Let $K$ be a finite extension of $\mathbb{Q}\_p$. Let $F$ be a Lubin–Tate formal group law defined over $K$ with endomorphism $f(T)$ corresponding to $\pi$ (a uniformizer of $K$). Then one can define the logarithm of $F$ to be $\lambda\_F(T) = \lim\_{n\rightarrow\infty}\pi^{-n}f^n(T)$. Then $\lambda\_F(T) = T + \text{higher-degree terms}$, so the logarithm is invertible under function composition. We define $\operatorname{exp}\_F(T)$ to be the inverse of $\lambda\_F$ under composition. Then $\operatorname{exp}\_F(\lambda\_F(T)) = T$. However I don't understand how this is possible if $\lambda\_F$ is not one to one. In particular $\lambda\_F$ sends all torsion points of $F$ to 0. Please let me know what is wrong here.



Question about log and exp of a formal group law
The radius of convergence of any formal group $F$ (one-dimensional, finite height) is $1$, in other words $L\_F$ will converge at all $z\in\Bbb C\_p$ with $v(z)>0$.


In particular, the logarithm is convergent at all torsion points of the formal group. What goes wrong, goes wrong with the exponential, whose series is not convergent far enough from the origin to reach any of the torsion points.


