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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.

Viktor Vaughn
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Sasha Mayer
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1 Answers1

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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 $g^{-1}$.

Viktor Vaughn
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  • thanks so much for help! Following your way, with $R_{g^{-1}}$ and $h=g$, the inverse inclusion is evident. But am i right, that you've mentioned tangent bundle for some extra information and never used it later? Also, i expected that $T_{e}(G) \cdot g$ is not just notation, but also means product of matrixes (since $G$ is subgroup of $GL$), but i can't see it from here.. – Sasha Mayer Dec 19 '16 at 02:45
  • If I'm looking at the same reference, the authors define both $g v$ and $v g$. My comment was simply meant to explain the notation: $v(gh) = (vg)h$, which gives a right action of $G$ on the set of all tangent vectors. The authors state that one can interpret this as matrix multiplication if the tangent spaces are embedded in $\text{End}(V)$. I'm on the road right now, but I'll try to update my answer explaining that assertion in a few days. – Viktor Vaughn Dec 19 '16 at 04:30
  • many thanks, it will be also very useful! – Sasha Mayer Dec 19 '16 at 17:54