Characterizing the tangent space of a Lie group at an arbitrary point

Question

Suppose $G$ is a subgroup of the general linear group $\mathrm{GL}(n)$. Is the following statement true for all such $G$ and $g \in G$? If so, can you provide a proof? $$T_{g}G = \{ g v : v \in T_{I} G\}$$

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P.S. This question and its answer are perhaps closely related, but as a beginner I couldn't connect the dots at the end (assuming the above statement is true). Here is my attempt:

If $L_g : x \mapsto g x$, then I know that, by definition, the differential of $L_g$ at the identity maps the tangent space at the identity (Lie algebra $T_{I} G$)) to the tangent space of $G$ at $g$ (i.e., $T_{g}G$). The differential is given by $dL_g : v \mapsto gv$ where $v \in T_IG$. Now to finish the proof, I have to show that $dL_g$ is surjective, but I'm not sure how to proceed. Here it is stated that the differential is an isomorphism (thus bijective) - can you explain why this is the case and provide a proof?

Thanks.

Answer 1

Consider the translation $L_g:G\to G$. Then ${\rm d}(L_g)_e : T_eG = \mathfrak{g}\to T_gG$ is a linear map. It is an isomorphism because $L_g$ is a diffeomorphism. The inverse of $L_g$ is $L_{g^{-1}}$ and so the chain rule gives $({\rm d}(L_g)_e)^{-1} = {\rm d}(L_{g^{-1}})_g$. This means that $$T_gG = \{{\rm d}(L_g)_e(X) \mid X \in \mathfrak{g}\}.$$

This holds for any Lie group. If $G\leq {\rm GL}(n,\Bbb R)$, then $L_g$ is the restriction of a linear map, so ${\rm d}(L_g)_e = L_g$ , leading to $$T_gG =\{gX \mid X \in \mathfrak{g}\}.$$

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Follow-up questions:

1. by definition of Lie group, the multiplication $$\mu:G\times G \ni (g,h)\mapsto gh\in G$$is smooth, so if you freeze one of the arguments, you get that $L_g$ and $R_h$ are smooth as well. More precisely, $L_g = \mu \circ \iota_g$ is a composition of smooth maps, hence smooth, where $\iota_g:G\to G\times G$ given by $\iota_g(h)=(g,h)$ is the inclusion. You know that set-theoretically $(L_g)^{-1} = L_{g^{-1}}$, but $L_{g^{-1}}$ is automatically smooth by the above argument, since it is a translation (no matter by what element).

2. Actually the differential at any point is an isomorphism: $({\rm d}(L_g)_h)^{-1} = {\rm d}(L_{g^{-1}})_{gh}$. For more details, see (3) below.

3. If $f:M\to N$ is a diffeomorphism between any manifolds with inverse $f^{-1}:N\to M$, differentiate $f^{-1}\circ f={\rm Id}_M$ at $p\in M$ to get ${\rm d}(f^{-1})_{f(p)} \circ {\rm d}f_p = {\rm Id}_{T_pM}$. Similarly, differentiate $f\circ f^{-1} = {\rm Id}_N$ at $f(p)\in N$ to get ${\rm d}f_p\circ {\rm d}(f^{-1})_{f(p)} = {\rm Id}_{T_{f(p)}N}$. This shows that $({\rm d}f_p)^{-1} = {\rm d}(f^{-1})_{f(p)}$. The slogan is "the inverse of the derivative is the derivative of the inverse" (but you need to get base point right, compare with Calculus 1: $(f^{-1})'(y)=1/f'(x)$, where $y=f(x)$).