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Adams operations on the virtual K-theory of (1,n)\mathbb{P}(1,n)

Takashi Kimura Department of Mathematics and Statistics; 111 Cummington Mall, Boston University; Boston, MA 02215, USA kimura@math.bu.edu  and  Ross Sweet Department of Mathematics and Statistics; 111 Cummington Mall, Boston University; Boston, MA 02215, USA rsweet@math.bu.edu
(Date: July 27, 2025)
Abstract.

We analyze the structure of the virtual (orbifold) K-theory ring of the complex orbifold (1,n)\mathbb{P}(1,n) and its virtual Adams (or power) operations, by using the non-Abelian localization theorem of Edidin-Graham [3]. In particular, we identify the group of virtual line elements and obtain a natural presentation for the virtual K-theory ring in terms of these virtual line elements. This yields a surjective homomorphism from the virtual K-theory ring of (1,n)\mathbb{P}(1,n) to the ordinary K-theory ring of a crepant resolution of the cotangent bundle of (1,n)\mathbb{P}(1,n) which respects the Adams operations. Furthermore, there is a natural subring of the virtual K-theory ring of (1,n)\mathbb{P}(1,n) which is isomorphic to the ordinary K-theory ring of the resolution. This generalizes the results of Edidin-Jarvis-Kimura [6] who proved the latter for n=2,3n=2,3.

1. Introduction

We begin by quickly recalling a few facts about the virtual K-theory ring and virtual Adams (or power) operations of a complex orbifold. We will focus upon the case of primary interest to us, (1,n)\mathbb{P}(1,n), for simplicity of exposition, keeping in mind that these notions can be defined quite generally, e.g. for toric orbifolds. We refer the interested reader to [5, 6] for the general case.

Let (1,n)\mathbb{P}(1,n) be the weighted projective line regarded as the toric orbifold [X/G][X/G] where the smooth manifold X=2{(0,0)}X=\mathbb{C}^{2}-\{(0,0)\} has the action of GG, the complex torus \mathbb{C}^{*}, given by the map G×XXG\times X\to X taking (α,(z1,z2))(αz1,αnz2)(\alpha,(z_{1},z_{2}))\mapsto(\alpha z_{1},\alpha^{n}z_{2}). The associated inertia orbifold I(1,n)I\mathbb{P}(1,n) is the quotient orbifold [(IGX)/G][(I_{G}X)/G] where the inertia manifold IGXG×XI_{G}X\subset G\times X, consisting of all points (m,x)(m,x) such that mx=xmx=x, is a GG-manifold with the induced action G×IGXIGXG\times I_{G}X\to I_{G}X given by (h,(g,x))(hgh1,hx)(h,(g,x))\mapsto(hgh^{-1},hx). Notice that I(1,n)I\mathbb{P}(1,n) contains (1,n)\mathbb{P}(1,n) via the the GG-equivariant embedding XIGXX\to I_{G}X taking x(1,x)x\mapsto(1,x).

Let K(I(1,n))K(I\mathbb{P}(1,n)) be the Grothendieck group of complex vector bundles on the orbifold I(1,n)I\mathbb{P}(1,n) or, equivalently, K(I(1,n))K(I\mathbb{P}(1,n)) is KG(IGX)K_{G}(I_{G}X), the Grothendieck group of GG-equivariant vector bundles on IGXI_{G}X. There are many so-called inertial products on K(I(1,n))K(I\mathbb{P}(1,n)) in the sense of [5] which make K(I(1,n))K(I\mathbb{P}(1,n)) into a commutative, unital ring containing the ordinary K-theory ring K((1,n))K(\mathbb{P}(1,n)), called the untwisted sector, as a subring. Recall that the motivating example of an inertial product is the orbifold product which, in K-theory, is due to Adem-Ruan-Zhang [1] and Jarvis-Kaufmann-Kimura [9] (and, in terms of a Lie group presentation, Edidin-Jarvis-Kimura [4]). This orbifold product is a K-theoretic version of an orbifold product originally introduced in cohomology theory by Chen-Ruan [2] as the genus 0, 33-pointed, degree 0 stable maps contribution to (orbifold) Gromov-Witten theory.

The virtual (orbifold) product is an inertial product introduced by Gonzalez-Lupercio-Segovia-Uribe-Xicotencatl [8]. Furthermore, in [6], it was shown that the virtual K-theory ring of any toric orbifold admits virtual Adams (or power) operations, a collection of ring homomorphisms ψ~k:K(I(1,n))K(I(1,n))\widetilde{\psi}^{k}:K(I\mathbb{P}(1,n))\to K(I\mathbb{P}(1,n)) which satisfy ψ1=id\psi^{1}=id, and the conditions ψ~kψ~=ψ~k\widetilde{\psi}^{k}\circ\widetilde{\psi}^{\ell}=\widetilde{\psi}^{k\ell} for all k,1k,\ell\geq 1, yielding the structure of a so-called ψ\psi-ring. In fact, the ψ\psi-ring structure on K(I(1,n))K(I\mathbb{P}(1,n)) induces a λ\lambda-ring structure on K(I(1,n))K(I\mathbb{P}(1,n))_{\mathbb{Q}}. In addition, on the untwisted sector K((1,n))K(\mathbb{P}(1,n)), these virtual Adams operations reduce to the ordinary Adams operations which satisfy

ψ~k()=k\widetilde{\psi}^{k}({\mathscr{L}})={\mathscr{L}}^{k} (1)

for all k1k\geq 1 when {\mathscr{L}} is the class of any line bundle.

The previous equation motivates the following definition: An element {\mathscr{L}} in the virtual ψ\psi-ring K(I(1,n))K(I\mathbb{P}(1,n))_{\mathbb{C}} is said to be a virtual line element if it is invertible and satisfies Equation (1) for all k1k\geq 1. The multiplicative group of virtual line elements 𝒫K(I(1,n))\mathcal{P}\subset K(I\mathbb{P}(1,n))_{\mathbb{C}} contains, in particular, classes of ordinary line bundles on (1,n)\mathbb{P}(1,n). Virtual line elements share many of the same properties as ordinary line bundles in ordinary equivariant K-theory. For example, they possess a virtual version of an Euler class constructed from the virtual λ\lambda-ring structure and a virtual version of dualization (see [6] for details).

In this paper, we apply non-Abelian localization [3] to K(I(1,n))K(I\mathbb{P}(1,n))_{\mathbb{C}} and express the virtual product and virtual Adams operations in terms of natural generators on the localization. We use these generators to calculate the group of virtual line elements 𝒫\mathcal{P}, show that 𝒫\mathcal{P} spans the vector space K(I(1,n))K(I\mathbb{P}(1,n))_{\mathbb{C}}, and then give a simple presentation of the virtual K-theory ring K(I(1,n))K(I\mathbb{P}(1,n))_{\mathbb{C}} in terms of these virtual line elements. Finally, we show that for a particular crepant resolution, ZnZ_{n}, of the total space of the cotangent bundle of (1,n)\mathbb{P}(1,n), there is a surjective homomorphism from the virtual K-theory ring K(I(1,n))K(I\mathbb{P}(1,n))_{\mathbb{C}} to ordinary K-theory ring K(Zn)K(Z_{n})_{\mathbb{C}} which respects the ψ\psi-ring structures in the spirit of the hyper-Kähler resolution conjecture. Indeed, there is a summand of the virtual K-theory ring K(I(1,n))K(I\mathbb{P}(1,n))_{\mathbb{C}}, isomorphic to the completion with respect to the augmentation ideal, which is isomorphic to K(Zn)K(Z_{n})_{\mathbb{C}} as ψ\psi-rings. These results generalize those of [6] where a different presentation of the virtual K-theory ring and the same results for the crepant resolution for n=2,3n=2,3 were obtained without using localization.

Finally, it is worth pointing out that these techniques should apply to general toric orbifolds. It would be very interesting to study these questions in greater generality.

Acknowledgments

We would like to thank Dan Edidin, Tyler Jarvis, and Tomoo Matsumura for useful discussions.

2. Review of inertial and virtual K-theory

This section provides a brief introduction to theory of inertial pairs on smooth Deligne-Mumford stacks developed in [5, 6]. An inertial pair defines a multiplication on the K-theory and Chow ring of the inertia stack. These rings have compatible power operations for a special case of inertial pairs, which will be of interest here. Let XX be a scheme or algebraic space (or smooth manifold) and GG a linear algebraic group acting properly on XX. Then the quotient [X/G][X/G] is a smooth Deligne-Mumford stack (or complex orbifold).

Definition 2.1.

Define the inertia space of XX to be

IGX={(g,x)|gx=x}G×X.I_{G}X=\{(g,x)\,|\,gx=x\}\subset G\times X.

Let Xm={(m,x)|mx=x}IGXX^{m}=\{(m,x)\,|\,mx=x\}\subseteq I_{G}X. In particular, X1=XX^{1}=X. The inertia space has an action of the group GG via h(g,x)=(hgh1,x)h\cdot(g,x)=(hgh^{-1},x). We may then define the double inertia space by

𝕀G2X={(g1,g2,x)|gix=x,i=1,2}G2×X.\mathbb{I}^{2}_{G}X=\{(g_{1},g_{2},x)\,|\,g_{i}x=x,\,i=1,2\}\subset G^{2}\times X.
Definition 2.2.

Let μ:𝕀G2XIGX\mu:\mathbb{I}^{2}_{G}X\rightarrow I_{G}X be the composition map μ(g1,g2,x)=(g1g2,x)\mu(g_{1},g_{2},x)=(g_{1}g_{2},x), and let ei:𝕀G2XIGXe_{i}:\mathbb{I}^{2}_{G}X\rightarrow I_{G}X for i=1,2i=1,2 be the evaluation maps ei(g1,g2,x)=(gi,x)e_{i}(g_{1},g_{2},x)=(g_{i},x).

An inertial product will be defined on KG(IGX)K_{G}(I_{G}X) and AG(IGX)A^{*}_{G}(I_{G}X) associated to an inertial pair (,𝒮)(\mathscr{R},\mathscr{S}) [5], where \mathscr{R} is a GG-equivariant vector bundle on 𝕀G2X\mathbb{I}^{2}_{G}X and 𝒮\mathscr{S} is a non-negative class in KG(IGX)K_{G}(I_{G}X)_{\mathbb{Q}}, as we will briefly review below. In a special case of inertial pairs associated to vector bundles, we will obtain the virtual orbifold product, which will be the focus of the remaining sections of this paper.

Definition 2.3.

For x,yKG(IGX)x,y\in K_{G}(I_{G}X), define the inertial product on KG(IGX)K_{G}(I_{G}X) by

xy=μ(e1(x)e2(y)eu()),x*_{\mathscr{R}}y=\mu_{*}(e_{1}^{*}(x)\cdot e_{2}^{*}(y)\cdot\operatorname{eu}(\mathscr{R})), (2)

where eu()\operatorname{eu}(\mathscr{R}) is the Euler class of \mathscr{R}. Recall that in K-theory, the Euler class is λ1()\lambda_{-1}(\mathscr{R}^{*}) and in Chow or cohomology is the top Chern class. The same formula gives the inertial product on AG(IGX)A^{*}_{G}(I_{G}X), and on HG(IGX)H_{G}^{*}(I_{G}X).

The compatibility condition on 𝒮\mathscr{S} giving an inertial pair is the identity

=e1(𝒮)+e2(𝒮)μ(𝒮)+Tμ,\mathscr{R}=e_{1}^{*}(\mathscr{S})+e_{2}^{*}(\mathscr{S})-\mu^{*}(\mathscr{S})+T_{\mu},

where TμT_{\mu} is the relative tangent bundle. We refer the reader to [5] §3 for additional details concerning the definition of an inertial pair.

Proposition 2.4 ([4],§3).

If (,𝒮)(\mathscr{R},\mathscr{S}) is an inertial pair, then the associated inertial product *_{\mathscr{R}} on KG(IGX)K_{G}(I_{G}X) is commutative and associative with identity 𝟏XKG(X)\mathbf{1}_{X}\in K_{G}(X). Further, KG(X)KG(IGX)K_{G}(X)\subseteq K_{G}(I_{G}X) is a subring where the inertial product reduces to the ordinary product on equivariant K-theory.

The same result holds for Chow and cohomology. Furthermore, there is an inertial Chern character map that gives a homomorphism between the inertial K-theory and inertial Chow rings.

Proposition 2.5 ([5],Prop 3.8).

If (,𝒮)(\mathscr{R},\mathscr{S}) is an inertial pair, then the inertial Chern character

𝒞h~:KG(IGX)AG(IGX),\widetilde{\mathscr{C}\kern-2.5pt{h}}:K_{G}(I_{G}X)_{\mathbb{Q}}\longrightarrow A^{*}_{G}(I_{G}X)_{\mathbb{Q}},

given by

𝒞h~(V)=Ch(V)Td(𝒮),\widetilde{\mathscr{C}\kern-2.5pt{h}}(V)=\mathrm{Ch}(V)\cdot\operatorname{Td}(-\mathscr{S}),

where Ch(V)\mathrm{Ch}(V) is the ordinary Chern character of VV, Td(V)\operatorname{Td}(V) is the Todd class, and \cdot is the ordinary product, is a homomorphism of rings with respect to the inertial products on KG(IGX)K_{G}(I_{G}X) and AG(IGX)A^{*}_{G}(I_{G}X) (or HG(IGX)H_{G}^{*}(I_{G}X)).

The class 𝒮\mathscr{S} is also used to define a new grading on the inertial Chow and K-theory.

Definition 2.6.

The 𝒮\mathscr{S}-age of a connected component [U/G][U/G] of IGXI_{G}X is defined to be age𝒮(U)=rk(𝒮|U).\operatorname{age}_{\mathscr{S}}(U)=\mathrm{rk}(\mathscr{S}|_{U}). For an element KG(IGX)\mathscr{F}\in K_{G}(I_{G}X) supported on UU, its 𝒮\mathscr{S}-degree is

deg𝒮=age𝒮(U)mod,\deg_{\mathscr{S}}\mathscr{F}=\operatorname{age}_{\mathscr{S}}(U)\mod\mathbb{Z},

giving a /\mathbb{Q}/\mathbb{Z}-grading on KG(IGX)K_{G}(I_{G}X). The 𝒮\mathscr{S}-grading of an element xAG(IGX)x\in A^{*}_{G}(I_{G}X) is the rational number

deg𝒮x|U=degx|U+age𝒮(U),\deg_{\mathscr{S}}x|_{U}=\deg x|_{U}+\operatorname{age}_{\mathscr{S}}(U),

which gives a \mathbb{Q}-grading on Chow. For a class in cohomology, xHG(IGX)x\in H_{G}^{*}(I_{G}X), the 𝒮\mathscr{S}-grading is the rational number

deg𝒮x|U=degx|U+2age𝒮(U).\deg_{\mathscr{S}}x|_{U}=\deg x|_{U}+2\operatorname{age}_{\mathscr{S}}(U).

Denote by AG{q}(IGX)A^{\{q\}}_{G}(I_{G}X) the subspace of AG(IGX)A^{*}_{G}(I_{G}X) of elements of 𝒮\mathscr{S}-degree qlq\in\mathbb{Q}^{l}, where ll is the number of of connected components of I𝒳I\mathscr{X}. The restriction of the inertial Chern character to AG{0}(IGX)A^{\{0\}}_{G}(I_{G}X), denoted 𝒞h~0\widetilde{\mathscr{C}\kern-2.5pt{h}}^{0}, is the inertial rank for 𝒮\mathscr{S}.

Proposition 3.11 in [5] states that the inertial Chern character homomorphism preserves the 𝒮\mathscr{S}-degree modulo \mathbb{Z}.

With respect to the usual product on K-theory, have the following definition.

Definition 2.7.

Let YY be a manifold, and GG a closed algebraic group. The augmentation homomorphism on equivariant K-theory ϵ:KG(Y)KG(Y)\epsilon:K_{G}(Y)\rightarrow K_{G}(Y) is given by

ϵ(|U)=Ch0(|U)𝒪U\epsilon(\mathscr{F}|_{U})=\mathrm{Ch}^{0}(\mathscr{F}|_{U})\mathscr{O}_{U}

where [U/G][U/G] is a connected component of [Y/G][Y/G]. If \mathscr{F} is a vector bundle on UU, then Ch0(|U)\mathrm{Ch}^{0}(\mathscr{F}|_{U}) is the rank of \mathscr{F} on UU. The kernel of ϵ\epsilon, 𝔞Y\mathfrak{a}_{Y}, is called the augmentation ideal.

Definition 2.8.

The completion of the ring KG(Y)K_{G}(Y) with respect to the augmentation ideal is called the augmentation completion, K^G(Y)\widehat{K}_{G}(Y)_{\mathbb{Q}}.

Definition 2.9.

An inertial augmentation homomorphism is ϵ~:KG(IGX)KG(IGX)\widetilde{\epsilon}:K_{G}(I_{G}X)\rightarrow K_{G}(I_{G}X), given by

ϵ~(|U)=𝒞h~0(|U)𝒪U,\widetilde{\epsilon}(\mathscr{F}|_{U})=\widetilde{\mathscr{C}\kern-2.5pt{h}}^{0}(\mathscr{F}|_{U})\mathscr{O}_{U},

where UU is as above. We call (KG(IGX),,1,ϵ~)(K_{G}(I_{G}X),*,1,\widetilde{\epsilon}) the augmented inertial ring.

Remark 2.10.

By the definition of 𝒞h~0\widetilde{\mathscr{C}\kern-2.5pt{h}}^{0}, we have that

ϵ~(|U)={ϵ(|U),ifage𝒮(U)=00,otherwise.\widetilde{\epsilon}(\mathscr{F}|_{U})=\begin{cases}{\epsilon}(\mathscr{F}|_{U}),\hskip 7.22743pt\textup{if}\,\,\operatorname{age}_{\mathscr{S}}(U)=0\\ 0,\hskip 7.22743pt\textup{otherwise}\end{cases}.

