4
$\begingroup$

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.

$\endgroup$
3
  • 5
    $\begingroup$ It's 1 to 1 in a neighborhood of 0, and so the identity is also. But as you say, it can't be true over all p-adic algebraic integers, even when restricted to positive valuation. Try to compute the radius of convergence of the log function. Note that all of this has an analogue over the complex numbers $\endgroup$ Oct 23 at 6:57
  • 2
    $\begingroup$ To echo something implicit in Dror's comment: the same issue arises when F is the multiplicative group law, so it might be best to understand that one first; in this case we have an explicit formula for the coefficients of log and exp, so the radius of convergence is easily computable. $\endgroup$ Oct 23 at 8:17
  • $\begingroup$ The part which is very confusing to me is that this means the torsion points of $F$ cannot be in the radius of convergence of log, but we can still evaluate log to be 0 at these points, so I must be missing something. Another somewhat related question I have is suppose $\alpha$ is in the radius of convergence of log. Then do we always have log$(\alpha)$ in the radius of convergence of exp$_F$? $\endgroup$ Oct 23 at 17:26
11
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ As Lubin has commented somewhere else on this site, it is a remarkable property that the well-behaved exponential (everywhere convergent, and homomorphism) and badly-behaved logarithm (finite radius of convergence, not a homomorphism) in complex analysis switch roles in $p$-adic analysis, where the logarithm is better behaved than the exponential in the sense of having a larger radius of convergence (although both are $p$-adically nice in being homomorphisms between their discs of convergence). $\endgroup$
    – KConrad
    Oct 30 at 22:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.