Let $k$ be a (non-archimedean) local field of positive characteristic $p$ and $\mathfrak{n}$ be any finite-dimensional nilpotent Lie algebra over $k$ with nilpotence length $l<p$. It is well-known that $\mathfrak{n}$ admits a faithful finite-dimensional linear representation by nilpotent matrices, which we can simultaneously triangularize. Hence, the Campbell-Hausdorff formula implies that $N:=\exp(\mathfrak{n})$ is a linear algebraic unipotent group defined over $k$ and its Lie algebra is $\mathfrak{n}$. Then the adjoint action $\text{Ad}:N\rightarrow \text{GL}(\mathfrak{n})$ is defined by \begin{align*} \text{Ad}(n)(X):=\log(n\exp(X)n^{-1})\ \text{for $n\in N$ and $X\in \mathfrak{n}$}, \end{align*} where $\log: N\rightarrow \mathfrak{n}$ denotes the inverse of $\exp$. Let $\mathfrak{n}^*$ denote the linear dual space of the underlying vector space of $\mathfrak{n}$ and let $\text{Ad}^*:N\rightarrow \text{GL}(\mathfrak{n}^*)$ be the co-adjoint action given by $\text{Ad}^*(n)(f)=f\circ \text{Ad}(n^{-1})$ for $n\in N$ and $f\in \mathfrak{n}^*$. If $N$ and $\mathfrak{n}$ have nilpotence length $l<p$,

Q1: Can we conclude that all $\text{Ad}^*(N)$-orbits are closed in Hausdorff topology from $\mathfrak{n}^*$? I think we already know that all $\text{Ad}^*(N)$-orbits are local closed in Hausdorff topology from $\mathfrak{n}^*$, because it is a $k$-rational action when $l<p$ see locally closed orbits in metric Hausdorff topology

Q:2 Can we conclude that all stabilizer groups of the co-adjoint action are smooth? By the way, affine algebraic groups in characteristic zero are always smooth.

A positive answer of Q2 implies a positive answer of Q1 by Corollary 3.1.3."On the topology of relative orbits for actions of algebraic groups over complete fields""MR2721858" by Bac and Thang

  • $\begingroup$ I am not sure what you mean by "Hausdorff topology from $\mathfrak n^*$". Is that the one induced by some isomorphism of vector spaces with $k^N$? Proposition 2.1 in math.stanford.edu/~conrad/249BW16Page/handouts/unipgp.pdf proves that the orbits are closed in the Zariski topology. If the Hausdorff topology that you have in mind is finer than the Zariski topology on $\mathfrak n^*$, this gives an affirmative answer to Q1. $\endgroup$ – user94041 Feb 14 '17 at 17:43
  • $\begingroup$ @user94041: Proposition 2.1 implies that U.x is closed in Zariski topology in X. I am asking for U(k).x is Hausdorff closed in X(k). They are different. An interesting example saying that: G.v is closed doesn't imply that G(k).v is closed is given in Section 7.1 content.algebraicgeometry.nl/2014-5/2014-5-025.pdf $\endgroup$ – m07kl Feb 14 '17 at 20:27
  • 1
    $\begingroup$ m07kl, as remarked above, over algebraically closed field it is known that the orbits are closed for a unipotent group action on an affine variety. If you knew that the first Galois cohomology of a unipotent group is trivial it will also follow that this is the case for the action of the group of points over a local field. The vanishing of this Galois cohomology is known in characteristic 0. I am sure you could find this result (as well as an explanation of the general theory, in case you are not familiar with it) in the monumental book by Paltonov and Rapinchuk. $\endgroup$ – Uri Bader Feb 15 '17 at 13:30
  • 2
    $\begingroup$ (cont.) I would work through the proof. My guess is that it works verbatim, assuming the characteristic is high enough. Good luck. $\endgroup$ – Uri Bader Feb 15 '17 at 13:30
  • 1
    $\begingroup$ Another approach which might work in all characteristics is to imitate the argument in the Bernstein-Zelevinsky appendix: change the algebraic structure in a way that does not change the $k$-points but makes things smoother. Unfortunately, not being an expert myself, I cannot advice you further without devoting too much time here. If something will come to mind I'll let you know. $\endgroup$ – Uri Bader Feb 17 '17 at 7:52

Your Answer

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

Browse other questions tagged or ask your own question.