Tangent spaces to the orbit of a Lie group

Question

Let $G\subset GL(n)$ be a Lie subgroup and denote $$ M:=G x_0 = \{ Ax_0\ :\ A\in G\}\subset \mathbb R^n,$$ where $x_0\ne 0$ is a fixed vector in $\mathbb R^n$. Then $M$ is a smooth submanifold of $\mathbb R^n$.

Question. Is it true that $$\tag{1}A(T_{x_0} M)=T_{Ax_0} M,\qquad \forall A\in G\ ?$$

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At first sight I would say that (1) is true. However, if the Lie algebra $\mathfrak g$ is given by $$\mathfrak g = \text{span}\ (g_1, g_2\ldots g_m), $$ then $$T_{x_0} M = \text{span}\, (g_1 x_0, \ldots ,g_mx_0),$$ and analogously $$ T_{Ax_0} M=\text{span}\, (g_1 Ax_0, \ldots, g_mAx_0), $$ while $$ A(T_{x_0} M) = \text{span}\, (Ag_1 x_0, \ldots ,Ag_mx_0),$$ and I don't see a reason why the last two vector spaces should coincide. I don't know how to handle the commutators $[g_j, A]$, where $g_j\in\mathfrak g$ and $A\in G$.

Answer 1

The restriction $f_A$ of the map $A$ to $M$ induces a diffeomorphism of $M$, remark $df_A=A$ since $A$ is linear. We have $f_A(x_0)=A(x_0)$, and $df_A(T_{x_0}(M))=A(T_{x_0}M)=T_{f_A(x_0)}M=T_{A(x_0)}M$.

Answer 2

I would like to complement Tsemo's answer, which is short and elegant, with a couple commentaries.

Tsemo's argument shows that $T_{Ax_0} M = A(T_{x_0} M)$, that is $$\tag{1} \text{span}\, ( g_1 A x_0, \ldots ,g_n Ax_0) = \text{span}\,(Ag_1 x_0,\ldots, Ag_n x_0). $$ This identity can be proved directly, by observing that $$\tag{2} A^{-1} g A \in \mathfrak g, \qquad \forall g\in\mathfrak g,\ \forall A\in G, $$ so, even if $g_jA\ne Ag_j$ in general, it is nonetheless true that $g_jA$ is a linear combination of $Ag_1,\ldots Ag_n$, and (1) follows from basic linear algebra.

Now, the identity (2) is a basic fact of abstract Lie group theory and it looks quite trivial. I have tried to prove (2) via the matrix exponential, that is, taking $A=e^{\lambda h}$ for a $h\in \mathfrak g$. Then (2) is equivalent to $$ \left(1-\lambda h +\frac{\lambda^2}{2}h^2 -\ldots\right) g \left(1+\lambda h +\frac{\lambda^2}{2} h^2+\ldots\right)\in \mathfrak g, $$ which looks much less obvious. I have computed some terms of the left-hand side; $$ g +\lambda(gh-hg)+\frac{\lambda^2}{2}(h^2g+gh^2-2hgh)+\ldots $$ and to my amazement I have discovered that they equal $$ g+\lambda [g, h] +\frac{\lambda^2}{2}[[g,h],h]+\ldots $$ which is in agreement with (2), since linear combinations of commutators in $\mathfrak g$ are in $\mathfrak g$. Indeed, this is the definition of Lie algebra.

The moral is that the modern viewpoint on Lie groups and Lie algebras is truly impressive, but it can also look a lot dryer than it actually is.