The augmented inertial ring above has compatible Adams operations, as shown in [6].

Definition 2.11.

Let RR be a commutative, unital ring together with a collection of ring homomorphisms ψn:RR\psi^{n}:R\rightarrow R for all n1n\geq 1. RR is called a ψ\psi-ring if for all xRx\in R and for all n1n\geq 1,

  1. (1)

    ψ1(x)=x\psi^{1}(x)=x

  2. (2)

    ψn(ψm(x))=ψnm(x)\psi^{n}(\psi^{m}(x))=\psi^{nm}(x)

The homomorphisms ψn\psi^{n} are called Adams (or power) operations.

We will also need to consider ψ\psi-rings with an augmentation.

Definition 2.12.

If (R,,1,ϵ)(R,\cdot,1,\epsilon) is an augmented ring, then (R,,1,ψ,ϵ)(R,\cdot,1,\psi,\epsilon) is an augmented ψ\psi-ring if

ϵ(ψk())=ψk(ϵ())=ϵ(),\epsilon(\psi^{k}(\mathscr{F}))=\psi^{k}(\epsilon(\mathscr{F}))=\epsilon(\mathscr{F}), (3)

for all k1k\geq 1. Set ψ0=ϵ\psi^{0}=\epsilon.

Recall that ordinary equivariant K-theory KG(X)K_{G}(X) is a ψ\psi-ring whose Adams operations ψk:KG(X)KG(X)\psi^{k}:K_{G}(X)\to K_{G}(X) are defined though the λ\lambda-ring structure of KG(X)K_{G}(X). That is, for all i0i\geq 0, define maps λi:KG(X)KG(X)\lambda^{i}:K_{G}(X)\to K_{G}(X) by demanding that for any GG-equivariant vector bundle VV, λi([V]):=[ΛiV]\lambda^{i}([V]):=[\Lambda^{i}V], the class of the ii-th exterior power of VV, and also by demanding that the series λt:=i0tiλi:KG(X)KG(X)[[t]]\lambda_{t}:=\sum_{i\geq 0}t^{i}\lambda^{i}:K_{G}(X)\to K_{G}(X)[[t]] satisfy the multiplicativity relation λt(1+2)=λt(1)λt(2)\lambda_{t}({\mathscr{F}}_{1}+{\mathscr{F}}_{2})=\lambda_{t}({\mathscr{F}}_{1})\lambda_{t}({\mathscr{F}}_{2}) for all 1,2{\mathscr{F}}_{1},{\mathscr{F}}_{2} in KG(X)K_{G}(X). These λ\lambda-operations make KG(X)K_{G}(X) into a so-called λ\lambda-ring. The Adams operations are defined through the equality of power series

λt()=exp(n1(1)n1tnnψn()).\lambda_{t}({\mathscr{F}})=\exp\left(\sum_{n\geq 1}(-1)^{n-1}\frac{t^{n}}{n}\psi^{n}({\mathscr{F}})\right). (4)

In the special case where \mathscr{L} is the class of a line bundle, it follows that

ψk()=k.\psi^{k}(\mathscr{L})=\mathscr{L}^{\otimes k}. (5)

If an element {\mathscr{F}} in KG(X)K_{G}(X) is the class of a rank nn vector bundle then λt()\lambda_{t}({\mathscr{F}}) is a degree nn polynomial in tt and λn()\lambda^{n}({\mathscr{F}}) is an invertible element of KG(X)K_{G}(X). An element {\mathscr{F}} in KG(X)K_{G}(X) which satisfies these properties is said to be a λ\lambda-positive element of rank nn. Although a rank nn λ\lambda-positive element {\mathscr{F}} need not be the class of a rank nn vector bundle, in general, they do share many of the properties of vector bundles, e.g. they possess an Euler class in K-theory, Chow and cohomology.

We will now define the inertial Adams operations. We begin by introducing Bott classes.

Definition 2.13.

Let YY be a manifold, and let \mathscr{L} be an equivariant line bundle in KG(Y)K_{G}(Y). For each j1j\geq 1, the j-th Bott class of \mathscr{L} is

θj()=1j1=i=0j1i.\theta^{j}(\mathscr{L})=\frac{1-\mathscr{L}^{j}}{1-\mathscr{L}}=\sum_{i=0}^{j-1}\mathscr{L}^{i}. (6)

The splitting principle is used to extend this definition to any equivariant vector bundle in KG(Y)K_{G}(Y). The result is a multiplicative class, where for any vector bundles 1\mathscr{F}_{1} and 2\mathscr{F}_{2},

θj(1+2)=θj(1)θj(2).\theta^{j}(\mathscr{F}_{1}+\mathscr{F}_{2})=\theta^{j}(\mathscr{F}_{1})\theta^{j}(\mathscr{F}_{2}).
Definition 2.14.

If 𝒮\mathscr{S} can be represented by a vector bundle, then the inertial pair is called strongly Gorenstein.

Definition 2.15.

Suppose the inertial pair (,𝒮)(\mathscr{R},\mathscr{S}) is strongly Gorenstein. Then for all k1k\geq 1, the inertial Adams operations ψ~k:KG(IGX)KG(IGX)\widetilde{\psi}^{k}:K_{G}(I_{G}X)\rightarrow K_{G}(I_{G}X) are given by

ψ~k()=ψk()θk(𝒮),\widetilde{\psi}^{k}(\mathscr{F})=\psi^{k}(\mathscr{F})\cdot\theta^{k}(\mathscr{S}^{*}), (7)

where \cdot is the usual product on KG(IGX)K_{G}(I_{G}X).

We are motivated by equation (5) to introduce the following analog of classes of line bundles in ordinary equivariant K-theory.

Definition 2.16.

An invertible element \mathscr{L} in the inertial K-theory ring KG(IGX)K_{G}(I_{G}X) satisfying equation (1),

ψ~k()=k,\widetilde{\psi}^{k}(\mathscr{L})=\mathscr{L}^{*k},

for all k1k\geq 1 is called an inertial line element.

Notice that after tensoring with \mathbb{Q}, an inertial line element in KG(IGX)K_{G}(I_{G}X)_{\mathbb{Q}} is nothing more than a λ\lambda-positive element of rank 11 with respect to the inertial λ\lambda-ring structure λ~t:KG(IGX)KG(IGX)\widetilde{\lambda}_{t}:K_{G}(I_{G}X)_{\mathbb{Q}}\to K_{G}(I_{G}X)_{\mathbb{Q}}, where λ~t:=i0tiλ~i\widetilde{\lambda}_{t}:=\sum_{i\geq 0}t^{i}\widetilde{\lambda}^{i} is defined from the inertial Adams operations ψ~k\widetilde{\psi}^{k} through Equation (4) but where ψn\psi^{n} is replaced by ψ~n\widetilde{\psi}^{n}, λt\lambda_{t} is replaced by λ~t\widetilde{\lambda}_{t}, and all products are understood to be the inertial product.

Theorem 2.17 ([6],Theorem 5.16).

Let GG be a diagonalizable group and let (,𝒮)(\mathscr{R},\mathscr{S}) be a strongly Gorenstein inertial pair on IGXI_{G}X. The ring (KG(IGX),,𝟏,ϵ~,ψ~)(K_{G}(I_{G}X),*_{\mathscr{R}},\mathbf{1},\widetilde{\epsilon},\widetilde{\psi}) is an augmented ψ\psi-ring.

The virtual orbifold product was defined in [8], and in [5] was shown to arise from a particular inertial pair.

Definition 2.18.

The tangent bundle of 𝒳\mathscr{X} gives an equivariant bundle 𝕋KG(X)\mathbb{T}\in K_{G}(X) on XX. Denote by 𝕋|𝕀G2X\mathbb{T}|_{\mathbb{I}^{2}_{G}X} the pullback of 𝕋\mathbb{T} to 𝕀G2X\mathbb{I}^{2}_{G}X via the projection map from 𝕀G2X\mathbb{I}^{2}_{G}X to XX. Further, let 𝕋IGX\mathbb{T}_{I_{G}X} be the tangent bundle of I𝒳I\mathscr{X} and 𝕋𝕀G2X\mathbb{T}_{\mathbb{I}^{2}_{G}X} be the tangent bundle of I2𝒳I^{2}\mathscr{X}. The virtual obstruction class =𝕋virtKG(𝕀G2X)\mathscr{R}=\mathbb{T}^{virt}\in K_{G}(\mathbb{I}^{2}_{G}X) is defined by

𝕋|𝕀G2X+𝕋𝕀G2Xe1𝕋IGXe2𝕋IGX.\mathbb{T}|_{\mathbb{I}^{2}_{G}X}+\mathbb{T}_{\mathbb{I}^{2}_{G}X}-e_{1}^{*}\mathbb{T}_{I_{G}X}-e_{2}^{*}\mathbb{T}_{I_{G}X}. (8)
Definition 2.19.

Let 𝐍\mathbf{N} be the normal bundle of the projection IGXXI_{G}X\rightarrow X

Proposition 2.20 ([5],Proposition 4.3.2).

The pair (𝕋virt,𝐍)(\mathbb{T}^{virt},\mathbf{N}) is a strongly Gorenstein inertial pair.

Corollary 2.21 ([5],Corollary 4.3.4).

The virtual orbifold Chow ring from [8] is isomorphic as a ring to the inertial Chow ring associated to the inertial pair (𝕋virt,𝐍)(\mathbb{T}^{virt},\mathbf{N}).

Definition 2.22.

For the virtual K-theory above, we will call ψ~k\widetilde{\psi}^{k} virtual Adams operations and ϵ~\widetilde{\epsilon} the virtual augmentation.

In [6], a variant of the hyper-Kähler resolution conjecture was introduced for inertial K-theory. Let 𝒳=[X/G]\mathscr{X}=[X/G] be an orbifold where GG is diagonalizable, and let ZZ be a hyper-Kähler resolution of 𝕋𝒳\mathbb{T}^{*}\mathscr{X}. There is an isomorphism between K^(I𝒳)\widehat{K}(I\mathscr{X})_{\mathbb{C}} with the virtual orbifold product and K(Z)K(Z)_{\mathbb{C}} with the usual product.

3. Virtual K-theory of (1,n)\mathbb{P}(1,n)

The statements and results in this section are a synopsis of the results from §7 of [6]. Let X=2{(0,0)}X=\mathbb{C}^{2}-\{(0,0)\} have an action of the group \mathbb{C}^{*} given by λ(z1,z2)=(λz1,λnz2)\lambda\cdot(z_{1},z_{2})=(\lambda z_{1},\lambda^{n}z_{2}), for a positive integer n2n\geq 2. Then the quotient space [X/][X/\mathbb{C}^{*}] is the global quotient orbifold (1,n)\mathbb{P}(1,n). As a vector space, the virtual orbifold K-theory is given by the equivariant K-theory of the inertia manifold IGX=m=0n1XmI_{G}X=\coprod_{m=0}^{n-1}X^{m}, the disjoint union of the fixed point set of XX under the action of an element of the group /n={0,,n1}\mathbb{Z}/n\mathbb{Z}=\{0,\ldots,n-1\} using additive notation. We view /n\mathbb{Z}/n\mathbb{Z} as a subgroup of \mathbb{C}^{*} under the identification mζmm\mapsto\zeta^{m}, for ζ=exp(2πi/n)\zeta=exp(2\pi i/n). With respect to the usual product, when m=1m=1, this gives the ring

KG(X0)[x0±1](x01)(x01)n,K_{G}(X^{0})\cong\frac{\mathbb{Z}[x_{0}^{\pm 1}]}{\langle(x_{0}-1)(x_{0}-1)^{n}\rangle},

and when m{1,,n1}m\in\{1,\ldots,n-1\},

KG(Xm)[xm±1]xmn1.K_{G}(X^{m})\cong\frac{\mathbb{Z}[x_{m}^{\pm 1}]}{\langle x_{m}^{n}-1\rangle}.

Denote the identities in each ring above by 1m1_{m} where m{0,,n1}m\in\{0,\ldots,n-1\}. Equation (8) gives the formula for \mathscr{R} and 𝒮m:=𝒮|Xm=xm\mathscr{S}_{m}:=\mathscr{S}|_{X^{m}}=x_{m}, so the virtual multiplication in this case can be written as

xm1a1xm2a2=xm1+m2a1+a2eu(𝒮m1+𝒮m2𝒮m1+m2),x_{m_{1}}^{a_{1}}*x_{m_{2}}^{a_{2}}=x_{m_{1}+m_{2}}^{a_{1}+a_{2}}\cdot\operatorname{eu}(\mathscr{S}_{m_{1}}+\mathscr{S}_{m_{2}}-\mathscr{S}_{m_{1}+m_{2}}), (9)

where the Euler class is

eu(𝒮m1+𝒮m2𝒮m1+m2)={1if either m1=0 or m2=0,1m1+m22xm1+m21+xm1+m22if m1+m2=nm10 and m20,1m1+m2xm1+m21otherwise.\operatorname{eu}(\mathscr{S}_{m_{1}}+\mathscr{S}_{m_{2}}-\mathscr{S}_{m_{1}+m_{2}})=\begin{cases}1&\mbox{if either $m_{1}=0$ or $m_{2}=0$,}\\ 1_{m_{1}+m_{2}}-2x_{m_{1}+m_{2}}^{-1}+x_{m_{1}+m_{2}}^{-2}&\mbox{if $m_{1}+m_{2}=n$, $m_{1}\not=0$ and $m_{2}\not=0$,}\\ 1_{m_{1}+m_{2}}-x_{m_{1}+m_{2}}^{-1}&\mbox{otherwise}.\end{cases}

The virtual augmentation is given by ϵ~(x0a)=10\widetilde{\epsilon}(x_{0}^{a})=1_{0} and ϵ~(xma)=0\widetilde{\epsilon}(x_{m}^{a})=0 for m{1,,n1}m\in\{1,\ldots,n-1\}. For the definition of the virtual Adams operations, see equation (7). Here, the Bott classes are given by

θj(𝒮m)=θj(xm1)=i=0j1xmi.\theta^{j}(\mathscr{S}_{m}^{*})=\theta^{j}(x_{m}^{-1})=\sum_{i=0}^{j-1}x_{m}^{-i}.

In Proposition 7.22 of [6], Edidin, Jarvis, and Kimura prove that a subring of the inertial K-theory of (1,2)\mathbb{P}(1,2) and (1,3)\mathbb{P}(1,3) is isomorphic to the ordinary K-theory of a toric crepant resolution of the cotangent bundle of (1,2)\mathbb{P}(1,2) and (1,3)\mathbb{P}(1,3), respectively. The cotangent bundle to (1,n)\mathbb{P}(1,n) is the quotient stack [(X×𝔸1)/][(X\times\mathbb{A}^{1})/\mathbb{C}^{*}] where the action of \mathbb{C}^{*} has weights (1,n,(n+1))(1,n,-(n+1)). There is a simplicial fan associated to this quotient, and a subdivision of this fan determines the toric crepant resolution of the cotangent bundle. See [6] §7.3 for more details. This resolution has ordinary K-theory

K(Zn)=[χ0±1,,χn1±1]eu(χ0),,eu(χn1)2=[χ0±1,,χn1±1](χ01),,(χn11)2.K(Z_{n})=\frac{\mathbb{Z}[\chi_{0}^{\pm 1},\ldots,\chi_{n-1}^{\pm 1}]}{\langle\operatorname{eu}(\chi_{0}),\ldots,\operatorname{eu}(\chi_{n-1})\rangle^{2}}=\frac{\mathbb{Z}[\chi_{0}^{\pm 1},\ldots,\chi_{n-1}^{\pm 1}]}{\langle(\chi_{0}-1),\ldots,(\chi_{n-1}-1)\rangle^{2}}. (10)

4. Localization

Henceforth, we will assume complex coefficients unless otherwise stated. Edidin and Graham [3] proved the following proposition.

Proposition 4.1.

Let GG be an algebraic group that acts on an algebraic space YY with finite stabilizers. Then, the localization maps give a direct sum decomposition

Γ:KG(Y)iKG(Y)𝔪Φi\Gamma:K_{G}(Y)\stackrel{{\scriptstyle\sim}}{{\longrightarrow}}\bigoplus_{i}K_{G}(Y)_{\mathfrak{m}_{\Phi_{i}}}

In the case of (1,n)\mathbb{P}(1,n), we take Y=IGXY=I_{G}X, and the maximal ideals 𝔪iR(G)[x±1]\mathfrak{m}_{i}\subseteq R(G)\cong\mathbb{C}[x^{\pm 1}], for i/ni\in\mathbb{Z}/n\mathbb{Z}, are the maximal ideals of characters vanishing on ζi\zeta^{i}. Thus, 𝔪0=x1\mathfrak{m}_{0}=\langle x-1\rangle and 𝔪i=xζi\mathfrak{m}_{i}=\langle x-\zeta^{i}\rangle where ζ=exp(2πi/n)\zeta=exp(2\pi i/n). We define

𝒦=l=0n1𝒦i,\mathcal{K}=\bigoplus_{l=0}^{n-1}\mathcal{K}_{i},

where the summands 𝒦l\mathcal{K}_{l} are the localizations above

𝒦l=KG(IX)𝔪l.\mathcal{K}_{l}=K_{G}(IX)_{\mathfrak{m}_{l}}.

Define generators of the direct sum decomposition of localizations by

xml:=Γ(xm)|KG(Xm)𝔪l.x_{ml}:=\Gamma(x_{m})|_{K_{G}(X^{m})_{\mathfrak{m}_{l}}}.

The result is a decomposition as a vector space of the K-theory of (1,n)\mathbb{P}(1,n), as in the diagram below.

