The mod 2 cohomology of the connective ko spectrum is known to be the module $\mathcal{A}\otimes_{\mathcal{A}_2} \mathbb{F}_{2}$, where $\mathcal{A}$ denotes the Steenrod algebra, and $\mathcal{A}_2$ denotes the subalgebra generated by $Sq^1$ and $Sq^2$.

Where can I find the original calculation for referencing it ? What is the structure as module over the steenrod algebra of the mod 2 cohomology of periodic KO theory?

  • 2
    $\begingroup$ Unless I'm mistaken $\pi_*(H\mathbb{F}_2\wedge KO)=0$ (it is a ring of positive characteristic containing an isomorphism between the additive and the multiplicative formal group law), so the answer to your second question is rather trivial $\endgroup$ – Denis Nardin Jul 24 '19 at 16:45
  • 1
    $\begingroup$ Indeed, as the periodicity generator ie the Bott element induces zero-map in ordinary homology! $\endgroup$ – Prasit Jul 24 '19 at 16:56
  • 4
    $\begingroup$ It is more usual to use $\mathcal{A}_1$ to denote the algebra generated by $\mathrm{Sq}^1$ and $\mathrm{Sq}^2$. $\endgroup$ – John Palmieri Jul 24 '19 at 19:57

Ravenel in his Complex Cobordism and Stable Homotopy Groups of Spheres attributes this result to Stong, in Determination of $H^*(BO(k,⋯,∞),Z_2)$ and $H^∗(BU(k,⋯,∞),Z_2)$, but looking at that paper (which is concerned mainly with the "unstable" cohomology of the various constituent spaces of $ko$), he attributes this further to

Adams, J. F., On Chern characters and the structure of the unitary group, Proc. Camb. Philos. Soc. 57, 189-199 (1961). ZBL0103.16001.

There you can find indeed the required result as Lemma 4.

Regarding your second question, the cohomology $H^*(KO;\mathbb{F}_2)$ is zero for chromatic reasons (the spectrum $KU\wedge H\mathbb{F}_2$ carries an isomorphism of the additive and multiplicative formal group law in characteristic two, so it must be trivial, and then you can use the fact that $KU=KO/\eta$ to conclude, since $\eta$ is nilpotent in $\pi_*KO$), hence the map $H^*(KO;\mathbb{F}_2)→H^*(ko;\mathbb{F}_2)$ is trivial.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.