If I define the inverse map in a Lie group $G$ as,
$$i: G \rightarrow G,\quad i(g) = g^{-1}, \forall g \in G \tag1$$
I think that the associated push-forward would be,
$$i_*: T_gG \rightarrow T_{g^{-1}}G, \quad i_*(X|_g) = X|_{g^{-1}} \equiv -X|_g, \forall X \in \mathfrak{X}(G) \tag2$$
Where, $\mathfrak{X}(G)$ is the set of tangent vector fields in $G$
Is Eq. (2) the action of $i_*$ or it is wrong? The part which I'm not very sure is $i_*(X|_g) = X|_{g^{-1}}$. I have this doubt because in other posts (e.g. Pushforward of Inverse Map around the identity? or Differential of the inversion of Lie group) it is handle $T_eG$ and not $T_gG$ in general.
Best Answer
In general the proposed equation makes no sense, as quantities on the two sides live in different tangent spaces: For $X \in T_g G$, we have $-X_g \in T_g G$ but $X_{g^{-1}} \in T_{g^{-1}} G$, and these spaces coincide only if $g^2 = e$.
On the other hand, unwinding definitions gives $$i = R_{g^{-1}} \circ i \circ L_{g^{-1}}$$ (here, $L_h$ and $R_h$ respectively denote left and right multiplication by $h$), so differentiating gives $$T_g i = T_e R_{g^{-1}} \circ T_e i \circ T_g L_{g^{-1}},$$ and using that $T_e i \cdot Y = - Y$ gives what you wanted in your equation (2): $$\boxed{T_g i \cdot X = T_e R_{g^{-1}} (-T_g L_{g^{-1}} \cdot X) = -T_g (L_{g^{-1}} \circ R_{g^{-1}}) \cdot X }.$$ For a general Lie group this is already fully simplified, but for a linear Lie group $G \leq GL(n, \Bbb R)$ the usual matrix identifications give $$\boxed{T_A i \cdot X = -A^{-1} X A^{-1}} .$$
Remark The proposed equation is wrong for a different reason, too, namely that the value of the vector field $X$ at one point on $G$ need not be related it its value at any other point. Such a relation is forced, however, if we restrict our attention to left- or right-invariant vector fields.