Let $\mathfrak{g}$ a real Lie algebra and $s$ an involutive automorphism of $\mathfrak{g}$. Let $\mathfrak{f}$ the set of fixed point of $s$ and let $G$ a Lie group with Lie algebra $\mathfrak{g}$ and $K$ a Lie subgroup of $G$ with Lie algebra $\mathfrak{f}$. Suppose $K$ is connected and closed in $G$. For any $G$-invariant metric the space $G/K$ is a Riemannian locally symmetric space (in particular, it is an homogeneous space). The subspace $\mathfrak{p}:=\{X \in G : sX=-X \}$ can be indentified as $(G/K)_{o=\pi(e)}$ by using the map $d\pi$, where $\pi$ is the natural projection.
Let $\tau_g : G/K \rightarrow G/K$ such that $\tau_g(xK)=gxK $and $T_X$ the restriction of $\left(\operatorname{ad}X\right)^2$ to $\mathfrak{p}$.
It is possible to prove (see the book of Helgason) that the differential of the (Riemmanian) exponential is given by: $$ d \operatorname{Exp}_X =(d \tau_{\exp(X)})_o \circ \sum_{n=0}^\infty \frac{(T_X)^n}{(2n+1)!}$$ where $\exp$ is the Lie exponential and $X \in \mathfrak{p}$.
I'm wondering if there is some nice formula for the inverse of such map. In particular, if $G$ is a matrix Lie group the inverse exponential is: $$ d \exp_X^{-1}=\sum_{n=0}^\infty B_j (\operatorname{ad}_X)^j $$ where $B_j$ are the bernoulli numbers.
Because $G$ is a matrix Lie group we have that $(d\tau_{\exp (X)})_o=\exp(X)$ so i'm expecting some similar formula to holds in this case, but i can't figure out how to espress it