LIFTING COMMUTATION RELATIONS IN CUNTZ ALGEBRAS 5 BRUCEBLACKADAR 1 0 2 t c O Abstract. Weexaminesplittingofthequotientmapfromthefullfreeprod- uctA∗B,ortheunitalfreeproductA∗CB,tothe(maximal)tensorproduct 3 A⊗B, forunital C*-algebras A andB. Such a splittingisvery rare,but we 1 showthereisoneifAandB areboththeCuntzalgebraO2 orO∞,andina fewothercases. Thesplittingisnotexplicit(andinprincipleprobablycannot ] be). WealsodescribesevereK-theoreticobstructions toasplitting. A O . h 1. Introduction t a m Lifting commutation relations from quotients of C*-algebras is a difficult and often unsolvable problem. We consider the essentially generic case where A and B [ are C*-algebras(to avoidunnecessary technicalities, we will only consider the case 3 ofunital AandB), andwehavethe quotientmapfromthe fullfree productorfull v unital free product to the tensor product. 3 8 Recall that the full free product A∗B is the universal C*-algebra generated by 1 copies of A and B with no relations (note that this free product is nonunital even 3 if A and B are unital), the unital free product A∗CB is the universal C*-algebra 0 generatedbycopiesofAandB withacommonunit,andthetensorproductA⊗B . 1 is the universal C*-algebra generated by commuting copies of A and B with a 0 common unit (all tensor products in this paper are maximal; our examples are 5 nuclear, so this is not much of an issue). There is a natural quotient map π from 1 : thefull freeproductA∗B, ortheunitalfreeproductA∗CB, tothe tensorproduct v A⊗B. i X We consider the question of whether there is a splitting (cross section) for this r quotient map, i.e. a *-homomorphism σ : A ⊗ B → A ∗ B (or to A ∗C B, not a necessarily unital) with π◦σ the identity on A⊗B. The short answer is “rarely”. Existence of a splitting for the quotient map from A∗B to A⊗B is equivalent to having a functorial procedure for beginning with two (not necessarily unital) *-homomorphismsφ:A→D and ψ :B →D of A and B into a C*-algebraD and manufacturing new *-homomorphisms φ˜: A → D and ψ˜: B → D such that φ˜(A) and φ˜(B) commute and have a common unit. “Functorial” means: (i) If f :D →D′ is a *-homomorphism, then f]◦φ=f ◦φ˜and f]◦ψ =f ◦ψ˜. (ii) If φ(A) and ψ(B) already commute and have a common unit, then φ˜= φ and ψ˜=ψ. Date:October 14,2015. 1 2 BRUCEBLACKADAR [Suchφandψdefinea*-homomorphismφ∗ψfromA∗BtoD;letφ˜andψ˜bedefined by (φ∗ψ)◦σ for a splitting σ]. Sucha functorialprocedureis automatically point- norm continuous in the sense that if φ :A→D and ψ :B →D converge point- n n norm to φ and ψ respectively, then φ˜ → φ˜ and ψ˜ → ψ˜ point-norm respectively n n [it suffices to note that φ ∗ψ →φ∗ψ point-norm]. n n Existence of a splitting is also equivalent to being able to always solve the fol- lowing lifting problem: given a C*-algebra D, a (closed) ideal J of D, and *- homomorphisms φ¯: A → D/J and ψ¯ : B → D/J such that φ¯(A) and ψ¯(B) com- mute and have a common unit, and φ¯ and ψ¯ separately lift to *-homomorphisms φ : A → D and ψ : B → D, find lifts φ˜ : A → D and ψ˜ : B → D such that φ˜(A) and φ˜(B) commute and have a common unit. We now give examples suggesting that splittings are “rare”. Example1.1. ConsiderthesimplestnontrivialC*-algebraC2. Thereisanexplicit description of C2∗CC2, which is the universal unital C*-algebra generated by two projections,asthe continuousfunctionsfrom[0,1]to M whicharediagonalatthe 2 endpoints(cf.