What is the universal property of associated graded?

Question

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?

Answer 1

$\DeclareMathOperator{\Rees}{Rees}\DeclareMathOperator{\gr}{gr}$ The associated graded of a filtered $\mathbf R$-module $M$ is the universal $\mathbf R$-module with a map of the Rees module of $M$ over $\mathbf R[t]$ to $\gr M$.

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

Answer 2

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 ⊕N_i to ⊕N_i/N_i-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.

Answer 3

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.

Answer 4

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 $\operatorname{coker}(0: B_{i-1} \to B_i) = \operatorname{id}_{B_i}$.