In this blog post, we look at some examples of extensions of C*-algebras and introduce the ${Ext}$ (semi-)group. We only scrape the surface of what is known at Brown-Douglas-Fillmore theory, which as far as I know was the first place where the dual of the topological K-theory group (the ${K^1}$ -group) got its additive structure.

We will follow the book of Jensen&Thosme almost word for word. The book is a great reference for this material, but somewhat ill suited to give the “big picture” and for some reason does make use of standard terminology (the words Corona algebra, pullback diagram, dilation are never mentioned in this book, thought they are used extensively), making it hard search for other references. Also it gives very little motivation and historical background for the material covered.

Other references for this material include , , and though the latter only covers extensions of the compact operators. But any book on operator theory is likely to cover this to some extent.

1. A historical note

The birthplace of BDF-theory is the study of compact perturbations. A compact perturbation of an operator ${T}$ is are operators of the form ${T + K}$ where ${K}$ is a compact operator. There are several reasons why compact perturbations are of interest.

First of all, as the compact operators are the closest relative to the finite dimensional operators, they are the nicer operators in operator theory, and if two operators differ by a compact operator, it is quite natural to consider them as essentially the same.

Secondly, in many cases one can obtain significantly simpler expressions for the operators up to compact perturbations. To mention some – unitaries and self-adjoint operators can be represented as diagonal operators (Weyl-von Neumann theorem); an operator is Fredholm if and only if it is invertible modulo compact operators (note the “s” in operators – a Fredholm operator is invertible modulo a fixed compact operator if and only if its index is zero (see the example on the Toeplitz operator below for more on this)).

Since the compact operators ${\mathbb{K}\subset B(H)}$ sit as a closed two-sided ideal of ${B(H)}$ , one can represent the equivalence classes of compact perturbations by elements in

$Q(H):= B(H)/\mathbb{K}(H)$

called the Calkin algebra. Denote by ${\pi: B(H)\rightarrow Q(H)}$ the quotient map. Now if ${T\in B(H)}$ is a normal operator, define ${E(T) : = \pi^{-1}(C^\star(\pi(T)) ) \subset B(H)}$ . It is straightforward to check that we have produced an exact sequence

$0\rightarrow \mathbb{K}\rightarrow E(T)\rightarrow C^\star (\pi(T)) \rightarrow 0.$

Two such exact sequences are equivalent (in the sense defined below) if and only if ${T}$ and ${T'}$ are essentially unitary equivalent, so one has translated the study of essential unitary equivalence to the study of equivalences of extensions.

2. The Busby invariant of an extension

Given a short exact sequence of C*-algebras

$0\rightarrow \bb \xrightarrow[]{\iota} \mathcal{E}\xrightarrow[]{p} \aa \rightarrow 0.$

One says that ${\mathcal{E}}$ is an extension of ${\bb}$ by ${\aa}$ (thought some authors prefer to say ${\mathcal{E}}$ is an extension of ${\aa}$ by ${\bb}$ ). Throughout this post, ${\iota}$ will always denote the map ${\iota: \bb \rightarrow \mathcal{E}}$ and ${p}$ the map ${p: \mathcal{E} \rightarrow \aa}$ in the above short exact sequence.

The most natural equivalence relation for two extensions is that of isomorphism. Two extensions

$0\rightarrow \bb \rightarrow \mathcal{E}\rightarrow \aa \rightarrow 0$

$0\rightarrow \bb' \rightarrow \mathcal{E}'\rightarrow \aa' \rightarrow 0$

are said to be isomorphic if ${\aa = \aa'}$ , ${\bb = \bb'}$ and there is an isomorphism ${\mathcal{E}\rightarrow \mathcal{E}'}$ such that all square in the following diagram commute

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There is a nifty way to represent the isomorphism class of extensions of ${\bb}$ through a certain map from ${\aa}$ to the Corona algebra of ${\aa}$ by ${\bb}$ , by a *-homomorphism

$\tau: \aa \rightarrow Q(\bb)= M(\bb)/\bb$

called the Busby invariant.

