Let $G \subset GL(V)$ be linear Lie group. And $T_e(G)$ be its tangent space at identity, g -- arbitrary element in $G$. Then, $T_g(G)=T_e(G)\cdot g$.
I found this statement as an exercise in Vinberg and as it is nearly at the begining, it shouldn't use any advanced theory of Lie groups.
Can someone make it clear? Thanks a lot.
For each $g \in G$, we have the (smooth) right translation map $R_g: x \mapsto xg$. For each $h \in G$ this induces a (linear) derivative map $d(R_g)_h: T_h(G) \to T_{hg}(G)$. Given a tangent vector $v \in T_h(G)$, the authors define the tangent vector $vg \in T_{hg}(G)$ by pushing forward by this map: $$ vg := d(R_g)_h(v) \in T_{hg}(G) \, . $$ (One can show that this is indeed a right action on the tangent bundle $TG$ since $d(R_{g_1 g_2}) = d(R_{g_2}) \circ d(R_{g_1})$.) In particular, for $h = e$ and $v \in T_e(G)$, we get a tangent vector $vg \in T_g(G)$, which shows that $T_e(G) g \subseteq T_g(G)$.
Do you see how to show the reverse containment? Hint: Use $g^{-1}$.