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Let $K=\mathbb{F}_2((t))$ be the field of formal Laurent series over the finite field $\mathbb{F}_2$. Now consider $K^3$ as an additive group and its dual group $\hat{K^3}$, which consists of all continuous group homomorphism $\chi$ from $K^3$ into the circle Group $\mathbb{T}$, i.e., $\chi$ is a unitary character of $K^3$.

It is well-known that for a fixed unitary character $\chi$ of $K$, distinct from the unit character. The mapping $\Phi: K^3\rightarrow \hat{K^3}$, $y\mapsto \chi_y$ is a topological group isomorphism, where $\chi_y$ is the character of $K^3$ defined by $\chi_y(x)=\chi(\sum_{i=1}^3 x_iy_i)$.

Define an action of $SL(2,K)$ on $K^3$ by

$g.x=(ax_1+bx_2,cx_1+dx_2,x_3+x_1x_2+(ax_1+bx_2)(cx_1+dx_2))$,

where $g=\begin{pmatrix} a&b\\ c&d\\ \end{pmatrix}$$ \in SL(2,K)$ and $x=(x_1,x_2,x_3)\in K^3$.

Since characteristic of $K$ is 2, we know that 1=-1. Moreover, $1=det(g)=ad-bc=ad+bc$. It follows that

$g.x=(ax_1+bx_2,cx_1+dx_2,x_3+acx_1^2+bdx_2^2)$.

Since $(x+y)^2=x^2+y^2$ for all $x,y\in K$, it is easy to see this action is well-defined. In some sense, you can regard the semidirect product $K^3\rtimes SL(2,K)$ with this action as the Jacobi group in characteristic 2.

Now we consider the dual action $\alpha: SL(2,K)\curvearrowright \hat{K^3}$ by $(g.\chi)(x)=\chi(g^{-1}x)$.

My question is that what is the corresponding action of $SL(2,K)$ on $K^3$ via the isomorphism $\Phi$? More precisely, find the unique element $y'$ (depending on $g$ and $y$) in $K^3$ such that $\chi_{y'}=\alpha(g)\chi_y$.

The isomorphism $\Phi:K^3\rightarrow \hat{K^3}$ is not canonical and we have to fix $\chi$. My favorite choice is that $\chi:\mathbb{F_2}((t))\rightarrow \mathbb{T}$, $\chi(\sum a_nt^n)=\exp(\frac{2\pi i a_{-1}}{2})$.

My motivation is to compute all the orbits of the dual action $\alpha: SL(2,K)\curvearrowright \hat{K^3}$.

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    $\begingroup$ If $V$ is a finite-dimensional $K$-vector space, then $\hat{V}$ is isomorphic (not canonically) to $V$ as a topological group. View $V$ as an $R$-module, where $R$ is the ring of $K$-linear endomorphism. Then $V$ is naturally a left $R$-module, which makes $\hat{V}$ naturally a (continuous) right-module, which is the commutant in the full endomorphism group (as topological group) of the multiplication by $t$. With a choice of basis, the action is just given by transpose matrices. (...) $\endgroup$
    – YCor
    Commented Dec 4, 2015 at 14:44
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    $\begingroup$ (...) by restriction we have a right group action of $\mathrm{GL}(V)$ (group of $K$-linear isomorphism, which composed with the inversion map yields a left action. So if $H$ acts on $K^d$ through a map $j:H\to\mathrm{GL}_d(K)$, then the Pontryagin dual of $K^d$ is $H$-equivariantly isomorphic to $K^d$ with the action of $H$ given by $h\mapsto (j(h)^t)^{-1}$, if I'm correct. In particular when $H\subset\mathrm{GL}_d(K)$ is stable under transposition, then its Pontryagin dual orbit can be identified to its ordinary orbits. $\endgroup$
    – YCor
    Commented Dec 4, 2015 at 14:46
  • $\begingroup$ Dear YCor: Thanks very much for your answer. Is the identification of its Pontryagin dual orbit with its ordinary orbits topological? because I would like to know whether all orbits are locally closed. $\endgroup$
    – m07kl
    Commented Dec 4, 2015 at 16:02
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    $\begingroup$ Yes it's topological, but I already wrote it. $\endgroup$
    – YCor
    Commented Dec 4, 2015 at 16:04
  • $\begingroup$ Thanks. So we are done, because the algebraic group SL(2,K) acts algebraically on the affine varety K^3. So all orbits are locally closed in Zariski topology, hence in particualr, in locally compact topology. $\endgroup$
    – m07kl
    Commented Dec 4, 2015 at 17:04

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