A question regarding cohomology Let $X=\mathbb{P}^n$ denote the complex projective space of complex dimension $n$. Let $G$ denote the Grassmannian of lines in $X$. There is a tautological bundle $P\subset X\times G,$ given by {$(x,l) : x\in l$}. Let $p_X:P\rightarrow X,p_G:P\rightarrow G$ be the restrictions of the projections to $P$. Let $[H]\in H^2(X,\mathbb{Z})$ be a hyperplane class, obtained by the fundamental class of a general hyperplane section of $X$ (which lives in $H_{2n}(X,\mathbb{Z})\cong H^2(X,\mathbb{Z})$ by Poincare duality).
If $m,m'$ denotes the complex dimension of $P , G$ respectively, notice that $p_X ^{*}([H])\in H^2(P,\mathbb{Z})\cong H_{2m-2}(P,\mathbb{Z}),$ and $2m-2 = 2m'$ since $P$ is a bundle of projective line over $G$. So $p_{G*}(p_X ^{*}([H]))\in H_{2m'}(G,\mathbb{Z}).$
Now, in a paper I am reading, they have written that taking the cap product of $p_{G*}(p_X ^{*}([H]))$ with some cohomology class gives back the same class (Here we have identified appropriate homology and cohomology classes using Poincare duality).

So my question is the following: does $p_{G*}(p_X ^{*}([H]))$ give the fundamental class of $G$ ? (taking G as a compact orientable manifold)

Any help would be appreciable.
 A: The first thing to note is that $p_X^*([H])$ is the Poincare dual of the class $[p_X^{-1}(H)]$. There are several ways to see this. One way is to realise $[H]$ as the euler class of $\mathcal{O}(1)$ on $\mathbb{P}^n$. So $p_X^*([H]) = e(p_X^*\mathcal{O}(1))$. The Euler class is given by the zero locus of a generic section. You can take the pullback of a linear form $L$ (you should check that this is 'generic', in the sense that it intersects transversally with the zero section), and evaluate its zero locus. Its zero locus in this case is $(x,l)$ such that $L(x)=0$. So you end up with $p_X^{-1}(H)$. (well, possibly with a different H which is linearly equivalent). 
Here's a MO thread discussing this (pullback of homology classes, transversality issues) in higher generality: https://mathoverflow.net/questions/115764/pullback-map-in-homology.
Another way is to observe that this $P\rightarrow X$ is a projective bundle (in particular, it is a flat morphism), and you can apply flat pullback for Chow groups. 
What remains is to compute the degree of the map $p_X^{-1}(H) \rightarrow G$. But observe that the pre-image of a point $[L] \in G$ is the unique intersection of $H$ and $L$ in $\mathbb{P}^n$. So this is a degree $1$ map, hence $p_{G*}(p_X^*([H])) = [G]$. 
