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?

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    $\begingroup$ 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
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    $\begingroup$ 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

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 (...)

  • $\begingroup$ 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

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