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For a finite CW-complex $X$, the K-theory group $K^{-1}(X)$ is isomorphic to the group of homotopy classes of maps $[X, U(\infty)]$. The group of isomorphism classes of line bundles on $X$, which I denote by $\text{Pic}(X)$, is isomorphic to $[X, \mathbb{CP}^\infty]$, and it acts on $K^{-1}(X)$ by tensor product.

My question: does anybody know how to write down explicitly the action of $\mathbb{CP}^\infty$ on $U(\infty)$ which induces the action of $\text{Pic}(X)$ on $K^{-1}(X)$?

Note: Of course the action of $\mathbb{CP}^\infty$ on $U(\infty)$ should be compatible with the H-space structure of $\mathbb{CP}^\infty$, which induces the tensor product of line bundles. For this H-space structure see this post, which inspires my question.

Alex Fok
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    I would (very naively!) have expected an action of $U(\infty)$ on $\mathbb{CP}^{\infty}$ (coming from the action of $U(n+1)$ on $\mathbb{C}^{n+1}$) rather than the other way around. Now I'm curious :) – Qfwfq May 07 '15 at 13:59

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Probably it works like this. Every element of $\mathbb {CP}^{\infty}$ corresponds to an injective map $\mathbb C \to \mathbb C^{\mathbb N}$ up to scaling. Tensoring with $\mathbb C^{\mathbb N}$, we get an injective map $\mathbb C^{\mathbb N} \to \mathbb C^{\mathbb N}$ up to scaling. We can send a unitary matrix $U$ acting on $\mathbb C^{\mathbb N}$ to the unitary matrix that acts on the image of the tensor product map as $U$ and acts on its orthogonal complement as $1$. This is scale-invariant, so it is a well-defined map $\mathbb {CP}^{\infty} \times U(\infty) \to U(\infty)$. It's compatible with the H-space structure, which can also be defined using tensor product and the $\mathbb C^{\mathbb N \times \mathbb N} = \mathbb C^{\mathbb N}$ trick in a similar way.

My only evidence that it's related to the K-theory map you describe is that it is a form of tensor product.

Will Sawin
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  • I would appreciate it if you could point out more explicitly why your action of $\mathbb{CP}^\infty$ on $U(\infty)$ induces the action of $\text{Pic}(X)$ on $K^{-1}(X)$. – Alex Fok May 11 '15 at 11:59