Let $X$ and $Y$ be compact spaces (or closed manifolds if you want). I have two questions relating the ring structure of topological $K$-theory. To motivate my questions let me give some background information. Consider the product $$K^0(X)\times K^0(Y)\to K^0(X\times Y)$$ given by external tensor product of complex vector bundles. Then in principle one can define a product $$K^{-n}(X)\times K^{-m}(Y)\to K^{-(n+m)}(X\times Y)$$ for any integers $n$ and $m$.

I want to see if one can write down the explicit formula of the map $$K^{-1}(X)\times K^{-1}(Y)\to K^0(X\times Y),$$ where $K^{-1}(X)$ is defined in terms of equivalence classes of $(E, U)$, where $E\to X$ is a complex vector bundles and $U:E\to E$ is an automorphism.

My questions:

  1. I am not sure how does one take the automorphism $U$ into the product structure? I tried to find the answer to this question by google and reading the books on topological $K$-theory I know of, but I cannot find a reference talking about this issue explicitly.

  2. Suppose we have a satisfactory answer to the above question. I guess it's reasonable to expect $$\textrm{ch}(x\cup y)=\textrm{ch}^{\textrm{odd}}(x)\cup\textrm{ch}^{\textrm{odd}}(y),$$ where $x\in K^{-1}(X)$, $y\in K^{-1}(Y)$, $\textrm{ch}:K^0\to H^{\textrm{even}}(\mathbb{Q})$ and $\textrm{ch}^{\textrm{odd}}:K^{-1}\to H^{\textrm{odd}}(\mathbb{Q})$ are the Chern character and the odd Chern character respectively. I don't even find this "claim" or "fact" so it could be wrong. However, if it is correct, may I know a reference for this?


1 Answer 1

  1. The product of $K^{−1}$ with itself can be defined by identifying $K^{−1}(X)$ with $K^0(X\wedge S^1)$ using the suspension isomorphism, multiplying the resulting two elements of $K^0(X\wedge S^1)$, obtaining an element of $K^0(X\wedge X\wedge S^2)$, and applying Bott periodicity to get to $K^0(X\wedge X)$. Choosing an explicit construction of the suspension isomorphism gives an explicit construction of the multiplication map.

  2. The Chern character is induced by a morphism of ring spectra and therefore is automatically multiplicative. See Reference for $E_{\infty}$-ness of the Chern Character for more information.

  • $\begingroup$ First of all thank you for your answers. After looking at the answers you gave at the post you cited I got a rough idea. Of course the $E_\infty$ things are well beyond my knowledge. Somehow I wonder whether one can prove the equality in Q.2 by elementary means, meaning that whether one can prove the equality in Q.2 at the differential form level by finding an explicit exact form (something similar to a transgression form) for which they differ by. $\endgroup$
    – Ho Man-Ho
    Commented Mar 8, 2019 at 4:32

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