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 $\displaystyle = d$

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?