It is defined implicitly as follows, for any ${e\in \mathcal{E}}$ let ${T(e)\in M(\bb)}$ be the multiplier for which

$\iota(T(e)(b)) = e \iota(b) \ \ \ \ \ (1)$

for all ${b\in \bb}$ . Since ${\iota}$ is injective, this uniquely determines ${T(e)}$ . Recall that the multiplier algebra consists of all bounded linear adjointable operators on ${\bb}$ , where ${\bb}$ is treated as a Hilbert ${\bb}$ -module over itself with the inner product

$\langle x, y \rangle := x^\star y \qquad \forall x, y \in \bb.$

The operator ${T(e)}$ is clearly linear and since

$\begin{align*} \iota(\langle T(e)(b), c\rangle) & = \iota(T(e)(b)^\star c)\\& = \iota(T(e)(b))^\star\iota(c) \\& = \iota(b)^\star e^\star \iota(c) \\& = \iota(b)^\star T(e^\star)(c) \\& = \langle b, T(e^\star)(b)\rangle \\\end{align*}$

${T(e)}$ is adjointable with adjoint ${T(e)^\star = T(e^\star)}$ . Boundedness of ${T(e)}$ is automatic, since all adjointable operators are automatically bounded (see for more on Hilbert C*-modules). Note that if ${b, b' \in \bb}$ then

$\iota (T( \iota(b') ) b) = \iota(b'b)$

i.e. ${T(\iota(b'))}$ is just the “multiplication by ${b'}$ ” map, so ${T}$ maps ${\iota(\bb)}$ bijectively onto ${\bb\subset M(\bb)}$ . It may not be injective on ${\mathcal{E}}$ though, but is certainly injective if ${\iota(\bb)}$ is an essential ideal (exercise! or see Prop. 3.12.8).

The last map we will need is a linear splitting

$s: \aa \rightarrow \mathcal{E} \qquad p\circ s = id_{\aa}.$

A priori we don’t know much about ${s}$ , other than the fact that it exists since any short exact sequence of vector spaces splits by some linear map. We can now define

Definition 1 (Busby map) – The Busby invariant (or map) is the *-homomorphism

$\tau = q_B\circ T \circ s$

where ${q_B: M(\bb)\rightarrow M(\bb)/\bb =: Q(\bb)}$ is the quotient map.

It should be emphasised that though ${T\circ s}$ depends on the choice of ${s}$ , the Busby map ${\tau}$ does not as any right inverse of ${p}$ will do, linear or not. To see this, let ${s}$ and ${s'}$ be two splittings of ${p}$ , then for any ${a\in \aa}$ we have ${s(a)- s'(a) \in \iota(\bb)}$ so ${T(s(a)) - T(s'(a)) \in \bb}$ hence ${q_B(T(s(a)) - T(s'(a))) = 0}$ .

The following theorem shows why we care about these Busby maps –

Theorem 2 – Let ${\aa}$ and ${\bb}$ be any C*-algebras. There is a 1-1 correspondence between isomorphism classes of extensions of ${\aa}$ by ${\bb}$ and *-homomrphisms ${\tau: \aa\rightarrow Q(\bb)}$ .

