Let $V$ be a $n$-dimensional $\mathbf{Q}_p$-vector space with a continuous action of $\operatorname{Gal}(\bar{L}/L)$, where $L$ is a complete discretely valued field of characteristic zero with perfect residue field of characteristic $p$.

Question: Is there one standard definition of what it means for $V$ to be ordinary, and if so, what is it?

The reason I ask is that I have seen a few different definitions that don't seem to quite coincide (and perhaps this is just the state of things).

For example, in Ralph Greenberg's Iwasawa Theory for $p$-adic Representations, he requires there to be a filtration

$$\cdots \subseteq F^{i+1}V\subseteq F^iV\subseteq\cdots$$

of $V$ by $G_L$-stable subspaces satisfying the following conditions:

(i) $F^iV=0$ for $i \gg 0$

(ii) $F^iV=V$ for $i \ll 0$

(iii) the inertia group of $G_K$ acts by $\chi_p^i$ on $F^iV/F^{i+1}V$, where $\chi_p^i$ is the $p$-adic cyclotomic character

I guess it is implicit in (iii) that any jump in the filtration gives a $1$-dimensional quotient. Greenberg proves in this paper (at least for $L=\mathbf{Q}_p$) that such a representation is Hodge-Tate, but in his proof, it seems that he is not requiring $F^iV/F^{i+1}V$ to be $\leq 1$ dimensional, because he calls this dimension $h_i$ and proves that this quotient, when tensored up to $\mathbf{C}_p$ (completion of $\bar{\mathbf{Q}}_p$) is isomorphic to $\mathbf{C}_p(i)^{h_i}$ (at least I think this is what he does).

This definition seems to me (unless I'm missing something which is entirely possible) to differ slightly from the one given in Tom Weston's Iwasawa Invariants of Galois Deformations (where he takes $L$ to be a finite extension of $\mathbf{Q}_p$). He calls $V$ nearly ordinary if there is a composition series

$$0=V^0\subsetneq V^1\subsetneq\cdots\subseteq V^n=V$$

of the $\mathbf{Q}_p[G_L]$-module. He says that if $V$ is Hodge-Tate, then for each $i$, there is an open subgroup of inertia and an integer $m_i$ such that the open subgroup acts on $V^i/V^{i-1}$ by $\chi_p^{m_i}$. He then calls $V$ ordinary if $m_1\geq m_2\geq\cdots\geq m_n$. It seems to me that if the Hodge-Tate weights (the $m_i$) are distinct, and I can take the open subgroup for each $i$ to be the entire inertia group, then Weston's definition of ordinary implies Greenberg's, but if there are the $m_i$ are not all distinct, then it doesn't seem to work. Does Greenberg's definition force the Hodge-Tate weights to all appear with multiplicity one?

Finally, I'm pretty certain I've seen a $2$-dimensional $V$ (at least when $V$ is attached to a $p$-ordinary modular form) called ordinary if it has a $1$-dimensional unramified $G_L$-quotient (if I'm not mistaken Greenberg's definition, in the $2$-dimensional case, reduces to the existence of one-dimensional $G_L$-quotient that is a Tate twist of an unramified character). This use of the term "ordinary" makes sense to me because it is satisfied by the $p$-adic representation attached to an elliptic curve over $L$ with good, ordinary reduction (and perhaps this is the origin of the term).

I apologize if there are mistakes in the above, or if I've failed to see some obvious equivalences. I'm sort of just learning some of this stuff.

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    $\begingroup$ One probably wants the definition of "ordinary" to be invariant under twisting, which rules out the definition in your last paragraph. E.g. should the $H^1_{et}$ of an ordinary elliptic curve, which is $T_p(E)(-1)$, be ordinary or not? $\endgroup$ Nov 25, 2011 at 8:23

2 Answers 2


Weston's definition is more general than Greenberg's.

  • If $V$ is 1-dimensional and corresponds to a ramified finite-order character, then $V$ is ordinary in Weston's sense, but not in Greenberg's.
  • If $V$ is "Greenberg ordinary", then I claim that [EDIT: after possibly extending the coefficients to some finite extension $E / \mathbf{Q}_p$] the filtration $F^i$ can be refined to a filtration satisfying Weston's conditions. For each $i$, $(F^i V / F^{i+1} V)(-i)$ is an unramified representation of $G_K$, so it's uniquely determined by where Frobenius goes. Since any endomorphism of a finite-dimensional space has an eigenvalue [EDIT: over some finite extension!], this allows us to split off a 1-dimensional piece. Continuing in this fashion, we can find a full flag of $G_K$-stable subspaces of $V$, with the Hodge-Tate weights of the quotients in the right order. So "Greenberg ordinary" representations are "Weston ordinary" after some finite extension of the coefficients.
  • If $V$ is Weston ordinary and semistable, then $V$ is Greenberg ordinary, because subquotients of semistable representations are semistable, and a semistable representation with all Hodge-Tate weights 0 must be unramified.
  • Conversely, if $V$ is Greenberg ordinary, then it's semistable. This is a fairly hard theorem, originally due to Perrin-Riou. It follows from the fact that for a semistable representation $W$ with all Hodge-Tate weights $\ge 1$, any extension of $W$ by the trivial rep is also semistable, i.e. $H^1_{st}(L, V) = H^1_g(L, V) = H^1(L, V)$.

