Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $R$ be a commutative ring, and, for $n\ge0$, ${\mathcal{A}}_n={\mathcal{A}}_n(R)$ the group of series $u(x)=\sum_0^\infty a_jx^{j+1}\in R[[x]]$ for which $a_0\in R^\times$ and $u(x)\equiv x\pmod{x^{n+1}}$. The group operation is composition of series. A series in ${\mathcal{A}}_n$ that’s not in ${\mathcal{A}}_{n+1}$ is said to have depth $n$. When $\kappa$ is a field with $p^s$ elements, ${\mathcal{A}}_1(\kappa)$ is often called the Nottingham group, but it seems to have been designated by different names in different times and places.

One sees that ${\mathcal{A}}_n$ is the projective limit of the groups ${\mathcal{A}}_n/{\mathcal{A}}_{n+r}$, and for $n>0$ these are unipotent algebraic groups of dimension $r$. So it's the proalgebraic group ${\mathcal{A}}_1$, defined over ${\mathbb{F}}_p$, that I'm asking about, and the existence of (finite-dimensional) algebraic subgroups in it.

There's one that's obvious, namely the image of ${\mathbf{G}}_{\mathrm a}\to {\mathcal{A}}_1$ by the homomorphism $$ t\mapsto \frac{x}{1-tx} =x+tx^2+t^2x^3+\cdots=u_t(x) $$ fractional-linear transformation leaving the origin fixed, expanded as series.

Almost as obvious is to apply the transformation $$ u(x)=x(1+g(x))\mapsto (u(x^m))^{1/m} =x(1+g(x^m))^{1/m} $$ clearly a group homomorphism, well defined when $m$ is prime to $p$, because you're raising a principal unit to an exponent that's in ${\mathbb{Z}}_p$. This takes $u_t$, a subgroup of Nottingham of depth one to a group of depth $m$. In particular, the Nottingham proalgebraic group has lots of one-dimensional algebraic subgroups.

There are no commutative connected algebraic subgroups of dimension greater than $1$: I've found an argument that depends on higher ramification theory and uses Hasse-Arf, but breaks down completely in any noncommutative case. I would like to settle the question of whether there are any algebraic subgroups of Nottingham of dimension greater than $1$, necessarily noncommutative: it will be enough to dispose of the two-dimensional case. So it occurs to me that this may already be known, or maybe someone out there can suggest an approach to me.

share|improve this question
    
Don't you mean $a_0=0$ instead of $a_0\in R^\times$? –  YCor Jan 28 '12 at 13:40
    
@Yves: No, he means $a_0\in R^{\times}$, because he's labeled his series as $a_0x+a_1x^2+a_2x^3+\cdots$. (Note that the $j$'th term in his sum is $a_jx^{j+1}$.) Although slightly nonstandard, it's a useful labeling for composition of unit power series. OTOH, for $n\ge1$, his congruence condition forces $a_0=1$ for all power series in $\mathcal{A}_n$. But $\mathcal{A}_0$ contains all series with $a_0\in R^{\times}$. –  Joe Silverman Jan 28 '12 at 20:16
    
The numbering is nonstandard for me, but seems very common usage among the group theorists. –  Lubin Jan 28 '12 at 20:41
    
Indeed, becasue (pro-$p$) group theorists usually take $a_0=1$. So if we would have numbered the usual way, then we would have to start with $a_2$. Also many times we write the elemnts as $x(1+g(x))$, so this way $g(x)$ has the "right" form. –  Yiftach Barnea Jan 28 '12 at 20:49
    
Regarding your question, several years ago I talked with Steve Donkin about the Nottingham group and as I rememebr he thought about it as a proalgebraic group. So you might want to ask him. –  Yiftach Barnea Jan 28 '12 at 20:52

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.