# What is the universal property of associated graded?

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?

• 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

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.

• 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.

• 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

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.

• The definition of Day convolution that I turned up, via ncatlab, is intimidating. Is it the same as declaring $(V \otimes W)_i = \sum_{j + k = i} V_j \otimes W_k$, and letting the map $(V \otimes W)_i$ to $(V \otimes W)_{i + 1}$ send $V_j \otimes W_k$ to $V_{j + 1} \otimes W_k + V_j \otimes W_{k + 1}$? I guess that it must be more than that, because the sums I am writing seem only to make sense when the tensor products live in some common space, which we are explicitly avoiding assuming. – LSpice Jan 3 '16 at 18:56

I always find it helpful to write down the unit and counit in order to understand an adjunction, so I'll just expand on Nicolas Schmidt's excellent answer.

From the point of view discussed by Nicolas, let us consider filtered and graded objects in an abelian category. Denote by $\operatorname{triv}$ the trivial (generalized) filtration, so that the adjunction is written $\operatorname{gr} \dashv \operatorname{triv}$. The unit of the adjunction $$\eta_A: A \to \operatorname{triv}\operatorname{gr} A$$ is given on the $i$th piece of a (generalized) filtration $a_i: A_{i-1} \to A_i$, $i \in \mathbb{Z}$ by the cokernel $\operatorname{coker}(a_i): A_i \to \operatorname{Coker}(a_i)=\operatorname{gr}_iA$. The counit $$\varepsilon_B : \operatorname{gr}\operatorname{triv}(B) \to B$$ is given in degree $i$ for a $\mathbb{Z}$-graded object $B$ by the identity $\operatorname{id}_{B_i}$. Indeed, this is because $\operatorname{coker}(0: B_{i-1} \to B_i) = \operatorname{id}_{B_i}$.