## Segre Classes of Cones

Last time, we talked about the Normal Cone.  We’re going to go back a bit and increase the generality before coming back to it.  Let $C$ be a cone over $X$, and let $P=P(C\oplus 1)$ be the projective closure.  We define the Segre class of the cone $C$, $s(C)$ in $A_*(X)$ to be $s(C)=q_*\left(\sum_{i\geq 0} c_1(\mathscr{O}(1))^i\cap [P]\right)$, where $q:P\to X$ is the projection.

So…that’s a fairly opaque definition.  What does it give us in a case we understand? Let $E$ be a vector bundle, so it’s a special cone.  Then $[P(E\oplus 1)]=q^*[X]$, and so the definition can be unraveled (annoying calculation, do it in the privacy of your office) to get $c(E\oplus 1)^{-1}\cap [X]$, but then $c(E\oplus 1)=c(E)$, so for a vector bundle, $s(E)=c(E)^{-1}\cap [X]$.

Additionally, you get geometric multiplicities of the cone out: $s(C)=\sum_{i=1}^t m_i s(C_i)$ where $C_i$ are the cone’s irreducible components.

So this Segre class tells us interesting things, and in the cases we’ve already looked at, gives us familiar things.  But what happens if we try to combine cones? Well, if $C$ and $D$ are cones defined by $S^*$ and $T^*$, we get a cone $C\times_X D$ given by $S^*\otimes_{\mathscr{O}_X} T^*$.  Now, if $D$ is actually a vector bundle $E$, we define $C\oplus E = C\times_X E$ and get $s(C\oplus E)=c(E)^{-1}\cap s(C)$.

This is actually a general pattern.  If we want to do anything interesting with cones, we need to include the special case of vector bundles in order to work it out.  For example, a morphism of cones $\phi:C\to C'$ is a morphism of their algebras $\phi^*:S^{*'}\to S^*$, and a short exact sequence is $0\to E\to C\to C'\to 0$, where $E$ is a vector bundle requires that the map on algebras for $C\to C'$ be injective adn the map $S^*\to \mathrm{Sym}(\mathscr{E})$ be surjective, and that there is $\tilde{\mathscr{E}}$ locally free subsheaf of $S^1$ such that $S^{*'}\otimes_{\mathscr{O}_X} \mathrm{Sym}(\tilde{\mathscr{E}})\to S^*$ is an isomorphism.  Yeah, it’s kind of a mess.

However, we do get the good result that if $0\to E\to C\to C'\to 0$ is a short exact sequence of cones, we have $s(C')=c(E)\cap s(C)$.  In particular, if we have a coherent sheaf $\mathscr{F}$, we get a cone $C(\mathscr{F})=\underline{\mathrm{Spec}}(\mathrm{Sym}(\mathscr{F}))$, and we define $s(\mathscr{F})=s(C(\mathscr{F}))$.  Then if we have $0\to \mathscr{F}'\to\mathscr{F}\to\mathscr{E}\to 0$ with $\mathscr{E}$ locally free, we get precisely the above back, and get the same formula.