Let $\mathbb{P}^3 = P(\mathbb{C}^4)$ and $\gamma:C\rightarrow \mathbb{P}^3$ be a smoothly embedded algebraic space curve. Then its total Chern class is $c(C) = c(TC) = 1+a\in H^*(C)$, where $a^2=0$. Since $C$ is compact and orientable we have Poincare duality and hence a well defined 'shriek – map' $$\gamma^!: H^*(C)\overset{PD_C}{\longrightarrow} H_{2-*}(M) \overset{\gamma_*}{\longrightarrow} H_{2-*}(\mathbb{P}^3)\overset{PD_{\mathbb{P}^3}^{-1}}{\longrightarrow}H^{*+4}(\mathbb{P}^3).$$
Let $g$ be the genus of $C$ and let $\chi(C):=2-2g$ and $d$ be the degree of $C$. Then $$\gamma^!(1) = d\cdot c_1(\mathcal{O}_{\mathbb{P}^3}(1))^2\quad\text{and }\quad \gamma^!(a) = \chi \cdot c_1(\mathcal{O}_{\mathbb{P}^3}(1))^3.$$
Does anyone know why these two relations hold and point me to some literature? My knowledge in algebraic geometry is very limited since I am more familiar with manifolds, homology/cohomolgy of topological spaces so Chern classes for me are always defined in that context. I know that there are Chern classes in algebraic geometry (i.e. for Chow rings) and that these notions are related, but my understanding is very superficial.
Still it would help me a lot if I could understand the relations above. Thanks!
Best Answer
You can understand these facts using just a bare minimum of algebraic geometry. All you need to know about $c_1(\mathcal{O}_{\mathbb{P}^3}(1))$ in this context is that it is the standard generator of $H^2(\mathbb{P}^3)$ (corresponding to the fundamental class of $\mathbb{P}^1=S^2$ when you restrict to $\mathbb{P}^1\subset\mathbb{P}^3$). This can also be described as the Poincaré dual of the fundamental class of $\mathbb{P}^2\subset\mathbb{P}^3$, or of any hyperplane in $\mathbb{P}^3$ (since the inclusion maps of different hyperplanes $\mathbb{P}^2\to\mathbb{P}^3$ are all homotopic). How you prove this depends on how you define Chern classes, but for some definitions it is essentially by definition: for instance the first Chern class of the tautological line bundle on $\mathbb{P}^\infty$ can be defined to be negative the standard generator of $H^2(\mathbb{P}^\infty)$, and then for other line bundles it can be defined by pulling back under the classifying map to $\mathbb{P}^\infty$. The bundle $\mathcal{O}_{\mathbb{P}^3}(1)$ is just the dual of the tautological bundle on $\mathbb{P}^3$, so its first Chern class is the negative of the first Chern class of the tautological bundle, which is just the pullback of the tautological bundle under the inclusion $\mathbb{P}^3\to\mathbb{P}^\infty$.
Let me now say a bit about Poincaré duality on $\mathbb{P}^3$. If you have two smooth oriented submanifolds $A$ and $B$ of an oriented manifold which intersect transversely, then the Poincaré dual of the fundamental class of the intersection $A\cap B$ is just the cup product of the Poincaré duals of the fundamental classes of $A$ and $B$. This plays very nicely with smooth subvarieties of $\mathbb{P}^3$ via Bézout's theorem. In particular, if $H\subset\mathbb{P}^3$ is a hyperplane, Bézout's theorem says that for any smooth curve $C\subset\mathbb{P}^3$ of degree $d$ which intersects $H$ transversely, $C$ and $H$ intersect at $d$ points. So, writing $t$ for the Poincaré dual of the fundamental class of $H$ (which, as mentioned above, is $c_1(\mathcal{O}_{\mathbb{P}^3}(1))$) and $c$ for the Poincaré dual of the fundamental class of $C$, $tc=dt^3$.
Now to compute $\gamma^!(1)$, we first take the Poincaré dual of $1\in H^0(C)$, which is just the fundamental class $[C]\in H_2(C)$. Then we push this forward to get the class $[C]\in H_2(\mathbb{P}^3)$, and take its Poincaré dual $c\in H^4(\mathbb{P}^3)$. Since $tc=dt^3$, this class $c=\gamma^!(1)$ must be $dt^2$.
To compute $\gamma^!(a)$, we first take the Poincaré dual of $a=c_1(TC)\in H^2(C)$. The top Chern class of a vector bundle is the same as its Euler class, and the Euler class of the tangent bundle of an oriented manifold is dual to the Euler characteristic of the manifold (that's why it's called the Euler class). So the Poincaré dual of $a$ is $\chi\in H_0(C)$ (when you identify $H_0(C)$ with $\mathbb{Z}$ in the canonical way). Pushing this forward to $\mathbb{P}^3$ we get $\chi\in H_0(\mathbb{P}^3)$, whose Poincaré dual is $\chi t^3$.