Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Suppose K is a local field , G is its galois group, V a fine dimensional Vector space over F, which is a sub field of K, and totally ramified over $Q_p$. Consdider the linear action of G on V (V is not just a $Z_p$ representation ), are there similar theories dealing with such situation like Fontaine's theory, someting like filtered $\varphi$ module with srong p-divisible peoperties and maybe with some extra structure? Are there any reference? Thank you!

share|improve this question
    
I have edited it to more easy case. –  TOM Sep 6 '12 at 7:09
    
You also changed the setting completely. Is $F$ an extension of $K$ or a subfield of $K$?? –  Laurent Berger Sep 6 '12 at 7:15
    
I am sorry, F should be a subfield of K. –  TOM Sep 6 '12 at 7:18
2  
If you're looking at linear representations of $G$ with coefficients, then everything works "the same". See for example 3.1 of Breuil-Mézard's 2002 Duke paper. –  Laurent Berger Sep 6 '12 at 11:50
    
It is helpful, thank you! –  TOM Sep 6 '12 at 12:08
add comment

1 Answer

If $F$ is not finite but rather equal to $C_p$ then this is really Sen's theory (see for instance Fontaine's course notes in Astérisque 295). If $F$ is merely a finite extension of $K$, then I'm not sure that you need to introduce a lot of machinery: restrict your representation to $G_F$ so that it's linear, do what you have to do, and then take in account the extra structure that you had. Alternatively, a semilinear representation is the same as an element of $H^1(G,GL_d(F))$, so you could use Galois-cohomological techniques, especially the inflation-restriction sequence with $G_F$ and $G_K$.

EDIT : this answered the question for $F$-semilinear representations of $G_K$ with $F$ an extension of $K$. Since then the OP has modified his question so my answer is not relevant anymore :(

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.