The topological $K$-theory of a complex compact manifold $X$ is the commutative monoid $K(X)$ of isomorphism classes of complex vector bundles. Two classes $[E]$ and $[F]$ are equivalent in $K$-theory if they are stably equivalent: there exists a trivial bundle $G$ with $$ (E \oplus G) \cong (F \oplus G). $$

In what sense is this the correct notion of equivalence in $K(X)$?

In other words, what does stable equivalence buy you?

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    $\begingroup$ The correct notion in order to do what? $\endgroup$ Nov 3, 2010 at 19:35
  • $\begingroup$ Clarified, I hope. $\endgroup$ Nov 3, 2010 at 19:42

2 Answers 2


It comes down to whether or not you want $K(X)$ to be a group. If all you care about is isomorphism classes of vector bundles, that's a perfectly good monoid. People wouldn't call the monoid $K$-theory because $K$-theory of stuff generally means you've performed some kind of group-completion of a monoid or a category or something.

So there's two ways to respond to your question. The simplest would be "why don't vector bundles have inverses?" which on the most superficial level has the answer not all vector bundles are $0$-dimensional.

Another way to respond would be, what if we performed the formal group-completion of the monoid of isomorphism classes of vector bundles? Well, you get the same thing. General nonsense says you have a map from the group completion to $K(X)$ defined your way, since bundles over reasonable spaces have complementary bundles.

Is this roughly an answer to your question?

edit in response to your comment: Say $X$ is a finite-dimensional CW-complex, and $\epsilon$ is a finite-dimensional vector bundle over $X$. Then there is a classifying map for $\epsilon$, this is a map $X \to G_{\infty,k}$ where $k$ is the dimension of $\epsilon$. $G_{\infty,k}$ is the Grassmannian of $k$-dimensional subspaces of $\mathbb R^\infty$. A basic theorem about $G_{\infty,k}$ is that any map of a finite-dimensional CW-complex into $G_{\infty,k}$ is homotopic to a map into $G_{m,k}$ for $m$ perhaps very large. $G_{m,k}$ is canonically isomorphic to $G_{m,m-k}$ (take orthogonal complements). The bundle classified by the corresponding map $X \to G_{m,m-k}$ is the complementary bundle to $\epsilon$, denote it $\epsilon'$. By design, $\epsilon \oplus \epsilon'$ is a trivial bundle as it's the pull-back of the tangent bundle to $\mathbb R^m$. The fact that maps $X \to G_{\infty,k}$ are homotopic to maps $X \to G_{m,k}$ has many proofs -- one is from the Schubert cell decomposition of $G_{m,k}$ together with cellular approximation.

That's the idea. I believe complementary bundles exist for more than bundles over finite-dimensional spaces but that's the most convenient argument that comes to mind (have to run off to teach a class now).

2nd edit: Let $\mathcal V(X)$ be the group completion of the monoid $\mathcal M(X)$ of isomorphism classes of finite-dimensional vector bundles over $X$. Let $K(X)$ be the stable isomorphism classes of vector bundles over $X$. $K(X)$ is a group for a fairly broad class of spaces, by the arguments above. So by group-completion formal nonsense there is a map $\mathcal V(X) \to K(X)$. The question is whether this map has an inverse or not. Given an element $[\alpha] \in K(X)$, $\alpha \in \mathcal M(X)$ is a vector bundle over $X$ and is well-defined up to stable equivalence. To $\alpha$ there is an associated element of $\mathcal V(X)$ and the question is, is the map

$$K(X) \to \mathcal V(X)$$

well-defined, where you map $[\alpha]_{K(X)} \longmapsto [\alpha]_{\mathcal V(X)}$?

Ah, okay, I see the problem now. This isn't well-defined. But things are off only by a tiny thing. $\mathcal V(X)$ contains a copy of the integers as a direct-summand. This is because $\mathcal M(X) \to \mathbb Z$ given by taking the dimension of a bundle is a homomorphism of monoids (using addition in $\mathbb Z$). There is a map back $\{0,1,2,\cdots \} \to \mathcal M(X)$ given by taking trivial bundles over $X$. So in $\mathcal V(X)$ this gives a splitting $\mathcal V(X) = \mathbb Z \oplus \overline{\mathcal V(X)}$ where $\overline{\mathcal V(X)}$ is the "reduced" group-completion of isomorphism types of vector bundles over $X$. There is a well-defined map $K(X) \to \overline{\mathcal V(X)}$ since you can think of $\overline{\mathcal V(X)}$ as being $\mathcal V(X)$ modulo trivial bundles.

Does that make sense now? I haven't thought about this stuff in much detail for years.

  • $\begingroup$ Are you saying that stable equivalence endows isomorphism classes with a formal inverse? That I don't see. $\endgroup$ Nov 3, 2010 at 19:44
  • $\begingroup$ Yes, provided $X$ is a reasonable space. The idea is summarized in my edit above. $\endgroup$ Nov 3, 2010 at 20:55
  • $\begingroup$ Hi Ryan, thanks for your clarification, but it's still murky for me. I don't see why $E \cong F \not \Leftrightarrow (E \oplus G) \cong (F \oplus G)$ $\endgroup$ Nov 4, 2010 at 13:57
  • $\begingroup$ I'm not sure what your asking in the comment above. But I think you're asking why is the map from stable bundles to the group-completion of isomorphism classes of bundles an isomorphism of groups? I have to prep to teach a class but later this afternoon I'll likely have time to respond. $\endgroup$ Nov 4, 2010 at 21:02
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    $\begingroup$ The tangent bundle to $S^2$ is non-trivial. So let $E$ be the tangent bundle to $S^2$ and $F$ the trivial $2$-dimensional bundle over $S^2$. Let $G$ be a trivial $1$-dimensional bundle over $S^2$, then $E+G$ and $F+G$ are isomorphic since the normal bundle to $S^2$ in $\mathbb R^3$ is trivial, so $E+G$ is the pull-back of $\mathbb R^3$'s tangent bundle, which is trivial. $\endgroup$ Nov 5, 2010 at 21:48

I know it's been a long time, but I figured that for the sake of people who may read this in the future, I should point out that the object $K(X)$ as you've defined it is NOT a group, which is clear already in the case that $X$ is a point.

You DO get a group (let's stick to compact spaces $X$) if you allow the trivial bundle $G$ appearing on either side of your equation to be different, say $G$ on the left and $G'$ on the right. The result is usually written $\tilde{K}(X)$, and the group $K(X)$ is $Z\oplus\tilde{K}(X)$.

Anyway, things that "stable equivalence" buys you is 1) makes calculations tractable (I think there are very few spaces for which the monoid of isomorphism classes of vector bundles is known) and 2) allows you to apply the well-oiled machinery of stable homotopy theory.

Also some evidence that "stable equivalence" is the "right" equivalence comes from the theory of characteristic classes, e.g. Stiefel-Whitney, Chern, Pontryagin, etc. since those classes are all "blind" to adding a trivial summand to your bundle.


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