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5:50 PM
Does anyone know what is $KO^{tC_2}[1/2]$? (with the trivial action)
 
 
4 hours later…
9:59 PM
@DenisNardin I think I got a proof that $π_*KO^{tC_2}[1/2]=\mathbb{Q}_2[v_1^{\pm2}]$, but confirmations are welcome...
 
10:55 PM
@DenisNardin I agree with your calculation. And you don't need to invert 2, already \pi_* KO^{t C_2} = Q_2[t^{\pm 1}]. (Where t comes from a Thom class for 4*sign.)
I don't know a quick proof, just a calculation with the KO-theory of projective spaces.
 

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