Proof

Let’s first show that the Busby invariant is an isomorphism invariant as the name suggests. Assume $\mathcal{E}$ and $\mathcal{E}'$ are two isomorphic extensions of $\bb$ by $\aa$ . Let ${\tau}$ and ${\tau'}$ be their associated Busby invariants. Then using equation 1 we find that

(eq 1) $\begin{align*} \iota((T'(\phi(e))(b)) &= (\phi^{-1} \circ \iota')(T'(\phi(e))(b)) \\ & = \phi^{-1}(\phi(e ) \iota'(b)) \\ & = e (\phi\circ\iota')(b) = e\iota(b) \end{align*}$

from which one can deduce that ${T' = T\circ \phi}$ (in the notation of Definition 1). Note also that if ${s: \aa \rightarrow \mathcal{E}}$ is a splitting of ${p}$ then ${s' = \phi\circ s}$ is a splitting of ${p'}$ , hence

$\begin{align*} \tau' &= q_B\circ T'\circ (\phi\circ s) \\ &= q_B\circ (T'\circ \phi) \circ s \\ &= q_B\circ T \circ s = \tau. \end{align*}$

Now given a morphism ${\tau : \aa \rightarrow Q(\bb)}$ , let’s show that there is an extension for which ${\tau}$ is the Busby invariant. To this end, let

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be the pullback diagram of ${\tau}$ and ${q_B}$ defined, as in , by

$\mathcal{E}_\tau := \{ (b, a) \in M(\bb)\oplus \aa ~|~ q_B(b) = \tau(a) \}.$

After some diagram chasing one can conclude that all squares in the above diagram commutes (and that all horizontal sequences are exact)

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where ${\iota(b) = (b,0)}$ , ${p(b,a) = a}$ and ${T(b,a) = b}$ . One can check that for any ${e\in \mathcal{E}_\tau}$ we have ${\iota(T(e)(\tilde{b})) = e\iota(b)}$ hence ${\tau}$ is indeed the Busby invariant of the above extension. Lastly, we need to show that if two extensions have the same Busby invariant, then they are isomorphic. For this, we will show that if ${\tau}$ is the Busby invariant associated to an extension

$0\rightarrow \bb \xrightarrow[]{\iota} \mathcal{E}\xrightarrow[]{p} \aa \rightarrow 0$

then the extensions is isomorphic to the previously defined extension

$0\rightarrow \bb \xrightarrow[]{\iota'} \mathcal{E}_{\tau}\xrightarrow[]{p'} \aa \rightarrow 0.$

We can use the universal properties of the pullback ${\mathcal{E}_\tau}$ to produce a commutative diagram

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Now since

$T'\circ \iota' = T\circ \iota = (T'\circ \phi) \circ \iota$

and ${T'}$ restricts to an isomorphism ${\iota'(\bb)\rightarrow \bb \subset M(\bb)}$ , we get that

$\iota' =\phi \circ \iota$

hence the two extensions are indeed isomorphic.

To summarize, for any ${\tau \in Hom(\aa, Q(\bb))}$ we have found a specific extension of ${\bb}$ by ${\aa}$ constructed as the pullback of ${Q(\bb)}$ by ${\tau}$ and ${q_B}$ . Then we have shown that an extensions has the Busby invariant ${\tau}$ if and only if it is isomorphic to this pullback extension. The benefit of using ${\tau}$ rather than explicit extensions is that it removes the dependence of ${\mathcal{E}}$ and it gives us an easier way to define the group operations on extensions later on. Additionally, one can deduce properties of the extension just by looking at its associated Busby invariant. For example, the set ${Hom({\mathbb C}, Q(\bb))}$ is in 1-1 correspondence with the projections in ${Q(\bb)}$ .

Other examples are

(Essential extensions) ${\mathcal{E}}$ is an essential extension (i.e. ${\iota(\bb)\subset \mathcal{E}}$ is an essential ideal) if and only if ${\tau}$ is injective.
(Trivial extensions) The extension is split by a ${\star}$ -homomorphism if and only if the Busby invariant lifts to *-homomorphism ${\tilde{\tau}: \aa \rightarrow M(\bb)}$ .
(Very trivial extension) The extension is isomorphic to ${0\rightarrow \aa\rightarrow \aa\oplus\bb\rightarrow \bb \rightarrow 0}$ if and only if ${\tau = 0}$ .
(Unital extensions) The extension is unital if and only if ${\tau}$ is unital.
(Invertible extensions) When ${\aa}$ is separable and ${\bb}$ is stable, the extension is invertible (meaning it defines an invertible element in the $Ext$ semigroup defined below) if and only if there is a contractive completely positive map ${s: \aa \rightarrow M(\bb)}$ such that ${\tau = q_\bb\circ s}$ .