[x0±1](x010)(x0n10)\textstyle{\frac{\mathbb{C}[x_{0}^{\pm 1}]}{\langle(x_{0}-1_{0})(x_{0}^{n}-1_{0})\rangle}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}Γ\scriptstyle{\Gamma}\textstyle{\oplus}[x1±1]x1n11\textstyle{\frac{\mathbb{C}[x_{1}^{\pm 1}]}{\langle x_{1}^{n}-1_{1}\rangle}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}Γ\scriptstyle{\Gamma}\textstyle{\oplus}\textstyle{\cdots}\textstyle{\oplus}[xn1±1]xn1n1n1\textstyle{\frac{\mathbb{C}[x_{n-1}^{\pm 1}]}{\langle x_{n-1}^{n}-1_{n-1}\rangle}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}Γ\scriptstyle{\Gamma}[x00±1]I00\textstyle{\frac{\mathbb{C}[x_{00}^{\pm 1}]}{I_{00}}}\textstyle{\oplus}[x10±1]I10\textstyle{\frac{\mathbb{C}[x_{10}^{\pm 1}]}{I_{10}}}\textstyle{\oplus}\textstyle{\cdots}\textstyle{\oplus}[x(n1)0±1]I(n1)0\textstyle{\frac{\mathbb{C}[x_{(n-1)0}^{\pm 1}]}{I_{(n-1)0}}}\textstyle{\oplus}\textstyle{\oplus}\textstyle{\oplus}[x01±1]I01\textstyle{\frac{\mathbb{C}[x_{01}^{\pm 1}]}{I_{01}}}\textstyle{\oplus}[x11±1]I11\textstyle{\frac{\mathbb{C}[x_{11}^{\pm 1}]}{I_{11}}}\textstyle{\oplus}\textstyle{\cdots}\textstyle{\oplus}[x(n1)1±1]I(n1)1\textstyle{\frac{\mathbb{C}[x_{(n-1)1}^{\pm 1}]}{I_{(n-1)1}}}\textstyle{\oplus}\textstyle{\oplus}\textstyle{\oplus}\textstyle{\vdots}\textstyle{\vdots}\textstyle{\vdots}\textstyle{\oplus}\textstyle{\oplus}\textstyle{\oplus}[x0(n1)±1]I0(n1)\textstyle{\frac{\mathbb{C}[x_{0(n-1)}^{\pm 1}]}{I_{0(n-1)}}}\textstyle{\oplus}[x1(n1)±1]I1(n1)\textstyle{\frac{\mathbb{C}[x_{1(n-1)}^{\pm 1}]}{I_{1(n-1)}}}\textstyle{\oplus}\textstyle{\cdots}\textstyle{\oplus}[x(n1)(n1)±1]I(n1)(n1)\textstyle{\frac{\mathbb{C}[x_{(n-1)(n-1)}^{\pm 1}]}{I_{(n-1)(n-1)}}}

where Iml=xmlζl1mlI_{ml}=\langle x_{ml}-\zeta^{l}1_{ml}\rangle for all mm and ll, except for I00=(x00100)2I_{00}=\langle(x_{00}-1_{00})^{2}\rangle.

Proposition 4.2.

The rows in 𝒦\mathcal{K} are the localizations 𝒦l\mathcal{K}_{l} for l=0,,n1l=0,\ldots,n-1, and the inverse images of the generators of 𝒦\mathcal{K} are

Γ1(100)\displaystyle\Gamma^{-1}(1_{00}) =12n((1n)x0+(1+n))x0n1x01\displaystyle=\frac{1}{2n}((1-n)x_{0}+(1+n))\frac{x_{0}^{n}-1}{x_{0}-1} (11)
Γ1(x00)\displaystyle\Gamma^{-1}(x_{00}) =12n((3n)x0+(n1))x0n1x01\displaystyle=\frac{1}{2n}((3-n)x_{0}+(n-1))\frac{x_{0}^{n}-1}{x_{0}-1} (12)
Γ1(10l)\displaystyle\Gamma^{-1}(1_{0l}) =ζln(ζl1)(x01)x0n1x0ζlforl0\displaystyle=\frac{\zeta^{l}}{n(\zeta^{l}-1)}(x_{0}-1)\frac{x_{0}^{n}-1}{x_{0}-\zeta^{l}}\hskip 14.45377pt\textup{for}\,\,l\neq 0 (13)
Γ1(1ml)\displaystyle\Gamma^{-1}(1_{ml}) =ζlnxmn1xmζlform0,\displaystyle=\frac{\zeta^{l}}{n}\frac{x_{m}^{n}-1}{x_{m}-\zeta^{l}}\hskip 14.45377pt\textup{for}\,\,m\neq 0, (14)

where all products above are with respect to the ordinary product on KG(I(1,n))K_{G}(I\mathbb{P}(1,n))_{\mathbb{C}} with complex coefficients, and where we regard un1uζl=il(uζi)\frac{u^{n}-1}{u-\zeta^{l}}=\prod_{i\neq l}(u-\zeta^{i}).

Proof.

We now calculate the map Γ1\Gamma^{-1} on the generators of the decomposition. Consider the element 1001_{00}. If Γ(f)=100\Gamma(f)=1_{00}, then f(x0)=p(x0)(x01)(x0n1)+q(x0)f(x_{0})=p(x_{0})(x_{0}-1)(x_{0}^{n}-1)+q(x_{0}), where pp and qq are polynomials, and the degree of qq is at most nn. The localization maps give that f0mod(x0ζl)f\equiv 0\bmod{(x_{0}-\zeta^{l})}, where l0l\neq 0. Thus, q(x0)=(ax0+b)l0(x0ζl)=(ax0+b)x0n1x01q(x_{0})=(ax_{0}+b)\prod_{l\neq 0}(x_{0}-\zeta^{l})=(ax_{0}+b)\frac{x_{0}^{n}-1}{x_{0}-1}, where a,ba,b\in\mathbb{C}. We also have that f1mod(x01)2f\equiv 1\bmod{(x_{0}-1)^{2}}, or equivalently, f(1)=1f(1)=1 and f(1)=0f^{\prime}(1)=0. Hence, we have

f(1)\displaystyle f(1) =limx01(ax0+b)x0n1x01=n(a+b)\displaystyle=\lim_{x_{0}\to 1}(ax_{0}+b)\frac{x_{0}^{n}-1}{x_{0}-1}=n(a+b)
1n\displaystyle\frac{1}{n} =a+b.\displaystyle=a+b.

Similarly,

f(1)\displaystyle f^{\prime}(1) =limx01(n1)x0nnx0n1+1(x01)2(ax0+b)+ax0n1x01\displaystyle=\lim_{x_{0}\to 1}\frac{(n-1)x_{0}^{n}-nx_{0}^{n-1}+1}{(x_{0}-1)^{2}}(ax_{0}+b)+a\frac{x_{0}^{n}-1}{x_{0}-1}
0\displaystyle 0 =n(n1)2(a+b)+na=n(n+1)2a+n(n1)2b.\displaystyle=\frac{n(n-1)}{2}(a+b)+na=\frac{n(n+1)}{2}a+\frac{n(n-1)}{2}b.

Combining these two expressions, we obtain

a=1n2nb=1+n2n.a=\frac{1-n}{2n}\hskip 14.45377ptb=\frac{1+n}{2n}.

The inverse of 1001_{00} can then be written as

Γ1(100)=12n((1n)x0+(1+n))x0n1x01.\Gamma^{-1}(1_{00})=\frac{1}{2n}((1-n)x_{0}+(1+n))\frac{x_{0}^{n}-1}{x_{0}-1}.

The computation for x00x_{00} is similar to that for 1001_{00}. The only difference is now we require that f(1)=1f^{\prime}(1)=1 instead of 0. This gives

1=n(n+1)2a+n(n1)2b.1=\frac{n(n+1)}{2}a+\frac{n(n-1)}{2}b.

Solving for aa and bb now gives

a=3n2nb=n12n.a=\frac{3-n}{2n}\hskip 14.45377ptb=\frac{n-1}{2n}.

Therefore, we obtain an expression for the inverse of x00x_{00} given by

Γ1(x00)=12n((3n)x0+(n1))x0n1x01.\Gamma^{-1}(x_{00})=\frac{1}{2n}((3-n)x_{0}+(n-1))\frac{x_{0}^{n}-1}{x_{0}-1}.

Now consider the element 10l1_{0l}, where l0l\neq 0. As before, we assume 10l=Γ(f(x0))=p(x0)(x01)(x0n1)+g(x0)1_{0l}=\Gamma(f(x_{0}))=p(x_{0})(x_{0}-1)(x_{0}^{n}-1)+g(x_{0}) where the degree of gg is at most nn. As 10l[x0l±1]x0lζl10l1_{0l}\in\frac{\mathbb{C}[x_{0l}^{\pm 1}]}{\langle x_{0l}-\zeta^{l}1_{0l}\rangle}, we have f0mod(xζk)f\equiv 0\,mod(x-\zeta^{k}) for klk\neq l. This allows us to write

g(x0)=r(x0)kl(x0ζk)=r(x0)x0n1x0ζl,g(x_{0})=r(x_{0})\prod_{k\neq l}(x_{0}-\zeta^{k})=r(x_{0})\frac{x_{0}^{n}-1}{x_{0}-\zeta^{l}},

for a polynomial rr of degree at most one. Note that f0mod(x01)2f\equiv 0\bmod{(x_{0}-1)^{2}}, so r(x0)=α(x01)r(x_{0})=\alpha(x_{0}-1) for some constant α\alpha. We also have f1mod(xζl)f\equiv 1\bmod{(x-\zeta^{l})}. So we have

f(ζl)=limx0ζlα(x01)x0n1x0ζl=limx0ζlα(x01)nx0n11=α(ζl1)nζl.f(\zeta^{l})=\lim_{x_{0}\to\zeta^{l}}\alpha(x_{0}-1)\frac{x_{0}^{n}-1}{x_{0}-\zeta^{l}}=\lim_{x_{0}\to\zeta^{l}}\alpha(x_{0}-1)\frac{nx_{0}^{n-1}}{1}=\alpha(\zeta^{l}-1)n\zeta^{-l}.

And therefore,

α=ζln(ζl1).\alpha=\frac{\zeta^{l}}{n(\zeta^{l}-1)}.

We now have a description of the inverse of 10l1_{0l}

Γ1(10l)=ζln(ζl1)(x01)x0n1x0ζl.\Gamma^{-1}(1_{0l})=\frac{\zeta^{l}}{n(\zeta^{l}-1)}(x_{0}-1)\frac{x_{0}^{n}-1}{x_{0}-\zeta^{l}}.

The last generators are 1ml1_{ml} where mm is nonzero. In this case, 1ml=Γ(f(xm))=p(xm)(xm1)+q(xm)1_{ml}=\Gamma(f(x_{m}))=p(x_{m})(x_{m}-1)+q(x_{m}) where the degree of qq is at most n1n-1. Since 1ml[xml±1]xmlζl1ml1_{ml}\in\frac{\mathbb{C}[x_{ml}^{\pm 1}]}{\langle x_{ml}-\zeta^{l}1_{ml}\rangle}, we have f0mod(xζk)f\equiv 0\bmod{(x-\zeta^{k})} for klk\neq l. Therefore,

g(xm)=αkl(xmζk)=αxmn1xmζl,g(x_{m})=\alpha\prod_{k\neq l}(x_{m}-\zeta^{k})=\alpha\frac{x_{m}^{n}-1}{x_{m}-\zeta^{l}},

where α\alpha is a constant. Furthermore, f1mod(xζl)f\equiv 1\bmod{(x-\zeta^{l})}, so we can solve for α\alpha,

f(ζl)=limxmζlαxmn1xmζl=limxmζlαnxmn11=αnζl.f(\zeta^{l})=\lim_{x_{m}\to\zeta^{l}}\alpha\frac{x_{m}^{n}-1}{x_{m}-\zeta^{l}}=\lim_{x_{m}\to\zeta^{l}}\alpha\frac{nx_{m}^{n-1}}{1}=\alpha n\zeta^{-l}.

We then get

α=ζln.\alpha=\frac{\zeta^{l}}{n}.

The inverse of 1ml1_{ml} is given by

Γ1(1ml)=ζlnxmn1xmζl.\Gamma^{-1}(1_{ml})=\frac{\zeta^{l}}{n}\cdot\frac{x_{m}^{n}-1}{x_{m}-\zeta^{l}}.

5. The virtual product on the localization

In the previous section, we constructed an inverse to the localization maps. Using the vector space isomorphism Γ\Gamma, and the virtual product on (1,n)\mathbb{P}(1,n) as given in equation (9), a virtual product can be constructed on the localization 𝒦=l=0n1𝒦l\mathcal{K}=\bigoplus_{l=0}^{n-1}\mathcal{K}_{l}.

Proposition 5.1.

The virtual product on 𝒦\mathcal{K} is given by linearly extending the virtual multiplications on the generators of 𝒦\mathcal{K} below.

100100\displaystyle 1_{00}*1_{00} =100\displaystyle=1_{00} 100x00\displaystyle 1_{00}*x_{00} =x00\displaystyle=x_{00}
1001m0\displaystyle 1_{00}*1_{m0} =1m0,if m0\displaystyle=1_{m0},\hskip 7.22743pt\textup{if $m\neq 0$} x00x00\displaystyle x_{00}*x_{00} =2x00100\displaystyle=2x_{00}-1_{00}
x001m0\displaystyle x_{00}*1_{m0} =1m0,if m0\displaystyle=1_{m0},\hskip 7.22743pt\textup{if $m\neq 0$} 1m101m20\displaystyle 1_{m_{1}0}*1_{m_{2}0} =0\displaystyle=0
10l1ml\displaystyle 1_{0l}*1_{ml} =1ml\displaystyle=1_{ml} x001ml\displaystyle x_{00}*1_{ml} =0,if l0\displaystyle=0,\hskip 7.22743pt\textup{if $l\neq 0$}
1m1l1m2l\displaystyle 1_{m_{1}l}*1_{m_{2}l} =(1ζl)1(m1+m2)l,if m1+m2n,m1,m20\displaystyle=(1-\zeta^{-l})1_{(m_{1}+m_{2})l},\hskip 7.22743pt\textup{if $m_{1}+m_{2}\neq n,\,\,m_{1},m_{2}\neq 0$}
1m1l1m2l\displaystyle 1_{m_{1}l}*1_{m_{2}l} =(1ζl)210l,if m1+m2=n,m1,m20\displaystyle=(1-\zeta^{-l})^{2}1_{0l},\hskip 7.22743pt\textup{if $m_{1}+m_{2}=n,\,\,m_{1},m_{2}\neq 0$}
1m1l11m2l2\displaystyle 1_{m_{1}l_{1}}*1_{m_{2}l_{2}} =0,if l1l2\displaystyle=0,\hskip 7.22743pt\textup{if $l_{1}\neq l_{2}$}
Proof.

We use the definition of virtual multiplication in equation (9) to multiply the inverse of elements in 𝒦\mathcal{K}, and then apply Γ\Gamma to obtain a multiplication in 𝒦\mathcal{K}.

Γ1(100)Γ1(100)\displaystyle\Gamma^{-1}(1_{00})*\Gamma^{-1}(1_{00}) =14n2((1n)x0n+2x0n1++2x0+(1+n))((1n)x0n+2x0n1\displaystyle=\frac{1}{4n^{2}}((1-n)x_{0}^{n}+2x_{0}^{n-1}+\cdots+2x_{0}+(1+n))*((1-n)x_{0}^{n}+2x_{0}^{n-1}
++2x0+(1+n))\displaystyle\quad+\cdots+2x_{0}+(1+n))
=14n2((12n+n2)x02n+k=1n1(4(1n)+4(k1))x02nk\displaystyle=\frac{1}{4n^{2}}((1-2n+n^{2})x_{0}^{2n}+\sum_{k=1}^{n-1}(4(1-n)+4(k-1))x_{0}^{2n-k}
+(2(1n2)+4(n1))x0n+k=1n1(4(1+n)+4(nk1))x0nk+(1+2n+n2))\displaystyle\quad+(2(1-n^{2})+4(n-1))x_{0}^{n}+\sum_{k=1}^{n-1}(4(1+n)+4(n-k-1))x_{0}^{n-k}+(1+2n+n^{2}))
=14n2((12n+n2)x02n+k=1n1(4k4n)(x0n+x0nk1)\displaystyle=\frac{1}{4n^{2}}((1-2n+n^{2})x_{0}^{2n}+\sum_{k=1}^{n-1}(4k-4n)(x_{0}^{n}+x_{0}^{n-k}-1)
+(2(1n2)+4(n1))x0n+k=1n1(8n4k)x0nk+(1+2n+n2))\displaystyle\quad+(2(1-n^{2})+4(n-1))x_{0}^{n}+\sum_{k=1}^{n-1}(8n-4k)x_{0}^{n-k}+(1+2n+n^{2}))
=14n2((2n2+2n)x0n+k=1n14nx0nk+(2n2+2n))\displaystyle=\frac{1}{4n^{2}}((-2n^{2}+2n)x_{0}^{n}+\sum_{k=1}^{n-1}4nx_{0}^{n-k}+(2n^{2}+2n))
=12n((1n)x0n+k=1n12x0nk+(1+n))\displaystyle=\frac{1}{2n}((1-n)x_{0}^{n}+\sum_{k=1}^{n-1}2x_{0}^{n-k}+(1+n))
=Γ1(100).\displaystyle=\Gamma^{-1}(1_{00}).

