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

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? $$T_{g}G = \{ g v : v \in T_{I} G\}$$

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.

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 $$T_gG =\{gX \mid X \in \mathfrak{g}\}.$$

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 $$y=f(x)$$).

• Thanks Ivo. Sorry in advance for my basic questions! Commented Sep 23, 2020 at 18:03
• You're good. If anything is unclear just ask. Commented Sep 23, 2020 at 18:04
• Thanks! As I said, I'm very new to manifolds and Lie groups and I'm not very comfortable with many definitions yet. 1) How to formally show that $L_g$ is a diffeomorphism? (It is of course bijective - then I have to show that its inverse is differentiable?) 2) Why $L_g$ being a diffeomorphism implies that the differential at identity is an isomorphism? 3) Can you please write the chain rule more explicitly? (I'm not sure how chain rule looks like when dealing with maps between manifolds). Commented Sep 23, 2020 at 18:23
• I added answers to all the questions in the original post. Commented Sep 23, 2020 at 18:42
• Thanks a lot, Ivo! Commented Sep 23, 2020 at 18:53