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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

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

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 $N$ has 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

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

Let $k$ be a (non-archimedean) local field of positive characteristic $p$ and $\mathfrak{n}$ be any finite-dimensional nilpotent Lie algebra over $k$. 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 $N$ has 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

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 $N$ has 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