Let $\mathcal{E} \rightarrow X$ be a complex vector bundle of rank $r+1$ and let $F=\mathbb{P}^r \rightarrow E = \mathbb{P}\mathcal{E}\rightarrow X$ be the associated projective bundle. We know that under the assumption that $H^{\bullet}(E) \rightarrow H^{\bullet}(F)$ is surjective, by the LeraySerre spectral sequence we have for any field $k$ $$H^{\bullet}(E, k) \simeq H^{\bullet}(X,k) \otimes H^{\bullet}(F,k)$$ Where the tensor product on the right is graduated. Can we say anything more about the algebra structure of left hand side when we take $\mathbb{Z}$ coefficients, maybe in terms of the Chern classes of the bundle, unravelling the definitions of the LeraySerre spectral sequence?
1 Answer
As you say, if the map $H(E)\to H(F)$ is surjective, the LeraySerre spectral sequence degenerates at the $E^2$page which is therefore isomorphic to the $E^\infty$page. Unwrapping the definitions, this means that there is a filtration on $H^*(E)$ such that the associated graded is $H^*(X)\otimes H^*(F)\cong H^*(X)[x]/x^{r+1}$. Choosing a representative of the class $x$, we obtain that $H^*(E)$ is generated as a $H^*(X)$algebra (with $H^*(X)$ in filtration degree $0$) by a single element $x$ in degree and filtration degree $2$ such that $x^{r+1}$ sits in degree less than $2r+2$. This can only happen if there is a relationship of the form $x^{r+1} = \sum_{k=0}^r p^*\gamma_{r+1k}x^k$ with $\gamma_k\in H^{2k}(X)$. Determining the concrete values of the $c_k$ amounts to solving the extension problem of the LeraySerre spectral sequence, which is in a sense precisely the part of the problem the spectral sequence doesn't capture. However, in this case we can solve this problem geometrically: By construction, the pullback of $\mathcal E$ to $E$ splits off a line bundle $L$, and we may take $x = c_1(L)$. Since $p^*\mathcal E/L$ is $r$dimensional, its $(r+1)$th Chern class vanishes, and the sum formula for the total Chern class yields $$ c(p^*\mathcal E/L) = c(p^*\mathcal E)(1x)^{1}. $$ Expanding the second factor as $\sum_{k=0}^r x^k$ and taking the resulting expression for the $(r+1)$th Chern class yields $\gamma_k = c_k(\mathcal E)$. In fact, this can be used to define Chern classes for general complex oriented cohomology theories.

$\begingroup$ When you say: "The pullback splits off a line bundle" you mean you take $L$ to be line subbundle of $p^{\ast}\mathcal{E}$ generated by the tautological section? $\endgroup$ Feb 18, 2019 at 16:25

$\begingroup$ More or less  a point in $E$ is a point $x\in X$, together with a onedimensional subspace $V$ of $\mathcal E_x$. The fiber of $L$ over $(x,V)$ is precisely the onedimensional subspace $V$ of $(p^*\mathcal E)_{(x,V)}\cong \mathcal E_x$. Note that $L$ is not generated by a nonvanishing section (otherwise it would be trivial, but by definition its restriction to every fiber is nontrivial), and there is no canonical section of $p^*\mathcal E$. $\endgroup$ Feb 18, 2019 at 20:04