The cohomology of projective hypersurfaces

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?

hyh Says:

March 6, 2012 at 12:28 am

Something can be said about the ring structure using Hodge theory.

E.g. for a quartic $X$ in $\mathbb{P}^3$, Hodge theory says $H^2(X)\otimes\mathbb{C}$ can be decomposed as the direct sum of 3 subspaces, $H^{2,0}, H^{1,1}, H^{0,2}$. Their dimensions are 1, 20, 1 respectively. $H^2(X;\mathbb{Z})$ sits inside the direct sum of them as a unimodular lattice by Poincare duality, and that lattice is even, by the adjunction formula for self-intersections of curves in it.

By some further properties of Hodge decomposition and Hodge index theorem on intersection of curves you can find it is a lattice with sign (3,19), and in fact it ($H^2(X;\mathbb{Z})$) as a unimodular lattice, is isomorphic to 3 copies of $U$ plus 2 copies of the lattice $E_8$, where $U$ is the lattice freely generated over $\mathbb{Z}$ by $a$ and $b$ with $\langle a,a\rangle=\langle b,b\rangle=0$ and $\langle a,b\rangle= 1$. Those are in general true for all K3 surfaces, not only quartic in $\mathbb{P}^3$.

For higher degree things may be more complicated. If you have a quintic in $\mathbb{P}^3$ then it’s second Betti number is 53 by the formula you gave above, and the Hodge number (i.e. the dimension of $H^{2,0}, H^{1,1}, H^{0,2}$) are 5, 43, 5 respectively. Seems like it is more complicated in this case. But similar analysis should give you a even unimodular lattice whose signature is (6, 47). ——These are some fresh calculation I just made, so I might be wrong here.

On the topology side for simply-connected 4-manifolds, the intersection form on H^2 almost determined its homeomorphism type. You can read more on the 4-manifold page on wikipedia.

Akhil Mathew Says:

March 6, 2012 at 8:50 am
Thanks a lot! This is pretty interesting — do you have a reference for these kinds of calculations?

To use TeX in WordPress comments, you should write “latex” after the first dollar sign, before the math.

1. Hyh Says:
  
  March 6, 2012 at 12:35 pm
  Emm, I guess you can read about this in a survey book on algebraic surface by Shafarevich. It is in the series of Russian math encyclopedia: http://www.amazon.com/Algebraic-Geometry-Cohomology-Encyclopaedia-Mathematical/dp/3540546804
  
  1. Akhil Mathew Says:
    
    March 8, 2012 at 12:17 am
    Thanks for this reference.

March 6, 2012 at 12:30 am

Sorry, seems like latex on WordPress shouldn’t be bounded by dollar signs? I’m too used to mathoverflow I guess. Or you don’t allow latex in comments?

sanathdevalapurkar Says:

July 9, 2014 at 5:53 pm
@hyh I know it’s probably too late to answer, but if you wish to enter, for example $ax^2+bx+c=0$ in WordPress, you enclose “latex ax^2+bx+c=0” (without the quotes) in dollar signs. Hope this helps,

Theo Says:

March 6, 2012 at 9:34 am

Incidentally, if you have any comments on http://mathoverflow.net/questions/89738/, I’d love to hear them 🙂

Qiaochu Yuan Says:

November 12, 2013 at 8:46 pm

Can you explain how you’re using the Lefschetz hyperplane theorem to conclude that $M$ is simply connected? This is probably quite basic but I don’t see it. Shouldn’t a hyperplane section of $M$ be some projective curve, which a priori could have nontrivial fundamental group?

Akhil Mathew Says:

November 12, 2013 at 9:17 pm
We’re treating $M$ as a hyperplane section in $\mathbb{P}^3$ (imbedded via Veronese in a larger projective space), so we can apply Lefschetz directly to $M$ (you’re right it would break down in one smaller dimension).

1. Qiaochu Yuan Says:
  
  November 12, 2013 at 10:30 pm
  I see. Thanks!

December 14, 2013 at 5:59 pm

There is no torsion in $H^2$ by universal coefficients; this is true more generally for any simply connected space.

Ozlem Says:

November 11, 2014 at 6:13 pm
sorry but I don’t see how you get there is no torsion by universal coefficients ?

June 14, 2014 at 3:47 pm

I found out how to get the ring structure on cohomology. I’m in the middle of writing a blog post about it.

Qiaochu Yuan Says:

June 14, 2014 at 3:49 pm
Sorry, I should have specified that I only really know how to get the ring structure in the case $n = 3$ , although I suspect the argument generalizes. The very short sketch is to use the Hirzebruch signature theorem to calculate the signature and then appeal to the classification of indefinite unimodular lattices.

Climbing Mount Bourbaki