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Let $A$ be a nonunital C*-algebra and let $I \subset A$ be a dense, $*$-closed, 2-sided ideal. I was under the impression that there existed some "obvious" argument proving that $I$ carries all the $K$-theory of $A$, but it does not seem so clear to me... For example, it's not hard to prove things like this:

Claim: For any projection $e \in M_n(A)$, there are projections in $e' \in M_n(I)$ arbitrarily near to $e$.

Sketch: Since $M_n(I)$ is dense in $M_n(A)$, we can find $x \in M_n(I)$ as near as we like to $e$. Without loss of generality, $x$ is self-adjoint or replace it by $\frac{x+x^*}{2}$ which is also close to $e$ and belongs to $M_n(I)$ since $I$ is self-adjoint. Define \begin{align*} f(t) = \begin{cases} 0 & \text{ if } t< 1/2 \\ 1/t & \text{ if } t > 1/2 \\ \end{cases} && && \chi(t) = \begin{cases} 0 & \text{ if } t < 1/2 \\ 1 & \text{ if } t > 1/2 \\ \end{cases} \end{align*} and notice that $\chi(t) = t f(t)$. Now, since $\operatorname{spec}(x) \subset \mathbb{R}$ is concentrated near $\{0,1\}$, we have that the projection $e' = \chi(x)$ is very close to $x$. Moreover, $e' = x f(x)$ belongs to $M_n(I)$ since the latter is an ideal in $M_n(A)$.

Now, at first glance, the above claim would seem to imply the map $K_0(I) \to K_0(A)$ induced by the containment $I \subset A$ is surjective, but there is a gap. By the usual definition, $K_0(A)$ is a subgroup of $K_0( \tilde A)$, where $\tilde A$ is the unitization of $A$. So, it would seem we actually need to prove the containment $\tilde I \subset \tilde A$ induces an isomorphism on $K$-theory (note that $\tilde I$ is not an ideal in $\tilde A$!). So, given any projection $e = a + e_0 \in M_n(\tilde A)$ where $a \in M_n(A)$ and $e_0$ is a projection in $M_n(\mathbb{C}$), how does one adapt the argument above to produce a nearby projection $e' \in M_n(\tilde I)$? One can go ahead and replace $a$ by $x \in M_n(I)$ close to $a$, and then show that $e' = \chi(x+e_0)$ is a projection close to $e$ with scalar part $e_0$, but I am uncertain how to show $e' - e_0$ belongs to $M_n(I)$, or indeed if this should hold in general.

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You need your dense subalgebra to be closed under holomorphic functional calculus if you want its K-groups to coincide with the K-groups of its completion. Such subalgebras are called local $C^\ast$-algebras. And you don't need it to be an ideal. –  AlexE May 1 at 21:22
    
@AlexE: So, going that route, I guess what I don't understand is the following point. Let $A$ be a nonunital $C^*$-algebra, and let $B \subset A$ be a dense $*$-subalgebra which is "local" in the sense that, for all $x \in M_n(B)$, for all $f$ holomorphic in a neighbourhood of $\operatorname{spec}_{M_n(A)}(x)$ with $f(0) =0$, one has $f(x) \in M_n(B)$. Does it then follow that $M_n(\tilde B) \subset M_n(\tilde A)$ is also closed under functional calculus? Roughly, if $B$ is local in $A$, why is $\tilde B$ local in $\tilde A$? –  Michael May 1 at 22:01
    
Also, I sort of expected that, for ideals, easier methods might suffice? –  Michael May 1 at 22:05
1  
Holomorphic functional calculus for a non-unital algebra A is defined inside its unitization $\tilde{A}$ and f(0)=0 ensures that we end up with something in A, not in $\tilde{A}$. So to show that $f(b + \lambda) \in \tilde{B}$, use functional calculus to write $b + \lambda = (id + \lambda)(b)$. Now you get $f(b + \lambda) = (f \circ (id + \lambda))(b)$ which is now in $\tilde{B}$, since B is closed under holomorphic functional calculus, but we have not necessarily $(f \circ (id + \lambda))(0) = 0$. –  AlexE May 2 at 8:23
    
+1, it is a nice trick writing $b+\lambda = (id + \lambda)(b)$. This isn't the whole story, though, since, if $b \in M_n(B)$ and $\lambda \in M_n(\mathbb{C})$, it is unclear whether $f(b + \lambda)$ can be related to a holomorphic function of $b$. –  Michael May 2 at 10:42

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