Remark 2: () (since ${\tau(1)}$ determins ${\tau}$ and is a projection). It turns out that ${Hom({\mathbb C}, Q(\bb)) = K_0(Q(\bb))\simeq K_1(\bb)}$ .

3. Defining Ext(A,B)

In the spirit of generalized homology theories we would like an equivalence relation on the collection of extensions that is invariant under some natural notion of homotopy of extensions of operator algebras and functorial. The isomorphism classes are in this sense too strong a relation to be of use here. After (presumably much) trial and error the following equivalence relation on extensions has been the most fruitful for the definition of the Ext group

Definition 3 (Unitary equivalence) – Two extensions ${\tau_1}$ and ${\tau_2}$ of ${\aa}$ by ${\bb}$ are said to be unitarily equivalent if there is a unitary operator ${u\in M(\bb)}$ such that ${q_B(u)^\star \tau_1(a) q_B(u) = \tau_2(a)}$ for all ${a\in \aa}$ .

Since ${q_B(u)}$ is a unitary in ${Q(\bb)}$ one may wonder why we didn’t simply pick ${u \in Q(\bb)}$ in the first place. This will remain mysterious for the time being, but we mention that the two equivalence relations are not the same as not all unitaries lift from quotients (unless they are in the connected component of the identity), and indeed they yield different group structures on the Cuntz algebras ${\mathcal{O}_n }$ (see section V.6). In the above is called “strong” unitary equivalence while the equivalent definition with ${u\in Q(\bb)}$ is called “weak” unitary equivalence.

The (strong) unitary equivalence is equivalent to the existence of an isomorphism ${\phi: \mathcal{E}_1\rightarrow \mathcal{E}_2}$ and unitary ${u\in M(\bb)}$ such that the following diagram commutes

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Assume from now on that ${\bb}$ is stable that is ${\bb\otimes \mathbb{K} \simeq \bb}$ where ${\mathbb{K}}$ are the compacts operators ${l^2({\mathbb N})}.$ We will need this assumption to define addition of two equivalence classes of extensions: Let ${H}$ be an infinite dimensional (separable) Hilbert space, then the Hilbert spaces

$H^n = \underbrace{ H\oplus...\oplus H}_{n}$

is unitarily equivalent to ${H}$ . Let ${U: H\rightarrow H^n}$ be the unitary determining the isomorphism. Then we can define an isomorphism ${\mathbb{K} \simeq M_n(\mathbb{K}) = \mathbb{K}\otimes M_n({\mathbb C})}$ by ${T\mapsto U T U^\star}$ . Similarly, for any stable C*-algebra ${\bb}$ we get an isomorphism

$\bb \simeq \bb \otimes \mathbb{K} \simeq \bb \otimes M_n(\mathbb{K}) \simeq M_n(\bb \otimes \mathbb{K}) \simeq M_n(\bb).$

The isomorphisms constructed in this way are called “standard isomorphisms”. Now let ${\langle \tau_1\rangle }$ and ${\langle\tau_2\rangle}$ denote the unitary equivalence class of two extensions. Then using the above identification with a fixed isomorphism as above, we define ${\langle \tau_1 \rangle \oplus \langle \tau_2\rangle}$ to be the unitary equivalence class of the extensions associated with map

$\aa \xrightarrow[]{\tau_1\oplus \tau_2} M_2(Q(\bb)) \rightarrow Q(\bb).$

We refer the reader to Lemma 3.2.3 (and the discussion right above) for the proof this is a well defined associative product on the unitary equivalence classes of extensions of ${\aa}$ by ${\bb}$ . To see it is abelian, just conjugate with (the unitary matrix) ${q_\bb\left( \begin{smallmatrix} 0 & 1 \\ 1 & 0 \end{smallmatrix}\right)}$ .

