K-Theory and the Stack of Vector Bundles - MathOverflow

K-Theory and the Stack of Vector Bundles

2010-04-21T04:40:12Z

I have some understanding that vector bundles provide a basic, familiar example of what I should call a stack. Namely, consider the functor $Vect$ that assigns to a space $X$ the set of isomorphism classes of vector bundles on $X$. This functor isn't local, in the sense that the isoclass of a vector bundle isn't determined by its restriction to an open cover, but rather by gluing data on overlapping sets in a cover. Since for any space $Y$ a map $X \to Y$ is determined by what it does when restricted to a cover of $X$, this tells us there is no space $Y$ that represents the functor $Vect$ in this fashion. However, I can also consider $Vect$ as a stack, which assigns to $X$ the groupoid of vector bundles on $X$. This gadget is fancy enough to understand how vector bundles glue together, and so recovers the locality missing from our earlier functor.

In K-theory, we attach to a space $X$ a ring $K(X)$ whose underlying group is the the free abelian group on the set of isoclasses of vector bundles on $X$, mod short exact sequences. It turns out that one can describe $K(X)$ as the set of homotopy classes of maps from $X$ to $\mathbb{Z} \times BU(\infty)$.

At this point my meager understanding of K-theory seems to be contradict what I said in the first paragraph. The fact that $K(X)$ has a classifying space seems at odds with the observation that vector bundles aren't determined by their restrictions to open covers, whereas maps to another space are. Is something wrong with what I've said so far? If not, perhaps there isn't a contradiction because either 1) $K(X)$ isn't quite the set of isoclasses of vector bundles, but rather a group completion thereof, or 2) we're looking at homotopy classes of maps to $\mathbb{Z} \times BU(\infty)$, so what I said in the first paragraph doesn't apply?

Answer by Kevin Lin for K-Theory and the Stack of Vector Bundles

2010-04-21T06:18:56Z

Read this

Answer by Jeffrey Giansiracusa for K-Theory and the Stack of Vector Bundles

2010-04-21T07:12:38Z

Johannes Ebert and I wrote about this in our paper http://arxiv.org/abs/0712.0702.

The groupoid of vector bundles doesn't quite give you topological K-theory directly - you have to first pass through the group completion of this with respect to the monoidal product given by direct sum of vector bundles. Alternatively, you can instead work with virtual bundles $E-F$, which can be made to form a groupoid in a reasonable way, and on compact bases this is enough, though for noncompact bases topological K-theory classes are non quite represented by virtual bundles.

Getting on to stacks here, a stack in the category of manifolds which admits an atlas (i.e. a representable surjective map from a topological space) has a homotopy type. An easy way to build the homotopy type is the take a Lie groupoid that presents the stack and then form the classifying space of this groupoid. Passing to homotopy types gives a well-defined functor to the homotopy category of topological spaces.

You can make a classifying stack for vector bundles, and its homotopy type will be $\coprod_n BU(n)$. Direct sum makes this stack into a monoid, and its homotopy type then has the usual monoidal structure given by $\oplus$. If you group complete the homotopy type then you get the representing space $Z\times BU$ for topological K-theory.

You can also avoid having to apply group completion by making a classifying stack for virtual bundles. Johannes and I did exactly this in our paper. (Technically, if you allow noncompact bases then it is only a pre-stack and needs to be stackified, which adds in the objects slightly more general than virtual bundles that I alluded to above.) It turns out that the homotopy type of the classifying stack of virtual bundles is exactly $Z\times BU$.

Thus, if you have a virtual bundle on $X$, this represents a class in $K(X)$, but it also is classified by a map to the classifying stack of virtual bundles, and taking homotopy types gives the corresponding map $X \to Z\times BU$.