# Linear isotropy group at the origin of a homogeneous space

Let $$G$$ be a Lie group and $$H$$ a compact subgroup. I am trying to understand Example 1.3 in Kobayashi & Nomizu's "Foundations..." , in Chapter IV.1. They say we consider the linear isotropy group $$\tilde H$$ at the origin $$o$$ of $$G/H$$, where $$o$$ is the coset $$H$$; they don't describe further how this is defined.

I am tempted to think that to define this, we start with the group $$G_H := \{ g \in G: gH = H \}$$. Then the differential of the multiplication with an element of $$G_H$$ maps the tangent space at any point of $$H$$ to the tangent space at another point of $$H$$, so intuitively it should indeed give a linear isomorphism on $$T_o (G/H)$$. But I'm having trouble defining it rigorously, first of all because I don't quite know how to describe elements of $$T_o (G/H)$$. I'm tempted to say a vector $$T_o (G/H)$$ is of the form $$X + TH$$, where $$X \in T_e G$$ ($$e$$ = the identity of $$G$$). But this doesn't even make sense to me, since the bundle $$TH$$ consists of vectors at other points than $$e$$...

• treat the tangent space as the lie algebra $g/h$? Commented Aug 1 at 14:30
• Your first sentence is wrong: $G/K$ typically is not a Lie group (is not even homeomorphic to one). Commented Aug 1 at 15:30
• @MoisheKohan This makes matters worse then. How can I make sense of Example 1.3, then? Commented Aug 2 at 15:49
• $G$ acts on the space $G/H$ from the left. The subgroup $H\subset G$ preserves the point $o=[H]$, i.e. for any $h\in H$ you have $ho=o$. Then the differential $d_0h:T_oG/H\to T_oG/H$ is a linear map, you can check that $d_o h' d_oh = d_o(h' h)$ for any $h',h\in H$. This gives you a group homomorphism $H\to GL(T_oG/H)$. The image of this homomorphism is what I would call the linear isotropy group of $o$ in $G/H$. Commented Aug 3 at 9:23