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While I was computing the cohomology ring $E^*(\mathbb{C}P^n)$ for $E$ an oriented ring spectrum, via AHSS I realised that orientation is not a necessary hypothesis in order to compute the cohomology of it.

In fact, for any cohomology theory $h^*$, we have that the second page of the AHSS is (by UCT) $$E^{s,t}_2\cong H^s(\mathbb{C}P^n)\otimes h^t(\text{pt.})$$ since we know that AHSS is a multiplicative spectral sequence, everything boils down to show that $$d_2(y\otimes \eta)=0$$ where $y$ is the "generator" of the cohomology ring $H^*(\mathbb{C}P^n)\cong \mathbb{Z}[y]/(y^{n+1})$

Let $i\colon \mathbb{C}P^1\to \mathbb{C}P^n$ be the inclusion, we know that it induces isomorphism in second cohomology. Therefore by naturality of the spectral sequence we know that $d_2(y\otimes \eta)=d(i^{-1}y'\otimes \eta)=i^{-1}d(y'\otimes \eta)$ where $y'$ is the "generator" of the cohomology ring of $\mathbb{C}P^1$. The latter differential is clearly zero since AHSS for $\mathbb{C}P^1$ has non zero column apart from the 0 and the second one. Therefore the AHSS collapses.

If we know that the spectral sequence is bounded below (i.e. for example induced by a bounded below spectrum), since the stable page is a free $h^*(\text{pt.})$-module we have that the stable page is isomorphic to $h^*(\mathbb{C}P^n)$, and we can even find the ring structure again by multiplicativity.

It seems that this is a powerful result, but I can't find it anywhere in the references. Maybe I'm missing some subtle point. Can someone provide me some references or finding the error?

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I'm pretty sure the condition of complex orientability is necessary for the collapse of the spectral sequence; consider for example the AHSS for real $K$-theory of $\mathbb{C}P^n$.

In your argument, you seem to be supposing that there is a class in $E^* \mathbb{C}P^\infty$ (or rather $E^* \mathbb{C}P^n$) represented by $y \otimes \eta$ on the $E_2$ page whose pullback along the inclusion $i$ represents the generator of $E^* \mathbb{C}P^1$, which is the definition of a complex orientation of $E$.

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  • $\begingroup$ yes I found this morning a flaw in my reasoning, but completely forget to update! The flaw is the chain of equalities $d_2(y\otimes \eta)=d(i^{-1}y'\otimes \eta)=i^{-1}d(y'\otimes \eta)$, since I'm not allowed to consider $i^{-1}$ in the last step, since it's no more invertible there $\endgroup$ – Riccardo Jun 15 '16 at 18:24

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