ALGEBRAIC K-THEORY OF POSET REPRESENTATIONS

Reinhard C. Laubenbacher

New Mexico State University Department of Mathematics Las Cruces, NM 88003

Abstract. Let S be a finite partially ordered set and k a field. This paper relates the algebraic K–theory ring of the category of k–representations of S to the Möbius ring of S.

Let S be a finite partially ordered set and k a field. There are two notions of a k-representation of S.

Definition 1. A k-representation of a poset S is a family (V, V_{s}) of finite dimensional k-vector spaces such that V_{s} ⊆ V for all s ∈ S and V_{s} ⊆ V_{t} if and only if s ≤ t in S.

Morphisms of such representations are linear transformations which preserve the subspace lattice. Denote the resulting category by R(S). It is worth remarking that R(S) is equivalent to a full subcategory of the category of finitely generated modules over a finite dimensional k-algebra, namely the k-incidence algebra of the poset S_{0}^{∞}, obtained by adjoining a least and greatest element to S [3].)

Definition 2. A k-representation of S is a functor S ⟶ Mod_{k}, where S is S viewed as a category in the usual way and Mod_{k} is the category of finite-dimensional k-vector spaces.

Morphisms are just natural transformations of functors. Denote the resulting category by A(S). Tensor product induces a product structure on both categories. Both categories are exact, with the obvious notion of exactness in each case, whereas only A(S) is abelian.

The first notion of a poset representation has been used in the study of certain classes of abelian groups (surveyed in [1]) while both notions appear in the representation theory of algebras, for instance (see e.g. [5]). The purpose of this note is to compute the algebraic K–theory of both categories in terms of the Möbius ring M[S] of S.

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The author would like to thank the referee for vastly simplifying and generalizing an earlier version of this paper. Thanks are also due to David Arnold and Keith Dennis for helpful conversations, as well as the Cornell University Mathematics Department for its hospitality while part of this work was being done.

First we need to recall some properties of the Möbius ring of a poset [6]. Let T be a poset, and μ its Möbius function. Recall that μ is defined as follows: μ(s, t) = 0, unless s ≤ t; if s ≤ t then μ(s, t) is defined recursively by the formula

∑_{s≤z≤t} μ(s, z) = δ_{s,t}.

Let M[T] be the free ℤ -module on the elements of T. For each pair of elements (s, t) in T define a function φ_{s,t} : T ⟶ ℤ by

φ_{s,t}(z) = ∑_{u≤s,t} μ(z, u).

Now we can define a product on M[T] by

st = ∑_{z∈T} φ_{s,t}(z)z,

and extend to all of M[T] by linearity. Then M[T] is a commutative ring, the Möbius ring of T.

One also observes that the elements

e_{t} = ∑_{s∈T} μ(s, t)s

are primitive idempotents in M[T], are mutually orthogonal, and

e = ∑_{t∈T} e_{t}

is the identity of M[T]. Hence, there is a ring isomorphism

M[T] ≅ ∏_{t∈T} ℤe_{t}.

Theorem 1. Let S be a finite poset and k a field. Let S′ be the poset obtained from S by adjoining a largest element. Then:

(1) There is a graded ring isomorphism K_{∗}(A(S)) ≅ K_{∗}(k) ⊗ M[S];

(2) There is a graded ring isomorphism K_{∗}(R(S)) ≅ K_{∗}(A(S′)).

Proof. For each s ∈ S there is an exact functor A(S) ⟶ Mod_{k}, sending a representation F to F (s). These functors preserve tensor product and thus assemble to an exact, product preserving functor

ϕ : A(S) ⟶ ∏_{s∈S} Mod_{k} .

There is an exact functor ψ going the other way, which sends (V_{s})_{s∈S} to the functor which takes the value V_{s} on s ∈ S and zero on morphisms. Then ϕ ∘ ψ = id.

