# The product of (first) chern classes

I was reading Fulton's Intersection Theory and came across this useful formula for the chern class of the tensor product: $$c_r(E\otimes L) = \sum_{i=0}^r c_1(L)^ic_{r-i}(E).$$ However, I couldn't find a definition of what exactly the product of two chern classes was. Is it purely formal or can it be simplifed?

In particular, for a line bundle $\mathcal{L}=\mathcal{O}(D)$ on a variety $X$, we know $c_1(\mathcal{L}) = D$. It's tempting to write $$c_1(\mathcal{L})^2 = D^2 = D.D$$ where $D.D$ is the self-intersection as usual, but I can't really justify this from the axioms or basic results I've found.

Is the above true? If so, how does one justify it, and if not, how should I think about products and power of (first) chern classes?

Thanks.

• It is just the cup product... – Alan Muniz Mar 11 '16 at 22:51
• By the way, as Fulton defines it in terms of Segre classes, he states in the definition of Chern class: "Here the $s_j(E)$ are regarded as endomorphisms of $A^{\ast}X$, with products denoting composition; " – Alan Muniz Mar 11 '16 at 22:56
• $c_i(E)$ is in $H^{2i}(X,\mathbb{Z})$ you can use the cup profuct in any compatible cohomology... – Alan Muniz Mar 12 '16 at 13:54
• And, yes! For a line bundle $L$ associated to a divisor $D$, $c_1(L)^r = D^r$. – Alan Muniz Mar 12 '16 at 13:56
• Ok. Later I'll post the answer. This mobile app is not very good :( The self intersection is defined somewhere in Fulton's book. I don't remember where now. – Alan Muniz Mar 12 '16 at 14:02