# Is there a good definition of (topological) K-Theory over arbitrary spaces?

Hi

(this is my very first question here, so please don't hurt me...)

for some time now i've been looking for a sufficiently aesthetical definition of (topological) K-theory of arbitrary spaces, yet been unable to find or come up with one. The definition I know goes as follows:

For X connected, compact, hausdorff one defines $V(X) = \text{ set of isoclasses of vectorbundles over} X$ which becomes a comm. monoid under direct sum and then $KO(X) = K(V(X))$ where the righthandside just means group completion of a comm monoid.

Here already the "isoclasses" of bundles bothers me, because this "set" is not really a set, is it? (?its elements being proper classes?). I guess this may be salvaged by instead looking at equi. classes of systems of transition functions taking values in $GL(\mathbb R)$, modulo some further restriction and relations !?

Anyway, this is something I might even live with, but one goes on to show $KO(X) \cong [X, \mathbb Z \times BGL(\mathbb R)]$ for such spaces, and for a CW-complex C sets $KO(C) = [C, \mathbb Z \times BGL]$. This is the only possible defintion (up to nat. iso), when trying to end up with a cohomology-like functor; right? Finally for a general space Z we pick (for every space simultaneously?!) a CW-substitue say C' and put $KO(Z) = [C', \mathbb Z \times BGL]$.

However, using this as defintion, there really is very little beauty left in K-Theory for me. I know that just putting $KO(X) = K(V(X))$ goes awry (bundles must be allowed to have varying dimension over different components and in turn must allow for a partition of unity and so on...)

So my question is: Is there a way of altering the definition of V(X) sufficiently so as to give a "correct" definition of K-Theory? Or some other way of producing these groups, nicely? Nicely should in particular mean, without homotopy theory oder cell complexes, so that for instance homotopy invariance is not directly built into the definition. And if so, what about relative groups, or even higher ones?

After all for singular cohomology and bordism there also are descriptions using homotopy theory (via Eilenberg-MacLane resp. Thom-Spaces) just as above, but for an arbtrary space there are entirely different (better?) descpriptions in terms of singular chains and manifolds.

Maybe I should add that passing to spectra makes everything even worse in my opinion.

• As I understand, the compact Hausdorff condition is needed to ensure stable inverses, ie for every bundle $\zeta$ there is a bundle $\xi$ such that $\zeta\oplus\xi$ is trivial. What if you only consider iso classes of stably invertible bundles in the definition of $V(X)$? Do you still end up with $[X,Z\times BGL]$? – Mark Grant Aug 29 '11 at 14:59
• Please, don't use the spellings "plz", "Thx". – Georges Elencwajg Aug 29 '11 at 19:13
• edited but may I ask why? – old account Aug 30 '11 at 11:52
• We like our vwls. – Donu Arapura Aug 30 '11 at 12:39
• @pudin: thanks for your editing. The reason abbreviations like "plz" are not welcome here is that they look childish, buddy-buddy, unprofessional and obviously serve no useful purpose.You are requesting help from (among others) some of the best mathematicians on earth, at least four Fields medalists, and from mature people many of whom are middle-aged or more. Since you have never met (most of) them, a modicum of reserve and civility is expected on this site. This little point settled, I'm sure you meant well and we are very happy to welcome you in our community. – Georges Elencwajg Aug 30 '11 at 15:21

1. You might want to look at 'Vector bundles over classifying spaces of compact Lie groups' by Jackowski and Oliver. They discuss a situation in which you can understand $K(V(X))$ quite explicitly and it is interestingly different from $[X,\mathbb{Z}\times BO]$.

2. Picking $C'$ for all $Z$ is not a terribly big deal, because you can do it functorially: just use the geometric realisation of the singular complex of $Z$.

3. Some things work better if you define $V(X)$ to be the set of isomorphism classes of numerable vector bundles, ie those for which there exists a trivialising cover with a subordinate partition of unity. For most spaces (compact spaces, CW complexes, metric spaces, ...) this makes no difference. In cases where it does make a difference, there is a good argument that numerability should be part of the 'right definition' of a vector bundle. I don't remember exactly how much this buys you in the foundations of $K$-theory, however. It will not affect the Jackowski-Oliver examples, as their base spaces are CW complexes.

