The following fact is asserted by wikipedia:

The $K$-theory of $BU(n)$ is the numerical symmetric polynomials, i.e the subring of $\mathbb{Z}[x_1, \ldots, x_n]$ that is preserved under the action of the symmetric groups.

I was wondering if anyone knows the proof of this/or a reference to it.


  • 1
    $\begingroup$ A reference can be found in Segal's paper "Equivariant K-theory". He proved this in quite detail. $\endgroup$ – Bombyx mori Nov 28 '12 at 1:13

Here's a sketch of one way of doing this (I haven't looked at Segal's argument, probably he has a better one)...

First one computes that the K-theory of the projectivization of a rank n complex vector bundle is a truncated polynomial algebra over the K-theory of the base on the element $\mathcal{O}(1)-1$ where $\mathcal{O}(1)$ is the canonical line bundle. This can be found, for example, in Atiyah's book on K-theory and is a formal consequence of the corresponding calculation for $K(\mathbb{C}P^n)$ which, in turn, follows from the calculation of the $K$-theory of spheres (Bott periodicity) and a Mayer-Vietoris sequence (I'm not sure if that's the argument Atiyah gives though).

From here we prove the splitting principle: Given a bundle, $E \rightarrow X$ notice that the map $\mathbb{P}E \rightarrow X$ induces an injection on $K$-theory and that the pullback of $E$ splits off a line bundle (the canonical one, in fact). Iterating this we get the existence of a space $Y$ so that $Y \rightarrow X$ induces an injection on $K$-theory and the given bundle splits as a sum of line bundles.

Applying this procedure to the universal bundle over $BU(n)$ (maybe you have to be careful about spaces being compact if you want to use the geometric definition of $K$-theory, but that's okay just look at the finite-dimensional Grassmannians and make a limiting argument) we get that the K-theory of $BU(n)$ injects into the $K$-theory of $(\mathbb{C}P^{\infty})^n$. It is clear that the image lives inside the symmetric polynomials, so we'll be done if we can calculate $K(BU(n))$ and see that it has the right size. For this, use the AHSS.

  • $\begingroup$ sorry for the late accept, Dylan but this does make sense! It's analogous (somewhat) to the ordinary cohomology computation (in terms of the maximal tori of $U(n)$) I suppose! $\endgroup$ – Elden Elmanto Jan 26 '13 at 5:02

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