8
$\begingroup$

Up until now, I've taken it for granted that the topological k-theory of a space $X$ is equal to the K-theory of vector bundles on $X$. $K_0$'s of both coincide (Serre-Swan) however, is it the case that $K_i(\text{Vect}(X)) \cong K_{top}^i(X)$? ($\text{Vect}(X)$ being the category of (real/complex) vector bundles over $X$.)

I'm doubting this now, however, and I can't seem to find anything written up explicitly on the relationship between the two.

In what way, if at all, is topological K-theory a special case of Quillen's K-theory? What is the relationship?

$\endgroup$
  • 1
    $\begingroup$ Topological $K$-theory is a special case of the $K$-theory of operator algebras. Algebraic $K$-theory doesn't have Bott periodicity, so the higher $K$-groups needn't agree. I don't know whether the algebraic and operator algebraic $K_1$s are the same. $\endgroup$ – Kevin Carlson Dec 16 '13 at 1:15
  • 2
    $\begingroup$ Here's a reference: looks like the algebraic $K$-theory of $C^*$-algebras is not well-understood beyond $K_0$ or finite coefficients. math.uiuc.edu/K-theory/0128/Kregularity.pdf $\endgroup$ – Kevin Carlson Dec 16 '13 at 1:29
  • $\begingroup$ Rephrased the question. $\endgroup$ – Joshua Seaton Dec 16 '13 at 1:43
  • $\begingroup$ Thanks, Kevin. We wouldn't need a Bott periodicity for algebraic K-theory in general, just one for the class of rings of functions on spaces. I couldn't initally why this couldn't be, a priori. In any case though, you're right that this doesn't work. As we can take $X$ to be a point. There's a theorem of Suslin (I think) that desrcibes the alg K-theory of an algebraically closed field of char 0 and it has that K_2 is uniquely divisible, which can't be the case for X=point, in which which we're comparing the algebraic K-theory of $\mathbb{C}$ with the top K-theory of a point. $\endgroup$ – Joshua Seaton Dec 16 '13 at 1:47
  • $\begingroup$ (Somewhat) related: Topological vs. Algebraic K-Theory $\endgroup$ – Grigory M Dec 26 '13 at 22:09
4
$\begingroup$

I've taken it for granted that the topological K-theory of a space X is equal to the K-theory of vector bundles on X

This is not true even for a point: $K_1^{alg}(\mathrm{Vect}(pt))=K_1^{alg}(\mathbb C)=\mathbb C ^\times$.

(Of course, there is a map from $GL(\mathbb C)$ with discrete topology to ‘ordinary’ $GL(\mathbb C)$ inducing a map $K^{\text{alg}}(\mathbb C)\to K^{\text{top}}(pt)$; but this map is far from being isomorphism. That’s how I think about the difference in general case: in algebraic K-theory we forget about non-trivial topology [on our group]…)


Looks like some results in the direction you want are discussed in Michael Paluch. Algebraic K-theory and topological spaces K-theory:

In this note we discuss the algebraic and topological K-theories of an admissible space X and demonstrate how one may recover the connective topological K-theory of X from the algebraic K-theory of a simplicial ring which encodes the topological structure of the Fréchet algebra of continuous functions on X (...)

$\endgroup$
  • $\begingroup$ Thank you for the response. I apologise for taking as long as I did to get back to you. School has been hectic and I probably won't be able to look into this for a while. But I certainly will revisit this answer. $\endgroup$ – Joshua Seaton Dec 27 '13 at 23:06

Your Answer

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

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