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If you have non-Arch. local field F and E its finite extension, I am just wondering if anybody has any idea about the action of $\operatorname{Gal}(E/F)= \operatorname{Aut}_F(E)$ on the lines in $k^2_E$, Where $k_E$ is a residue field of the extension?

Note:

${ \text{Adjacent points} } \simeq \mathbb{P}^1_k$
$\mathbb{P}_{1}(k):=$ set of all dim 1 subspaces of $k\mathcal{P}^{1}$ a 2-dim k-vector space.
$\mathbb{E}\mathcal{P^{1}}=$ lines in $\mathcal{k}_{E}^{2}$.

edited Oct 12 2010 at 11:14

asked Oct 6 2010 at 16:30

Dragon
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Perhaps you should explain the notation ? – Chandan Singh Dalawat Oct 6 2010 at 16:48

am working on base change for Gl(2), and this mean if i have 2 trees, one for GL(2,F) and another for GL(2,E) where E is the extension of F. So I Should describ the base change over these trees. – Dragon Oct 6 2010 at 17:41

I tried to fix your LaTeX in a way that preserved the meaning of what you wrote, but there are some factual errors in your question. – S. Carnahan♦ Oct 7 2010 at 2:54

I don't know if this special case helps : Assume that the residue field k of F is finite with q=p^f elements, and that the degree d=[l:k] of the residue field l of E over k divides q-1. Then the normal basis theorem implies that for every character x:Gal(l|k)-->k^*, the x-eigenspace l(x) is a k-line. – Chandan Singh Dalawat Oct 7 2010 at 3:17

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The tree of $SL(2,E)$ has vertices given by homothety classes of lattices, and edges given by lattice inclusions with quotient isomorphic to $k_E$. This is not isomorphic to the $E$ points of the projective line, and it does not parametrize lines in $k^2_E$. The $E$ points of the projective line are identified with the ends of the tree, and the lines in $k^2_E$ can be identified with the adjacent points of a fixed lattice defined over $\mathcal{O}_F$. – S. Carnahan♦ Oct 7 2010 at 3:19

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S. Carnahan · Accepted Answer · 2010-10-23 11:59:05Z UTC

The action is fairly straightforward. Any line in $k_E^2$ can be described by an equation of the form $ax+by=c$, where $a,b,c \in k_E$, and $a$ and $b$ are not both zero. You can describe the action of $\operatorname{Gal}(E/F)$ on $k_E$ by taking reduction modulo the maximal ideal. Alternatively, you can look at the action on the (prime-to-residue-characteristic) roots of unity in $E$ living over units in $k_E$. You can then use the action on $k_E$ to get the diagonal action on the coefficients $a,b,c$ in the equation $ax+by=c$.

If you want to restrict your view to lines through the origin, you get an action on pairs $(a,b)$, and the equivalence relation that sends proportional pairs to the same point in $\mathbb{P}^1$ is Galois-stable.

Galois Action on the lines in the k^{2}_{E}

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