This is from the book "Notes on the Brown-Douglas-Fillmore Theorem". The page describes how the addition of $C^*$ extension works (in terms of Busby invariants). It says that $\rho$ is an injective *-homomorphism that is determined by the given diagram. And also it is stated that $Q(H_{1}) \oplus Q(H_{2})$ is a subalgebra of $Q(H_{1} \oplus H_{2})$.
My questions are:
- Does this $\rho$ have any special form? Is it somehow obvious?
- What is the goal of identifying $Q(H_{1}) \oplus Q(H_{2})$ as the subalgebra of $Q(H_{1} \oplus H_{2})$? As far as I see it we don't get any explicit summation of extensions this way anyway? Why can we identify it as a subalgebra?
- How do I check the last part of the proof of Lemma 2.6.1 that $[\tau]$ depends only on the class of $\tau_1$ and $\tau_2$?
I am having kind of a hard time to understand this addition operation in such abstract terms. $Q$ is the Calkin Algebra by the way.
The paper does not say "a" $*$-homomorphism but "the" $*$-homomorphism determined by the diagram. There is a canonical embedding of $L(H_1)\oplus L(H_2)$ into $L(H_1\oplus H_2)$ where you consider $(T_1\oplus T_2)(x\oplus y)=T_1x\oplus T_2y$. In other words, you consider $T_1\oplus T_2$ as the diagonal matrix $\begin{bmatrix}T_1&0\\0&T_2\end{bmatrix}$. So $\rho$ is the map $$ \rho(\pi_1(T_1)\oplus\pi_2(T_2))=\pi_{1,2}(T_1\oplus T_2), $$ where $\pi_{1,2}$ is the quotient map $L(H_1\oplus H_2)\to L(H_1\oplus H_2)/K(H_1\oplus H_2)$. This is well-defined and injective because $\pi_1(T_1)\oplus\pi_2(T_2)=0$ if and only if $T_1$ and $T_2$ are compact.
By the equivalent definition, $\def\ext{\operatorname{Ext}}\ext(X)$ is the set of equivalence classes of $*$-homomorphisms $\tau:C(X)\to Q(H_\tau)$. For reasons, you want to define a group operator on $\ext(X)$, so you need to give a meaning to $\tau_1+\tau_2$. These maps into Calkin algebras over different Hilbert spaces, so an actual sum makes no sense. The next best thing is a direct sum, but $\tau_1\oplus\tau_2$ maps $C(X)$ into $Q(H_1)\oplus Q(H_2)$. But this direct sum is not a Calkin algebra, so $\tau_1\oplus\tau_2$ is not in $\ext(X)$. But this is easily solve by using the canonical embedding of $Q(H_1)\oplus Q(H_2)$ inside $Q(H_1\oplus H_2)$, which is $\rho$. That's why $\tau_1+\tau_2$ is defined to be the $*$-homomorphism $C(X)\to Q(H_1\oplus H_2)$ given by $$ (\tau_1+\tau_2)f=\rho(\tau_1(f)\oplus\tau_2(f)). $$
To show that $[\tau_1+\tau_2]$ is well-defined, we need to show that if $[\tau_1]=[\tau_3]$ and $[\tau_2]=[\tau_4]$ then $[\tau_1+\tau_2]=[\tau_3+\tau_4]$. So by hypothesis there exist unitaries $U_1,U_2$ with $\tau_3=U_1\tau_1 U_1^*$ and $\tau_4=U_2\tau_2U_2^*$. Then you check that $U_1\oplus U_2$ is a unitary the implements $[\tau_1+\tau_2]=[\tau_3+\tau_4]$.