[Bla06,IV.1.4.2]). Fromthisdescriptionitiseasytoseethatthereis nosplittingforthequotientmapfromC2∗CC2 toC2⊗C2 (whichisjustevaluation attheendpointsof[0,1]). Thereisa fortiorinosplittingforthequotientmapfrom C2∗C2 to C2⊗C2, since this quotient map factors through C2∗CC2. There is also no splitting for the quotient map from C2∗C2 to C2∗CC2 (4.1). Example 1.2. Hereis aneasierexample. Letn>1. Then the quotientmapfrom Mn∗CMn toMn⊗Mn cannotsplit(unitally). ForMn∗CMn hasMn asaquotient, but there is no nonzero homomorphismfrom Mn⊗Mn ∼=Mn2 to Mn. (Actually it cannot split nonunitally either; cf. Theorem 3.10, §4.) Similarly, if m and n (>1) are not relatively prime, there is no splitting for the quotient map from Mm ∗C Mn to Mm ⊗Mn ∼= Mmn, since Mm ∗CMn has Mp as a quotient, where p is the least common multiple of m and n, and p < mn. If m andn arerelativelyprime,the questionis moredelicate,but itseems unlikelythat Mm∗CMn contains a unital copy of Mmn (there is no such copy if either m or n is prime [RV98]). These obstructions to a splitting are K-theoretic; see section 3 for a discussion. Thereissomewhatmorehope ingeneralofgettingasplitting ifthe freeproduct is replaced by a “soft tensor product” A⊛ B where the copies of A and B are ǫ required to approximately commute (there are various ways to make this precise; cf.2.1). InthecaseofC2⊗C2,wewanttorestricttothecasewherethecommutator pq−qp of the two generating projections has small norm, say ≤ ǫ for a specified ǫ > 0. In fact, in this case, as soon as ǫ < 1 there is a splitting, as is easily seen. 2 A similar result holds for M ⊛ M , where the matrix units of the two matrix m ǫ n algebrasǫ-commute,andmoregenerallyforA⊛ B forAandB finite-dimensional, ǫ for small enough ǫ. But in only slightly more general settings there is no splitting even in the soft tensor product case: Example 1.3. If weregardC(T)⊗C(T) asthe universalC*-algebrageneratedby twocommutingunitaries,andweletC(T)⊛ C(T)bethe“softtorus,”theuniversal ǫ C*-algebrageneratedbytwounitariesuandvwithkuv−vuk≤ǫ([Exe93],[EEL91], [EE02]), then there is no splitting for the naturalquotientmapfromC(T)⊛ C(T) ǫ LIFTING COMMUTATION RELATIONS 3 to C(T)⊗C(T) for any ǫ > 0 (the Voiculescu matrices ([Voi83], [EL89]) can be used to show this is impossible). Thus a fortiori there can be no splitting for the quotient map from C(T)∗CC(T) to C(T)⊗C(T), since this quotient map factors throughC(T)⊛ C(T). Thisexamplecanbeessentiallysummarizedbysayingthat ǫ C(T2)∼=C(T)⊗C(T) is not semiprojective (2.3). Example 1.4. Not all obstructions to a splitting are K-theoretic. For example, there is no splitting for the quotient map from C([0,1])⊛ C([0,1]) to C([0,1])⊗ ǫ C([0,1])∼= C([0,1]2) for any ǫ > 0 since C([0,1]2) is not semiprojective (2.3), and hence no splitting for C([0,1])∗CC([0,1])→C([0,1])⊗C([0,1]) eventhoughthere is a splitting onthe K-theorylevel. (There is, however,a splitting for the quotient mapfromC([0,1])∗C([0,1])toC([0,1])∗CC([0,1]);infact,thereisasimpleexplicit cross section for the map from A∗C([0,1]) to A∗CC([0,1]) for any unital A.) One can ask whether the quotient map from A∗C B to A⊗B ever splits (if neither A nor B is C). Perhaps surprisingly,the answer is yes: it splits if A and B are certain Kirchberg algebras (but far from all pairs of Kirchberg algebras). For example,inoneofourmainresultsweshow(Corollary5.8)itsplitsifA=B =O 2 or if A is any semiprojective Kirchberg algebra and B = O∞. In fact, even the quotient map from A∗B to A⊗B splits in these cases. It has been known that in each of these cases, the quotient map from A⊛ B to A⊗B splits if ǫ > 0 ǫ is