Consider a smooth surface {M \subset \mathop{\mathbb P}^3(\mathbb{C})} of degree {d}. We are interested in determining its cohomology.

1. A fibration argument

A key observation is that all such {M}‘s are diffeomorphic. (When {\mathop{\mathbb P}^3} is replaced by {\mathop{\mathbb P}^2}, then this is just the observation that the genus is determined by the degree, in the case of a plane curve.) In fact, consider the space {V} of all degree {d} homogeneous equations, so that {\mathop{\mathbb P}(V)} is the space of all smooth surfaces of degree {d}. There is a universal hypersurface {H \subset \mathop{\mathbb P}^3 \times \mathop{\mathbb P}(V)} consisting of pairs {(p, M)} where {p} is a point lying on the hypersurface {M}. This admits a map

\displaystyle \pi: H \rightarrow \mathop{\mathbb P}(V)

which is (at least intuitively) a fiber bundle over the locus of smooth hypersurfaces. Consequently, if {U \subset \mathop{\mathbb P}(V)} corresponds to smooth hypersurfaces, we get an honest fiber bundle

\displaystyle \pi^{-1}(U) \rightarrow U .

But {U} is connected, since we have thrown away a complex codimension {\geq 1} subset to get {U} from {\mathop{\mathbb P}(V)}; this means that the fibers are all diffeomorphic.

This argument fails when one considers only the real points of a variety, because a codimension one subset of a real variety may disconnect the variety.

2. Three-fourths of the cohomology

OK, so let’s go back to the original situation. We are interested in determining the cohomology of {M}. The first thing to note is that, by the Lefschetz hyperplane theorem, {M} is simply connected. We consequently have

\displaystyle H^0(M; \mathbb{Z}) = \mathbb{Z}; \quad H_1(M; \mathbb{Z}) = H^1(M; \mathbb{Z}) = 0 .

By Poincaré duality, we’ll also have

\displaystyle H^4(M; \mathbb{Z}) = \mathbb{Z}; \quad H^3(M; \mathbb{Z}) = 0.

The main step now is to determine the second Betti number. There is a nice argument for this in a paper by Milnor “On simply connected 4-manifolds” which I’d like to describe here.

The key point is that to determine the second Betti number, we may as well determine the Euler characteristic of {M}; we can do this by computing the top Chern class of {TM} and evaluating on a fundamental class.

Unfortunately, we don’t really have a good grasp of the fundamental class of {M}. We do, however, have a completely good grasp of the cohomology of the ambient space {\mathop{\mathbb P}^3(\mathbb{C})}: the cohomology here is generated by the hyperplane class {\alpha \in H^2(\mathop{\mathbb P}^3(\mathbb{C}))}. If {i: M \hookrightarrow \mathop{\mathbb P}^3(\mathbb{C})} is the inclusion, we then observe that

\displaystyle <i^* \alpha^2, [M]> = d

because the intersection of {\alpha} with itself in {M} has {d} points (by Bezout’s theorem). As a result, if we run into something which is a multiple of {\alpha^2}, we can easily evaluate it on the fundamental class.

3. Computing Chern classes

The Chern classes of {\mathop{\mathbb P}^3(\mathbb{C})} are known: the total Chern class is {(1 + \alpha)^4 = 1 + 4 \alpha + 6 \alpha^2 + 4 \alpha^3}. That is,

\displaystyle c(T\mathop{\mathbb P}^3(\mathbb{C})) = 1 + 4 \alpha + 6 \alpha^2 + 4 \alpha^3

which means that if we pull back along {i: M \hookrightarrow \mathop{\mathbb P}^3(\mathbb{C})}, we have

\displaystyle c(i^* T\mathop{\mathbb P}^3(\mathbb{C})) = 1 + 4 i^* \alpha + 6i^* \alpha^2.

To get the Chern classes of {TM}, we’ll use the fact that the normal bundle of {M} is easily computed via the adjunction formula.

Proposition 1 (Adjunction formula) Let {D \subset X} be an inclusion of a smooth divisor in a smooth complex manifold. Then the normal bundle of {D} is just the associated line bundle {\O(D)} restricted to {D}.

Consequently, we have

\displaystyle TM \oplus i^* \mathcal{O}(d) = i^* T\mathop{\mathbb P}^3(\mathbb{C})

by adjunction. If we take Chern classes, we get that

\displaystyle (1 + c_1(TM) + c_2(TM)) (1 + d i^* \alpha) = (1 + 4i^* \alpha + 6i^* \alpha^2).

It follows that {c_1(TM) = (4 - d)i^*\alpha} and

\displaystyle c_2(TM) = 6i^*\alpha^2 - d i^* \alpha c_1(TM).

In other words,

\displaystyle c_2(TM) = (6 + d(d-4)) i^*\alpha^2

4. Finishing the computation

Now that we have the Chern class {c(TM)}, we can obtain the Euler characteristic easily by evaluating the top Chern class on {[M]}.

If we evaluate on the fundamental class of {M}, we get

\displaystyle \chi(M) = d( 6 + d(d-4)),

which means that the second Betti number of {M} is {d(6 + d(d-4)) - 2}. In Milnor’s paper, the case {d = 4} is handled.

This method seems to work to handle smooth hypersurfaces in {\mathop{\mathbb P}^n} for any {n}, but it only gives the cohomology mod torsion; also, it does not give the ring structure. Does anyone know how to obtain this extra information?