Let $U$ denote the limiting group of the chain $U(1) \to U(2) \to U(3) \to \cdots$

I wish to compute the group $K^{-1}\mathbb{C}/\mathbb{Z}(BU \times \mathbb{Z})$. For this, we have the long exact sequence

$\cdots \to K^{-1}(M) \to H^{odd}(M;\mathbb{C}) \to K^{-1}\mathbb{C}/\mathbb{Z}(M) \to K(M) \xrightarrow{ch \otimes \mathbb{C}} H^{even}(M;\mathbb{C}) \to \cdots$ .

It therefore helps to know the groups $K^{\pm 1}(BU \times \mathbb{Z})$ and $H^*(BU \times \mathbb{Z};\mathbb{C})$.

However I am unable to compute these groups. Could anyone please suggest some references where these groups may have been explicitly computed ? Or may be some hints as to how I may proceed ? Thanks so much !

  • 10
    $\begingroup$ $H^\ast(BU;\mathbb Z) = \mathbb Z[c_1,c_2,\dots]$ is given by the Chern classes. A great place to learn about this is Milnor and Stasheff's Characteristic Classes. I should know something about $K^\ast(BU)$ off the top of my head, but I don't. But googling "K-theory of BU" leads to this wikipedia page, which calculates both your groups in section 3 and 4 respectively, referring to Adams' Stable Homotopy and Generalized Homology for the $K$-theory computation. I'm still confused by the notation $K^n\mathbb C / \mathbb Z$... $\endgroup$
    – Tim Campion
    Oct 14, 2020 at 19:10
  • 6
    $\begingroup$ Ah, and as wikipedia notes, since $BU$ has even cells, its odd-dimensional $K$-theory vanishes. $\endgroup$
    – Tim Campion
    Oct 14, 2020 at 19:13
  • 5
    $\begingroup$ The odd cohomology vanishes, and then since $KU_\ast$ is also concentrated in even degrees the Atiyah-Hirzebruch spectral sequence collapses (everything is in even bidegree so there is no room for differentials) -- then since everything is in even bidegree, it's in even total degree, so the $K$-theory is also concentrated in even degrees. $\endgroup$
    – Tim Campion
    Oct 14, 2020 at 19:41
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    $\begingroup$ I feel like anything I know about this stuff I learned through some weird osmosis at some point, but I think the Adams book mentioned above might be a good resource. Just try not to get too hung up on whatever point-set formalism he's using. $\endgroup$
    – Tim Campion
    Oct 14, 2020 at 19:47
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    $\begingroup$ Yes Adams book is going to be one of the best places to learn this. Be warned: his book is written in reverse order of how one learning the subject should read it. You should read what he has to say about the AH spectral sequence (I think in the third part of the book), and then look at his computations of complex oriented cohomology using it (K-theory is complex oriented). This will be the second part of the book. $\endgroup$ Oct 14, 2020 at 19:59

1 Answer 1


First, you just have $K^n(\mathbb{Z}\times BU)=\text{Map}(\mathbb{Z},K^n(BU))$, so you can work with $K^*(BU)$, which is technically more convenient. In particular, this is the inverse limit of the rings $K^*(BU(n))$. If $V$ is a complex vector bundle of dimension $n$ over a base space $X$, we can consider the polynomial $g_V(u)=\sum_{k=0}^n[\lambda^k(V)]u^{n-k}\in K^0(X)[u]$. We can then define $c^{KU}_k(V)\in K^0(X)$ to be the coefficient of $t^{n-k}$ in $g_V(t-1)$. By considering the universal case, we get classes $c^{KU}_0=1,c^{KU}_1,\dotsc,c^{KU}_n\in K^0(BU(n))$. It is a standard fact that $K^0(BU(n))=\mathbb{Z}[[c^{KU}_1,\dotsc,c^{KU}_n]]$, with $c^{KU}_n$ mapping to zero in $K^0(BU(n-1))$. One way to prove this is to use the general framework of complex orientable cohomology theories. Another is to use the Atiyah-Hirzebruch spectral sequence, which is in (even,even) bidegrees and so collapses. By passing to the limit we get $$ K^0(BU) = \mathbb{Z}[[c^{KU}_1,c^{KU}_2,c^{KU}_3,\dotsc]]. $$ We also have $$ H^*(BU) = \mathbb{Z}[[c^{H}_1,c^{H}_2,c^{H}_3,\dotsc]], $$ with $c_k^H\in H^{2k}(BU)$ being the Chern classes in ordinary cohomology.


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