sufficiently small, where A⊛ B denotes the universal C*-algebra in which the ǫ standardgeneratorsandtheiradjointsinthetwoalgebrasǫ-commute(thisiseasily provedfrom the definition of semiprojectivity, cf. 2.3); but it came as a surprise to the author that there is a splitting even when no commutation condition (or any other relation) between the algebras is assumed. SeeSection2fordefinitionsandabriefdiscussionofsemiprojectivityandKirch- berg algebras, and [Bla06] for general information about C*-algebras. [Lor97] contains an extensive discussion of lifting problems and semiprojectivity for C*- algebras. Iamindebtedtotherefereeforpointingoutagapinthe proofofTheorem3.10, and for comments leading to Theorem 5.2 and an improvement in Theorem 5.3. I also thank Mikael Rørdam and Wilhelm Winter for helpful conversations. 2. Preliminaries 2.1. If A and B are separable and unital, and G and H are sets of generators for A and B respectively which are finite or sequences converging to 0, and ǫ > 0, we define A⊛ B to be the universal unital C*-algebra generated by unital copies of ǫ A and B such that k[a,b]k≤ ǫ for all a ∈G, b ∈ H. The notation does not reflect thedependence onG andH,whichisnottooimportantqualitatively;ifweinstead write A⊛ǫ,G,HB, and G′ and H′ are different sequences of generators converging to 0, then for any ǫ>0 there is an ǫ′ >0 such that the natural quotient map from A⊛ǫ,G,HB to A⊗B factors through A⊛ǫ′,G′,H′ B. We will thus fix G and H and just write A⊛ B. ǫ The natural quotient map from A⊛ǫ B to A⊗B factors through A⊛ǫ′ B for any ǫ′ <ǫ. 2.2. Recall the definition of semiprojectivity ([Bla85], [Bla06, II.8.3.7]): A separa- bleC*-algebraAissemiprojectiveif,wheneverDisaC*-algebra,(J )anincreasing n 4 BRUCEBLACKADAR sequence of closed (two-sided) ideals of D, and J = [∪J ]−, then any homomor- n phism φ: A→D/J can be partially lifted to a homomorphism ψ :A→D/J for n some sufficiently large n. IfAisunital,thenincheckingsemiprojectivityfromthedefinitionD andφmay be chosen unital, and then the partial lift ψ can be required to be unital (in fact, ψ will automatically become unital if n is sufficiently increased). There aremany knownexamples ofsemiprojectiveC*-algebras,but semiprojec- tivity is quite restrictive. For example, if X is a compact metrizable space, C(X) is semiprojective if and only if X is an ANR of dimension ≤ 1 [ST12]; the dimen- sionrestrictionis essentiallyexactlythe factthatcommutationrelationscannotbe partially lifted in general. See also 2.5. Proposition 2.3. Let A and B be unital semiprojective C*-algebras. Then the quotient map from A⊛ B to A⊗B splits (unitally) for some ǫ>0 (hence for all ǫ sufficiently small ǫ>0) if and only if A⊗B is semiprojective. Proof. Suppose A⊗B is semiprojective. Set D =A∗CB and Jn the kernel of the natural quotient map from D onto A⊛ B. Then (J ) is an increasing sequence 1/n n of closed ideals of D, and if J =[∪ J ]−, then D/J ∼=A⊗B. By semiprojectivity n n there is a (unital) partial lift of the identity map on A⊗B for some n. Conversely, suppose there is a splitting σ for the quotient map from A⊛ B to ǫ A⊗B for some ǫ > 0. Let D, (J ), φ be as in the definition of semiprojectivity, n with D and φ unital. For some sufficiently large n there are unital partial lifts ψ A of φ|A⊗1 and ψB of φ|1⊗B, which give a unital homomorphism ψ from A∗CB to D/Jn which is a lift of φ◦π, where π is the quotient map from A∗CB to A⊗B (this is essentially the argument that shows that A∗CB is semiprojective). Since