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?


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 $E$.

  • $\begingroup$ 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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