Concrete pull-back calculation along H-space map

Question

I am trying to calculate the pull-back of a cohomology class on the loopspace of the algebraic $K$-theory space $\Omega K(\mathbb{C})$ along the H-space map of $K(\mathbb{C}).$

Let $b_k\in \tilde{H}^{2k}(\Omega K(\mathbb{C});\mathbb{R})$ be a reduced cohomology class on the loopspace of the complex K theory space (it is in fact derived from the Borel regulator classes, but this should not matter). The H space structure on $K(\mathbb{C})$ gives a map $$ \Omega K(\mathbb{C})\wedge \Omega K(\mathbb{C})\to \Omega K(\mathbb{C})$$ (By forming sums of chain complexes.) The Kunneth formula tells us that $$\tilde{H}^{2k}(\Omega K(\mathbb{C})\wedge \Omega K(\mathbb{C}); \mathbb{R})\cong \bigoplus_{i+j=2k}\tilde{H}^{i}(\Omega K(\mathbb{C});\mathbb{R})\otimes \tilde{H}^{j}(\Omega K(\mathbb{C});\mathbb{R}).$$

In this scenario I want to calculate $f^*b_k\in\tilde{H}^{2k}(\Omega K(\mathbb{C})\wedge \Omega K(\mathbb{C});\mathbb{R}).$ I am hoping that $f^*b_k=b_k\otimes 1+1\otimes b_k.$ Has someone done this or a similar calculation before?

Answer 1

I will assume that by $\wedge$ you meant $\times$, and did not mean to write reduced cohomology (because I don't think the $H$-space structure gives you a map out of the smash product, and $b_k \otimes 1$ is not a tensor product of reduced classes).

Let $X$ be a unital $H$-space with multiplication $\mu : X \times X \to X$. By an analogue of the Eckmann---Hilton argument, the map $\Omega \mu$ is homotopic to the loop concatenation map $$c : \Omega X \times \Omega X \to \Omega X.$$

If $b \in \tilde{H}^k(X;R)$ is represented by a pointed map $f : X \to K(R,k)$, then $\Omega f : \Omega X \to K(R,k-1)$ represents a class $\tau(b) \in \tilde{H}^{k-1}(\Omega X;R) \subset {H}^{k-1}(\Omega X;R)$. This always satisfies $$c^*(\tau(b)) = \tau(b) \otimes 1 + 1 \otimes \tau(b).$$ Dualising Proposition 16.19 of Switzer's book gives a proof.