1
$\begingroup$

Suppose $K$ is an unramified extension of $\mathbb Q_p$ of degree $m$, and $\sigma$ is the $p$ power frobenius on $K$. Suppose $V$ is a $2$ dimensional admissible filtered $\phi$ module over $K$.

I want to calculate the set $$Z(\phi)=\{f:K\text{-linear endomorphism of }V\text{ such that }f\phi=\phi f\}\subset M_2(K)\subset M_{2m}(\mathbb Q_p)$$ as a subvariety. (The second containment is obtained up to choosing a basis of $K/\mathbb Q_p$.)

We know from Lemma 2.1 of LV the dimension is at most $4$. But I'm still not clear about the structure of $Z(\phi)$. (For "structure" I mean some properties like "matrices commutes with a Jordan block/ canonical normal form is a polynomial of this matrix".)

Improve this question
$\endgroup$
4
  • $\begingroup$ Have you tried working out the $m = 1$ case first? In this case you're just looking for everything that commutes with a given linear endomorphism; and there are three cases according to whether $\phi$ is (a) diagonalisable with distinct eigenvalues, (b) scalar, or (c) a single 2x2 Jordan block, all of which you can easily bash out by hand. $\endgroup$
    – David Loeffler
    Commented Nov 1, 2023 at 15:08
  • $\begingroup$ @DavidLoeffler Yes $m=1$ is trivial. When $m>1$ $\phi$ is not linear and that is the hard part. $\endgroup$
    – Richard
    Commented Nov 2, 2023 at 6:29
  • $\begingroup$ I heard you the first time – you don't need to "@" tag someone who is already part of the comment exchange, they'll already get the notification. (It's sometimes done anyway when there are lots of different people involved in the conversation but that's not the case here.) $\endgroup$
    – David Loeffler
    Commented Nov 2, 2023 at 6:30
  • $\begingroup$ Can you be a little more precise what you mean by "structure"? $\endgroup$
    – David Loeffler
    Commented Nov 2, 2023 at 6:30

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .