Idempotent completeness of KK (and why we care)

Let’s talk about idempotent completeness!

This blog will deal with idempotent completeness, as it relates to the UCT class in KK-theory. For an outside observer (i.e. a non category theorist) like myself working mostly with operator algebras it may seem abelian categories as quite elusive. The usual categories of Banach/C*-algebras is not even additive even if we use completely positive maps as morphisms; the category of vector bundles, though additive, does not admit kernels; even kasparovs KK-category whose objects are separable C*-algebras, and whose morphisms ${Hom (A, B) = KK(A, B)}$ are the KK-groups, may lack kernels and cokernels for an arbitrary morphism.

However the category KK does have many nice structural features (it is additive and monoidal, it admit a triangulation, it is Karoubian…) which often is enough for practical purposes.

1. Kernels in KK

It may be instructive to see what goes wrong in the case of the category KK when we want to find a kernel for a morphism. First recall

Definition 1 Let ${KK}$ denote the category whose objects are separable C*-algebras, and whose morphisms ${Hom(A, B) := KK(A, B)}$ are the Kasparov bivariant K-groups

By a result of Higson, the functor $KK: C^*alg \to KK$ can also be characterized by universal properties, namely, it is the the universal functor which is homotopy invariant, split exact and stable.

Since the Hom-sets are ${\Z}$ -modules ${KK}$ is an additive category, however it fails to be abelian as not every morphism has a kernel or a cokernel.

First recall that a categorical kernel a morphism ${f\in Hom(X, Y)}$ , is an object ${ker(f)}$ and a morphism (necessarily a monomorphism) ${k: Ker(f) \to X}$ for which ${f\circ k = 0}$ , satisfying the following universal property: If ${h: Z\to X}$ is any other morphism satisfying ${f\circ h}= 0$ , then it factors uniquely through ${k}$ in the sense that the following diagram commutes –

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Remark) A little bit of thought will show that we get a bijection

$\{h \in KK(Z, X)~|~ f\circ h = 0 \} ~ \leftrightarrow ~ KK(Z,ker(f))$

which is natural in ${Z}$ . More precisely, we have by Yoneda’s lemma, that an object ${ker(f)}$ is a kernel if and only if it represents the functor

$Z\mapsto \{ h\in K(Z, X) ~|~ f\circ h = 0\}$

i.e. if ${KK(Z, ker(f)) = \{ h\in K(Z, X) ~|~ f\circ h = 0\}}$ naturally in ${Z}$ . Alternatively, that ${KK(-, ker(f))}$ represents the kernel of the natural transformation

$f_*: KK(-, A) \to KK(-, B)$

given by the Kasparov product with ${f}$ . Note that the kernel ${ker(f)}$ is precisely the kernel of ${KK(B, X)\to KK(B, Y)}$ induced by ${f}$ as maps of ${\Z}$ -modules, meaning we have a short exact sequence

$0\to KK(B, ker(f)) \to KK(B, X) \to KK(B, Y) \to 0.$

It readily follows that if $Y$ is contractible (hence KK-equivalent to $0$ ) we would then

$ker(f) \sim_{kk} X.$

1.1. A map without a kernel

To show not all maps in ${KK}$ admit a kernel we use a semi-standard example from homological algebra of a non-representable functor. Set ${Y = X = \C}$ and let ${f = n \cdot id_\C}$ the multiplication by ${n}$ on the generator ${id \in KK(\C, \C)}$ . Finding a kernel for ${f}$ is equivalent to showing the functor

$Z \mapsto \{h \in KK(Z, \C) ~|~ n\cdot h = 0\}$

i.e. to the n-torsion functor, is representable. This functor is well known in the category of $\Z$ -modules to be non-representable from which one can show it is not representable in KK.

2. Idempotent completeness of KK

Assuming a morphism ${f\in KK(A, B)}$ in an arbitrary additive category admits both a kernel and a cokernel for which the sequence

$0 \to ker(f) \to A \to coker(f) \to 0$

splits. Then

$A \sim_{KK} ker(f) \oplus coker(f).$

So split kernels/cokernels are KK-equivalent to direct summands of ${A}$ . This direct sum decomposition implies (or rather, is equivalent to) the existence of an idempotent ${p_k \in KK(A, A)}$ (thought of as projection onto ${ker(f)}$ ) for which ${p_k}$ and ${(id_A - p_i)}$ split in the sense of Definition 2 below.

