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.
Log and Exp of Formal Group Law – Common Questions
formal-groupsnt.number-theoryp-adic
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I don't think you should be trying to interpret this stuff over the (real or) complex numbers. Discs do not have nice algebraic properties in the archimedean world as they do $p$-adically. Even $p$-adically, one never talks about an actual radius of convergence but just makes sure things converge on the maximal ideal and work there. For example, the Lubin--Tate series could even be a polynomial, with infinite radius of convergence, but still one just focuses on what it does inside the unit disc ($p$-adically).
Okay, here's a few words about the relation between the $L$-series and the formal group. In general, if $F(X,Y)$ is the formal group law for $\hat G$, then there is an associated formal invariant differential $\omega(T)=P(T)dT$ given by $P(T)=F_X(0,T)^{-1}$. Formally integrating the power series $\omega(T)$ gives the formal logarithm $\ell(T)=\int_0^T\omega(T)$. The logarithm maps $\hat G$ to the additive formal group, so we can recover the formal group as $$ F(X,Y) = \ell^{-1}(\ell(X)+\ell(Y))$$. (See, e.g., Chapter IV of Arithmetic of Elliptic Curves for details.)
Now let $E$ be an elliptic curve and $\omega=dx/(2y+a_1x+a_3)$ be an invariant differential on $E$. If $E$ is modular, say corresponding to the cusp form $g(q)$, then we have (maybe up to a constant scaling factor) $\omega = g(q) dq/q = \sum_{n=1}^{\infty} a_nq^{n-1}$. Eichler-Shimura tell us that the coefficients of $g(q)$ are the coefficients of the $L$-series $L(s)=\sum_{n=1}^\infty a_n n^{-s}$. Integrating $\omega$ gives the elliptic logarithm, which is the function you denoted by $f$, i.e., $f(q)=\sum_{n=1}^\infty a_nq^n/n$, and then the formal group law on $E$ is $F(X,Y)=f^{-1}(f(X)+f(Y))$.
To me, the amazing thing here is that the Mellin transform of the invariant differential gives the $L$-series. Going from the invariant differential to the formal group law via the logarithm is quite natural.
Best Answer
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.