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Polynomials in ℤ[t] are categorified by considering Euler characteristics of complexes of finite-dimensional graded vector spaces. Now, given a rational function that has a power series expansion with integer coefficients, it seems natural to consider complexes of (locally finite-dimensional) graded vector spaces.

Are there nice examples of this in nature?

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Could you clarify what you mean by the Euler characteristic of a complex of finite-dimensional graded vector spaces? I just need to make sure I know where t comes from. – Qiaochu Yuan Oct 20 '09 at 19:15
@Qiaochu. The graded dimension of a graded vector space V=\oplus_i V_i is dim_t(V) = \sum_i t^i dim(V_i). The graded Euler characteristic of a complex of graded vector spaces is just the alternating sum of the graded dimensions of the terms of the complex. – Scott Morrison Oct 20 '09 at 21:06
up vote 16 down vote accepted

Yes, the particular equation you wrote is categorified by the free resolution of k as module over k[x] by the complex $k[x] \overset{x}\longrightarrow k[x]$ given by multiplication by x. It also appears in the numerical criterion for Koszulity of k[x] (see the paper of Beilinson, Ginzburg and Soergel).

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I think the following does what you want:

Let R be a polynomial ring in one variable x over a field with its usual grading; i.e. x has degree 1. Next consider the graded complex that is R in degree 0 and R[1] in degree -1 --- here by R[1] we mean
R[1]_ i=R_{i-1}, and with the non-trivial map given by multiplication by x.

I guess that you would understand the Euler characteristic of this complex to be

(1-t)(1+t+t^2+t^3+...) since you might consider the "graded dimension" of the degree 0 part to be 1+t+t^2+... and the "graded dimension" of the degree -1 part to be (t+t^2+...).

However when you take homology you get 0 in degree -1 and just k in degree 0 and this k has "graded dimension" 1.

You do get things like this in nature when you try to compute Euler characteristics of p-torsion modules for Iwasawa algebras: see for some more calculations in this context. The notion of graded Brauer character there replaces the simpler notion of graded dimension.

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Aaah. To slow. That's what comes from writing at length... – Simon Wadsley Oct 20 '09 at 20:19

I don't know whether or not this satisfies the criterion of "categorification" (what that?), but the equation (1 - t)(1 + t + t2 + ...) = 1 and its relation to vector spaces is well-known to differential topologists and geometers. We use it all the time in K-theory and index theory where it becomes the identity Λ-1V ⊗ S1V = ℂ. Here, Λ-1V denotes the alternating sum of the exterior powers of V whilst S1V is the sum of the symmetric powers.

Of course, the sum of the symmetric powers isn't a class in K-theory as it is an infinite sum. To get round this, we work in K[[t]] and allow parameters, whereupon the equation becomes Λ-tV ⊗ StV = ℂ. Here, the t means formally multiply the kth exterior or symmetric power by tk.

Where this breaks out of mere formalism and becomes very powerful is in equivariant K-theory. Then the parameter becomes a way of measuring the action of the group, which is usually S1 or a finite cyclic group for index theory calculations. In particular, in Witten's original adaptation of index theory to loop spaces, we end up with positive energy representations of S1 which become vector bundles over the original (finite dimensional) manifold with circle actions preserving the fibres. One can decompose these according to the circle action whereupon one has a vector bundle for each k ∈ ℤ. The positive energy criterion means that these are trivial below a certain integer and are always finite dimensional. However, as there are an infinite number of them then the total dimension can be infinite dimensional. Then the identity Λ-tV ⊗ StV = ℂ has real meaning as the power of the t parameter indicates how the circle acts on that component of the vector bundle. That is, if the circle action on V is the standard action then it is multiplication by tk on Λk V and on Sk V.

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To tie it up with Ben's answer, the proof of this formula is through the Koszul complex of which Ben's resolution is a simple example. – Torsten Ekedahl Feb 18 '11 at 7:25

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