This is one of many good feature of such kernel/cokernel pairs. In our case however, where kernels/cokernels rarely exist, we may still wonder if split kernel/cokernel exists for some restricted family of morphisms.

Definition 2 A morphism ${e\in Hom (A, A)}$ in an arbitrary category is called idempotent if ${e^2 = e}$ . It is called split if there exists an object ${Y}$ and morphisms

$r: Y \to A \qquad s: A\to Y$

such that

$s\circ r = id_A \qquad r\circ s = e.$

Definition 3 An additive category is called Karoubian, or idempotent complete, if for any idempotent ${e \in Hom(A, A)}$ the following equivalent conditions hold

${e}$ is split by some maps ${r: R\to A}$ , ${ s: A \to R}$ (i.e. ${r\circ s = e}$ and ${s\circ r = id_R}$ )
${e}$ admits a kernel (necessarily isomorphic to ${R}$ )
${e}$ admits a cokernel (necessarily isomorphic to ${ker(1-e)}$ ).
${e}$ determines a splitting ${A \simeq Ker(e) \oplus ker(1-e)}$

The equivalence of the above is the content of [2, Proposition 6.5.4] and these notes. Karoubian categories are strictly weaker than abelian categories as we only require idempotents to admit kernels/cokernels, however we do require these kernels to be split, which is not always the case for arbitrary kernels in abelian categories. However, for idempotents, this is always the case. Let us state the following:

Corollary 4 Any abelian category is automatically Karoubian.

Proof

The only non-trivial thing to check is the existence of a splitting of the sequence

$0 \to ker(e) \xrightarrow[]{\iota} A \to Im(e) \xrightarrow{p} 0.$

for an arbitrary idempotent ${e \in Hom(A, A)}$ . Recall that for abelian categories left and right and two sided splittings are equivalent (by the splitting lemma), so we only need to prove the map ${p: A \to Im(e)}$ has a natural splitting ${r: Im(e) \to A}$ . But since ${e^2 = e}$ , the map ${e}$ itself factors through ${Im(e)}$ and obviously determines a right splitting.

Now, let us state some important tools to determine when a category is Karoubian, at least for triangulated categories (we adopt the convention of [1] throughout this post).

Proposition 5 [1,Prop. 5] A triangulated category is Karoubian if it is closed under countable coproducts.

Since the category KK admits the structure of a triangulated category with countable coproducts (see [3]), the following corollary is immediate

Corollary 6 The category KK is idempotent complete!

3. Consequences for the universal coefficient theorem

The universal coefficient theorem class (UCT class for short) or bootstrap class, is the class of C*-algebras which satisfy the KK-theoretic universal coefficient theorem class formulated by Rosenberg and Schochet. Not all separable C*-algebras are in the UCT class. It is still unclear if every separable nuclear C*-algebra is in the UCT class.

The UCT class is closed closed under direct sums, KK-equivalence, crossed products by ${\Z}$ and ${\R}$ , inductive limits, suspensions and satisfies the two out of three property, meaning if

$0\to A \to C\to B \to 0$

is an extension of C*-algebras, with 2 of the algebras in the UCT class, then so is the third. In particular if

$A\oplus B \quad \text{and } \quad B$

is in the UCT class, then so is ${A}$ . But we can do better!

Recall that a retract of a C*-algebra ${A}$ to a C*-algebra ${B}$ is a *-homomorphism ${f: A\to B}$ which admits a right inverse, meaning there is a *-homomorphism ${r: B \to A}$ such that

$f\circ r = id_B.$

Proposition 7 If ${A}$ is in the UCT class, and there are elements ${f\in KK(B,A)}$ and ${g \in KK(B, A)}$ such that ${f\circ g = id_B \in KK(B, B)}$ . Then ${B}$ is in the UCT class.

