7
$\begingroup$

There are several ways in which one can define $K$-theory for $C^*$-algebras: for $K_0(A)$ group two aproaches: algebraic (using idempotents) and topological (using projections, i.e. self-adjoint idempotents) are equivalent. However recently I read that $K_0(A)$ may be defined as $\pi_1(GL_{\infty}(A))$ (the fundamental group of invertible matrices of arbitrary size over $A$). How to prove that fact?

Improve this question
$\endgroup$
6
  • 3
    $\begingroup$ Here's a sketch. $GL_n(A)$ deformation retracts onto $U_n(A)$, so $K_1(A)$ (homotopy classes of unitaries over $A$) is isomorphic to $\pi_0(GL_\infty(A))$. But $\pi_1(GL_\infty(A)) \cong \pi_0(\Sigma GL_\infty(A)) \cong \pi_0(GL_\infty(\Sigma A)) \cong K_1(\Sigma A)$, where $\Sigma$ means suspension. Now apply Bott periodicity. (I'm being fast and loose with basepoints / unitalizations here, but that's the general idea.) $\endgroup$
    – Paul Siegel
    Commented Apr 23, 2016 at 19:12
  • 4
    $\begingroup$ Come to think of it, I find it a bit strange to think of this as a "definition" of $K_0(A)$ - Bott probably would have called this an example of "that old French trick of turning a theorem into a definition". Funny that somebody did it to Bott's own theorem. $\endgroup$
    – Paul Siegel
    Commented Apr 23, 2016 at 19:16
  • $\begingroup$ You can also recover the natural pre-ordering on K${}_0$ by looking at the set images of loops of the form $\exp 2\pi i t p$ where $p$ is a projection (in some matrix algebra over the C*-algebra). $\endgroup$
    – David Handelman
    Commented Apr 23, 2016 at 22:00
  • $\begingroup$ @Paul Siegel thank you for your comment. Looks very reasonable however let me ask: is this first isomorphism obvious? Namely suspension is defined as $\Sigma A=C_0(\mathbb{R}) \otimes A \cong C_0(\mathbb{R} \to A)$ and the fundamental group is defined in terms of loops so does these two object coincide? $\endgroup$
    – truebaran
    Commented Apr 24, 2016 at 0:51
  • 1
    $\begingroup$ @truebaran In the first isomorphism we suspend $GL_\infty(A)$ as a topological space, not a C*-algebra (I don't see any natural C*-algebra structure on it). In the second step we pass to the C*-algebra suspension of $A$, but it is here where we should be careful about basepoints and such: I think the right thing to do is use the unitalization of $\Sigma A$ instead of $\Sigma A$, so that we get $C(S^1) \otimes A$ in the case where $A$ is unital. If $A$ is non-unital then you have to be more careful, but the definition of $K_1$ is rigged to make everything work out. $\endgroup$
    – Paul Siegel
    Commented Apr 24, 2016 at 15:49

0

Reset to default

You must log in to answer this question.