Taking Γ\Gamma of both sides of the equation above gives the multiplication

100100=100.1_{00}*1_{00}=1_{00}.
Γ1(100)Γ1(x00)\displaystyle\Gamma^{-1}(1_{00})*\Gamma^{-1}(x_{00}) =14n2((1n)x0n+2x0n1++2x0+(1+n))((3n)x0n+2x0n1\displaystyle=\frac{1}{4n^{2}}((1-n)x_{0}^{n}+2x_{0}^{n-1}+\cdots+2x_{0}+(1+n))*((3-n)x_{0}^{n}+2x_{0}^{n-1}
++2x0+(n1))\displaystyle\quad+\cdots+2x_{0}+(n-1))
=14n2((34n+n2)x02n+k=1n1(44n+4k)x02nk+(2n2+8n2)x0n\displaystyle=\frac{1}{4n^{2}}((3-4n+n^{2})x_{0}^{2n}+\sum_{k=1}^{n-1}(4-4n+4k)x_{0}^{2n-k}+(-2n^{2}+8n-2)x_{0}^{n}
+k=1n1(8n44k)x0nk+(n21))\displaystyle\quad+\sum_{k=1}^{n-1}(8n-4-4k)x_{0}^{n-k}+(n^{2}-1))
=14n2((34n+n2)x02n+k=1n1(44n+4k)(x0n+x0nk1)+(2n2+8n2)x0n\displaystyle=\frac{1}{4n^{2}}((3-4n+n^{2})x_{0}^{2n}+\sum_{k=1}^{n-1}(4-4n+4k)(x_{0}^{n}+x_{0}^{n-k}-1)+(-2n^{2}+8n-2)x_{0}^{n}
+k=1n1(8n44k)x0nk+(n21))\displaystyle\quad+\sum_{k=1}^{n-1}(8n-4-4k)x_{0}^{n-k}+(n^{2}-1))
=14n2(((68n+2n2)+k=1n1(44n4k)+(2n2+8n2))x0n\displaystyle=\frac{1}{4n^{2}}(((6-8n+2n^{2})+\sum_{k=1}^{n-1}(4-4n-4k)+(-2n^{2}+8n-2))x_{0}^{n}
+k=1n1(4n)x0nk+((34n+n2)k=1n1(44n4k)+(n21)))\displaystyle\quad+\sum_{k=1}^{n-1}(4n)x_{0}^{n-k}+(-(3-4n+n^{2})-\sum_{k=1}^{n-1}(4-4n-4k)+(n^{2}-1)))
=14n2((2n2+6n)x0n+k=1n1(4n)x0nk+(2n22n))\displaystyle=\frac{1}{4n^{2}}((-2n^{2}+6n)x_{0}^{n}+\sum_{k=1}^{n-1}(4n)x_{0}^{n-k}+(2n^{2}-2n))
=12n((3n)x0n+2x0n1++2x0+(n1))\displaystyle=\frac{1}{2n}((3-n)x_{0}^{n}+2x_{0}^{n-1}+\cdots+2x_{0}+(n-1))
=Γ1(x00).\displaystyle=\Gamma^{-1}(x_{00}).

This gives the multiplication

100x00=x00.1_{00}*x_{00}=x_{00}.
Γ1(100)Γ1(1m0)\displaystyle\Gamma^{-1}(1_{00})*\Gamma^{-1}(1_{m0}) =12n2((1n)x0n+2x0n1++2x0+(1+n))(xmn1++xm+1)\displaystyle=\frac{1}{2n^{2}}((1-n)x_{0}^{n}+2x_{0}^{n-1}+\cdots+2x_{0}+(1+n))*(x_{m}^{n-1}+\cdots+x_{m}+1)
=12n2(k=1n((1n)+(2(k1)))xm2nk+k=1n(2(nk)+(1+n))xmnk)\displaystyle=\frac{1}{2n^{2}}(\sum_{k=1}^{n}((1-n)+(2(k-1)))x_{m}^{2n-k}+\sum_{k=1}^{n}(2(n-k)+(1+n))x_{m}^{n-k})
=12n2(k=1n(2kn1)xmnk+k=1n(3n2k+1)xmnk)\displaystyle=\frac{1}{2n^{2}}(\sum_{k=1}^{n}(2k-n-1)x_{m}^{n-k}+\sum_{k=1}^{n}(3n-2k+1)x_{m}^{n-k})
=12n2(k=1n2nxmnk)\displaystyle=\frac{1}{2n^{2}}(\sum_{k=1}^{n}2nx_{m}^{n-k})
=1n(xmn1++xm+1)\displaystyle=\frac{1}{n}(x_{m}^{n-1}+\cdots+x_{m}+1)
=Γ1(1m0).\displaystyle=\Gamma^{-1}(1_{m0}).

Thus, we have

1001m0=1m0.1_{00}*1_{m0}=1_{m0}.
Γ1(x00)Γ1(x00)\displaystyle\Gamma^{-1}(x_{00})*\Gamma^{-1}(x_{00}) =14n2((3n)x0n+2x0n1++2x0+(n1))((3n)x0n+2x0n1\displaystyle=\frac{1}{4n^{2}}((3-n)x_{0}^{n}+2x_{0}^{n-1}+\cdots+2x_{0}+(n-1))*((3-n)x_{0}^{n}+2x_{0}^{n-1}
++2x0+(n1))\displaystyle\quad+\cdots+2x_{0}+(n-1))
=14n2((96n+n2)x02n+k=1n1(4(3n)+4(k+1))x02nk\displaystyle=\frac{1}{4n^{2}}((9-6n+n^{2})x_{0}^{2n}+\sum_{k=1}^{n-1}(4(3-n)+4(k+1))x_{0}^{2n-k}
+(2(n2+4n3)+4(n1))x0n+k=1n1(4(n1)+4(n1k))x0nk\displaystyle\quad+(2(-n^{2}+4n-3)+4(n-1))x_{0}^{n}+\sum_{k=1}^{n-1}(4(n-1)+4(n-1-k))x_{0}^{n-k}
+(n22n+1))\displaystyle\quad+(n^{2}-2n+1))
=14n2((96n+n2)(2x0n1)+k=1n1(4(k+2)4n)(x0n+x0nk1)\displaystyle=\frac{1}{4n^{2}}((9-6n+n^{2})(2x_{0}^{n}-1)+\sum_{k=1}^{n-1}(4(k+2)-4n)(x_{0}^{n}+x_{0}^{n-k}-1)
+(2n2+12n10)x0n+k=1n1(8n84k)x0nk+(n22n+1))\displaystyle\quad+(-2n^{2}+12n-10)x_{0}^{n}+\sum_{k=1}^{n-1}(8n-8-4k)x_{0}^{n-k}+(n^{2}-2n+1))
=14n2((2n2+10n)x0n+k=1n14nx0nk+(2n26n))\displaystyle=\frac{1}{4n^{2}}((-2n^{2}+10n)x_{0}^{n}+\sum_{k=1}^{n-1}4nx_{0}^{n-k}+(2n^{2}-6n))
=12n((5n)x0n+2x0n1++2x0+(n3))\displaystyle=\frac{1}{2n}((5-n)x_{0}^{n}+2x_{0}^{n-1}+\cdots+2x_{0}+(n-3))
=2Γ1(x00)Γ1(100).\displaystyle=2\Gamma^{-1}(x_{00})-\Gamma^{-1}(1_{00}).

Therefore,

x00x00=2x00100.x_{00}*x_{00}=2x_{00}-1_{00}.

and more generally,

x00k=kx00(k1)100.x_{00}^{k}=kx_{00}-(k-1)1_{00}.
Γ1(x00)Γ1(1m0)\displaystyle\Gamma^{-1}(x_{00})*\Gamma^{-1}(1_{m0}) =12n2((3n)x0n+2x0n1++2x0+(n1))(xmn1++xm+1)\displaystyle=\frac{1}{2n^{2}}((3-n)x_{0}^{n}+2x_{0}^{n-1}+\cdots+2x_{0}+(n-1))*(x_{m}^{n-1}+\cdots+x_{m}+1)
=12n2(k=1n((3n)+2(k1))xm2nk+k=1n(2(nk)+(n1))xmnk)\displaystyle=\frac{1}{2n^{2}}(\sum_{k=1}^{n}((3-n)+2(k-1))x_{m}^{2n-k}+\sum_{k=1}^{n}(2(n-k)+(n-1))x_{m}^{n-k})
=12n2(k=1n(2k+1n)xmnk+k=1n(3n2k1)xmnk)\displaystyle=\frac{1}{2n^{2}}(\sum_{k=1}^{n}(2k+1-n)x_{m}^{n-k}+\sum_{k=1}^{n}(3n-2k-1)x_{m}^{n-k})
=12n2(k=1n2nxmnk)\displaystyle=\frac{1}{2n^{2}}(\sum_{k=1}^{n}2nx_{m}^{n-k})
=1n(xmn1++xm+1)\displaystyle=\frac{1}{n}(x_{m}^{n-1}+\cdots+x_{m}+1)
=Γ1(1m0).\displaystyle=\Gamma^{-1}(1_{m0}).

This calculation shows that

x001m0=1m0.x_{00}*1_{m0}=1_{m0}.

Suppose that m1+m2nm_{1}+m_{2}\neq n.

Γ1(1m10)Γ1(1m20)\displaystyle\Gamma^{-1}(1_{m_{1}0})*\Gamma^{-1}(1_{m_{2}0}) =1n2(xm1n1++xm1+1)(xm2n1++xm2+1)\displaystyle=\frac{1}{n^{2}}(x_{m_{1}}^{n-1}+\cdots+x_{m_{1}}+1)*(x_{m_{2}}^{n-1}+\cdots+x_{m_{2}}+1)
=1n2(k=1nkxm1+m22nk1(1xm1+m2n1)+k=1n1(nk)xm1+m2nk1(1xm1+m2n1)\displaystyle=\frac{1}{n^{2}}(\sum_{k=1}^{n}kx_{m_{1}+m_{2}}^{2n-k-1}(1-x_{m_{1}+m_{2}}^{n-1})+\sum_{k=1}^{n-1}(n-k)x_{m_{1}+m_{2}}^{n-k-1}(1-x_{m_{1}+m_{2}}^{n-1})
=1n2(k=1nk(xm1+m22nk1xm1+m23nk2)+k=1n1(nk)(xm1+m2nk1xm1+m22nk2))\displaystyle=\frac{1}{n^{2}}(\sum_{k=1}^{n}k(x_{m_{1}+m_{2}}^{2n-k-1}-x_{m_{1}+m_{2}}^{3n-k-2})+\sum_{k=1}^{n-1}(n-k)(x_{m_{1}+m_{2}}^{n-k-1}-x_{m_{1}+m_{2}}^{2n-k-2}))
=1n2(k=1nk(xm1+m2nk1xm1+m2nk2)+k=1n1(nk)(xm1+m2nk1xm1+m2nk2))\displaystyle=\frac{1}{n^{2}}(\sum_{k=1}^{n}k(x_{m_{1}+m_{2}}^{n-k-1}-x_{m_{1}+m_{2}}^{n-k-2})+\sum_{k=1}^{n-1}(n-k)(x_{m_{1}+m_{2}}^{n-k-1}-x_{m_{1}+m_{2}}^{n-k-2}))
=1n2(k=1nn(xm1+m2nkxm1+m2nk1))\displaystyle=\frac{1}{n^{2}}(\sum_{k=1}^{n}n(x_{m_{1}+m_{2}}^{n-k}-x_{m_{1}+m_{2}}^{n-k-1}))
=0.\displaystyle=0.

This identity gives

1m101m20=0,ifm1+m2n.1_{m_{1}0}*1_{m_{2}0}=0,\hskip 7.22743pt\textup{if}\,\,m_{1}+m_{2}\neq n.

Now suppose that m1+m2=nm_{1}+m_{2}=n, then

Γ1(1m10)Γ1(1m20)\displaystyle\Gamma^{-1}(1_{m_{1}0})*\Gamma^{-1}(1_{m_{2}0}) =1n2(xm1n1++xm1+1)(xm2n1++xm2+1)\displaystyle=\frac{1}{n^{2}}(x_{m_{1}}^{n-1}+\cdots+x_{m_{1}}+1)*(x_{m_{2}}^{n-1}+\cdots+x_{m_{2}}+1)
=1n2(k=1nkx02nk1(x0n22x0n1+x0n)+k=1n1(nk)x0nk1(x0n22x0n1+x0n))\displaystyle=\frac{1}{n^{2}}(\sum_{k=1}^{n}kx_{0}^{2n-k-1}(x_{0}^{n-2}-2x_{0}^{n-1}+x_{0}^{n})+\sum_{k=1}^{n-1}(n-k)x_{0}^{n-k-1}(x_{0}^{n-2}-2x_{0}^{n-1}+x_{0}^{n}))
=1n2(k=1nk(x03nk32x03nk2+x03nk1)+k=1n1(nk)(x02nk32x02nk2+x02nk1))\displaystyle=\frac{1}{n^{2}}(\sum_{k=1}^{n}k(x_{0}^{3n-k-3}-2x_{0}^{3n-k-2}+x_{0}^{3n-k-1})+\sum_{k=1}^{n-1}(n-k)(x_{0}^{2n-k-3}-2x_{0}^{2n-k-2}+x_{0}^{2n-k-1}))
=1n2(k=1nk(x0nk32x0nk2+x0nk1)+k=1n1(nk)(x0nk32x0nk2+x0nk1))\displaystyle=\frac{1}{n^{2}}(\sum_{k=1}^{n}k(x_{0}^{n-k-3}-2x_{0}^{n-k-2}+x_{0}^{n-k-1})+\sum_{k=1}^{n-1}(n-k)(x_{0}^{n-k-3}-2x_{0}^{n-k-2}+x_{0}^{n-k-1}))
=1n(k=1n(x0nk32x0nk2+x0nk1))\displaystyle=\frac{1}{n}(\sum_{k=1}^{n}(x_{0}^{n-k-3}-2x_{0}^{n-k-2}+x_{0}^{n-k-1}))
=0.\displaystyle=0.

Here, the fourth equality uses the relations x02nk=x0n+x0nk1x_{0}^{2n-k}=x_{0}^{n}+x_{0}^{n-k}-1 and x03nk=2x0n+x0nk2x_{0}^{3n-k}=2x_{0}^{n}+x_{0}^{n-k}-2 for 0kn0\leq k\leq n. The conclusion is that the elements 1m01_{m0} are nilpotent, i.e.

1m101m20=0,ifm1+m2=n.1_{m_{1}0}*1_{m_{2}0}=0,\hskip 7.22743pt\textup{if}\,\,m_{1}+m_{2}=n.

This concludes the calculation of the virtual product on 𝒦0\mathcal{K}_{0}.

We will calculate the virtual product on 𝒦l\mathcal{K}_{l} similarly. Henceforth, assume that l0l\neq 0. First, suppose that m1+m2nm_{1}+m_{2}\neq n.

Γ1(1m1l)Γ1(1m2l)\displaystyle\Gamma^{-1}(1_{m_{1}l})*\Gamma^{-1}(1_{m_{2}l}) =1n2(ζl(n1)xm1n1++ζlxm1+1)(ζl(n1)xm2n1++ζlxm2+1)\displaystyle=\frac{1}{n^{2}}(\zeta^{-l(n-1)}x_{m_{1}}^{n-1}+\cdots+\zeta^{-l}x_{m_{1}}+1)*(\zeta^{-l(n-1)}x_{m_{2}}^{n-1}+\cdots+\zeta^{-l}x_{m_{2}}+1)
=1n2(k=1nkζl(2nk1)xm1+m22nk1(1xm1+m2n1)+k=1n1(nk)ζl(nk1)xm1+m2nk1(1xm1+m2n1))\displaystyle=\frac{1}{n^{2}}(\sum_{k=1}^{n}k\zeta^{-l(2n-k-1)}x_{m_{1}+m_{2}}^{2n-k-1}(1-x_{m_{1}+m_{2}}^{n-1})+\sum_{k=1}^{n-1}(n-k)\zeta^{-l(n-k-1)}x_{m_{1}+m_{2}}^{n-k-1}(1-x_{m_{1}+m_{2}}^{n-1}))
=1n2(k=1nkζl(2nk1)(xm1+m22nk1xm1+m23nk2)+k=1n1(nk)ζl(nk1)(xm1+m2nk1xm1+m22nk2))\displaystyle=\frac{1}{n^{2}}(\sum_{k=1}^{n}k\zeta^{-l(2n-k-1)}(x_{m_{1}+m_{2}}^{2n-k-1}-x_{m_{1}+m_{2}}^{3n-k-2})+\sum_{k=1}^{n-1}(n-k)\zeta^{-l(n-k-1)}(x_{m_{1}+m_{2}}^{n-k-1}-x_{m_{1}+m_{2}}^{2n-k-2}))
=1n2(k=1nkζl(nk1)(xm1+m2nk1xm1+m2nk2)+k=1n1(nk)(ζl(nk1)(xm1+m2nk1xm1+m2nk2))\displaystyle=\frac{1}{n^{2}}(\sum_{k=1}^{n}k\zeta^{-l(n-k-1)}(x_{m_{1}+m_{2}}^{n-k-1}-x_{m_{1}+m_{2}}^{n-k-2})+\sum_{k=1}^{n-1}(n-k)(\zeta^{-l(n-k-1)}(x_{m_{1}+m_{2}}^{n-k-1}-x_{m_{1}+m_{2}}^{n-k-2}))
=1nk=1nζl(nk1)(xm1+m2nk1xm1+m2nk2)\displaystyle=\frac{1}{n}\sum_{k=1}^{n}\zeta^{-l(n-k-1)}(x_{m_{1}+m_{2}}^{n-k-1}-x_{m_{1}+m_{2}}^{n-k-2})
=1nk=1n(ζl(nk1)ζl(nk2))xm1+m2nk1\displaystyle=\frac{1}{n}\sum_{k=1}^{n}(\zeta^{-l(n-k-1)}-\zeta^{-l(n-k-2)})x_{m_{1}+m_{2}}^{n-k-1}
=1n(1ζl)k=1nζl(nk1)xm1+m2nk1\displaystyle=\frac{1}{n}(1-\zeta^{-l})\sum_{k=1}^{n}\zeta^{-l(n-k-1)}x_{m_{1}+m_{2}}^{n-k-1}
=(1ζl)Γ1(1m1+m20).\displaystyle=(1-\zeta^{-l})\Gamma^{-1}(1_{m_{1}+m_{2}0}).

Thus, we obtain

1m1l1m2l=(1ζl)1m1+m2l.1_{m_{1}l}*1_{m_{2}l}=(1-\zeta^{-l})1_{m_{1}+m_{2}l}.

