On the Grothendieck groups of toric stacks

In this note, we prove that the Grothendieck group of a smooth complete toric Deligne-Mumford stack is torsion free. This statement holds when the generic point is stacky. We also construct an example of open toric stack with torsion in K-theory. This is a part of the author's Ph.D thesis. A similar result has been proved by Goldin, Harada, Holm, Kimura and Knutson in [GHHKK] using symplectic methods.

Let N be a free abelian group of rank d and N R = N ⊗ R. Given a complete simplicial fan Σ in N R , one chooses an integral element v i in each of the one-dimensional cones of Σ. This defines a stacky fan Σ in the sense of [BCS]. We denote the corresponding toric Deligne-Mumford stack by X Σ . Recall the Grothendieck group K 0 (X Σ ) is defined to be the free abelian group generated by all formal combinations of coherent sheaves on X Σ modding out by the short exact sequences. Each element v i corresponds to a toric invariant divisor E i . This divisor E i determines an invertible sheaf O(E i ). We denote its equivalent class in K 0 (X Σ ) by R i . The ring structure of K 0 (X Σ ) is given by tensor product of coherent sheaves. K-theory of smooth toric stacks has been studied in [BH]. In particular they computed K 0 (X Σ ) explicitly by writing out its generators and relations.

Theorem 2.1. [BH] Let B be the quotient of the Laurent polynomial ring Z[x 1 , x -1 1 , . . . , x n , x -1 n ] by the ideal generated by the relations

Then the map from B to K 0 (X Σ ) which sends x i to R i is an isomorphism of rings.

Our main result is the following.

Theorem 2.2. The Grothendieck group K 0 (X Σ ) of a complete smooth toric Deligne-Mumford stack X Σ is a free Z module.

Proof. We denote the Laurent polynomial ring Z[x 1 , x -1 1 , . . . , x n , x -1 n ] by R. Let A = R/I, where I is generated by i∈S (1 -x i ) = 0 for any set S ⊆ [1, . . . , n] such that {v i |i ∈ S} are not contained in any cone of Σ. And B = A/J, where J is generated by n Laurent polynomials

where m j is an integral basis of M.

First we want to replace h j by

. They generate the same ideal J but this collection avoids negative powers. To prove B is a free Z module we need to show that the multiplication map B → pB is an injection for any prime p. Let K(g 1 , . . . , g d ) and K(g 1 , . . . , g d , p) be the Koszul complexes for sequences g 1 , . . . , g d and g 1 , . . . , g d , p of elements of the ring A. These two Koszul complexes are related by the following lemma.

Here Cone means mapping cone of complexes.

Proof. See corollary 17.11. of [E].

According to this lemma, we get a long exact sequence of cohomology groups:

We will show that all the cohomology groups of K(g 1 , . . . , g d ) and K(g 1 , . . . , g d , p) vanish except at one position. More precisely, the only non vanishing piece of (2.1) is:

To prove this we need a result about Cohen-Macaulay properties of Stanley-Reisner rings.

Theorem 2.4.

Proof. If we make a change of variables x i to 1 -x i , then we see that A ′ is nothing but the Stanley-Reisner ring associated to supporting polytope of Σ.

It is a general fact that the Stanley-Reisner ring of polytopes are CM over any field(See Chapter 5 of [BrHe]). Furthermore one can show it is actually CM over Z(See Exercise 5.1.25 of [BrHe]). We will sketch the solution of this exercise in the following remark.

Remark 2.5. Consider the flat morphism Z → A ′ . For any maximal ideal q ⊂ A ′ , we have q ∩ Z = (p). In order to show A ′ is CM it suffices to check it for each fiber, i.e. A ′ q /pA ′ q is CM for all the maximal ideal q. If (p) is not (0) then A ′ q /pA ′ q = (A ′ ⊗ Z/pZ) q . It is CM because Stanley-Reisner ring over the field is CM. So we just need to show that for any maximal ideal q, the restriction q ∩ Z is not (0). Suppose this is the case, we will have an inclusion Z → A ′ /q. However, since we assume q ∩ Z = (0), the field A ′ /q must have characteristic zero. But this contradicts the fact that A ′ is finitely generated over Z because Q is not finitely generated over Z.

Corollary 2.6. The ring A is Cohen-Macaulay.

Proof. Because A is a localization of A ′ and being CM ring is a local property, A is CM by Theorem 2.4.

Remark 2.7. It follows from the general theory of Stanley Reisner ring (Theorem 5.1.16 of [BrHe]) that A ′ has Krull dimension d + 1.

Lemma 2.8. [E] Suppose M is a finitely generated module over ring A and

Lemma 2.9. The quotient A/J is a finitely generated abelian group.

Proof. Let k be any field and f be an arbitrary map from A/J to k. Maximal ideals of A/J are in one to one correspondence with such map f . We want to solve for (x 1 , . . . , x n ) that satisfy equations in ideal I and J in the field k. Recall elements of ideal I are in form of i∈S (1 -x i ) for any subset S ⊆ [1, . . . , n] such that one dimensional rays v i , i ∈ S are not contained in any cone of Σ. So x i equals 1 outside some cone σ. Then equations in J reduce to v i ∈σ x m,v i i =1. We can choose the dual vector m such

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