the partial lifts of the generators of A and B asymptotically commute as n → ∞, if n is sufficiently increased, the map ψ factors through A⊛ B. Then ψ◦σ is a ǫ partial lift of φ, so A⊗B is semiprojective. (cid:3) 2.4. A Kirchberg algebra is a separable nuclear purely infinite (simple unital) C*- algebra. ItisnotknownwhethersuchaC*-algebraisautomaticallyinthebootstrap class for the Universal Coefficient Theorem [Bla98, 22.3.4]; a Kirchberg algebra in this bootstrap class is called a UCT Kirchberg algebra. The first Kirchberg algebras to be studied were the Cuntz algebras O , 2 ≤ n n ≤ ∞. If 2 ≤ n < ∞, O is the universal (unital) C*-algebra generated by n n isometries with mutually orthogonal range projections adding up to the identity. O∞ is the universal(unital) C*-algebrageneratedby a sequence of isometrieswith mutually orthogonal range projections. The next class was the (simple) Cuntz- Krieger algebras O . These are all UCT Kirchberg algebras. A Kirchberg (in part done also independently by Phillips) showed that the UCT Kirchberg algebras are classified by their K-theory. If A is a unital C*-algebra, write L(A) for the triple (K (A),K (A),[1 ]); L(A) is partof the Elliott invariant 0 1 A of A, and is the entire Elliott invariant of A if A is a Kirchberg algebra. A unital *-homomorphism φ from A to B induces a morphism from L(A) to L(B), i.e. a homomorphismφ∗0 :K0(A)→K0(B) with φ∗0([1A])=[1B] and a homomorphism φ∗1 : K1(A) → K1(B). The following facts are known (see [Rør02] for a good exposition): (i) If A and B are UCT Kirchberg algebras and α : L(A) → L(B) is a mor- phism, then there is a unital *-homomorphism φ: A→B with φ∗ =α. If LIFTING COMMUTATION RELATIONS 5 α is an isomorphism, φ can be chosen to be an isomorphism. In particu- lar, L(A) is a complete isomorphisminvariantof A among UCT Kirchberg algebras. (ii) If G and G are any countable abelian groups (recall that the K-groups 0 1 of any separable C*-algebra are countable), and u is any element of G , 0 then thereis a UCT KirchbergalgebraA(unique upto isomorphism)with L(A)∼=(G ,G ,u). 0 1 Cuntz showed that L(On) = (Zn−1,0,1) if n < ∞ and L(O∞) = (Z,0,1). As technical results toward the above classification, Kirchberg ([KP00]; cf. [Rør02]) showed the following facts about tensor products: (iii) For any Kirchberg algebra A (UCT or not), O ⊗A ∼= O . In particular, 2 2 O ⊗O ∼=O . 2 2 2 (iv) For any Kirchberg algebra A (UCT or not), O∞⊗A ∼= A. In particular, O∞⊗O∞ ∼=O∞. (v) The infinite tensor products O2∞ and O∞∞ are isomorphic to O2 and O∞ respectively. The isomorphism O ⊗O ∼= O (which was first shown by Elliott; cf. [Rør94]) 2 2 2 and the others are quite deep results. They are highly nonconstructive; in fact, it isinprincipleessentiallyimpossibletogiveanexplicitisomorphismofO ⊗O and 2 2 O by the results of [AC13]. 2 2.5. Ithasrecentlybeenshown[End15]thatifAisaUCTKirchbergalgebra,then A is semiprojective if and only if K∗(A) is finitely generated (in fact, this problem was the original motivation for the work of the present paper). Special cases were previously known: it is easy to show that O (n<∞) and, more generally, O for n A any A is semiprojective [Bla85]. O∞ was shown to be semiprojective in [Bla04], and more general results were obtained in [Szy02] and [Spi09]. 