Now suppose that m1+m2=nm_{1}+m_{2}=n, then

Γ1(1m1l)Γ1(1m2l)\displaystyle\Gamma^{-1}(1_{m_{1}l})*\Gamma^{-1}(1_{m_{2}l}) =1n2(ζl(n1)xm1n1++ζlxm1+1)(ζl(n1)xm2n1++ζlxm2+1)\displaystyle=\frac{1}{n^{2}}(\zeta^{-l(n-1)}x_{m_{1}}^{n-1}+\cdots+\zeta^{-l}x_{m_{1}}+1)*(\zeta^{-l(n-1)}x_{m_{2}}^{n-1}+\cdots+\zeta^{-l}x_{m_{2}}+1)
=1n2(k=1nkζl(2nk1)x02nk1(x0n22x0n1+x0n)+\displaystyle=\frac{1}{n^{2}}(\sum_{k=1}^{n}k\zeta^{-l(2n-k-1)}x_{0}^{2n-k-1}(x_{0}^{n-2}-2x_{0}^{n-1}+x_{0}^{n})+
+k=1n1(nk)ζl(nk1)x0nk1(x0n22x0n1+x0n))\displaystyle\quad+\sum_{k=1}^{n-1}(n-k)\zeta^{-l(n-k-1)}x_{0}^{n-k-1}(x_{0}^{n-2}-2x_{0}^{n-1}+x_{0}^{n}))
=1n2(k=1nkζl(2nk1)(x03nk32x03nk2+x03nk1)+\displaystyle=\frac{1}{n^{2}}(\sum_{k=1}^{n}k\zeta^{-l(2n-k-1)}(x_{0}^{3n-k-3}-2x_{0}^{3n-k-2}+x_{0}^{3n-k-1})+
+k=1n1(nk)ζl(nk1)(x02nk32x02nk2+x02nk1))\displaystyle\quad+\sum_{k=1}^{n-1}(n-k)\zeta^{-l(n-k-1)}(x_{0}^{2n-k-3}-2x_{0}^{2n-k-2}+x_{0}^{2n-k-1}))
=1n2(k=1nkζl(nk1)(x0nk32x0nk2+x0nk1)+\displaystyle=\frac{1}{n^{2}}(\sum_{k=1}^{n}k\zeta^{-l(n-k-1)}(x_{0}^{n-k-3}-2x_{0}^{n-k-2}+x_{0}^{n-k-1})+
+k=1n1(nk)ζl(nk1)(x0nk32x0nk2+x0nk1))\displaystyle\quad+\sum_{k=1}^{n-1}(n-k)\zeta^{-l(n-k-1)}(x_{0}^{n-k-3}-2x_{0}^{n-k-2}+x_{0}^{n-k-1}))
=1nk=1nζl(nk1)(x0nk32x0nk2+x0nk1)\displaystyle=\frac{1}{n}\sum_{k=1}^{n}\zeta^{-l(n-k-1)}(x_{0}^{n-k-3}-2x_{0}^{n-k-2}+x_{0}^{n-k-1})
=1nk=1n(ζl(nk3)2ζl(nk2)+ζl(nk1))x0nk1\displaystyle=\frac{1}{n}\sum_{k=1}^{n}(\zeta^{-l(n-k-3)}-2\zeta^{-l(n-k-2)}+\zeta^{-l(n-k-1)})x_{0}^{n-k-1}
=1n(1ζl)2k=1nζl(nk1)x0nk1\displaystyle=\frac{1}{n}(1-\zeta^{-l})^{2}\sum_{k=1}^{n}\zeta^{-l(n-k-1)}x_{0}^{n-k-1}
=(1ζl)2Γ1(10l).\displaystyle=(1-\zeta^{-l})^{2}\Gamma^{-1}(1_{0l}).

Taking Γ\Gamma of both sides of the equation gives

1m1l1m2l=(1ζl)210l.1_{m_{1}l}*1_{m_{2}l}=(1-\zeta^{-l})^{2}1_{0l}.

Suppose that m0m\neq 0, then

Γ1(10l)Γ1(1ml)\displaystyle\Gamma^{-1}(1_{0l})*\Gamma^{-1}(1_{ml}) =1n(ζl1)(ζl(n1)x0n+(ζl(n2)ζl(n1))x0n1\displaystyle=\frac{1}{n(\zeta^{l}-1)}(\zeta^{-l(n-1)}x_{0}^{n}+(\zeta^{-l(n-2)}-\zeta^{-l(n-1)})x_{0}^{n-1}
++(1ζl)1)(ζl(n1)xmn1++ζlxm+1)\displaystyle\quad+\cdots+(1-\zeta^{-l})-1)*(\zeta^{-l(n-1)}x_{m}^{n-1}+\cdots+\zeta^{-l}x_{m}+1)
=1n2(ζl1)(k=1n(kζl(nk1)(k1)ζl(nk))xm2nk+\displaystyle=\frac{1}{n^{2}(\zeta^{l}-1)}(\sum_{k=1}^{n}(k\zeta^{-l(n-k-1)}-(k-1)\zeta^{-l(n-k)})x_{m}^{2n-k}+
+k=1n((nk)ζl(nk1)(nk+1)ζl(nk))xmnk)\displaystyle\quad+\sum_{k=1}^{n}((n-k)\zeta^{-l(n-k-1)}-(n-k+1)\zeta^{-l(n-k)})x_{m}^{n-k})
=1n(ζl1)k=1n(ζl(nk1)ζl(nk))xmnk\displaystyle=\frac{1}{n(\zeta^{l}-1)}\sum_{k=1}^{n}(\zeta^{-l(n-k-1)}-\zeta^{-l(n-k)})x_{m}^{n-k}
=1nk=1nζl(nk)xmnk\displaystyle=\frac{1}{n}\sum_{k=1}^{n}\zeta^{-l(n-k)}x_{m}^{n-k}
=Γ1(1ml).\displaystyle=\Gamma^{-1}(1_{ml}).

This calculation shows that

10l1ml=1ml.1_{0l}*1_{ml}=1_{ml}.

Combining the previous three identities and using associativity of virtual multiplication, we also have

10l10l=10l.1_{0l}*1_{0l}=1_{0l}.

Suppose l1l2l_{1}\neq l_{2}, let η=l1l2\eta=l_{1}-l_{2}, and without loss of generality, take m1=1=m2m_{1}=1=m_{2}. Consider the product

x1n1x1ζl1x2n1x2ζl2\displaystyle\frac{x_{1}^{n}-1}{x_{1}-\zeta^{l_{1}}}*\frac{x_{2}^{n}-1}{x_{2}-\zeta^{l_{2}}} =(ζl1(n1)x1n1++1)(ζl2(n1)x2n1++1)\displaystyle=(\zeta^{-l_{1}(n-1)}x_{1}^{n-1}+\cdots+1)*(\zeta^{-l_{2}(n-1)}x_{2}^{n-1}+\cdots+1)
=k=2n(i=1k1ζl1(ni)l2(nk+i))x22nk(1x2n1)\displaystyle=\sum_{k=2}^{n}\Big{(}\sum_{i=1}^{k-1}\zeta^{-l_{1}(n-i)-l_{2}(n-k+i)}\Big{)}x_{2}^{2n-k}(1-x_{2}^{n-1})
+k=1n(i=0nkζl1(nki)l2i)x2nk(1x2n1)\displaystyle\quad+\sum_{k=1}^{n}\Big{(}\sum_{i=0}^{n-k}\zeta^{-l_{1}(n-k-i)-l_{2}i}\Big{)}x_{2}^{n-k}(1-x_{2}^{n-1})
=k=2n(i=1k1ζl1(ni)l2(nk+i))(x2nkx2nk1)\displaystyle=\sum_{k=2}^{n}\Big{(}\sum_{i=1}^{k-1}\zeta^{-l_{1}(n-i)-l_{2}(n-k+i)}\Big{)}(x_{2}^{n-k}-x_{2}^{n-k-1})
+k=1n(i=0nkζl1(nki)l2i)(x2nkx2nk1)\displaystyle\quad+\sum_{k=1}^{n}\Big{(}\sum_{i=0}^{n-k}\zeta^{-l_{1}(n-k-i)-l_{2}i}\Big{)}(x_{2}^{n-k}-x_{2}^{n-k-1})
=k=2n(i=1k1ζiη+kl2)(x2nkx2nk1)\displaystyle=\sum_{k=2}^{n}\Big{(}\sum_{i=1}^{k-1}\zeta^{i\eta+kl_{2}}\Big{)}(x_{2}^{n-k}-x_{2}^{n-k-1})
+k=1n(i=knζiη+kl2)(x2nkx2nk1)\displaystyle\quad+\sum_{k=1}^{n}\Big{(}\sum_{i=k}^{n}\zeta^{i\eta+kl_{2}}\Big{)}(x_{2}^{n-k}-x_{2}^{n-k-1})
=0,\displaystyle=0,

using the identity 1+ζη++ζ(n1)η=01+\zeta^{\eta}+\cdots+\zeta^{(n-1)\eta}=0, which holds since η0\eta\neq 0. As all of the inverse images of the generators of 𝒦\mathcal{K} contain a factor of the form xmn1xmζl\frac{x_{m}^{n}-1}{x_{m}-\zeta^{l}} for some mm and some ll, we see that whenever l1l2l_{1}\neq l_{2},

1m1l11m2l2=0.1_{m_{1}l_{1}}*1_{m_{2}l_{2}}=0.

6. The virtual Adams operations on the localization

The virtual K-theory ring has extra structure, given by the virtual Adams (or ψ\psi-) operations. As with the virtual product, we can use the isomorphism Γ\Gamma to induce virtual Adams operations on the direct sum decomposition. Let d=gcd(k,n)d=\gcd(k,n) for all integers k1k\geq 1.

Proposition 6.1.

Given k1k\geq 1 and l=0,,n1l=0,\ldots,n-1, let the dd solutions to the equivalence kylmodnky\equiv l\bmod{n} be given by y=siy=s_{i}, where i=1,,di=1,\ldots,d, so that when l=0l=0, s1=0s_{1}=0. The virtual Adams operations ψ~k:𝒦𝒦\widetilde{\psi}^{k}:\mathcal{K}\rightarrow\mathcal{K} are given by

ψ~k(10l)\displaystyle\widetilde{\psi}^{k}(1_{0l}) =i=1d10si,ifd|l,l0\displaystyle=\sum_{i=1}^{d}1_{0s_{i}},\hskip 7.22743pt\textup{if}\,\,d\,|\,l,\,l\neq 0 ψ~k(10l)\displaystyle\widetilde{\psi}^{k}(1_{0l}) =0,ifdl,l0\displaystyle=0,\hskip 7.22743pt\textup{if}\,\,d\nmid l,\,l\neq 0
ψ~k(1ml)\displaystyle\widetilde{\psi}^{k}(1_{ml}) =i=1dζl1ζsi11msi,ifd|l,m0l\displaystyle=\sum_{i=1}^{d}\frac{\zeta^{-l}-1}{\zeta^{-s_{i}}-1}1_{ms_{i}},\hskip 7.22743pt\textup{if}\,\,d\,|\,l,\,m\neq 0\neq l ψ~k(1ml)\displaystyle\widetilde{\psi}^{k}(1_{ml}) =0,ifdl,m0l\displaystyle=0,\hskip 7.22743pt\textup{if}\,\,d\nmid l,\,m\neq 0\neq l
ψ~k(1m0)\displaystyle\widetilde{\psi}^{k}(1_{m0}) =k1m0,ifm0\displaystyle=k1_{m0},\hskip 7.22743pt\textup{if}\,\,m\neq 0 ψ~k(100)\displaystyle\widetilde{\psi}^{k}(1_{00}) =i=1d10si\displaystyle=\sum_{i=1}^{d}1_{0s_{i}}
ψ~k(x00)\displaystyle\widetilde{\psi}^{k}(x_{00}) =kx00(k1)100+i=2d10si.\displaystyle=kx_{00}-(k-1)1_{00}+\sum_{i=2}^{d}1_{0s_{i}}.
Proof.

Suppose l0l\neq 0 and m0m\neq 0, and recall the expression for the inverse of 1ml1_{ml} in equation (14). Let

h(xm)\displaystyle h(x_{m}) =ψ~k(Γ1(1ml))\displaystyle=\widetilde{\psi}^{k}(\Gamma^{-1}(1_{ml}))
=ψk(Γ1(1ml))θk(xm1)\displaystyle=\psi^{k}(\Gamma^{-1}(1_{ml}))\theta^{k}(x_{m}^{-1})
=ζlnxmkn1xmkζl(1+xm1++xmk+1)\displaystyle=\frac{\zeta^{l}}{n}\cdot\frac{x_{m}^{kn}-1}{x_{m}^{k}-\zeta^{l}}\cdot(1+x_{m}^{-1}+\cdots+x_{m}^{-k+1})
=f(xm)(xmn1)+g(xm).\displaystyle=f(x_{m})(x_{m}^{n}-1)+g(x_{m}).

where the degree of gg is at most n1n-1. Suppose that dd does not divide ll, then there are no solutions to the equivalence kylmodnky\equiv l\bmod{n}. Therefore, h(ζy)=0h(\zeta^{y})=0 for all 0yn10\leq y\leq n-1. But the degree of gg is at most n1n-1, so we must have g(xm)=0g(x_{m})=0. In this case, we have

ψ~k(1ml)=0.\widetilde{\psi}^{k}(1_{ml})=0.

Now consider the case where dd divides ll. If yy does not satisfy the equivalence kylmodnky\equiv l\bmod{n}, then h(ζy)=0h(\zeta^{y})=0, so we can write

g(xm)=p(xm)kylmodn(xmζy)g(x_{m})=p(x_{m})\prod_{ky\not\equiv l\bmod{n}}(x_{m}-\zeta^{y})

Let sis_{i}, for 1id1\leq i\leq d, be the solutions to the above equivalence, then

h(ζsi)\displaystyle h(\zeta^{s_{i}}) =limxmζsiζlnxmkn1xmkζl(1+xm1++xmk+1)\displaystyle=\lim_{x_{m}\to\zeta^{s_{i}}}\frac{\zeta^{l}}{n}\cdot\frac{x_{m}^{kn}-1}{x_{m}^{k}-\zeta^{l}}\cdot(1+x_{m}^{-1}+\cdots+x_{m}^{-k+1})
=limxmζsiζlnxmkn1xmkζlxmk1xm11\displaystyle=\lim_{x_{m}\to\zeta^{s_{i}}}\frac{\zeta^{l}}{n}\cdot\frac{x_{m}^{kn}-1}{x_{m}^{k}-\zeta^{l}}\cdot\frac{x_{m}^{-k}-1}{x_{m}^{-1}-1}
=limxmζsiζlnknxmkn1kxmk1xmk1xm11\displaystyle=\lim_{x_{m}\to\zeta^{s_{i}}}\frac{\zeta^{l}}{n}\cdot\frac{knx_{m}^{kn-1}}{kx_{m}^{k-1}}\cdot\frac{x_{m}^{-k}-1}{x_{m}^{-1}-1}
=ζlnkn(ζsi)kn1k(ζsi)k1(ζsi)k1(ζsi)11\displaystyle=\frac{\zeta^{l}}{n}\cdot\frac{kn(\zeta^{s_{i}})^{kn-1}}{k(\zeta^{s_{i}})^{k-1}}\cdot\frac{(\zeta^{s_{i}})^{-k}-1}{(\zeta^{s_{i}})^{-1}-1}
=ζlnknζsikζlζsiζl1ζsi1\displaystyle=\frac{\zeta^{l}}{n}\cdot\frac{kn\zeta^{-s_{i}}}{k\zeta^{l}\zeta^{-s_{i}}}\cdot\frac{\zeta^{-l}-1}{\zeta^{-s_{i}}-1}
=ζl1ζsi1.\displaystyle=\frac{\zeta^{-l}-1}{\zeta^{-s_{i}}-1}.

On the other hand,

g(ζsi)\displaystyle g(\zeta^{s_{i}}) =limxmζsip(xm)kylmodn(xmζy)\displaystyle=\lim_{x_{m}\to\zeta^{s_{i}}}p(x_{m})\prod_{ky\not\equiv l\bmod{n}}(x_{m}-\zeta^{y})
=limxmζsip(xm)xmn1(xmζs1)(xmζsd)\displaystyle=\lim_{x_{m}\to\zeta^{s_{i}}}p(x_{m})\frac{x_{m}^{n}-1}{(x_{m}-\zeta^{s_{1}})\cdots(x_{m}-\zeta^{s_{d}})}
=limxmζsip(xm)nxmn1j=1dtj(xmζst)\displaystyle=\lim_{x_{m}\to\zeta^{s_{i}}}p(x_{m})\frac{nx_{m}^{n-1}}{\sum_{j=1}^{d}\prod_{t\neq j}(x_{m}-\zeta^{s_{t}})}
=p(ζsi)nζsiti(ζsiζst).\displaystyle=p(\zeta^{s_{i}})\frac{n\zeta^{-s_{i}}}{\prod_{t\neq i}(\zeta^{s_{i}}-\zeta^{s_{t}})}.

We can then solve for p(ζsi)p(\zeta^{s_{i}}).

h(ζsi)\displaystyle h(\zeta^{s_{i}}) =g(ζsi)\displaystyle=g(\zeta^{s_{i}})
ζl1ζsi1\displaystyle\frac{\zeta^{-l}-1}{\zeta^{-s_{i}}-1} =p(ζsi)nζsiti(ζsiζst)\displaystyle=p(\zeta^{s_{i}})\frac{n\zeta^{-s_{i}}}{\prod_{t\neq i}(\zeta^{s_{i}}-\zeta^{s_{t}})}
p(ζsi)\displaystyle p(\zeta^{s_{i}}) =ζsinti(ζsiζst)ζl1ζsi1.\displaystyle=\frac{\zeta^{s_{i}}}{n}\prod_{t\neq i}(\zeta^{s_{i}}-\zeta^{s_{t}})\frac{\zeta^{-l}-1}{\zeta^{-s_{i}}-1}.

Thus, we have an expression for the virtual Adams operations.