• ad 1. oh nice, included in there is a characterization of the "bundle part" of K(BG). ad 2. but would you really define the n-th singular cohomology as homotopy classes of maps from the geometric realization of the singular complex of Z to an appropriate EilenbergMaclane space? ad 3. if again i'm not mistaken, maps to BG correspond exactly to numerable G principle bundles even for non cw-complexes, which means even if i included this (which one certainly must to get homotopy invariance of pullbacks) the definition wouldn't "work" for non CW-complexes since K-Theory uses CW-substitutes – old account Aug 30 '11 at 14:33

This is a response only to a tangential part of your question, on set theory, and not to the bulk. It continues some of the discussion in the comments to Alain Valette's answer.

I guess the thing to emphasize is that in some technical sense you are correct: standard (e.g. ZFC) approaches to set theory often do not give meaning to sentences like "The set of isomorphism classes of all finite-dimensional vector bundles". Certainly if you think that an "isomorphism class" is itself necessarily a "set" of things, and that the elements of a "set" should themselves be "sets", as is often done, that is.

But from a modern perspective, this is a perverse way to go about doing mathematics. The notion of "the collection of isomorphism classes of finite-dimensional vector bundles" is perfectly good. Moreover, there are many sets-with-structure that represent this notion. So pick one --- for example, choose your favorite copy of $\mathbb R$, your favorite construction of $\mathrm{GL}(\mathbb R,n)$, etc., and form some set of transition functions and so on. At the end of the day, whatever honest set you actually construct is just as good as any other. The most important moral from category theory is that what you should mean by "the X" is "any object satisfying the distinguished properties of X, along with the instructions of how to make it isomorphic to any other object with the same distinguished properties".

Since you should not expect the Axiom of Choice, and other difficult things, to work for very large collections, it is important to worry about the "size" of a collection. Most mathematicians only have two "sizes". One size is called "is a set", and the other is called "is larger than a set". Things like Choice are assumed to work without complication for anything that "is a set", and you may be more suspicious for "things larger than a set". For this notion of "is a set", it certainly is true that the collection of isomorphism classes of finite-dimensional vector bundles "is a set". Proof: represent it by a set in some way, I don't care how.

Anyway, in the comments to Alain's answer, it looks like you understand all of this perfectly well. So don't sweat this part of the language!

My understanding of your question is that you are after the so-called "representable $K$-theory"; see e.g. Karoubi's 1970 paper: http://www.math.jussieu.fr/~karoubi/Publications/06.pdf in which he defines $K$-theory for paracompact spaces.

Note that the set of isomorphism classes of vector bundles over a compact $X$, is indeed a set; to see it (there might be a shorter argument) by Serre-Swan the category of vector bundles over $X$ is equivalent to the category of projective finite-type modules over the ring $R=C(X)$ of continuous functions on $X$. And projective modules over $n$ generators are (up to isomorphism) images of idempotent matrices in the ring $M_n(R)$. Sets all over.

• hmm doesn't the "up to isomorphism" destroy your argument? one may restrict to equivalence classes of some set of bundles such that any bundle is isomorphic to one in there (or in your case the idempotents) but the set of ALL isoclasses? – old account Aug 31 '11 at 14:10
• @ pudin. Sorry, I don't understand your comment. What is an isoclass, maybe? Can you explain what you mean when $X$ is a point, i.e. we deal with finite-dimensional vector spaces over a given field? – Alain Valette Aug 31 '11 at 15:08
• All vectorbundles do not form a set right? (if i am wrong here than just ignore me) Just like all vectorspaces don't or all sets don't. At least there is no a priori reason why they should. Now saying "Set of isomorphism classes of vectorbundles" is without meaning somehow, isn't it? The only way I know out of this is to pick a SET $S$ of vectorbundles with the property that any vectorbundle is isomorphic to one in $S$. (for a compact space X the set $S$ might be the set of finitedim. subbundles of $X \times \mathbb R^\infty$). Here "set of isoclasses" does make sense and... – old account Aug 31 '11 at 15:27
• @pudin: I quit, by lack of interest for the direction taken by the discussion. (We started from K-theory, remember?). I suggest you pass the question to a set-theorist. – Alain Valette Sep 1 '11 at 18:13
• @pudin: Here is a very adequate quote from John von Neumann: "Young man, in mathematics you don't understand things. You just get used to them." – Alain Valette Sep 1 '11 at 19:45