We are now ready to define the ${Ext}$ -semigroup

Definition 4 (The Ext semigroup) – The semigroup ${\operatorname{Ext}(\aa, \bb)}$ is defined to be the collection of equivalence classes of extensions determined by ${\tau\sim \tau'}$ if there are trivial extensions ${\lambda, \lambda'}$ such that

$\langle \tau \oplus\lambda\rangle = \langle \tau'\oplus \lambda'\rangle$

with addition defined as above.

Remark (Why quotienting out the split extensions?) The reason we quotient out the split extensions in the above definition is to ensure that the abelian semigroup ${\operatorname{Ext}(\aa, \bb)}$ will at the very least be a monoid (i.e. have zero element). As mentioned in the introduction, the origins of extensions theory was the study of a single essentially normal operator ${T\in B(H)}$ , yielding extensions ${\mathbb{K}}$ by ${C(\sigma_e(T))}$ . It turned out that all split extensions of ${\mathbb{K}}$ by a commutative C*-algebra are unitarily equivalent. For this reason many references that only cover extensions of ${\mathbb{K}}$ by ${C(X)}$ introduce ${Ext}$ solely as unitary equivalence classes of extensions.

It may be helpful to have a concrete extensions corresponding to the element ${[\tau\oplus \tau']}$ . Given two equivalence classes of extension ${[\tau], [\tau']\in \operatorname{Ext}(\aa, \bb)}$ one can represent ${[\tau \oplus \tau ']}$ by the extension

$0\rightarrow M_2(\bb)\xrightarrow[\hat{\iota}]{} \widehat{\mathcal{E}} \xrightarrow[\hat{p}]{}\aa \rightarrow 0$

where

$\widehat{\mathcal{E}} = \left\{ \left. \begin{bmatrix} e_1 & b_1 \\ b_2 & e_2 \end{bmatrix} \right| e_i \in \mathcal{E}_i, b_i \in \bb, p_1(e_1) = p_2(e_2) \right\}$

and ${\hat{p} = p_1\oplus p_2: \begin{bmatrix} e_1 & b_1 \\ b_2 & e_2 \end{bmatrix} \mapsto p_i(e_i)}$ .

4. Examples

Let’s try to work through some examples to get a better grip of these group. As above, let

$0\rightarrow \bb \xrightarrow[]{\iota} \mathcal{E}\xrightarrow[]{p} \aa \rightarrow 0$

be and an extension with ${\bb}$ stable.

4.1. Trivial and useless extensions

The absolute simplest cases are the following

Example 1 – Let ${\bb}$ be unital, then ${Q(\bb) = \{0\}}$ , so ${\tau \in Hom(\aa, Q(\bb))}$ must be trivial and ${\operatorname{Ext}(\bb, \aa)= 0}$ .

A slightly less trivial, but significantly more useless example is the following

Example 2 (Silly example) – If ${M(\bb)}$ is projective then since every Busby map lifts to a *-homomorphism to ${M(\bb)}$ , ${\operatorname{Ext}(\aa, \bb)}$ is always trivial.

This is because an extension is split if and only if the Busby invariant lifts to a *-homomorphism ${\aa\rightarrow M(\bb)}$ , which it does since ${M(\bb)}$ is projective. Exercise: A projective C*-algebra is contractible. Can a contractible C*-algebra have projections? What can we deduce about the preceding example.

