Let $K$ be a local field of characteristic $p$, where $p$ is a prime number greater than 2. In particular, $(x+y)^p=x^p+y^p$ for $x,y\in K$.

The fake Heisenberg group is defined to be $$ G=\{\begin{pmatrix} 1&a&b\\ 0&1&a^p\\ 0&0&1\\ \end{pmatrix}: a,b\in K\}$$ It is a non-abelian, $K$-split connected algebraic group, which is also unipotent with nilpotent length two. Consider its Lie ring $$ \mathfrak{g}=\text{Log}(G)=\{\begin{pmatrix} 0&x&y\\ 0&0&x^p\\ 0&0&0\\ \end{pmatrix}: x,y\in K\}$$ Since $K$ is not perfect, $\mathfrak{g}$ is not stable under scalar multiplication. In particular, $\mathfrak{g}$ is not a vector Space and is different from Lie(G), the Lie algebra associated to G.

Q.1: Do we know the unitary dual $\hat{G}$ of $G$? Notice that $G$ is a central non-split extension of the additive group $K$: $$0\rightarrow K\rightarrow G \rightarrow K \rightarrow 0.$$ I can image to apply the Machey Machine to this group extension similar to the real Heisenberg groups.

Q.2: Is $G$ CCR? or type I?

Q.3: Let $\hat{\mathfrak{g}}$ be the Pontrjagin dual of the additive group $\mathfrak{g}$ and let $G$ act on $\hat{\mathfrak{g}}$ by the coadjoint action. Are all orbits closed or locally closed in $\hat{\mathfrak{g}}$? Recall that the adjoint action of $G$ on $\mathfrak{g}$ is defined by $Ad(g)(X)=gXg^{-1}$ for $g\in G$ and $X\in \mathfrak{g}$ and the coadjoint action of $G$ on $\hat{\mathfrak{g}}$ is defined by $Ad^*(g)(f)=f\circ Ad(g^{-1})$ for $f\in \hat{\mathfrak{g}}$.

Dear all, my motivation is about the Kirillov orbit method for unipotent Groups over a local field of positive characteristic (see Corollary 7.3 and Exapmple 8.3 in A general Kirillov Theory for locally compact nilpotent groups). It describes the unitary dual of $G$ by its coadjoint orbits. In order to do this, we need to know Q.3 has an affirmative answer according to Corollary 7.3 and Exapmple 8.3 in op. cit.

So if the associated Lie ring $\mathfrak{g}$ is a vector space (this is always the case if char(K)=0), then we can identify its Pontrygagin dual with its linear dual, on which the unipotent group $G$ acts algebraiclly by the coadjoint action. Since unipotent Groups act algebraiclly on an affine variety, all its orbits are closed. Then the orbit method (Corollary 7.3 in op. cit.) implies that the group $G$ is CCR and the Kirillov-orbit map is a homeomorphism. Hence, we have affirmative answers for all my questions.

For unipotent Groups over a local field $K$ with char(K)=p. I find the above obstruction. However, if we start with a Lie algebra $\mathfrak{g}$ (in particular, a vector space) with nilpotent length less that $p$ and consider the group $G=\exp(\mathfrak{g})$ obtained by exponentiating it, then we have affirmative answers for all my questions by exactly same argyments.

Remark that $\exp: \mathfrak{g}\rightarrow G, \exp(X)=\sum_{l=0}^{3}\frac{X^l}{l!}$ with the inverse $\text{Log}: G\rightarrow \mathfrak{g}: \text{Log}(g)=\sum_{l=1}^{3} \frac{(-1)^l}{l+1}(1-g)^l$.

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    $\begingroup$ What you call the "Lie ring" of $G$ is not its Lie algebra (the Lie algebra is given by replacing $x^p$ with 0, and Lie algebras of $K$-group schemes are naturally $K$-vector spaces), so what do you mean by "Lie ring"? (Coadjoint action of $G$ on that $\widehat{\mathfrak{g}}$ doesn't make sense.) Also, to make sense of exp($\mathfrak{g}$) as a smooth unipotent $K$-group using the BCH group law in a useful way, I think you need a torus action with very tightly controlled weights, but I see no such action for this example, so be clearer about this "exp" (definition, properties). $\endgroup$ – grghxy Jul 5 '15 at 14:04
  • $\begingroup$ Dear grghxy: thanks for comments. You are right $\mathfrak{g}$ is not the Lie algebra associated to G. Its Lie algebra is always vector Space and you can find the definition of a Lie ring in Definition 4.1 in op. cit. "exp" and "log" are the formal power series and you can find their origin in "The Fourier transform for nilpotent locally compact Groups: I " by Roger Howe projecteuclid.org/euclid.pjm/1102810614. The coadjoint action of $G$ is the dual action of the conjugation action of $G$ on $\mathfrak{g}$ see page 12 in op. cit. I don't think G is smooth. $\endgroup$ – m07kl Jul 5 '15 at 15:26
  • $\begingroup$ Dear grghxy: I have added more in my post. Please try Again. $\endgroup$ – m07kl Jul 6 '15 at 11:52
  • $\begingroup$ Since $\mathfrak{g}$ is not the Lie algebra of $G$, the motivation for this questions escapes me and so I prefer not to dwell more on it; good luck. $\endgroup$ – grghxy Jul 6 '15 at 14:10

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