rharron
commented
11 years ago I forgot to mention the simple solution: each element has, in addition to an ordp, an M. I've brought up this possibility a couple times for non-families, but then once I understood the "magic" that was going on, I saw that it was unnecessary because you can safely conflate the p-adic precision with the filtration precision. But for families you can't.
If we do this, we could store the zero distribution in D / Fil^M as (p^(M_c_p), M, ()), i.e. valuation p^(M_c_p), known mod Fil^M, and no moments. Similarly, an element of (p^rD) / Fil^M could be stored as (p^r, M, (a_0 mod p^(Mc_p - r), ..., a_i mod p^((M-i) * cp - r), ..., a{M-1-r} mod p^((r+1) \ c_p - r)) (where I've dropped the ceiling from the notation for legibility).
rpollack9974
The way we do relative precision for overconvergent distributions relies on the basic isomorphism (p^r D) / Fil^M \cong p^r (D / Fil^{M-r}), i.e. if mu is an element of D / Fil^M that happens to be divisible by p, then we can "factor out" the p and drop a moment and consider mu to be an element of D / Fil^{M-1}, but shifted up. Conversely, if we have an element in p (D / Fil^{M-1}), we can also simply think of it as an element of (pD) / Fil^M since the new moment will just be O(p). This is used throughout our code, in particular we want to be able to do this in solving the difference equation (which introduces denominators).
But this doesn't work in families. Take, for instance, p = 5 and M = 4. If mu is in D / Fil^4, then its moments are known to precision 3,3,2,1, respectively, but if nu is in D / Fil^3, its precisions are 3,2,1. So, if mu has moments (p a0, p a1, p a2, p a3), its first moment is known up to p^3. Factoring out p and dropping the last moment yields moments (a0, a1, a2) where a0 is known up to p^2! Crap! The total measure of an element of D / Fil^3 must be known up to p^3. In particular, this means we can't store the 0 distribution in the same way. In our non-families code, we use that we can keep factoring out p and dropping a moment to store 0 as a distribution with no moments. So, the zero distribution in D / Fil^4 (for non-families) is the same as a distribution in p (D / Fil^3), or p^2 (D / Fil^2), p^3 (D / Fil^1), p^4 (D / D), and so is stored as p^4 * (). But for families, if mu = 0 mod Fil^4, then it is not known to be p times something that is zero mod Fil^3. Simply put: p * Fil^3 is not Fil^4.
I'll read carefully what Rob (P) has written in sections 8 and 9 of our paper and see if I can make sense of a good way of dealing with this. But if anyone else has comments, go right ahead!