3. K-Theoretic Obstructions, Unital Free Product Case It turns out that there are rather severe K-theoretic obstructions to a splitting for the quotient map from A ∗C B to A ⊗ B. In this section, we will restrict attentiontoseparablenuclearC*-algebrasinthe UCTclass[Bla98, 22.3.4],sothat theKu¨nnethTheoremforTensorProducts[Bla98,23.1.3]andtheresultsof[Ger97] hold, although some of what we do here works in greater generality. The analysis is largely an elementary (but moderately complicated) exercise in group theory, so some details are omitted. Let A and B be unital C*-algebras of the above form. We recall the results of the Ku¨nneth theorem and of [Ger97]: K0(A⊗B)∼=[K0(A)⊗ZK0(B)]⊕[K1(A)⊗ZK1(B)]⊕TorZ1(K0(A),K1(B))⊕TorZ1(K1(A),K0(B)) K1(A⊗B)∼=[K0(A)⊗ZK1(B)]⊕[K1(A)⊗ZK0(B)]⊕TorZ1(K0(A),K0(B))⊕TorZ1(K1(A),K1(B)) Z where Tor denotes the Tor-functor of homological algebra. 1 From the exact sequence of [Ger97], we obtain: K (A∗B)∼=K (A)⊕K (B) 0 0 0 K (A∗B)∼=K (A)⊕K (B) 1 1 1 6 BRUCEBLACKADAR K0(A∗CB)∼=[K0(A)⊕K0(B)]/h([1A],−[1B])i K1(A∗CB)∼=K1(A)⊕K1(B)⊕Z if [1 ] and [1 ] are torsion elements of K (A) and K (B) respectively, and A B 0 0 K1(A∗CB)∼=K1(A)⊕K1(B) otherwise. In K1(A∗CB), the first two summands are well defined as subsets of the group, but the extra Z summand is not. The quotient map from A∗B to A∗C B induces the obvious maps on the K- groups. The quotient map π from A∗CB to A⊗B induces the following maps: on K0, π∗0(x,y)=x⊗[1B]+[1A]⊗y in the K0(A)⊗ZK0(B) summand; on the first two summands of K1, π∗1(x,y) = ([1A]⊗y)⊕(x⊗[1B]) in the [K0(A)⊗Z K1(B)]⊕ [K1(A)⊗ZK0(B)] summands. If there is a splitting, there must be crosssections for these maps. There will be a K-theoretic obstruction almost any time both A and B have nontrivial K , or if 1 the rank of K of both A and B is at least 2. 0 We first dispose of the extra summand of Z in K when it occurs. One can give 1 a more precise and detailed analysis of this summand, but the following argument suffices for the purposes of this paper. Lemma 3.1. If [1A] and [1B] are torsion, the extrasummand Z in K1(A∗CB) can be chosen so that the restriction φ of the quotient map to this summand maps into Z Tor (K (A),K (B)). 1 0 0 Proof. Suppose m[1 ] = 0 in K (A) and n[1 ] = 0 in K (B) for natural numbers A 0 B 0 m,n. Sincethe exactsequenceof[Ger97]isnatural(functorial),itdoesnotchange if we replace A and B by A⊗O∞ and B ⊗O∞ respectively, i.e. we may assume A and B are properly infinite. Thus A and B contain unital copies of O and m+1 On+1 respectively. There is a corresponding summand of Z in K1(Om+1∗COn+1), which is well defined as the entire group K1(Om+1∗COn+1), and the Z summand in K1(A∗C B) can be taken to be the image of this summand, which maps into TorZ(K (O ),K (O )) since this is all of K (O ⊗O ). (cid:3) 1 0 m+1 0 n+1 1 m+1 n+1 Corollary 3.2. There can be a splitting at the K-theory level only if φ is zero, i.e. the image of K1(A∗CB) in K1(A⊗B) is the same as the image of the subgroup K (A)⊕K (B). 1 1 Z Proof. Tor (K (A),K (B))isatorsiongroup,soφcannotbeinjective. Aquotient 1 0 0 map of Z can only split if it is zero or injective. (cid:3) Thus we are reduced to considering only the summands K (A) ⊕ K (B) of 1 1 K1(A∗CB). Proposition 3.3. A splitting for the quotient map from A∗CB to A⊗B is im- possible under any of the following conditions: (i) K1(A)⊗ZK1(B)6=0. (ii) (rank(K (A)))(rank(K (B))) > rank(K (A)) + rank(K (B)), or 0 0 0 0 rank(K (A))+rank(K (B))−1 if [1 ] or [1 ] has infinite order. 0 0 A B LIFTING COMMUTATION RELATIONS 7 Z (iii) Tor1(K∗(A),K∗(B))6=0. This is only the beginning of the K-theoretic obstructions. We will analyze the case where K∗(A) and K∗(B) are finitely generated. Then we have K (A)=Za⊕G , K (A)=Zn⊕G 0 0 1 1 K (B)=Zb⊕H , K (B)=Zm⊕H 0 0 1 1 for nonnegative integers a,b,n,m and finite abelian groups G ,G ,H ,H . 