ψ~k(1ml)=i=1dζl1ζsi11msi.\widetilde{\psi}^{k}(1_{ml})=\sum_{i=1}^{d}\frac{\zeta^{-l}-1}{\zeta^{-s_{i}}-1}1_{ms_{i}}.

Suppose l0l\neq 0, and recall equation (13) for Γ1(10l)\Gamma^{-1}(1_{0l}). Set

h(x0)\displaystyle h(x_{0}) =ψ~k(Γ1(10l))\displaystyle=\widetilde{\psi}^{k}(\Gamma^{-1}(1_{0l}))
=ψk(Γ1(10l))\displaystyle=\psi^{k}(\Gamma^{-1}(1_{0l}))
=ζln(ζl1)(x0k1)x0kn1x0kζl\displaystyle=\frac{\zeta^{l}}{n(\zeta^{l}-1)}\cdot(x_{0}^{k}-1)\frac{x_{0}^{kn}-1}{x_{0}^{k}-\zeta^{l}}
=f(x0)(x01)(x0n1)+g(x0),\displaystyle=f(x_{0})(x_{0}-1)(x_{0}^{n}-1)+g(x_{0}),

where the degree of gg is at most nn. As before, if dd does not divide ll, then kylmodnky\equiv l\bmod{n} has no solutions. Then every ζy\zeta^{y} is a root of gg, and 1 is a double root. This contradicts the degree condition of gg, so we must have g(x0)=0g(x_{0})=0, and

ψ~k(10l)=0.\widetilde{\psi}^{k}(1_{0l})=0.

Assume that dd divides ll, so that kylmodnky\equiv l\bmod{n} has dd solutions, denoted sis_{i} as above. If yy is not a solution, then h(ζy)=0h(\zeta^{y})=0. In particular, we have h(1)=0=h(1)h(1)=0=h^{\prime}(1). We can then write

g(x0)=p(x0)(x01)kylmodn(x0ζy).g(x_{0})=p(x_{0})(x_{0}-1)\prod_{ky\not\equiv l\bmod{n}}(x_{0}-\zeta^{y}).

Now assume that sis_{i} is a solution.

h(ζsi)\displaystyle h(\zeta^{s_{i}}) =limx0ζsiζln(ζl1)(x0k1)x0kn1x0kζl\displaystyle=\lim_{x_{0}\to\zeta^{s_{i}}}\frac{\zeta^{l}}{n(\zeta^{l}-1)}\cdot(x_{0}^{k}-1)\frac{x_{0}^{kn}-1}{x_{0}^{k}-\zeta^{l}}
=limx0ζsiζln(ζl1)(x0k1)knx0kn1kx0k1\displaystyle=\lim_{x_{0}\to\zeta^{s_{i}}}\frac{\zeta^{l}}{n(\zeta^{l}-1)}\cdot(x_{0}^{k}-1)\frac{knx_{0}^{kn-1}}{kx_{0}^{k-1}}
=ζln(ζl1)((ζsi)k1)kn(ζsi)kn1k(ζsi)k1\displaystyle=\frac{\zeta^{l}}{n(\zeta^{l}-1)}\cdot((\zeta^{s_{i}})^{k}-1)\frac{kn(\zeta^{s_{i}})^{kn-1}}{k(\zeta^{s_{i}})^{k-1}}
=ζln(ζl1)(ζl1)knζsikζlζsi\displaystyle=\frac{\zeta^{l}}{n(\zeta^{l}-1)}\cdot(\zeta^{l}-1)\frac{kn\zeta^{-s_{i}}}{k\zeta^{l}\zeta^{-s_{i}}}
=1.\displaystyle=1.

On the other hand,

g(ζsi)\displaystyle g(\zeta^{s_{i}}) =limx0ζsip(x0)(x01)kylmodn(x0ζy)\displaystyle=\lim_{x_{0}\to\zeta^{s_{i}}}p(x_{0})(x_{0}-1)\prod_{ky\not\equiv l\bmod{n}}(x_{0}-\zeta^{y})
=limx0ζsip(x0)(x01)x0n1(x0ζs1)(x0ζsd)\displaystyle=\lim_{x_{0}\to\zeta^{s_{i}}}p(x_{0})(x_{0}-1)\frac{x_{0}^{n}-1}{(x_{0}-\zeta^{s_{1}})\cdots(x_{0}-\zeta^{s_{d}})}
=limx0ζsip(x0)(x01)nx0n1j=1dtj(x0ζst)\displaystyle=\lim_{x_{0}\to\zeta^{s_{i}}}p(x_{0})(x_{0}-1)\frac{nx_{0}^{n-1}}{\sum_{j=1}^{d}\prod_{t\neq j}(x_{0}-\zeta^{s_{t}})}
=p(ζsi)(ζsi1)nζsiti(ζsiζst).\displaystyle=p(\zeta^{s_{i}})(\zeta^{s_{i}}-1)\frac{n\zeta^{-s_{i}}}{\prod_{t\neq i}(\zeta^{s_{i}}-\zeta^{s_{t}})}.

Since

h(ζsi)=g(ζsi),h(\zeta^{s_{i}})=g(\zeta^{s_{i}}),

we obtain

1=p(ζsi)(ζsi1)nζsiti(ζsiζst),1=p(\zeta^{s_{i}})(\zeta^{s_{i}}-1)\frac{n\zeta^{-s_{i}}}{\prod_{t\neq i}(\zeta^{s_{i}}-\zeta^{s_{t}})},

and thus,

p(ζsi)=ζsin(ζsi1)ti(ζsiζst).p(\zeta^{s_{i}})=\frac{\zeta^{s_{i}}}{n(\zeta^{s_{i}}-1)}\prod_{t\neq i}(\zeta^{s_{i}}-\zeta^{s_{t}}).

We can then conclude that

ψ~k(10l)=i=1d10si.\widetilde{\psi}^{k}(1_{0l})=\sum_{i=1}^{d}1_{0s_{i}}.

Equation (14) gives Γ1(1m0)\Gamma^{-1}(1_{m0}) for m0m\neq 0. Define

h(xm)\displaystyle h(x_{m}) =ψ~k(Γ1(1m0))\displaystyle=\widetilde{\psi}^{k}(\Gamma^{-1}(1_{m0}))
=ψk(Γ1(1m0))θk(xm1)\displaystyle=\psi^{k}(\Gamma^{-1}(1_{m0}))\theta^{k}(x_{m}^{-1})
=1nxmkn1xmk1xmk1xm11\displaystyle=\frac{1}{n}\cdot\frac{x_{m}^{kn}-1}{x_{m}^{k}-1}\cdot\frac{x_{m}^{-k}-1}{x_{m}^{-1}-1}
=f(xm)(xmn1)+g(xm).\displaystyle=f(x_{m})(x_{m}^{n}-1)+g(x_{m}).

for gg with degree at most n1n-1. If yy is not a solution to ky0modnky\equiv 0\bmod{n}, then h(ζy)=0h(\zeta^{y})=0 so that

g(xm)=p(xm)ky0modn(x0ζy).g(x_{m})=p(x_{m})\prod_{ky\not\equiv 0\bmod{n}}(x_{0}-\zeta^{y}).

Further, if si0s_{i}\neq 0 is a solution to the equivalence, then ((ζy)k1)/((ζy)11)=0((\zeta^{y})^{-k}-1)/((\zeta^{y})^{-1}-1)=0, so we have

g(xm)=p~(xm)y0(x0ζy)=p~(xm)xmn1xm1,g(x_{m})=\tilde{p}(x_{m})\prod_{y\neq 0}(x_{0}-\zeta^{y})=\tilde{p}(x_{m})\frac{x_{m}^{n}-1}{x_{m}-1},

for some polynomial p~(xm)\tilde{p}(x_{m}). When y=0y=0,

h(1)\displaystyle h(1) =limxm11nxmkn1xmk1xmk1xm11\displaystyle=\lim_{x_{m}\to 1}\frac{1}{n}\cdot\frac{x_{m}^{kn}-1}{x_{m}^{k}-1}\cdot\frac{x_{m}^{-k}-1}{x_{m}^{-1}-1}
=limxm11nknxmkn1kxmk1kxmk1xm2\displaystyle=\lim_{x_{m}\to 1}\frac{1}{n}\cdot\frac{knx_{m}^{kn-1}}{kx_{m}^{k-1}}\cdot\frac{-kx_{m}^{-k-1}}{-x_{m}^{-2}}
=1nknkk1\displaystyle=\frac{1}{n}\cdot\frac{kn}{k}\cdot\frac{-k}{-1}
=k.\displaystyle=k.

Evaluating gg at 1 gives

g(1)\displaystyle g(1) =limxm1p~(xm)xmn1xm1\displaystyle=\lim_{x_{m}\to 1}\tilde{p}(x_{m})\frac{x_{m}^{n}-1}{x_{m}-1}
=limxm1p~(xm)nxmn11\displaystyle=\lim_{x_{m}\to 1}\tilde{p}(x_{m})\frac{nx_{m}^{n-1}}{1}
=p~(1)n.\displaystyle=\tilde{p}(1)n.

Therefore, we have g(xm)=knxmn1xm1=1m0g(x_{m})=\frac{k}{n}\frac{x_{m}^{n}-1}{x_{m}-1}=1_{m0}, and so

ψ~k(1m0)=k1m0.\widetilde{\psi}^{k}(1_{m0})=k1_{m0}.

The polynomial defining Γ1(100)\Gamma^{-1}(1_{00}) is given by equation (11). The Adams operations are

h(x0)\displaystyle h(x_{0}) =ψ~k(Γ1(100))\displaystyle=\widetilde{\psi}^{k}(\Gamma^{-1}(1_{00}))
=ψk(Γ1(100))\displaystyle=\psi^{k}(\Gamma^{-1}(1_{00}))
=12n((1n)x0k+(1+n))x0kn1x0k1\displaystyle=\frac{1}{2n}((1-n)x_{0}^{k}+(1+n))\frac{x_{0}^{kn}-1}{x_{0}^{k}-1}
=f(x0)(x01)(x0n1)+g(x0).\displaystyle=f(x_{0})(x_{0}-1)(x_{0}^{n}-1)+g(x_{0}).

where gg is a polynomial of degree at most nn. As above, if yy is not a solution to ky0modnky\equiv 0\bmod{n}, then h(ζy)=0h(\zeta^{y})=0, and we can write

g(x0)=p(x0)ky0modn(x0ζy)=p(x0)x0n1(x0ζs1)(x0ζsd).g(x_{0})=p(x_{0})\prod_{ky\not\equiv 0\bmod{n}}(x_{0}-\zeta^{y})=p(x_{0})\frac{x_{0}^{n}-1}{(x_{0}-\zeta^{s_{1}})\cdots(x_{0}-\zeta^{s_{d}})}.

Now suppose sis_{i} is a solution.

h(ζsi)\displaystyle h(\zeta^{s_{i}}) =limx0ζsi12n((1n)x0k+(1+n))x0kn1x0k1\displaystyle=\lim_{x_{0}\to\zeta^{s_{i}}}\frac{1}{2n}((1-n)x_{0}^{k}+(1+n))\frac{x_{0}^{kn}-1}{x_{0}^{k}-1}
=limx0ζsi12n((1n)x0k+(1+n))knx0kn1kx0k1\displaystyle=\lim_{x_{0}\to\zeta^{s_{i}}}\frac{1}{2n}((1-n)x_{0}^{k}+(1+n))\frac{knx_{0}^{kn-1}}{kx_{0}^{k-1}}
=12n((1n)(ζsi)k+(1+n))kn(ζsi)kn1k(ζsi)k1\displaystyle=\frac{1}{2n}((1-n)(\zeta^{s_{i}})^{k}+(1+n))\frac{kn(\zeta^{s_{i}})^{kn-1}}{k(\zeta^{s_{i}})^{k-1}}
=12n((1n)+(1+n))knζsikζsi\displaystyle=\frac{1}{2n}((1-n)+(1+n))\frac{kn\zeta^{-s_{i}}}{k\zeta^{-s_{i}}}
=1.\displaystyle=1.

Evaluating gg gives

g(ζsi)\displaystyle g(\zeta^{s_{i}}) =limx0ζsip(x0)x0n1(x0ζs1)(x0ζsd)\displaystyle=\lim_{x_{0}\to\zeta^{s_{i}}}p(x_{0})\frac{x_{0}^{n}-1}{(x_{0}-\zeta^{s_{1}})\cdots(x_{0}-\zeta^{s_{d}})}
=limx0ζsip(x0)nx0n1j=1dtj(x0ζst)\displaystyle=\lim_{x_{0}\to\zeta^{s_{i}}}p(x_{0})\frac{nx_{0}^{n-1}}{\sum_{j=1}^{d}\prod_{t\neq j}(x_{0}-\zeta^{s_{t}})}
=p(ζsi)nζsiti(ζsiζst).\displaystyle=p(\zeta^{s_{i}})\frac{n\zeta^{-s_{i}}}{\prod_{t\neq i}(\zeta^{s_{i}}-\zeta^{s_{t}})}.

Setting these expressions equal and solving for p(ζsi)p(\zeta^{s_{i}}) yields

p(ζsi)=ζsinti(ζsiζst).p(\zeta^{s_{i}})=\frac{\zeta^{s_{i}}}{n}\prod_{t\neq i}(\zeta^{s_{i}}-\zeta^{s_{t}}).

Since it is possible for 1 to be a double root, we take derivatives and evaluate at 1.

h(x0)\displaystyle h^{\prime}(x_{0}) =12n(k(1n)x0k1)x0kn1x0k1\displaystyle=\frac{1}{2n}(k(1-n)x_{0}^{k-1})\frac{x_{0}^{kn}-1}{x_{0}^{k}-1}
+12n((1n)x0k+(1+n))(x0k1)knx0kn1(x0kn1)kx0k1(x0k1)2\displaystyle\quad+\frac{1}{2n}((1-n)x_{0}^{k}+(1+n))\frac{(x_{0}^{k}-1)knx_{0}^{kn-1}-(x_{0}^{kn}-1)kx_{0}^{k-1}}{(x_{0}^{k}-1)^{2}}
h(1)\displaystyle h^{\prime}(1) =12nk(1n)knk+12n(2)kn(n1)2\displaystyle=\frac{1}{2n}k(1-n)\frac{kn}{k}+\frac{1}{2n}(2)\frac{kn(n-1)}{2}
=0.\displaystyle=0.

Recall that s1=0s_{1}=0 in this case.

g(x0)\displaystyle g^{\prime}(x_{0}) =p(x0)x0n1t=1d(x0ζst)\displaystyle=p^{\prime}(x_{0})\frac{x_{0}^{n}-1}{\prod_{t=1}^{d}(x_{0}-\zeta^{s_{t}})}
+p(x0)nx0n1t=1d(x0ζst)(x0n1)j=1dtj(x0ζst)t=1d(x0ζst)2\displaystyle\quad+p(x_{0})\frac{nx_{0}^{n-1}\prod_{t=1}^{d}(x_{0}-\zeta^{s_{t}})-(x_{0}^{n}-1)\sum_{j=1}^{d}\prod_{t\neq j}(x_{0}-\zeta^{s_{t}})}{\prod_{t=1}^{d}(x_{0}-\zeta^{s_{t}})^{2}}
g(1)\displaystyle g^{\prime}(1) =limx01[p(x0)x0n1t=1d(x0ζst)\displaystyle=\lim_{x_{0}\to 1}\Big{[}p^{\prime}(x_{0})\frac{x_{0}^{n}-1}{\prod_{t=1}^{d}(x_{0}-\zeta^{s_{t}})}
+p(x0)nx0n1t=1d(x0ζst)(x0n1)j=1dtj(x0ζst)t=1d(x0ζst)2]\displaystyle\quad+p(x_{0})\frac{nx_{0}^{n-1}\prod_{t=1}^{d}(x_{0}-\zeta^{s_{t}})-(x_{0}^{n}-1)\sum_{j=1}^{d}\prod_{t\neq j}(x_{0}-\zeta^{s_{t}})}{\prod_{t=1}^{d}(x_{0}-\zeta^{s_{t}})^{2}}\Big{]}
=p(1)nt=2d(1ζst)\displaystyle=p^{\prime}(1)\frac{n}{\prod_{t=2}^{d}(1-\zeta^{s_{t}})}
+1n[t=2d(1ζst)]n(n1)t=2d(1ζst)2nj=2dtjor 1(1ζst)2t=2d(1ζst)2.\displaystyle\quad+\frac{1}{n}\Big{[}\prod_{t=2}^{d}(1-\zeta^{s_{t}})\Big{]}\frac{n(n-1)\prod_{t=2}^{d}(1-\zeta^{s_{t}})-2n\sum_{j=2}^{d}\prod_{t\neq j\,\textup{or}\,1}(1-\zeta^{s_{t}})}{2\prod_{t=2}^{d}(1-\zeta^{s_{t}})^{2}}.

So we have

0\displaystyle 0 =p(1)nt=2d(1ζst)\displaystyle=p^{\prime}(1)\frac{n}{\prod_{t=2}^{d}(1-\zeta^{s_{t}})}
+1n[t=2d(1ζst)]n(n1)t=2d(1ζst)2nj=2dtjor 1(1ζst)2t=2d(1ζst)2\displaystyle\quad+\frac{1}{n}\Big{[}\prod_{t=2}^{d}(1-\zeta^{s_{t}})\Big{]}\frac{n(n-1)\prod_{t=2}^{d}(1-\zeta^{s_{t}})-2n\sum_{j=2}^{d}\prod_{t\neq j\,\textup{or}\,1}(1-\zeta^{s_{t}})}{2\prod_{t=2}^{d}(1-\zeta^{s_{t}})^{2}}
0\displaystyle 0 =2np(1)+(n1)t=2d(1ζst)2j=2dtjor 1(1ζst)\displaystyle=2np^{\prime}(1)+(n-1)\prod_{t=2}^{d}(1-\zeta^{s_{t}})-2\sum_{j=2}^{d}\prod_{t\neq j\,\textup{or}\,1}(1-\zeta^{s_{t}})
p(1)\displaystyle p^{\prime}(1) =12n(2j=2dtjor 1(1ζst)(n1)t=2d(1ζst)).\displaystyle=\frac{1}{2n}\Big{(}2\sum_{j=2}^{d}\prod_{t\neq j\,\textup{or}\,1}(1-\zeta^{s_{t}})-(n-1)\prod_{t=2}^{d}(1-\zeta^{s_{t}})\Big{)}.

