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I'm trying to figure out why there are 27 lines on a smooth cubic hypersurface in $\mathbb{P}^3$ using Chern classes, without looking up the proof in 3264 and All That. One thing confusing me is the following:

We want to find the top Chern class of the bundle $E$ over $\mathbb{G}(1,3)$ whose fiber over a line $L$ is the space of cubic forms on $\mathbb{P}^3$, restricted to $L$. After all, a smooth cubic hypersurface gives a section of this vector bundle, and the top Chern class will hopefully tell us that this section needs to vanish 27 times.

The first way I thought of to study $E$ is to note that it is a quotient of the trivial bundle $\mathbb{G}(1,3)\times\mathbb{C}^{20}\to\mathbb{G}(1,3),$ where $\mathbb{C}^{20}$ parametrizes cubic forms on $\mathbb{P}^3$ and the map $$\mathbb{G}(1,3)\times\mathbb{C}^{20}\to E$$ of bundles maps $(L,f)\mapsto(L,f|_L)\in E.$ Then these fit in the short exact sequence $$0\to F\to\mathbb{G}(1,3)\times\mathbb{C}^{20}\to E\to0,$$ where $F$ is the bundle whose fiber over $L$ is the cubic forms of $\mathbb{P}^3$ that vanish on $L$.

Now the part I'm getting confused about. Properties of Chern classes tell us that the total Chern classes satisfy $$c(\mathbb{G}(1,3)\times\mathbb{C}^{20})=c(E)c(F).$$ Since $\mathbb{G}(1,3)\times\mathbb{C}^{20}$ is a trivial bundle, I'm pretty sure it has trivial total Chern class $c(\mathbb{G}(1,3)\times\mathbb{C}^{20})=1.$ But this would mean that $c(E)$ and $c(F)$ are units, i.e. trivial themselves, right? Which is very hopefully not the case. Help? (I could be way off base on much of this, since I came up with it right as I was falling asleep last night.)

(Edited to fix minor errors.)

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up vote 2 down vote accepted

The cohomology ring $H^*(G(1,3),\mathbb Z)$ has many units and to say that $c(E)\in H^*(G(1,3),\mathbb Z)$ is one of them is far from implying that $c(E)=1$ and even further from implying that $E$ is trivial.

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Thanks - I am now back on the right track and it was kind of a silly issue! –  Rob Silversmith Nov 16 '12 at 20:18

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