0 1 0 1 In order to have a splitting, from 3.3(iii) we need the ordersof G and G to be 0 1 relatively prime to the orders of H and H , so we will assume this from now on. 0 1 We also havethatK1(A)⊗ZK1(B)=0 is necessaryfor a splitting by 3.3(i), which impliesthatnandmcannotbothbe nonzero;withoutlossofgeneralityweassume m = 0. We must have Zn⊗ZH1 ∼= H1n = 0, so either n = 0 or H1 = 0. We then obtain: (1) K0(A∗CB)=(Za+b⊕G0⊕H0)/h[1A],−[1B]i (2) K (A⊗B)=Zab⊕Gb ⊕Ha 0 0 0 (3) K1(A∗CB)=Zn⊕G1⊕H1 (4) K (A⊗B)=Znb⊕Gb ⊕Ha⊕Hn 1 1 1 0 since Gi⊗ZHj =0 for all i,j. We need for the induced maps from Ki(A∗CB) to Ki(A⊗B) to be surjective and have a cross section. For this, we need Hn =0, so n=0 or H =0. 0 0 3.4. There are two trivial cases: K∗(A) is either (0,0), or (Z,0) and [1A] = 1, in whichcasethereisnorestrictiononK∗(B). (Thesituationis,ofcourse,symmetric in A and B.) 3.5. Next we can easily dispose of the case where n>0. In this case we observed earlier that H0 = H1 = 0, and from (4) we see that b ≤ 1. Thus K∗(B) is either (0,0) or (Z,0). Suppose K0(B) = Z. The map π∗1 from K1(A∗C B) ∼= Zn⊕G1 to K (A⊗B)∼=Z⊕G is multiplication by [1 ]; for this to be surjective, we need 1 1 B for [1 ] to be a generator of K (B). Thus we are in one of the trivial cases. B 0 3.6. From now on we assume n=0, i.e. K (A) and K (B) are finite groups. The 1 1 situation is now symmetric in A and B. The map from K0(A∗CB) to K0(A⊗B) must send Za+b/h[1 ],−[1 ]i onto Zab, and thus we must have ab ≤ a+b, and if A B either the torsion-free part of [1 ] or [1 ] is nonzero, we must have ab≤a+b−1. A B Thus we have that either a or b is ≤ 1, or a = b = 2 and the torsion-free part of both[1A]and[1B]is0. Butinthe casea=b=2,the mapsπ∗0 onthe torsion-free partofK arezeroandcannotbesurjective. Thusthereisnosplittinginthiscase. 0 3.7. Nowassumewithoutlossofgeneralitythata=0 or1. We firstdisposeofthe cases where [1 ] or [1 ] has finite order (which includes the case a=0). Suppose A B (K (A),K (A),[1 ])=(Z⊕G ,G ,(u,r)) 0 1 A 0 1 (K (B),K (B),[1 ])=(Zb⊕H ,H ,(v,s)) 0 1 B 0 1 8 BRUCEBLACKADAR Then K∗(A∗CB)=((Z⊕G0⊕Zb⊕H0)/h(u,r,v,s)i,G1⊕H1) K∗(A⊗B)=(Zb⊕Gb0⊕H0,Gb1⊕H1) Ifu=0, thenπ∗0 maps Zb⊕H0 to 0, sofor the mapto be surjective wemust have b≤ 1 and H0 = 0; if b =1, for π∗0 to map Z⊕G0 onto K0(A⊗B) we must have v = 1. Similarly, π∗1 maps the H1 to 0, so H1 = 0. Thus we are in a trivial case (for B). If v = 0, we argue similarly that G = G = 0 and u = 1, so we are again in a 0 1 trivial case (for A). Now consider the case (K (A),K (A),[1 ])=(G ,G ,r) 0 1 A 0 1 (K (B),K (B),[1 ])=(Zb⊕H ,H ,(v,s)) 0 1 B 0 1 where b ≥ 1 and v = 0. Then as before G = G = 0 and we are in a trivial case 0 1 (for A). The last case is where both K∗(A) and K∗(B) are torsion (finite) groups. Then K∗(A⊗B)=(0,0) and there is no K-theoretic restriction. 3.8. The remaining case is where a = 1, b ≥ 1, and [1 ] and [1 ] have infinite A B order. This is the most delicate case. Write u and v for the components of [1 ] A and[1 ]inthetorsion-freepartsofK (A)andK (B)respectively;wemayassume B 0 0 without loss of generality that u>0 and that v =(w,0,...,0) with w >0. First consider the case b=1. We have K0(A∗CB)∼=(Z⊕G0⊕Z⊕H0)/h(u,r,−w,−s)i∼=Z2/h(u,−w)i⊕G0⊕H0 and the map π∗0 to K0(A ⊗ B) = Z ⊕ Gb0 ⊕ H0 must send Z2/h(u,−w)i onto the first coordinate. This map is multiplication by w on the first coordinate plus multiplication by u on the second; thus it can only be surjective if u and w are relatively prime. In this case, it is both injective and surjective on this piece. The map π∗0 maps the G0 to the G0 and is multiplication by w. For this to be surjective (hence bijective), we must have wG = G , i.e. w must be relatively 0 0 prime to the order of G . Similarly, u must be relatively prime to the order of H . 