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.
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
$$\left(\pi^{*}(g)(1_{ {\mathbb C^{\times}}^{*} })\right)(1_{C^{\times}})=(\text{formula 1})=... $$