4.2. The Toeplitz extension

Example 3 (Unilateral shift and the Toeplitz algebra) – It is easy to see check that the unilateral shift operator ${T: l^2({\mathbb N})\rightarrow l^2({\mathbb N})}$ given by

$T(x_1,x_2, ...) = (0, x_1, x_2, ...)$

is neither invertible nor normal. It is a good exercise to show that

there is no compact ${k\in \mathbb{K}(l^2({\mathbb N}))}$ for which the operator ${T + k}$ is invertible (hint, it suffices to check this for finite rank operators)
there is no compact ${k\in \mathbb{K}(l^2({\mathbb N}))}$ for which the operator ${T + k}$ is normal
the quotient ${\pi(T)\in Q(l^2({\mathbb N}))}$ (the Calkin algebra) is both normal and invertible.

The operator ${T}$ is the prototypical example of an injective operator that is not surjective. It is also a Fredholm operator of index ${-1}$ . It turns out that a Fredholm operator is invertible up to a (fixed) compact operator if and only if its Fredholm index is zero, hence our unilateral shift has no such lift. The Toeplitz algebra ${\mathcal{T}}$ has already appeared in this blog (TODO). It is the algebra generated by ${I}$ and ${T}$ and we have seen it contains the compact operators ${\mathbb{K} = \mathbb{K}(l^2({\mathbb N}))}$ as an ideal from which we produced an extension

$0\rightarrow \mathbb{K}\xrightarrow[\iota]{} \mathcal{T}\xrightarrow[p]{} C(S^1)\rightarrow 0.$

This extension is semi-split and actually generates the group ${Ext(C(S^1), \mathbb{K})}$ . The Fredholm index of ${F}$ is then the negative of the winding number of the corresponding function ${p(F) \in C(S^1)}$ by the so called Noether–Gohberg–Krein Index Theorem.

4.3. Unitization

Example 4 (The unital extension) – The unitization of a non-unital C*-algebra ${\aa}$ is the essential trivial extension

$0 \rightarrow \aa \rightarrow \aa^1\rightarrow {\mathbb C}\rightarrow 0$

4.4. Algebras with the local lifting property

One says that a C*-algebra ${\aa}$ has the lifting property if for every C*-algebra ${\mathcal{C}}$ and completely positive contraction

$\phi: \aa \rightarrow \mathcal{C}/J$

with ${J}$ an ideal of ${\mathcal{C}}$ , ${\phi}$ lifts to a completely positive contraction ${\hat{\phi}: \aa \rightarrow \mathcal{C}}$ . The following is immediate –

Example 5 – If ${\aa}$ is separable and has the lifting property, then ${\operatorname{Ext}(\aa, \bb)}$ is a group.

Proof

Since ${\aa}$ is separable, ${\tau}$ is invertible if and only if it is semisplit (by a result of Arveson). However if ${\aa = \mathcal{E}/\bb}$ is separable and nuclear, then any map ${\phi: \aa\rightarrow \mathcal{E}/\bb}$ lift to a completely positive contractive map.

We get the immediate corollary

Example 6 – If ${\aa}$ is separable and nuclear, then ${\operatorname{Ext}(\aa, \bb)}$ is a group.

Proof

This follows by Choi-Effros lifting theorem which asserts that all separable nuclear C*-algebras have lifting property.

4.5. Commutative extensions

Here are some examples of extensions of commutative C*-algebras. As we have seen in the previous example, if we assume the algebras are separable they are automatically invertible extensions.

Example 7 – If ${\aa}$ and ${\bb}$ are commutative the ${\mathcal{E}}$ is commutative.

Proof

Since ${\mathcal{E}}$ is isomorphic to the pullback of ${Q(\bb)}$ by ${\tau: \aa\rightarrow Q(\bb)}$ and ${q_b: M(\bb)\rightarrow Q(\bb)}$ it sits as a subalgebra of ${ M(\bb)\oplus \aa}$ , which is commutative.