0 0 The same argument for π∗1 shows that u and w are relatively prime to the orders of H and G respectively. 1 1 We cannot rule out the case where u and v are greater than 1 and relatively prime. For example, consider the case u = 2, v = 3. For simplicity suppose the K-groups are torsion-free, i.e. K∗(A) = (Z,0,2) and K∗(B) = (Z,0,3). Then K0(A∗CB)=Z2/h(2,−3)i with order unit [(2,0)]=[(0,3)]. K0(A⊗B)=Z with orderunit 6,and the map π∗0 sends [(x,y)] to 3x+2y, andis an isomorphism;the inverse map sends n to [(n,−n)]. It is possible that this inverse map is induced by a homomorphism on the algebra level in some cases (e.g. possibly in the case A = M2(O∞) and B = M3(O∞)) (3.12), although it is not in the case A = M2, B =M (3.11). 3 Thus the possibilities are L(A)=(Z⊕G ,G ,(u,r)), L(B)=(Z⊕H ,H ,(w,s)) 0 1 0 1 whereuisrelativelyprimetow andtotheordersofH andH ,andw isrelatively 0 1 prime to the orders of G and G . 0 1 LIFTING COMMUTATION RELATIONS 9 3.9. Now consider the case b>1. We have K0(A∗CB)∼=(Z⊕G0⊕Zb⊕H0)/h(u,r,−w,0,...,0,−s)i ∼=Z2/h(u,−w)i⊕Zb−1⊕G ⊕H 0 0 and the map π∗0 to K0(A⊗ B) = Zb ⊕Gb0 ⊕ H0 must send Z2/h(u,−w)i onto the first coordinate. This map is multiplication by w on the first coordinate plus multiplication by u on the second; thus it can only be surjective if u and w are relatively prime. In this case, it is both injective and surjective on this piece. The map π∗0 sends G0 into Gb0 and sends x to (wx,0,...,0). This can only be surjective if G0 = 0. Similarly, π∗1 sends G1 into Gb1 by the same formula, so G1 =0. Thus K∗(A)=(Z,0). However, we do not need to have u = 1. But there is a restriction: the map π∗0 sends H0 into H0 by multiplication by u; this needs to be surjective, i.e. u is relatively prime to the order of H . Similarly, uH = H , so u is relatively prime 0 1 1 to the order of H . Thus, for any b>1, we have the possibilities 1 L(A)=(Z,0,u), L(B)=(Zb⊕H ,H ,(w,0,...,0,s)) 0 1 where u is relatively prime to w and to the orders of H and H . 0 1 We summarize: Theorem 3.10. Let A and B be separable nuclear unital C*-algebras in the UCT class withfinitely generatedK-theory. Then therecan bea splittingfor the quotient map from A∗CB to A⊗B only in the following situations: (i) L(A)=(0,0,0)or (Z,0,1), L(B)arbitrary (or vice versa). Inthefirstcase L(A⊗B)=(0,0,0), and in the second L(A⊗B)∼=L(B). (ii) L(A)=(G ,G ,r), L(B)=(H ,H ,s). In this case L(A⊗B)=(0,0,0). 0 1 0 1 (iii) L(A) = (Z⊕G ,G ,(u,r)), L(B) = (Z⊕H ,H ,(w,s)), where u is rela- 0 1 0 1 tively prime to w and to the orders of H and H , and w is relatively prime 0 1 to the orders of G and G . 0 1 (iv) L(A)=(Z,0,u), L(B)=(Zb⊕H ,H ,(w,0,...,0,s)), where b>1, and u 0 1 is relatively prime to w and to the orders of H and H . 0 1 In (ii)–(iv), G ,G ,H ,H are any finite abelian groups with the orders of G and 0 1 0 1 0 G relatively prime to the orders of H and H . 