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I'm reading through Michael Taylor's notes from this PDF.

The author starts saying that given an $N$-dimensional Lie group $G$ and some covector $\omega_e \in \bigwedge^N T^*_e G$, there is a unique differential $N$-form $\omega_\ell$ that is left-invariant, meaning $\omega_\ell(e)=\omega_e$ and $L_g^*\omega_\ell=\omega_\ell$ with $L_g(h)\equiv gh$. Similarly, there is a unique right-invariant $\omega_r(e)=\omega_r$, $R_g^* \omega_r =\omega_r$. They then observe that these must also satisfy $$R_h^* \omega_\ell = \alpha(h)\omega_\ell, \qquad L_h^* \omega_r = \beta(g) \omega_r,$$ for some pair of homomorphisms $\alpha,\beta:G\to(0,\infty)$. The group is said to be unimodular iff $\alpha=1$ or $\beta=1$.

Shortly thereafter, the authors discuss the adjoint representation of the group, by defining $$K_g:G\to G, \qquad K_g(h)\equiv ghg^{-1}$$ and defining $\operatorname{Ad}(g)=DK_g(e):T_e G\to T_e G$, with $D$ denoting the differential. After a few standard observations about the adjoint representation, they say that comparing the two equations above, we find that $$\alpha(g) = \det(\operatorname{Ad}(g)).$$ Is there a more explicit way to see where this relation comes from?

I can see some connection: I can write $K_g=L_g \circ R_g^{-1}$ and thus $$\operatorname{Ad}(g)=DK_g(e) = (DL_g(g^{-1}))\circ (D R_g^{-1}(e)).$$ Still, I think I'm missing something, because I don't how to link this with the statements about left- or right-invariant differential forms made previously, as well as whether we should make a choice of $\omega_e$ for this to work, etc.

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The reason why the choice of $\omega_e$ does not matter is that the space $\bigwedge^NT_e^*G$ is one dimensional. Hence different choices of $\omega_e$ are related by a constant factor and hence lead to the same maps $\alpha$ and $\beta$.

This also is the reason why $R_h^*\omega_\ell$ must be a constant multiple of $\omega_\ell$. Indeed it suffices to evaluate $R_h^*\omega_\ell$ in $e$ and compare this to $\omega_e$. By definition, $R_h^*\omega_\ell(e):(T_eG)^N\to\mathbb R$ is given by $\omega_\ell(h)\circ (DR_h(e))^N$. But in turn $\omega_\ell(h):(T_hG)^N\to\mathbb R$ is given by $\omega_e\circ (DL_{h^{-1}}(h))^N$. So overall, you see that you get that $$R_h^*\omega_\ell(e)=\omega_e\circ (DL_{h^{-1}}(h)\circ DR_h(e))^N=\omega_e\circ(\text{Ad}(h))^N=\det(\text{Ad}(h))\omega_e.$$

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