TITLE: Calculating the extraordinary cohomology of $\mathbb{C}P^n$
QUESTION [6 upvotes]: Let $E$ be a ring spectrum with an orientation. Now I want to calculate  $E^*(\mathbb{C}P^n)$.
The definition of orientation I am using is: There is an element $x \in E^*(\mathbb{C}P^{\infty})$ such that it's restriction to $\mathbb{C}P^1$ is a generator over $\pi_*(E)$.
To calculate this I want to use the Atiyah-Hirzebruch spectral sequence. But I am unable to figure out why all the differentials vanish. I believe one has to some how use the fact that $E$ has an orientation but I am unable to figure out how. Thank you.

REPLY [7 votes]: Your condition should be that the orientation is an element of the reduced $E$ cohomology of $CP^\infty$, restricting to a generator of $\pi_0(E)$.
Look at the reduced AHSS. The elements with bidegree $(2,0)$ are perm cycles in the SS for $CP^1$ for degree reasons. The information about the orientation tells you they have to be perm cycles in the SS for $CP^\infty$ too and, in particular, in the SS for $CP^n$.
Since we have a splitting $* \to CP^n \to *$ the reduced AHSS sits inside the unreduced SS, and in the unreduced SS,
elements with $s=0$ are perm cycles.
Now multiply.