# Cohomology of projective bundle only depends on base and fiber?

Let $P\to X$ be a $\mathbb P^n$ bundle. Is it true that all the (co)homology group only depends on $X$ and $n$ (and independent of the transform funcions) ?

• Yes for the cohomology groups, no for the ring structure. You have $H^*(P)=H^*(X)[\xi]/(\xi^{n+1}+c_1\xi^n+...+c_{n+1}\xi^0)$ where $c_1,...,c_n$ are the Chern classes and $\xi$ is of degree 1. – Roland Feb 10 '18 at 18:43
• @Roland The last $x_{n+1}$ should be $c_{n+1}$? and could you write a little more about how you get this? – Akatsuki Feb 10 '18 at 18:50
• Yes, that's a typo. Look for projective bundle theorem (it is stated here en.wikipedia.org/wiki/Projective_bundle). – Roland Feb 10 '18 at 18:53
• @Roland Thanks! You can post as an answer if you want. – Akatsuki Feb 10 '18 at 18:57
• @Akatsuki This is a nontrivial result, see this answer : math.stackexchange.com/questions/2461550/… (this is even true for Hodge numbers so of course it implies the version).But in fact, what I wrote was a bit misleading : we only have isomorphism as a group and not in general as a ring (as Roland already said). Edit : and I don't know a reference, but googling "Grothendieck ring of varieties" and also "Bittner's theorem" should give you some references. – Nicolas Hemelsoet Feb 12 '18 at 8:03

But note however that even if the cohomology groups $H^*(P)$ only depends on the base scheme $X$, this is not true for the ring structure. In fact we have : $$H^*(P)=H^*(X)[\xi]/(\xi^n+c_1\xi^{n-1}+...+c_{n+1}\xi^0)$$ where $\xi\in H^2(P)$ is of degree 2 and is the class $c_1(\mathcal{O}_{P}(1))$ of the tautological bundle, and $c_1,...,c_{n+1}\in H^*(X)$ are the Chern classes of $P$.
• I am sure this is true if the $\mathbb{P}^n$ bundle comes from a vector bundle because you can play around with a tautological bundle, the Leray-Hirsch Theorem etc. But do all $\mathbb P^n$ bundles come from vector bundles? Do you have a reference? – Thomas Rot Feb 13 '18 at 9:19