Example 8 – For a manifold with boundary ${\overline{X}}$ we have get a short exact sequence

$0\rightarrow C_0(X) \rightarrow C(\overline{X})\rightarrow C(\partial X) \rightarrow 0$

In the case where ${X}$ is the Poincare disk one can produce an explicit expression for a semi-splitting using classical harmonic analysis. Explicitly, for a function ${f\in C(\partial X) = C(S^1)}$ define a function

$F_f\in C(X)$

$F_f(x) = \int_{S^1}P(x, \xi)f(\xi) d\lambda$

where ${P(x, \xi)}$ is the Poisson kernel. The function ${F_f}$ can then extended to ${\overline{X}}$ , and it can be shown that ${F_f|_{\partial X} = f}$ , hence the assignment ${f\mapsto F_f}$ is a splitting of this extension. Similar explicit splittings can be constructed for arbitrary symmetric spaces of non-compact type of rank 1 using their geodesic boundary (=visual boundary = Gromov boundary = sphrere at infinity). The process is described at the final section of

Example 9 – Let $X = \R^n$ for $n\geq 2$ , then the spectrum of the Corona algebra $Q(X) = C(\beta X \backslash X)$ consists of exactly one connected component (Corollary 3.7), hence there is exactly two extension $Ext(C(X), \C)$ , one trivial and one unital, corresponding to the unital morphism $\tau: \C \to Q(X)$ . In case $n =1$ there are two connected components, hence by the same logic, four such isomorphism classes of extensions, one trivial, one unital and two corresponding to which component one sends the unit in $\C$ .

Example 10 –

5. A note on the case where $B$ is not stable

The reader may have paused at the previous extensions of commutative C*-algebras, as these are never stable. In this section we adress this issue and define what is ment by “a class in $Ext(\aa, \bb)$ representing the extension” in the case where $\bb$ is not stable.

Assuming $\aa$ and $\bb$ are separable C*-algebra, but $\bb$ not stable, then it is customary to define $Ext(\aa, \bb) := Ext(\aa, \bb\otimes \mathbb{K})$ , but this is rather uninformative. If $\tau$ is the Busby invariant of an extension of $\aa$ by $\bb$ representing an isomorphism class of such extensions, it is not clear what extension of $\aa$ by $\bb\otimes \mathbb{K}$ corresponds to $\tau$ .

The short answer is given by Remark 2 section 7 of where it is shown that there is an imbedding $Q(\bb)\to Q(\bb\otimes \mathbb{K})$ defined by sending $x\mapsto x\otimes p_1$ for a projection $p_1\in \mathbb{K}$ . This gives us a new Busby function

$\tau_K : \aa \to Q(\bb\otimes \mathbb{K})$

given by composition with the above inclusion. The extension corresponding to $\tau_K$ is the one which represents the extension $\tau$ in $Ext(\aa, \bb\otimes \mathbb{K})$ .

6. Relations to KK-theory

There is yet another way to describe the group of invertible extensions $Ext^{-1}(A, B)$ , namely, using the Kasparov $KK_1$ -group. Explicitly, let $A$ be a separable C*-algebra and $B$ a $\sigma$ -unital and stable C*-algebra, then we have an isomorphism

$KK_1(A, B) = Ext^{-1}(A, B)$

This is Corollary 18.5.4 of , though the proof is somewhat left to the imagination. Let us here sketch how to construct the map that yields the isomorphism. The main technique employed here is the KSGNS theorem, which is Kasparovs version of Stinesprings dilation theorem (Theorem 5.6 of ).

We have already mentioned that for separable $A$ , an extension $[\tau]\in Ext(A, B)$ is invertible if and only if there is a completely positive lift $s: A\to M(B)$ of $\tau$ , i.e. $s$ satisfies $\tau = q_B\circ s: A\to Q(B)$ . This is equivalent to having a completely positive splitting of the quotient map $p: E\to A$ . The reader is encuraged to show that the map $a\mapsto (s(a), a)$ is the needed splitting if we put $E$ in the standard form as described above. Alternatively, here is a complete proof

