What is the universal property of associated graded? - MathOverflow

What is the universal property of associated graded?

2009-10-10T20:09:51Z

Given a filtered vector space (or module over a ring) 0=V₀⊆V₁⊆...⊆V, you can construct the associated graded vector space gr(V)=⊕_iV_i+1/V_i. Does gr(V) 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→V_i+1/V_i should be, but what would the map ∪V_i→⊕_iV_i+1/V_i be?

Answer by Ben Webster for What is the universal property of associated graded?

2009-10-10T20:25:18Z

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.

Answer by Anton Geraschenko for What is the universal property of associated graded?

2009-10-12T03:33:33Z

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.