# differential geometry – A formula for Lie groups

Let $$G$$ be a Lie group and $$H$$ a closed subgroup. We denote by $$mathfrak g$$ and $$mathfrak h$$ by the Lie algebras of $$G$$ and $$H$$ respectively. For $$xin G,$$ denote $$tau(x):G/Hto G/H$$ by $$gHmapsto xgH.$$ Denote $$pi:Gto G/H$$ to be the quotient map and put $$o=pi(e)$$ where $$e$$ is the identity of $$G.$$ For $$hin H$$ denote $$dtau(h):=dtau(h)(o)$$ and $$dpi:=dpi(e).$$ I want to understand the following Lemma (lem 1.7) of Helgason’s book `Groups, Geometry and Analysis’

$$det(dtau(h))=frac{det Ad_G(h)}{det Ad_H(h)}.$$

Helgason proves that if $$mathfrak m$$ is any subspace so that $$mathfrak g=mathfrak hoplus mathfrak m$$, then for any $$hin H$$ we have $$dpi|_{mathfrak m}circ A_h=dtau(h)circ dpi|_{mathfrak m},$$ where $$A_h(X)$$ is the projection of $$A_G(h)X$$ onto $$mathfrak m$$ for $$Xinmathfrak m.$$ Therefore, $$det A_h=det(dpi(h))$$. Also $$Ad_G(h)X=Ad_H(h)X$$ for $$Xinmathfrak h.$$ I understand upto his point. Now Helgason directly draws the conclusion that $$det Ad_G(h)=det(dtau(h))det(Ad_H(h)).$$ I do not understand this. It seems that the problem will be solved if we can show that $$Ad_G(h)=Ad_H(h)oplus A_h$$. But this seems unlikely.