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For a compact Hausdorff topological space $X$, its K-theory $K^0(X)$ is defined to be the Grothendieck group of the isomorphism classes of finite dimensional vector spaces on $X$. For example $K^0(\text{pt})=\mathbb{Z}$.

My question is: is there a categorification of K-theory, just as we can categofify natural numbers to vector spaces?

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  • $\begingroup$ I've wondered idly sometimes if the answer to this is something like "the derived category of vector bundles on $X$", in the sense that the Grothendieck group accomplishes subtraction, and a complex of vector spaces being exact means that the alternating sum of their dimensions is zero. $\endgroup$
    – Ryan Reich
    Commented Aug 4, 2012 at 20:37
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    $\begingroup$ I don't know about topological K-theory, but Toën has developed a categorification of algebraic K-theory, which he calls secondary K-theory. I have a set of notes (math.uchicago.edu/~ejenkins/nwtft.html) from lectures he gave a few years ago. (Apologies for my poor handwriting.) $\endgroup$
    – Evan Jenkins
    Commented Aug 4, 2012 at 20:59
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    $\begingroup$ Isn't the category of vector bundles on $X$ already a categorification of $K^0$ in the sense that $K^0$ decategorifies it? $\endgroup$
    – Qiaochu Yuan
    Commented Aug 5, 2012 at 3:23
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    $\begingroup$ @Qiaochu: The category of vector bundles is very useful sometimes. However a major drawback (for me) is that in the category of vector bundles there is no "negative vector bundles". In other words, we can only take direct sum of vector bundles but cannot take "direct difference" of vector bundles. In K-theory, taking difference is very important, especially in its connection with index theory. So that's why I ask for a better categorification. $\endgroup$
    – Zhaoting Wei
    Commented Aug 6, 2012 at 4:30
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    $\begingroup$ @Zhaoting: then you can take $\mathbb{Z}_2$-graded vector bundles. $\endgroup$
    – Qiaochu Yuan
    Commented Mar 2, 2015 at 6:13

3 Answers 3

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The abelian group $K^0(X)$ has a natural $(\infty,0)$-categorification'', meaning a spectrum $K(X)$ whose $\pi_0$ recovers $K^0(X)$: take $K(X)$ to be the spectrum of maps from (the suspension spectrum of) $X$ to the (complex or real, depending) K-theory spectrum. Or we could try a different variant, the spectrum given as the group completion of the topological groupoid of vector bundles over X under direct sum; I think this variant would give the connective cover of $K(X)$.

To be clear, this is an analog of natural numbers categorifying to finite-dimensional vector spaces only if you force yourself to neglect the non-invertible maps between vector spaces.

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  • $\begingroup$ @Dustin: Thank you for your answer! I think it is what I am looking for. Could you give more details? For example, is the "complex K-theory spectrum" just the set of Fredholm operators in a seperable Hilbert space? $\endgroup$
    – Zhaoting Wei
    Commented Aug 4, 2012 at 21:59
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    $\begingroup$ A spectrum is a collection of pointed spaces $E_n$ ($n\in\mathbb Z$), along with a collection of homotopy equivalences from of $E_n$ to the loop space of $E_{n+1}$. The zeroth space of the $K$-theory spectrum is the space of Fredholm operators, but the $K$-theory spectrum contains more data than just its zeroth space. $\endgroup$
    – André Henriques
    Commented Aug 4, 2012 at 22:18
  • $\begingroup$ Hi Zhaoting - that's a good first approximation, but as Andre says, there's more data involved in the definition of a spectrum. One part is that the whole sequence of spaces, each looping to the previous, makes the zeroth space a loop space to arbitrarily high order, which encodes a sort of abelian group law on the zeroth space, enhancing the abelian group law on its $\pi_0$ that we know and love. But moreover the low dimensional homotopy groups of those higher spaces have potential to contribute to negative-dimensional homotopy groups of the spectrum, making spectra a strange -- (contd) $\endgroup$
    – Dustin Clausen
    Commented Aug 5, 2012 at 3:05
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    $\begingroup$ -- though, as evidence bears out, very natural -- categorification of abelian groups, where the data doesn't just go up but also down. In the case at hand, the K-theory spectrum is defined using Bott periodicity: for instance for complex K-theory, the n^{th} space is Z x BU for n even and U for n odd. And then the spectrum K(X) I was talking about is such that its n^{th} homotopy group is the topological K-group K^{-n}(X), for n an arbitrary integer. I'd recommend Segal's paper "categories and cohomology theories" for getting a grip on this sort of thing. $\endgroup$
    – Dustin Clausen
    Commented Aug 5, 2012 at 3:11
  • $\begingroup$ @ Dustin: Thank you very much for your explanation! It is very illustrative and I will read that paper you mentioned. $\endgroup$
    – Zhaoting Wei
    Commented Aug 5, 2012 at 3:48
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Maybe I am missing something, but is there a reason no one has mentioned the work on 2-vector spaces. One place to start is http://hopf.math.purdue.edu//Baas-Dundas-Rognes/segal60.pdf . They state a conjecture that the algebraic K-theory of the category of 2-vector spaces (over $\mathbb{C}$) is the algebraic K-theory of $ku$. The conjecture is proved in http://arxiv.org/pdf/0706.0531.pdf by Baas, Dundas, Richter and Rognes.

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One answer these days is to think of $K$-theory as represented by a commutative ring spectrum (alias $E_{\infty}$-ring spectrum) $K$. Then there is a perfectly good theory of $K$-module spectra, to which we can apply Waldhausen's approach to algebraic $K$-theory to construct the $K$-theory of $K$-theory. This first appeared in EKMM (Elmendorf-Kriz-Mandell-May. Rings, modules, and algebras in stable homotopy theory. AMS 1997) and has been much studied since, for example by Blumberg and Mandell. The localization sequence for the algebraic $K$-theory of topological $K$-theory.

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