Take the 2-minute tour ×

Given a filtered vector space (or module over a ring) $0=V_{0}\subseteq V_{1}\subseteq\cdots\subseteq V$, you can construct the associated graded vector space $\mathrm{gr}\left(V\right)=\oplus_{i}V_{i+1}/V_{i}$. Does $\mathrm{gr}\left(V\right)$ satisfy a universal property? What is it?

Before anybody hastily says, "it's the universal graded vector space with a filtered map from $V$," let me point out that it's not so simple. A map of filtered vector spaces is a map of vector spaces which respects the filtration. It's clear what the map $V_{i+1}\rightarrow V_{i+1}/V_{i}$ should be, but what would the map $\cup_{i}V_{i+1}\rightarrow\oplus_{i}V_{i+1}/V_{i}$ be?

share|improve this question
Not an answer, so I'll put it as a comment: Should there even be canonical maps between V and gr V? If V is locally finite over a field, say, then any good map between V and gr V should be an iso, so if you have it in one direction, then you'd have it in the other, right? My first thought was to try to send v \mapsto \sum_i \phi_i(v), where \phi_{i+1} is the map you describe, except that \phi_i(v) is not defined if i < \deg(v), and v \mapsto \sum_{i \geq \deg(v)} \phi_i(v) is not linear. (By \deg(v) I mean the smallest i so that v\in V_i.) –  Theo Johnson-Freyd Oct 11 '09 at 0:43

3 Answers 3

up vote 11 down vote accepted

The associated graded of a filtered R-module M is the universal R-module with a map of the Rees module of M over R[t] to gr M.

Let me explain what the Rees module Rees(M) is: it's the submodule of M[t,t-1] which is generated as a R[t] module by tiM_i. Give this the obvious grading by degree of t. So Rees(M)/tRees(M)=gr M, whereas Rees(M)/(t-1)Rees(M)=M with the induced filtration. This is the thing that has a map to gr M.

share|improve this answer
I'm not convinced. It sounds like you're saying I got the arrow the wrong way in my preemptive defense. But if it had the universal property you're describing, wouldn't there be a canonical arrow from \oplus V_{i+1}/V_i to V? That seems even more unlikely. –  Anton Geraschenko Oct 10 '09 at 20:38
Well, now I have a hopefully more convincing answer. –  Ben Webster Oct 10 '09 at 21:21
Sorry, what is "a" in "Rees(M)/(t-a)Rees(M)"? –  Theo Johnson-Freyd Oct 11 '09 at 0:37
@Theo: I think "a" was meant to be 1, so that an element m gets identified with tm. –  Anton Geraschenko Oct 11 '09 at 3:17
Well, I'd had any non-zero element of the base field in mind, but maybe it's simpler to stick with 1. –  Ben Webster Oct 11 '09 at 4:03

A universal property comes from an adjunction. From this point of view, associated graded has no universal property because it is not left or right adjoint.

Proof. If gr(-) were left (right) adjoint, then it would respect cokernels (kernels). Consider the morphism of filtered vector spaces (0⊆0⊆V)→(0⊆V⊆V) (the three pieces are the 0-, 1-, and 2-filtered parts) which is just the identity map on V. It's kernel and cokernel are trivial. But the induced map gr(0⊆0⊆V)→gr(0⊆V⊆V) is the zero map from V (in degree 2) to V (in degree 1), which has non-trivial kernel and cokernel. So the associated graded of the (co)kernel is not the (co)kernel of the associated graded map.

Ben's solution is to write this poorly behaved functor as a composition of two nicer functors. The first functor is Rees:R-filmod→R[t]-grmod (from the category of filtered R-modules to the category of graded R[t]-modules). I think this functor is right adjoint to R[t]/(t-1)⊗-.

The second is R[t]/(t)⊗-:R[t]-grmod→R-grmod, the functor that takes ⊕Ni to ⊕Ni/Ni-1. R[t]/(t)⊗- is left adjoint to the functor that takes a graded R-module to the same graded module, regarded as an R[t]-module by letting t act by 0.

Upshot: associated graded is not an adjoint functor, so it doesn't have a nice universal property by itself, but it is the composition of a right adjoint functor and a left adjoint functor, which do have universal properties.

share|improve this answer
Minor corrections: the morphism is of filtered vector spaces, and the degrees of the associated graded map should be shifted up. –  S. Carnahan Oct 12 '09 at 4:33
Corrected. Thanks Scott. –  Anton Geraschenko Oct 12 '09 at 4:36

The associated graded functor has an obvious universal property if you use a sufficiently nice definition of the notion of "being filtered". A good notion of the category of filtered objects over a category $\mathcal{C}$ consists of the functor category $\text{Fun}((\mathbb{Z},\leq), \mathcal{C})$, where the poset $(\mathbb{Z},\leq)$ is viewed as a category. In other words you just give yourself the filtration pieces $V_i$ and arbitrary maps $V_i \rightarrow V_{i+1}$ instead of just monomorphisms.

This category has a tensor product via Day convolution, and the dualizable objects (if $\mathcal{C}$ is say the category of vector spaces over a field) essentially correspond to the classical filtered vector spaces, via $V = \text{colim}_i V_i$

The associated graded functor then simply is the left adjoint to the "trivial filtration functor", sending a graded vector space $(V_i)_{i \in \mathbb{Z}}$ to the filtered vector space with $V_i \rightarrow V_{i+1}$ being the zero map.

share|improve this answer

Your Answer


