# Why is Lie group unimodularity related to the determinant of the adjoint?

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.

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 $$R_h^*\omega_\ell(e)=\omega_e\circ (DL_{h^{-1}}(h)\circ DR_h(e))^N=\omega_e\circ(\text{Ad}(h))^N=\det(\text{Ad}(h))\omega_e.$$