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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?

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    $\begingroup$ 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
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    $\begingroup$ Indeed, as the periodicity generator ie the Bott element induces zero-map in ordinary homology! $\endgroup$ – Prasit Jul 24 '19 at 16:56
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    $\begingroup$ 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
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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.

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