3
$\begingroup$

Thank you for answering my question.

I have another question about the $K_1$ group. As you may know, some books define the $K_1$ group like below:

enter image description here

Also, it defines the $K_0$ group for an arbitrary C*-algebra like below:

enter image description here

Where $K_{00}(A)$ is the Grothendieck group for the abelian semigroup $V(A)$.

My question is how can I define $K_1(A)$ like the way of $K_0(A)$ more precisely how can I define $K_1(A)$ like below:

enter image description here

$\endgroup$
3
  • 1
    $\begingroup$ It seems when we define $K_0$ like above, we add some 0s to the matrices and when we calculate the $Ker$, we nicely remove those 0s. $\endgroup$ – Peg Leg Jonathan Jul 11 '20 at 7:41
  • $\begingroup$ anyone can help? $\endgroup$ – Peg Leg Jonathan Jul 11 '20 at 10:16
  • 1
    $\begingroup$ This is how $K_1$ is already defined. Since $K_1(\mathbb C) = 0$ you have $\ker (K_1(A^+) \to K_1(\mathbb C)) = K_1(A^+) = K_1(A)$. $\endgroup$ – Jamie Gabe Jul 11 '20 at 10:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.