Furthermore, id_{A(S)} and ψ ∘ ϕ are homotopy equivalent, thus ϕ and ψ induce inverse isomorphisms in K-theory. To see this we will construct a cofiltration of the identity functor by exact quotient functors

id_{A(S)} = F^{0} ↠ F^{1} ↠ ⋯ ↠ F^{n} = 0

as follows. Suppose that S has n elements s_{1}, …, s_{n}, ordered in such a way that if j ≦ i and s_{i} and s_{j} are comparable, then s_{j} ≥ s_{i}. Now define F^{1} : A(S) ⟶ A(S) by F^{1}(F)(t) = F (t) if t ≠ s_{1} and F^{1}(F)(s_{1}) = 0. Let id_{A(S)} ↠ F^{1} be the obvious projection. Then its kernel is the functor P^{1}(F)(t) = 0 if t ≠ s_{1} and P^{1}(F)(s_{1}) = F (s_{1}), so that we have a short exact sequence of exact functors

0 ⟶ P^{1} ⟶ F^{0} ⟶ F^{1} ⟶ 0 .

Inductively define F^{i}(F)(s_{j}) = 0 if j ≦ i and F^{i}(F)(s_{j}) = F (s_{j}) otherwise. Then, for each i, we have a short exact sequence of functors

0 ⟶ P^{i} ⟶ F^{i} ⟶ F^{i+1} ⟶ 0 .

It follows that id_{A(S)} and ∑ P^{i} induce the same map on K-groups [4, Cor. 3.1]. Thus, we have an isomorphism of graded rings:

K_{∗}(A(S)) ⟶ ∏_{s∈S} K_{∗}(k) .

Now observe that

∏_{s∈S} K_{∗}(k) ≅ (∏_{s∈S} ℤ) ⊗ K_{∗}(k) ≅ M[S] ⊗ K_{∗}(k)

as graded rings. This proves part (1) of the theorem.

To prove (2), consider the functor ϕ : R(S) ⟶ A(S′), which takes a representation (V, V_{s}) to the functor on S′ taking the value V on the largest element and V_{s} on the other elements s ∈ S, with the appropriate inclusions as values on the morphisms. In order to show that this (exact) functor induces an isomorphism of K-groups, observe first that it is an exact equivalence between R(S) and its image in A(S′), with an exact inverse. To complete the proof of the theorem we will construct a finite resolution of an object in A(S′) by objects in the image of ϕ, and then appeal to the Resolution Theorem [4, Theorem 4.3]. So, let F be in A(S′). For each minimal element s ∈ S′ define the functor Q_{s} in A(S′) by Q_{s}(t) = F (s) for all t ≥ s and Q_{s}(t) = 0 otherwise, with identities, resp. inclusions as morphisms. Define a natural transformation Q_{s} ⟶ F by setting Q_{s}(t) ⟶ F (t) equal to F (s) ⟶ F (t) for t ≥ s and zero otherwise. Taking sums we obtain a functor ⊕_{s min}Q_{s} ⟶ F. Now let t be a minimal element such that (⊕_{s min}Q_{s})(t) ⟶ F (t) is not onto. Choose a complement of the image, say V_{t}, and define Q_{t}(w) = V_{t} for all t ≤ w. Proceeding in this fashion and taking the direct sum over all s ∈ S, we obtain a functor P = ⊕_{s∈S}Q_{s} in the image of ϕ and a surjection P ⟶ F.

Therefore, in order to construct a resolution of F by objects in im(ϕ), first find a surjection P_{0} ⟶ F ⟶ 0, and then construct P_{1} ⟶ P_{0} via a surjection onto ker(P_{0} ⟶ F). In this way we obtain a resolution which terminates after a finite number of steps, since S is finite and the number of nonzero values of P_{i} decreases as i increases. Thus, the conditions of the Resolution Theorem are satisfied, and the proof of the theorem is complete. □

