\lambda, v \rangle = \langle \lambda, \chi(\det g)^{-1} v \rangle = \lambda(\chi(\det g)^{-1} v) = \chi(\det g)^{-1} \lambda(v)$. We have
$$
\langle \pi^*(g)\lambda, v \rangle = \pi^*(g)(\lambda(v)). \quad (1)
$$
Therefore $\pi^*(g)=\chi(\det g)^{-1}$. Is the proof correct? Is (1) correct? Thank you very much.

 A: Formula (1) is correct and is defined, in general, for
$$ \pi^{*}: G\rightarrow \operatorname{Aut}_\mathbb C(V^{*}),$$
where $G$ is any given group and $V^{*}$ is the linear dual of the vector space $V$ over $\mathbb C$.
All you need it to consider formula (1) in the case
$$\lambda\in {\mathbb C^{\times}}^{*},~~ \lambda=1_{ {\mathbb C^{\times}}^{*} }$$
and 
$$v=1_{\mathbb C^{\times}},  $$
as the notes implicitly assume the identifications
$$\operatorname{Aut}_\mathbb C(\mathbb C^{\times})\simeq \mathbb C^{\times} $$
via  $\operatorname{Aut}_\mathbb C(\mathbb C^{\times})\ni\rho\mapsto \rho(1_{\mathbb C^{\times}}  ),$
and
$$\operatorname{Aut}_\mathbb C({\mathbb C^{\times}}^{*})\simeq 
{\mathbb C^{\times}}^{*} $$
via $\mathcal I:\operatorname{Aut}_\mathbb C({\mathbb C^{\times}}^{*})
\rightarrow {\mathbb C^{\times}}^{*}$, $\Phi\mapsto
\mathcal I(\Phi),$ $\mathcal I(\Phi):=\Phi(1_{ {\mathbb C^{\times}}^{*} })$.
In other words, to arrive at $\pi^{*}(g)=\chi(det g)^{−1}$ one needs to use the composition  
$$ G\stackrel{\pi^*}{\rightarrow} \operatorname{Aut}_\mathbb C({\mathbb C^{\times}}^{*})\stackrel{\mathcal I}{\rightarrow} {\mathbb C^{\times}}^{*}$$
in formula (1).
EDIT 
More explicitly:
$$(\mathcal I\circ \pi^{*})(g)=
\mathcal I(\pi^{*}(g))=\pi^{*}(g)(1_{ {\mathbb C^{\times}}^{*} })\in 
{\mathbb C^{\times}}^{*}; $$
then
$$\left(\pi^{*}(g)(1_{ {\mathbb C^{\times}}^{*} })\right)(1_{C^{\times}})=(\text{formula 1})=...
 $$
