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The sum of two extension is defined in the following way:

Let $\mathcal{B}$ and $\mathcal{A}$ be two separable C*-algebras with $\mathcal{B}$ stable and $\phi,\psi: \mathcal{A} \rightarrow \mathcal{M(B)/B}$ be two extensions of $\mathcal{B}$ by $\mathcal{A}$. Then the sum of extension is defined in the following way: $$(\phi \oplus \psi)(a)=\pi(S)\phi(a)\pi(S^*)+\pi(T)\psi(a)\pi(T^*)$$ where $S,T \in \mathcal{M(B)}$ are isometries such that $SS^*+TT^*=1_{\mathcal{M(B)}}$.

I do not follow the motivation why we define the sum in this way and what is the guarantee that we find two such isometries $S$ and $T$ such that $SS^*+TT^*=1_{\mathcal{M(B)}}$.

Can you refer me some notes or papers where I can find the details of this BDF sum?

Also I do not get the idea why we define the BDF equivalance of extensions as follows:

Two extensions $\phi$ and $\psi$ are equivalent if there exists $\rho,\sigma: \mathcal{A} \rightarrow \mathcal{M(B)/B}$ trivial extensions such that $$\phi \oplus \rho \sim_u \psi \oplus \sigma$$

Any reference of notes and paper will be very helpful. Thanks in advance.

Arindam
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  • By extension, you mean the corresponding Busby invariant? – PStheman Nov 12 '19 at 22:26
  • How are you thinking of $\mathcal{M}(B)$?

    In "Elements of KK-theory" by Jensen and Thomsen, the multiplier algebra is thought of as the space of adjointable operators on $B$ as a right Hilbert $B$-module. In the first couple of sections, it is shown that $\mathcal{L}_B(\mathcal{H}_B) \simeq \mathcal{M}(\mathbb{K} \otimes B)$, which is in turn isomorphic to $\mathcal{M}(B)$ by stability. $\mathcal{H}_B$ is a lot like a countable infinite-dimensional Hilbert space, so proceed how you would in $\mathcal{B}(H)$ to get two isometries $S,T$ such that $SS^* + TT^* = 1$.

     – PStheman Nov 12 '19 at 22:47
  • For $\mathcal{M(B)}$, I am more used to the definition of Idealizer. Let $\phi: \mathcal{B} \rightarrow B(\mathcal{H})$ be a non-degenerate representation. Then idealizer of $\mathcal{B}$ =${T \in B(\mathcal{H})| T\mathcal{\phi(B)} \subseteq \mathcal{B} \text{ and } \mathcal{\phi(B)}T \subseteq \mathcal{\phi(B)}}$. we can show $\mathcal{\phi(B)}$ sits inside Idealizer of $\mathcal{B}$ as an essential ideal and Idealizer satisfies the universal property of multiplier algebra. Can you write the idea in the comment as an answer? – Arindam Nov 13 '19 at 01:58

1 Answers1

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Here's one way you can obtain the two isometries:

For any $C^*$-algebras $A$ and $B$, $M(A)\otimes M(B)$ always embeds unitally in $M(A\otimes B)$ (tensor products here are minimal). Thus if $B$ is stable, $M(B)\otimes\mathbb B$ embeds in $M(B\otimes\mathbb K)\cong M(B)$. So take two isometries $s,t\in\mathbb B$ such that $ss^*+tt^*=1_\mathbb B$, and put $S=1_{M(B)}\otimes s$, $T=1_{M(B)}\otimes t$.

For motivation, I would recommend consulting the original papers of Brown, Douglas, and Fillmore.

Aweygan
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  • Thanks for this answer, but I also don't get how this condition on summing both isometries to 1 is necessary. – craaaft Nov 15 '23 at 22:27
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    @craaaft The condition $ss^+tt^=1_{\mathbb{B}}$ implies that $SS^+TT^=1_{\mathcal{M(B)}}$. Intuitively, this means that $s$ and $t$ cut the Hilbert space into orthogonal subspaces, whose direct sum is the whole space, and such that each subspace is isomorphic to the whole space. – Aweygan Nov 16 '23 at 02:50
  • Ah ok, I wasn't aware of that. I have a more fundamental question - why can't we define the sum as just $(\phi \oplus \psi)(a)=\phi(a)+\psi(a)$? – craaaft Nov 16 '23 at 08:27
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    @craaaft The map you've defined isn't a $*$-homomorphism – Aweygan Nov 16 '23 at 17:05
  • Ok I tried to check if it is a $$-homomorphism and I end up with this: $\pi(S)\phi(ab)\pi(S)^{} + \pi(T)\psi(ab)\pi(T)^{} = \pi(S)\phi(a)\pi(S)^{}\pi(S)\phi(b)\pi(S)^{} + \pi(T)\psi(a)\pi(T)^{}\pi(S)\phi(b)\pi(S)^{} + \pi(S)\phi(a)\pi(S)^{}\pi(T)\psi(b)\pi(T)^{} + \pi(T)\psi(a)\pi(T)^{}\pi(T)\psi(b)\pi(T)^{}$ which would mean that $T^{}S=0$ and $S^{}T=0$ for the equality to be correct. Is that right? Also is there a more general lemma about how those isometries give rise to a $$-homomorphism in general? Thank you! – craaaft Nov 17 '23 at 11:49
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    @craaaft I'll answer your first question here, and suggest you make a new post on this website regarding the second. We have $T^S=0$ by the $C^$-identity: \begin{align} |T^S|^2 =|S^TT^S| =|S^(1-SS^)S| =0. \end{align*} – Aweygan Nov 17 '23 at 18:15
  • ah I see. thank you, will do! – craaaft Nov 17 '23 at 19:24
  • @craaaft You're welcome, glad to help! – Aweygan Nov 17 '23 at 22:55
  • Sorry, another question that I have is I am wondering why for example in the original paper (and other places) they don't talk at all how the BDF sum comes to be but just define it as a direct sum. Is saying that something is a direct sum imply that the sum looks in that specific way? – craaaft Dec 20 '23 at 10:22