Consider the sum

r(x0)\displaystyle r(x_{0}) =i=1dΓ1(10si)\displaystyle=\sum_{i=1}^{d}\Gamma^{-1}(1_{0s_{i}})
=Γ1(100)+i=2dΓ1(10si)\displaystyle=\Gamma^{-1}(1_{00})+\sum_{i=2}^{d}\Gamma^{-1}(1_{0s_{i}})
=12n((1n)x0+(1+n))x0n1x01+i=2dζsin(ζsi1)(x01)x0n1x0ζsi.\displaystyle=\frac{1}{2n}((1-n)x_{0}+(1+n))\frac{x_{0}^{n}-1}{x_{0}-1}+\sum_{i=2}^{d}\frac{\zeta^{s_{i}}}{n(\zeta^{s_{i}}-1)}(x_{0}-1)\frac{x_{0}^{n}-1}{x_{0}-\zeta^{s_{i}}}.

Notice that r(ζy)r(\zeta^{y}) agrees with the expression above for all values of yy and has r(1)=h(1)r^{\prime}(1)=h^{\prime}(1), so by taking Γ\Gamma of each side, we obtain

ψ~k(100)=i=1d10si.\widetilde{\psi}^{k}(1_{00})=\sum_{i=1}^{d}1_{0s_{i}}.

Lastly, we consider the Adams operations on x00x_{00}. Recall that Γ1(x00)\Gamma^{-1}(x_{00}) is given in equation (12). Let

h(x0)\displaystyle h(x_{0}) =ψ~k(Γ1(x00))\displaystyle=\widetilde{\psi}^{k}(\Gamma^{-1}(x_{00}))
=ψk(Γ1(x00))\displaystyle=\psi^{k}(\Gamma^{-1}(x_{00}))
=12n((3n)x0k+(n1))x0kn1x0k1\displaystyle=\frac{1}{2n}((3-n)x_{0}^{k}+(n-1))\frac{x_{0}^{kn}-1}{x_{0}^{k}-1}
=f(x0)(x01)(x0n1)+g(x0).\displaystyle=f(x_{0})(x_{0}-1)(x_{0}^{n}-1)+g(x_{0}).

where the degree of gg is at most nn. When yy is not a solution to ky0modnky\equiv 0\bmod{n}, then h(ζy)=0h(\zeta^{y})=0, so that

g(x0)=p(x0)ky0modn(x0ζy)=p(x0)x0n1t=1d(x0ζst).g(x_{0})=p(x_{0})\prod_{ky\not\equiv 0\bmod{n}}(x_{0}-\zeta^{y})=p(x_{0})\frac{x_{0}^{n}-1}{\prod_{t=1}^{d}(x_{0}-\zeta^{s_{t}})}.

Assume sis_{i} is a solution to the equivalence ky0modnky\equiv 0\bmod{n}.

h(ζsi)\displaystyle h(\zeta^{s_{i}}) =limx0ζsi12n((1n)x0k+(1+n))x0kn1x0k1\displaystyle=\lim_{x_{0}\to\zeta^{s_{i}}}\frac{1}{2n}((1-n)x_{0}^{k}+(1+n))\frac{x_{0}^{kn}-1}{x_{0}^{k}-1}
=limx0ζsi12n((3n)x0k+(n1))knx0kn1kx0k1\displaystyle=\lim_{x_{0}\to\zeta^{s_{i}}}\frac{1}{2n}((3-n)x_{0}^{k}+(n-1))\frac{knx_{0}^{kn-1}}{kx_{0}^{k-1}}
=12n((3n)(ζsi)k+(n1))kn(ζsi)kn1k(ζsi)k1\displaystyle=\frac{1}{2n}((3-n)(\zeta^{s_{i}})^{k}+(n-1))\frac{kn(\zeta^{s_{i}})^{kn-1}}{k(\zeta^{s_{i}})^{k-1}}
=12n((3n)+(n1))knζsikζsi\displaystyle=\frac{1}{2n}((3-n)+(n-1))\frac{kn\zeta^{-s_{i}}}{k\zeta^{-s_{i}}}
=1\displaystyle=1
g(ζsi)\displaystyle g(\zeta^{s_{i}}) =limx0ζsip(x0)x0n1t=1d(x0ζst)\displaystyle=\lim_{x_{0}\to\zeta^{s_{i}}}p(x_{0})\frac{x_{0}^{n}-1}{\prod_{t=1}^{d}(x_{0}-\zeta^{s_{t}})}
=limx0ζsip(x0)nx0n1j=1dtj(x0ζst)\displaystyle=\lim_{x_{0}\to\zeta^{s_{i}}}p(x_{0})\frac{nx_{0}^{n-1}}{\sum_{j=1}^{d}\prod_{t\neq j}(x_{0}-\zeta^{s_{t}})}
=p(ζsi)nζsiti(ζsiζst).\displaystyle=p(\zeta^{s_{i}})\frac{n\zeta^{-s_{i}}}{\prod_{t\neq i}(\zeta^{s_{i}}-\zeta^{s_{t}})}.

Equating these two expressions and solving for p(ζsi)p(\zeta^{s_{i}}) gives

p(ζsi)=ζsinti(ζsiζst).p(\zeta^{s_{i}})=\frac{\zeta^{s_{i}}}{n}\prod_{t\neq i}(\zeta^{s_{i}}-\zeta^{s_{t}}).

As in the case of 1001_{00}, 1 may be a double root, so we take derivatives and evaluate to obtain

h(x0)\displaystyle h^{\prime}(x_{0}) =12n(k(3n)x0k1)x0kn1x0k1\displaystyle=\frac{1}{2n}(k(3-n)x_{0}^{k-1})\frac{x_{0}^{kn}-1}{x_{0}^{k}-1}
+12n((3n)x0k+(n1))(x0k1)knx0kn1(x0kn1)kx0k1(x0k1)2\displaystyle\quad+\frac{1}{2n}((3-n)x_{0}^{k}+(n-1))\frac{(x_{0}^{k}-1)knx_{0}^{kn-1}-(x_{0}^{kn}-1)kx_{0}^{k-1}}{(x_{0}^{k}-1)^{2}}
h(1)\displaystyle h^{\prime}(1) =12nk(3n)knk+12n(2)kn(n1)2\displaystyle=\frac{1}{2n}k(3-n)\frac{kn}{k}+\frac{1}{2n}(2)\frac{kn(n-1)}{2}
=k.\displaystyle=k.

The same calculation as for 1001_{00} holds for g(1)g^{\prime}(1), so we find that

p(1)=12n(2k+2j=2dtjor 1(1ζst)(n1)t=2d(1ζst)).p^{\prime}(1)=\frac{1}{2n}\Big{(}2k+2\sum_{j=2}^{d}\prod_{t\neq j\,\textup{or}\,1}(1-\zeta^{s_{t}})-(n-1)\prod_{t=2}^{d}(1-\zeta^{s_{t}})\Big{)}.

Consider the expression

r(x0)\displaystyle r(x_{0}) =kΓ1(x00)(k1)Γ1(100)+i=2dΓ1(10si)\displaystyle=k\Gamma^{-1}(x_{00})-(k-1)\Gamma^{-1}(1_{00})+\sum_{i=2}^{d}\Gamma^{-1}(1_{0s_{i}})
=k12n((3n)x0+(n1))x0n1x01+(k1)12n((1n)x0+(1+n))x0n1x01\displaystyle=k\frac{1}{2n}((3-n)x_{0}+(n-1))\frac{x_{0}^{n}-1}{x_{0}-1}+(k-1)\frac{1}{2n}((1-n)x_{0}+(1+n))\frac{x_{0}^{n}-1}{x_{0}-1}
+i=2dζsin(ζsi1)(x01)x0n1x0ζsi.\displaystyle\quad+\sum_{i=2}^{d}\frac{\zeta^{s_{i}}}{n(\zeta^{s_{i}}-1)}(x_{0}-1)\frac{x_{0}^{n}-1}{x_{0}-\zeta^{s_{i}}}.

Since r(ζy)=h(ζy)r(\zeta^{y})=h(\zeta^{y}) for all yy, and r(1)=h(1)r^{\prime}(1)=h^{\prime}(1), we must have r(x0)=h(x0)r(x_{0})=h(x_{0}). Taking Γ\Gamma of both sides gives

ψ~k(x00)=kx00(k1)100+i=2d10si.\widetilde{\psi}^{k}(x_{00})=kx_{00}-(k-1)1_{00}+\sum_{i=2}^{d}1_{0s_{i}}.

It will be useful to renormalize the generators of KlK_{l} for l0l\neq 0 and m0m\neq 0 to make the virtual multiplication and virtual Adams operations more concise.

Definition 6.2.

Let

1^ml:=11ζl1ml\hat{1}_{ml}:=\frac{1}{1-\zeta^{-l}}1_{ml}

for l0l\neq 0 and m0m\neq 0. When l=0l=0, let 1^m0=1m0\hat{1}_{m0}=1_{m0} for all mm. Similarly, let 1^0l=10l\hat{1}_{0l}=1_{0l} for all ll.

With these generators, it is immediate that whenever l0l\neq 0, the localization 𝒦l\mathcal{K}_{l} is isomorphic as a ring to the group ring [/n]\mathbb{C}[\mathbb{Z}/n\mathbb{Z}]. Since the group ring has a semisimple basis, we can construct a semsimple basis for 𝒦l\mathcal{K}_{l} as follows.

Definition 6.3.

For l0l\neq 0, set

ulq=1ni=0n1ζiq1^ilu_{l}^{q}=\frac{1}{n}\sum_{i=0}^{n-1}\zeta^{-iq}\hat{1}_{il}

for q=0,,n1q=0,\ldots,n-1. Define u00=x00100u_{0}^{0}=x_{00}-1_{00} and u0m=1m0u_{0}^{m}=1_{m0} for m0m\neq 0.

With these generators, the virtual multiplication becomes

ul1q1ul2q2=δl1l2δq1q2ul1q1,u_{l_{1}}^{q_{1}}*u_{l_{2}}^{q_{2}}=\delta_{l_{1}l_{2}}\delta_{q_{1}q_{2}}u_{l_{1}}^{q_{1}},

where δ\delta is the Kronecker delta. Note that the identity element 𝟏𝒦\mathbf{1}\in\mathcal{K} is

𝟏=l=0n110l.\mathbf{1}=\sum_{l=0}^{n-1}1_{0l}.

Given these generators with the virtual multiplication, K((1,n))K(\mathbb{P}(1,n))_{\mathbb{C}} has the following presentation:

[ulq,10l]l,q=0n1/I.\mathbb{C}[u_{l}^{q},1_{0l}]_{l,q=0}^{n-1}/I.

where the ideal II is generated by the polynomials below.

u0qu0q\displaystyle u_{0}^{q}u_{0}^{q^{\prime}} ulqulqδllδqqulq,\displaystyle u_{l}^{q}u_{l^{\prime}}^{q^{\prime}}-\delta_{ll^{\prime}}\delta_{qq^{\prime}}u_{l}^{q}, if ll and ll^{\prime} are not both 0
l=0n110l𝟏\displaystyle\sum_{l=0}^{n-1}1_{0l}-\mathbf{1} q=0n1ulq10l,\displaystyle\sum_{q=0}^{n-1}u_{l}^{q}-1_{0l}, ifl0\displaystyle\hskip 7.22743pt\textup{if}\,\,\,l\neq 0
10l10lδll10l,\displaystyle 1_{0l}1_{0l^{\prime}}-\delta_{ll^{\prime}}1_{0l}, for l=0,,n1l=0,\ldots,n-1 ulq10lδllulq,\displaystyle u_{l}^{q}1_{0l^{\prime}}-\delta_{ll^{\prime}}u_{l}^{q}, for l=0,,n1.\displaystyle\hskip 7.22743pt\textup{for $l=0,\ldots,n-1$}.

Proposition 6.1 can be restated in terms of these new generators.

Proposition 6.4.

With respect to the generators ulqu_{l}^{q} on the localization, the virtual Adams operations have the following forms for all k1k\geq 1.

ψ~k(u0q)\displaystyle\widetilde{\psi}^{k}(u_{0}^{q}) =ku0q,ifq=0,,n1\displaystyle=ku_{0}^{q},\hskip 7.22743pt\textup{if}\,\,q=0,\ldots,n-1
ψ~k(ulq)\displaystyle\widetilde{\psi}^{k}(u_{l}^{q}) ={i=1dusiq,for allq=0,,n1ifd|landl00,ifdl\displaystyle=\begin{cases}\sum_{i=1}^{d}u_{s_{i}}^{q},\hskip 7.22743pt\textup{for all}\,\,q=0,\ldots,n-1\,\,\textup{if}\,\,d\,|\,l\,\,\textup{and}\,\,l\neq 0\\ 0,\hskip 7.22743pt\textup{if}\,\,d\nmid l\end{cases}
ψ~k(𝟏)\displaystyle\widetilde{\psi}^{k}(\mathbf{1}) =𝟏.\displaystyle=\mathbf{1}.

7. The virtual line elements

Consider the augmented virtual K-ring (K(I(1,n)),,𝟏,ψ~k,ϵ~)(K(I\mathbb{P}(1,n))_{\mathbb{C}},*,\mathbf{1},\widetilde{\psi}^{k},\widetilde{\epsilon}) with \mathbb{C}-linear extensions of ψ~k\widetilde{\psi}^{k} and ϵ~\widetilde{\epsilon}.

Proposition 7.1.

The group of virtual line elements, 𝒫\mathcal{P}, is isomorphic to (/n)n×n(\mathbb{Z}/n\mathbb{Z})^{n}\times\mathbb{C}^{n} via the isomorphism

Φ:(/n)n×n𝒫\Phi:(\mathbb{Z}/n\mathbb{Z})^{n}\times\mathbb{C}^{n}\longrightarrow\mathcal{P}
(f0,,fn1;β00,,β0n1)𝟏+q=0n1l=1n1(ζlfq1)ulq+q=0n1β0qu0q.(f_{0},\ldots,f_{n-1};\beta^{0}_{0},\ldots,\beta_{0}^{n-1})\mapsto\mathbf{1}+\sum_{q=0}^{n-1}\sum_{l=1}^{n-1}(\zeta^{lf_{q}}-1)u_{l}^{q}+\sum_{q=0}^{n-1}\beta_{0}^{q}u_{0}^{q}.
Proof.

Every element in K((1,n))K(\mathbb{P}(1,n))_{\mathbb{C}} can be written as

α𝟏+l,q=0n1βlqulq\alpha\mathbf{1}+\sum_{l,q=0}^{n-1}\beta_{l}^{q}u_{l}^{q}

We would like to find the virtual line elements with respect to these generators. Recall that the virtual line elements are determined by the equation

k=ψ~k()\mathscr{L}^{k}=\widetilde{\psi}^{k}(\mathscr{L})

with respect to the virtual multiplication and Adams operations, for all positive integers kk. First, consider the case where α=0\alpha=0, so that =l,q=0n1βlqulq\mathscr{L}=\sum_{l,q=0}^{n-1}\beta_{l}^{q}u_{l}^{q}. Then if k=nk=n, we have

ψ~n()=q=0n1nβ0qu0q\widetilde{\psi}^{n}(\mathscr{L})=\sum_{q=0}^{n-1}n\beta_{0}^{q}u_{0}^{q}

and by the relations above,

n=l=1n1q=0n1(βlq)nulq.\mathscr{L}^{n}=\sum_{l=1}^{n-1}\sum_{q=0}^{n-1}(\beta_{l}^{q})^{n}u_{l}^{q}.

Thus, it is immediate that βlq=0\beta_{l}^{q}=0 for all ll and qq, which contradicts the invertibility of \mathscr{L}. Thus, we may assume that α0\alpha\neq 0. Write β~lq=βlq/α\widetilde{\beta}_{l}^{q}=\beta_{l}^{q}/\alpha and l=ndb+rl=\frac{n}{d}b+r for some bb and rr, and d=gcd(k,n)d=\gcd(k,n). In this case,

ψ~k()\displaystyle\widetilde{\psi}^{k}(\mathscr{L}) =α𝟏+q=0n1i=1dβlqusiq+q=0p1β0qku0q\displaystyle=\alpha\mathbf{1}+\sum_{q=0}^{n-1}\sum_{i=1}^{d}\beta_{l}^{q}u_{s_{i}}^{q}+\sum_{q=0}^{p-1}\beta_{0}^{q}ku_{0}^{q}
=α𝟏+q=0n1r=0nd1b=0d1βrkmodnqulq+q=0n1β0qku0q,\displaystyle=\alpha\mathbf{1}+\sum_{q=0}^{n-1}\sum_{r=0}^{\frac{n}{d}-1}\sum_{b=0}^{d-1}\beta_{rk\bmod{n}}^{q}u_{l}^{q}+\sum_{q=0}^{n-1}\beta_{0}^{q}ku_{0}^{q},

for values of k0modnk\not\equiv 0\bmod{n}. If k0modnk\equiv 0\bmod{n}, then

ψ~k()=α𝟏+q=0n1β0qku0q.\widetilde{\psi}^{k}(\mathscr{L})=\alpha\mathbf{1}+\sum_{q=0}^{n-1}\beta_{0}^{q}ku_{0}^{q}.

