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I came across the space $BU_\otimes$ when struggling with twisted K-theory. Segal proved that this is an H-space, right? I have read a dozen times by now that the group $[X, BU_\otimes]$ consists of the vector bundles of virtual dimension 1. I can clearly see the semi-group structure there, but what does the inverse of such a bundle look like?

Best regards, Ulrich Pennig

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up vote 13 down vote accepted

I'll write $U(X)=[X,BU_\otimes]$. So $U(X)\subset K(X)=[X,Z\times BU]$ is the multiplicatively closed subset corresponding to "virtual bundles of rank 1". Likewise, I'll write $I(X)=[X,BU]\subset K(X)$ for the ideal of "virtual bundles of rank 0". There's a bijection $a\mapsto 1+a$ from $I(X)$ to $U(X)$.

Here's the claim: if $X$ is compact (let's say it's a finite CW-complex of dimension $n$), then $I(X)$ is a nilpotent ideal (in fact, $I(X)^{n+1}=0$). This is a generic fact about multiplicative cohomology theories, and it can be proved in a number of ways; you can think about it in terms of the multiplicative properties of the Atiyah-Hirzebruch spectral sequence, for instance.

For such $X$, it is clear that elements of $1+a\in U(X)$ are invertible, given by the series $(1+a)^{-1}=1+a+a^2+\cdots$ which terminates since $a\in I(X)$ and $I(X)$ is nilpotent.

For infinite dimensional $X$, you need to argue a little harder. If $X=\lim_{\to} X_i$ where the $X_i$ are finite CW-complexes, then there is a surjection $K(X)\to \lim_{\leftarrow} K(X_i)$, which restricts to surjections for $I(X)$ and $U(X)$. The kernel is a $\lim^1$-term. If the $\lim^1$-term vanishes, then it's clear that elements of $U(X)$ are invertible, since their images in the $U(X_i)$ are invertible. It's enough to check that $\lim^1$-vanishes in the case that $X=BU_\otimes$, which is a standard calculation.

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