So we have Greenberg ordinary $\Leftrightarrow$ semistable, and Weston ordinary [EDIT: after some finite coefficient extension]. One can check, incidentally, that a semistable representation is Greenberg-ordinary if and only if the Newton polygon and the Hodge polygon on $\mathbf{D}_{st}(V)$ coincide (recall that the Newton is always on or above the Hodge).

My impression is that the more general definition used by Weston is more standard nowadays, since it covers a lot more interesting objects. E.g. the representation attached by Hida to a p-adic ordinary modular form of non-integer weight will be Weston ordinary, but it won't be Hodge-Tate, let alone semistable.

  • $\begingroup$ Thank you very much, David. I was mistaken in my belief that the non-trivial quotients in Greenberg's filtration had to be one-dimensional. It makes a lot more sense now. $\endgroup$ Nov 25, 2011 at 14:35
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    $\begingroup$ Yes. Greenberg's filtration could even be trivial (e.g. if V is unramified). I think Greenberg does it this way because his filtration is uniquely determined if it exists, while refining it to a full flag can be done in many ways (possibly infinitely many) if there are repeated Hodge-Tate weights. $\endgroup$ Nov 25, 2011 at 14:54
  • $\begingroup$ Dear @David Loeffler, I was looking at this answer recently because it was relevant to something I was doing, and I realize I don't quite understand you're argument for why an ordinary filtration in the sense of Greenberg can be refined to a filtration with one-dimensional quotients. I understand that $(\mathrm{Fil}^iV/\mathrm{Fil}^{i+1}V)(-i)$ is unramified, and hence determined by where Frobenius goes. But why does it necessarily have an eigenvalue in the field of coefficients, i.e., why can't its characteristic polynomial have no roots in the base field? Your argument seems to be to take $\endgroup$ Jul 21, 2012 at 20:12
  • $\begingroup$ an eigenvalue of Frobenius and take the corresponding eigenspace (which is $G_K$-stable) to refine the filtration further...but I don't see how to do this if we have to pass to, say, an extension of the base field in order to have eigenvalues. $\endgroup$ Jul 21, 2012 at 20:13
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    $\begingroup$ @KeenanKidwell: You are quite right, sorry! If $A$ is the $(p-1)\times(p-1)$ companion matrix of the $p$-th cyclotomic polynomial, then there is a well-defined unramified $p$-adic representation of $G_{\mathbf{Q}_p}$ which sends arithmetic Frobenius to $A$, and it's ordinary in Greenberg's sense but not in Weston's. The correct statement is that any Greenberg-ordinary representation becomes Weston-ordinary after some finite extension of the coefficient field -- I've edited my answer to explain this. $\endgroup$ Jul 23, 2012 at 8:54

David Loeffler is correct.

In general, ordinary seems to have several interpretations, which are related but distinct.

One of them is simply: fully reducible, i.e. admits a Galois invariant filtration with one-dimensional graded pieces. This is even more general than Tom Weston's definition, and is also what some people might call nearly ordinary (but is more general than what other people might call nearly ordinary!).

One of them is Tom Weston's nearly ordinary, and another is his ordinary.

Then there is Greenberg's (but again, this is what some people might just call nearly ordinary). This is also used by Perrin-Riou and Fontaine, I think.

Then there is the third one you mention, in which one requires that the top graded piece be unramified, rather than just a cyclotomic power on inertia.

Each of these definitions arises in some way that seemed natural to the author at the time. E.g. the third, most restrictive, condition, is related to ordinary elliptic curves, as you note, and is also related to what happens in the Galois reps. of Hida families of ordinary $p$-adic modular forms.

On the other hand, a general (i.e. non-integral weight) member of a Hida family will give a two-dimensional representation which (at least for one choice of normalization) has an unramified quotient as in your third definition, but its Galois invariant subrep. will not be an integral power of the cyclotomic character. Thus, if one wants to call these ordinary, one is led to some version of my first definition, or of Weston's definition.

The nearly ordinary terminology arose (as far as I know) in Hida's work in order to get a twist-invariant notion (because when generalizing Hida theory from modular forms to more general automorphic forms, the requirement of having an unramified quotient starts to look less natural --- because it is not twist-invariant, it breaks some of the natural symmetry).

The Fontaine--Greenberg--Perrin-Riou definition combines the notion of being fully reducible with being semistable (as David Loeffler notes), which is natural if one is focussing on semistable or crystalline reps., but which again is not so natural if one is studying $p$-adic families of reps. (say attached to Hida families), since it won't be preserved under interpolation.

Speaking as someone who works in the field of $p$-adic Galois reps., I personally use ordinary in the most general sense, i.e. simply to mean fully reducible, and I don't use the term nearly ordinary at all (except to communicate with other people who do use it); on the other hand, when using ordinary with this very general meaning, I expect to sometimes have to clarify my meaning to those I'm talking to. Similarly, if anyone else uses ordinary in a talk or a paper, and then seems to begin making unwarranted deductions, I am aware that they may be using a more restrictive definition, and will ask them precisely what they mean (or look carefully in the text for the definition they are using).

My hope is that over time, the most general meaning will be the one that takes over; only time will tell if this really happens, though!

  • $\begingroup$ Thanks for elaborating on the different uses of the term, Matt...very helpful as always! $\endgroup$ Nov 25, 2011 at 14:32

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