Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $A$ be a C* algebra, $J$ an ideal, $\pi: A \to A/J$ the quotient map. Recall that the relative K theory group $K_0(A, A/J)$ consists of equivalence classes of triples $(p,q,x)$ where $p$ and $q$ are projections over $A$ and $\pi(x)$ implements a Murray-Von-Neumann equivalence between $\pi(p)$ and $\pi(q)$. One has the standard equivalence relations involving homotopies and direct sums, together with the relation $[p,q,x] = 0$ if $x$ implements a Murray-Von-Neumann equivalence between $p$ and $q$.

One can check that the sequence $K(A, A/J) \to K(A) \to K(A/J)$ is exact in the middle, where the first map is given by $[p,q,x] \mapsto [p] - [q]$. I am trying to understand how the excision theorem for K-theory of C* algebras works, and it would help if I had a direct proof of the following lemma: if the exact sequence $0 \to J \to A \to A/J \to 0$ splits then the map $K(A, A/J) \to K(A)$ is injective. It would even be helpful to understand what is going on in the simple case where $A$ is the unitalization of a nonunital C* algebra $J$ (where the relevant short exact sequence necessarily splits).

Can anyone help?

share|cite|improve this question
If the sequence $0 \rightarrow J \rightarrow A \rightarrow A/J \rightarrow 0$ splits and $A$ is unital, then $K_0(A,A/J) \cong \ker \left(K_0(A) \rightarrow K_0(A/J) \right)$. To show this w/o excision is exercise 4.10.7 on p. 114 here: – Steve Huntsman May 20 '10 at 16:02
Arg... you're right, but unfortunately they don't give any hints. I'll poke through that chapter and see if there are any useful tools, but otherwise I'm just as stuck. – Paul Siegel May 21 '10 at 19:42

3 Answers 3

The unitalization case $A=J^+$ is treated in Blackadar's $K$-theory for Operator Algebras. In Proposition 5.4.1 he directly shows that $$ K_0(J^+,J)\cong\ker(K_0(J^+)\to K_0({J^+}/J)). $$

share|cite|improve this answer

For the case where $A = C(X)$ for some compact Hausdorff space and $J$ is an ideal in $A$, the lemma you state is Theorem 2.2.10 in my book "Complex Topological K-Theory." As the title suggests, I only consider K-theory of topological spaces in my book, but it would not be too hard for you to take the material in Section 2.2 and adapt it to the K-theory of $C^*$-algebras.

share|cite|improve this answer

Although this question was posted and answered a very long time ago, I just stumbled upon it and I thought it might be worthwhile if an answer is provided near the post (it actually seems like a full answer is not yet available).

First I will assume $A$ is untital. Second I will use the fact that given $[(p',q',x')]\in K_0(A,A/J)$ I can write it as $[(\ell_k,q,\ell_k)]$ where $\pi(q)=\pi(\ell_k)=\ell_k$ where $\ell_k$ is the standard (bigger than $k\times k$) matrix over the complex numbers with ones on the first $k$ entries of the diagonal and zeros elsewhere. A sketch of the proof is:

-note that $$[(p',q',x')]=[(p'\oplus 1-p',q'\oplus 1-p', x'\oplus 1-p')]$$ -note that $$[(p'\oplus 1-p',q'\oplus 1-p', x'\oplus 1-p')]=[(\ell_k,q'',x'')]$$ - Use the fact that given a partial isometry $y$ we can find a path of unitaries in matrices of quadruple the size from the identity to $V$ such that $V^*yy^*V=y^*y$ and $yy^*Vy^*y=y$ (where we include partial isometries and projections in the upper left hand corner in bigger matrices).

-So let $u_t$ this path of unitaries for the partial isometry $\pi(x)$ and let $v_t$ a lift starting at the identity.

-Note that $$[(\ell_k,q'',x'')]=[(\ell_k, q, x)],$$ where $\ell_k=\pi(\ell_k)=\pi(q)=\pi(x)$ (use $(\ell_k,v_t^*q''v_t,v_t^*x'')$).

-Note that $$[(\ell_k, q, x)]=[(\ell_k,q,\ell_k)]$$ by a linear path $t\ell_k+(1-t)x$ since $\pi(x)=\pi(\ell_k)$.

Denote by $\sigma:A/J\rightarrow A$ the section which exists by assumption. Now suppose $r=[(\ell_k, q, \ell_k)]$ is a general element in $K_0(A,A/J)$ such that $[\ell_k]-[q]=0$ in $K_0(A)$. Then there must exist a path of unitaries $u_t$ starting at the identity such that $u_1^*qu_1=\ell_k$. So we have $r=[(\ell_k,\ell_k, u_1^*\ell_k)]$. Now denote $U:=\sigma(\pi(u_1))$ then $\pi(U)=\pi(u_1)$ and thus by a linear path $r=[(\ell_k,\ell_k, U^*\ell_k)]$. Now note that $U^*\ell_k U=\sigma\pi(u_1^*\ell_ku_1)=\ell_k$ which shows that $(\ell_k,\ell_k,U^*\ell_k)$ is degenerate thus $r=0$. So we find that the kernel of $K_0(A,A/J)$ is trivial. Which was to be shown.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.