# $K$-Theory of $BU(n)$

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.

Thanks!

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

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.
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! – Elden Elmanto Jan 26 '13 at 5:02