1 0 1 In all cases except (ii), K∗(A∗CB) is the same as K∗(A⊗B) and the maps π∗0 and π∗1 are isomorphisms. In case (ii), we have L(A∗CB)=((G0/rG0)⊕(H0/sH0),G1⊕H1,0) i.e. the identity always has class 0. Theorem 3.10 does not guarantee a splitting in these cases; there often is not. For one thing, there may be more obstructions coming from ordered K-theory: Example 3.11. (cf. Example 1.2) Suppose K∗(A) = K∗(B) = (Z,0) but [1A] = m, [1B] = n with m,n > 1. Then K∗(A ⊗ B) ∼= (Z,0) and K∗(A ∗C B) ∼= (Z2/h(m,−n)i,0), so there is potentially a splitting on the group level. But de- pending on the actual map π∗ there may not be one at the scaled ordered group level, or even at the group level. Supposen=m. ThenK0(A∗CB)∼=Z⊕Zn,andtheorderunitinK0(A∗CB)is (n,0). TheorderunitinK0(A⊗B)isn2. Themapπ∗ :K0(A∗CB)→K0(A⊗B)is 10 BRUCEBLACKADAR multiplication by n, which is not surjective, so there canbe no crosssection. More generally, in n and m are not relatively prime, and d is their greatest common divisor and p is their least common multiple, then K0(A∗CB) ∼= Z⊕Zd and the order unit in K0(A ∗C B) is (p,0). The order unit in K0(A ⊗ B) is mn; π∗ is multiplication by mn =d and is not surjective. p Inthesecasesthereisnotasplittingatthegrouplevel. Butsupposemandnare relativelyprime. Thenthereisasplittingatthegrouplevel(π∗ isanisomorphism). ButthepositiveconeinK (A⊗B)canbelargerthanthepositiveconeintheunital 0 freeproductK0(A∗CB),forexampleifA=Mm,B =Mn;inthiscasethepositive coneinK (A⊗B)∼=Zistheusualone,butitisshownin[RV98]that(atleastifm 0 or n is prime) the positive cone in K0(A∗CB) is the subsemigroup of Z generated by m and n. Example3.12. ThesituationcanbenicerwithKirchbergalgebras: theunitalfree product M2(O∞)∗CM3(O∞) contains a unital copy of M6. Let {eij :1≤i,j ≤3} be the standard matrix units in the M3(O∞). There are mutually orthogonal subprojectionsf1,f2,f3,g1,g2ofe11inM3(O∞)addinguptoe11suchthatf1,f2,f3 areequivalenttoe andg andg areequivalent(take[f ]=[e ]and[g ]=−[e ] 11 1 2 k 11 k 11 for all k in K0(O∞)∼=Z). Then f +f +f ∼1 . 1 2 3 M3(O∞) We have that 1 =1 since the free product is unital, and 1 is M3(O∞) M2(O∞) M2(O∞) the sum of two equivalent projections in the M2(O∞), so f1+f2+f3 is the sum of two equivalent projections r ,r . Set p = r +g and p = r +g ; then 1 2 11 1 1 12 2 2 e =p +p ,andp ∼p . Forj =2,3setp =e p e andp =e p e ; 11 11 12 11 12 j1 j1 11 1j j2 j1 12 1j then {p :1≤j ≤3,1≤k≤2} jk are six mutually orthogonal equivalent projections in M2(O∞)∗CM3(O∞) adding up to the identity. This argument generalizes to show that if m and n are relatively prime, then Mm(O∞)∗CMn(O∞)containsa unitalcopyofMmn. Thus thereis no K-theoretic obstruction to a splitting in this case. Example 3.13. IfAorB doesnothavefinitely generatedK-theory,the situation can be much more complicated. For example, if A is the CAR algebra, there does not appear to be any K-theoretic obstruction to a cross section for the quotient map from A∗CA to A⊗A, but it is questionable that there is a splitting in this case. One unresolved question is: if there is a splitting for the quotient map from A∗C B to A⊗B, is there necessarily a unital splitting (and is a splitting even necessarilyunital)? It follows fromTheorem3.10and the comment afterwardthat in the UCT case with finitely generated K-theory, any splitting must necessarily at least send the identity of A⊗B to a projection in A∗CB whose K0-class is the same as the K -class of the identity. 0 4. K-Theoretic Obstructions, General Free Product Case We can do an essentially identical analysis of obstructions to splitting of the quotient map from A∗B to A⊗B. But this case can also be handled more easily: ifthereisasplittingforthismap,thereisalsoasplittingforthequotientmapfrom