Proof

We can assume that

$\begin{align*} E &= \{(x, y ) \in M(B)\oplus A ~|~ q_B(x) = \tau(y) = q_B(s(y))\} \\ & = \{(x, y ) \in M(B)\oplus A ~|~ q_B(x - s(y))= 0\} \end{align*}$

Hence we have a well defined bijective linear map $E\to B\oplus A$ given by

$(x, y) \mapsto (x-s(y), y)$

The inverse of this map

$h: B\oplus A \to E \qquad h(w, z) = (w + s(z), z)$

is a linear combination of completely positive maps. Composing the inclusion $a \mapsto (0, a) \in B\oplus A$ with $h$ gives the desired completely positive splitting of $q$ .
.

Now the KSGNS theorem assigns to $s: A\to M(B)$ a (uniquely determined up to unitary equivalence) triple $(\pi, S, v)$ , where $S$ is a Hilbert $B$ -module, $\pi: A\to L(S)$ a $\star$ -homomorphism, $v\in L(B, S)$ such that

$s(a) = v^* \pi(a) v$

Note that this forces $v^*v = 1$ as $s$ is a splitting, hence $v$ is an isometry (on $\pi(A)S$ , but this can always be assumed to be the whole of $S$ after passing to a homotopic cycle). The Hilbert module $S$ is the completion of the algebraic tensor product $S_0 = A\otimes B$ with respect to the (degenerate) inner product

$\langle \sum_i a_i\otimes b_i, \sum_j a_j'\otimes b_j'\rangle : = \langle \sum_{i,j} b_i, s(a_ia_j') b_j \rangle_B = \sum_{i,j} b_i^* s(a_ia_j') b_j$

and the map $v$ is simply the inclusion $v(x) = 1\otimes x$ if $A$ is unital, or $v(x) = \lim_i e_i\otimes x$ (the strict limit) if $A$ is not unital, where $e_i$ is an approximate unit for $A$ . The adjoint is then $v^*(a\otimes b ) = s(a)b$ and so $v^*v = id$ and $vv^*(a\otimes b) = 1 \otimes s(a) b$ is a projection on $S$ , since $s(1)=1$ . Note that complete positivity of $s$ is precisely what is needed to ensure the above inner product is positive.

Having this at our disposal, one can define a Kasparov cycle

$[S, \pi, F] \in KK_1(A, B)$

where $F = 2vv^* - 1$ . The proof that this induces an isomorphism $Ext(A, B)^{-1} = KK_1(A, B)$ is due to Kasparov and can be found in .

Conversely, if we are given a cycle $[S_B, \pi, F] \in KK^1(A, B)$ we can produce an element in $Ext^{-1}$ by assigning to it the sequence corresponding to the map

$0\to \mathbb{K}(S_B) \to E \to A \to 0$

with corresponding Busby map $\tau = \pi: A\to M(\mathbb{K}(S_B)) = L(S_B)$ . Explicitly, we have

$E = \{ (a, h) \in A\oplus L(S_B) ~|~ [h] = [\pi(a)] \text{ in } L(S_B)/\mathbb{K}(S_B)\}$

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Theorem of the month – The Ext semi-group for C*-algebras

1. A historical note

2. The Busby invariant of an extension

3. Defining Ext(A,B)

4. Examples

4.1. Trivial and useless extensions

4.2. The Toeplitz extension

4.3. Unitization

4.4. Algebras with the local lifting property

4.5. Commutative extensions

5. A note on the case where $B$ is not stable

6. Relations to KK-theory

References

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1. A historical note

2. The Busby invariant of an extension

3. Defining Ext(A,B)

4. Examples

4.1. Trivial and useless extensions

4.2. The Toeplitz extension

4.3. Unitization

4.4. Algebras with the local lifting property

4.5. Commutative extensions

5. A note on the case where is not stable

6. Relations to KK-theory

References

Leave a Reply Cancel reply

5. A note on the case where $B$ is not stable