A question on the ring structure of topological K-theory and Chern character

Question

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?

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.