Using a variation of the isomorphisms in Theorem 1, one can give an explicit formula for the multiplication in K_{0}(R(S)). The indecomposable projectives in R(S) can be described explicitly as follows. They are all of the form {k, U(s)}, where either U(s) = 0 for all s ∈ S or there exists a t ∈ S such that U(s) = k for all s ≥ t and U(s) = 0 otherwise. (For details see [2].) They form a poset if we use inclusion as the order relation. It is then easy to see that this poset is isomorphic to the poset S_{0}^{op} obtained from S by inverting the order relation and adjoining a least element 0. In particular, if S contains n elements, then there are n + 1 indecomposable projectives, up to isomorphism. Using a construction similar to the one in the proof of the theorem, it is easy to see that every object of R(S) has a finite projective resolution, so that the indecomposable projectives furnish a generating set for K_{0}(R(S)). For s ∈ S_{0}^{op} denote the indecomposable projective corresponding to it by P_{s}.

Theorem 2. There are ring isomorphisms K_{0}(R(S)) ≅ M[S_{0}^{op}] ≅ M[S′], and the product structure in K_{0}(R(S)) is as follows: If s, t ∈ S_{0}^{op}, then

[P_{s} ⊗ P_{t}] = ∑_{u∈S_{0}^{op}} φ_{s,t}(u)[P_{u}] .

Proof. The first isomorphism is induced by sending the generator [P_{s}] of K_{0}(R(S)) to the generator s of M[S_{0}^{op}], and [P_{0}] to 0; the second isomorphism is obvious from earlier remarks. Since R(S) is a Krull-Schmidt category [2] the first map is an additive isomorphism. It remains to show that it is multiplicative, for which it is sufficient to consider the generators. First consider [P_{0}]. If [P_{s}] is another generator, then

[P_{0}] · [P_{s}] = [P_{0} ⊗ P_{s}] = [P_{0}],

which can be immediately seen from the definition of the tensor product in R(S).

If s, t ∈ S, then [P_{s} ⊗ P_{t}] = [{V, V (t)}], where V = k and V (u) = k if u ≥ s, t and 0 otherwise.

Denote the order relation in S_{0}^{op} by ⪯. Let [P_{s} ⊗ P_{t}] = ∑_{u∈S_{0}^{op}} α_{u}[P_{u}]. Then the coefficient α_{u} has the property that

∑_{v⪯u} α_{v} = 1.

We need to show that α_{u} = φ_{s,t}(u). First observe that this is true for all maximal elements u ⪯ s, t. Furthermore,

∑_{u⪯v⪯s,t} φ_{s,t}(v) = ∑_{u⪯v⪯s,t} ∑_{u⪯v⪯z⪯s,t} μ(v, z)

= ∑_{u⪯z⪯s,t} ∑_{u⪯v⪯z⪯s,t} μ(v, z)

= ∑_{u⪯z⪯t} δ_{u,z}

= δ_{u,u}

= 1.

An inductive argument now shows that

α_{u} = φ_{s,t}(u)

for all u. □

REFERENCES

[1] D. M. Arnold, Butler Groups, Representations of Finite Posets, and Modules over Multiple Pullback Rings, to appear.

[2] D. M. Arnold, C. I. Vinsonhaler, Invariants for Classes of Indecomposable Representations of Finite Posets, to appear in Trans. AMS.

[3] R. Bautista and R. Martinez, Representations of Partially Ordered Sets and 1-Gorenstein Artin Algebras, Proc. of 1978 Antwerp Ring Theory Conference, Marcel Dekker, 1979, pp. 385-433.

[4] D. Quillen, Higher Algebraic K-theory: I, in Algebraic K-theory I – Higher K-Theories, Lecture Notes in Mathematics 341, Springer Verlag, New York, 1972.

[5] C. M. Ringel, Tame Algebras and Integral Quadratic Forms, Lecture Notes in Mathematics 1099, Springer Verlag, New York, 1984.

[6] L. Solomon, The Burnside Algebra of a Finite Group, Journal of Combinatorial Theory 2 (1967) 603-615.