On the other hand,

k\displaystyle\mathscr{L}^{k} =(α𝟏+l,q=0n1βlqulq)k\displaystyle=\Big{(}\alpha\mathbf{1}+\sum_{l,q=0}^{n-1}\beta_{l}^{q}u_{l}^{q}\Big{)}^{k}
=αk(𝟏+l=1n1q=0n1β~lqulq+q=0n1β~0qu0q)k\displaystyle=\alpha^{k}\Big{(}\mathbf{1}+\sum_{l=1}^{n-1}\sum_{q=0}^{n-1}\widetilde{\beta}_{l}^{q}u_{l}^{q}+\sum_{q=0}^{n-1}\widetilde{\beta}_{0}^{q}u_{0}^{q}\Big{)}^{k}
=αk(𝟏+l=1n1q=0n1((1+β~lq)k1)ulq+kq=0n1β~0q).\displaystyle=\alpha^{k}\Big{(}\mathbf{1}+\sum_{l=1}^{n-1}\sum_{q=0}^{n-1}((1+\widetilde{\beta}_{l}^{q})^{k}-1)u_{l}^{q}+k\sum_{q=0}^{n-1}\widetilde{\beta}_{0}^{q}\Big{)}.

In order for equation (7) to be satisfied, the following three conditions must hold

αk\displaystyle\alpha^{k} =α,\displaystyle=\alpha, αk((1+β~lq)k1)\displaystyle\alpha^{k}((1+\widetilde{\beta}_{l}^{q})^{k}-1) =βrkmodnq,\displaystyle=\beta_{rk\bmod{n}}^{q}, αkkβ~0q=kβ0q.\displaystyle\alpha^{k}k\widetilde{\beta}_{0}^{q}=k\beta_{0}^{q}.

Since these equations must hold for all positive integers kk, the first condition implies that α=1\alpha=1. Thus, the second and third conditions reduce to

(1+βlq)k1\displaystyle(1+\beta_{l}^{q})^{k}-1 =βrkmodnq,\displaystyle=\beta_{rk\bmod{n}}^{q}, kβ0q\displaystyle k\beta_{0}^{q} =kβ0q.\displaystyle=k\beta_{0}^{q}.

The third condition is satisfied for any choice of β0q\beta_{0}^{q}\in\mathbb{C}. If r=1r=1 and k=nk=n, then the second equation states that

(1+βnb/d+1q)n\displaystyle(1+\beta_{nb/d+1}^{q})^{n} =0\displaystyle=0
βnb/d+1q\displaystyle\beta_{nb/d+1}^{q} =ζfq1,\displaystyle=\zeta^{f_{q}}-1,

for some fq=0,,n1f_{q}=0,\ldots,n-1. Further, each βkq\beta_{k}^{q} is determined by βnb/d+1q\beta_{nb/d+1}^{q} as follows:

(1+βnb/d+1q)k1\displaystyle(1+\beta_{nb/d+1}^{q})^{k}-1 =βkq\displaystyle=\beta_{k}^{q}
(ζfq)k1\displaystyle(\zeta^{f_{q}})^{k}-1 =βkq\displaystyle=\beta_{k}^{q}
ζkfq1\displaystyle\zeta^{kf_{q}}-1 =βkq.\displaystyle=\beta_{k}^{q}.

Therefore, a general line element in terms of the generators above has the form

=𝟏+q=0n1l=1n1(ζlfq1)ulq+q=0n1β0qu0q,\mathscr{L}=\mathbf{1}+\sum_{q=0}^{n-1}\sum_{l=1}^{n-1}(\zeta^{lf_{q}}-1)u_{l}^{q}+\sum_{q=0}^{n-1}\beta_{0}^{q}u_{0}^{q},

where fq{0,,n1}f_{q}\in\{0,\ldots,n-1\} and β0q\beta_{0}^{q}\in\mathbb{C}. Write this virtual line element as (f0,,fn1;β00,,β0n1)\mathscr{L}(f_{0},\ldots,f_{n-1};\beta_{0}^{0},\ldots,\beta_{0}^{n-1}). Let \mathscr{L} and \mathscr{L}^{\prime} be virtual line elements, then

\displaystyle\mathscr{L}\mathscr{L}^{\prime} =𝟏+q=0n1l=1n1((ζlfq1)+(ζlfq1)(ζlfq1)+(ζlfq1))ulq+q=0n1(β0q+β0q)u0q\displaystyle=\mathbf{1}+\sum_{q=0}^{n-1}\sum_{l=1}^{n-1}((\zeta^{lf_{q}}-1)+(\zeta^{lf_{q}}-1)(\zeta^{lf^{\prime}_{q}}-1)+(\zeta^{lf^{\prime}_{q}}-1))u_{l}^{q}+\sum_{q=0}^{n-1}(\beta_{0}^{q}+{\beta^{\prime}}_{0}^{q})u_{0}^{q}
=𝟏+q=0n1l=1n1(ζl(fq+fq)1)ulq+q=0n1(β0q+β0q)u0q.\displaystyle=\mathbf{1}+\sum_{q=0}^{n-1}\sum_{l=1}^{n-1}(\zeta^{l(f_{q}+f_{q}^{\prime})}-1)u_{l}^{q}+\sum_{q=0}^{n-1}(\beta_{0}^{q}+{\beta^{\prime}}_{0}^{q})u_{0}^{q}.

Thus, multiplication of virtual line elements has the following property:

=(f0+f0,,fn1+fn1;β00+β00,,β0n1+β0n1).\mathscr{L}\mathscr{L}^{\prime}=\mathscr{L}(f_{0}+f^{\prime}_{0},\ldots,f_{n-1}+f^{\prime}_{n-1};\beta_{0}^{0}+{\beta^{\prime}}_{0}^{0},\ldots,\beta_{0}^{n-1}+{\beta^{\prime}}_{0}^{n-1}). (15)

Note that this implies that (f0,,fp1;β00,,β0p1)1=(f0,,fp1;β00,,β0n1)\mathscr{L}(f_{0},\ldots,f_{p-1};\beta_{0}^{0},\ldots,\beta_{0}^{p-1})^{-1}=\mathscr{L}(-f_{0},\ldots,-f_{p-1};-\beta_{0}^{0},\ldots,-\beta_{0}^{n-1}). Thus, the map Φ\Phi as defined above is an isomorphism of groups. ∎

Proposition 7.2.

The virtual line elements 𝒫\mathcal{P} span K(I(1,n))K(I\mathbb{P}(1,n))_{\mathbb{C}} as a complex vector space.

Proof.

We show that every generator ulqu_{l}^{q} and 𝟏\mathbf{1} can be written as a linear combination of line elements. Note that 𝟏=(0,,0;0,,0)\mathbf{1}=\mathscr{L}(0,\ldots,0;0,\ldots,0). By subtracting the constant term from the line element (0,,0;0,,0,1,0,,0)\mathscr{L}(0,\ldots,0;0,\ldots,0,1,0,\ldots,0), we obtain u0qu_{0}^{q}. Let MM be the n(n1)×nnn(n-1)\times n^{n} matrix whose rows are given by pairs (l,q)(l,q) and columns by nn-tuples (f0,,fn1)(f_{0},\ldots,f_{n-1}), where an entry of MM is (ζlfq1)(\zeta^{lf_{q}}-1). Construct a submatrix AA of MM whose columns are the columns of MM determined by the nn-tuples (α,0,,0),(0,α,0,,0),,(0,,0,α)(\alpha,0,\ldots,0),\,(0,\alpha,0,\ldots,0),\ldots,\,(0,\ldots,0,\alpha) for each α/n\alpha\in\mathbb{Z}/n\mathbb{Z}. The matrix AA is a block diagonal matrix, whose diagonal blocks, BB, are the (n1)×(n1)(n-1)\times(n-1) matrices

B=(ζ1ζ21ζn11ζ21ζ2(2)1ζ2(n1)1ζn11ζ2(n1)1ζ(n1)(n1)1).B=\begin{pmatrix}\zeta-1&\zeta^{2}-1&\cdots&\zeta^{n-1}-1\\ \zeta^{2}-1&\zeta^{2(2)}-1&\cdots&\zeta^{2(n-1)}-1\\ \vdots&\vdots&\ddots&\vdots\\ \zeta^{n-1}-1&\zeta^{2(n-1)}-1&\cdots&\zeta^{(n-1)(n-1)}-1\end{pmatrix}.

Squaring the matrix BB yields the real matrix

B2=(nn2nn2nn2nnn).B^{2}=\begin{pmatrix}n&n&\cdots&2n\\ \vdots&\vdots&\iddots&\vdots\\ n&2n&\cdots&n\\ 2n&n&\cdots&n\end{pmatrix}.

This matrix has a linearly independent set of n1n-1 eigenvectors

(1111),(1100),(1010),,(1001).\begin{pmatrix}1\\ 1\\ 1\\ \vdots\\ 1\end{pmatrix},\hskip 7.22743pt\begin{pmatrix}1\\ -1\\ 0\\ \vdots\\ 0\end{pmatrix},\hskip 7.22743pt\begin{pmatrix}1\\ 0\\ -1\\ \vdots\\ 0\end{pmatrix},\cdots,\begin{pmatrix}1\\ 0\\ 0\\ \vdots\\ -1\end{pmatrix}.

As AA is block diagonal, A2A^{2} is also block diagonal with diagonal block entries equal to B2B^{2}. Thus, AA has a linearly independent set of n(n1)n(n-1) eigenvectors given by placing the eigenvectors above in a column vector with the rest of the entries equal to zero. Thus, the rank of A2A^{2} is n(n1)n(n-1), and so the determinant is nonzero. This implies the determinant of AA is nonzero, and so we conclude that the rank of AA is n(n1)n(n-1). Therefore, the rank of MM is n(n1)n(n-1) as well. ∎

8. Comparing the virtual K-theory of (1,n)\mathbb{P}(1,n) with the K-theory of its crepant resolution

To obtain the result from [6] for (1,n)\mathbb{P}(1,n), we need to show that the localization 𝒦0{\mathcal{K}_{0}}_{\mathbb{C}} with its virtual product is isomorphic as a ring to the K-theory of a resolution of the cotangent bundle of (1,n)\mathbb{P}(1,n) with the ordinary product. Recall that the K-theory ring of the resolution tensored with \mathbb{C} is isomorphic as a ring to

K(Zn)=[χ0±1,,χn1±1](χi1)(χj1),K(Z_{n})_{\mathbb{C}}=\frac{\mathbb{C}[\chi_{0}^{\pm 1},\ldots,\chi_{n-1}^{\pm 1}]}{\langle(\chi_{i}-1)(\chi_{j}-1)\rangle},

where ZnZ_{n} is the resolution of [6], discussed in Section 3 of this paper.

There are natural generators for 𝒦\mathcal{K} given by the line elements

σi\displaystyle\sigma_{i} =(0,,0,1,0,,0;0,,0)\displaystyle=\mathscr{L}(0,\ldots,0,1,0,\ldots,0;0,\ldots,0)
νj\displaystyle\nu_{j} =𝟏+δjwhere\displaystyle=\mathbf{1}+\delta_{j}\hskip 7.22743pt\textup{where}
δj\displaystyle\delta_{j} =(0,,0;0,,0,1,0,,0)(0,,0;0,,0)\displaystyle=\mathscr{L}(0,\ldots,0;0,\ldots,0,1,0,\ldots,0)-\mathscr{L}(0,\ldots,0;0,\ldots,0)
=(0,,0;0,,0,1,0,,0)𝟏,\displaystyle=\mathscr{L}(0,\ldots,0;0,\ldots,0,1,0,\ldots,0)-\mathbf{1},

where the 1 is in the iith and jjth slots, respectively. Note that in the proof of Proposition 7.2, we show that the line elements are spanned as a vector space by (0,,0,α,0,,0;0,,0)\mathscr{L}(0,\ldots,0,\alpha,0,\ldots,0;0,\ldots,0) and (0,,0;0,,0,α,0,,0)\mathscr{L}(0,\ldots,0;0,\ldots,0,\alpha,0,\ldots,0), for α/n\alpha\in\mathbb{Z}/n\mathbb{Z}. Since multiplication of line elements is additive in the entries of the \mathscr{L}’s, we see that σi\sigma_{i} and νj\nu_{j} indeed generate 𝒦\mathcal{K} as a ring.

Theorem 8.1.

The virtual K-theory ring K(I(1,n))K(I\mathbb{P}(1,n))_{\mathbb{C}} admits the following presentation:

[σ0±,,σn1±,ν0±,,νn1±]/I\mathbb{C}[\sigma_{0}^{\pm},\ldots,\sigma_{n-1}^{\pm},\nu_{0}^{\pm},\ldots,\nu_{n-1}^{\pm}]/I

where the ideal II is generated by the polynomials

σin𝟏\displaystyle\sigma_{i}^{n}-\mathbf{1} (νi𝟏)(νj𝟏)\displaystyle(\nu_{i}-\mathbf{1})(\nu_{j}-\mathbf{1})
σi(νj𝟏)(νj𝟏)\displaystyle\sigma_{i}(\nu_{j}-\mathbf{1})-(\nu_{j}-\mathbf{1}) (σi𝟏)(σj𝟏),\displaystyle(\sigma_{i}-\mathbf{1})(\sigma_{j}-\mathbf{1}), ifij,\displaystyle\hskip 7.22743pt\textup{if}\,\,i\neq j,

for the virtual line elements σi\sigma_{i} and νj\nu_{j} defined above.

Proof.

The virtual line elements σi\sigma_{i} and νj\nu_{j} generate 𝒫\mathcal{P} by the multiplication property (15) of virtual line elements. Applying this property again, the relations in the ideal II follow easily. ∎

We will now use Theorem 8.1 to find a presentation for the localization 𝒦0\mathcal{K}_{0} with complex coefficients. This will be isomorphic as a ψ\psi-ring to the ordinary K-theory of the resolution ZnZ_{n}, and by establishing an isomorphism as ψ\psi-rings between 𝒦0{\mathcal{K}_{0}}_{\mathbb{C}} and the virtual augmentation completion K^(I(1,n))\widehat{K}(I\mathbb{P}(1,n))_{\mathbb{C}} we will have a generalization of Proposition 7.22 of [6].

By the definition of the virtual line elements \mathscr{L}, the map Γ0:=Γ|𝒦0\Gamma_{0}:=\Gamma|_{\mathcal{K}_{0}} takes

𝒦\textstyle{\mathcal{K}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}Γ0\scriptstyle{\Gamma_{0}}𝒦0\textstyle{\mathcal{K}_{0}}   𝟏+l=1p1q=0p1(ζlfq1)ulq+q=0p1β0qu0q\textstyle{\mathbf{1}+\displaystyle\sum_{l=1}^{p-1}\sum_{q=0}^{p-1}(\zeta^{lf_{q}}-1)u_{l}^{q}+\displaystyle\sum_{q=0}^{p-1}\beta_{0}^{q}u_{0}^{q}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}100+q=0p1β0qu0q\textstyle{1_{00}+\displaystyle\sum_{q=0}^{p-1}\beta_{0}^{q}u_{0}^{q}}

Note that this implies that all the generators σi\sigma_{i} will be mapped to the identity 1001_{00}. In terms of the presentation of 𝒦\mathcal{K} above, we can interpret Γ0\Gamma_{0} by the canonical projection

𝒦[σ0±1,,σn1±1,ν0±1,,νn1±1]/I\textstyle{{\mathcal{K}}_{\mathbb{C}}\cong\mathbb{C}[\sigma_{0}^{\pm 1},\ldots,\sigma_{n-1}^{\pm 1},\nu_{0}^{\pm 1},\ldots,\nu_{n-1}^{\pm 1}]/I\ignorespaces\ignorespaces\ignorespaces\ignorespaces}𝒦0[σ0±1,,σn1±1,ν0±1,,νn1±1]/I\textstyle{{\mathcal{K}_{0}}_{\mathbb{C}}\cong\mathbb{C}[\sigma_{0}^{\pm 1},\ldots,\sigma_{n-1}^{\pm 1},\nu_{0}^{\pm 1},\ldots,\nu_{n-1}^{\pm 1}]/I^{\prime}}

where II^{\prime} is generated by the same polynomials as II with the addition of the polynomials σi𝟏\sigma_{i}-\mathbf{1}. Further, the target ring is isomorphic as a ring to

[ν^0±1,,ν^n1±1](ν^i1)(ν^j1)\frac{\mathbb{C}[\widehat{\nu}_{0}^{\pm 1},\ldots,\widehat{\nu}_{n-1}^{\pm 1}]}{\langle(\widehat{\nu}_{i}-1)(\widehat{\nu}_{j}-1)\rangle}

via a ring isomorphism sending νiν^i\nu_{i}\mapsto\widehat{\nu}_{i} and σj1\sigma_{j}\mapsto 1. In the proof of Proposition 7.22 of [6], it is noted that since the virtual K-theory is the quotient of the coordinate ring of a torus, the virtual augmentation completion K^(I(1,n))\widehat{K}(I\mathbb{P}(1,n))_{\mathbb{C}} is the localization of K(I(1,n))K(I\mathbb{P}(1,n))_{\mathbb{C}} at the maximal ideal corresponding to the identity of the torus. But this is exactly the localization 𝒦0\mathcal{K}_{0}. Thus, we have that

K^(I(1,n))[ν^0±1,,ν^n1±1](ν^i1)(ν^j1).\widehat{K}(I\mathbb{P}(1,n))_{\mathbb{C}}\cong\frac{\mathbb{C}[\widehat{\nu}_{0}^{\pm 1},\ldots,\widehat{\nu}_{n-1}^{\pm 1}]}{\langle(\widehat{\nu}_{i}-1)(\widehat{\nu}_{j}-1)\rangle}.

This is clearly isomorphic to the K-theory of the resolution, so we have proved the following theorem.

Theorem 8.2.

For all integers n2n\geq 2, there is an isomorphism of ψ\psi-rings

K^(I(1,n))K(Zn),\widehat{K}(I\mathbb{P}(1,n))_{\mathbb{C}}\cong K(Z_{n})_{\mathbb{C}},

with respect to the virtual product on the virtual augmentation completion and the ordinary product on the K-theory of ZnZ_{n}, where ZnZ_{n} is the toric crepant resolution of (1,n)\mathbb{P}(1,n) from [7].

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