I try to find the automorphisms $\sigma$ of $\mathbb F_q((\frac1T))$ with the following properties: $\sigma$ is an isometry,and $\sigma(\mathbb F_q[T])\subseteq\mathbb F_q[T]$. Almong the automorphisms $$\mathcal N=\{\sigma(\frac1T)=\sum_{n\ge1} \alpha_i\frac1{T^i}\mid\alpha_i\in\mathbb F_q \}$$ the Nottingham group of $\mathbb F_q((\frac1T))$, I only found the trivial automorphisms $T\mapsto \alpha T$. Does there exist other ones (maybe not in $\mathcal N$)?

  • $\begingroup$ You left out the condition $\alpha_1 \not= 0$. $\endgroup$ – KConrad Feb 9 '15 at 17:29

There are a few more. Suppose $\sigma(T) =a_n T^n + \dots + a_0$. Then

$$ \sigma \left(\frac{1}{T}\right)= \frac{1}{\sigma(T)} = a_n^{-1} T^{-n} \left(1- \frac{a_{n-1}}{a_n} T^{-1} + \frac{a_{n-1}^2 - a_n a_{n-2}}{a_n^2}T^{-2} + \dots \right)$$

This is not an automorphism unless $n=1$. In that case the formula looks like:

$$ \sigma\left(\frac{1}{T} \right) = \frac{1}{a_1} \frac{1}{T} - \frac{a_0}{a_1^2} \frac{1}{T^2} + \frac{a_0^2}{a_1^3} \frac{1}{T^3} + \dots $$

or more elegantly:

$$ \sigma\left(\frac{1}{T} \right) = \alpha \frac{1}{T} + \alpha \beta \frac{1}{T^2} + \alpha \beta^2 \frac{1}{T^3} + \dots $$

Theser are all.


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