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Let $F \subset K$ be a field extension of degree $n$, then $F^n \cong K$ as $F$-vectorspaces. Now $K^\times$ acts on $F^n$, by multiplication on $K$, and so $K^\times$ embeds into $GL_n(F)$, and every Galois element gives an automorphism of $K^\times$.

Question: Under which conditions can it be extended to an automorphism of $GL_n(F)$? How?

Cyclic extension, abelian extension, solvalabe extension, general extension?

I am mostly interested in the case, where $F$ is a local field.

Example $\mathbb{R} \subset \mathbb{C}$: We fix an $\mathbb{R}$-basis $\\{ 1,i \\}$. The multipliaction by $a+ib$ correspond to the matrix $$ \begin{pmatrix} a & -b \newline b & a \end{pmatrix}.$$ The Galois element is complex conjugation and corresponds to transpose on the above matrices. Can it be extended to the group $GL_2(\mathbb{C})$?

Motivation: I am actually hoping for an explanation of the Caley transform introduced here on page 2: http://www.ams.org/mathscinet/search/publdoc.html?pg1=IID&s1=162025&vfpref=html&r=28&mx-pid=237707

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    Small typo - $K^\times$ embeds into $GL_n(F)$.2012-04-24
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    I think I am misunderstanding the question. $Gal(K/F)$ respects the $F$-space structure on $K$, and so lives in $GL_n(F)$, where it acts on $K^\times$ via conjugation. Clearly this extends to all of $GL_n(F)$.2012-04-24
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    Could you include one worked example in the question so it is clearer exactly what you're looking for? Are you assuming the $K/F$ is a Galois extension? You don't say that in the question, but just mention a "Galois element", so it's not clear if you mean an $F$-automorphism of $K$, whether or not $K$ is actually Galois over $F$.2012-04-25
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    For $n=2$, when $K/F$ is quadratic (and the embedding is obvious), if I understand the question correctly, "conjugation by the longest Weyl element" extends the action of the non-trivial element of the Galois group to $GL_2(F)$.2012-04-25
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    What @SteveD said is correct. Read his comment carefully again. Then work it out in the example of $\mathbb{C}/\mathbb{R}$: transposing $\left(\begin{smallmatrix}a&b\\-b&a\end{smallmatrix}\right)$ is the same as conjugating by $\left(\begin{smallmatrix}1&0\\0&-1\end{smallmatrix}\right)$, which is precisely how Galois acts with respect to your chosen basis.2012-04-25
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    My bad, now I get it. thanks, Alex B. and Steve D.2012-04-25
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    You are welcome! As a meta remark: if you delete comments that others have responded to, then you leave other peoples' comments orphaned, which is not very nice. E.g. now, the second sentence of my previous comment seems to come out of nowhere.2012-04-25

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Maybe I am wrong, but consider rather the more canonical group $\mathrm{Aut}_F(K)$ of $F$-linear automorphisms on $K$. The Galois group naturally embedds and acts as inner automorphisms. The elements of $K^\times$ correspond to homothecies and the (inner) Galos action on the homothecies correspond to the natural action on $K$.