The proof of this statement can be found in [3] Section 6.2 where the authors show that the UCT-class is the smallest localizing subcategory containing ${\C}$ (see also this these). Such subcategories are automatically “thick” triangulated category in the sense of [1, Def. 2.1.6] (meaning closed under retracts).

In the above proposition the map ${f}$ is sometimes referred to as being a KK-retract as it mirrors the retractions in the category of C*-algebras. Note that any retract of C*-algebras automatically induces a KK-retract, but the converse definitely not true as the following examples suggests

Example 1) Let ${\iota : B\to A}$ be the inclusion of a subalgebra of ${A}$ and let ${E: A\to B}$ be a conditional expectation with finite Watatani index (see this overview). Recall that a contractive completely postive map such that ${E|_B = id_B}$ is called a conditional expectation. Finite index just ensures that the action of $A$ on the Hilbert $B$ -module given by “localization” $A^E$ of the Hilbert $A$ module $A$ at $E$ , are compact operators. Then the induced class ${[E] \in KK(A, B)}$ is represented by the Fredholm module

$(A^E, M_B, 0)$

where $A^E$ is $A$ treated as a Hilbert $B$ -module wrt the inner product $\langle x, y \rangle_E := E(x^*y)$ , and $M_B(x) := b x$ is the multiplication by $b$ map. The element $[E]$ is a KK-retract with right inverse ${[\iota] \in KK(B, A)}$ . Note that we have a change in order when we represent composition in KK using Kasparov products, so $id = [\iota]\otimes [E] = [E] \circ [\iota]$ . Thus $E$ is the KK-retract in this example not $[\iota]$ .

Examples 2) Let ${f: C(M) \to C(N)}$ be a K-oriented map between two smooth manifolds. Then ${f}$ induces a shriek map (or wrong-way map) ${f^! \in KK(C(N), C(M))}$ such that

$(f\otimes f^!) = Id_B \in KK(C(M), C(M))$

If follows that ${f \in KK(C(M), C(N))}$ is a KK-retract.

Example 3) If every there was a ranking of idempotents in KK-theory, Kasparov’s gamma element $\gamma \in KK_G(\C, \C)$ would probably “bring home the bacon” as they say. Its importance comes from the fact that the Baum-Connes assembly map factors through the image of the “multiplication by $\gamma$ ” map on $K_0(C_r^*(G))$ . There is also a $\gamma$ -element living in $KK(A\rtimes_rG, A \rtimes_rG)$ for a C*-algebra, obtained by first taking the exterior product of $\gamma$ with $id\in KK_G(A, A)$ , and then using the descent map. One example where this $\gamma$ element appears is in Kasparov’s Thom isomorphism theorem which succinctly states the following: Let $G$ be connected (say) Lie group and $K\subset G$ a maximal compact subgroup, then there are elements

$\alpha \in KK(A\rtimes_r G, A\otimes C_0(V) \rtimes_r K) \qquad \beta \in KK(A\otimes C_0(V) \rtimes_r K, A\rtimes_r G)$

such that

$\alpha \otimes \beta = \gamma \qquad \beta \otimes \alpha = id$

In certain cases we know $\gamma = id$ hence the above turns into a KK-equivalence. In general however, we only have a KK-retract. By idempotent completeness we thus get

$A\otimes C_0(V) \rtimes_r K \sim_{KK} A\rtimes_r G \oplus ker(\gamma).$

If we know $G$ satisfies the Baum-Connes conjecture with coefficients (if for instance $G$ is amenable), then multiplication by $\gamma$ on $K_0(A\rtimes_r G)$ is an isomorphism. We leave it as an exercise to check that $ker(\gamma) \subset KK(A\rtimes_r G, A\rtimes_r G)$ must have trivial K-theory. If we know $ker(\gamma)$ was in the UCT class it would have to be zero, but there are examples of C*-algebras with trivial K-theory which are not KK-equivalent to $0$ . So one just has to deal with that summand. In several cases $\gamma$ has been shown to be $1$ though. Thus we recover Connes Thom isomorphism which gives a KK-equivalence $A\rtimes \R \sim_{KK_{+1}} A$ as $\R$ has no non-trivial compact subgroups.